Linear Circuit Analysis- DeCarlo-Lin 3e

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LINEAR

CIRCUITS

TIME DOMAIN, PHASOR, AND LAPLACE TRANSrORM APPROACHES THIRD

EDITION

Raymond A. DeCarlo Purdue University Pen-Min Lin Purdue University

Kendall Hunt

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Cover image (^^Jîkiaui

^ Used under license from Shutterstock, Inc.

Kendall Hunft p u b l i s h i n g

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www.kendallhunt.cpm Send all inquiries to: 4050 Westmark Drive Dubuque, lA 52004-1840 Copyright © 2001, 2009 Raymond A. DeCarlo and Pen-Min Lin Copyright © 1995 Prentice-Hall, Inc. ISBN 978-0-7575-6499-4 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the copyright owner.

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Printed in the United States of America 10 9 8 7 6 5 4 3 O

TABLE OF CO N TEN TS

Preface......................................................................................................................................................................vii Chapter 1 • Charge, Current, Voltage and Ohm’s Law ............................................................................ 1 Chapter 2 • Kirchhoff’s Current & Voltage Laws and Series-Parallel Resistive C ircu its..............51 Chapter 3 • Nodal and Loop Analyses....................................................................................................... 107 Chapter 4 • T he Operational Amplifier..................................................................................................... 155 Chapter 5 * Linearity, Superposition, and Source Transform ation................................................... 191 Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems.................................... 227 Chapter 7 • Inductors and C apacitors....................................................................................................... 269 Chapter 8 • First Order RL and RC Circuits...........................................................................................321 Chapter 9 • Second Order Linear Circuits................................................................................................379 Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods .................................................431 Chapter 11 • Sinusoidal State State Power Calculations.......................................................................499 Chapter 12 • Laplace Transform Analysis L Basics................................................................................. 543 Chapter 13 • Laplace Transform Analysis II: Circuit Applications................................................... 603 Chapter 14 • Laplace Transform Analysis III; Transfer Function Applications.............................683 Chapter 15 * Time Domain Circuit Response Computations: The Convolution M ethod...... 763 Chapter 16 • Band-Pass Circuits and Resonance....................................................................................811 Chapter 17 * Magnetically Coupled Circuits and Transformers........................................................ 883 Chapter 18 • Tw o-Ports...................................................................................................................................959 Chapter 19 • Principles o f Basic Filtering ............................................................................................. 1031 Chapter 20 • Brief Introduction to Fourier Series .............................................................................. 1085 In d ex................................................................................................................................................................... 1119

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PREFACE For the last several decades, EE/ECE departments o f US universities have typically required two semesters o f linear circuits during the sophomore year for EE majors and one semester for other engineering majors. Over the same time period discrete time system concepts and computer engi neering principles have become required fare for EE undergraduates. Thus we continue to use Laplace transforms as a vehicle for understanding basic concepts such as impedance, admittance, fdtering, and magnetic circuits. Further, software programs such as PSpice, MATLAB and its tool boxes, Mathematica, Maple, and a host o f other tools have streamlined the computational drudg ery o f engineering analysis and design. MATLAB remains a working tool in this 3'''^ edition o f Linear Circuits. In addition to a continuing extensive use o f MATLAB, we have removed much o f the more com plex material from the book and rewritten much o f the remaining book in an attempt to make the text and the examples more illustrative and accessible. More importantly, many o f the more diffi cult homework exercises have been replaced with more routine problems often with numerical answers or checks. Our hope is that we have made the text more readable and understandable by today’s engineering undergraduates.

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Charge, Current, Voltage and Ohm’s Law CHAPTER O U TLIN E 1. 2. 3. 4. 5. 6. 7. 8.

Role and Importance o f Circuits in Engineering Charge and Current Voltage Circuit Elements Voltage, Current, Power, Energy, Relationships Ideal Voltage and Current Sources Resistance, Ohm’s Law, and Power (a Reprise) V-I Characteristics o f Ideal Resistors, Constant Voltage, and Constant Current Sources Summary Terms and Concepts Problems

CHAPTER O B jEC TIV ES 1.

Introduce and investigate three basic electrical quantities: charge, current, and voltage, and the conventions for their reference directions.

2. 3.

Define a two-terminal circuit element. Define and investigate power and energy conversion in electric circuits, and demonstrate

4.

that these quantities are conserved. Define independent and dependent voltage and current sources that act as energy or sig

5. 6. 7. 8.

nal generators in a circuit. Define Ohm’s law, v{t) = R i{t), for a resistor with resistance R. Investigate power dissipation in a resistor. Classify memoryless circuit elements by dieir terminal voltage-current relationships. Explain the difference between a device and its circuit model.

ch ap ter 1 • Charge, Current, Voltage and O hm ’s Law

1. ROLE AND IM PORTANCE OF CIRCUITS IN ENGINEERING Are you curious about how fuses blow? About the meaning o f different wattages on Hght bulbs? About the heating elements in an oven? And how is the presence o f your car sensed at a stoplight? Circuit theory, the focus o f this text, provides answers to all these questions. W hen you learn basic circuit theory, you learn how to harness the power o f electricity, as is done, for example, in •

an electric motor that runs the compressor in an air conditioner or the pump in a dish washer;

• • • •

a microwave oven; a radio, TV, or stereo; an iPod; a car heater.

In this text, we define and analyze common circuit elements and describe their interaction. Our aim is to create a modular framework for analyzing circuit behavior, while simultaneously devel oping a set o f tools essential for circuit design. These skills are, o f course, crucial to every electri cal engineer. But they also have broad applicability in other fields. For instance, disciplines such as bioengineering and mechanical engineering have similar patterns o f analysis and often utilize circuit analogies.

W H A T IS A C IR C U IT ? A circuit is an energy or signal/information processor. Each circuit consists o f interconnections o f “simple” circuit elements, or devices. Each circuit element can, in turn, be thought o f as an ener gy or signal/information processor. For example, a circuit element called a “source” produces a voltage or a current signal. This signal may serve as a power source for the circuit, or it may rep resent information. Information in the form o f voltage or current signals can be processed by the circuit to produce new signals or new/different information. In a radio transmitter, electricity powers the circuits that convert pictures, voices, or music (that is, information) into electromag netic energy. This energy then radi ates into the atmosphere or into space from a transmitting antenna. A satellite in space can pick up this electromagnetic energy and trans mit it to locations all over the world. Similarly, a T V reception antenna or a satellite dish can pick up and direct this energy to a T V set. T h e T V contains circuits (Figure 1.1) that reconvert the information within the received signal back into pictures with sound. FIG U RE 1.1 Cathode ray tube with surrounding circuitry for converting electrical signals into pictures.

Chapter 1 • Charge, Current, Voltage and O hm ’s Law

2. CH A RGE AND CU RREN T CH A R G E Charge is an electrical property o f matter. Matter consists o f atoms. Roughly speaking, an atom contains a nucleus that is made up o f positively charged protons and neutrons (which have no charge). T he nucleus is surrounded by a cloud o f negatively charged electrons. Th e accumulated charge on 6.2415 x 10’^ electrons equals -1 coulomb (C). Thus, the charge on an electron is -1 .6 0 2 1 7 6 X 10-19 C. Particles with opposite charges attract each other, whereas those with similar charges repel. The force o f attraction or repulsion between two charged bodies is inversely proportional to the square o f the distance between them, assuming the dimensions o f the bodies are very small compared with the distance o f separation. Two equally charged particles 1 meter (m) apart in free space have charges o f 1 C each if they repel each other with a force o f 10“^ c^ Newtons (N), where c = 3 x 10^ m/s is the speed o f light, by definition. The force is attractive if the particles have opposite charges. Notationally, Q will denote a fixed charge, and q or q{t), a time-varying charge.

Exercise. How many electrons have a combined charge o f -5 3 .4 0 6 x 10

C?

AN SW ER; 333,3 9 1 ,5 9 7

Exercise. Sketch the time-dependent charge profile q{t) = 3 (l-^ ^ 0 C, ? > 0, present on a metal plate. M ATLAB is a good tool for such sketches.

A conductor refers to a material in which electrons can move to neighboring atoms with relative ease. Metals, carbon, and acids are common conductors. Copper wire is probably the most com mon conductor. An ideal conductor offers zero resistance to electron movement. Wires are assumed to be ideal conductors, unless otherwise indicated. Insulators oppose electron movement. Common insulators include dry air, dry wood, ceramic, glass, and plastic. An ideal insulator offers infinite opposition to electron movement.

C U R R EN T Current refers to the net flow o f charge across any cross section o f a conductor. T he net move ment o f 1 coulomb (1 C) o f charge through a cross section o f a conductor in 1 second (1 sec) produces an electric current o f 1 ampere (1 A). The ampere is the basic unit o f electric current and equals 1 C/s. The direction o f current flow is taken by convention as opposite to the direction o f electron flow, as illustrated in Figure 1.2. This is because early in the history o f electricity, scientists erroneously believed that current was the movement o f only positive charges, as illustrated in Figure 1.3. In metallic conductors, current consists solely o f the movement o f electrons. However, as our under standing o f device physics advanced, scientists learned that in ionized gases, in electrolytic solu

c h ap ter 1 • Charge, Current, Voltage and O hm ’s Law

tions, and in some semiconductor materials, movement o f positive charges constitutes part or all o f the total current flow.

One Ampere of Current "

One ; ; Cloud o f \ se co n d ^ ....... |---- 6.24x10’® 1 later i ; k electrons

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Boundary FIG U RE 1.2 A cloud o f negative charge moves past a cross section of an ideal conductor from right to left. By convention, the positive current direction is taken as left to right.

One Ampere of Current One Coulom b

One

of positive

'second later

charge

Boundary

FIGURE 1.3 In the late nineteenth cenmry, current was thought to be the movement of a positive charge past a cross section of a conduaor, giving rise to the conventional reference “direction of positive current flow.” Both Figures 1.2 and 1.3 depict a current o f 1 A flowing from left to right. In circuit analysis, we do not distinguish between these two cases: each is represented symbolically, as in Figure 1.4(a). The arrowhead serves as a reference for determining the true direction o f the current. A positive value o f current means the current flows in the same direction as the arrow. A current o f negative value implies flow is in the opposite direction o f the arrow. For example, in both Figures 1.4a and b, a current o f 1 A flows from left to right.

1A

-1A

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(a) (b) FIG U RE 1.4 1 A of current flows from left to right through a general circuit element.


In Figure 1.4, the current is constant. The wall socket in a typical home is a source o f alternating current, which changes its sign periodically, as we will describe shortly. In addition, a current direc tion may not be known a priori. These situations require the notion o f a negative current. E X A M P L E 1.1. Figure 1.5 shows a slab o f material in which the following is true: 1. Positive charge carriers move from left to right at the rate o f 0.2 C/s. 2. Negative charge carriers move from right to left at the rate o f 0.48 C/s. Given these conditions, a) Find and /^; b)

Describe the charge movement on the wire at the boundaries A and B. B

A

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,

Connecting

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Connecting wire

wire Sem iconductor iVlaterial

F IG U R E 1.5 Material through which positive and negative charges move. S o lu tio n

a)

The current from left to right, due to the movement o f the positive charges, is 0.2 A. The current from left to right, due to the movement o f the negative charges, is 0.48 A. Therefore, /^, the total current from left to right, is 0.2 + 0.48 = 0.68 A. Since ly is the current from right to left, its value is then -0 .6 8 A.

b)

T he wire is a metallic conductor in which only electrons move. Therefore, at boundaries A and B, negative charges (carried by electrons) move from right to left at the rate o f 0.68 C/s.

Exercise. In Example 1.1, suppose positive-charge carriers move from right to left at the rate o f 0.5 C/s, and negative carriers move from left to right at the rate o f 0.4 C/s. Find

and

AN SW ER: /, = - 0 .9 A; ^ = 0.9 A

If a net charge

crosses a boundary in a short time frame o f At (in seconds), then the approxi

mate current flow is /=

Aq At ( 1 . 1)

where I, in this case, is a constant. The instantaneous (time-dependent) current flow is the limit ing case o f Equation 1.1, i.e.,


dq{t) dt

( 1. 2)

Here q{t) is the amount o f charge that has crossed the boundary in the time interval [tQ, t] . The equivalent integral counterpart o f Equation 1.2 is

q{t) = J i{r)dr (1.3) E X A M P L E 1.2 The charge crossing a boundary in a wire is given in Figure 1.6(a) for ? > 0. Plot the current i{t) through the wire.

(a)

(b)

FIG U RE 1.6 (a) Charge crossing a hypothetical boundary; (b) current flow associated with the charge plot o f (a).


S o lutio n

As per Equation 1.2, the current is the time derivative o f q{t). The slopes o f the straight-Une seg ments o f q{f} in Figure 1.6(a) determine the piecewise constant current plotted in Figure 1.6(b).

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Exercise. The charge crossing a boundary in a wire varies as q[t) = ---------------- C, for t >Q. Compute the current flow. A N SW ER: sin(cof) A, for f > 0

Exercise. Repeat the preceding exercise if q{t) = 5e A N SW ER:

C, for t > 0 .

A, for f > 0

E X A M PLE 1.3 Find q{t), the charge transported through a cross section o f a conductor over [0, f], and also the total charge Q transported, if the current dirough the conductor is given by die waveform o f Figure 1.7(a).

-l-*-t(se c)

FIG U RE 1.7 (a) Square-wave current signal; (b) q{t) equal to the integral of i{t) given in (a).


S o lutio n

From Equation 1.3, for t>Q, q{t)=p{T)clT

Thus, q{t) is the running area under the i{t) versus t curve. Since i{t) is piecevv'ise constant, the integral is piecewise linear because the area either increases or decreases linearly with time, as shown in Figure 1.7(b). Since q{t) is constant for ^ > 3, the total charge transported is Q = q{5) = 3 C.

Exercise. If the current flow through a cross section o f conductor is i{t) = cos(120jtf) A for ? > 0 and 0 otherwise, find q{t) for t>Qi. AN SW ER: q{t)

‘

120jt

C for r > 0

Exercise. Suppose the current through a cross section o f conductor is given in Figure 1.8. Find q{t) for t > 0 .

FIGURE 1.8 AN SW ER; q(t) =

C for 0 <

1; q{t) = IC for r > I

T Y P ES OF C U R R EN T There are two very important current types: direct current (do) and alternating current (ac). Constant current (i.e., dqldt = / is constant) is called direct current, which is illustrated graphi cally m Figure 1.9(a). Figure 1.9(b) shows an alternating current, generally meaning a sinusoidal waveform, i.e., current o f the form y4sin(w? + ()>), where A is the peak magnitude, co is the angu lar frequency, and (|) is the phase angle o f the sine wave. W ith alternating current, the instanta neous value o f the waveform changes periodically through negative and positive values, i.e., the

ch a p ter 1 • Charge, Current, Voltage and O h m s Law

direction o f the current flow changes regularly as indicated by the + and - values in Figure 1.9(b). Household current is ac. Lastly, Figure 1.9(c) shows a current that is neither dc nor ac, but that nevertheless will appear in later circuit analyses. There are many other types o f waveforms. Interestingly, currents inside com puters, C D players, TV s, and other entertainment devices are typically neither dc nor ac.

i(t) (A)

t(sec) -H ----------------------1 -

-I-----►

3 (a)

F IG U R E 1.9 (a) Direct current, or dc; i{t) = Iq\ (b) alternating current, or ac;

i{t) = 1 2 0-^ sin (1 2 0 ?) A; (c) neither ac nor dc.

10


Because the value o f an ac waveform changes with time, ac is measured in different ways. Suppose the instantaneous value o f the current at time t is A!sin(ci)i- + (j>). The term peak value refers to K in K sin(co? + (j)). The peak-to-peak value is 2K. Another measure o f the alternating current, indicative o f its heating effect, is the root mean square (rms), or effective value. The rms or effec tive value is related to the peak value by the formula

rms =

X peak-value = Q .lO llK

(i.4)

A derivation o f Equation 1.4 with an explanation o f its meaning will be given in Chapter 11. A special instrument called an ammeter measures current. Some ammeters read the peak value, whereas some others read the rms value. One type o f ammeter, based on the interaction between the current and a permanent magnet, reads the average value o f a current. From calculus, Fave! the average value o f any function y(^), over the time interval [0, 7] is given by

(1.5) For a general ac waveform, the average value is zero. However, ac signals are often rectified, i.e., converted to their absolute values, in power-supply circuits. For such circuits, the average value o f the rectified signal is important. From Equation 1.5, the average value o f the absolute value o f an ac waveform over one complete cycle with T = 2jt/co, is

K ^ 2.K Average Value = —^\s,m{wt)\dt = ----- J 0

2K -cos{(ot) T

(O

0.5T

sin(cot)clt

^ 0

2,K — = 0.636K jt

( 1.6)

i.e., 0 .636 X peak value.

Exercise. Suppose i{t) - 169.7 sin(50jtr) A. Find the peak value, the peak-to-peak value, the rms value o f i{t), and the average value o f AN SW ER: 169.7, 339.4, 120, and 107.93 A, respectively

3. VO LTAG E W hat causes current to flow? An analogous question might be. W hat causes water to flow in a pipe or a hose? W ithout pressure from either a pump or gravity, water in a pipe is still. Pressure from a water tower, a pressured bug sprayer tank, or a pump on a fire truck will force water flow In electrical circuits, the “pressure” that forces electrons to flow, i.e., produces a current in a wire or a device, is called voltage. Strictly speaking, water flows from a point o f higher pressure— say, p o in ts — to a point o f lower pressure— say, point 5 — along a pipe. Between the two points and B, there is said to be a pressure drop. In electrical circuits, a voltage drop from point A to point B

11


along a conductor will force current to flow from point A to point B; there is said to be a voltage drop from point A to point B in such cases. Gravity forces the water to flow from a higher elevation to a lower elevation. An analogous phe nomenon occurs in an electric field, as illustrated in Figure 1.10(a). Figure 1.10(a) shows two con ducting plates separated by a vacuum. O n the top plate is a fixed amount o f positive static charge. On the bottom plate is an equal amount o f negative static charge. Suppose a small positive charge were placed between the plates. This small charge would experience a force directed toward the negatively charged bottom plate. Part o f the force is due to repulsion by the positive charges on the top plate, and part is due to the attraction by the negative charges on the bottom plate. This repulsion and attraction marks the presence o f an electric field produced by the opposite sets o f static charges on the plates. The electric field indicated in Figure 1.10 sets up an “electric pressure” or voltage drop from the top plate to the bottom plate, which forces positive charges to flow “downhill” in the way that water flows from a water tower to your faucet. Unlike water flow, negative charges are forced “uphill” from the negatively charged bottom plate to the positively charged top plate. As men tioned in the previous section, this constitutes a net current flow caused by the bilateral flow o f positive and negative charges. The point is that current flow is induced by an electric pressure called a voltage drop.

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© © © © © © © © A 0

A

Positive

negative

charge, q Electric Field

Force on

Electric Field

charge q

charge Negative

B © charge,-q

B © © © 0

Force on

© © © © (a)

© © © © © © © © (b)

FIGURE 1.10 (a) Positive charge in a (uniform) electric field; (b) negative charge in a uniform elearic field. As mentioned, in Figure 1.10, the positive charge ^ at ^ tends to move toward B. We say, quali tatively, that point A in the electric field is at a higher potential than point B. Equivalently, point 5 is at a lower potential than point A. An analogy is now evident: a positive charge in an electric field “falls” from a higher potential point to a lower potential point, just as a ball falls from a high er elevation to a lower elevation in a gravitational field. Note, however, that if we turn the whole setup o f Figure 1.10(a) upside down, the positive charge q still moves from point A to point B, an upward spatial movement. Similarly, if a negative charge - q is placed at B, as in Figure 1.10(b), then the negative charge experiences an upward-pulling force, moving from the lower potential, point B, to the higher potential, point A.

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Chapter 1 ® Charge, Current, Voltage and Ohms Law

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Again, consider Figure 1. 10(a). As the charge q moves from point ^ toward B, it picks up veloci ty and gains kinetic energy. Just before q hits the bottom plate, the kinetic energy gained equals the (constant) force acting on q multiplied by the distance traveled in the direction o f the force. The kinetic energy is proportional to q and to

the “distance traveled.” Therefore,

energy converted = kinetic energy gained oc q

n

The missing proportionality constant in this relationship is defined as the potential difference or voltage between A and B, The term “voltage” is synonymous with “potential difference.” Mathematically, , . , voltage = potential difference =

energy converted magnitude of charge

( 1.8)

The standard unit for measuring potential difference or voltage is the volt (V). According to

Equation 1.8, i f 1 joule {]) o f energy is convertedfrom one form to another when moving 1 C o f charge from point K to point B, then the potential difference, or voltage, between A and B w i VTIn equation form, with standard units of V, J, and C, we have 1V = 1 ^

(1.9)

O

The use of terms such as “elevation diflFerence,” “energy converted,” “potential difference,” or “voltage” implies that they all have positive values. If the word “difference” is changed to “drop” (or to “rise”), then potential drop and elevation drop have either positive or negative values, as the case may be. The following four statements illustrate this point in the context of Figure 1.10: The voltage between (or across) A a n d 5 is 2 V. The voltage between (or across) B and A is 2 V. ' The voltage drop from A to B is 2W.

{

•The voltage drop from B to A Is - 2 V.

.

.

.

.

’

This discussion describes the phenomena of “voltage.” Voltage causes current flow. But what pro duces voltage or electric pressure? Voltage can be generated by chemical action, as in batteries. In a battery, chemical action causes an excess of positive charge to reside at a terminal marked with a plus sign and an equal amount of negative charge to reside at a terminal marked with a negative

^

sign. When a device such as a headlight is connected between the terminals, the voltage causes a current to flow through the headlight, heating up the tiny wire and making it “Ught up.” Another source of voltage/current is an electric generator in which mechanical energy used to rotate the shaft of the generator is converted to electrical energy using properties of electro-magnetic fields.

^

All types of circuit analysis require knowledge of the potential difference between two points, say

^ ^

A and B, and specifically whether point A or point 5 is at a higher potential. To this end, we speak of the voltage drop from point A to point B, conveniently denoted by a double-subscript, as Vj^. If the value of is positive, then point ^ is at a higher potential than point B. On the other hand, if is negative, then point 5 is at a higher potential than point A. Since stands for the voltage drop from point B to point A,

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The double-subscript convention is one o f three methods commonly used to unambiguously specify a voltage drop. Using this convention requires labeling all points o f interest with letters or integers so that

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sense. A second, more-common convention uses + and - markings

on two points, together with a variable or numerical labeling o f the voltage drop from the point marked + to the point marked - . Figure 1.11 illustrates this second convention, where Vq denotes the voltage drop from A (marked +) to B (marked - ) . If Vq is positive, then ^4 is at a higher potential than

B. O n the other hand, if Vq is negative, then 5 is at a higher potential than A. The value o f Vq, togeth er with the markings + and

stipulates which terminal is at a higher potential; neither alone can do

this. For a general circuit element, the (+, —) markings— that is, the reference directions— can be assigned arbitrarily. A third method for specifying a voltage drop, using a single subscript, will be dis cussed in Chapter 2. B -I-

V„

FIGURE 1.11 The + and - markings establish a reference direction for voltage drop. For accuracy, always place the (+, - ) markings reasonably close to the circuit element to avoid uncertainty. The following example illustrates the use o f the double subscript and the (+, - ) markings for des ignating voltage drops. E X A M P L E 1.4 Figure 1.12 shows a circuit consisting o f four general circuit elements, with voltage drops as indi cated. Suppose we know that

= 4 V, and

= 9 V. Find the values o f

V^q and

CD-

V

-I-

3V

FIG U RE 1.12 Arbitrary circuit elements for exploring the use of (+, - ) for specifying a voltage drop.


14

S o lutio n

T he meaiiing o f the double subscript notation and the (+, - ) markings for a voltage imply that

'DA

^ 5 C = 3I V ^CZ> = - ^ Z )C = -(-2 ) = 2 V

Exercise. In Figure 1.12, find

and Vp.^-

A N S W E R :- 3 V ; - 2 V

Exercise. T he convention o f the (+, - ) markings is commonly used as described. Figure 1.13 shows an old 12-V automobile battery whose (+, - ) markings cannot be seen because o f the corrosion o f the terminals. A digital voltmeter (DVM ) is connected across the terminals, as shown. The display reads -1 2 V. Figure out the (+, - ) marking o f the battery terminals. A N SW ER: left terminal, right terminal, +

DVM

12V battery

FIG U RE 1.13 Digital voltmeter connected to a 12-V (car) battery whose plus and minus markings have corroded away.

One final note: As with current, there are different types o f voltages— dc voltage, ac voltage, and general voltage waveforms. Figure 1.9, with the vertical axis relabeled as v{t), illustrates different voltage types.

4. C IR CU IT ELEM EN TS Circuits consist o f interconnections o f circuit elements. The most basic circuit element has two terminals, and is called a two-terminal circuit element, as illustrated in Figure 1.14. A circuit eie-

Chapter 1 • Charge, Current, Voltage and O h m s Law

15

ment called a source provides either voltage, current, or both. The battery is a very common source, providing nearly constant voltage and the usually small current needed to operate small electronic devices. Car batteries, for example, are typically 12 volts and can produce large currents during starting. The wall outlet in a home can be thought o f as a 110 -volt ac source. Figure 1.14(a) shows a (battery) voltage across a general undefined circuit element. A current z(r) flows through the element. Recall from our earlier intuitive discussion that voltage is analogous to water pressure: pressure causes water to flow through pipes; voltage causes current to flow through cir cuit elements. Total water into a pipe equals total water out o f the pipe. Analogously, the current

entering a two-terminal device must, by definition, equal the current leaving the two-terminal device.

Current

FIGURE 1.14 (a) General circuit element (connected to a battery) as an energy or signal processor: v(i) is the voltage developed across the circuit element, and z'(r) is the current flowing through the circuit element; (b) practical example of a general circuit element (car headlight) connected to a car battery. The circuit element o f Figure 1.14(a) has a specific labeling: the current i(f) flows from the plus terminal to the minus terminal through the circuit element. Such a labeling o f the voltage-current reference directions is called the passive sign convention. In contrast, the current iij) flows from the minus terminal to the plus terminal through the battery; this labeling is conventional for sources but not for non-source circuit elements. In addition to sources, there are other common two-terminal circuit elements: • The resistor •

The capacitor

•

The inductor

For a resistor, the amount o f current flow depends on a property called resistance; the smaller the resistance, the larger the current flow for a fixed voltage across the resistor. A small-diameter pipe offers more resistance to water flow than a large-diameter pipe. Similarly, different types o f con ductors offer different resistances to current flow. A conductor that is designed to have a specific resistance is called a resistor. If the device is an ideal resistor, then v(f) = Ri{i), where i? is a con stant o f resistance. More on this shortly. The circuit elements called the capacitor and the inductor will be described later in the text. Also, future chapters will describe the operational amplifier and the transformer that are circuit elements having more than two terminals.

16


5. VO LTAG E, CURRENT, POW ER, ENERGY, RELATIONSHIPS The relationship between voltage across and current through a two-terminal element determines whether power (and, thus, energy) is delivered or absorbed. The heating element in an electric oven can be thought o f as a resistor. The heating element absorbs electric energy and converts it into heat energy that cooks, among other things, turkey dinners. In Figure 1.14(a), a battery is connected to a circuit element. Figure 1.14(b) concretely illustrates this with a 12-V car battery connected to a headlight. W ith reference to Figures 1.14(a) and 1.14(b), suppose v{t) = 12 V, and i{t) = 5 A: 5 A o f current flows through the headlight. The head light converts electrical energy into heat and light. Power (in watts) is the rate at which the ener gy is converted. At each instant o f time, the electrical power delivered to (absorbed by) the head light is pit) = v[t)i{t) - 12 X 5 = 60 watts. Similarly, at each instant o f time, the battery can be viewed as delivering 60 watts o f power to the headlight. Inside the battery, the stored potential energy o f the chemicals and metals undergoes a chemical reaction that produces the electrical potential difference and the current flow to the headlight: chemical energy is converted into elec trical energy that is converted into light and heat. Figure 1.15 depicts a more general scenario: a circuit element is connected to its surrounding cir cuit at points A and B. (One, o f course, could imagine that the “remainder o f circuit” is a battery, and circuit element 1 is a headlight.) Suppose there is a constant voltage drop from A to B, denot ed by

Also assume that a constant current

flows from terminal A to terminal B through

circuit element 1, as shown.

FIG U RE 1.15 A general circuit in which a two-element circuit element is extracted and labeled according to the passive sign convention. For discussion purposes, assume

> 0 and

> 0. During a time interval o f T s, (V^g x T) C o f

charge moves through circuit element 1 from A to B. In “falling” from a higher potential, point A, to a lower potential, point B, the charge loses electric potential energy. The lost potential energy is con verted within element 1 into some other form o f energy— heat or light being two o f several possibil ities. According to Equation 1.8, the amount o f energy converted {absorbed by the element) is y. T) >Q. The power absorbedhj element 1 is, by definition, the rate at which it converts or absorbs energy. This rate equals

â b (âb ^ T) ■^Vab I a b > 0 . T


17

Exercise. In Figure 1.15, the current ÂB - 5 niA, and

= 400 V. W hat is the energy absorbed by circuit element 1 in one minute? W hat is the power absorbed by circuit element 1? AN SW ER: W = 120 J; P = 2 watts

W ith respect to Figure 1.15, for constant (direct) voltages and currents, we arrive at a very simple relationship:

P\-V ab I ab where

0 -1 0 )

is the power (in W ) absorbed by the circuit element. Consequently, the energy, W , (in

J), absorbed during the time interval Tis

W^=P\xT

(1-11)

Now, let us reconsider Figure 1.15. One can think o f-/ ^ g as flowing from A w B through the remainder o f the circuit. In this case,

-1 ^ ^ ^ ^ < 0 . This means that the remain

der o f the circuit absorbs negative power or equivalently delivers |^ 5 (—

| = â ÂB

circuit

element 1. As such, the remainder o f the circuit is said to generate electric energy. By definition, the electric power generated by the remainder o f the circuit is the rate at which it generates elec tric energy. From Equation 1.8, this rate equals --------- ---------- - ^ ab âb Observe that the rate at which the remainder o f the circuit generates power precisely equals the rate at which circuit element 1 absorbs power. This equality is called the principle o f conserva tion o f power: total power generated equals total power absorbed. Equivalently, the sum o f the powers absorbed by all the circuit elements must add to zero,

Exercise. In Figure 1.15, -Pg =

+ Pq = y^gl^B

ÂB^^ÂB^ ~

watts, i.e., the remainder o f the circuit absorbs - 1 0 watts o f

power. How much power does circuit element 1 absorb? A N SW ER: 10 watts

In general, whenever a two-terminal general circuit element is labeled according to the passive sign convention, as in Figure 1.15, then P = whereas

P = V^b âb ^

> 0 means the element absorbs (positive) power,

absorbs negative power or delivers (positive) power to whatever it

is connected. As a general convention, non-source circuit elements are labeled according to the passive sign convention. Usually, sources are labeled with the current leaving the terminal labeled with “+”. For such labeling o f sources, if the product o f the source voltage and the current leaving the “+” terminal is positive, then the source is delivering power to the network.

Chapter 1 • Charge, Current, Vohage and O hm ’s Law

RULE FOR C A LC U LA TIN G A B SO R B ED PO W ER The power absorbed by any circuit element (Figure 1.16) with terminals labeled A and B is equal to the voltage drop from A m B multiplied by the current through the element from A to 5,

+

V AB FIGURE 1.16

Exercise. Compute the power absorbed by each o f the elements in Figure 1.17. -1A

_____________

-2A

_____________

2A

>

< Z3

10V

10V

10V

(a)

(b)

(c)

FIGURE 1.17 AN SW ER: (a) 10 W; (b) - 2 0 W; (c) 20 W

As mentioned, power is the rate o f change o f work per unit o f time. T he ability to determine the power absorbed by each circuit element is highly important because using a circuit element or some device beyond its power-handling capability could damage the device, cause a fire, or result in a serious disaster. This is why households use circuit breakers to make sure electrical wiring is not overloaded.

Exercise. In Figure 1.18, a car heater is attached to a 12-volt D C voltage source. How much power can the car heater absorb before the 20 -amp fuse blows. 20 Amp Fuse

FIGURE 1.18 Car heater connected to a 12-volt car battery through a 20-amp fuse. A N SW ER: 240 watts

19


As mentioned earlier, the calculated value o f absorbed power P may be negative. If the absorbed power P is negative, then the circuit element actually generates power or, equivalently, delivers power to the remainder o f the circuit. In any circuit, some elements will have positive absorbed powers, whereas some others will have negative absorbed powers. If one adds up the absorbed powers o f ALL elements, the sum is zero! This is a universal property called conservation o f power.

PRIN C IPLE OF CO N SERVA TIO N OF PO W ER The sum o f the powers absorbed by all elements in a circuit is zero at any instant o f time. Equivalently, the sum o f the absorbed powers equals the sum o f the generated powers at each instant o f time.

The 2"*^ edition o f this text contains a rigorous proof o f this principle. For the present, we will simply use it to solve various problems. The following example will help clarify the sign conven tions and illustrate the principle o f conservation o f power. E X A M P L E 1.5 Light bulbs come in all sorts o f shapes, sizes, and wattages. W a t t l e measures the power consumed by a bulb. Typical wattages include 15, 25, 40, 60, 75, and 100 W. Power consumptions differ because the current required to light a higher-wattage (and brighter) bulb is larger for a fixed out let voltage: a higher-wattage bulb converts more electric energy into light energy. In Figure 1.19, the source delivers 215 watts o f power. W hat is the wattage o f the unlabeled bulb?

7?

100V

’ watts

watts

watts

FIG U RE 1.19. Three bulbs connected to a 100-V battery. S o lutio n

From conservation o f power, the total power delivered by the battery equals the total power absorbed by all the bulbs. Therefore, the power absorbed by the unknown bulb is 215 - 4 0 - 100 = 75 watts

Exercise. Determine the current / leaving the battery in Example 1.5. AN SW ER: 2.15 amps

20


EXA M PLE 1.6 An electroplating apparatus uses electrical current to coat materials with metals such as copper or silver. In Figure 1.20, suppose a 2 2 0 -V electrical source supplies 10 A dc to the electroplating apparatus.

10A Electroplating Apparatus

FIGURE 1.20 Electrical source operating an electroplating apparatus. a) b)

W hat is the power consumed by the apparatus? If electric energy costs 10 cents per kilowatt-hour (kW h), what will it cost to operate the apparatus for a single 12 -h day?

S o lutio n

Step 1. From Equation 1.10, the power consumed is /> = 220

X

10 = 2 2 0 0 W, or 2.2 kW

Step 2. According to Equation 1.11, the energy consumed per 12-h period is 2.2

X

12 = 26.4 kWh

Step 3. Therefore, the cost to operate is 26 .4

X

.01 = $ 2.64 / day

Exercise. Suppose the electroplating apparatus o f Example 1.6 draws 12 A D C at the same volt age. W hat is the cost o f operation for a single 12-h day? W hat is the cost o f operating for a 20 workday month? AN SW ER: $3,168; $63.36

E X A M PLE 1.7 Each box in the circuit o f Figure 1.21 is a two-terminal element. Compute the power absorbed by each circuit element. W hich elements are delivering power? Verify the conservation o f power prin ciple for this circuit.

Chapter 1 • Charge, Current, Voltage and O hm ’s

21

FIG U RE 1.21 Circuit containing several general circuit elements. S o lutio n Step 1. Compute power absorbed by each element. Using either Equation 1.10 or the power con

sumption rule, the power absorbed by each element is

a)

For element 1 P i = 4 X 1 = 4 W

b)

For element 2 P l = 8 x 2 = 1 6 W

c)

For element 3 ^ 3 = 10 X 1 = 10 W

d)

For element 4

e)

For element 5 P 5 = 2

0

For element 6 Pe = 1 0 X ( - 2 ) = - 2 0 W

14 x

x

(-1)=-14W

2 = 4W

Step 2 . Verify conservation o f power. Since P 4 and Pg are negative, element 4 delivers 14 W, and

element 6 delivers 20 W o f power. T he remaining four elements absorb power. Observe that the sum o f the six absorbed powers, 4 + 16 + 10 - 14 + 4 - 2 0 = 0, as expected from the principle o f conservation o f power. Equivalently, the total positive generated power, (14 + 20) = 34 W, equals the total positive absorbed power, (4 + 16 + 10 + 4) = 34 W.

Exercise. In Figure 1.22, find the powers absorbed by elements 1, 2, and 3.

FIG U RE 1.22 AN SW ER: 8 W, 20 W, - 2 8 W; element 3 equivalently delivers 28 W


22

Exercise. In Figure 1.22, suppose the current 2 A were changed to - 4 A. W hat is the new power absorbed by element 3? A N SW ER: 56 watts

If the power absorbed by a circuit element is positive, the exact nature o f the element determines the type o f energy conversion that takes place. For example, a circuit element called a resistor (to be dis cussed shortly) converts electric energy into heat. If the circuit element is a battery that is being charged, then electric energy is converted into chemical energy within the battery. If the circuit ele ment is a dc motor turning a fan, then electrical energy is converted into mechanical energy.

N O N -D C PO W ER A N D EN ER G Y C A LC U LA TIO N S Consider Figure 1.23, where i{t) is an arbitrary time-varying current entering a general two-ter minal circuit element, and v{t) is the time-varying voltage across the element. Because voltage and current are functions o f time, the power p{t) = v{t)i{t) is also a function o f time. For any specific value o f ^ = ?j, the value p{t^) indicates the power absorbed by the element at that particular time— hence, the terminology instantaneous power for p{t). i(t)

Circuit Elem ent Absorbing Power p(t)

FIGURE 1.23 Calculation of absorbed power for time-varying voltages and currents for circuit ele ments labeled with the passive sign convention; here, power is p{t) = v{t)i{t). Equation 1.12 extends Equation 1.10 in the obvious way.

p{t) = v{t)i{t)

( 1. 12)

i.e., the instantaneous (absorbed) power p{t), in W, is the product o f the voltage v{t), in V, and the current i{t), in A, with labeling according to the passive sign convention. This product also makes sense from a dimensional point o f view; , volts X amps =

joules coulombs joules ;— - x = coulomb second second

Knowing the power p{t) absorbed by a circuit element as a function o f t allows one to compute the energy W{tQ, t) absorbed by the element during the time interval [^q, t > Iq], W[tQ, t) (J) is the integral o f p{t) (W) with respect to t over [?q, t], i.e..


23

W(to,t)^ r p ir ) d r

where

the lower limit o f the integral, could possibly be -oo. For the dc case, p{t) = P (a con

stant). From Equation 1.13,

t t W(tQ,t) = f p ( r ) d T = P f d r = P(t-tQ) = P x T

where T = t - t^, as given in Equation 1.11. If, in Equation 1.13, tg = -oo, then W (-co, t) becomes a function only o f t which, for convenience, is denoted by

t W{t)= f p ( r ) d r L

(1.14)

W{t) = W{—00, t), in joules, represents the total energy absorbed by the circuit element from the beginning o f time to the present time rwhen p{t) is in watts.

Exercise, a) Suppose the power absorbed by a circuit element over [0,oo) is p{i) = W (0, oo). b) Now suppose the absorbed power o f the circuit element is p{t) =

j

j >0

watts. Find

for t > 0 .

•

A N SW ER: 4 J; (4+t) J

Since energy is the integral o f power, power is the rate o f change (derivative) o f energy. Differentiating both sides o f Equation 1.14 yields the expected equation for instantaneous power.

dW(t) v m o = P ( o = ^

(1.15a)

or, equivalently, for t > (q,

Exercise. Suppose that for t > 0 , the work done by an electronic device satisfies W{t) = 10(1 —

J- If the voltage supplied by the device is 10 V, then for t > 0, find the power and current supplied by the device, assuming standard labeling, i.e., the passive sign convention. AN SW ER: p{t) = \0e-‘ watts; i(f) = e'‘ A

24


EXA M PLE 1.8 In the circuit o f Figure 1.23, the current i{t) and vokage v{t) have the waveforms graphed in Figure 1.24. Sicetch p{t), the instantaneous power absorbed by the circuit element, and then sketch W(0,

t), the energy absorbed over the interval [0 , i\.

FIGURE 1.24 (a) Current and (b) voltage profdes with respect to t for circuit o f Figure 1.23. S o lutio n

A simple graphical multiplication o f Figures 1.24(a) and (b) yields the sketch o f the curves in instantaneous power shown in Figure 1.25(a). From Equation 1.13 with = 0, we have, for 0 <

t< % ,2

p(T)dr - J — c/t = — 0

and for t> 5, t

5

t

t

W ( 0 , t ) - J p(r)dT = J p{x)dT + J p(r)dT - 5 + J ' d T - 5 + ( t - 5 ) = l 0

0

Figure 1.25(b) presents the resulting graph.

5

5


25

(b) FIGURE 1.25 (a) Profile of the instantaneous power p{t) = v{t)i{t) for the current and voltage wave forms of Figure 1.24; (b) associated profde of energy versus time.

6. IDEAL VO LTAG E AND CU RREN T SO URCES Two-terminal circuit elements may be classified according to their terminal voltage-current rela tionships. The goal o f this section is to define ideal voltage and current sources via their termi nal voltage-current relationships. The wall socket o f a typical home represents a practical voltage source. After flipping the switch on an appliance plugged into a wall socket, a current flows through the internal circuitry o f the appliance, which, for a vacuum cleaner or dishwasher, converts electrical energy into mechanical energy. For modest amounts o f current draw (below the fuse setting), the voltage nearly maintains its nominal pattern o f 120 / 2 sin(120 lit) = 169.7 sin(120 nt) V. This practical situation is ide

26


alized in circuit analysis by the ideal voltage source symbol shown in Figure 1.26(a), a circle with a ± reference inside. The symbol is more commonly referred to as independent voltage source.

FIGURE 1.26 Equivalent representations of ideal voltage source attached to a hypothetical circuit. The waveform or signal v{t) in Figure 1.26 represents the voltage produced by the source at each time t. The plus and minus (+, —), on the source define a reference polarity. T he reference polari ty is a labeling or reference frame for standardized voltage measurement. T he reference polarity does not mean that v(t) is positive. Rather, the reference polarity (+, - ) means that the voltage drop from + to - is v{t), whatever its value/sign. Finally, the voltage source is ideal because it maintains the given voltage v{t), regardless o f the current drawn from the source by the attached circuit.

voltage (V) V,

1(A)

(b)

FIGURE 1.27 (a) Ideal battery representation of ideal voltage source; (b) v-i characteristic of ideal battery. Figure 1.2 7 (a) shows a source symbol for an ideal battery. The voltage drop from the long-dash side to the short-dash side is Vg, with Vjj > 0. In commercial products, the terminal marked with a + sign corresponds to the long-dash side o f Figure 1.27(a). An ideal battery produces a constant voltage under all operating conditions, i.e., regardless o f current drawn from an attached circuit or circuit element, as indicated by the v-i characteristic o f Figure 1.27(b). Real batteries are not ideal but approximate the ideal case over a manufacturer-specified range o f current requirements. Practical sources (i.e., non-ideal); voltage sources, such as commercial dc and ac generators; and real batteries deviate from the ideal in many respects. One important respect is that the terminal voltage depends on the current delivered by the source. The most common generators convert mechanical energy into electrical energy, while batteries convert chemical energy into electrical


27

energy. There are two general battery categories: nonrechargeable and rechargeable. A discussion o f the dramatically advancing battery technology is beyond the scope o f this text. Besides batteries and ideal voltage sources, devices called ideal or independent current sources maintain fixed current waveforms into a circuit, as illustrated in Figure 1.28. T he symbol o f an ideal current source is a circle with an arrow inside, indicating a reference current direction. An ideal current source produces and maintains the current i{t) under all operating conditions. O f course, the current i{t) flowing from the source can be a constant (dc), sinusoidal (ac), or any other time-varying function.

FIGURE 1.28 Equivalent ideal current sources whose current i{t) is maintained under all operating conditions o f the circuit. In nature, lightning is an example o f an approximately ideal current source. W hen lightning strikes a lightning rod, the path to the ground is almost a short circuit, and very little voltage is developed between the top o f the rod and the ground. However, if lightning strikes a tree, the path o f the current to the ground is impeded by the trunk o f the tree. A large voltage then develops from the top o f the tree to the ground. Independent sources have conventional labeling, as shown in Figure 1.29, which is different from that o f the passive sign convention. Here the source delivers power if p{t) = v{t)i{t) > 0 and would absorb power i f p{t) = < 0. A complicated circuit called a battery charger can deliver ener gy to a drained car battery. T he car battery, although usually a source delivering power, exempli fies a source absorbing power from the charger.

FIG U RE 1.29 Common voltage and current source labeling.

28


Another type o f ideal source is a dependent source. A dependent source or a controlled source produces a current or voltage that depends on a current through or voltage across some other ele ment in the circuit. Such sources model real-world devices that are used in real circuits. In the text, the symbol for a dependent source is a diamond. If a ± appears inside the diamond, it is a depend ent voltage source, as illustrated in Figure 1.30. If an arrow appears inside the diamond, it is a dependent current source, as illustrated in Figure 1.31. In Figure 1.30, the voltage across the dia mond-shaped source, v{t), depends either on a current, labeled through some other circuit device, or on the voltage across it. If the voltage across the source depends on the voltage v^, i.e., v{t) = p then the source is called a voltage-controlled voltage source (VCVS). If the volt age across the source depends on the current z^, i.e., v{t) = then the source is called a cur rent-controlled voltage source (CCVS).

FIGURE 1.30 The right element is a voltage-controlled voltage source (VCVS) if v{t) = (p is here dimensionless), or a current-controlled voltage source (CCVS) if v(t) = rî (r^ here has units of ohm).

Exercise. The voltage across a particular circuit element is element is

= 5 V, and the current through the

0.5 A, using the standard labeling.

a)

If a V CV S (Figure 1.30) with p = 0.4 were associated with the controlled-source branch, fmd vit).

b)

If a CCV S (Figure 1.30) with

= 3 £2 were associated with the controlled branch, fmd

v{t). ANSW ER: a) 2 V; b) 1.5 V

There is dual terminology for dependent current sources. The configuration o f Figure 1.31 shows a voltage-controlled current source (VCCS), i.e., i{t) = g^v^, or a current-controlled current source (CCCS), for which i{t) =

r~\

29


i(t) =

or Pi

Q V

or

X

Pi: -o FIG U RE 1.31 The right element is a voltage-controlled current source (VCCS) if i{t) =

(g^ has

units o f siemens) or a current-controlled current source (CCCS) if i{t) = |3/^ ((3 is dimensionless). Source voltages or currents are called excitations, inputs, or input signals. A constant voltage will nor mally be denoted by an uppercase letter, such as V, Vq, cally be denoted by /, /g, /p

V^, and so on. A constant current will typi

and so on. The units are volts, amperes, and so on. Smaller and larger

quantities are expressed by the use o f prefixes, as defined in Standard Engineering Notation Table 1.1.

Exercise. The voltage across a particular circuit element is ment is

= 5, and the current through the ele

= 0.5 A using the standard labeling.

a)

If a VCCS (Figure 1.31) with ^^ = 0.1 S were associated with the controlled-source branch, find

b)

If a CCCS (Figure 1.31) with P = 0.5 were associated with the controlled-source branch, find i{i).

i{i). AN SW ER; a) 0.5 A; b) 0.25 A

TABLE 1.1. Engineering Notation for Large and Small Quantities

i

w

Name

Prefix

Value

femto

f

10-15

pico

P

10-12

nano

n

10-9

micro

P

10-6

milli

m

10-3

kilo

k

103

mega

M

lO^’

g‘ga

G

109

tera

T

1012

30


7. RESISTANCE, O HM 'S LAW, AND POW ER (A REPRISE) Different materials allow electrons to move from atom to atom with different levels o f ease. Suppose the same dc voltage is applied to two conductors, one carbon and one copper, o f the same size and shape. Two different currents will flow. T he current flow depends on a property o f the conductor called resistance: the smaller the resistance, the larger the current flow for a fixed volt age. The idea is similar to water flow through different-diameter pipes (analogous to electrical con ductors): for a given pressure, a larger-diameter pipe allows a larger volume o f water to flow and, therefore, has a smaller resistance than a pipe with, say, half the diameter. A conductor designed to have a specific resistance is called a resistor. Hence, a resistor is a device that impedes current flow. Just as dams impede water flow and provide flood control for rivers, resistors provide a means to control current flow in a circuit. Further, resistors are a good approx imate model to a wide assortment o f electric devices such as light bulbs and heating elements in ovens. Figure 1.32(a) shows the standard symbol for a resistor, where the voltage and current ref erence directions are marked in accordance with xhie.passive sign convention. Figure 1.32(b) pic tures a resistor connected to an ideal battery.

I +

R V

-

(a) FIG U RE 1.32 (a) Symbol for a resistor with reference voltage polarity and current direction consistent with the passive sign convention; (b) resistor connected to an ideal battery. In 1827, Ohm observed that for a connection like that o f Figure 1.32(b), the direct current through the conductor/resistor is proportional to the voltage across the conductor/resistor, i.e., I = V. Inserting a proportionality constant, one can write

or, equivalently,

1 = — V —GV ^

(1 .1 6a)

V = R1 The proportionality constant R is the resistance o f the conductor in ohms. The resistance R meas ures the degree to which the device impedes current flow. For conductors/resistors, the ohm (Q) is the basic unit o f resistance. A two-terminal device has a 1-Q resistance i f a 1-V excitation causes

1-A o f current to flow. In Equation 1.16(a), the proportionality constant is the reciprocal o f R, i.e., G = HR, which is called the conductance o f the device. T he unit for conductance according to the International System o f Units (SI) system is the siemen, S. In the United States, the older term for the unit o f conductance is the mho ^5, that is, ohm spelled backward, which is still widely used. In this text, we try to adhere to the SI system. If a device or wire has zero resistance {R = 0) or infinite conductance {G = t»), it is termed a short circuit. On the other hand, if a device or wire has infinite resistance (zero conductance), it is called an open circuit. Technically speaking,


31

a resistor means a real physical device, with resistance being the essential property o f the device. In most o f the literature on electronic circuits, resistor and resistance are used synonymously, and we will continue this practice.

O H M 'S LAW Ohm’s law, as observed for constant voltages and currents, is given by Equation 1.16(b), with its equivalent form in Equation 1.16(a). However, it is true for all time-dependent waveforms exciting a linear resistor. Thus, we can generalize Equation 1.16 as v (0 = « W

or

i{t) = —v{t) = Gv{t)

(1.17b)

R

according to Figure 1.33, whose voltage-current labeling is consistent with the passive sign con vention. i(t) AO +

^ v(t)

----- OB -

FIG U R E 1.33 If either the voltage or the current direction is reversed, but not both, then Ohm’s law becomes v(t) = -Ri{t). As an aid in writing the correct v-i relationship for a resistor. Ohm’s law is stated here in words:

For a resistor connected between terminals A and B, the voltage drop from A to B is equal to the resistance multiplied by the current flowing from A to B through the resistor.

Exercise. Find the resistance R for each o f the resistor configurations in Figure 1.34. AN SW ER: (a) 12 Q ; (b) 3 Q; (c) 6 Q

-1A +

R 12V (a)

4A -

+

R 12V

-2A - -

(b)

R 12V

+

(c)

FIG U RE 1.34

Once the voltage and the current associated with a resistor are known, the power absorbed by the resistor is easily calculated. Assuming the passive sign convention, then combining Equation 1.12 for

32


power and Ohms law (Equation 1.17), the instantaneous absorbed power is 9

p{t) = v{t)i{t) = i' ^ { t ) R ^ ^ ^ R

(1.18a)

which for the dc case reduces to

Exercise. Find the power absorbed by each o f the resistors in Figure 1.35.

80 +

12V (a)

4A -

R 100 (b)

90 -

12V

+

(c)

FIGURE 1.35 AN SW ER: (a) 18 W; (b) 160 W; (c) 16 W

Equations 1.18(a) and (b) bring out a very important property; a resistor always absorbs power, dissipating it as heat. Intuitively speaking, electrons that flow through the resistor collide with other particles along the way. The process resembles the action in a pinball game: the pinball suecessively collides with various pegs as it rolls from a higher to a lower elevation. W ith each colli sion, part o f the electron’s kinetic energy is converted into heat as the voltage pressure continues to reaccelerate the electron. Electrical energy that is converted to heat or used to overcome friction is usually called a loss. Such losses are termed /-squared-i? {f-R) losses because o f the form o f Equation 1.18. On the other hand, a stove’s heating element purposely converts to heat as much electric energy as possible, in which case, the P-R loss is desirable. This heating effect also proves useful as the basis for the oper ation o f fuses. A fuse is a short piece o f inexpensive conductor with a very low resistance and a predetermined current-carrying capacity. When inserted in a circuit, it carries the current o f the equipment or appliances it must protect. W hen the current rises above the fuse rating, the gener ated heat melts the conducting metal inside the fuse, opening the circuit and preventing damage to the more-expensive appliance. Oversized fuses or solid-wire jumpers circumvent safe fuse oper ation by permitting unsafe operation at overload currents, with consequent electrical damage to the appliance that may cause overheating and fire. Resistance o f a conductor depends on the material and its geometrical structure. For a specific temperature, R is proportional to the length I o f a conductor and inversely proportional to its cross-sectional area A,

R= p^

(1.19)

^

33


where the proportionaUty constant p is the resistivity in ohm-meters (Q • m). T he resistivity o f copper at 2 0 °C is 1.7 x 10“^ Q •m. Table 1.2 lists the relative resistivities o f various materials with respect to copper. Table 1.2 Resistivities of Various Materials Relative to Copper. Silver

0.94

Chromium

1.8

Tin

6.7

Copper

1.00

Zinc

3.4

Carbon

2 .4 X 10^

Gold

1.4

Nickel

5.1

Aluminum

1.6

*The resistivity of copper at 20 degrees C is 1.7 x 10"^ Qxm. EXA M PLE 1.9 Sixteen-gauge (16 AWG) copper wire has a resistance o f 4 .0 9 4 Q for every 1,000 feet o f wire. Find the resistance o f 100 feet o f 16 AWG aluminum wire and 100 feet o f 16 AWG nickel wire. Then find the voltage across each wire and the power absorbed (given off as heat) by each wire if a 10A direct current flows through 100 feet o f each wire. S o lutio n

The resistivities o f aluminum and nickel wire relative to copper are 1.6 and 5.1, respectively. Hence, 100 feet o f aluminum/nickel wire has a resistance o f (aluminum) 1.6 x 0.4094 = 0.655 Q (nickel) 5.1 X 0.4094 = 2.088 Q Given a 10-A current flowing through 100 feet o f copper, aluminum, and nickel wire, Ohm’s law implies (copper) V = /?/ = 0 .4 0 9 4 x 10 = 4 .0 9 4 V (aluminum) V = RI = 0 .655 x 10 = 6.55 V (nickel) V = RI = 2 .088 x 10 - 2 0.88 V Finally, from Equation 1.18(b), the absorbed power given off as heat is (copper) P = V I ^ R I - = 0 .4 0 9 4 x 100 = 4 0 .9 4 W (aluminum) p = VI = RI^^ 0 .655 x 100 = 6 5.5 W (nickel) P = VI = RI~ = 2 .088 x 100 = 208.8 W Notice that every 100 feet o f 16 AWG aluminum wire would absorb 65.5 - 4 0.9 = 2 4 .6 W more power than copper. And nickel wire absorbs even more power: ^ ^

208.8

■ “ 4 0.94 times more power than copper per unit length. This absorbed power, given off as heat, is why nickel wire is used for heating elements in toasters and ovens.

34

Chapter 1 • Charge, Current, Vokage and O hm ’s Law

Exercise, (a) If a constant current o f 10 A flows through 1,000 feet o f (16 AWG) copper wire, how many watts o f heat are generated by the wire? (b) If the wire o f part (a) were changed to (16 AWG) aluminum, how many watts o f heat would be generated? AN SW ER; (a) 409.4 watts; (b) 65 5 .0 4 watts

Temperature also affects resistance. For example, light bulbs have a “cold” resistance and a “hot” resistance o f more importance during lighting. For most metallic conductors, resistance increases with increasing temperature— except carbon, which has a decrease in resistance as temperature rises. Since resistors absorb power dissipated as heat, they should have adequate physical dimen sions to better radiate the heat or there must be some external cooling to prevent overheating. EXA M PLE 1.10 T he hot resistance o f a light bulb is 120 Q. Find the current through and the power absorbed by the bulb if it is connected across a constant 90-V source, as illustrated in Figure 1.36. -O 90V —

90V

R =120Q -O -

FIG U RE 1.36 Light bulb and equivalent resistive circuit model. S o lutio n

Step 1. From Ohm’s law. Equation 1.16(a),

V 90 / = - = ----- = 0.75 A R

120

Step 2 . By Equation 1.18(b), the power absorbed by the lamp is P = 0.752

X

120 = 67.5 W

Step 3 . C/?eck conservation o f power. T he power delivered by the source is 90 x 0.75 = 67.5 W. Therefore, the power delivered by the source equals the power absorbed by the resistor. This ver ifies conservation o f power for the circuit.

Exercise. In Example 1.10, suppose the battery voltage is cut in half to 60 V. W hat is the power absorbed by the lamp? W hat is the power delivered by the battery? Repeat with the battery volt age changed to 120 V. AN SW ER; 30 watts; 120 watts

35


The following example illustrates power consumption for a parallel connection o f light bulbs. EXA M PLE 1.11 Figure 1.37 shows four automobile halogen Hght bulbs connected in parallel across a 12-V bat tery. Find the following; (a) The effective “hot” resistance o f each bulb (b) T he total power delivered by the source (c) After 700 hours o f operation, the current supplied by the source drops to 11.417 A. Discover which light bulb has burned out.

ijt )

27 watts

35 watts

50 watts

60 watts

12V

F IG U R E 1.37 Parallel connection of light bulbs. S o lutio n

(a) From Equation 1.18(b), P - VÎR,

12^ Rxiw -

27 144

^50W =

50

= 5.33D

= 2 .88 Q

144 ^ 3 5 W = ^ - 4 .1 1 4 Q 144

- ' 60

= 2.4 Q

(b) The power delivered by the source equals the sum o f the powers consumed by each bulb, which is 172 W. (c) Since the current supplied by the source has dropped to 11.417 A, then the power delivered by the source drops to P;„urcenew ~

^ 11.417 = 137 watts, which is 35 watts less than the ear

lier-delivered power o f 172 watts. Hence, the 35-watt bulb has gone dark.

Exercise. Repeat Example 1.11 (a) with the battery voltage changed to 48 V and a new set o f light bulbs whose operating voltage is 48 V. AN SW ER; 85.333 Q; R^c,^= 65.83 Q; R^q^ = 4 6 .0 8 Q; R(^^^= 38.4 Q..

E X A M PLE 1.12 W hen connected to a 120-volt source, halogen light bulb number 1 uses 40 watts o f power. W hen similarly connected, halogen light bulb 2 uses 60 watts o f power. (a) (b)

Find the hot resistance o f each bulb. If the two bulbs are connected in a series, as in Figure 1.38 and placed across the 120-V source, find the power absorbed by each bulb and the power delivered by the source, assuming the hot resistances computed in part (a) do not change.

36


(c)

Find the voltage

and V2 across each bulb.

b u ib l

120V

120 V ^ = i-

bulb 2

(a)

(b)

FIGURE 1.38 Series connection o f two light bulbs and equivalent resistive circuit model. S o lu tio n

Step 1. Find the hot resistances. The hot resistances o f each bulb are given by

120^

Vt

40

‘ bulb\

= 360 and R 2 =

120 "

= 240

Step 2. Find the current through each bulb, the power absorbed by each bulb, and the power delivered by the source. The circuit o f Figure 1.38(a) has the equivalent representation in terms o f resistanc es in Figure 1.38(b). By definition, in a nvo-terminal circuit element, the current entering each resistor equals the current leaving. Therefore, the current through each resistor in the series con nection is the same, and is denoted /. So the new power dissipated by each bulb/resistor is

^l,new —

Snd P2^new ~ ^2^

To calculate these values, we need to know I. By conservation o f power, the power delivered by the source is the sum o f the absorbed powers, i.e., ^ sou rce ~

X I — P\^new

^ 1,new ~

+ -^2^

Hence, dividing through by /, 120 = RÎ + R 2I = (Ri + R 2)I = 600/ Therefore,

;=™=o.2A 600

Hence, = 360 X 0 .2 ^ = 14.4 W ,

= 240 x 0.2^ = 9.6 W,

= 24 W

37


Step 3. Find voltages across each bulb. From Ohm’s law, Vl =7?,7 = 72 V and V2 = ^ 2 ^ = 48

Although involved, the solution o f this problem uses the definition o f a two-terminal circuit ele ment and conservation o f power to arrive at the result in a roundabout way. In Chapter 2, we can more directly arrive at the answers by using Kirchhoff’s voltage and current laws. A potential problem with series connections o f light bulbs is circuit failure. If one bulb burns out, i.e., the filament in the bulb open-circuits, then all other lights are extinguished. Parallel circuits continue to operate in the presence o f open-circuit failures and are easier to fix: only the unlit bulb must be replaced.

8. V-l CH A RA CTERISTICS OF IDEAL RESISTOR, CO N STA N T VO LTAG E, AN D CO N STA N T CU RREN T SO URCES The ideal (linear) resistor is a device that satisfies Ohm’s law. Ohm’s law is a relationship between the current through the linear resistor and the volt age across it. A graph o f this relation ship is known as the v-i characteristic o f the resistor. The ideal resistor stud ied in this chapter has the v-i charac

v(V )

teristic given in Figure 1.39. The slope o f the line in the v-i plane is the value o f the resistance. Recall that an ideal voltage source maintains a given voltage, irrespective o f the current demands o f the attached circuit. For constant-voltage sources, as shown in Figure 1.40(a), this property is depicted graphically by a constant hori

FIGURE 1.39 Linear resistor characteristic in which voltage is the constant times the current through the resistor.

zontal line (slope equals 0 ) in the v-i plane (Figure 1.40(b)). This means that the “internal” resistance o f an ideal voltage source is zero. Further, if

= 0, the voltage source looks like a short circuit because the current flow, generated

by the remaining circuit, will induce no voltage across the source. For now, we must be content with this brief discussion. Chapter 2 will reiterate and expand on these ideas.

38


V

+ V ( + )

•V-

/ Circuit

V

-►I (a)

(b)

FIG U RE 1.40 (a) Constant source

attached to circuit; (b) v-i characteristic

is a constant horizontal line in the v-i plane. Analogously, an ideal current source maintains the given current, irrespective o f the voltage requirements o f the attached circuit. For constant-current sources, as in Figure 1.41(a), this prop erty is depicted by a constant vertical line (infinite slope) in the v-i plane (Figure 1.41(b)). This means that an ideal current source has infinite “internal” resistance. Further, if

= 0, the current

source looks like an open circuit because no current will flow, regardless o f any voltage generated by the rest o f the circuit. Again, we must be content with this brief discussion until Chapter 2 reit erates and expands on the ideas.

V

+ Y

/ Circuit /

(a) FIG U RE 1.41 (a) Constant source

(b) attached to circuit;

(b) v-i characteristic is a constant vertical line in the v-i plane.

9. SUM M ARY Building on a simplified physics o f charge (coulombs), electric fields, and charge movement, this chapter set forth the notions o f current, i{t) or / for dc, and voltage, v{t) or V for constant volt ages. A rigorous treatment would require field theory and quantum electronics. More specifically, the notions o f current, current direction, voltage, and voltage polarity, a two-terminal circuit ele ment (the current entering equals the current leaving), the passive sign convention, power con sumption [pit) = v{t)i{t) assuming the passive sign convention], and dissipated energy (the inte gral o f power) were all defined. In general, we can say that every circuit element does one o f the following: • • • •

Absorbs energy Stores energy Delivers energy, or Converts energy from one form to another


39

T he chapter subsequently introduced ideal independent and dependent voltage and current sources: the voltage-controlled voltage source (VCVS), the current-controlled voltage source (CC VS), the voltage-controlled current source (VCCS), and the current-controlled current source (C C C S). A dependent source produces a voltage or current proportional to a voltage across or a current through some other element o f the circuit. The various types o f dependent sources are summarized in Table 1.4. TABLE 1.4 Summary of the Four Possible Dependent Sources.

VCVS (Voltage-Controlled Voltage Source, p is dimensionless)

ccvs (Current-Controlled Current Source,

is in

ohms)

V CC S (Voltage-Controlled Voltage Source, is in S)

CCCS (Current-Controlled Current Source, P is dimensionless)

-I-

40


The chapter keynoted a special two-terminal element, called a resistor, whose terminal voltage and current satisfied Ohms law, v(t) = Ri{t), where v{t) is the voltage in volts, R is the resistance in ohms, and i{t) is the current in amperes. The resistor, as defined in this chapter, is a passive ele ment, meaning that it always absorbs power,/>{;■) = v{t)i{t) = i?-{t)IR = Rp-{t) > 0 since R>Q. This absorbed power is dissipated as heat. Hence, the (passive) resistor models the heating elements in a stove or toaster oven quite well. In addition, the resistor models the hot resistance o f a light bulb. Throughout the text, the resistor will often represent a fixed electrical load. In a later chapter, we will discover that it is possible to construct a device with a negative resistance, R
Current

Coulomb Ampere (A) (C)

Voltage

Resistance

Conductance

Volt (V)

Ohm (Q)

S (Siemens) mhof3

Power

w a tt

= volt X

am p

Energy

Joule (J)

Throughout this chapter, a number o f examples illustrated the various concepts that were intro duced. Some simple resisrive circuits were analyzed. To analyze more complex circuits, one needs Kirchhoff’s voltage and current laws, which specify how circuit elements interact in a complex cir cuit. These basic laws o f circuit theory are set forth in the next chapter.

10. TERM S AND C O N C EPTS Alternating current: a sinusoidally time-varying current signal having the form A'sin(co?+(j)). Battery: a device that converts chemical energy into electrical energy, and maintains approxi mately a constant voltage between its terminals. Charge: an electric property o f matter, measured in Coulombs. Like charges repel, and unlike charges attract each other. Each electron carries the smallest known indivisible amount o f charge equal to - 1.6 x 10“ '^ Coulomb. Conductance: reciprocal o f resistance, with siemens (S) (or formerly, mhos) as its unit. Conductor: a material, usually a metal, in which electrons can move to neighboring atoms with relative ease. Conservation o f power (energy): the sum o f powers generated by a group o f circuit elements is equal to the sum o f powers absorbed by the remaining circuit elements. Current: the movement o f charges constitutes an electric current. Current is measured in Amperes. One Ampere means movement o f charges through a surface at the rate o f 1 Coulomb per second. Current source: a device that generates electrical current. Dependent (controlled) current source: a current source whose output current depends on the voltage or current o f some other element in the circuit.

Chapter 1 ®Charge, Current, Voltage and Ohms Law

41

'w '

Dependent (controlled) voltage source: a voltage source whose output voltage depends on the voltage or current of some other element in the circuit.

Direct current: a current constant with time. Ideal conductor: offers zero resistance to electron movement. Ideal insulator: offers infinite resistance to electron movement. Independent (ideal) current source: an ideal device that delivers current as a prescribed function of time, e.g., {2 cos(/) + 12}A, no matter what circuit element is connected across its ter minals.

Independent (ideal) voltage source: an ideal device whose terminal voltage is a prescribed func tion of time, e.g., {2 cos{t) + 12}V, no matter what current goes through the device. Instantaneous power: the value of p{t) = at a particular time instant. Insulator: a material that opposes easy electron movement. Mho: historical unit of conductance equal to the reciprocal of an ohm. Ohm: unit of resistance. One ohm equals the ratio of IV to lA. Ohm’s law: for a linear conductor, the current through the conductor at any time t is proportional to the voltage across the conductor at the same time.

Open circuit: connection of infinite resistance or zero conductance. Passive sign convention: voltage and current reference directions, indicated by +, - , and an arrow, which conform to that shown in Figure 1.15.

Peak-to-peak value: equals 2 K 'm K sin(co^ + (()) of the ac waveform. Peak value: refers to K m K sin(cor + (|)) of the ac waveform. Power: rate of change of work per unit of time. Resistance: for a resistor, v{t) a i{t). The proportionality constant R is called the resistance, i.e., v{i) = Ri{t). Resistance is measured in ohms: 1 ohm means the voltage is 1 V when the current is 1 A.

Resistivity: the resistance of a conductor is proportional to its length and inversely proportional to its cross-sectional area. The proportionality constant p is called the resistivity of the material. The resistivity of copper at 2 0 ^C is 1.7 x 10~^ ohm-meters.

Resistor: physical device that obeys Ohms law. There are commercially available nonlinear resis tors that do not obey Ohms law. Resistors convert electric energy into heat.

Root mean square (rms) or eflfective value: measure of ac current, which is related to the peak value by the formula rms = 0.7071if, where K sin(o)^ + (|)) is the ac waveform. Short circuit: connection of zero resistance or infinite conductance. Siemens: unit of conductance (formerly, mho) or inverse ohms. v-i characteristic: graphical or functional representation of a memoryless circuit element. Voltage (potential difference): positive charge, without obstruction, will move from a higher potential point to a lower potential point, accompanied by a conversion of energy. Voltage is measured in volts; 1 volt between two points A and B means that the energy converted when moving 1 Coulomb of charge between A and B is 1 joule.

V oltê source: device that generates an electric voltage or potential difference. Wattage: measure of power consumption.

42


PROBLEMS

AN SW ER: (a) -0 .1 2 1 3 C; (b) 121,3 A; (c) 3.75 X

10^'; (d) 1 - exp(-5^) for ? > 0 and 0 for ^ <

0 , from left to right; (e) line segments joining

C H A R G E A N D C U R R EN T PRO BLEM S

(0,0), (0,1), (2,1), (2,-1), (5,-1), (5,1), (6,1), (6,2), (7,2), (7,-2), (8,-2).

1.Consider the diagram o f Figure P I.la . (a) (b)

(c)

(d)

Determine the charge on 7 .5 7 3 X

10 ^^ electrons. If this number o f electrons moves uni formly from the left end o f a wire to the right in 1 ms (milli second), what current flows through the wire? How many electrons must pass a given point in 1 minute to produce a current o f 10 Amperes? If the charge profile across the cross-sec tion o f a conductor from left: to riglit is given by q{t) = t+ 0.2e'5^- 0.2 C for t > 0 and zero for ? < 0 , plot the profile o f the current that flows across the botmdary. In

2. For the following questions, draw diagrams whenever necessary. (a) (b)

Determine the charge on 6 .023 X lO '^ electrons. If this number o f electrons moves uni formly from the left end o f a wire to the right in 1 ms (milli second), what current flows through the wire?

(c)

(d)

How many electrons must pass a given point in 1 minute to produce a current o f 5 Amperes? The charge profile residing in a vol ume V = 10 cm^ is given by q(i) = t + 0.5 sin( 7i:^) C for t> 0 and zero for t <

what direction would the current flow?

0. Plot the current that flows across the boundary o f the volume for 0 < ? < 2 sec. In what direction would the current flow at ? = 1 second? Explain. 3. Reconsider Figure 1.5 in the text in which changed to ijyf) and

is

is changed to z'^(?). Suppose

(i)

Positive charge carriers move from left

(ii)

Negative charge carriers move from

to right at the rate o f 2cos(10z-) C/s right to left at the rate o f 6 cos( 10 ?) C/s Repeat part (d) for the charge wave

(a)

Find ij^t) and i^(?) as functions o f time.

form (in coulombs) sketched in Figure P .l.lb .

(b)

Describe the charge movement on the wire at the boundaries A and B.

4. (a)

Suppose the charge transported across the cross section o f a conductor for t > 0 is q(t) = e'' sin( 12071?) C. Find the current, z(r), t> 0 ,flowing in the con ductor.

(b)

The charge crossing a boundary in a wire is given in Figure P I.4 for t > 0. Plot the current i(t) through the wire. See Example 1.2.

43


AN SW ER: Q = integral o f current. Hence, Q = 0.4 - 0.2 = 0.2. 7. (a)

(b)

Figure P 1.4 Charge crossing a hypothetical boundary. (c) The current in an ideal conductor is

5. (a)

(b)

given by i{t) = 5 - 3e'^^ - 2e‘^^ A for t > 0. Determine the charge transferred, q{t), as a function o f time for t > 0 . Repeat part (a) for the current plot sketched in Figure P I .5.

The current in an ideal conductor is given by i{t) = 2 - cos(2?) A for t > 0 and 0 for t < 0. Determine the charge transferred, q{i), as a function o f time for ? > 0 . Now suppose the charge transferred across some surface for ^ > 0 from left to right is q{i) = 2 - cos(2?) C. Find the current i(t) through the sur face for ^ > 0 from left to right. Repeat part (a) for the current plot sketched in Figure P I .7. Again, the current is zero for t< 0.

i(t) (A)

2--

-t (secs)

-

1- -

Figure P I.7 -

2- -

CHECK: (c) For ,6 > ? > 3 , g (0 = j - 4 t + 12

Figure P I.5

6 . A plot o f the current flowing past point A is shown on the graph o f Figure P I . 6 . Find the net positive charge transferred in the direction o f the current arrow during the interval 0 < ^<

8 . Find i(t) when the charge transported across a surface cutting a conductor is shown in Figure P I.8.

6 sec, in Coulombs. i (amps) 0.1 -t (sec) 10

-0.1--

t-*-t (sec)

Figure P I .6

Figure P I .8

44


9. (a)

Find the average value o f the voltage,

12.(a)

2k T = -

(b)

(0

Hint: See Equation 1.6. (b)

W hich elements in Figure P I. 12 are labeled with the passive sign conven tion'

v{t) = K cos(cof) over one period,

Find the average value o f the absolute

In the circuit o f Figure P I. 12, volt ages, currents, and powers o f some ele ments have been measured and indi cated in the diagram.

value o f the voltage, v(f) = K cos(cot) over one period, T = 2u/cl).

5A + 3V -

2V -

+ 5V -

-cz>

V O LTA G E, CURREN T, POW ER, EN ER G Y

“

10. In Figure P I. 10, suppose we know that

A

4A

E

+ 4V

Vab = 8 V and Vad = 18 V. Find the values of

il'

7A

3A

0 '

Figure P I. 12 (i)

If element A generates 28 power, find Va (ii) Find the power absorbed by ment B. (iii) I f element C generates 6 power, find Vq (iv) I f element D absorbs 2 7 power, find //).

6 V

+

- 4 V

-

W ele W W

(v) If element E absorbs 4 W power, Figure P I. 10 11. (a)

(b)

2A

W hich o f the three elements in Figure P I. 11 is labeled with the passive sign convention?

find I e (vi) Find the power absorbed by ele ment F. 13. Consider the circuit o f Figure P I. 13.

Find the absorbed powers for each cir cuit element in Figure P 1.11.

____________

2A

(a)

___________

20 V

lO V

(i)

(ii)

__ . ■

Find the power absorbed by the circuit elements 1 and 2 .

(b) Show that the algebraic sum o f the absorbed powers is zero. Be careful o f sign. 4A.

3A

2A

C _ 3 20 V

(iii) Figure PI. 11 A N SW ERS (b): (i) -40 W; (ii) 20 W; (iii) 60 W

6V

1A 3A

lOV

circuit element 1

+

circuit 6V element 2

16v Q ) 2 A lOV

Figure P I. 13

45


14. (a)

Determine the power absorbed by

+ 80V -

-25A

each o f the circuit elements in Figures (b)

P 1.14a below. Show that the algebraic sum o f the absorbed powers is zero. Be careftil o f sign.

(c)

Repeat parts (a) and (b) for die circuit of Figure P l.l4 b , where v^{t) = 3 for t>Q, Vjit) = 1 +

^^for t>Q, z'j(?)

for t>Q, and ijit) =

=2 +

for t

> 0. 9A

Figure P I. 15

14A|

+

circuit 15V element 1

4A 5A ( ^ ^ 2 0 V

lOA

6

+ 10V-

circuit element 2

5V

lOV

- 5V + 5A

(a)

100(1 - e * ) mA for t > 0. (a) How much energy does the element A absorb for the interval [0 , t] ? (b) I f element B is a 5 Q resistor, deter

circuit element 2 -v,(t) +

2A

3.334, Rfj = 20 in ohms.

16. In the circuit shown in Figure P I. 16, i(t) =

Figure P I. 14a

I

30

CHECK: Re

(c)

mine the power absorbed at time t, and the energy absorbed for the inter val [0 , t], W hat is the energy delivered by the source over the interval [0 , i\?

4V (b)

Figure P 1.14b 15. In the circuit o f Figure P I. 15, there are three independent sources and five ordinary resistors. (a) Determine which o f the circuit ele

(b)

ments are sources and which are resis tors. Determine the value o f the resistance for each o f the resistors.

25V

©

ti

Figure P I. 16 C H E C K : (b) 125 W, 1 2 5 tJ 17. Suppose energy cost in Indiana is 10 cents per kwh. (a) How much does it cost to run a 100watt T V set 8 hours per day for 30 days? (b) How many 100-watt light bulbs run for 6 hours a day are needed to use $9.00 o f energy every 30 days? A N SW ER: (a) 8 cents per day; $2.40 per month; (b) 5 bulbs

46

Chapter 1 • Charge, Current, Vohage and O hm ’s Law

RESISTA N CE 18. Using Equation 1.19 and Table 1.2 , find the resistance o f a nickel ribbon having these dimensions: length: width: thickness: 19. (a) (b) (c)

40 m 1.5 cm

0.1 cm

Compute the resistance o f800 feet o f 14gauge copper wire (2.575 Q /1000 ft). Repeat (a) for 200 feet o f 14-gauge nickel wire. If one end o f the copper wire is soldered to one end o f the nickel wire, find the total resistance o f the 1000 feet o f wire. Can you justify your answer?

20. The resistance o f a conductor is function of the temperature T (in °C ). Over a range o f temperature that is not too distant from 2 0 °C , the relationship between R{ T) and T is linear

Figure P I.21 22. For the circuit o f Figure P 1.22,

= 1Q , and

the input current to the circuit o f is i(^t) = 400sin(207i^) mA for ? > 0 and zero for t
the

instantaneous

power delivered by the source. Using a graphing program, graph the power delivered as a function o f time for 0 < ? < 0.5 s. (b) Now compute and graph an expres sion for the energy dissipated in the resistor as a function o f for 0 < ? <

0.5 s.

and can be expressed as /?(7) = i?(20)[l + a{T-

20 )] where a is called the temperature coeffi cient o f the conducting material. For copper a = 0.0039 per °C . If the resistance o f a coil of wire is 21 Ohms at -1 0 ° C, what is the resist ance when the wire is operating at 10° C? AN SW ER: 22.85 Q

A VERA G E VALU E, PO W ER, A N D EN ER G Y C A LC U LA TIO N S 2 1 . The current through a 500 £2 resistor is given in Figure P I .21 where /^ = 6 mA. (a)

(b)

How much total charge is transferred over the time interval ^ = 0 to ^ = 2 sec onds? How much total energy must a source deliver over the time interval t = 0 X.Ot

= 2 seconds? (c)

If /(?) in Figure P I.21 is periodic, with period equal to 2 seconds, i.e., the indi cated waveform is replicated every two seconds, find the average power absorbed by the resistor. Use intuitive reasoning.

Figure P I.22 23. The switch S in Figure P I .23 is assumed to be ideal, i.e., it behaves as a short circuit when closed, and as an open circuit when open. Suppose the switch is repeatedly closed for 1 ms and opened for 1ms. (a) W hat is the average value o f z(^)? (b) W hat is the average power delivered by the source?

Kt)>'

Figure P I.23 A N SW ERS: 0.25 mA. 1.25m W

47


24. Repeat problem 23 when the switch is repeatedly closed for 3 ms and opened for 1 ms.

28. In Figure P1.28, F q = 125 V.

(a)

Suppose bulb A and bulb B each use 100 watts o f power. Find /q and the

CH ECK:

= 1.875 m W

hot resistance o f each bulb. (b)

Suppose bulb A uses 40 watts o f power

25. In Figure P1.25, Vq = 10 V, and the switch S

and bulb B 60 watts o f power. Find /q

alternately stays at position A for 4 ms and at

and the hot resistance o f each bulb.

position B for 1 ms. Find the average value o f i(tj. 5kn

Figure P I.25 C H E C K : Average current is 1.8 mA.

A PPLIC A TIO N S OF O H M 'S LAW

ANSW ER: (b) /g = 0.8 A,

26 (a)

93.75 Q

W hat is the safe maximum current o f

a 0.25 W, 2 77 k fi metal film resistor used in a radio receiver? (b) W hat is the safe maximum current o f a 1 W, 130 £2 resistor? (c) W hat is the safe maximum current o f a 2 kW, 2 £2 resistor used in an electric power station? A N SW ERS: (a) 0.95 mA, (b) 87.7 niA. (c) 3 1.6 A

= 62.5 Q ,

=

29. T he power delivered by the source in the circuit o f Figure P I .29 is 750 watts.

(a)

If /^ = 5 A, determine the value o f R.

(b)

Suppose now that R = 11 Q. Find Hint: W hat is the power consumed in each resistor as a function o f /^?

27. In Figure P I .27, Kq = 120 V. (a)

T he power absorbed by the bulb in the circuit shown in Figure P I.2 7 a is Figure P I.29

60 watts. Find the value o f the hot resistance o f the bulb. (b)

The power delivered by the source to the parallel connection o f two identi cal bulbs in Figure 1.27b is 150 watts. Find the hot resistance o f each bulb.

AN SW ER: (a) 6 Q, (b) 4.33 A 30. Consider the circuit o f Figure P I.30. The power consumed by each resistor is known to be

^2Q ^

watts, PjQ = 48 watts, = 64 watts, PjQ = 3075.2 watts, and = 1944 watts.

8

(a) (b)

(a)

(b) Figure P I.27

AN SW ER: (b) 192 D.

For each resistor, determine the indi cated voltage or current. Determine the total power delivered by the two sources.

48


(ii) Determine the total power deliv ered by the battery.

20 + V, -

(iii) Determine the current, + V,-

3n

(iv) Assuming that each bulb behaves as a resistor, determine the hot

40

''" 6

60 <

>50

resistance o f each bulb. (b)

Figure P I . 3 0

C H E C K : (a)

blow the 15-amp fuse in the circuit o f

= 28 V, Vg = 108 V, (b) 5523.2

Figure P I.3 2 b .

(c)

31. The power absorbed by the resistor R in the circuit o f Figure P I.31 is 100 watts and

(a)

Find the value o f R.

(b)

Find the value o f the current flow ing through R and determine its direction as per the passive sign convention. Find the power absorbed by the 20 Q resistor.

Repeat (b) for C C bulbs. Ignition Switch

Ko = 2 0 V

(d)

Determine the number o f AA bulbs in parallel that would be required to

watts

(c)

deliv

ered by the battery.

12V

Find the power delivered by the source and the value o f /q.

Figure PI.32a

12V

Figure PI.32b 200

Figure P1.31. C H E C K : /q = 6 A. 32. (a)

Consider the circuit o f Figure P I.32a, which shows three lamps, AA, BB, and C C in a parallel circuit. This is a simplified example o f a light circuit on a car, in your house, or possibly on a Christmas tree. Halogen bulb AA uses 35 watts when lit, the Halogen Xenon bulb BB uses 36 watts when lit, and the incandescent bulb C C bulb uses 25 watts when lit. (i) Determine the current through each bulb.

AN SW ERS: (a) (i) 2 .9 167, 3, 2.08; (ii) 6.2 A; (iii) 96; (iv) 4 .1 1 ,4 , 5.76; (b) n > 6 ; (c) n > 8 33. An automobile battery has a terminal volt age o f approximately 12 V when the engine is not running and the starter motor is not engaged. A car with such a battery is parked at a picnic. For music, the car stereo is playing, using 240 watts, and some o f the lights are on using 120 watts. W ith this load, the battery will supply approximately 3 M J o f energy before it will have insufficient stored energy to start the car. (a) W hat power does the battery supply (b)

to the load? W hat current does the battery supply to the load?

49


(c)

Approximately how long can the car remain parked with the stereo and lights on and still start the car?

C H E C K : (c) 2.31 hours 34. In Figure P I .34, Vq = 24 V, i?, = 4 Q, the

Figure PI.35b

unknown resistance, Rj, consumes 20 watts o f power. Find

and Rj- (How many possible

D EP EN D EN T SO U RC E PRO BLEM S 36. Consider the circuit in Figure P I .36.

solutions are there?)

(a)

-A /V ^ R,

If 1/ = 6 V, find

and the power in

watts absorbed by the load R^. W hat is the power delivered by each source?

I'.

(b)

If the power absorbed by the load resistor is Pj^ = 80 watts, then find V^, /( and I/.

Figure P I.34 C H EC K : 1 A, 20 Q or ?????

R, = 20n

35. In Figure P I .35, Vq = 48 V. (a)

Determine the value(s) o f the current in the D C resistive circuit o f Figure P I.3 5 a

(b)

(c)

given

that

the

unknown

Figure P I.36

devices absorb the powers indicated. W hat value o f Vq results in a unique

C H EC K : (a)

solution for I J

(b) 1/ = 4 V

= 60 V,

= 3.6 watts;

If the circuit is modified as shown in Figure P I.35b , determine the rwo new values o f the current 1^.

37. For the circuit in Figure P I .37, determine and in terms o f /-^, , R2 and p.

AN SW ER: (a) 10, 2 A; (b) 35.77 V

Figure P1.37 38. Consider the circuit o f Figure P I .38. (a) Determine an expression for F igure P I . 3 5 a

and

the voltage gain

V G y = ^ Vin terms o f R^, R2 , a , and V-^^. (b)

If R.ÂQ. and a = 0.8, determine the

50

Chapter 1 « Charge, Current, Voltage and Ohms Law

(c)

value of J?2 so that the voltage gain Gy

41. For the circuit o f Figure P 1.41, suppose

= 4. Given your answer to (b), determine the power gain, which is the ratio of the power delivered to divided by

= 1 0 V. (a) (b)

the power delivered by the source.

Find the output voltage and output current. Find the voltage gain

Gv =

Gp =

ôut

, and the power gain

^ Pin ■

(c) (d)

Figure P I.38

Find the power delivered by each source. Suppose the power absorbed by the 2 k Q output resistor were 80 watts. Find the power delivered by the input source, and the voltage

39. For the circuit of Figure P I .39, suppose

- 100 mA,

50 Q,

10 Q, and

100 Q. (a)

Find the output voltage and out

(b)

put current. Find the current gain, G/ =

0.1V.

0.21

v,^

Rb

, the v oltê gain Figure P I.41

Gy =

, and the power gain C H EC K : (b)

(c)

10

G p = ^ P-

42. For the circuit of Figure P 1.42, suppose

Findthepowerabsorbedbyeachresistor.

^ 3=

.

I

10i?i. Find the resistor values so that Gy = —^

I

= 1000

Vin 2001

1000

2kO

2kn

Figure P I.39 40. In problem 39, suppose R^= 1 k Q, 7?2=

C ty

10 Q, and R^= 20 Q and P ^ = 80 watts. Find ^2. Figure P I.42

CHECK; 4 . 1 mA C H EC K : i?,= 5 k Q

n

C H A P KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits A CAR HEATER FAN SPEED -CO N TRO L APPLICATION One use o f resistors in electronic circuits is to control current flow, just as dams control water flow along rivers. Ohm’s law, V = RI, gauges the ability o f resistors to control this current flow: for a fixed voltage, high values o f resistance lead to small currents, whereas low values o f resistance lead to higher currents. This property underlies the adjustment o f the blower (fan) speed for ventila tion in a typical car, as represented in the following diagram.

In this diagram, three resistors are connected in series, and their connecting points are attached to a switch. As we will learn in this chapter, the resistance o f a series connection is the sum o f the resistances. So with the switch in the low position, the 12-V car battery sees three resistors in series with the motor. Th e series connection o f three resistors represents a “large” resistance and heavily restricts the current through the motor. W ith less current, there is less power, and the fan motor speed is slow. W hen the switch moves to the Med-1 position, a resistor is bypassed, producing less

52

Chapter 2 • K irchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits

resistance in the series circuit and allowing more current to flow. More current flow increases the fan motor speed. Each successive switch position removes resistance from the circuit, and the fan motor speed increases accordingly. Analysis o f such practical circuits builds on the principles set forth in this chapter.

CHAPTER O U TLIN E 1.

2. 3. 4. 5.

6. 7. 8. 9.

Introduction and Terminology: Parallel, Series, Node, Branch, and so on KirchhofFs Current Law Kirchhoff’s Voltage Law Series Resistances and Voltage Division Parallel Resistances and Current Division Series-Parallel Interconnections Dependent Sources Revisited Model for a Non-ideal Battery Non-ideal Sources Summary Terms and Concepts Problems

CHAPTER OBJECTIVES Define and utilize Kirchhoff’s current law (KCL), which governs the distribution o f cur rents into or out o f a node. Define and utilize Kirchhoff’s voltage law (KVL), which governs the distribution o f volt ages in a circuit. Introduce series and parallel resistive circuits. Develop a voltage division formula that specifies how voltages distribute across series con nections o f resistors. Develop a current division formula that specifies how currents distribute through a par allel connection o f resistors. ' Show that a series connection o f resistors has an equivalent resistance equal to the sum o f the resistances in the series connection. Show that a parallel connection o f resistors has an equivalent conductance equal to the sum o f the conductances in the parallel connection. Explore the calculation o f the equivalent resistance/conductance o f a series-parallel con nection o f resistances, i.e., a circuit having a mixed connection o f series and parallel con nections o f resistors. Explore the calculation o f voltages, currents, and power in a series-parallel connection o f resistances. Revisit the notion o f a dependent source and use a V C C S to model an amplifier circuit. Describe a practical battery source and look at a general practical source model.

Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits

53

1. IN TRO D U CTIO N AND TER M IN O LO G Y: PARALLEL, SERIES, N OD E, BRANCH, AND SO ON The circuits studied in Chapter 1 were interconnections o f resistors and sources that were two-terminal circuit elements. This chapter sets forth K irchhoff’s voltage law (KVL) and Kirchhoff’s current law (KCL). These laws govern the voltage relationships and the current relationships, respectively, o f interconnections o f twoterminal circuit elements. Some new terminology underpins the statements o f KVL and KCL. Figure 2.2a shows a series circuit con sisting o f a sequential connection o f two-terminal cir cuit elements (resistors) end to end. The common con

FIG U RE 2.1

nection point between any elements is called a node. In general, a node is the connection point o f one or more circuit elements. Figure 2.2b shows a par allel circuit, in which the top terminals and the bottom terminals o f each resistor are wired together. The common connection point o f the top terminals is a node, as is the common con nection point o f the bottom terminals. An important property o f the series connection o f Figure 2.2a is that all the rwo-terminal elements carry the same current, in this case

because the input current for each two-terminal element

must equal the exit current. Similarly, in a parallel connection, such as Figure 2.2b, the same volt age, in this case, Vj^, appears across every circuit element.

+ node

d-

0 -1-

node 1

Vr -

b

node 2

(a) (b) FIG U RE 2.2 (a) Series connection of resistors with the property that each resistor carries the same current; (b) parallel connection of resistors with the property that the same voltage appears across each resistor. Sources interconnected with circuit elements produce currents through the elements and voltages across the elements. For example, a voltage source connected across Figure 2.2a would generate a current 2^ and the voltages

through v^. K irchhoff’s voltage law (KVL) governs the distribu

tion o f voltages around loops o f circuit elements, as shown in Figure 2.2a. Similarly, a current source connected across the circuit o f Figure 2.2b would produce the voltage and the currents Z] through K irchhoff’s current law (KCL) governs the flow o f currents into and out o f a com-

54

Chapter 2 • KirchhofPs Current & Voltage Laws and Series-Parallel Resistive Circuits

mon connection point or node, as in the top and bottom connections o f Figure 2.2b. This chap ter sets forth precise statements o f these laws and illustrates their application. A proper statement o f KVL and KCL requires the additional notion o f branch. A branch o f a cir cuit is a generic name for a two-terminal circuit element and is denoted by a line segment, as in Figure 2.3. T he endpoints o f a branch (the terminals o f the circuit element) are called nodes, as in Figure 2.3a. Ordinarily, however, node means a common connection point o f two or more cir cuit elements (branches), as shown in Figure 2.3b.

node A

(a) (b) FIG U RE 2.3 (a) Single branch representing a circuit element with terminals labeled as nodes A and B; (b) interconnection of branches (circuit elements) with common connection points labeled as nodes A through D. The voltage polarity and current direction for the branches in Figures 2.2 and 2.3 are labeled in accor dance with the passive sign convention; the arrowhead on a branch denotes the reference current direc tion, which is from plus to minus. Recall that the + to - does not mean that the voltage is always posi tive if measured from the plus-sign to the minus-sign. In general, reference directions can be assigned arbitrarily. The conventional assignment o f voltage polarity and current direction to voltage and current soiurces is given in Figure 2.4, which is different from the passive sign convention. Note that with these conventional assignments, the (instantaneous) power delivered by a source is/^/^) = power absorbed by a source is

Circuit

Circuit

' 'J O ©

(a)

th^

= -pjeff)-

(b)

FIG U RE 2.4 Conventional labeling o f (a) voltage, and (b) current sources.


55

2. KIRCH HOFF'S CU RREN T LAW Imagine a number o f branches connected at a common point, as at node A o f Figure 2.3b. The current through each branch has a reference direction indicated by an arrow. If the arrow points toward the node, the reference direction o f the current is entering the node; if the arrow points away from the node, the reference direction o f the current is leaving. If a current is referenced as leaving a node, then the negative o f the current enters the node, and conversely.

KIR C H H O FF'S C U R R EN T LAW (K C L ) S tatem ent 1: The algebraic sum o f the currents entering a node is zero for every instant o f

time. S tatem ent 2 : Equivalently, the algebraic sum o f the currents leaving a node is zero for every

instant o f time.

The two statements o f KCL are equivalent because the negative o f the sum o f the currents enter ing a node corresponds to the sum o f the currents leaving the node. Further, from physics we know that charge is neither created nor destroyed. Thus, the charge transported into the node must equal the charge leaving the node because charge cannot accumulate at a node. KCL expresses the con servation o f charge law in terms o f branch currents. Moreover, KCL specifies how branch currents interact at a node, regardless o f the type o f element connected to the node. Referring to Figure 2.3b, KCL at nodey4 requires that i^{t) + - iîi) = 0 for all t. KCL at node Finally, KCL at node D requires that 25(f) =

B requires that

E X A M P L E 2.1 For the node shown in Figure 2.5, find

FIG U RE 2.5 Connection o f five circuit elements at a single node.

56


S o lutio n

By KCL, the sum o f the currents entering the node must be zero. Hence, the current z^(^) = 9cos(2r) - 3cos(2?) - cos(2r) - 2cos(2z) = 3cos(2^) A.

Exercise. 1. Suppose the current through the voltage source in Figure 2.5 is changed to - 2 cos(2?). Find AN SW ER: - 4 cos(2^) A. 2. Three branches connect at a node. All branch currents have reference directions leaving the node. If /j = /2 = 2 A, then find ly A N SW ER; - 4 A

Two implications o f KCL are o f immediate interest. First, as a general rule, KCL forbids the series connection o f current sources. Figure 2.6a shows an invalid connection o f two arbitrary current sources i,(?) and iî), where z,(?)

i^t). It is invalid because KCL requires that i^{t) = i2 {t). On

the other hand, a parallel connection o f two current sources can be combined to form an equiva lent source, as in Figure 2.6b, where

= i^{t) + i2 {t).

(a)

-O

(b)

FIG U RE 2.6 (a) Invalid connection of two arbitrary current sources when Z](z)

ijii)- Avoid this violation o f KCL; (b) equivalent representation o f a parallel connection o f rwo current sources in which = i]{t) +

A second immediate consequence o f KCL is that a current source supplying zero current [i{t) = 0 in Figure 2.7] is equivalent to an open circuit because the current through an open circuit is zero. An open circuit has infinite resistance, or zero conductance. This means that a current source has infinite internal resistance. From another angle, a constant current source is represented by a ver tical line in the iv plane (see Figure 1.4 lb ). The slope o f the vertical line, which is infinite, deter mines the internal resistance o f the source.


57

-O +

v(t) -O FIG U RE 2.7 Ideal current source with i{t) = 0 is an open circuit. A typical application o f KCL is given in the following example. E X A M P L E 2.2 In the parallel resistive circuit o f Figure 2.8, the voltage across each resistor is 6 cos(z) V. Find the current through each resistor and the current,

supplied by the voltage source.

'm

6cos(t) V

IQ

<20

<3Q

FIG U RE 2.8 Parallel resistive circuit for Example 2.2.

S o lu tio n

By Ohm ’s law,

/^ l ( 0 = 6 c o s ( 0 A

.

6 c o s (/)

'R l W -------- = 3 c o s ( 0

A

, „ 3 , „ - 5 i 2 5 W = 2 c o s ( ,) A By KCL,

iinif) = '« l ( 0 + '« 2 (0 +

= 6 cos(r) + 3cos(/) + 2cos(/) = llco s(/ ) A

Exercise. 1. In Figure 2.8, suppose the source voltage is changed to a constant, labeled Iin in terms o f V-^. AN SW ER:

Find

58

Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits

2. Suppose the source voltage in the circuit o f Figure 2.8 were changed to - 1 2 cos(2?) V. Find AN SW ER: - 2 2 cos(2?) A

Kirchhoff’s current law holds for closed curves or surfaces, called Gaussian curves or surfaces. A Gaussian curve or surface is a closed curve (such as a circle in a plane) or a closed surface (such as a sphere or ellipsoid in three dimensions). A Gaussian curve or surface has a well-defined inside and outside. Figure 2.9 illustrates the idea o f a Gaussian curve for three (planar) situations.

.............

Two Terminal Circuit Elem ent / i,(t)

i,(t) -A '/'' ............

V

(a)

o

(b)

FIGURE 2.9 Illustrations of Gaussian curves: (a) enclosure of a two-terminal element; (b) enclosure o f a three-terminal device, such as a transistor; (c) enclosure of a three-node interconnection with an arbitrary circuit. For the two-terminal circuit element o f Figure 2.9a, KCL for Gaussian curves implies that i^{t) =

i2 {t), which is precisely the definition o f a two-terminal circuit element. For the three-terminal device o f Figure 2.9b, KCL for Gaussian curves implies that i^{t) = i^{t) + i2 {t). Finally, for Figure 2.9c, i^ —i^ + if^ = 0. From these illustrations, one might imagine that the use o f Gaussian surfaces might simplify or provide a short cut to certain branch current computations. T he general state ment o f KCL for Gaussian surfaces is next followed by an example that demonstrates its use for computing branch currents.

59


K C L FOR G A U SSIA N CU R V ES O R SURFACES The algebraic sum o f the currents leaving (or entering) a Gaussian curve (or surface) is zero for every instant o f time.

E X A M P L E 2 .3 This example shows how the use o f a Gaussian curve or surface can sometimes simplify a calcula tion. Figure 2.10 portrays a complicated circuit whose branch currents and voltages are not solv able by methods learned so far. Our objective is to find the current

without having to solve a

set o f complex circuit equations.

FIG U RE 2.10 Circuit for Example 2.3, showing a Gaussian surface to compute

directly.

S o lutio n

Using KCL for the indicated Gaussian curve, - 1 .1 5 + / ^ - 0.3 + 0.95 = 0. Equivalently, /^ = 1.15 + 0.3 - 0.95 = 0.5 A. In the next chapter, circuits such as the one in Figure 2 .10 are analyzed using a technique called nodal analysis.

Exercise. 1. Draw a Gaussian surface on the circuit in Figure 2.10 that is different from the sur face given but still allows one to compute /^. AN SW ER; One choice is a circle enclosing the bottom node.

60

Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits

2. Draw an appropriate Gaussian curve to find / in the graphical circuit representation in Figure

2 . 11. AN SW ER: 2 A

FIG U RE 2.11 Graph representation of a circuit.

3. KIRCHHOFF'S VO LTAG E LAW Kirchhoff’s voltage law (KVL) specifies how voltages distribute across the elements o f a circuit. Before conveying four equivalent versions o f KVL, we first set forth several necessary background concepts. The first is the notion o f a closed path. In a circuit, a closed path is a connection o f two-terminal elements that ends and begins at the same node and which traverses each node in the connection only once. Figure 2.12 illustrates several closed paths. One closed path is A-B-CD-E-A, i.e., it begins at node A, moves to node B, drops to node C, moves through element 4 to node D, down through element 6 to reference node E, and back through the voltage source to A. A second closed path is A-B-C-E-A, and a third is B-D-C-B. A second concept pertinent to our KVL statements is that o f a node voltage with respect to a ref erence. A node voltage o f a circuit is the voltage drop from a given node to a reference node. The reference node is usually indicated on the circuit or is taken as ground. The circuit o f Figure 2.12 has branches labeled 1 through 6 and nodes labeled A through E, with node E taken as the refer

Vg, Vq and v^. The voltage denotes the voltage drop from node A to node E-, denotes the voltage drop from node D to node E, and similarly for the remaining node voltages. Node E, being the reference node, has zero as its node ence node. The associated node voltages are denoted by

voltage.

61


+

V.

v„

BD

FIG U RE 2.12 Circuit diagram illustrating (i) three closed paths {A-B-C-D-E-A)\ (ii) the concept of node voltages with respect to a given reference node E, (iii) the concept of branch voltages

and Vj^-,

and

The concept o f a closed path and the concept o f a node voltage allow us to state our first two ver sions o f Kirchhoff’s voltage law.

KIRCH HOFF'S V O LTA G E LAW (K V L) Kirchhoff’s voltage law can be stated in different ways. Following are two equivalent state ments o f the law. Statem ent 1: The algebraic sum o f the voltage drops around any closed path is zero at every

instant o f time. Statem ent 2 : For any pair o f nodes j and k, the voltage drop Vjj^ from node j to node k is

given by

at every instant o f time, where Vj is the voltage at node j with respect to the reference and is the voltage at node ^ with respect to reference. Herey and k stand for arbitrary node indices. For example, in Figure 2.12, j, k can be any o f the nodes A, B, C, D, or E.

62


Referring back to Figure 2.12, for the closed

A-B-C-D-E-A, statement 1 o f KVL requires that

ÂB ^ ^BC ^CD + ^DE ÊA ^ O' for Figure 2.12, from statement 2 o f KVL, the branch voltages = v^ - Vg and Vqj^ = ^D- Hence, = -v^. Thus, by knowing the node voltages o f a circuit, one can easily compute the branch voltages.

Exercise. 1. Find Vy^g, VgQ and V^^-for the circuit o f Figure 2.1 3 in which we have introduced the ground symbol at node E to identify the reference node. AN SW ERS: = - 3 V, Vg(^=2\ V, = 18 V 2. Again, with reference to Figure 2.13, find the node voltages V^, Vg, Vq and AN SW ERS: 2 V, 5 V, - 1 6 V, - 6 V 3. In Figure 2.13, suppose the branch labeled 6 V is now labeled - 1 2 V. Find AN SW ER: - 3 V

D

FIG U RE 2.13

A third concept needed for two further equivalent statements o f KVL is that o f a closed node sequence. A closed node sequence is a finite sequence o f nodes that begins and ends at the same node. A closed node sequence generalizes the notion o f a closed path. Finally, we define the notion o f a connected circuit. In a connected circuit, each node can be reached from any other node by some path through the circuit elements. Figures 2 .12 and 2.14 show connected circuits. However, in Figure 2.14, the sequence o f nodes A-B-C-D-E-A is a closed node sequence but not a closed path because there is no circuit element between nodes B and C.

+ 2.5V -

+ 10V D +

V,

F IG U R E 2 .1 4 Simple dependent source circuit for illustrating the concepts o f a connected circuit and a closed node sequence.

63


This brings us to our last two equivalent statements o f KVL.

KIRCH HOFF'S V O LTA G E LAW (K V L) Following are two additional equivalent statements o f KVL. Statem ent 3 ; For connected circuits and any node sequence, say A -D -B-... -G-P, the volt

age drop

ÂP = ÂD + ^DB + - +

GP

at every instant o f time. S tatem ent 4 : For connected circuits, the algebraic sum o f the node-to-node voltages for any

closed node sequence is zero for every instant o f time.

Referring back to Figure 2.12, statement 3 o f KVL impHes that

Vab +

=

Referring to Figure 2.14, for the closed node sequence E-A-B-E, V-^ = 10 =

+ Vg = 2.5 + Vg and Vg = 7.5 V. Now, consider the closed node sequence E-C-D-E. For this sequence,

~

^CD

Equivalently, 30 = 10 + Vj^, in which case

= 20 V. Finally,

consider the closed node sequence, E-B-C-E, which is not a closed path because there is no cir cuit element between nodes B and C. Nevertheless, by statement 4 o f KVL, - Vg+ Vb £+ ^ c ~ ^ or equivalently that

= Vg -

= 7.5 - 30 = - 22.5 V.

Exercise. 1. In Figure 2.14, suppose = 20 V, V^g = 5 V, and and Vj^. AN SW ER: Kg = 15 V, 60 V, V^g = 45 V and = 40 V

10 V and Vj^ = - 3 V. Find v^ q

2. (a) In the circuit o f Figure 2.15, suppose (b) Suppose Vg = 1 2 0

cos( 1 2 0 tc?),

Vg^ =

18

= 20 V. Find Vg, Vq

cos( 1 2 0 ti?)

and

3 2 cos(1207t^). Find

at ? =

0 .5 s.

(c) Find

when v^ = 1 0 0 V,

= - 1 0 V and

25V.

SC RA M BLED A N SW ER: 85 V, - 1 3 V, - 7 0 V

D

FIG U RE 2.15 Circuit with nodes labeled A through E. Node E is taken as the reference node.

64

Chapter 2 • K irchhoff s Current & Voltage Laws and Series-Parallel Resistive Circuits

Two further implications o f the KVL are o f immediate interest. First, as a general rule, KVL for bids the parallel connection o f two voltage sources— say, and V2 (^)— for which Vj(?) as illustrated in Figure 2.16a. O n the other hand, two voltage sources in series can be combined to form a single source, as illustrated in Figure 2.16b, where

{t) = v-^{t) + V2 {t).

FIGURE 2.16 (a) An improper connecdon of voltage sources when v^{t) ^ i>2 (i); (b) an equivalent representation of two voltage sources connected in series in which

= Vj{i) +

Second, a voltage source supplying 0 V is equivalent to a short circuit, as illustrated in Figure 2.17. Also, the internal resistance o f a voltage source is zero. One can see this by referring to the fact that in the iv plane, an ideal dc voltage source is represented by a horizontal line, as was illus trated in Figure 1.40, The slope o f the line is zero and represents the resistance o f the source. These ideas are dual to those expressed for current sources earlier.

-O

-o +

ov

ov

FIG U RE 2.17 A 0 V voltage source is equivalent to a short circuit. Finally, note that all four KVL statements can be justified using the definition and the notation for “voltage” drop presented in Chapter 1. The justification is more readily comprehended via the analogy o f the gravitational field, also developed in that section. Also, observe that KVL holds for all closed node sequences, independent o f the device represented by each branch o f the connect ed circuit. The distribution o f voltages around closed paths can be viewed as a special case o f this general statement.

4. SERIES RESISTANCES AND VO LTAG E DIVISION During holidays, one often sees strings o f lights hanging between poles or trees. Sometimes these strings consist o f a series connection o f light bulbs. Each light bulb contains a filament, a coil of


65

wire, that gives o ff an intense light when hot. In a circuit’s perspective, the filament acts as a resis tor and has an equivalent hot resistance. The series connection o f bulbs can be modeled by a series connection o f resistors, with each resistor paired with a specific bulb. Computing the voltage across each light (a very important type o f calculation) would then be equivalent to finding the voltage across each o f the resistors in the equivalent circuit model. It is quite common to model electrical loads, such as a light, by resistors. E X A M P L E 2 .4 Figure 2.18a shows a voltage source v-JJ) connected to three resistors in series. The objectives o f this example are to compute

the voltages Vj{t),j = 1, 2, 3, across each resistor, and

the

equivalent resistance seen by the voltage source.

ijt )

(a) (b) FIGURE 2.18 (a) Three series resistors connected across a voltage source. By the definition o f a two-terminal resistor or by the KCL, the current through each resistor is (b) equivalent resistance

= R^ ■¥Rj + R^ seen by the source, i.e., v-^{t) = Reqii„{t)-

So l u t io n Step 1. Express the voltage across each resistor in terms o f the input current. For the circuit o f Figure 2.18a, the current through each resistor is i^JJ) by KCL. From Ohm’s law, the voltage across each resistor is

fo t j = 1, 2, 3. Step 2. Express v-J^t) in terms ofi-JJ), solvefor i-^<^t), and then compute an expression for v^t) in terms

o f the Rj and v-J^t). By KVL, the source voltage equals the sum o f the resistor voltages, i.e.. (2 . 1)

where we have substituted Rjii„{t) = Vj{t). Dividing Equation 2.1 by (i?j + Rj + -^3) yields

Since v-{t) = RjiiJyt) for j = 1, 2, 3,

66


V j{t) =

= --- ---------5

^

—

,2 2^)

Equation 2.2a is a voltê division formula for a three-resistor series circuit. This formula imphes that if a resistance R. is small relative to the other resistances in the series circuit, then only a small portion o f the source voltage develops across it. O n the other hand, if a resistance R. is large rela tive to the other resistances, then a larger portion o f the source voltage will develop across it. One concludes that the voltage distributes around a loop o f resistors in proportion to the value o f each resistance. The proportion is simply the ratio o f the branch resistance R- to the total series resist ance.

Step 3. Compute the equivalent resistance R^^ seen by the voltage source. The equivalent resistance seen by the voltage source for a resistive circuit is implicidy defined by Ohm s law, i.e., Viî) = êqhri^^nonzero currents, the equivalent resistance is defined as

Figure 2.18b illustrates the idea o f the equivalence. By Equation 2.1, v^î) = R^,îj„{t) = (7?^ + +

implies that the equivalent resistance is R^^ = R-^ + R^ + Ry This means that from the

perspective o f the voltage source, the series connection o f resistors is equivalent to a single resistor o f value equal to the sum o f the resistances. A formal discussion o f equivalent resistance and its generalization (the Thevenin resistance) is taken up in Chapter 6 .

Exercise. In Figure 2.18, suppose R^ = 5R^ and R^ = 2Ry Find R AN SW ER: R^^ = GR^, ^ ^rid

,

and

Example 2.4 suggests some generalizations. Consider Figure 2.19. The first is that the equivalent resistance R^^ seen by the source is the sum o f the resistors. This means that resistances in series add, i.e., resistors in series can be combined into a single resistor whose resistance is the sum o f the indi vidual resistances.

Req = R \ + R 2+ "' + Rn Further, since vi^t) = Rjii„{t), a general voltage division formula can be derived as

Ri R\ +

+ ■■' + Rn (2 .2 b)

fory = 1, ... , n.

67


FIG U RE 2.19 Series circuit o f n resistors driven by a voltage source.

Exercise. In Figure 2.19, suppose each resistor has value Rq. Find the equivalent resistance seen by the source and the voltage across each resistor in terms o f the source voltages. A N SW ER: nR^^, v jt)ln

EXA M PLE 2.5 Find the equivalent resistance seen by the source and the voltages 2.20. W hat is the power dissipated in the 14-Q resistor if

and Vj for the circuit o f Figure = 2 V?

FIG U R E 2.20 Series circuit containing a dependent voltage source.

So l u t io n From the preceding discussion, R^^ is defined by Ohm’s law, i.e., v-J^t) = Rgî-Js)Step 1. Express v-^ in terms o f the remaining branch voltages. From KVL,

în = Vi + V2 + 2vj = 3vi + V2

(2.3)

68


Step 2 . Express the branch voltages in terms o f

and substitute into Equation 2.3. To express

and ^2 in

terms of i^^, observe that i^^ is the current through each resistor (KCL or definition o f two-terminal circuit element) and use Ohms law:

= 2i-^ and Vj = 14/^^. Subsrimting into Equation 2.3 yields ^ in =

Therefore,

20//„ = R e q iin

= 20

Notice that the dependent source increases the resistance o f the two series resistors by 4 Q. Dependent sources can increase or decrease the resistance o f the circuit. W ith dependent sources, it is even possible to make the equivalent resistance negative. Step 3. Find the power absorbed by the 14-Q. resistor. To find the power absorbed by the l4-£2 resis tor when = 2 V, first compute i^^ via Ohm s law; i^^ = v J R = = 0 .0 1 x 1 4 = 0.14 W.

= 2/20 = 0.1 A. It follows that P

Exercise. Suppose the dependent source in the circuit o f Figure 2 .20 has its value changed to 2[v-^ + V2 ). Find R^q. AN SW ER: 48 Q

5. PARALLEL RESISTANCES AND CU RREN T DIVISION Many o f the electrical outlets in the average home are connected in parallel. W hen too many appliances are connected to the same outlet or set o f oudets on the same fused circuit, a fuse will blow or a circuit breaker will open. Although each appliance uses only a portion o f the maximum allowable current for the (fused) circuit, together, the total current exceeds the allowable limit. Because o f this common occurrence, an engineering student ought to know how current distrib utes through a parallel connection o f loads (resistors). To keep the analysis simple, consider a set o f three parallel resistors driven by a current source. E X A M P L E 2 .6 Figure 2.21a shows a circuit o f three parallel resistors driven by a current source. Our objectives are to find expressions for v-Jyf), ij^t) in terms o f the input current i-JJ) and the circuit conductances (the recip rocal o f the circuit resistances) and the equivalent resistance seen by the current source.

-o ijt )

0

-I-

- f-

v jt )

ijt )

v Jt)

-O

(a)

(b)

FIG U RE 2.21 (a) Three parallel resistors driven by a current source; (b) equivalent resistive circuit as seen from source.


69

So l u t io n Step 1. Find expressions for ij{t) in terms o f v-J^t). The variable that Hnks the branch current ij{t) to the input current i^J^t) is the voltage which by KVL appears across each resistor. Since v-J^t) appears across each o f the resistors, Ohm’s law implies that each resistor current is

ij{t) = ^ ^

(2.4)

where Gj = HRj is the conductance in siemens andy = 1, 2, 3. Step 2. Compute v^î) in terms ofi-J^t). Applying KCL to the top node o f the circuit yields

ii„{t) = /j(?) + z'2 (?) + i^{t) Using Equation 2.4 to substitute for each ij{t) and then solving for v-^ yields

(j\ + Cj2

■

R]

Rn

êq

^3

R^ (2.5)

Step 3. Compute ijyi) in terms ofi-J^t). To obtain a relationship between i-J,t) and ij{t), substitute Equation 2.5 into Equation 2.4 to obtain

1 G,

.

^

± + + Rl R2 R3

.

G, ,

G 1 + G 2 + G3 (2 .6)

Equation 2.6 is called a current division formula. It says that currents distribute through the branches o f a parallel resistive circuit in proportion to the conductance o f the particular branch

G. relative to the total conductance o f the circuit G^^ = G^ + G2 + Gy The greater the conduc tance, i.e., the smaller the resistance, the larger the proportion o f current flow through the associ ated branch. Step 4. Compute the equivalent resistance

= G^ v-^, defines G^^ or, equivalently,

seen by the source. As in Example 2.5, Ohm’s law, i-^ From Equation 2.5, G^^ = G j + G2 + G3 is the equiva

lent conductance o f the parallel circuit, and the equivalent resistance is

R.

1 ± +

R,

R2

The idea is illustrated in Figure 2.21b.

R3

1

1

G 1 + G2 + G 3

G ,,

70


Exercise. In Figure 2.21a, suppose through if = 1Oe~‘ A.

= 1 Q,

= 0-5

= 0.5 O.. Find the current

and

A N SW ER: z, (?) = 2e“Â

T he Example 2.6 suggests a very important property. Since

hn ={G\ + G2 + G 3 )v,„ = G Vj„ in addition to implying that G

= G^ + G2 + G^ is the equivalent conductance seen by the source

o f the parallel circuit, one can further interpret this to mean that conductances in parallel add to

form equivalent conductances. This parallels the property that resistors in series add to form equiv alent resistances. O n the other hand, resistances in parallel do not add, and conductances in series do not add. We can conclude that from the perspective o f the source, the parallel circuit o f Figure 2.21a has the equivalent representations given in Figure 2.21b. These ideas generalize to n resistors in parallel, as illustrated in Figure 2.22. In particular, the equivalent resistance R

o f the parallel set o f resistors in Figure 2 .22 is

Ren = - 1

1

R,

R2

R., (2.7)

êq = + *^2 + ••• + equivalent conductance. Further, the current through each branch satisfies the general current division formula

R,

G/

1 1

1

+ ---- -!-••• + R„ ^1 ^2

rj A A-•■•j-+ rG „ G +. r G2 +

.

G, .

~ Gr^ q (2 .8 )

■ o -Iijt )

( f

)

V,

(t)

-o FIG U RE 2.22 Parallel connection o f n resistors driven by current source.


Exercise. Consider Figure 2.22. Suppose ten 10 Q resistors are in parallel. Find

71

and the cur

rent through each resistor. AN SW ER: \ Q. and each current is Q.\i-J^t)

E X A M P L E 2 .7 Consider the circuit o f Figure 2.23 exhibiting a current source driving two parallel resistors. Show that

R

Ry

■o

© -O FIG U RE 2.23 Two resistors in parallel driven by a current source.

So l u t io n Step 1. Find the equivalent resistance seen by the current source. From Equation 2.7, with « = 2, it follows that

R^R2

1

Ri ^ Ri This formula, called the product over sum rule, is quite useful in many calculations. Step 2. Find i^{t) and iît)- From Equation 2.8, with w = 2, it follows that

R^

Gi ■

X 7X

■

Ry and

/2 (0 =

Gi 4- G 2

Ri

,__ Ri + R'^ R2

■

72


Exercise. In Figure 2.23, suppose

= 12 A and

Find Rj so that

= 10

= 4 A.

AN SW ER: /?, = 20 Q

E X A M P L E 2 .8 the current ijit) through i?2> and the

For the circuit o f Figure 2.24, find the input voltage instantaneous power absorbed by Rj when

5e ‘ ^ 0

t< 0

o

y r i,(t) ^2

ijt )

U i3 (t) SG 3

l|i,( t )
> = 0 .0 5 U > = 0 .1 5 U > = 0 .0 2 u V = 0 . 0 3 0 -O FIG U RE 2.2 4 Parallel connection of four resistors.

So l u t io n Step 1. Compute the equivalent conductance and equivalent resistance o f the circuit. Since conduc tances in parallel add,

^ e q - G j + G 2 + G 3 + G 4 = 0.25 S

and

Step 2 . Compute v^JJ). From Ohm’s law, the voltage across the current source is

2 0 e~ 'V t^ 0 0

t< 0

Step 3. Compute the current i2 (t). Using the current division formula o f Equation 2.8 yields

3e~'A t > 0 Cjeq

0.25

0

r<0

73

Chapter 2 • K irchhoff’s Current & Voltage Laws and Series-Parallel Resistive Circuits

Step 4. Compute the power absorbed by Rj- To compute the power absorbed by P 2 (0 = v,„(O x i2 {t) = 1'2 (0 r ^2 =

for t >0,

W

Exercises. 1. For the circuit o f Figure 2.24, find i/^(t) and the power absorbed by i?4 . A N SW ER: 0.6^-^ A, 12

W

2. In the circuit o f Figure 2.24, suppose each conductance is doubled and i-J^t) = 100 mA. Find

R , V- (?), and the power absorbed by the new Gy AN SW ER; 2 Q, 200 mV, and 1.6 mW

6. SERIES-PARALLEL IN TERCO N N ECTIO N S The last two sections covered series and parallel resistive networks. Suppose we take a series circuit and connect it in parallel with another series circuit; this is a parallel connection o f two series cir cuits. Alternately, we could take two parallel circuits and connect them in series. This would result in a series connection o f parallel circuits. We could also put a series connection o f two parallel sub circuits in parallel with a replica o f itself or some other series or parallel circuit. Many other inter connections are possible. Arbitrary series and parallel connections o f such subcircuits are called series-parallel circuits. This section explores the calculation o f the equivalent resistance o f seriesparallel circuits by repeated use o f formulas for series and parallel resistance computation. Related voltage and current computation is also explored. Example 2.11 presents a practical application o f series-parallel concepts. EXA M PLE 2 .9 Find the equivalent resistance, R^^, and the voltage across the source,

the voltages

V2 , the

power absorbed by the 6 kQ resistor, and the power delivered by the source for the circuit o f Figure 2.25, when

= 20 mA.

FIGURE 2.25 Series-parallel resistive circuit.

74


So l u t io n Step 1. Compute

first compute R^^y and

To compute

8 x 4.8

Kq\ = ------= — 1+ 4 .8

38.4 = 3 kQ 12.e

and

^^■^2

1

1

1 ^ 6

12

6

4

---- 1---- 1--- ------12

The resistance in parallel with the 2k£2 resistor is, say,

Reqh ~ 1000 + R^qX + R^q2 ~ 6 Finally,

R,„ = IkQ. /!R,„. =

2 +6

= 1.5 kQ

Step 2. Compute V-^ . From Ohm’s law.

= Reqhn = 20 X 1.5 = 30 V Step 3. Compute Vj and V^- By voltage division.

'

Req-i

6

= >5 V and Vj = - ^ v ; „ = ?

6

Step 4. Compute the power absorbed by the 6

resistor.

^ (V 2 )^ ^ 1 0 0 6000

6000

1 60

Step 5. Compute the power delivered by the source.

^source ~ înhn — 30 x 0 .0 2 = 0 .6 W

Exercise. 1. W hat is the current through the 2 k£2 resistor firom top to bottom? AN SW ER: 50 mA 2 . In Example 2.9, suppose the resistance o f each resistor is doubled. Find the new R^ and the power delivered by the source. AN SW ER: 3 k n , 1.2 watts

This example points out a very interesting fact: finding the equivalent resistance o f a series-paral lel connection o f resistors requires only two types o f arithmetic operations no matter the network complexity: adding two numbers and taking the reciprocal o f a number. A hand calculator easily executes both operations. Such is not the case with a non-series-parallel network. To find the equivalent resistance o f a non-series-parallel network, one usually must write simultaneous equa tions and evaluate determinants, a topic detailed in Chapter 3 .


75

It is then important to recognize when a problem belongs to the series-parallel category in order to take advantage o f the simple arithmetic operations. In the previous series-parallel examples, one— and only one— independent source was specified on the circuit diagram. This is part o f the definition o f a seriesparallel network. The independent source must be indicated, or, equivalendy, the pair o f input termi nals to which the source is connected must be specified. The specification o f the input terminals deter mines whether or not a network is series-parallel. The following example illustrates the effect of differ ent input terminal designations on the computation o f equivalent resistance. E X A M P L E 2 .1 0 For the circuit o f Figure 2.26a, determine whether or not the network is series-parallel as seen from each o f the following terminal pairs: 1. 2.

C a s e l;( A , B) Case 2; (A, C)

3.

Case 3: (C, D)

If the answer is affirmative, give an expression and compute the numerical value for the equivalent resistance, using the notation // (double slash) for combining resistances in parallel, i.e., and

are

parallel, and RÎIR2 llR^ means

is in parallel with

means

vvhich is in parallel with Ry

(a)

>R,

A<

R1

D'

(0 FIG U RE 2.26 (a) From terminals (C, D) the network is not series-parallel. However, from terminals

{A, B) the network is a series-parallel one. (b) Redrawing of the network of (a) as seen from terminals (A, Q; the resulting network is series-parallel.(c) Non-series-parallel network seen from (C, D).

76


So l u t io n Case 1. Find equivalent resistance seen at (A, B). W ith an independent source connected to nodes A and B, the source sees a series-parallel network. By inspection o f Figure 2.26a, the equivalent

^

resistance is i?eq = ^ i //[(^2 +

+ ^ 5)] = 20//[(4 + 6)//(2 + 8 )] = 4

^ I

Case 2. Find equivalent resistance seen at (A, Q. W ith (A, Q as the input terminal pair, the net work is again series-parallel. This is made apparent by redrawing the network, as shown in Figure 2.26b, from which i?eq =

+ [(i ?4 +

= 4//{6 + [(2 + 8)//20]} = 3.0 4 Q

Case 3. Find equivalent resistance seen at (C, U). W ith (C, D) as the input terminal pair, the net work is not series-parallel, as can be garnered from Figure 2.26c. T he calculation o f

for this

case requires methods to be discussed in Chapter 3 and is omitted.

Exercise. 1. In Figure 2.26b, suppose AN SW ER: 3.11 Q

is changed to 40 Q. Find

In electrical engineering laboratories, a student often uses a meter to measure voltages associated with a piece o f electronic equipment. In older laboratories, or when using an inexpensive meter, the voltage reading will sometimes differ from what the student calculated or expected to meas ure. Typically, this results from the loading effect o f the meter. Using the concept o f series-paral lel resistances, the following example explores the phenomenon o f loading. E X A M P L E 2 .il Suppose the circuit in Figure 2.27a is part o f a laboratory experiment to verify voltage division. In this experiment, you calculate the expected voltage Vq and then measure the circuit voltage using an inexpensive voltmeter. (a)

Calculate the expected voltage Vq in Figure 2.27a.

(b)

A voltmeter with a 1-kQ/V sensitivity is used to measure

V q.You

use a 0 -10-V range. In

this range, the meter is represented by a 10-kQ resistance, i.e., 10 kD = full-scale reading meter sensitivity = 10 V x 1 kQ/V. W hat voltage will the meter read?

X

(c)

A better-quality voltmeter with a 2 0 - k H / V sensitivity is used to measure the same volt-

^0’ ^ 0 -10-V scale. This better-quality meter is represented by a 2 0 0 -k tl resist ance. W hat new voltage will the meter read?

_

77


—

_

" h

10 kQ

— r-O +

15V^

10 kO

20 ko

f-O— -1-

—

IS v X >10 kO

20 kO

lOkQ

— r-O-----1-

20 kQ

-o FIG U RE 2.27 Three circuits for exploring the effect of loading on a circuit: (a) circuit for validating voltage division; (b) circuit o f (a) with an attached voltmeter having an internal resistance o f 10 k£2; (c) circuit o f (a) with an attached voltmeter having an internal resistance of 200 k^2.

So l u t io n (a) Voltage division on the circuit o f Figure 2.27a yields 20 Vo =

(b)

-15 = 10 V

O n the 0-10-V range, the voltmeter internal resistance between the probes is 10 kD, as stated. This represents a 1G-Id2 load connected in parallel with the 20-kQ resistance, as shown in Figure 2.27b. The voltage Kg will now change because the 15-V source no longer sees 10 ld2 in series with 20 kO. Rather, the source sees 10 kQ in series with 6 .67 k ii = 20 kX2//10 kQ. By voltage division, 6 .67 -15 = 6 V Vo = 10 -H6 .67 This is a 4 0 % deviation from the true answer, V q = 10 V, as calculated in part (a).

(c)

O n the 0 -10-V range with the better voltmeter, the internal resistance between the probes is 200 kD. As before, this represents a 200-kQ load connected in parallel with the 2 0 -k 0 resistance, as shown in Figure 2.27c. 20 kQ//200 klQ = 18.18 kO. By voltage division, this yields 18.18 Vb =

-15 = 9 .6 7 7 V

10 + 18.18

This 3.23% deviation is within a reasonable tolerance o f the precise answer o f 10 V. Example 2.11 demonstrates the effect o f loading due to a measuring instrument, emphasizing the importance o f choosing a good voltmeter with adequate sensitivity. Although modern-day volt meters typically have sensitivities better than 20 kQ/V, a meter with a sensitivity o f 1 kQ./Y is used in the example to dramatize the effect o f loading.

Exercise. 1. Repeat Example 2.11 if the 20-kQ resistance is changed to 40 kO. AN.SWER: 12 V, 6.667 V, 1 1.538 V

78


2. Th e circuit o f Figure 2.28 shows a voltage divider whose voltage Kq is to be measured by a volt meter having an internal resistance o f 80 kO. Find Kq without the meter attached, and then find the value o f Vq measured by the meter.

20 V

FIG U RE 2.28 Voltage divider circuit. AN SW ER: 15 V, 13.71 V

7. D EPEN D EN T SO U RCES REVISITED Chapter 1 introduced the notion o f a dependent or controlled source whose voltage or current depends on the voltage or current in another branch o f the network, i.e., each source has a con trolling voltage or current and an output voltage or current. Figure 2.29 depicts the four types o f controlled sources designated by a diamond containing either a ± or an arrow:

1. 2.

Voltage-controlled voltage source

(VCVS)

Voltage-controlled current source

(VCCS)

3.

Current-controlled voltage source (CCVS)

4.

Current-controlled current source (CCCS)

An arrow inside the diamond indicates a controlled current source having the reference current direction given by the arrow. A ± inside the diamond specifies a controlled voltage source, with the reference voltage polarity given by the ± sign. A parameter value completes the specification o f a linear controlled source. In Figure 2.2 9 the (constant) parameters are fx, g^, r^, and |3. These symbols are common to many electronic circuit texts and have useful physical interpretations to practicing engineers and technicians. For consistency, a ^^-type controlled source is a V C C S and a jO,-type source is a V C V S, and so on.

79


O +

a ................... o (a) VCVS or |j-type

O

o

(b) VCVS or g -type

o

O Pi,

>r

6-

■o

(c )C C V S o rr -type

<;

o - ......... .........o (d) CCCS or P-type

F IG U R E 2.29 Designations for the various controlled sources. In practical controllecl sources, the controUing voltage (t>j in Figure 2.29a and b) or current (z'j in Figure 2.29c and d) is ordinarily associated with a particular circuit element, but not always. For generality, the controlling voltage

in Figure 2.29a and b is shown across a pair o f nodes. Also,

in Figure 2.29c and d, the controlling current Zj is shown to flow through a short circuit. (Strictly speaking, neither an open circuit nor a short circuit is a circuit element.) In a real circuit, the cur rent may be flowing through an actual circuit element, such as a resistor or even a source. In Figure 2.29b, once the controlling voltage v-^ is known, the right-hand source behaves as an independent current source o f value

Since the unit for

is volt, the unit for^^ is amperes per volt, or siemens. Since

is amperes and the unit for has units o f conductance, and the

controlling and controlled variables belong to two different network branches,

is called a trans

fer conductance, or transconductance. The other controlled sources have a similar interpretation. The parameter

has the unit o f resistance, ohms, and is called a transfer resistance. The param

eter |i is dimensionless because the controlling voltage has units o f volts and the output vari able must have units o f volts. Similarly, the parameter (3 is dimensionless. The units and asso ciation are set forth in Table 2.1.

80


Table 2.1 Units and Association. Type

vcvs VCCS

ccvs cccs

Unit

Appellation

dimensionless

Voltage gain

Parameter

siemens

Sm

P

Transfer conductance

ohm

Transfer resistance

dimensionless

Current gain

Figure 2.29 portrays each controlled source as a four-terminal device. In practical circuits, the great majority o f controlled sources have one terminal or node in common, making them threeterminal devices. The dashed lines joining the two bottom nodes in Figure 2 .29 suggest this quite common configuration. The controlled sources as defined in Figure 2.29 have linear v-i relationships. Controlled sources may also have a nonlinear v-i relationship. In such cases, the element will be called a nonlinear controlled source. This text deals only with linear controlled sources. The next few examples describe some o f the unique features o f controlled sources.

Exercise. Find v^,

and the power delivered by each source in Figure 2.30.

FIG U RE 2.30 AN SW ER: 4 V, 0.05 A, 1.6 W, 0.05 W

E X A M P L E 2 .1 2 This example analyzes the circuit o f Figure 2.31. The independent voltage source in series with the 3-Q. resistor represents a practical source discussed at greater length later in this chapter. The circuit within the box o f Figure 2.30 approximates a simplified amplifier circuit by a V C C S. The

8 -Q resistor is considered a load and could, for example, model a loudspeaker. Two important quantities o f an amplifier circuit are voltage gain and power gain, which are computed here along with various other quantities. (a) (b)

Find the equivalent resistance Compute / .

seen by the independent voltage source.


(c)

Compute /out'

(d)

Compute

(e)

Compute the voltage gain

(f)

Compute the power

(g)

Compute the power delivered by the dependent current source.

(h)

Compute

81

delivered to the amplifier.

the power absorbed by the 8-D. resistor.

Compute the power gain, PgJP;„-

(i)

FIG U RE 2.31 Practical source (ideal independent voltage source in series with a resistor) driving a simplified VCCS approximation of an amplifier circuit loaded by an 8 -Q resistor.

So l u t io n (a)

Since resistances in series add,

(b) (c)

By Ohm’s law, /•„ = = 0.8 mA. To compute one must first compute K j. Here one can use Ohm’s law directly, since

= 3 + 47 = 50 Q.

we know /-^, or one can use voltage division. Doing the calculation by voltage division, V = — 4 0 X 10'^ = 3 7 .6 X 10'^ V ‘ 50 Using this value o f

and current division on the right half o f the circuit yields 0.125

79.8 X 3 7.6 X 1 0 '^ = 2 A

“ 0.125 + 0 .0 6 2 5

(d)

V

follows by Ohm’s law K,«.= 2

(e)

x

8 = 16V

The voltage gain with respect to the input signal is

^

= — = 400 0.0 4

82


(f)

By Equation 1.18, the power delivered to the amplifier circuit is

p.^ = (g)

4

= 47

= 47 X 0.82 >< io-<5 = 30.08

The power delivered by the dependent current source is ^VCCS = "ôut ^ 79 .8 K j = 16(79.8 x 0.0376) = 48.01 W

(h)

(i)

Pout is simply the product o f voltage and current delivered to the load

The resulting amplifier power gain is the ratio o f the power absorbed by the 8 -Q load to the power delivered to the amplifier, P-^,

^

Pin

= 1.064 X 10^ 3 0 ,0 8

Exercise. Suppose the 8-f2 load resistor in Figure 2.30 is changed to 16 Q. Compute and the power gain. AN SW ER: 1.5 A, 24 V, 1.197 X 10<^

The analysis in Example 2.12 required only KCL, KVL, and simple voltage divider and/or cur rent divider formulas. More complicated linear circuits necessitate a more systematic approach. To see this need, add a resistor between the top o f the 47-Q resistor and the top o f the dependent current source in Figure 2.31. The methods o f solution used in the example immediately break down because the circuit is no longer series-parallel; hence, one cannot use voltage division to compute V j. Chapter 3 will explain more systematic methods called nodal and loop analysis. Unlike a passive element such as a resistor, which always dissipates power as heat, a controlled source may generate power as computed in part (g) o f Example 2 . 12, or may dissipate power in other cases. Since a controlled source has the potential o f generating power, it is called an active element. In Example 2.12, the practical voltage source delivers 30.08 pW o f power to the circuit, which is easy to accept because the source could have been a small battery On the other hand, the con trolled source generates 48 W. This seems a litde puzzling. Where does the power come from? W hy not purchase a controlled source at a local electronics store and use it to power, say, a lamp? Here it is important to recognize that a controlled source is not a stand-alone component picked o ff the shelf like a resistor. A controlled source is usually constructed from one or more semicon ductor devices and requires a dc power supply for its operation. The power delivered by the con trolled source actually comes from the power supply. Here, we use the controlled source to math ematically model an amplifier and facilitate analysis o f the circuit.


83

W ith simple series-parallel connections o f resistors, the equivalent resistance is always positive. When controlled sources are present, a strange result may happen, as illustrated in the next exam ple. E X A M P L E 2 .1 3 Find the equivalent resistance = 2.

for the circuit o f Figure 2.32 when (a) p = 0.5 and (b) p

•O^ FIG U RE 2.32 Calculation of

for a circuit with controlled source for two values o f p.

So l u t io n W ith p unspecified, we can apply KVL to the single loop, noting that

Consequently, (1 - p)V^ =

= V^. Here,

and

i-x

For p = 0.5, R^q = 2R, which means that the dependent source acts like a resistor o f R Q.. In this case, it absorbs power. O n the other hand, for p = 2, R,^ = -R , a negative equivalent resistance. In this case, the dependent source acts like a -2R-Q. resistor and, in fact, delivers power to the inde pendent source. An important conclusion can be drawn from this example: in the study o f linear circuit analysis, controlled sources allow the possibility o f negative resistances. Since a negative resistance generates power, it is also an active element.

Exercise. In Figure 2.32, find the values o f p so that R^q - 0 .5 R and R^q - 2R. A N S W E R :-!, 0.5

84


Exercise. For Figure 2.33, find

for the following three values of^^: 0.5 mS, 1 mS, and 2 mS.

FIG U RE 2.33 AN SW ER: 2 k£l, open circuit, -1 kQ

8. M O D EL FOR A NON -IDEAL BATTERY The ideal battery o f Figure 1.30, repeated in Figure 2.34a, delivers a constant voltage regardless o f the current drawn by a load. T he i-v plane characteristic is a horizontal line through V^, as shown in Figure 1.40b and repeated in Figure 2.34b. Ideal batteries do not exist in the real world. The terminal voltage always depends on the supplied current. A more accurate representation o f a practical battery, but by no means a fully realistic one, is an ideal battery in series with a resist ance, say, R^, as shown in Figure 2.34c. R^ is termed an internal resistance, which crudely models the effects o f chemical action and electrodes inside the battery.

> Vs

(a)

(b)

FIG U RE 2.34 (a) Ideal battery; (b) i-v battery char acteristic; (c) battery model with internal resistance to crudely approximate effects of chemical action and presence o f electrodes; (d) nickel-cadmium battery.

(c)


85

E X A M P L E 2 .1 4 This example shows the effect o f the internal resistance o f a battery on the terminal voltage. Suppose a nickel-cadmium battery has an open circuit terminal voltage o f 6 volts. W hen con nected across a 2-Q. resistor, the voltage drops to 5.97 V. Find the internal resistance o f the bat tery.

So l u t io n Figure 2.35 illustrates the situation. Here, the dashed box represents the battery model with inter nal resistance R^. In Figure 2.35a, no load is connected to the battery. Hence, no current flows through the internal resistance, in which case, the terminal battery voltage is 6 V.

(a)

(b)

FIG U RE 2.35 Battery model with internal resistance; (a) open circuited (Is = 0); and (b) connected to a 2 -0 load. Figure 2.35b shows the battery connected to the 2-Q resistive load. The measured voltage is 5.97 V. By KVL, the voltage across the internal resistance, law, the current through

is

is Vj^ = 6 - 5-97 = 0.03 V. From Ohm’s

= (5.97/2) = 2 .985 A. Again, by Ohm’s law. 0.03 2.985

= 0 .0 1 0 0 5 Q

Exercise. In Example 2.14, suppose the internal resistance is known to be R^ = 0.005 Q and although the load resistance is unknown, the load current is 4 A. W hat is the voltage across the load resistance, and what is the load resistance? AN SW ER; 5.98 V and 1.495 ^

9. NON -IDEAL SO URCES Ideal voltage sources have zero internal resistance. Real voltage sources, such as batteries, have an internal resistance. The value o f this resistance may change with the current load. There may also


86

be other effects. However, for our purposes, a more realistic model o f a voltage source contains a series internal resistance, as illustrated in Figure 2.36a.

L .(t)

(b)

FIG U RE 2.36 (a) A non-ideal voltage source as an ideal voltage source with an internal series resist ance; (b) a non-ideal current source as an ideal current source with a parallel internal resistance. Ideal current sources have infinite internal resistance. Real current sources have a finite, typically large, internal resistance. Figure 2.36b depicts a more realistic current source model where the internal resistance is in parallel with the ideal current source. In the case o f constant voltage and current sources, ideal and non-ideal source models have a graphical interpretation. The i-v (current-voltage) characteristic o f an ideal constant voltage source {v^{t) = 1^) is a horizontal straight line. This means that the voltage supplied by the source is fixed for all possible current loads. An ideal constant current source (z^(z) =

has a vertical

straight line characteristic, which means that the current is constant for all possible voltages across the source. Figure 2.3 7 illustrates these relationships graphically.

Vout

V

(a)

FIG U RE 2.37 v-i characteristics o f (a) an ideal constant voltage source, and (b) an ideal constant current source.


87

The non-ideal case is quite different. Because o f the internal resistance a non-ideal constant volt age source i-v characteristic satisfies the linear relationship

ôut

^sôut

Ks'

and for a non-ideal constant current source in which

(2 . 10)

= l/R^,

hut — ^s'ôul

( 2 . 11 )

Equations 2.10 and 2.11 are illustrated by the graphs in Figure 2.38 when v^{t) = ideal voltage source and i^{t) =

for the non

for the non-ideal current source. For a voltage source, if the value

o f R; is very small in comparison with potential load resistances, as ordinarily expected, then the hne in Figure 2.38a approximates a horizontal line, the ideal case. O n the other hand, for a cur rent source, the line in Figure 2.38b approximates a vertical line whenever

is much much larg

er than a potential load resistance. This would then approximate the ideal current source case.

FIGURE 2.38 v-i characteristics of (a) non-ideal constant voltage source, and (b) non-ideal current source. In a similar way, non-ideal dependent voltage sources are a connection o f an ideal dependent source with a series resistance. A non-ideal dependent current source is a connection o f an ideal dependent current source with a parallel resistance. E X A M PLE 2 .1 6 Figure 2.39 shows the measured voltages o f a dc power supply found in an old laboratory. Assuming a non-ideal model o f Figure 2.38a, find the power supply.

and the internal resistance R^ of

88

Chapter 2 • K irch h offs Current & Voltage Laws and Series-Parallel Resistive Circuits

Vout (V) ' '

FIG U RE 2.39 Graph of measured voltages and currents for a dc power supply.

So l u t io n = -R^

From Equation 2.10, we know that

+ V^. From the graph, when

= 0,

10 V = I/. Further, R^ = - (9.8 - 10)/(0.5 - 0.0) = 0.4 Q.

SUM M ARY This chapter has presented the essential building blocks o f linear lumped circuit theory, beginning with the two fundamental laws for interconnected circuit elements: KVL and KCL. KVL states that for lumped circuits, the algebraic sum o f the voltages around any closed node sequence o f a circuit is zero. Similarly, KCL says that for lumped circuits, the algebraic sum o f the currents enter ing (or leaving) a node is zero. These laws in conjunction with Ohm’s law allowed us to develop a voltage division and a current division formula. The voltage division formula applies to series-resistive circuits driven by a volt age source. The voltage developed across each resistor was found to be proportional to the resist ance o f the particular element relative to the equivalent resistance seen by the source. For exam ple, in a two-resistor series circuit, Rj in series with we found that Vi =

7^1 + /?2

T he current division formula applies to parallel-resistive circuits driven by a current source. Here, the current through each resistor with conductance Gi was found to be proportional to G/ divid ed by the equivalent conductance seen by the source. Since conductance is the reciprocal o f resist ance, the idea can also be expressed in terms o f the resistances o f the circuit. For example, in a tworesistor parallel circuit, Rj is parallel with /?2> /i =

G, G 1 + G2

R, +Rn

In deriving the voltage division formula, we learned that the resistances o f a series connection o f resistors may be added together to obtain an equivalent resistance, prompting the phrase “resistors in series add.” Analogously, the derivation o f the current division formula for parallel circuits led

Chapter 2 • Kirchhofif’s Current & Voltage Laws and Series-Parallel Resistive Circuits

US

89

to conclude that a parallel connection o f resistors has an equivalent conductance equal to the

sum o f conductances. This is sometimes expressed in terms o f resistances as the inverse o f the sum o f reciprocal, i.e., for n resistors in parallel, p . 7^1

R„

which leads to the very special formula for two resistors in parallel,

R

=

often referred to as the product over sum rule. Dependent sources, first introduced in Chapter 1, were re-examined in greater detail. Some prac tical points were described. All o f the above ideas were applied to the analysis o f series-parallel networks that are interconnec tions o f series and parallel groupings o f resistors. Our analysis showed us how to compute the equivalent resistance o f series-parallel circuits. An example was given that described the applica tion o f these ideas to voltage measurement. This was followed by a discussion o f battery models and battery usage. Finally, battery modeling ideas were used to describe non-ideal source models.

12. TERM S AND C O N C EPTS Branch: a two-terminal circuit element denoted by a line segment. Closed node sequence: a finite sequence o f nodes that begins and ends with the same node. Closed path: a connection o f devices or branches through a sequence o f nodes so that the con nection ends on the node where it began and traverses each node in the connection only once. Connected circuit: one for which any node can be reached from any other node by some path through the circuit elements. Current division: the current in a branch o f a parallel-resistive circuit is equal to the input cur rent times the conductance o f the particular resistor, Gj, divided by the total parallel con ductance o f the circuit,

= G^ + ... + G^.

Dependent (controlled) current source: a current source whose output current depends on the voltage or current o f some other element in the circuit. Dependent (controlled) voltage source: a voltage source whose output voltage depends on the voltage or current o f some other element in the circuit. Gaussian surface: a closed curve in the plane or a closed surface in three dimensions. A Gaussian surface has a well-defined inside and outside. Kirchhoff'’* current law (KCL): the algebraic sum o f the currents entering a node o f a circuit consisting o f lumped elements is zero for every instant o f time. In general, for lumped circuits, the algebraic sum o f the currents entering (leaving) a Gaussian surface is zero at every instant o f time.

90

Chapter 2 ° KirchhoflF’s Current & Volt:«e Laws and Series-Parallel Resistive Circuits

Kirchho£F’$ voltage law (KVL); for lumped circuits, the algebraic sum of the voltage drops around any closed path in a network is zero at every instant of time. In general, for lumped connected circuits, the algebraic sum of the node-to-node voltages for any closed node sequence is zero for every instant of time.

Node: the common connection point between each element; in general, a node is a connection point of one or more circuit elements.

Node voltage: the voltage drop from a given node to the reference node. Parallel circuit: a side-by-side connection of two-terminal circuit elements whose top terminals are wired together and whose bottom terminals are wired together.

Series circuit: a sequential connection o f two terminal circuit elements, end-to-end. Voltage division: each resistor voltage in a series connection is a fraction o f the input voltage equal to the ratio of the branch resistance to the total series resistance.

// (double-slash): notation for combining resistances in parallel, i.e., RÛR2 means and i ?2 are in parallel, and RÎIR2 llRj^ means R^ is in parallel with which is in parallel with Ry r\ r\ . r\

o n

o

n n

n

91

Chapter 2 • K irchh off s Current & Voltage Laws and Series-Parallel Resistive Circuits

PROBLEMS KIR C H H O FF'S C U R R EN T LAW 1. (a)

Find the value o f /j for each o f the node connections in Figure P2.1a and P2.1b given that 1^ = 2 A,

Figure P2.3

1^ = 3 A, and (b)

= 4 A. Repeat part (a) when l 2 = I^ = 1^ = 2 A.

A N SW ER: ( b ) 4 A 4. (a)

Find the value o f /j in the circuit o f Figure P2.4.

(b)

Find the value o f

in the circuit o f

Figure P 2.4 by a single application o f KCL.

Figure P2.1 A N SW ERS: (b) 0, 2 A 2.

In the circuit o f Figure P2.2, each shaded

box is a general circuit element. = 20 mA, 1-^2 = ^0 mA,

(a)

Suppose

(b)

I ini = 100 mA> and /-^4 = 0.05 A. Apply KCL to find /j, Ij, ly and 1^. Repeat part (a) when = hni =

Figure P2.4 A N SW ER: (a) 6 A

KIR C H H O FF'S V O LTA G E LAW

100 mA.

5. (a)

Consider the circuit o f Figure P2.5a where each branch represents a circuit element. Find Vj and 1^2-

(b)

Find Kj and unspecified

for Figure P2.5b. Each branch

represents

an

unknown circuit element. .1 0 0

40V

lOOV Figure P2.2

+ V, -

6

3. (a)

For the circuit o f Figure P2.3a,

V,

-

+ VI

Reference Node

5A

Reference Node

(a)

find the value o f the current /j using

N

lO O v C ”^ + 40V

AN SW ERS: (b) (scrambled) 200, -300, -200, -300 mA

,

(b)

Figure P2.5

only a single application o f KCL. (Hint: Construct a Gaussian surface.) (b)

(c)

Repeat part (a) for 1^ True-False: can be uniquely deter mined as in part (a) and part (b).

6 . (a)

For the circuit o f Figure P2.6, deter mine the voltages f j, ... , ^4 and the power absorbed by each resistor.

92


(b)

Now determine the node voltages V^,

Vg, Vq and

with respect to the

reference

indicated

node

by

the

ground symbol.

(c)

Compute

50 V

SC RA M BLED AN SW ERS: (a) 2 V, - 2 V

K C L A N D KVL 9. (a)

Consider the circuit o f Figure P2.9a. Use KCL and KVL to find the voltage across each current source from the arrow head

Figure P2.6

to the arrow tail and the current through

A N SW ERS: (a) 180 V, 50 V, -110 V, 10 V

each voltage source from minus to plus. Finally, find the power delivered by each

7. (a)

Find the values o f the voltages Vp Kj,

source and verify conservation o f power,

and Vj in the circuit graph o f Figure

(b)

P2.7, where each branch represents a

For the circuit o f Figure P2.9b, find the voltages

and V^.

circuit element. (b)

Now determine the node voltages

K l’ ^B’

respect to the reference node indicated by

in iUV

6^

4A

30V

3A

10

the ground symbol.

(c)

Compute

and (a)

(b)

Figure P2.9 AN SW ERS: ( b ) - 1 7 V 14 V 10.

For the circuit in Figure P 2.10, calculate

the power delivered by each o f the eight inde pendent sources. Verify the principle o f con servation o f power. Figure P2.7 SC RA M BLED AN SW ERS: (a) -65 V, 15 V, 15 V; (b) 35 V, - 4 5 V, 10 V, - 5 V

8 . (a)

Use KVL to determine the voltages

(b)

1/ and V; in the circuit o f Figure P2.8. Now compute V^g. Figure P2.10 A N SW ERS: - 4 , - 9 , - 3 6 . 35, 10, 0 , 10, - 6 W

93


11.

(b)

Four circuit elements and a dependent

Supposing that /j = 4 A, /, = 2 A, and and the power

a = 0.25, determine delivered to Rj^.

voltage source are shown in the circuit o f Figure P 2 .11. The current through and the voltage across each element are identified on the dia gram. However, one— and only one— voltage (or current) value is labeled incorrectly. Mark the incorrect voltage (or current) on the circuit diagram and give the correct value for this volt age (or current).

Figure P2.14 A N SW ERS: (b) 80 Q, 20 W 15.

Consider the circuit in Figure P2.15 in = 1 A and R^ = 84 Q. Find the value

which

+

o f R^ for each o f the following cases:

25V

(a)

The power delivered by the source is

(b)

T he power absorbed by

(c)

watts. The power absorbed by R^ is 13.44

13.44 watts. Figure P 2 .ll Find the currents and voltages /^, V^, 1^,

12. and

is 13.44

watts.

in the circuit o f Figure P2.12.

Figure P2.15 SC RA M BLED A N SW ERS: 21, 16, 336, 56 Figure P2.12

KCL, KVL, A N D O H M 'S LAW 13.(a)

16 . Consider the circuit o f Figure P2.16. (a) (b)

If /? = 5 Q, find Vjf Find the value o f R when

For the circuit o f figure P2.13, sup = 875 mA and Rj_ = 80Q .

pose

Find V-^, /j, /j, and ly (b)

= 40 V.

Now suppose that = 7 A and -^2 ~ ^ A. Find V-^, ly Rj^, and the power

120V

o

4on

delivered to the load Rj^.

50V

o

O

20V

Figure P2.16 + V.

AN SW ER: (b) 25 Q 200

,4on

Figure P2.13 C H E C K : (b) P^= 160 watts I4.(a)

For the circuit o f Figure P2.13, deter mine in terms o f /j, a and Rj^.

17. For the circuit o f Figure 2.17, find (a) the voltage Vj and the power absorbed by the 10 Q resistor; (b) (c)

the voltage V2 , the power delivered by each source.

94


R >J .

10 k n ,

6kO Figure P 2.17 AN SW ER: (c) 12.5 watts and - 7 .5 watts

'+ ■V,

(a) Figure P2.20

18. Find the power absorbed by the unknown circuit element x and the voltage

in the cir

cuit o f Figure P2.18.

21.

500 50 V

6x 10" ANSWERS: (b) Vi = 3V 'in 4 8 V 4 kQ i 6xl 0' -i-/?-i-6axl0 The circuit o f in Figure P2.21 is a blower

motor control for a typical car heater. In this circuit,

6

+

0.8 A <----

resistors are used to control the current through a motor, thereby controlling the fan speed.

Figure P2.18 19. (a) (b)

Ignition Switch

JUA

Find the current 7/j and the voltage Kuf circuit o f Figure P 2 .19. If a resistor o f 7? Q is placed across the output terminals, determine the current

and the voltage

and

12V

Chassis . Ground

the power delivered by the 10 V 14V

44 V

6

'r

©

3R

10V

(a)

W ith the switch in the Lo position, the current supplied by the battery is

source.

2.5 A. The voltage drops across the

Figure P2.19 C H EC K : (b) K^„,= 4 V 20. (a)

^BC= 1-5 V, Vc/j = 0.625 V, and Vq = 3 .125 V. Consider the motor as rep

In Figure P2.20a, Vj = 32 V and the

resented by a load resistance.

power delivered by the source is 80

(i)

mW. Compute (b)

resistors and motor are Vjg = 6.75 V,

, V-^, and R.

In Figure P2.20b a dependent voltage source has been added to the circuit o f Figure P2.20a. Suppose Determine

= 40 V.

in terms o f a and R. If

= 0.8 mA and a = 5, find K, and R. (c)

Determine the value o f each resist ance and the value o f the equivalent resistance representing the motor.

(ii) Determine the power dissipated in each resistor and the power used by the motor. (iii) Determine the relative efficiency o f

For each circuit o f Figure P 2.20,

the circuit, which is the ratio o f the

determine the resistance seen by the voltage source,

power used by the motor to the power delivered by the battery.

95


(b)

W ith the switch in Med-1 position,

delivered is 1250 W. How many possible medi

determine:

um wattages are there and what are they?

(i)

C H EC K : 10 ohms, 40 ohms

T h e voltage drop across each resistor.

(ii) The current delivered by the battery. (iii) The relative efficiency of the circuit. (c)

(a)

Suppose /? = 20 Q, find the power

(b)

Suppose the power delivered by the

delivered by the current source.

Repeat part (b) with the switch in position Med-2.

(d)

24. Consider the circuit o f Figure P2.24.

The switch is in the high position. A

current source is 120 watts. Find the

winding in the motor shorts out. The

value o f R.

fuse blows. W hat is the largest equiva lent resistance o f the motor that will cause the fuse to blow? A N SW ERS: (a)

(i)

Rj^g = 2.7 Q.,

=0.6 0., RcD

“ (t)

^^^lOOV

= 0 .2 5 Q , 1.25 a

Figure P2.24

= 16.875 W Pbc = 3.75 W 1.5625 W P , , „ , , = 7.8125

(ii)

C H E C K : (b) 8
W (b)

(iii) 26% (i)K^5=0> ^s c = 3 -4 3 V , 1-43

25. Given that 4 W is absorbed by the 100-Q =

= 7.14 V

resistor, find V} and the power delivered by the voltage source in the circuit o f Figure P2.25.

(ii) 5.71 A (iii) 59.5% (c)

(i)

=0.

150 Q

2o on

Vcn = 2V,

Vmotor = 10 V

300n'

lo o n

(ii) B A (d)

(iii) 83.3% 0.4 Q

22. Suppose one has two resistors /?j = 20 Q and i ('2 = 20 Q that can be conected to a source,

= 100 V. By connecting the resistors to the

Figure P2.25 26. In the circuit o f Figure P2.26, suppose V2 = 60 V. Find

/^, and the power delivered by

the source.

source in different ways, what are the different i8 o n

wattages that can be delivered by the source?

6on'

The different types o f connections represent what might occur in an electric space heater having a low, medium, and high setting.

''• 6

4o n 9o n

1 800

C H EC K : There are three possible connections with medium using 500 watts. 23. In Problem 22, find the values o f R^ and so that the lowest wattage delivered by the 100 V source is 200 W and the highest wattage

Figure P2.26 SCRA M BLED AN SW ERS: 3 360, 840, 4

96

Chapter 2 • Kirchhofif’s Current & Voltage Laws and Series-Parallel Resistive Circuits

27. Find the power delivered by each independ ent source and the power absorbed by each resis tor in the circuit o f figure P2.27. (Check; Total of delivered power = total o f absorbed power.) 0.7 A

lOon

500 0.8 A 20V

6

C H EC K : (a) 45 < (c) 8 0 < i? ^ < 125

200

< 65; (b) -85 < 1^2 < '6 5 ;

Figure P2.27 SC RA M BLED A N SW ERS: 59.5, 9, 8, 49.


0.5, 45, 18

(a)

28.

(b)

Write an equation for

in terms o f a

and 4 For the circuit o f Figure P2.28 with the

indicated currents and voltages, find (a) (b) (c)

If 1/ = 40 V and a = 0.5, find the value o f the current

(c)

Currents /j through

How much power is delivered by the

Voltages Vj through

independent

Power delivered by each independent

power is delivered by the dependent

source?

How

much

current source

source? Verify the principle o f conser vation o f power for this circuit. 5 mA

500 0

2000 Figure P2.30 C H EC K : 4 = 0.05 A

Figure P2.28 C H E C K : ^2 = 12 V, /^ = 1 mA,

= 60m W

29. For the circuit o f Figure P2.29, find

V O LTA G E A N D C U R R EN T D IVISIO N 31. Consider the circuit o f Figure P2.31 in

(a)

Voltage drop V j, and

which

(b)

Voltage drop V2

V2 for each o f the following cases:

(c)

The value o f the unspecified resist

(a)

ance, R

= 30 V and

= 20 V. Find

and

Switch 5j is closed and switch S2 is open.

(b)

Switch

is open and switch S2 is

closed. (c)

Switch closed.

is closed and switch S2 is

97


34. For the circuit o f Figure P2.33,

(a)

= 120 V

= 120 Q. Find the value o f

Suppose

that is necessary to achieve V-^ = 90 V. Compute (b)

Find the values o f R^ and Rj that are necessary to achieve

Figure P2.31

= 100 V and

1/2 = 80 V. 32. Construct a series voltage divider circuit wiiose total resistance is 2400 Q as illustrated in Figure 2.32. (a)

Suppose Vj = 0 . 7 5 and Vj = 0 .2 5 V^. Find the values o f R-^,

(b)

^s'

Suppose V"j = 0 . 8 and V2 = 0.5V^. Find the values o f

(c)

6on

R2 , and R^.

Suppose K, = 0.81/ and Kj = 0.5 Find the values o f 7?,, R2 , and R^.

Figure P2.34 C H EC K ; (b) 60 Q, 240 Q

35 . Figure P2.35 shows a Wheatstone bridge circuit that is commonly used in a variety o f measurement equipment. The bridge circuit is said to be balanced if R J i j = Ri,R^- In this case, the voltage = 0 for any voltage V-^^. (a) Use voltage division to compute the voltages

Figure P2.32

and V^. Check:

AN SW ER: (c) R^=R^ = 960 Q, R^ = 4 80 Q

Rc + Rci 33. For the circuit o f Figure P 2.33, suppose = 48 V. (a) Find

(b)

about

with the switch in position A,

i.e., the switch is open. (b)

Find

with the switch in position B,

= 0, then what must be true and VJ. Show that =0 if and only if R^R^ = Rh^cSuppose that RJi^ = Rf^R^ and a 0.5D. resistor is connected across the

If

(c)

i.e., the switch is closed.

terminals. Then 6Q (d)

find the current

through the 0.5-^^ resistor. Suppose = 18 V, = 3 Q, = 2 Q, an d R ^ = 2^ - Find

Figure P2.33 C H E C K : (b) 25 <

< 30 V

AN SW ER: (d) 3 V

=6

98


36. Find V^, V , and

for the circuit o f Figure

P2.36 when 1/^ = 50 V and V^ 2 = 25 V. > ;i,

lokn

60kn

Figure P2.39 lokn

C H EC K :

= 6.4 watts

40. Find /p /2>l y V-^, and the power delivered by the source in the circuit o f Figure P2.40 when I-^ = 120 mA.

Figure P2.36 SC RA M BLED A N S W E R S :- 5 , 10, 15 37. Consider the circuit o f Figure P 2.37 in

-o

which /.^ = 0.1 A, Gj = 2.5 mS, (a)

i',

W ith the switch in position A, find "^^d the power delivered by the source.

(b)

’4kn

'• ©

Kq' h

6kn

9kn

Repeat part (a) if the switch is in posi tion B.

J 8 kO

-O Figure P2.40 C H E C K : V.^ = 360 V 41. Find /[ and I 2 for the circuit o f figure P2.41 when

= 10 mA, 1^^ = ^

and

= 14 mA.

Figure P2.37 SC RA M BLED AN SW ERS: (a) 50, 0 . 5 , 5, 50, 20

'■ 7

38. In the circuit o f Figure P2.38, it is required that /j = 0.81/^-^. Find R (in Q ), in terms o f /• , and V- in terms o f /■ .

r ’’ '“0f

>© Y

30mS>

Figure P2.41 42. For the circuit o f Figure P2.42, find the cur rents /p /2, ly and

-O V

'

when

= 300 mA.

I

j"

3oon> 6 o o n /l

I >i 2on

-o Figure P2.38

i 2on >4on CH ECK:

= 40/,„ and

39. In the circuit o f Figure P 2.39,' o

1

in

= 10 mA.

Find /p /j, V and the power delivered by the dependent source.

Figure P2.42 SC RA M BLED AN SW ERS: - 5 0 , 40, 80 (in mA)

N.


99

Rpo AND RELATED CALCULATIONS OF SERIES-PARALLEL CIRCUITS 43.

3000

For each o f the circuits o f Figure P 2.43,

find the value o f

3000 5000,

and the power delivered „

^

" V

if a 10-V source were connected.

5000 . 1.5kO<

J 7500

i6kn

lkO<

Ik0<

Figure P2.45 (b) SC RA M BLED AN SW ERS: (b) 29.63, 675, 0.1185 46.

Find

for each o f the circuits in Figure

P2.46. Notice that the circuit o f (b) is a modifica tion o f (a) and that o f (c) a modification o f (b). 2kn

15kn

Figure P2.43 A N SW ERS: 0.5R , 5 kQ, 2.6 kQ 44. Find the value o f R-^ for each o f the circuits o f Figure P2.44. O-

Figure P2.46 SC RA M BLED A N SW ERS: 60 kn ,

Figure 2.44 (a)

22.5 kO, 135 k n

1.5R

47. Find R^^ in the circuit o f Figure P2.47 (a)

W hen the switch is open

(b)

W hen the switch is closed

Figure 2.44 (b) 45. For each o f the circuits in Figure P2.45, compute the equivalent resistance R seen by the source, the input current delivered by each source, and

the power when

=

80 V. C H E C K : Answers are the same.

100


48. This is a conceptual problem and requires 8000

no calculations for the answer. Consider cir

>5000

cuits 1 and 2 o f Figure P2.48. All resistors are

2kO

greater than or equal to 1 Q.. We wish to deterO-

mine the relationship between R^^^ and R^ ^ 2 the presence o f the finite positive R-Q. resistor

(e) Figure P2.49

between points a and b. W hich o f the following

SCRAM BLED ANSWERS: 4 0 0 ,7 0 0 , 500, 1500

statements is true? > K ql

(a) (b)

50. Consider the circuit o f Figure P2.50. (a)

Kill = êq2 (d)

There

is no

general

relationship

between R , and R^^jrelation ship depends on the value o f R.

(b)

Explain your reasoning.

Suppose = 320 V, = 256 V, R^ = R^ = 800 n . Find R, and the resulting R^^. Suppose = 320 V, V^= 192 V, R^ = 400 Q, R^ = 800 Q. Find R, and the resulting R

C ircuit 1

Figure P2.50 SC RA M BLED AN SW ERS: 500, 1000, 1600, 400, 170.67, 128

C ircuit 2

5 L For the circuit of Figure P2.51: (a) Calculate R^^q the equivalent resist ance seen at terminals A and C, which would be the reading on an ohmmeter if the two probes were connected to A

Figure P2.48

and C, respectively. ♦ 49. For each o f the circuits o f Figure P2.49, find the value o f R that makes = 1000 Q,.

(b)

Calculate R^q the equivalent resist ance seen at terminals B and C, which would be the reading on an ohmmeter

O-

if the two probes were connected to B and C, respectively.

5000

>3kO O-

■7500

O-

be

mulas? State your reasons without per (b)

1 .2 kn ■

forming any calculations.

52Sn

o-

(c)

Can the equivalent resistance

calculated using the series-parallel for

(a)

O-

(c)

(d)


(a)

101

For Figure P2.53a, how many bulbs can be put in parallel before the 15 A fiise blows? Given the maximum number that can be put in parallel without blow

(b)

ing the fuse, find R and V^. In Figure 2.53b, bulbs BB and C C are 24 watt and 36 watt, respectively, at approximately 12 volts. Find the inter

52. Some physical problems have models that

nal resistances o f each bulb. How

are infinite ladders o f resistors, as illustrated in

many C C bulbs can be present before

Figure P2.52. (a) Find the equivalent resistance

the 15 amp fuse blows? Given this at the

number o f C C bulbs, find R^^ and

terminals a-b in figure P2.52a. (Hint: Since the resistive network is infinite, the equivalent resistance seen at termi nals a-b is the same as the equivalent resistance to the right o f terminals c-d\ this means that the network to the right o f c-d can be replaced by what???) Evaluate if

=1Q

and = 100 Q. This type of problem is useful for represent ing series and shunt conductance

(b)

in transmission lines. (b)

Find

Figure P2.53

at terminals a-b for the ladder

network o f Figure P2.52b.

C H E C K ; (a) n = 16; (b) n = 4

- 0 --------------->

>^1

54. In the circuit o f Figure P 2.54, V and

= 40 Q. The switch

= 330

closes at ? = 5

s, S2 closes at f = 10 s, 5^ closes at ? = 15 sec, bO -

and 54 closes at ^ = 20 sec. Plot

and cal

culate RAi) for 0 < /■< 25 s.

Figure P2.52 N U M ERICA L AN SW ERS: 10.512, 14.177 Figure P2.54 53.

Consider the circuit o f Figure 2.53a in

which = 0.5 £^. Suppose each AA-bulb rep resents a 12-watt fluorescent bulb at approxi mately 12 volts, having an internal resistance o f 12 a

55. Consider the circuit o f Figure P2.55. (a) Find max[ and the average (b)

value o f Zj(^). Find ijit), max[ i 2 {t)], and the average value o f i2 {t).

102

Chapter 2 • K irch h off’s Current & Voltage Laws and Series-Parallel Resistive Circuits

i,(t)

o

58. W ith the car engine turned off, you have

9kn 3kO <

been listening to the car radio. While the radio

6kn<

is on, you turn the ignition to start the engine.

3cost(2t)V •2kn

You noticed a momentary silence o f the radio. The following circuit analysis explains this

12V

effect quantitatively. Assume that with the car Figure P2.55

engine not running, the 12-V car battery is rep

56. Consider the circuits o f Figure P2.56. In

The load due to the car radio is represented by

resented by the model shown in Figure P2.58. Figure 2.56, (a) Find (b)

= 120 sin(377?) V and ou f

out'

an equivalent resistance o f 240 Q. The starter

=5

and the instantaneous

motor draws 150 A o f current when the igni

power absorbed by 30 D, resistor.

tion is turned on and before the engine starts.

If

Find

is replaced by a current source, = 120 sin(377?) mA,

up, find

pointing

and the instanta

neous power absorbed by 30 Q, resis tor. Does

affect the current through

at the moment when the ignition

switch is turned on. Compare this to the volt age

before the ignition switch is turned on.

W hy do you think the radio goes silent momentarily?

the other resistors in the circuit?

ignition model for car battery with engine not running

Chasis ground

Figure P2.58

C H EC K : /*3oq = 37.97sin^(377r) watts 57. The circuit o f Figure P 2.57 shows a simple scheme to determine Rq, the internal resistance o f the battery model. The loading effect due to the digital voltmeter may be neglected (consid er that the meter is represented by an infinite resistance). W ith the switch open, the meter reads 12 V. W ith the switch (briefly) closed, the reading drops to 11.96 V. Find the value o f R^.

CH ECK:

= 900 watts

59. The volume o f a car radio is not much affected by the on/off state o f the headlights. The following circuit analysis explains this phe nomenon quantitatively. Assume that with the car engine running, the 12-V car battery is rep resented by the model shown in Figure P2.59. Notice that the effective voltage o f the car battery increases due to the effect

model for a 12V battery

o f the alternator while the engine is running. The load due to the car radio

15Q

Figure P2.57 AN SW ER: 0.0376 Q

(a)

represented by an equivalent resist ance o f 240 £2. At 12 V dc, each head light consumes 35 W on low beam and 65 W on high beam. Find the equivalent resistance o f each headlight on low beam.

103


(b)

Find the equivalent resistance

v(V)

o f each headlight on high beam. (c)

Find

(d)

are turned off. Find when the low beams

when the headlights

are turned on. (e)

Find

when the high beams

are turned on. (f)

v (V) 60 ■ 40 -

How much power does each high beam consume given your

20 -

answer to part (e)? W hy is this

^--------- 1-----------

value different from 65 watts? (g)

0.5

1

1 V3

>i(A)

i(A)

(b)

How much power must the Figure P2.61

battery deliver to overcome its internal losses and operate the high beams and radio. C H E C K ; (a) 4.11 Q ; (f) 195.58 watts

DEPENDENT SOURCE PROBLEMS 62. In the circuit o f Figure P 2.62, determine so that the power delivered to the 5-kQ load resistor is lOOPy^, where P^^ is the mstantaneous

power

con

sumed by the 8-kQ resistor. Equivalently,

is the power

delivered by the non-ideal voltage source. Figure P2.59 60. A 50-cell lead acid storage battery has an open-circuit voltage o f 102 V and a total internal resistance o f 0.2 Q,. I f the battery delivers 40 A to a load (a)

Load

resistor, what is the terminal voltage? (b)

W hat is the terminal voltage when the battery is being charged at a 50 A rate?

(c)

Figure P2.62 AN SW ER: 6.25 mS

W hat is the power delivered by the charger in part (b)? How much o f the power is lost in the battery as heat?

63. Find the equivalent conductance G and then the equivalent resistance R “seen” by the

SCRAMBLED ANSWERS: 500, 112, 5600, 94

current source 1^ in the circuit o f Figure P2.63

61. A non-ideal constant voltage source, an ordinary resistor, and a non-ideal constant cur

when

in terms o f the literals R\,

rent source have the v-i characteristics given in Figure P 2.61. Determine the values o f the source voltage or current, the value o f the source internal resistance, and, finally, the value o f the resistance for Figure P2.61.

3

= 1 kQ, R2 =

gm = 0-2 mS.

and g^. Evaluate

104


MATLAB PROBLEMS (a) Find the output voltage,

the output

current (what is its direction), and the power absorbed by the load (8-Q resis tor) for the circuit o f Figure P2.66.

C H EC K :

Figure P2.63 = 10 kO.

64. For the circuit o f Figure P2.64, write a node equation that allows you to find

in terms o f

Figure 2.66

Then find

I

(b)

Using MATLAB or equivalent, com pute and PLO T with appropriate labels the power absorbed by the load, denot ed by R^, as Rj^ varies from 8 to 64 Q in increments o f 1 Q. Also plot the current, again using MATLAB, as a function o f

R^. At what value o f

is the absorbed

power a maximum? Knowing this is

Figure P2.64

important, for example, when matching

C H EC K : 0.75 S.

loudspeaker resistances to the output 65. In the circuit o f Figure P2.65 r^ =

resistance o f your stereo. For this prob

12.5 kQ and^^ = 12.5 mS:

lem, you should use MATLAB. You will

(a)

Compute the output voltage and out

need to turn in an original printout (no

put current in terms o f (b)

Compute the voltage gain, Gy=

copies permitted) o f your code and plots.

l/-^.

Hint: Begin your program

2kn

©

SmV,

8kQ'

2kn<

with the commands listed

5kO

below. ?? indicates that you

r I,

should

insert

the

proper

number or formula.

8kO<

Figure P2.65

RL = 8:1:64; % This command generates an array o f

numbers for RL beginning at 8 and ending at 64 in increments o f 1. I f you do not end it with a semicolon, it will list every entry o f the array. V2 = ?? % This value should be precomputed I L = ??;

PL = RL .* I L . ^ 2 ;


% Note that because IL and RL are

arrays o f numbers . ^ means to square each number in the array IL and . * means to multiply each number in IL by the corresponding number in the array for RL.

Beginning your MATLAB solution: % Define element values R l= 15; R2= 4; R3= 9; R4= 2; R5=8; R6=18; % To fin d Req start from right side. Ra= R4 + R5; Ga= 1/Ra;

plot(RL, PL)

Gb = Ga + 1/Rl;

grid % Plot IL in mA

Rb = 1/Gb; % Continue these additions and reciprocals until

plot(RL, IL *]0 0 0 )

obtaining Req.

grid % typing grid adds a grid to your

% To fin d Vout requires repeated use o f voltage and current division formulas.

plot. Always add a grid. % You can put both plots on the same graph as follows: plot(RL,IL*WOO,RL,PL) % The motivated student might investigate using the “hold” command instead.

Geq = 1/Req; IRc = 20*Gc/Geq; V Rb = IRc*Rb; % Now write down the MATLAB expression for

finding Vout. AN SW ERS: (a) 3 Q; (b) 24 V

►67. The analysis o f series-parallel circuits with numerical element values can be done with only two types arithmetic operations: adding two numbers and taking the reciprocal o f a number. As such, MATLAB is an extremely convenient tool for finding the equivalent resistances and the voltages and currents throughout a series-parallel circuit. This prob lem illustrates such a use o f MATLAB. For the circuit o f Figure P 2.67 (a)

Find R^^_

(b)

Find Vouf R^= 180

--R =20

R =90 20A

105

©

R,= 150 R =40

Figure P2.67

R =80

106


68. Use MATLAB to find R-^ and

for the

circuit o f Figure P2.68. Turn in your MATLAB code with your answers. Hint: Label the equiv alent seen at each node to facilitate computa-

“on of

K uf

-Cr 200 mA

--- — 2kO

—

-- ---------1,2kO

3kO

^3,2kn

(D IkO

2.2kO

3kO

^

1.6kn -

Figure P2.68 A N SW ERS: 591.2 Q, 8.869 V ► 69. Use MATLAB to find R^^,

and /j for

the circuit o f Figure P2.69.

300n

300

2on

ion eon i3on

10V

'4

6

4on

i5on

2oon >

135Q

> V 500*

-OFigure P2.69 ► 70. Use MATLAB to find Ri„ in the circuit o f Figure 2P.70. 3on

2on

8000

100

6000

600 1300 1500 100V

4000 <

400

1350

-o Figure P2.70 A N SW ER: 50.53 II, 133.8 mA

5000

< 500

C H A P Nodal and Loop Analyses

H ISTO RICAL NOTE For a network consisting o f resistors and independent voltage sources, one can apply KCL to the nodes, KVL to the various loops, and Ohm’s law to the elements to construct a large set o f simul taneous equations whose solution yields all currents and voltages in the circuit. In theory, this approach completely solves the basic analysis problem. In practice, this approach proves imprac tical because large numbers o f equations are required even for a small network. For example, a 6branch, 4-node network, with each node connected to the other nodes through a single element, leads to a set o f 12 equations in 12 unknowns: 3 equations from KCL, 3 equations from KVL, and 6 equations from the element v -i relationships. The 12 unknowns are the 6 branch currents and 6 branch voltages. Before the advent o f digital computers, engineers solved simultaneous equations manually, possi bly with the aid o f a slide rule, or some primitive mechanical calculating machines. Any technique or trick that reduced the number o f equations was highly treasured. In such an environment. Maxwell’s mesh analysis technique (1881) received much acclaim and credit. Through the use o f a fictitious circulating current, called a mesh current, Maxwell was able to greatly reduce the num ber o f equations. For the above-mentioned network, the number o f equations drops from 12 to 3 equations in the unknown mesh currents. An alternate KCL-based technique (now called nodal analysis) appeared in literature as early as 1901. The method did not gain momentum until the late 1940s, because most problems in the early days o f electrical engineering could be solved efficiendy using mesh equations in conjunc tion with some network theorems. W ith the invention o f multi-element vacuum tubes having interelectrode capacitances, some compelling reasons to use the node method appeared; primari ly, the node method accounts for the presence o f capacitances without introducing more equa tions, and secondly, those vacuum tubes that behave very much like current sources are more eas ily accommodated with nodal equations. By the late 1950s, almost all circuit texts presented both the mesh and node methods. Since the 1960s, many digital computer software programs (SPICE being the most ubiquitous) have been developed for the simulation o f electronic circuits that otherwise would defy hand cal

108

Chapter 3 * Nodal and Loop Analyses

culation. These software packages use a node equation method over the mesh equation approach. One o f several reasons is that a node is easily identifiable, whereas a set o f proper meshes is diffi cult for a computer to recognize. For resistive networks driven by current sources, writing node equations is straightforward. Certain difficulties arise in writing node equations for circuits containing independent and dependent voltage sources. During the 1970s, a modification o f the conventional node method by a research group at IBM resulted in the “modified nodal analysis” (MNA) technique. W ith the M N A method, the formulation o f network equations, even in the presence o f voltage sources and all types o f dependent sources, becomes very systematic. This chapter discusses the writing and solution o f equations to find pertinent voltages and cur rents for linear resistive networks.

CHAPTER O U TLIN E 1. 2. 3. 4. 5. 6. 7. 8.

Introduction, Review, and Terminology The Concepts o f Nodal and Loop Analysis Nodal Analysis I: Grounded Voltage Sources Nodal Analysis II: Floating Voltage Sources Loop Analysis Summary Terms and Concepts Problems

CH APTER O BJECTIVES 1.

2.

Describe and illustrate the method o f node analysis for the computation o f node voltages in a circuit. Knowledge o f the node voltages o f a circuit allows one to compute all the branch voltages and, thus, with knowledge o f the element values, all the branch currents. Define the notion o f a mesh or loop current and describe and illustrate the method o f mesh or, more generally, loop analysis for the computation o f loop currents in a circuit. Knowledge o f all the loop currents o f a circuit allows one to compute all the branch cur rents. Thus, in conjunction with the knowledge o f the branch element information, one can compute all the branch voltages.

3.

Formulate the node analysis and loop analysis equations as matrix equations and use matrix methods in their solution emphasizing the use o f existing software for the gener al solution.

4.

Describe and illustrate the modified nodal approach to circuit analysis. This method underlies the general software algorithms available for computer simulation o f circuits.

Chapter 3 • Nodal and Loop Analyses

109

1. IN TRO D U CTIO N , REVIEW, AND TER M IN O LO G Y Chapter 1 introduced basic circuit elements, Ohm’s law, and power calculations. Chapter 2 intro duced the important laws o f circuit theory, K Y L and KCL, and investigated series, parallel, and series-parallel circuits. Recall from Chapter 2 that a node voltage is the voltage drop from a given node to a reference node. As a brief review, consider Figure 3.1, which portrays a circuit labeled with nodes A through D having associated node voltages, V^, Vg, Vq V^, and eight branches, one for the current source and one for each o f the seven conductances,

... , Gj. (Since this

chapter deals almost exclusively with dc, the uppercase notation for voltages and currents is com monplace.)

FIG U RE 3.1. Diagram of a circuit with labeled node voltages, V^, Vg, Vq V^, with respect to the given reference node. KVL states that every branch voltage is the difference o f the node voltages present at the terminals o f the branch: for circuits in this text and all pairs o f nodes, j and k, the voltage drop from n o d ej to node k, is

at every instant o f time, where VJ- is the voltage at node j with respect to the reference and

is

the voltage at node k with respect to reference. Here, j and k stand for arbitrary indices and could be any o f the nodes. A, B, C, or D , in Figure 3.1. These statements mean that knowledge o f all node voltages in conjunction with device information paints a rather complete picture o f the cir cuit’s behavior. This chapter develops techniques for a systematic construction o f equations that characterize a circuit’s behavior. One last introductory point: Throughout this chapter and in many subsequent chapters, software programs such as MATLAB facilitate calculations. Constructing sets o f equations that character

110


ize the voltages and currents in a circuit is often a challenge. Solving such sets o f equations with out the use o f software tools presents a much greater challenge. Yet facilitated by MATLAB or equivalent, the calculations reduce to a hit o f the return key. MATLAB and the circuit simulation program called PSpice or Spice (utilized in Chapter 4) are but two o f the many modern and important software tools available to engineers.

2. TH E C O N C EPTS OF N O D AL AND LOOP ANALYSIS Nodal analysis is an organized means for computing ALL node voltages o f a circuit. Nodal analy sis builds around KCL, i.e., at each node o f the circuit, the sum o f the currents leaving (entering) the node is zero. Each current in the sum enters or leaves a node through a branch. Each branch current generally depends on the branch conductance, a subset o f the circuit node voltages, and possibly source values. After substituting this branch information for each current in a node’s KCL equation, one obtains a nodal equation. As an example, the nodal equation at node A in Figure 3.1 is /^-^ = /j + /y = G j (V ^- V^) + Gy (Vj - V^). The nodal equation at node C is -/2 + I^ + 1^^- Ij = Vj- + G j ( V ^ + Gy ( K ( j- V^) = 0. Writing such an equation at each circuit node (except the reference node) pro duces a set o f independent equations. O f course, one can substitute a KCL equation at the refer ence node for any o f the other equations and still obtain an independent set o f nodal equations. T he solution o f such a set o f nodal equations yields all circuit node voltages. Knowing all node volt ages permits us to compute all branch voltages. Knowing each branch voltage and each branch con ductance allows us to compute each branch current using Ohm’s law. The reference node may be chosen arbitrarily and can sometimes be chosen to greatly simplify the analysis. A set o f nodal equations has a matrix representation. The matrix representation permits easy solu tion for the node voltages using MATLAB or an equivalent software package. A variation o f the nodal analysis method, termed modified nodal analysis, relies heavily on matrix methods for constructing and solving the circuit equations. The basic principles o f this widespread analysis technique are illustrated in Section 4. Because computer-based circuit analysis packages build on a matrix formulation o f the circuit equations and because o f the widespread use o f matrices in circuits, systems, and control, we will stress a matrix formulation o f equations throughout this chapter. The student unfamiliar with matrix methods might look through a calculus text or a linear algebra text for a good explanation o f their basic properties and uses. T he counterpart to nodal analysis is loop analysis. In loop analysis, the counterpart o f a node voltage is a loop current, which circulates around a closed path in a circuit. A loop or closed path in a circuit is a contiguous sequence o f branches that begins and ends on the same node and touch es no other node more than once. For each loop in the circuit, one defines a loop current, as illus trated in Figure 3.2, that depicts three loops or closed paths having corresponding loop currents /p Ij, and ly O f course, one can draw other closed paths or loops for this circuit and define other loop currents.

111


90

FIG U RE 3.2. Simple resistive circuit showing three closed paths (dotted lines) that represent three loop currents, /j, Ij, and 1^; the branch current which is a difference o f the two loop currents through the resistor. Using a fluid flow analogy, one can think o f loop currents as fluid circulating through closed sec tions o f pipe. The fluid in different closed paths may share a segment o f pipe. This segment is anal ogous to a branch o f a circuit on which two or more loop currents are incident. The net current in the branch is analogous to the net fluid flow. Note that each branch current can be expressed as a sum o f loop currents with due regard to direction. For example, in Figure 3.2, the branch cur rent 7^3 = ^\ -

Using loop currents, element resistance values, and source values, it is possible

by KVL and Ohm’s law to express the sum o f the voltages around each loop in terms o f the loop currents. For example, the first loop, labeled

in Figure 3.2, has the loop equation

^ « = 9 / i + 3(/i -/ 2 )+ 6 (/ ^ -/ 3 ) We will explore this concept more thoroughly in Section 5. Here we see that loop analysis builds on KVL, whereas node analysis builds on KCL.

3. N O D AL ANALYSIS I: G RO U N D ED VO LTAG E SO URCES As mentioned earlier, nodal analysis is a technique for finding all node voltages in a circuit. W ith knowledge o f all the node voltages and all the element values, one can compute all branch volt ages and currents, and thus the power absorbed or delivered by each branch. This section describes nodal analysis for circuits containing dependent and independent current sources, resistances, and independent voltage sources that are grounded to the reference node (see Figure 3.3). Floating independent or dependent voltage sources (those not directly connected to the reference node) are covered in Section 4. For the class o f circuits discussed in this section, it is possible to write a nodal (KCL) equation at each node not connected to a voltage source. A node connected to a voltage source grounded to the reference node has a node voltage equal to the source voltage. The other node voltages must

112


be computed from the set o f nodal equations. Each nodal equation will sum the currents leaving a node. Each current in the sum will be expressed in terms o f dependent or independent current sources or branch conductances and node voltages. The set o f these equations will have a solution that yields all the pertinent node voltages o f the circuit. Examples 3.1 and 3.2 illustrate the basic techniques o f nodal analysis. EX A M P L E 3 .1 . The circuit o f Figure 3.3a contains an independent voltage source, an independent current source, and five resistances whose conductances in S are G j through Gy The nodes other than the refer ence are labeled with the node voltages V^,

and V^, which respectively denote nodes a, b, and

c. T he analysis o f this circuit illustrates the process o f nodal analysis to find the node voltages V^, y,,andK =

FIG U RE 3.3A. Resistive circuit for Example 3.1. Note that node voltage is specified by the voltage source.

So l u t io n . Step 1. Consider node c. A voltage source ties node c to the reference node. Hence, the node volt age is fixed at V-^, i.e., Because it is not necessary to apply KCL to this node unless the current through the voltage source is required, for example, when determining the power delivered by the source. Step 2. Sum the currents leaving node a. From KCL, the sum o f the currents leaving node a is zero. As per the partial circuit in Figure 3.3b, this requires that

Grouping the coefficients o f

and

tion yields our first nodal equation

and moving the source values to the right side o f the equa


113

G 5J 'V -aV .in')

(3.1) Step 3. Sum the currents leaving node b. Applying KCL to node b, reproduced in Figure 3.3c, yields the equation

G 2 ^y b- y a) ^G ,V b^ G, {V ,- VJ = Q After regrouping terms, one obtains our second nodal equation: G4(v ,- v

j

V

G 3V,

G .(V ,-V )

FIG U RE 3.3C

-G ^ V ^ + {G ^ + G ,+ G ,)V ,= G,V.„

(3.2)

Step 4 . Write set o f nodal equations in matrix form. Equations 3.1 and 3.2 in matrix form are Gi + G

2+

G5

-G 2

-G 2 G 2 -I- G 3 -I- G 4

•

(3.3)

Matrix equations organize relevant data into a unified framework. Because many calculators do matrix arithmetic, because o f the widespread availability o f matrix software packages such as MATLAB, and because equation solution techniques in circuits, systems, and control heavily uti lize matrix methods, the matrix equation formulation has widespread and critical importance. Step 5 . Solve the matrix equation 3.3: For this part, suppose that the conductance values in S are Gj = 0.2, G2 = 0.2, G3 = 0.3, G 4 = 0.1, G 5 = 0.4, that = 2.8 A, and that = 24 V. After sub stitution, equation 3.3 simpUfies to

114


■0.8

-0 .2 - ■K,'

- 0 .2

0 .6

■12.4' (3.4)

2.4

Solving using the inverse matrix method leads to the node voltages (in volts): ■ 0.8 - 0 .2

- 0 .2 ' 0.6

-1

T 2 .4 -

1

2.4

0 .6

0.4 4 0.2

0 .2 ' ■12.4'

18-

0.8

10

2.4

V

Alternately, one could have solved equation 3.4 via MATLAB, its equivalent, or the age-old hand method o f adding and subtracting equations. For example, in MATLAB »M =[0.8 -0.2;-0.2 0.6]; »b= [12.4 2.4]'; >>NodeV = M\b NodeV = 1.8000e+01 l.OOOOe+01 »% O R EQU IVA LEN TLY »NodeV = inv(M )*b NodeV = 18

10 Step 6. Compute

The branch voltage V^ = V^ -

18 - 10 = 8 V.

Exercises. 1. Utilize the solution o f Example 3.1 to compute the current leaving and the power delivered by the independent voltage source. AN SW ER: 3.8 A and 91.2 watts 2. Referring to Figure 3.3a and the values set forth in Step 5 o f Example 3.1, suppose the value o f is cut in half, the value o f V-^ is 24 V, and the value o f each o f the conductances is also cut in half W hat are the new values o f the node voltages? AN SW ER: All node voltages are the same. 3. By what single factor must the values o f

and V-^ be

multiplied so that the node voltages are doubled? AN SW ER: 2 L 4. Construct a node equation for A N SW ER: (G, + G ,)V ^ ^ 1 -

in Figure 3.4. F IG U R E 3.4.

115


EX A M PLE 3.2. Consider the circuit o f Figure 3.5a. Similar to Example 3.1, the objective is to find the node volt ages V^,Vf^, and

. However, in the circuit o f Figure 3.5a, an independent current source has

replaced the independent voltage source o f Figure 3.3a. This change unfreezes the constraint on the value o f

present in the circuit Figure 3.3a. There will result three nodal equations in the

three unknowns

, and V^.

0.4 U

FIG U RE 3.5A. Circuit containing two independent current sources and three unknown node voltages

, and V^.

So l u t io n . Step 1. Sum currents leaving node a. This step is the same as Step 2 o f Example 3.1. By inspec tion o f node a, 0 . 2 + 0.2(K^ - V^) + 0.4(V^ - V J - 2 = 0 which upon regrouping terms yields

0.8V ^ -0.2V ^ -0.4V ^ = 2

(3.5)

Step 2. Sum currents leaving node b. This step is the same as Step 3 o f Example 3.1. Again, by inspection, 0.2

- VJ + 0.3

+ 0.1 (V^ - V;) = 0

Simplification yields -0 .2 7 ^ + 0.6V ^ ^ -0.1K ^ = 0

(3.6)

Step 3. Sum currents leaving node c. Because a current source

o. 4 ( v - v :

drives node c, the similarity to example 3.1 ends, and we must write a third node equation. Summing the currents leaving node c, as shown in Figure 3.5b, yields 0 . 4 ( 1 / - K J + 0.1 ( K ^ - K ^ ) - 1 = 0 Upon simplification, we have -0.4V ^^-0.1K ^ + 0 .5 V ;= 1

(3.7)

116


Step 4. Write equations 3 .5 -3 .7 as a matrix equation and solve. The matrix form o f our nodal equations 3 .5 -3 .7 is

0.8

- 0.2

- 0 . 4 ' ■v;'

- 0.2

0.6

- 0.1

- 0 .4

- 0 .1

0.5

'T (3.8a)

= 0

1

Solving equation 3.8 using MATLAB or equivalent, using a calculator that does matrix operations, or solving via some form o f row reduction, one obtains the solution (in volts) ■0.8

= - 0.2 Vc

- 0 .4

- 0.2

- 0 .4 '

0.6 - 0.1

-1

■2 ‘

'2 .9

1.4

2.6

■2 ‘

■8.4'

- 0.1

0 = 1.4

2.4

1.6

0 = 4 .4

0.5

1

2.6

1.6

4 .4

1

V

(3.8b)

9.6

Specifically, in MATLAB » M = [0.8 -0.2 -0.4;-0.2 0.6 -0.1; -0.4 - 0.1 0.5]; >>b = [2 0 1]'; »NodeV = M\b NodeV = 8.4000e+00 4.4000e+00 9.6000e+00

Exercises. 1. Suppose the values o f the current sources in Figure 3.5a are doubled. W hat are the new values o f the node voltages? Hint: Consider the effect on equation 3.8. AN SW ER: All node voltages are doubled.

2 . Suppose the conductances in the circuit o f Figure 3.5a are cut in half, i.e., the resistances are doubled. W hat are the new node voltages? A N SW ER: Node voltages are doubled. 3. Suppose the conductances in the circuit o f Figure 3.5a are cut in half W hat happens to the magnitudes o f the branch currents? Hint: Express the branch current in terms o f the branch con ductance and its terminal node voltages. AN SW ER: The magnitudes o f the branch currents stay the same. I 4. Find two node equations characterizing the cir cuit o f Figure 3.6. AN SW ER: (G j + G^) -V„~G^^Vy =

and

FIG U RE 3.6


117

The matrices in equations 3.3, 3.4, and 3.8a are symmetric. A symmetric matrix, say A, is one

= A; this means that ifA = [a-^ is an n x n matrix whose i-j entry is a-, then A is symmetric if a-j = a^j^. In words, the off-diagonal entries are mirror images o f whose transpose equals itself, i.e., each other. For example. ■0.8

- 0.2

- 0 .4 '

A = - 0.2

0.6

- 0.1

- 0 .4

- 0.1

0.5

present in the circuit., as in Examples 3.1 and 3.2, the coefficient matrix o f the node equations (as exemplified in equations 3.3, 3.4, and 3.8) is always sym metric, provided the equations are written in the natural order. When only resistances, independent current sources, and grounded independent voltage sources are present in the circuit, the value o f the entries in the coefficient matrix o f the nodal equations can be computed by inspection. The 1-1 entry o f the matrix is the sum o f the conductances at node a (or 1); the 2 - 2 entry is the sum o f the conductances at node b (or 2). In general, the i-i entry o f the coefficient matrix is the sum o f the conductances incident at node i. Further, the 1 -2 entry o f the matrix is the negative o f the sum o f the conductances between nodes a and b (or between nodes 1 and 2), and the 2 -1 entry has the same value. In Example 3.2, the 1 -2 entry o f - 0 .2 S is the nega tive o f the sum o f the conductances between nodes a and b; the 1 -3 entry o f - 0 .4 S is the negative o f the sum o f the conductances between nodes a and c (or between 1 and 3, if the nodes were so numbered). Thus, whenever the circuit contains no dependent sources, the node equations can be written by inspection. Further, if independent voltage sources are absent, then the right-hand side o f the nodal matrix equation can also be written by inspection: the i-th entry is simply the sum o f the independent source currents injected into the i—th node at which KCL is applied. W hen controlled sources are present in the circuit, the resultant nodal matrix is generally not sym metric, as illustrated in the following two examples.

EXA M PLE 3.3. The circuit o f Figure 3.7 represents a small-signal low-frequency equivalent circuit o f an amplifi er in which the input signal V-^ is “amplified” at the output,

= V'2. Small-signal means that

the input signal should have a relatively small magnitude so that a LIN EA R circuit will adequately represent the amplifier. Similarly, low-frequency means that the frequency o f any sinusoidal input must be relatively low for the (resistive) circuit model o f the amplifier to remain valid. The amplifier circuit model contains a current-controlled current source (CC CS) and a voltagecontrolled current source (VCCS). These two dependent sources have currents that depend on other circuit parameters and require some special handling when constructing node equations. Our objective is to set forth the methodology for writing the node equations when dependent cur rent sources are part o f the circuit and to compute the magnitude o f the voltage gain, |

V-^ |

= \y 2 |y^n\■ Note that the source voltage, V-^, specifies the voltage at the node at the bottom o f Gp hence, a nodal equation at this node is unnecessary. Nodal equations must be written at the remaining


118

nodes, which are labeled with the voltages Vj, Kj (=

and V^. (Numbering and labeling is

often a matter o f personal preference. In this example, we have chosen 1, 2, and 3 as node labels, in contrast to the previous two examples, where we used a, b, and c.)

G,

FIG U RE 3.7. An equivalent circuit model o f an amplifier.

So l u t io n . Step 1. Sum the currents leaving node 1. Summing the current leaving node 1 leads to (^1 -

+ ^2 (^1 - ^ 3) + ^3 (^1 -

+ P

0

or, equivalently, after grouping like terms, (Gi + G2 + G3)Ki - G3 K2 -

(3.9)

Step 2. Substitute fo r i^ in equation 3 .9 and simplify. In equation 3.9, |3 i^ accounts for the effect o f the C C C S at node 1 and is not given in terms o f the circuit node voltages. To specify this term in terms o f the circuit node voltages, observe that in Figure 3.7, i^ is the current from node 2 to node 3 through G^. Hence,

I3t -I3G^{V2 - K3) = I3G^V^ - I3G^V^

(3.10)

Substituting equation 3.10 into 3.9, again grouping like terms, one obtains the first nodal equation,

{G, + G2 + G3) Vi + il3G^ - G3) K2 - (G 2 + PG^) K3 =

(3.11)

Step 3. Sum the currents leaving node 2. By the usual methods, G,{V^ - V,) +

+ G,{V^ - K3) +

0

which, after regrouping terms, reduces to - G 3 K1 + (G 3 + Gg 4. G,)V^ - G4 K3 +

= 0

( 3 . 12)

119


Step 4, Specify

in terms o f node voltages, substitute into equation 3 . 1 2 , an d simplify.

Inspecting the circuit o f Figure 3.7 shows that

is the voltage across

from node 1 to node 3.

Hence,

gm^. = & j y x - y , ) - g n y x - g r r y , Substituting

(3-13)

o f equation 3.13 into equation 3.12 leads to our second nodal equation, ^ 3)^ , + (G 3 + G4 + G ,)V ^ -{G , + ^ J K 3 = 0

(3.14)

Step 5. Sum the currents leaving node 3. Applying KCL to node 3 yields, G 2 ( V ,- V , ) ^ G , i V , - V , ) ^ G , V , - p i ^ - g ^ v ^ = 0

(3.15a)

Using equations 3.10 and 3.13 for , z and g v respectively, we have 0

= G , { V , - V ,) + G , { V , - V ,) + G ^ V ^ - ^ G , { V , - V , ) - g ^ { V , - V , )

Grouping like terms leads to our third equation in the three unknowns V^, V^, and V^:

-(G i + kJ V

x-

+ ^ 4)^2 + (^2 + G4 + I3G^ + G 5 + gJV ^ = 0

(3.15b)

Step 6. Put nodal equations in matrix form. The three nodal equations 3.11, 3.14, and 3.15b have the matrix form Gi + G 2 + G 3

Sm - G

PG^ —G 3

~ ^ 2

G 3 + Gg + G 4

3

“ ^2 “ 8 m

~Â -

~

-G 4 -

G 2 + G 4 + fiG^ + G 5 + g„

■y,'

G.V^v/

^2 =

0

^^3

0

Step 7. Substitute values and solve. Suppose that the various circuit conductances have the fol lowing values in [xS: G j = 1,000, G j = 2.0, G j = 1.0, G^ = 10, G j = 2 0 ,100, and Gg = 200 . Suppose further that

= 2.1 V, (3 = 4/1010 and^^ = 21,112 [xS. This allows us to generate the

following M ATLAB code for the solution: »G 1 = 1000e-6;G 2 = 2e-6; G 3 = le - 6 ; G 4 = lOe-6 ; »G5 = 20100e-6; G 6 = 200e-6; Vin = 2 . 1; »beta = 4/1010; gm = 2 1 1 12 e-6 ; »M =[G1+G 2+G 3 gm-G3 -G2-gm

beta*G 4-G 3

G 3+G 6+G 4 -G 4-beta*G 4

» b = [G l* V in »NodeV = M\b NodeV =

2 . 0000 e +00 -l.OOOOe+02 l.OOOOe+00

0

0]';

-G 2-beta*G 4;

-G4-gm ; G2+G4+beta*G4+G5+gm ];


120

in which case, ■V,-

■ 2

■

V^2 = - 1 0 0

V

1

V3

Step 8. Compute the voltage gain. The voltage gain o f the amplifier is given by Kut

V2

-1 0 0

Vin

Vin

2.1

= 4 7 .6 2

Exercises. 1. Suppose V-^ in the circuit o f Figure 3 .7 is doubled. W hat are the new node voltages? Hint: Consider the matrix equation o f Step 6. A N SW ER: Node voltages are doubled. 2. Suppose all conductances in the circuit o f Figure 3.7 are cut in half (resistances are doubled) and (3 is held constant. How must^^ change for the node voltages to remain at their same values? AN SW ER:

must double.

Realistic problems do not permit hand solutions. For hand solutions, the smallest number o f equations is generally desired. For matrix solutions using software packages such as MATLAB, more variables with more equations may often be easier to construct and may often result in more reliable numerical calculations. This can be illustrated using the equations o f Example 3.3. All the pertinent basic equations o f the circuit o f Figure 3 .7 can be written down as follows: from equa tions 3.9 and 3.10 we have

and

However, in contrast to the example, we do not substitute 3.10 into 3.9 to obtain 3.11. Rather, we just let them be two independent equations. Further, from equations 3.12, 3.13, and 3.15a, we have - G 3 K, + (G 3 + G4 + G ,) K2 - G 4 K3 + g^v^ = 0 and

By not substituting for and v^, we avoid unnecessary hand calculation, and if there is an error, it is easier to find. The resulting equations have the matrix form where i and now appear as additional unknowns, easily handled by a software program:

121


G] 4- G 2 + G 3

-G 3

-G 2

13

0

0

G4

-G 4

-I

0

-G 3

G 3 -1- G 4 -HGg

-G 4

0

8m

1

0

-1

0

-1

-G 2

-G 4

G2 + G 4 + G5

-/?

~Sm

1

•V,-

0

^2

= iy

0 0 0

As a general rule, we would reorder the equations so that rows 1, 3, and 5 came first, as they cor respond to the three nodal equations at Kj, V2 , and equations for i and

. Then we would write the constraint . Such a reordering leads to certain symmetry properties discussed earlier.

Exercise. Solve the above matrix equation in MATLAB or equivalent, using the numbers o f A and „ - 1 V. Example 3.3 to verify that = -1 .0 1 x 10^-3■ A

Matrix methods as used in the above examples and in the ones to follow necessitate the power o f a calculator or a software program such as MATLAB for easy solution. Such programs permit a straightforward calculation o f the required answers and are not prone to arithmetic errors. The next example illustrates how to write node equations for circuits containing a voltage-con trolled voltage source (VCVS) grounded to the reference node. The analysis o f CCVSs grounded to the reference node is similar. The more challenging analysis o f circuits containing floating dependent or independent voltage sources is taken up in the next section. EX A M P L E 3 .4 . The circuit o f Figure 3.8 models a poor operational amplifier circuit' in which the output voltage 1/^^ = V2 approximates

For the analysis, let |i = 70. The adjective “poor” arises because ^

should have a value much larger than 70. R3= lO k O

V. =

F IG U R E 3.8. A two-node (amplifier) circuit containing a grounded VCVS with jx = 70.

So l u t io n . The circuit contains two nodes labeled

and

(equivalently nodes 1 and 2) not constrained

by voltage sources. The goal o f our analysis is to find these node voltages by writing two equations in these voltages and solving. As is commonly the case, resistances are in ohms and will be con verted to conductances in S for convenience in writing the node equations.

/~N,

122


Step 1. Compute conductance values in S. Conductances are the reciprocal o f resistances, i.e., G-

= ÎRj- Hence, = 3.3 3 3 3 3 lQ-5, G^ = 10“^, G^ = 0.01, and G^ = lO'^

G j = 2.0 10-3,

Step 2 . Write a node equation at node 1. Summing the currents leaving node 1 yields

Grouping like terms leads to

{G ,^ G ^ ^ G ,)V ,-G ,V ^ = G,V^n Inserting numerical quantities yields the first node equation (3.16) Step 3. Sum currents leaving node

2

. Summing the currents leaving node 2 yields

^3 (^^2 - ^l) + ^5 ^2 + G4 {I /2 +

Kj) = 0

The dimensionless coefficient |i is placed with the conductance, obtain (HG4 -

while grouping like terms to

+ (G 3 + G4+ ^ 5)^2 = 0

Inserting the numerical values produces the second node equation 0.699 9 Vj + 0.0111 ^2 = 0

(3.17)

Step 4. Write equations 3 .1 6 and 3 .1 7 in matrix form and solve. In matrix form ■3.3333

-1

■ -V f

0 .6 9 9 9

O .O Ill

.^2.

■2' 0

Using the formula for the inverse o f a 2 x 2 matrix (interchange the diagonal entries, change the sign on the o ff diagonals, and divide by the determinant), one obtains •Vf

1

0.0111

1

0 .0 3 0 1 2 6

.^2.

0.7369

-0 .6 9 9 9

3.3333

-1 .8 9 9 6

In Example 3.4, observe that

=

^^ 2

= -1 -8 9 9 6 , which approximates - 2 V-^ since V- = 1 V.

Exercises. 1. Write MATLAB code to solve the above example. Check that your code works. Hint: See Example 3.3. 2. If R2 is changed to 100 k£2 in Example 3.4, show that V2 = -1 .9 0 6 3 V.


123

4. N O D AL AN ALYSIS II: FLOATING VO LTAG E SO URCES A floating voltage source means that neither node o f the source is connected to the reference node. When a floating dependent or independent voltage source is present with respect to a given reference node, a direct application o f KCL to either terminal node o f the voltage source is unfruit ful. There are several ways to handle this situation. One fruitful method is to enclose the source and its terminal nodes by a Gaussian surface, i.e., a closed curve, to create what is commonly called a supernode, as illustrated in Figure 3.9. One would then write KCL for the supernode as is done in a number o f circuit texts. However, there is a conceptually more straightforward approach, which is often called the modified nodal analysis, or MNA. In MNA, we add an addi tional current label to each floating voltage source. In Figure 3.9, we have added the current label

l^y. This additional current becomes an unknown in a set o f nodal equations generated by apply ing KCL to each node. At this point, further explanation is best done by an example, but the con cept is similar to the discussion following Example 3.3. E X A M P L E 3.5. Find the node voltages V^, Vy,

and the unknown current

in the circuit o f Figure 3.9, when

the bottom node is taken as reference.

FIG U RE 3.9. Resistive circuit containing a floating voltage source for the given reference; generally, the reference node may be chosen arbitrarily.

So l u t io n . Step 1. Write a node equation at node a. Summing the currents leaving node a yields 8 + 0.15

+ 3 + 0.2 ( K ^ -K .) = 0

After grouping terms appropriately, we have (0.15 + 0 .2 )K ^ - 0.15^ -^ - 0 .2 V ;+ 8 + 3 = 0 or, equivalently, 0.351^ ^ -0.15K ^ -0.2V ^ ^ = - 1 1 This provides our first equation in four unknowns.

(3.18)

124


Step 2. Write a nodal equation at node b. Here, =0

- 3 + 0 .1 5 ( K ^ - K ; + 0.05 or equivalently, - 0 .1 5

(0 .1 5 + 0.05) n - / ^ ^ = 3

Simplifying this expression leads to - 0 .1 5 K , + 0.2K^-/^^ = 3

(3.19)

Step 3. Write a nodal equation at node c. Here, ^.^ + 0-25 V ^-25 +

0

2

{V^- K J = 0

or equivalently. - 0 .2 K ,+ 0.45K^ + /,, = 25

(3.20)

Step 4. Write the node voltage relationship fo r the terminal nodes o f the floating voltage source, i.e., between the voltages

and V^. The voltages

Mathematically, this constraint is 1/ -

and

are constrained by the voltage source.

= 440, i.e., K ^ = 440

(3.21)

Step 5. Write thefour equations 3.18, 3.19, 3.20, and 3.21 in matrixform and solve. In matrix form, 0.35

- 0 .1 5

- 0 .2

0

-1 1

- 0 .1 5

0.2

0

-1

3

- 0 .2

0

0.45

1

25

0

-1

1

0

hb

(3.22)

440

Because o f the extra variable, the equations become too large for hand calculation. Hence, we use MATLAB as follows: >>M = [0.35 - 0 .1 5 - 0 .2 0 ;-0 .1 5 0.2 0 - 1 ; - 0 .2 0 0.45 1 ; 0 - 1 1 0]; »b = [-11 3 25 440]'; »x = M\b X =

-9.0000e+ 01 -3 .1 0 0 0 e + 0 2 1.3000e+02 -5.1500 e+ 01 Hence, = - 90 V,

= - 3 1 0 V,

= 130 V,

= -5 1 .5 A

Chapter 3 “ Nodal and Loop Analyses

12 5

In a conventional nodal analysis, all unknowns are node voltages. Here we have the additional unknown current,

. Because o f this additional unknown current, the method is called a mod

ified nodal analysis. Also, in this example, node d was taken as the reference node. However, one could just as easily take node b as the reference node, in which case, the voltage source would not have been floating. A home problem investigates this choice o f reference node.

Exercise. 1. For Example 3.5, compute the voltages

and

2. For Example 3.5, compute the power absorbed by the 0.15 S resistor. 3. Compute the power delivered by the floating voltage source. AN SW ERS in random order: 220 V, 22.6 6 kw, 310 V, 440 V, 7260 watts

The next example investigates a circuit having floating independent and dependent voltage sources. By convention, the reference node o f this circuit, figure 3.10, and all subsequent circuits, will be the bottom node o f the circuit unless stated otherwise. E X A M P L E 3.6. The circuit o f Figure 3.10 contains a floating independent and a floating dependent voltage source. Find the node voltages V^, Vy, V^, and the unknown currents

and

Then find the

power delivered by the 30 V source and the dependent source. 500

1

FIG U RE 3.10. Resistive circuit containing a floating dependent voltage source and a floating independent voltage when node d is chosen as the reference node.

So l u t io n . Step 1. Sum currents leaving node a. Here,

126


Equivalently,

100

(3.23)

100

“

Step 2. Sum currents leaving node b. Here,

100

500

Equivalently, (3.24) Step 3. Sum currents leaving node c. Here,

1 (3.25)

800

Step 4. Write an equation relating the terminal voltages o f the independent voltage source. Here, K ,-n = 3 0

(3.26)

Step 5. Write an equation relating the terminal voltages o f the dependent voltage source. Here, 40 K ,- K = 4 0 ., = — Equivalently, (3.27)

0.6V;, + 0 .4 y ^ - V ^ = 0

Step 6. Write equations 3.23 through 3 .2 7 in matrix form an d solve in MATLAB. Combining the above equations into a matrix produces 0 .03 -

0.01

-0 .0 1

0

1

0

0.012

0

0

-1

0.0 0 1 2 5

-1

1

0

0

âc

30

0

0

Icb,

0

0

0

0

-1

0 .6

0.4

________ -1

■2.2' 0

V'i

=

0

Again, this matrix equation is too large for hand computation. Hence in MATLAB, »M = [0.03 - 0 .0 1 0 1 0; -

0.01 0 .012 0 0

-

0 0 0.00125 - 1 1; 0 - 1 1 0 0; 0.6 0 . 4 - 1 0 0];

1;


127

>.b= [2.2 0 0 30 0]'; »x = M\b X =

l.OOOOe+02 5.0000e+01 8.0000e+01 -3 .0 0 0 0 e -0 1 -4 .0 0 0 0 e -0 1 Hence, 100 ■ 50 =

80

âc

- 0 .3

Icb_

- 0 .4

Step 4. Compute the power delivered by the 3 0 Vsource. The power delivered by the 30-V source is ^^./ = - 3 0 / ,^ = 3 0 x 0.4 = 1 2 W Step 5. Compute the power delivered by the dependent source. The power delivered by the depend ent source is Pdel =

=

- 4 0

X

^ ( - 0 . 3 ) = 0.12(V^ - V^) = 6 W 100

Exercises. 1. For Example 3.6, compute the voltages

and the power absorbed by the 800

Q resistor. A N SW ERS in random order: 8 watts, 20 V, - 8 0 V 2. Suppose the two independent voltage source values in Example 3.6 are doubled. W hat are the new node voltages? W hat are the new branch currents? A N SW ERS: Node voltages are doubled and branch currents are doubled. 3. Suppose all resistances in the circuit o f Figure 3.10 are doubled and the value o f the parameter on the dependent source is also doubled. W hat are the new branch currents? AN SW ER: All branch currents are cut in half.

The above example increases the number o f unknowns beyond the node voltages to include the two currents through the floating voltage sources. However, we could have included additional currents to the set o f equations making the dimension even higher. W ith a tool like MATLAB, this poses no difficulty. However, it does make hand computation a challenge. For example, we

128


= l^y as an additional variable with a corresponding increase in the num

could have included

ber o f equations. By adding addirional unknowns we would simplify the writing o f the individual node equations but increase the dimension o f the matrix equation. Specifically, the node equation at “a” becomes _

t i

I

50

IV

50

and the resulting larger matrix equation is 0.02

0

0

1

0

2.2

0

0.002

0

0

-1

0

0

0

0.00 1 2 5

-1

1

0

0

-1

1

0

0

0

0

-1

0

0

-4 0

-1

0

0

0

-100

0 âc

30

Ic h

0 0

This completes our discussion o f the standard nodal equation method o f circuit analysis. T he next section takes up a discussion o f an alternative analysis method entided loop analysis.

5. LOOP ANALYSIS Loop analysis is a second general analysis technique for computing the voltages and currents in a circuit. Mesh analysis is a special type o f loop analysis for planar circuits, i.e., circuits that can be drawn on a plane without branch crossings. For planar circuits, loops can be chosen as mesh es, as illustrated in Figure 3.2, or as in 3.11 below. Associated with each loop is a loop current. Loop currents circulate around closed paths (loops) in the circuit. Similarly, for planar circuits, the term mesh current is used traditionally for loop current. By KVL, the sum o f the voltages across each branch in a loop is zero. By expressing each o f these branch voltages in terms o f the designated loop currents, one can write an equation in the loop currents for each designated loop in the circuit. For branches that are often common to two or more designated loops, the branch current equals the net flow o f the loop currents incident on the branch. Writing an equation for each loop produces a set o f equations called loop equations. If sufficient independent loops are defined, one can solve the loop equations for the loop currents. Once the loop currents are known, we can easily compute the branch currents and then the branch voltages in the circuit. Then we can compute any other quantities o f interest, such as power absorbed, power delivered, voltage gain, etc. EXA M PLE 3.7. Consider the planar circuit o f Figure 3.11 with the three specified loops, which are also called meshes. Denote the “loop” currents for each loop by /j, 1^, and /j. The objective is to write three equations in the currents /j, and /j using KVL and solve these equations for their values. Then we will compute the power absorbed by the 2 -Q resistor marked with the voltage v. Suppose the source voltages are = 4 0 V andV^2 = 20 V.


129

FIG U RE 3.11. Resistive circuit containing only independent voltage sources for the loop analysis of Example 3.7.

So l u t io n . Step 1. Write a KVL equation based on loop 1 by summing voltages around this loop. Summing the voltages around loop 1 using Ohm’s law and the defined loop currents produces K., = 40 = /i + 4 (/; - /j) + Va + ih - ^3) = 6 /, - 4 /2 - 7 3 + 20

(3.25a)

Here, observe that the 4 Q resistor is incident on two loops; the net current flowing from top to bottom, i.e., with respect to the direction o f loop 1, is /j - l 2 - The idea is analogous to a pair o f distinct water pipes that share a common length. The common length is analogous to the 4 -Q resistor. The flow rate in each pipe is analogous to the currents /] and I 2 , which in fact, are rates at which charge flows past a cross sectional area o f the conductor. It follows that the net flow through the common length o f pipe with respect to the direction o f loop 1 is the difference in the net flow rates o f pipes 1 and 2, respectively. This is precisely the meaning o f /j - /2 . A similar explanation can be made for the 1 -Q resistor common to loops 1 and 3 for which the net flow rate with respect to the direction o f loop 1 is 7j - ly Simplifying equation 3.25a yields

6/1 - AI^ - 73 = 20

(3.25b)

Step 2. Write a KVL equation based on loop 2 by summing the voltages around this loop. Applying Ohm’s law and KVL to loop 2 produces 0 = 4(72 - 7,) + 272 + 2(^2 -

(^.26)

h'

Notice that with respect to the direction of loop 2, the net flow rate through the 4 Q resistor is

Step 3. Finally, write a KVL equation based on bop 3. Stmiming the voltages around loop 3 yields

V^2 = 20 = 2(73 - 72) + 73 + (73 - 7,) = - 7i - 272 + 473

(3.27)


130

Step 4 . Write eqtiations 3.25b, 3.26, an d 3 .2 7 in matrix form an d solve. Writing the above three loop equations in matrix form yields '6

-4

-r

W

-4

8

-2

h

-1

-2

4

■20'

-

0

(3.28)

20

h

Solving Equation 3.28 by the matrix inverse method (by a numerical algorithm or by Cramer’s rule) yields the loop currents in amps as

h'

6

-4

-r

-1

0 .35

'2 0 '

h

= -4

8

-2

0

h

-1

-2

4

20

= 0 .2 2 5

0 .225

0 .2

■20'

0 .2 8 7 5

0 .2

0

0 .2

0 .4

20

0 .2

11 ■ =

8.5 12

Step 5. Compute the power consumed by the 2 Q resistor. Knowledge o f the loop currents makes it possible to compute all voltages and currents in the circuit. For our purpose, the voltage

and the power absorbed by the 2 Q resistor is

/ 2 = 24.5 watts.

Exercises. All exercises are for the circuit o f Figure 3.11. 1. Compute the power delivered by the 20 V source. AN SW ER: 20 watts 2. Compute the power absorbed by the 4 Q resistor. AN SW ER: 25 watts 3. Suppose the source values are doubled. W hat are the new values o f the loop currents? AN SW ER: loop currents are doubled 4. Suppose the resistance values are multiplied by 4. W hat are the new loop currents? W hat are the new node voltages? AN SW ERS: Loop currents are 0.25 times their original values, and node voltages are unchanged.

Observe that there are no dependent current or voltage sources in the circuit. Similar to the nodal analysis case, whenever dependent sources are absent and the equations are written in the natural order, the loop (or mesh) equations are symmetric, as illustrated by the coefficient matrix o f equa tion 3.28 where, for example, the 1 -2 and 2 -1 entries coincide, as do the 1 -3 and 3 -1 entries, etc. Also, the value o f all entries can be computed by inspection. The 1-1 entry o f the matrix is the sum o f the resistances in loop 1; the 2 - 2 entry is the sum o f the resistances in loop 2, etc. In general, the i- i entry is the sum o f the resistances in loop i. T he 1 -2 entry o f the matrix is 'L{±R^ (the large sigma means summation), where each

is a resistance common to both loops 1 and 2.

Use the + sign when both loop currents circulate through

in the same direction, and use the -


131

sign otherwise. Further, if independent current sources are absent, then the right-hand side o f the loop equations can also be written by inspection. The i-th entry is simply the net voltage o f the sources in the i—th loop that tends to deliver a current in the direction o f the loop current.

Exercises. 1. Use the inspection rules described above to write two mesh equations for the circuit o f Figure 3.12, when both mesh currents are assigned clockwise direction.

FIG U RE 3.12. 2. Use the inspection rules described above to write two mesh equations for the circuit o f Figure 3.12, when the left mesh current is clockwise and the right mesh current is counterclockwise. 3. Use the inspection rules described above to determine the right-hand side o f the mesh equation for the circuit o f Figure 3.13.

AN SW ERS: 8, 0, 10

A simplifying reduction to the set o f loop equations occurs if an independent current source coin cides with a single loop current. The analysis becomes simpler because that loop current is no longer an unknown; rather it is equal to the value o f the source current if their directions coin cide, or to the negative value if their directions are opposing. Because the associated loop current is known, there are fewer loop equations to write and solve. One would apply KVL to such a loop only if it were necessary to compute the voltage across the independent current source, which might be necessary for determining the power delivered by the source. Th e entire situation is anal ogous to an independent voltage source tied between a node and the reference in nodal analysis. The following example illustrates the details o f this discussion.


132

EXA M PLE 3.8. The circuit o f Figure 3.1 4 is a modification o f the one o f 3.11 in which (i) a 1 ohm resistor on the perimeter o f the circuit is replaced by an 8 A independent current source, and (ii) the values o f the voltage sources are doubled. The currents for each loop are again denoted by /j, I 2 , and ly Our objective is to find all the loop currents, the voltage V^, and the power delivered by the 8 A source.

FIG U RE 3.14. A resistive circuit containing an independent current source on the perimeter o f loop 3 forcing /^ = 8 A.

So l u t io n . Step 1. So/ve fo r

by inspection. Because

is the only loop current circulating through the

branch containing the independent 8 A current source, /j = 8 A. This phenomena is similar to the fact that in nodal analysis, the node voltage o f a grounded voltage source is fixed at the voltage source value.

Step 2. Write a KVL equation fo r loop 1 by summing voltages around this loop. Summing the voltages around loop 1 using Ohm’s law and the designated loop currents produces 28 = / j + 4 ( /j - /2) + 12 + ( /j - 8) = 6 /, - 4 /2 + 4 Hence, 6 /, - 4 /2 = 24

(3.29)

Step 3. Write a KVL equation fo r loop 2 by summing the voltages around this loop. Applying KVL and Ohm’s law to loop 2 produces 0 = 4 (/2- /i) + 2 /2 + 2 (/2 - 8 ) = - 4 /, + 8 /2 - 16 Equivalently, - 4 /1 - 8 /2 = 16

(3.30)


133

Step 4 . Write above loop equations in matrix form and solve. The matrix form o f equations 3.29 and 3.30 is

' 6 -A

'i r

’24‘

-4

l2

16

8

Using the inverse matrix technique to compute the solution, we have II"

■6 -4

I2

-4

-1

8

24 '

1

16

“ 32

' 24' 4 6 16

'8

4

's' 6

Step 5. Compute V^. By KVL, v; = 2 (/2 - 8) + 12 + (/j - 8) = 8 V Step 6. Compute power delivered by 8 A source. Observe that the 8 A current source is labeled according to the passive sign convention, in which case,

Pdel = -

= - (8

8) = - 64 watts

Hence, the source actually absorbs 64 watts.

Exercise. In the circuit o f Figure 3.15, nvo o f the three mesh currents coincide with independent source currents. By writing and solving just one mesh equation, find /j.

A N SW ER: 3 A

Not only do independent current sources constrain loop currents, but dependent currents sources do also. This situation is illustrated in Example 3.9. E X A M PLE 3 .9 . This example illustrates the writing o f loop equations for a simplified small signal equivalent cir cuit, Figure 3.16, o f a two-stage amplifier that contains a current-controlled current source (C C C S) and a current-controlled voltage source (CC VS). This process extends the techniques o f Examples 3.7 and 3.8 to find some important characteristics o f the amplifier. Specifically, find (a) the input resistance seen by the source, i.e.,

= v-Jij^,

134


(b) the voltage gain, v

j and

(c) the voltage v across the dependent current source.

+

V

FIG U RE 3.16. Small signal equivalent circuit for a two-stage amplifier. Signals in amplifiers are usually time dependent, so we adopt the lowercase notation for voltages and currents.

So l u t io n . The circuit o f Figure 3.16 contains three loop or mesh currents. The direction o f the loops is a user-chosen preference. For convenience, we have chosen mesh current z'2 to be consistent with the direction o f the arrow in the dependent current source. Because this dependent current source lies on the perimeter o f the circuit, it constrains the value o f /'2> i-e-) ?2 ^ P ^b-

the control

ling current, , z'2 = . ih ~ ^ h' relationship implies that the mesh current o f loop 2 depends directly on the mesh current o f loop 1. This observation allows us to skip constructing a mesh equation for loop 2. Only equations for loops 1 and 3 are needed, thereby reducing the number o f simultaneous equations from three (because there are three loops) to two. Step 1. Apply KVL to loop/mesh 1. Here, by KVL and the observation that

în = h h +

('1 + P ' 1) =

+ (1 + p)

12

= P zp

i\

From this equation, we can immediately compute the input resistance V•

in

V•

in

(3.31)

Step 2. Apply KVL to loop/mesh 3. In this case, observe that z^ = — (z^ + z'g) = - (|3 z'j + Zj). By BCVL,

'3 -

K + K ('3 + ' 2) =

h +

(P h + ' 3) + K ('3 +P ' 1) = 0

Combining like terms, it follows that

K +

ph +

'3 = 0

(3.32)

135


Step 3. Write equations 3.31 and 3.32 in matrix form an d solve. The matrix form o f these equa tions

IS

/?^ + (P + l ) ^ ,

0

h

în

'3

0

(3.33)

)

V.

Because the solution is desired in terms o f the hteral variables, we solve equation 3.33 using Cramer’s rule, which utilizes determinants. In this task, first define

Using the notation A for the determinant, Cramers rule provides the solution for i^ according to the formula

det

în

0

0

R^ + R ^+r„

(3.34)

‘I = Applying Cramer’s rule for the solution o f Zj, yields

det

Step 4. Compute

/?^ + (|3 + l)7 ? e

V •

m

(3.35)

in terms o f v-^ and then the voltage gain vjv^^. As per the circuit o f Figure

3.11 and equations 3.34 and 3.35,

Vo =

V ;„ = r

ic = '•w(PM + '3) = r,i

After substituting for A, the voltage gain is

o

---V.

in

A

P/? I— d r= y

^

—r

Step 4 . Compute v. To compute the voltage across the dependent current source, apply KVL to mesh 2 to obtain n

= R^ (z'2 + /'i) + R, («2 + ^3) = [P

h +

h

136


Exercise. Find a simplified loop equation for

in the circuit o f Figure 3.17.

FIG U RE 3.17. c' '1

To see the importance o f the calculations o f the amplifier circuit o f Example 3.9, suppose two amplifiers are available for use with a non-ideal voltage source. The non-ideal voltage source is modeled by an ideal one-volt source in series with a 100 Q source resistance. Suppose amplifier 1 has a voltage gain, vjv^^ = 1 0 and

= 100 k£2. Suppose amplifier 2 has a voltage gain o f 100

and R -^ 2 = 5 £3. If amplifier 1 is attached to the non-ideal source, then by voltage division,

=

100,000/(100,000 + 100) = 0.999 V, whereas in the case o f amplifier 2, v -^ 2 = 5/(100 + 5) = 0 .0 4 7 6 V. In the first case, the gain from to is 10, yielding = 9.99 V. In the second case, the same gain is 100, yielding v^ 2 =

V. One concludes that amplifier 1 is better suited to this

particular application, although it has a lower voltage gain than amplifier 2. Hence, Example 3.9 illustrates the need to know both the voltage gain and the input resistance to determine the out put voltage in practical applications. Further, using the literal solution to the example allows us to apply the formulas to different sets o f parameter values without repeating the complete analysis. In the previous two examples, there were current sources on the perimeter o f the circuit. Such cur rent sources were incident to only one loop. It often happens that independent and dependent current sources can be common to two or more loops. When this happens, a situation analogous to floating voltage sources in nodal analysis occurs. To handle such cases, many texts define some thing called a supermesh and write a special loop equation for this supermesh. Supermeshes often confuse the beginner. There is an easier way. Example 3.10 below illustrates how to write “loop” equations when current sources are common to two or more loops. In such cases, we introduce auxiliary voltage variables across current sources common to two or more loops. The resulting set o f simultaneous equations will contain not only the loop currents as unknowns, but also the auxiliary voltages as unknowns. Because the resulting set o f equations contains both loop currents and additional (auxiliary) voltage variables, the equations are called modified loop equations. T he process o f writing modified loop equa tions is extremely systematic and straightforward. Further, it allows us to avoid explaining the very confusing concept o f a supermesh. On the other hand, the presence o f auxiliary voltage variables increases the number o f “unknowns,” i.e., the number o f simultaneous equations increases. Because o f the availability o f software packages such as MATLAB, M ATH EM ATICA, and MAPLE, this increased dimension is not a hindrance.

137


E X A M P L E 3 .1 0 . Consider the circuit o f Figure 3-18 in which

= 28 V and 7^2 = 0 .06 A. Note that the inde

pendent current source is common to loops 1 and 3 and a voltage-controlled current source is common to loops 1 and 2. Find values for the loop currents 7p Ij, l y and the power delivered by each independent source.

200Q

FIG U RE 3.18. Circuit containing a current source between loops.

So l u t io n . To begin the solution, we introduce two auxiliary voltage variables Vj and Vj associated with the current sources common to two (or multiple) loops. The purpose o f these variables is to facilitate the application o f KVL for constructing the loop equations. This will require that we obtain three KVL equations, one for each loop, and two constraint equations, one for each current source. Step 1. Apply KVL to loop 1. By a clear-cut applicanon o f KVL, 28 = 2007] - V] - V2

(3.36)

Step 2. Apply KVL and Ohm’s law to loop 2. Again applying KVL and Ohm’s law to loop 2, we obtain 100 7j + 200 (Jr^ —7^) + V2 = 0. After grouping like terms. 3 0 0 7 , - 20 0 /3 + V2 = 0

(3.37)

Step 3. Apply KVL to loop 3. Applying KVL to loop 3 yields 150 7^ + Vj + 200 (/j - Tj) = 0. Equivalendy, - 200/2 + 350/3 + V| = 0 Step 4. Write a constraint equation determined by the independent current source. Here, loops 1 and 3 are incident on the independent current source so that 0 .0 6 = / , -/ 3

(3.39)

138


Step 5. Write a constraint equation determined by the dependent current source. In a straightfor ward manner, we have / , - / , = 0 .0 2 V, = 0 .02 r2 0 0 (/3 -

12

) = 4 7 , - 47,

After simplification, /i + 3/2 - 4 /3 - 0

(3.40)

Step 6. Write equations 3 .3 6 to 3.40 in matrix form an d solve. The matrix form o f these equations is 0

0

-1

-r

w

300

-2 0 0

0

1

h

-2 0 0

350

1

0

h

0

-1

0

0

''I

0 .0 6

3

-4

0

0

V2

0

■ 28 0 =

0

(3.41)

Solving equation 3.41 by the matrix inverse method or by an available software package yields the solution (currents in A and voltages in V) given by equation 3 .42 below:

w

200

0

0

-1

h

0

300

-2 0 0

0

0

-2 0 0

350

1

0

-1

I

3

-4

Li ,''2.

=

j

-r -1 ■ 28

0 300 -200 0 1 0 -200 350 1 0 1 0 -1 0 0 1 3 -4 0 0]; »b = [28 0 0 0.06 0 ]’; »LoopIplus= M\b Looplplus = l.OOOOe-01 2,0000e-02 4,0000e-02 -l.OOOOe+01 2.0000e+00

■0.1 ■

1

0

0 .02

0

0

= 0 .0 4

0

0

0 .06

-To

0

0

0

2

1

.

which can be obtained using the following MATLAB code: M = [ 2 0 0 0 0 -1 -1

■

(3.42)

139


Step 7. Compute the powers delivered by the independent sources. First, the power delivered by the independent voltage source is

Py-source = 28 /j = 2.8 watts The power delivered by the independent current source is

Pl-source =

V^ = - 0 .6 watts

This last value indicates that the independent current source actually absorbs power from the cir cuit.

Exercise. For the circuit o f Figure 3.19, write the modified loop equations having two unknowns /j and V, following the procedure described in Example 3.10. Solve the equations and find the power absorbed by the 2 -Q resistor.

4A

AN SW ER: 18 watts

One final point before closing our discussion o f loop analysis. Loops can be chosen in different ways. Cleverly choosing loops can sometimes simplify the solution o f the associated equations. For example, by choosing a loop that passes through a current source so that no other loop is com mon to the source, the loop current is automatically specified by that current source.

6. SUM M ARY This chapter introduced the technique o f nodal analysis. Nodal analysis is a technique for writing a set o f equations whose solution yields all node voltages in a circuit. With knowledge o f all the node volt ages and all the element values, one can compute all branch voltages and currents. As mentioned, when ever there are no dependent sources present, the coefficient matrix o f the node equations is always sym metric. Hence, whenever dependent sources are absent, it is possible to write the nodal equation coef ficient matrix by inspection. Further, if independent voltage sources are absent, then the right-hand side o f the matrix form o f the nodal equations can also be written by inspection: the entry is simply the sum o f the independent source currents injected into the node at which KCL is applied. When VCCSs are present, the steps for writing nodal equations are the same as illustrated in Example 3.3. Generally, in such cases, the resultant coefficient matrix is not symmetric.

140


W hen floating dependent or independent voltage sources are present with respect to a given ref erence node, we introduce new current variable through these floating sources as unknowns. The node equations then incorporate these additional unknown currents, as was illustrated in Examples 3.5 and 3.6. This method increases the number o f equations but simplifies the con struction o f the individual equations. W ith a tool like MATLAB to compute solutions, there is no difficulty, although hand computation may become more difficult. This concept is the basis o f the modified nodal analysis method used in circuit simulation programs like SPICE. Loop/mesh analysis, an approach dual to nodal analysis, was introduced in Section 5. Mesh analy sis is a special case o f loop analysis for planar circuits when the loops are chosen to be the obvious meshes, similar in geometry to a fish net. In loop analysis, one sums the voltages around a loop or mesh to zero. Each o f the branch voltages in the loop is expressed as a product o f resistances and (fictitious) loop currents that circulate through the branch resistance, as illustrated in Figures 3.10, 3.14, and 3.16. The branch current o f the circuit are equal to the net flow o f the loop currents incident on a particular branch, meaning that each branch current is expressible as a sum o f loop currents. The desired set o f loop equations is produced by summing the voltages around each loop, expressing these voltages either as source values or as resistances times loop currents. One solves the loop equations for the loop currents. Once the loop currents are known, we can then compute the individual branch currents and then the branch voltages, and thus any other pertinent current, voltage, or power. Whenever there are no dependent sources present, the coefficient matrix o f the loop equations is always symmetric. Whenever dependent sources are absent, it is possible to eas ily write the loop matrix by inspection. As the size o f an arbitrary circuit grows larger, there are two good reasons for choosing the nodal method over the loop method; (i) the number o f nodal equations is usually smaller than the num ber o f loop equations, and (ii) the formulation o f nodal equations for computer solution is easier than methods based on loop equations. Writing nodal equations is particularly easy if the circuit contains only resistances, independent current sources, and VCCSs — for short, an R— I— g„ net work. For an

network, one simply applies KVL to every node (except the reference node)

and obtains a set o f node equations directly. For floating independent or dependent voltage sources, the task is more complex. Examples 3.5 and 3.6 illustrate cases where, besides the node voltages, additional unknown auxiliary currents are added. By adding additional auxiliary variables to the formulation o f the nodal equations, we described the concept behind the modified nodal analysis (MNA) method. The MNA method retains the simplicity o f the nodal method while removing its limitations and is the most commonly used method in present-day computer-aided circuit analysis programs.

7. TERM S AND C O N C EPTS Connected circuit: every pair o f nodes in the circuit is joined by some set o f branches. Cram er’s rule: a method for solving a linear matrix equation for the unknowns, one by one, through the use o f determinants; the method has serious numerical problems when implemented on a computer, but is often convenient for small, 2 x 2 or 3 x 3, hand cal culations. Floating source: neither node o f the source is connected to the reference node.

Chapter 3 ®Nodal and Loop Analyses

141

Gaussian surface: a closed curve or a closed surface surrounding two or more nodes. Linear matrix equation: an equation of the form Ax =b, where A i s z n x n matrix, x is an n-vector of unknowns, and b is an n-vector of constants. Loop (closed path): a contiguous sequence of branches that begins and ends on the same node and touches no node more than once.

Loop analysis: an organized method of circuit analysis for computing loop currents in a circuit. Knowledge of the loop currents allows one to compute the individual element currents and, consequently, the element voltages.

Loop current: a (fictitious) current circulating around a closed path in a circuit. Matrix inverse: the inverse, if it exists, of an n n matrix yl, denoted b y ^ “ ^ satisfies the equation A A~^ = A~^ ^ = I,, where / is the w x « identity matrix; the solution of the linear matrix equation, ^ is given hy x = A~^b, Mesh: After drawing a planar graph without branch crossing, the boundary of any region with finite area is called a mesh. Intuitively, meshes resemble the openings of a fish net.

Mesh analysis: the special case of loop analysis for planar circuits in which the loops are chosen to be the meshes.

Mesh current: a fictitious current circulating around a mesh in a planar circuit. Modified nodal analysis: a modification of the basic nodal analysis method in which the unknowns are the usual nodal voltages plus some naturally occurring auxiliary currents.

Nodal analysis: an organized method of circuit analysis built around KCL for computing all node voltages of a circuit.

Node voltage: the voltage drop from a given node to a reference node. Symmetric matrix: a matrix whose transpose is itself I f ^ = is a « x « matrix whose i- j entry is then A is symmetric if a - = a-.

142


3. For the circuit o f Figure P3.3, suppose = 1.2 A. Write a single node equation in the volt

PROBLEMS

age V and solve.

SIN G LE N ODE PROBLEM S 1. For the circuit o f Figure P3.1, write a single node equation in , Gj > <^3 > For a fixed K > 0 , R-^ = R , R2 = 2R , Compute Kj

^ 2-

= 2R. Figure P3.3

in terms o f R and V^j A N SW ER: - 6 V

4 Ksi-

M ULTIPLE N ODE PROBLEM S 4. The purpose o f this problem is to write the nodal equations direcdy by inspection o f the cir cuit diagram o f Figure P3.4. Recall that when the

Figure P3.1

network has only independent current sources and resistors, the nodal equation matrix is symmetric

A N SW ER: K, = l.5V^^

and the entries can be written down by inspection

2 . The battery o f your car has been dealt a sud

as per the discussion following Example 3.2.

den death by the sub-zero North wind and a

Construct the nodal equations in matrix form for the circuit o f Figure 3.4 by inspection.

faulty alternator. Unable to fight the elements, you wait a few days hoping for a thaw, which comes. You replace the alternator. Then, using your roommate’s car, you attempt a jump-start. Nothing happens. You let it sit for a while with your roommates car running juice into your battery for 20 minutes. Still, nothing happens. W hy won’t your car start? Consider the circuit

reference node

o f Figure P3.2. Notice that your “dead” batter

Figure P3.4

is labeled as Vq. Your roommate’s battery is

A FEW A N SW ERS: 'Fhe 3-3 entry is G^ + G^

labeled 12 V. Each battery has an internal

+ G^ + Gg, and the 2-1 entry is -G y

resistance o f 0.02 Q and the starter, an internal resistance o f 0.2 Q. The starter motor requires

5. Consider the circuit o f Figure P3.5 in which

50 A to crank the engine. Find the minimum

/^, = 0.5 A and V^ 2 = 40 V. Furtiier, let G, = 5 mS, G2 = 2.5 mS, G3 = 2.5 mS, and G^ = 12.5 mS. (a) By inspection, what is the

value o f voltage Vq needed before the starter can draw 50 A and work.

value o f 50 A

>0.020

7 0.20 ''load < v

I Live Battery

Write a minimum number o f node equations and put in matrix form.

0.02Q V

12V

(b) (c)

Solve the node equations for the voltages and

using MATLAB or the for

mula for the inverse o f a 2 x 2 matrix: Dead Battery

Figure P3.2

Starter Motor

a b c d

_

1

'd -b'

ad - be

-c a


143

(d)

Find 1 ^ ,7 ^ and

(e)

Find the power delivered by each source and the power absorbed by each resistor. Verify the principle of conservation of power.

8 . The circuit of Figure P3.8 is an experimental

reference node d

Figure P3.5

measurement circuit for determining tempera

In the circuit of Figure P3.6,

ture inside a cavern underneath the Polar ice 6 . (a)

= 8

cap. The cavern is heated by a fissure leading to

V* Further, = 5 kQ, R l = R^ = R^ = 20 k£2, and R^ = 10 kfl. Find the node voltages, and

some volcanic activity deep in the earth. The

V^, and also the voltage, V^. Compute

the range -2 5 ° C to +25°C . The nominal tem

the power absorbed by R^ and the

perature of the cavern is 0°C. In this type of cir

power delivered by each of the sources.

cuit, the voltage

It is suggested that you write your

the temperature changes. Suppose that

equations in matrix form and solve

V, and in kQ, R-^ = 20, i?2 ” 44, R^ - 20, and

using MATLAB or the formula for a 2

R^ = 12.5. Note that the 4 4

mA, V^2 -

X

(b)

2 inverse given in problem 5.

resistor

changes its value linearly from 15

IdQ to 65 k n as a fiinction of temperature over

- Vg is a measure of how = 50

resistor is a

result of manufacturing tolerances that often

Repeat part (a) when all resistances are

permit deviations from a nominal of, say, 40

cut in half

kflt, by as much as 2 0 %. As usual, it is cost ver sus precision. (a)

Write a set of nodal equations in the

(b)

Assuming

----------^/S/^-------

variables

and Vq = 40 kQ at 0°C, put

the nodal equations in matrix form and solve for the node voltages, and Vq (c)

Determine the power delivered by the source.

Figure P3.6 (d)

Use MATLAB to solve for all the node

7. In the circuit of Figure P3.7, ^2 = 4 V, and - 1 mA. Further, in mS, G| =

voltages as

0.4, G2 = 2, G3 = 3, and

out. Then find the linear equation

analysis to find

= 5. Use nodal

and V^. Then compute the

power delivered by the independent sources and the power absorbed by G2 .

varies from 15 to 65

ki2 in 1 kfi increments. Do not print relating to temperature. Plot Vq - y c as a W c tio n of temperature, i.e., over the range - 2 5 ° C to +25°C . Over what range of temperatures about 0 degrees would the sensor be reason ably accurate? W hy and why not?

144


II.

Consider the circuit o f Figure P 3 .l l .

Choose node D as the reference node. This choice eliminates the floating voltage source and hence the nodal equations can be written without the need o f a so-called supernode. Let = 0.0 8 S, Gj = 0.08 S, 0.0 2 S, G 5 = 0 .0 2 S, = 0.3 A, and

= 0.01 S, G^ =

= 0.3 A, 7^2 = 0 .2 A, 7^3

= 50 V. Write and solve a set

o f nodal equations for the voltages

Figure P3.8

and Vg = Vg0 . Then compute the powers 9. In the circuit o f Figure P3.9, ail resistances are 1 ItQ, except

= 500 Q. Suppose

delivered by each o f the sources.

= G,

100 V and 1 ^ 2 = 0-3 A. Compute all tiie node voltages o f the circuit. You may want to use MATLAB or a calculator that inverts matrices

G,

to compute the answer. Compute the power

© '

delivered by the independent sources.

Figure P 3 .1 1 12. In the circuit o f Figure P 3.12, = 20 kQ, T?3 = 20 kQ, = 5 mA. Find Figure P3.9 1,7,0.6,7

10. In the circuit o f Figure P3.10,

= 30 V,

= 0.6 A. Use nodal analysis

on the circuit below, as indicated: (a)

Figure P3.12

Write a nodal equation at node A.

(b)

Write a nodal equation at node B.

(c)

Write a third nodal equation at node C.

(d)

Solve the 3 equations in 3 unknowns

13. Consider Figure P3.13. (a)

M ATLAB, or using some other soft ware program to obtain all the node

this, let G-=\ I R -. (b)

W ith K ^ = 150V ,7?^ = lk Q ,7?, = 5 k Q ,

(c)

= 15 , Vg the power delivered by , and the power absorbed by R^ . Compute I 2 , the current through Rj ^2 = 10 kQ, T?3 = 10 kQ, and

voltages. Show ALL work/procedures.

mS, find

Find the power delivered by the inde pendent voltage source.

ion

100

Write the nodal equations and place in matrix form prior to solving. In doing

by hand, with your calculator, using

(e)

and the power delivered by

the dependent current source.

ANSWERS IN RANDOM O R D E R

7^2 = 1-2 A, and

= 5 kQ,

= 0.55 x 10“3, 7-„

from left to right.

L

ion V. ion.

©

Reference node

Figure P 3 .1 0

Figure P 3 .13


145

14. Consider the circuit of Figure P 3.14 (a)

17. Use nodal analysis to find the voltages

Write two node equations in terms of the literal variables in Figure P 3.14

(c)

(e)

in the circuit of Figure P 3.17.

= 20 Q, i?2 = 10 Q, = 10 Q, and I^=

= 4 Q,

=

A. Note that in

and put in matrix form.

0.1 S,

Solve the node equations for the volt

solving this problem, you are to generate three

and Vq when

6

= 0.1 A, 7^2 =

(nodal) equations in which the unknowns are

0-2 A, = 7 mS; = 2 mS, = 500 Q ; i?2 = 333.33 £2; and /?3 = 1 fl. Determine Kq.

you could eliminate the equation

ages

(d)

, and Suppose

for

but this problem is to illustrate that

such elimination is not necessary. Finally, deter

Determine the power delivered by

mine the equivalent resistance seen by the inde

each source. (Be careful of sign.)

pendent current source.

. 9.iVa

'4 Figure P3.14

Figure P3.17

15. Consider the circuit of Figure P3.15 in which = 20 V and (a)

= 0 ; node voltage C is

.

Write the two nodal equations in terms of the literal variables.

(b)

Suppose 11 = 6 and the following in S are given: = 0.5, G2 ” = 4, 6*5 = 1. Solve for

and

Check: 20 and 10 volts. (c)

Find

and then find the equivalent

resistance seen by the independent voltage source. (d)

"■

18. Consider the circuit of Figure P3.18. By choosing node C as the reference node, we elim inate a floating voltage source. Write an appro priate set of nodal equations, with node C as the reference node. Solve the nodal equations, speci fy the voltages V^q Vbo V^Q and V

and the

power delivered by the sources. Finally, find the equivalent resistance seen by the current source. Leti?i = 9 k Q ,7 ? 2 = l S k i 2 ’ = 6 kQ, kQ, = 3000 Q, and = 20 mA.

= 9

Find the power delivered by the inde pendent source and the dependent source.

(e)

W hat is the power absorbed by the output resistor?

-------- --------------

Figure P3.18. By choosing node C as the ref erence node, it is possible to simplify the con struction of the node equations. 19. Consider the circuit of Figure P 3.19 in

Figure P3.15 16. Redo problem 15 with

\ ‘ J

= 60V .

0.25 S and

which 60 ^2.

- 20 Q, /?2 ~ 20 Q, = 2 0 Q,

= 30 Q, R^ =

= 12 V, and

=

146


FLOATING VO LTAG E SO URCE PROBLEM S

0.6 A. The point o f this problem is to illustrate how a good choice o f reference node may sim plify the calculation o f node voltages, whereas a poor choice may lead to a complicated formu

22. For example 3.5 suppose all resistance values

lation o f the node equations.

are doubled, the floaring voltage source remains Choose a reference node so that there the same at 440 V, and all current sources are are no floating voltage sources. Write scaled down to one-half o f their original values. three equations in the unknown volt (a) Compute all node voltages and the ages. Solve for the node voltages. current C H EC K : i^ = 2 A k and iy = - 3 0 A. (b) Compute the voltages and Determine the power delivered by (b) (c) Compute the power absorbed by the each source. 0.075 S resistor. (c) Determine the power absorbed by (d) Compute the power delivered by the each resistor. floating voltage source. (d) Verify conservation o f power using the (a)

results o f parts (b) and (c).

23. For the circuit o f Figure 3.10 in Example 3.6, suppose the 110 V source is changed to 200 V and the 50 Q resistor is changed to 500 £2. Find the node voltages rents

and

Vy,

and the unknown cur

Then find the powers delivered by

the 30 V source and the dependent source. 24. Consider the circuit o f Figure P 3.24 in which = 2 0 0 V, V^2 = 5 0 V, R^ = 5 0 Q, R.^ = 20 Q, R^ = 50 Q, and R^ = 40 Q. (a)

Identify the floating voltage source and

(b)

W rite

add a current label through the source. 20. The nodal equations for the circuit in Figure P3.20 are 0.03 0.09

modified

nodal

equations,

which include both node voltages and - 0 .0 r ■Vf 0.04

unknown currents through any float

'A-.'

ing voltage sources.

0

.^0.

Compute the values o f R^, Rj ,

(c)

,a n d .

Solve the equations for the node voltages Kg and Vq and the current through the 50 V source. C H ECK : Vg = 50 V and

Figure P3.20

= 100 V.

(d)

Find the power consumed by R^ .

(e)

Determine the power delivered by each o f the sources.

21. Consider the circuit o f Figure P3.15, which has nodal equations {R^ ■ 0 .0 0 8

0 .0 1 9

- 0 .0 0 5 '

-0 .0 0 1

0 .0 0 5 5

- 0 .0 0 2

-0 .0 0 5

- 0 .0 0 2

0 .1 0 7

Compute the values of /?j,

0) given by ■ 0 =

0 O .n/

■

0 =

■

0 G A.

>-^3 >-^4’ -^5 ’ ^nd ,u.

Figure P 3 .2 4


147

25. Consider the circuit o f Figure P3.25 in

27. T he modified nodal equations for the cir

which l/j = 2 50 V,

cuit o f Figure P 3.27 are

: 20 Q, (a) (b)

= 5 0 Q,

= 5 0 V,

= 50 Q ,

= 40 Q , and R^ = 10 Q.

■0.004

-0 .0 0 1

-0 .0 0 2

add a current label through the source.

-0 .0 0 1

0.001

0

-1

W rite

-0 .0 0 2

0

0.004

1

2

-1

-1

0

Identify the floating voltage source and modified

nodal

equations,

which include both node voltages and

0 •T 4 '

O'

yB 0

^CB

0

unknown currents through any float (c)

ing voltage sources.

Compute all four resistor values and (3. Hint:

Solve the equations for the node volt-

Find all the conductances first and then convert

ages Vg and Vq and the current

to resistances.

through the 50 V source. C H EC K : 5 0 V a n d V^= 100 V. (d) (e)

Find the power consumed by R^ . Determine the power delivered by each o f the sources. Rc Figure P3.27

= 0.02 = 10 Q., =

28. For the circuit o f Figure P 3.28, S, ^2 = 0.025 S, Gg = 0.2 S,

0.4 A, and V^ 2 ~ 12 V. Use nodal analysis to find all node voltages, the current , the power absorbed by

and the power delivered by

the two sources. Figure P3.25 26. Consider the circuit o f Figure P3.26. R^ = 10 Q ,

= 100 Q, R^ = 100 Q,

= 50 Q ,

= 100 V, K^2 = 60 V, V^3 = 100 V, = 14 A. Label appropriate currents and I j^q through the floating voltage sources. (a)

Write the modified nodal equations for the three unknown node voltages and two unknown currents.

(b)

Solve for the five unknowns (in MATLAB).

(c)

Find the power delivered by each o f

(a)

Determine Vq

(b)

Label the current /ABUsing V^, Vg, and 4 as unknowns, write a 4 x 4 matrix set o f

(c)

nodal equations.

(d)

Solve the nodal equations for V^, Vg,

ÂB’ and 4 . (e)

(0

Determine the power absorbed by G2 . Determine the power delivered by all

sources. AN SW ERS (D) IN R A N D O M O R D E R : 12, 0.24, 9.6, 0.16

the sources.

29. Repeat Problem 28, except this time write Reference node

Figure P 3 .2 6

only three nodal equations in the variables V^,

Vg, and l^g. Notice that you must express

in

148


terms o f

and the appropriate conductance.

One can even reduce the number o f equations

AN SW ERS (R A N D O M IZ ED ): 250, 325, 1 2 5 ,7 5 ,2 5 0 , 25

to two using the so-called supernode approach,

32 Consider the circuit o f Figure P3.32, where

which is the subject o f other texts.

= 4 kQ, 30. For the circuit o f Figure P3.30,

= 100 Q,

mS,fj. = 4 S,

= 1 = 4/3 kQ, = 0.75 = 160 V, and = 40 mA.

= 300

(a)

Specify V^.

£2 A, = 2 A, Vj2 ~ nodal analysis to find all node voltages, the current , the

(b)

Write modified nodal equations.

= 20 Q,

= 20 Q, G4 = 0.09 S,

(c)

nodes B and C, and the power delivered by the

(d)

independent sources as follows:

Find the power delivered by each o f the sources.

(a)

Determine Vq

(b)

Write a set o f modified nodal equa

(e)

tions that contain extra current vari ables including

Solve the modified nodal equations in MATLAB.

power absorbed by the 20 Q resistor between

and

Compute the power absorbed by each resistor.

(f)

Verify conservation o f power.

.

(c)

Solve your nodal equations for the

(d)

Compute the power delivered by the independent sources.

unknowns. C H EC K : Vg= 180 V.

(e)

Compute the power absorbed by the 20

resistor between nodes B and C. Figure P 3.32 33. Consider the circuit in Figure P 3.33 in which V-^ = 60 V, G j = 0.1 S, G2 = 0.1 S,

= 0.3 S, G4 = 0.4 S, G 5 = 0.1 S, Gg = 0.1 S, Gj = 7/480 S, ^ = 3, and (3 = 2 . (a)

Figure P3.30

using

31. Use nodal analysis on the circuit o f Figure P3.31 as indicated. all resistors are 10 Q. (a)

W rite

= 100 V,

modified

1^2

nodal

= 1 A, and equations

including the extra variable IgQ (b) (c)

Write the modified nodal equations

(b) (c)

Vg Vq and /^âs unknowns.

Solve the modified nodal equations in MATLAB. Find

the equivalent

resistance seen by the independent voltage source.

Solve the modified nodal equations in MATLAB. Find the power delivered by each of the sources.

Reference node

Figure P3.31

Figure P3.33 AN SW ERS T O (C) IN RA N D O M O R D E R : 40, -2 5 ,3 8 .7 5 ;/ ?

= 12 Q .


149

SIN G LE LOOP-EQUATION PROBLEM S 34. In the circuit o f Figure P3.34, = 50 Q, and

= 0.5. If

= 400 Q,

= 50 V, find 4 ,

the power delivered by the independent and

37. Consider the circuit o f Figure P3.37 (a) Suppose

sources. R,

dependent voltage sources, and the equivalent resistance,

= 2 0 0 Q, R^ = 300 Q, R2 = 500

Q, /jj = 750 mA and I ^ 2 = 100 ^lAh the power delivered by each o f the independent

--------- ----------- ---

, seen by the independent source.

Figure P3.37 C H EC K : /j = 100 mA. (b) C H EC K S: R^^ = 2 50 Q, and

3 5 . In the circuit o f Figure P 3.35, = 5 0 Q, and = 50 V, find

= 8 watts. = 4 00 £2,

= 50 Q, and jj. = 0.5. If , the power delivered by the

independent and dependent voltage sources, and the equivalent resistance, R^^ , seen by the independent source.

Now suppose /^j = 4 0 0 mA and 1 ^ 2 = 100 mA and the loop equation for /j written in the standard way directly yields 2000/|= 60. Find R-^ and R2 if R^ = 600 fil. Note: If the equations are not written in the standard way, the solution is not unique. For example, multiplying both sides o f the above equation by 0.5 yields a different answer in which R-^ = 140 £2 , as opposed to the correct answer o f R-^ = 400 Q.

38. In the circuit o f Figure P 3.38, /j2 = 100 mA, 7^3 = 2 0 0 mA, and mA. Find Figure P3.35 CH ECKS: R^^ = 200 Q, and 36. In the circuit o f Figure P3.36,

= 56 V, = 100 mA

and the power delivered by each

independent source. 12.5 watts. = 200 V

and 1 ^ 2 = 20 mA. Find Then find the power delivered by each o f the independent sources. Finally, find the power absorbed by each resistor and verify conservation o f power for this circuit.

Figure P3.38 Figure P3.36 C H EC K : 7, = 4 mA.

C H EC K : Sum o f powers delivered by the sources is 15.68 watts.

150


with internal resistances R


(b)

(c)

Suppose

= 2 50 Q,

= 5 0 0 Q,

■20 Q and R2 = 80

Q (faulty connection) respectively connected in

= 100 V, and /S = 0.5. Use loop analy

parallel to supply power to a load o f 7?^ = 80 Q.

sis to find /j and R^^ .

Compute the power absorbed by the load R^

Compute the power dehvered by each

and the power delivered by each independent

source and absorbed by each resistor.

source. W hich battery supplies more current to

Verify conservation o f power.

R^ and hence more power to the load? How

Compute R^^ as a function o f R^,

much power is wasted by the internal resistanc

and 13 . Suppose 5 00 Q , plot R

s 13 s

2

= 250 Q and

=

as a function o f

,0

es o f the battery?

.

Pi,

0 ' C H EC K : P.s\

3.15 watts and P ,2 = 1 .8 watts.

Figure P3.39 C H EC K : Power absorbed by resistors is 15 watts and R^^ > 4 50 Q .

42. Reconsider the circuit o f Problem 3.41, redrawn with different loop currents in Figure 3.42a and 3.42b. Th e point o f this problem is

40. (a) For the circuit o f Figure P3.40, R^ = 1 kQ,

= 5 kQ, /?3 = 4 kQ,

= 100

mA and g„ = ‘i xlO “'^ S. Find /j and by writing two equations in the two unknowns /j and . The first

to verify that different sets o f independent loop equations produce the same element currents and branch voltages. (a)

/j and

equation is the usual loop equation

(b)

(c)

.

for the circuit o f Figure

3.42a, and then find the voltage across

and the second determines the rela tionship o f /j and

Write the new loop equations and find

and the power consumed by R^ . (b)

Write the new loop equations and find

Given your answer to (a), find the

/j and I 2 for the circuit o f Figure

equivalent resistance, R^^, seen by the independent source.

3.42b, and then find the voltage across and the power consumed by R^ .

Find the power delivered by the dependent source.

Figure 3.42 Figure P3.40

43. The matrix loop equation o f the circuit o f Figure P3.43 is

M ULTIPLE LOOP PROBLEM S 41. The circuit o f Figure P3.41 represents two non-ideal batteries = 21 V and = 24 V

• 150

-4 0

-1 0 0 '

-4 0

140

0

h

= -2 0

-1 0 0

0

150

h

20

h'

TOO

Chapter 3 ®Nodal and Loop Analyses

151

Find the value of each resistance and each source in the circuit.

(b)

the current in the locomotive motor and

(c)

repeat parts (a) and (b) when the loco

the power absorbed by the locomotive motive is 1/3 distant from either station

CHECKS: V^2 = ^0 V,

= 40 Q.

44. The mesh equations for the circuit of Figure P 3.44 are ■40

-8 0

- 1 0 ‘ \h'

-3 0

130

-5 0

h

-1 0

-5 0

70

[*3.

■n ■ =

-V2 0 Figure P3.45

Find

* 46. Reconsider Problem 3.45. Let

= 590 V

and R^ = 1.296 Q. This time, suppose there are two locomotives on the track. One is 1/3 dis tant from the East side station, and the other is 1/3 distant from the West side station. (a) Determine the resistance R in Figure P 3.46. (b)

Using the indicated currents, write a set of three mesh equations and solve

(c) 45. Figure P 3.45a shows an electric locomotive

(d)

propelled by a dc motor. The locomotive pulls a train of 12 cars. The motor behaves like a 590 V battery in series with a

for , ^2 ’ H• Determine the two motor currents. Determine the power delivered by each of the 660 V sources.

---------------------- -------- ------------- 1--------- -----------------R __ I ^ R A

1.296 Q resistor. Suppose the train is midway between stations, West side 660 V and East side, where 66 0 V dc

T

sources provide electricity. The resistance of the rails affects the cur rent received by the locomotive. The equivalent circuit diagram is given by Figure P3.45b, where R

660VVf R

R Figure P3.46

0.15 Q. Using m e ^ analy4 7 . Reconsider the Problem 3.5 and the circuit

sis find (a)

© T

l=

the currents

and I 2

of Figure P3.5. Draw two loop currents and

Chapter 3 ® Nodal and Loop Analyses

152

solve for these currents. Then compute the

50. Consider the circuit o f Figure P 3.50 in

node voltages

which = 40 V and V^ 2 = 20 V. Write a set o f three loop equations by inspection. Refer to

n\

Example 3.7 and the discussion following the

o

> and

48. In Figure P 3.48, let = 6 kQ,

= 9 kQ,

.

= 9 kQ,

= 18 kQ, =

example. Solve the loop equations using matrix

and /2.

software program. Compute the voltage v. Note

= 3000 Q, and

methods via your calculator or an appropriate

20 mA. (a)

Write two mesh equations in

that /| and I 2 should have values identical with

Put in matrix form and solve.

those in example 3.7. Finally, find the power

(b)

Specify the voltages 1^^,

,

(c)

^BO ^CD’ K iD' Find the power delivered by each o f

O

delivered by each o f the sources.

IQ

the sources.

49. Consider the circuit o f Figure P 3.49 in which 1/j = 250 V, V^ 2 = 50, V, = 50 Q, = 20 Q, = 50 Q, = 40 Q, and = 10 Q. (a)

51 . Consider the circuit o f Figure P3.51. (a)

Write three standard loop equations

matrix form in terms o f the literal

and put in matrix form. (b)

Write two mesh equations and put in

o

parameters.

Solve the equations for the loop cur

(b)

rents and determine the node voltages

Solve the mesh equations for the unknown currents assuming

and

Q, = 40 Q, A, and

.

= 20 Q, V.

= 100 = 80 Q,

(c)

Find the power consumed by

(d)

Determine the power delivered by

(c)

Find

each o f the sources.

(d)

Find the power delivered by the inde

r^

and V^.

pendent sources. (e)

Find the power delivered by the dependent source.

r )

n Figure P 3 .4 9

Figure P 3.51


153

52. Repeat Problem 51 when 7?, = 100 £3, ^2 = 40 Q, = 80 Q, = 60 £2, = 1 A, and

K2 =

V-


Write two mesh equations in terms o f the literal parameter values.

(b)

Solve the mesh equations assuming = 100 Q, ^2 = 40 Q,

= 60 f i,

80 Q,

1^2

= 80 Q,

=

= 1 A, and V^j =

40 V.

.

(c)

Compute

(d)

Find the power delivered by all the

and

sources in the circuit.

CHECK: /, = 0.1 A, K- = 50 V, and V = 20 V. 56. Consider the circuit o f Figure P3.56. (a)

Write the modified loop equations (using the indicated loops) in matrix form.

(b)

Figure P3.53

(c)

If I/, = 200 V, 7^2 = 0.3 A, /?! = 7^3 = 100 £2, T?2 = 400 £2, and /i = 0.5, find 7j , 72 , and V^ 2 ■ W hat is the power delivered by the three sources?

54 . Repeat Problem 53 when = 40 Q,

30 £3,

0.25 A, and

= 40 Q, 7?2 =

= 20 £2,

= 10 £2, 7^2 =

= 60 V. >mv,-

M O D IFIED LOOP ANALYSIS PROBLEM S 55 . Consider the circuit o f Figure P3.55. The objective o f this example is to illustrate a numerical approach to loop analysis where the number o f variables to be found is quite large,

57. Consider the circuit o f Figure P3.57. (a) Write the modified loop equations in matrix form.

but the equations are quite easy to write and do (b)

not require multiple substitutions. (a)

A, 7?, = 200 £2, 7?2 = 400 Q,

If the loop equation matrix is o f the

Q,

form below, compute the undeter (c) 1

1 + ??'

??

-1

0

R3

????

-1

(b)

If

y s -R ih n

A' =

0 0

= 500 £2, 7^2 = 100 Q,

Q, 7?4 = 100 £2,

= 100

= 500 Q, and TJj = 1150 Q,

compute 7j, 72 , 73 , and

mined entries. '7?2 + ? ?

If 1/, = 210 V, K^2 = 150 V, 7^3 = 0.1

= 400

= 150 V, /•„ = 0.5

A, find the three unknowns.

Compute the power delivered by the independent sources.

154

Chapter 3 ° Nodal and Loop Analyses

C H EC K : /j = 0.4 A and

= 50 V.

58. Consider the circuit of Figure P3.58. (a)

Write the modified loop equations in matrix form in terms of the literal values.

(b)

If

= 4 0 0 ^ ^ ^ 2 = 2 0 0 V,i?i = 3 0 Q , = 20 Q,

= 270 Q,

= 80 Q, T?5

= 140 Q, and compute /p /2 , 73, and (c)

Compute the power delivered by all sources.

(d)

Compute the power absorbed by each resistor and verify conservation of power. Ri

R3

R. -VESA 's ■<-----—N/N^ Rs Figure P3.58 CHECKS: V, = 265 V and watts.

= -3 9 7 .5

C H A P '

‘

"

/

f

L in D lif ie r The Operational Amplifier

Amplification o f voice allows announcers at sports events to convey their comments on the playby-play action to the crowd. At concerts, high-powered amplifiers project a singer’s voice and the instrumental music into a crowded auditorium. Electronic amplifiers make this possible. One o f the simplest and most common amplifiers is the operational amplifier, the subject o f this chapter. The word “operational,” though, suggests a purpose beyond simple amplification. Often one must sum signals to produce a new signal, or take the difference o f two signals. Sometimes one must decide whether one dc signal is larger than another. The operational amplifier is operational pre cisely because it can be configured to do these things and many other tasks, as we will see later in the text.

CH APTER O BjEC TIV ES 1.

Introduce the notion o f an ideal operational amplifier, called an op amp.

2. 3. 4.

Describe and analyze basic op amp circuits. Describe and illustrate a simple method for designing a general summing amplifier. Describe and illustrate the phenomenon o f saturation in op amp circuits and describe cir cuits that utilize saturation for their operation.

SECTIO N HEADIN GS 1. 2. 3. 4. 5.

Introduction The Idealized Operational Amplifier: Definition and Circuit Analysis The Design of General Summing Amplifiers Saturation and the Active Region of the Op Amp Summary Terms and Concepts Problems

156

Chapter 4 • T h e Operational Amplifier

1. IN TRO D U CTIO N Chapters 1 and 2 defined and discussed independent and dependent voltage and current sources. Chapter 3 investigated the nodal and loop analysis o f resistive circuits containing such sources. Ofi:en, dependent sources supply energy and power to a circuit, making them so-called active ele ments. O n the other hand, resistors are passive elements because they only absorb energy. Circuit models o f real amplifiers (see Examples 3.3 and 3.4 with associated Figures 3 .7 and 3.8, respec tively) contain controlled sources that underlie their analysis and performance evaluation. Indeed, the VCVS is the core component o f the operational amplifier (op amp), the main focus o f this chapter. Thus, the op amp is an active circuit element whose analysis is done with the techniques o f Chapters 1 through 3. A real op amp is a semiconductor device consisting o f nearly two dozen transistors and a dozen resistors sealed in a package from which a small number o f terminals protrude, as shown in Figure 4.1(a). Despite its apparent internal complexity, advances in integrated circuit manufacturing technology have made the op amp only slightly more expensive than a single discrete transistor. Its simplicity, utility, reliability, and low cost have made the op amp an essential basic building block in communication, control, and the instrumentation circuits that can be found in all under graduate EE laboratories. Top View Balance 1 [

TO-5

Dual-in-line

Inverting input

^

Noninverting ^ [ input

m

E- 4 [

3 6 Output H 5 Balance

(a) Inverting E+«-

input —

V-

Output

Reference node Noninverting

E-

input (c) (d) FIG U RE 4.1 (a) Typical op amp packages; (b) typical terminal arrangement of an op amp package; (c) dual power supply notation; (d) essential terminals for circuit analysis. Figure 4.1 (b) shows a typical arrangement o f terminals for a dual-in-line op amp package. Th e ter minal markings and the symbol shown in Figure 4.1(b) do not appear on the actual device, but

Chapter 4 • T h e Operational AmpUfier

157

are included here for reference. In Figure 4.1(b), the terminal labeled “N C ” (no connection) is not used. The E+ and E- terminals (Figure 4.1(b)) are connected to a dual power supply, illustrated in Figure 4.1(c), where

typically ranges between 3 V and 15 V, depending on the application;

adequate voltage is required for proper operation. The three terminals in Figure 4.1(b) marked “inverting input,” “non-inverting input,” and “output” interact with a surrounding circuit, and correspond to V,

and Vq in Figure 4.1(d). The two terminals labeled balance or ojf-set have

importance only when the op am is part o f a larger circuit: resistors o f appropriate values are con nected to these terminals to make sure the output voltage is zero when the input voltage is zero. This “balancing process” is best discussed in a laboratory session. This chapter sketches the basic properties o f the op amp: just enough to understand some o f the interesting applications. The ideal op amp model and the saturation model are described. Using these models and the principles o f analysis covered in Chapters 1 through 3, we then analyze the behavior o f some widely used op amp configurations. These application examples hint at the importance o f the op amp and furnish motivation for the study o f electronic circuits. Several o f the examples include a SPIC E simulanon o f the circuit being analyzed. SPIC E is a sophisticated circuit simulation program. Behind the user-interface, SPIC E uses complex models o f the real operational amplifier. Our purpose in using SPIC E simulation is to verify or test the theoretical analysis set forth in the examples. W hat we show is that the simplified theoretical analysis provides a very good approximation to the actual circuit behavior represented in the SPIC E simulation results. Industrial circuit designers often use SPIC E to visualize the expected behavior o f very complex circuits. Later chapters cover some o f the more complex op amp appli cations.

2. TH E ID EALIZED O PERATIO N AL AM PLIFIER This section analyzes resistive circuits containing an operational amplifier. Figure 4 .2 explicitly shows an op amp embedded in a surrounding resistive circuit.

FIG U RE 4.2 One possibility for analyzing op amp circuits is to represent the op amp by one o f the simplified models shown in Figure 4.3 that do not account for saturation effects. The first model o f Figure 4.3(a) consists o f an input resistor,

an output resistor,

and a V CV S with finite gain A.

O f practical import is the idealization o f this model (Figure 4.3(a)) to the one o f Figure 4.3(b) by

158


(1) letting R-^ become infinite, setting up an open circuit condition at the input terminals; (2) let ting

become zero, making the output voltage o f the op amp equal to that o f the V C V S; and

(3) letting the gain A approach infinity. These conditions are idealizations because (1) with R-^ infinite, there is no loading to a circuit attached to the input; (2) with R^^^ = 0, the full output voltage appears across any circuit connected to the output; and {5) A ^

leads to a simplifica

tion o f the associated analysis. These conditions, stated below as equation 4.1, define the so-called ideal operational amplifier; -> 00 (infinite gain)

R-^ Rout

(4.1a)

00 (infinite input resistance)

(4.1b)

0 (zero output resistance)

(4.1c)

Rest of Circuit

Rout +

+ V

A(v - V

(a)

—

FIG U RE 4.3 To see how this idealization simplifies op amp circuit analysis, consider an equivalent set o f conditions for the ideal op amp, called the virtual short circuit model:

(4.2a) (4.2b)

v^ = v_

(4.2c)

From Figure 4.3(b), the conditions that = 0 and i_ = Q follow directly from the open circuit con dition at the input terminals. The condition that = v_ (hence, the term “virtual short circuit”) will be discussed later, but occurs because A ^ co, forcing {v^ —v j 0.

159


The recommended way to analyze circuits containing op amps is to replace any ideal op amp by the model o f Figure 4.3{b), the virtual short circuit model o f equation 4.2. Th e following examples illus trate the use o f the virtual short circuit model. EX A M P L E 4 .1 . This example investigates the inverting am plifier o f Figure 4.4, which is used in a wide range o f commercial circuits. The objective is to compute

Rjs and

in terms o f

+

V

Rf

FIG U RE 4 .4 Inverting amplifier, assuming an ideal op amp in which V„„, = ---- —V:„ .


and v_. Since the + terminal is grounded,

= 0. From the virtual short prop

erty o f the ideal op amp, v_= v^ = Q. Step 2. Compute i^-^. Since v_ = 0, the voltage across R-^ is v-^. From Ohm’s law, v,„

R Step 3. Compute iy. Again, since v_ = 0, the voltage across Rf is

From Ohm’s law.

if = R Step 4. Relate the currents i-^ and ip and substitute the results o f Steps 2 and 3. From KCL, i^^ —i_ + ir= 0. From the properties o f the ideal op amp, i_ = 0, in which case, ir= -i-^. This imphes that

R* V Hence, the voltage gain relationship o f the inverting op am circuit is

Rf V ou t= --^ ^ in Rin

(4.3)

Equation 4.3 shows that the input and output voltages are always o f opposite polarity, hence the name inverting am plifier. One also observes that by choosing proper values for Rjr and R^ a volt age gain o f any magnitude is possible, in theory. In practice, other factors limit the range o f obtain able gains.

160


Exercises. 1. Find

for the circuit o f Figure 4.5.

lOOkO 25kQ -s/ W '

50 mV

FIG U RE 4.5 Inverting amplifier. 2. Find

for the circuit o f Figure 4.6.

lookn

FIG U RE 4.6 Inverting amplifier with additional resistor. A N SW ER F O R BO T H :

= - 200 mV.

A few remarks are in order. Op amp configurations in which one o f the input terminals is ground ed, as is the non-inverting terminal in Figure 4.4, are said to operate in the single-ended mode. The input terminal can be grounded directly or through a resistor, as in Exercise 2 above. Also, since

v_ =

or v_-v^ = 0, the terminals are virtually short circuited even though there is no hard-wired

direct connection between them. This condition is called a virtual short circuit. Further, if one o f the terminals is grounded, then the other terminal is said to be virtually grounded, as is the case in Figures 4.4, 4.5, and 4.6. Specifically, in Figure 4.4, there is a virtual ground at the invert ing input terminal. The next example continues the investigation o f the ideal inverting amplifier for the two-input, single-output op amp circuit o f Figure 4.7. The solution again makes use o f the virtual ground and virtual short circuit properties o f the ideal op amp.

161


E X A M P L E 4 .2 . For the circuit o f Figure 4.7 , our objective is to compute and the two input voltages

in terms o f R^, ^2>

and v^2 Rf

+ V

FIG U RE 4.7 Inverting (ideal) amplifier with rwo inputs for which Rf Rj v„„,=-^v,|V ,2.

So l u t io n . A s in Example 4.1, and by the same reasoning described there, v_ = v^ = 0. Step 1. Compute ij and

Since v_ = v^ = 0, the voltage across

is

and the voltage across Rj

is v^2 - From Ohm’s law,

R^ Step 2. Compute ij-. Again, since v_ = 0, the voltage across Rj- is

and from Ohm’s law.

'/ = ■R / Step 3. Relate the currents /j, ^2

h

and then substitute the results o f Steps 1 and 2. From KCL,

H —i_ + ijr = ^- From the properties o f the ideal op amp, i_ = 0, in which case, ijr = -(/j +

This implies that

Hence, Rf

Rj (4.4)

162


Exercise. In Figure 4.7, suppose Rjr= 100 IcD. Find AN SW ER:

= 25 kD and

and ^2 so that

= 50 kD.

E X A M P L E 4 .3 . This example analyzes the non-inverting operational am plifier circuit o f Figure 4.8. As in Examples 4.1 and 4.2, the objective is to compute in terms o f R^, R2 , and v-^. We show that

''out

V.

FIG U R E 4.8 A non-inverting op amp circuit.


and v_. Since the + terminal is connected to the input voltage source,

= v-^.

From the virtual short property o f the ideal op amp, v_ = v^ = Vj^. Step 2. Compute zj . Since v_ =

the voltage across the resistor

is

Observe that the current,

Zj, has reference direction different from the passive sign convention. Hence, from Ohm’s law,

h=Step 3. Compute ip Again, since v_ = v-^, the voltage across R^-is

lf =

From O hm s law,

Vout - Vin

Step 4 . Relate the currents z'j and ip and substitute the results o f Steps 1 and 2. From KCL, z'j - z_ + Zyr= 0. From the ideal op amp property o f equation 4.2, z_ = 0, forcing z^= -Zj. This implies that

Rf

/?,

163

Chapter 4 • T h e Operational AmpHfier

Hence, the input-output voltage relationship

Rf\

(4.5)

1+ ^

R i) / R \ From equation 4.5, the voltage gain is greater than 1, i.e., i 1 + - i - i > 1, and I } have the same polarity; the circuit is naturally called a non-inverting amplifier.

Exercise. For the non-inverting amplifier o f Figure 4.8, find

and

and

always

so that the gain is 2, and

when v-^ = 5 V, the power absorbed by R^ is 5 mW. i,n AN SW ER: Rf = 5\
EXA M PLE 4 .4 . This example analyzes the ideal general di^Ference amplifier circuit o f Figure 4.9. We show that

Kf ^’s 2 - - r ^ s \ R\

'’" " ' I

In a basic difference amplifier, the output is the difference o f two input voltages. For the gener al difference amplifier o f this example, the output is a difference o f the scaled input voltages,

=

for appropriate positive

and a^.

So l u t io n . From the ideal op amp property o f equations 4.2, v_ =

and no current enters the

inverting and non-inverting op amp input terminals. Step 1. Write a node equation at the non-inverting input terminal o f the op amp. Summing the cur rents leaving the + node o f the op amp yields

G 2 { v + - v , 2 ) + GgV+=Q

164


Solving for

leads to

(4.6a)

Step 2. Write a node equation at the inverting input terminal o f the op amp. Recall v_ = v^. T he sum o f the currents leaving the - node satisfies G i ( v+ - v, ] ) + G ^ ( v+ - v„„,) = 0 Thus, Cl

ôut =

+

1+

(4.6b)

Step 3. Combine Steps 1 and 2. Substituting equation 4.6a into 4.6b yields (4.7a) G t + Go or, in terms o f resistances,

[R g + R2 j

(4.7b)

^s2

Equations 4 .7 have the desired form:

appropriate positive constants
a ^, which can be obtained by proper choices o f the resistors.

Two special cases o f Example 4 .4 are o f practical importance. First, if

and R^ = /?2> then

equation 4.7b reduces to the classical difference amplifier equation.

out

= K {v , 2 -V si)

with K = I and for an arbitrary K > 0 , Rj- = KR^ and R^ = KR2 fits the bi bill.

Exercises. 1. In Figure 4.9, if

= 7?2 = 5 kQ and K = 2 , find R p and R^.

ANSW ER: R j - = R^= \0 kQ 2. Using the circuit o f Figure 4.9, design a difference amplifier so that

= 4(t^^2 “ ^ji) ^^d the

feedback resistance R^ = 20 kQ. A N SW ER: R, = R , ^ 5 kQ and R^ = 20 kQ 3. In Exercise 1, suppose R^ and /?yare scaled by a positive constant A'j, i.e., R^^^ = K^R^y and R2 and R are scaled by a positive constant K^. Determine the new input-output relationship. o

A N SW ER:

w ith

K

the sam e as in Exercise 1

165


The point o f Exercise 3 is that the group

, Rj] can be independently scaled by

and the group

{Rj, R^ independently scaled by K 2 without affecting the gain o f equation 4.7b. E X A M P L E 4 .5 . This example analyzes a special case o f the non-inverting amplifier called the buffer or isolation amplifier, shown in Figure 4.10, where = v-^. W hen connected between two circuits, the buffer amplifier prevents one circuit from having a loading effect on the other.

FIG U RE 4.10 The buffer or isolation amplifier for which

So l u t io n . From the connection shown in Figure 4 .1 0 , v-^ = ties o f the ideal op amp,

= v_, in which case

and

= = v_. From the proper

= v-^.

Exercise. Compute the power delivered by the source in Figure 4 .1 0 and the power delivered to the load R^. ^ AN SW ER: 0 and

R,

The circuit o f Figure 4.10 is called an isolation or buffer amplifier, because no current is drawn from the source maintaining

However, the op amp does supply current (and power) directly to the load by under the condition that

put current rating. Since Vg„f{t) =

not exceed the manufacturer’s maximum out

the circuit is also called a voltage follower.

Figure 4.11 shows a SPIC E simulation that verifies the behavior arrived at in Example 4.5. Here a dc voltage sweep, Q
166


buffer -DC Transfer-2

(V)

+0 .0006+000

+2 .000

+4.000

+6.000

V2 +8.000

+10.000

+12.000

FIG U RE 4.11 Spice simulation of voltage-follower circuit.

Exercise. Find

for the circuit o f Figure 4.12, the power supplied by the source

power supplied to the 12 kQ load. AN SW ERS:

=v,.0. 1 2 x 10-’

FIG U RE 4.12 Isolation of load from source using buffer amplifier.

and the


167

3. TH E DESIGN OF G EN ERAL SUM M IN G AMPLIFIERS^ Often data acquisition equipment and active fdters require multi-input single-output amplifiers having a more general summing characteristic, such as

ôut -

+ « 2 'a 2 ) +

)

(4.8)

where the constants ay>0 and |3>0. The inverting and non-inverting amplifier configurations (Examples 4.1, 4.2, and 4.3), as well as the difference amplifier configuration o f Example 4.4, are special cases o f equation 4.8. W ith a little cleverness, it is possible to design by inspection an op amp circuit whose input-output characteristic is precisely equation 4.8. The op amp circuit o f Figure 4.13 having the four inputs V^2 > ^hv ^bl accomplishes this. The circuit looks ordi nary except for the presence o f one additional conductance, AG, incident on the inverting termi nal o f the op amp. T he dashed lines in Figure 4.13 are present because this conductance may or may not be needed. Computation o f the values o f AG and

are explained in design Step 2,

below.

b2 0 — s / \ / V G

FIG U RE 4.13 A general op amp circuit that realizes equation 4.8.

Design Choices for the General Summing Circuit o f Figure 4.13 The first two design steps constitute a preliminary or prototype design, meaning that the feedback resistor is normalized to 1 Q , or equivalently, 1 S. After completing the prototype design, an engi neer would scale the resistances to more practical values without changing the gain characteristics. T he scaling procedure is explained in Step 3. D esign Step 1. Prototype design. Set G ^ = l S , G , , = a , S ,G ,2 = « 2 S, G^, = P i S , and CJ^2 = P2 S. For the design to remain simple, the total conductance incident on the inverting terminal must equal the total conductance incident on the non-inverting terminal. This is achieved by proper choice o f AG and/or G . T he proper choices are given in Step 2.

168


Design Step 2. Prototype design continued: Computation ofG^ andhr tS.G so that the total conduc

tance incident at the inverting terminal o f the op amp equals the total conductance incident at the non inverting terminal. To achieve this equality, recall that in design Step 1, Cy= 1 S,

S, and

S. Define

a numerical quantity ^ = (1 + a i +

) - (^1 +

)

The sign o f 6 leads to two cases: Case 1: If 8 > 0, then set (7 = 8 and AG = 0. .5

Case 2. If 8 < 0, set G to some value, for example, G = 1 S and AG = |8| + G . o & & Design Step 3. Scaling to achieve practical element values. Multiply all the resistances (divide all conductances) incident at the inverting input terminal o f the op amp by a constant

Similarly,

multiply all resistances (divide all conductances) incident at the non-inverting terminal o f the op amp by

It is permissible to choose

but this is not necessary.

EXA M PLE 4 .6 . Design an op amp circuit having the input-output relationship >ô»/ = - 7 V a l - 3 v ^ 2 + 2 v i , + 4 v^2 (4.9)

So l u t io n Step 1. Prototype design. Using Figure 4.13, choose Gy- = 1 S, G^j = 7 S, G^2 = 3 S, G^, = 2 S, and G^2 = 4 S . Step 2. Equalization o f total conductances at inverting and non-inverting terminals. Since 8 = (1 + 7 + 3) - (2 + 4) = 5 > 0, set AG = 0 and G^ = 8 = 5 S. The circuit in Figure 4.14(a) exemplifies the prototype design. Step 3. Scaling. To have practical element values, let us choose to a design with resistances Rjr= 100 kQ, 7?^, = 14.28 kQ, 25 kD and R„ = 20 kQ.

= 10^. This scaling leads = 3 3.33 kQ,

= 50 kQ, R^^ =

169


14.28 kQ V3, o------

lOOkO

25 kQ

(a)

(b)

FIGURE 4.14. (a) Prototype design of equation 4.9; (b) final design after scaling with

= lO^.

EXA M PLE 4 .7 . Design an op amp circuit to have the input-output relationship: I'ow/ = - 2 v „ i - 4 v^2 + 7 v/,| + 5 v^2

So l u t io n Step 1. Prototype design. Again, using Figure 4.13, choose

(4.10)

1S,G „=2S,G ,2 = 4S,G^,=7

S, and G^2 = 5 S. Step 2. Equalization o f total conductances at inverting and non-inverting terminals. 8 = (1 + 2 + 4) - (7 + 5) = - 5 < • 0; set C = 1 S, AG = |8| + G = 5 + 1 = 6 S. This prototype design is given in A o Figure 4.15(a). Step 3. Scaling. To have practical element values, let us again choose

= 10^. This scal

ing leads to a design with resistances Rjr = 100 kQ, R^j = 50 k fl, R ^2 = 25 kQ, = 14.28 kQ, R^j = 20 kQ, - 100 kQ, AR = 16.67 kQ. The final design is set forth in Figure 4.15(b). 16.67kn 50 kQ V3, o------v X / X . -

lOOkQ

25 kQ

'â2 O------- s/\v^v^

14.28 kQ lOOkQ

'^b2 O--20 kQ (a)

(b)

FIGURE 4.15 (a) Prototype design of equation 4.10; (b) final design after scaling with

= 10^.

170


Exercise. 1. Obtain an alternative design for Example 4 .7 such that

= 0, implying the saving

o f one resistor. AN SW ER: In prototype design, AG = 5 mho. 2. Design a difference amplifier so that

- v^-^, with

= 10 kD.

AN SW ER: See Figure 4.16. 10 kQ

At this point, the reader may wonder how this simple procedure is derived. The derivation o f this procedure is beyond the scope o f the Hght edition^ Th e interested reader is directed to the 2nd edition o f this text.

Exercise. 1. Find AN SW ER: „

the G- for the circuit in Figure 4.17(a).

in terms o f _

' out -

G,

G,

Gj

G3

,v2

2. Find t',,

in terms o f for the circuit in Figure 4.17(b).

AN SW ER:

-

- 7z>2 + 6K U

(a) FIG U RE 4 .17


171

4. SATURATION AND TH E A CTIVE REGION OF TH E OP AMP In the previous sections, we assumed the op am functioned ideally:

= v_ and

= i_=Q. For the

inverting amplifier o f Example 4.1, this led to the very simple gain formula,

'ôut

_ Rh,

Thus, as the input voltage increases, the output voltage increases proportionately. For real circuits, this proportional relationship holds only when

< V^^^for some value o f

that is associat

ed with the power supply voltage. Intuitively speaking, an op amp cannot generate an output volt age beyond that o f its power supply voltage, typically less than or equal to 15 V. W hen the V^^limit is reached, further increases in the magnitude o f v-^ produce no change in the value o f This behavior is called saturation. To explain this saturation behavior, we refer to Figure 4 .1 8 . In Figure 4.18,/'(v^ - v j represents a nonlinear controlled voltage source, as opposed to the linear relationship A(v^ - p_), shown in Figure 4.4(a). However, because the op amp functions more or less linearly until reaching its sat uration limits, we can approximate f(v^ - v_) by the three-segment piecewise linear relationship shown in Figure 4.19(a), wherein the saturation effects are captured by segments II and III. One observes that when

A

, the voltage/(f^ - z 'J clamps at

the voltage/(z'^ - v j clamps a saturation occurs are

A

. If

t

and when

^

A

j

As observed, the critical threshold voltages o f t^ât which = 15 V and A= 10^, the critical threshold voltages are ±0.15

mV; if A is infinite, as in Figure 4.19(b), then saturation occurs when \v^ > 0.

FIG U RE 4.18 Practical op amp model with a nonlinear controlled voltage source. The linear r e l a t i o n s h i p , - v^=A{v^ - v J , holds for segment 1 in Figure 4.19(a), which is said to be the linear region or active region o f the op amp, denoted by


172

Typical values o f finite A range from 10'^ to 10*^. The active region is the ordinary region o f oper ation. In the active region, the op amp provides a very high (open loop) voltage gain A, the slope o f segment I. The phrase “open loop” gain means that there is no connection through a wire, a resistor, or some other device back to the input terminals. Models o f the three operating regions o f the op amp are summarized in Table 4.1.

A f(v^-v) V

\

Positive Saturation

/ Active / Region

d

= V -V

Negative Saturation

\

-V

(b) FIG U RE 4.19 A piecewise linear (three-segment) curve for the op amp that specifies the active and positive/negative saturation regions of operation: (a) finite gain A, and (b) (ideal) infinite gain A. TABLE 4.1 Operating Regions o f the Op Amp with Associated Models

C urve Seg m ent

N ame

of

B

R e g io n D efining E quations Vcl =

Active

f(Vd) A

and

*^sar

II

Positive saturation

Vsa, A and

sat

III

Negative saturation

I dealized C ircuit M od el

Chapter 4 • T h e Operational Amphfier

173

The use o f a three-segment curve in Figure 4 .1 9 is different from the techniques o f earlier chap ters. The operating point,

determines the proper segment to be used for analysis. If the

input is small, one reasonably assumes the operation is in the active region, segment I. However, when the input magnitude is large, one must “guess and check” to determine the appropriate oper ating region. For example, should the guess be incorrect, then the model for one o f the other regions must be used and the analysis repeated until a valid solution (and operating region) is obtained. The following example illustrates the approach. E X A M P L E 4 .8 . The purpose o f this example is to illustrate that an op amp may operate in any o f three regions and also to illustrate that the determination o f the region o f operation using the “guess and check” method. Recall the inverting am plifier o f Figure 4.5. Suppose = 50 kQ, V; (b)

= 10 kQ and = 4 V; and (c)

= 15 V. Find

and

is infinite,

for the following three cases: (a)

= 0.5

= - 5 V. Finally, verify the theoretical analysis using SPICE.

So l u t io n (a) Assume the op amp operates in its active region. From equation 4.3 in Example 4.1, the out put voltage is vout „ = - ^ v , m, = - ^JQx 0 . 5 = - 2 . 5 V Since |-2.5| <

= 15 V the op amp operates in its active region; the answers

= - 2 .5 V and

v^ = 0 are valid. (b) W ith v-^ = 4 V, assuming operadon in the active region,

However, since |-20| > ôut

~

^

^ 50 = -----= -------------- x 4 = - 2 0 V.

10

Rl

= 15 V, the op amp does not operate in its active region. Therefore,

invalid, but does suggest operation in the negative saturation region. The negative

saturation model o f Table 4.1 yields the circuit o f Figure 4 .20 in which

= - 1 5 V.

FIG U RE 4.2 0 Op amp operating in negative saturation region. By writing and solving a single node equation at the inverting input terminal designated by the minus sign in Figure 4.20, we obtain Vj = - 0 .8 3 V. (c) With v-^ = - 5 V, assuming operation in the active region,

R'

~

50

^

= 25 V

This result suggests that the op amp is really operating in the positive saturation region. Using the positive saturation model o f Table 4.1, Figure 4.21 shows the proper circuit configuration with

174


ôut ~ 15 V. As in the previous case, by writing and solving a single node equation at the invert = 1.667 V. In this case and in case

ing terminal designated by the minus sign in Figure 4 .2 1 , (b) above,

0, as we were not in the active region o f operation, and it was necessary to change

the guessed region o f operation to obtain valid results.

FIG U RE 4.21 Op amp operating in positive saturation region. A SPIC E simulation was used to validate the theoretical analysis^. A D C sweep, - 4 <

< 4 V,

is an adequate input to demonstrate the saturation effects. In the SPIC E simulation, an accurate model for a 741 op amp manufactured by Burr Brown was used. T he resulting dc transfer curve is shown below in Figure 4.22. Lin/Decarlo E xI-D C Transfer-4 Output voltage (V) -4.000

- 1.000

+ 0 .0006+000

Vin +

1.000

+3.000

+4.000

V(IVM)

From this curve, one can see that the op amp saturates for input voltages v-^ such that

> 3,

and the op amp operates in its linear region whenever \v-^ < 3. As hoped, the simplified three-seg ment model in Table 4.1 yields very good results in all regions o f operation relative to the realis tic SPIC E simulation.

O ne can conclude from the above example that for the purpose o f faithfully amplifying an input signal, the input should not be so large as to drive the op amp into saturation. Driving an op amp into saturation distorts the output signal relative to the input. O n the other hand, for some spe cial applications, such as the com parator, saturation is precisely the property to be utilized. Figure


17 5

4.23 shows two comparator circuits. A com parator circuit compares tlie input voltage v-^ with a reference voltage Vygjr{or some multiple o f

Only two different output voltages are produced,

and the other for v-^ <

one for

EXA M PLE 4 .9 . For the com parator circuits shown in Figure 4.23, each op amp has infinite gain and a saturation voltage

= 1 5 V"^: v -^ relationship for the comparator o f Figure 4.23(a).

(a)

Find the

(b)

Repeat part (a) for Figure 4.23(b). Note that in both circuits, there is no connection between the output and inverting input terminals, a departure from all the previous circuit configurations. Because o f this, for ^ v_, and

almost all voltages,

= Vin O------s / W 2 0 kQ ''re f O --------

= -2 0 V

80 kn

1 . 1 .

(a)

(b)

FIG U RE 4 .23 Two comparator circuits that are used to determine when an input voltage is above or below a reference voltage. S

o l u t io n

(a) For

> - 5 V, the voltage v^=

v_ = -5 - v-^ < 0. Referring to Figure 4.19(b),

ôut ^ ~^sat = - 1 5 V. Similarly, for v-^ < - 5 V, the voltage v^ = v^ - v_ = -5 - v-^ > 0, and hence ôut = ^sat = 15 V. (b) By the fact that no current flows into the input terminals o f Figure 4.23(b), using nodal analy sis, we have that V

- V

20x10^

80x10^

m which case, - 4 + 0.8v,„

For

> - 5 V, the voltage v^ =

ration curve o f Figure 4.19(b),

- v_ = 0 - {-A +

Here, referring again to the satu

= - 1 5 V. Similarly, when v-^ < 5 V, the voltage

v^ = v ^ -v _ = Q -{-A + 0.8V.J > 0; hence,

= 15 V.

176


To verify this analysis, the circuit o f Figure 4.23(b) was simulated in B2Spice using a Burr Brown 741 op amp model. T he results o f the simulation are given in Figure 4.24. The theoretical analy sis based on the simplified models o f Table 4 .1 shows a very good match with the more realistic SPIC E simulation results. example4.8-DC Transfer-6 Output voltage (V ) -1 0 0 °

+2.000

+3.000

V(IVM)

FIG U RE 4.2 4 B2Spice simulation o f the comparator circuit of Figure 4.23(b).

Exercise. For the circuit o f Figure 4.25, suppose

= 12 V, find the range o f v^2 for which the op

amp is in positive saturation. Then find the range o f v^ 2 fof negative saturation. AN SW ER:

when

< -^ V, and

=-

when

V , C3------- v N / \ - -

75 kO

''s2 o-----s /s y \/-

-o +

25 kQ

1 . FIG U RE 4.25


177

5. SUM M ARY This chapter has introduced the operational amplifier and a number o f practical circuits that uti lize this new device. These circuits include the inverting and non-inverting amplifiers, the buffer amplifier, the difference amplifier, and the general summing amplifier. W ith regard to the gener al summing amplifier, a simple design algorithm is described and exemplified. The analysis o f these circuits builds on the definition o f an ideal op amp, meaning that, when properly config ured, no current enters the input terminals and the voltage across the input terminals is zero; these properties are referred to as the virtual short circuit model o f the op amp, i.e., the ideal op amp has infinite input resistance, zero output resistance, and an infinite internal gain, A. (See equations 4.1 and 4.2.) Practically speaking, the gain A, is not infinite, but ranges between lO'^ and 10*^. After exploring properties o f the ideal op amp, we discussed the phenomena o f output voltage sat uration. By introducing output saturation, the ideal model o f the op amp gives way to a more real istic one, characterized by three regions o f operation, each having its own “ideal” model, as set forth in Table 4.1. In practical design and applications, output saturation is either to be avoided or utilized to some advantage, as in the case o f the comparator circuit studied in Example 4.8. For a faithful amplification o f an input signal, saturation is to be avoided.

6. TERM S AND C O N C EPTS Active element: A circuit element that requires an outside power supply for proper operation and has the capability o f delivering net power to a circuit such as is the case for an op amp or negative resistance. Buffer: A circuit designed to prevent the loading effect in a multistage amplifier. It isolates two successive amplifier stages. Characteristics o f an ideal buffer are infinite input impedance, zero output impedance, and constant voltage gain. Com parator: an op amp circuit that compares the input voltage (or some multiple o f

with a reference voltage

only two different output voltages are produced, one for

<

v-^, and the other for v-^ < v-^. Difference amplifier: given two inputs,

and

a difference amplifier produces the output

ôut ^ appropriate constant k, often taken as 1. General summing amplifier: an op amp circuit having the input-output relationship

=

constant a - and p .. Ideal op amp: An operational amplifier with infinite input resistance and infinite open-loop gain. -

+ ^ n ^ a r) +

+ - + ^ m ^ bn )

p o s itiy c

Inverting amplifier: An operational amplifier connected to provide a negative voltage gain at dc. Linear active region: In the op amp output vs. input transfer characteristic, the region where the curve is essentially a straight line through the origin is called the linear active region. Non-inverting amplifier: An operational amplifier connected to provide a positive voltage gain at dc. Open-loop gain: The ratio o f the output voltage (loaded, but without any feedback connection) to the voltage across the two input terminals o f an op amp. The slope, p, o f the straight line in the active region o f an op amp is the open loop gain under no load condition. When a load

is present, the open loop gain is reduced to

the output resistance o f the op amp.

+ R^, where R^ is

178

Chapter 4 ®The Operational Amplifier

Operational amplifier (abbreviated op amp): A multi-stage amplifier with very high voltage gain (exceeding 10"^) used as a single circuit element.

Passive elements: a circuit element that cannot deliver net power to a circuit such as a resistor. Saturation regions: In the op amp output vs. input transfer characteristic, the region where the curve is essentially a horizontal line is called the saturation region. There are two such regions: one for positive input voltage, and the other for negative input voltage.

SPICE: Acronym for Simulation Program with Integrated Circuit Emphasis. It is a very sophisti cated software tool for simulating electronic circuit behavior.

Virtual ground: When an ideal op amp has one of its input terminals grounded, and is operat ing in the active region, then the other input terminal is also held at the ground potential because of the virtual short effect (see below). Such a condition is called a virtual

^

ground (in contrast to a physical ground).

Virtual short circuit: When an ideal op amp is operating in the active region, the voltage across the two input terminals is zero, even though the two terminals are not hard-wired togeth er. Such a condition is called a virtual short circuit (in contrast to a physical short circuit). Voltage follower: A voltage-controlled voltage source with gain equal to 1, often utilized to sep arate stages of amplification in a multi-stage amplifier device.

^The circuit proposed in this section is a modification of one proposed in W. J. Kerwin, L. P. Huesman, and R. W. Newcomb, “State-Variable Synthesis for Insensitive Integrated Circuit Transfer Functions,” IEEE Jr. of Solid State Circuits, Vol. SC-2, pp. 87-92, Sept. 1967. The modification consists of an additional resistor, which greatly simplifies the design calculations and was published by P. M. Lin as “Simple Design Procedure for a General Summer,” Electron. Eng., vol. 57, no. 708, pp. 37-38, Dec. 1985.

/—

'

^ See Linear Circuit Analysis by DeCarlo and Lin, 2nd edition, New York: Oxford University Press, 2002. ^ Any of the SPICE or PSPICE software programs available by a variety of vendors will suffice to obtain the indi cated curve. ^ An op amp and a comparator as seen in a parts catalog are essentially the same, except that the comparator device has a modified output stage that makes it compatible with digital circuits.

o

o

o

179


Problems ANALYSIS USING IDEAL OP AM P M O D EL 1. Consider the inverting amplifier circuit o f Figure P 4 .1, in which = 4 V. (a)

If

= 2 kD, find i ?2

dehvered to (b)

that the power

= 100

Figure 4.3

is 4 W.

Now suppose T(’2 = 12 kQ. Find

so

that the power delivered to Rj^ = 2 Id l is 450 mW. Then find the power con sumed in i?, and Rj-

4. In the circuits o f Figure P4.4, a source is rep resented by an ideal voltage source, v-J^t) = 4 V, in series with R^ = 10 Q resistor. The loadin both cases is (a)

= 40 Q.

W ith the load connected directly to the source, as shown in Figure P4.4(a), find the load voltage, the load current, the source current, and the power delivered to the load.

(b) Figure P 4.1 Check: (b) 1500
As in Figure 4.4(b), a buffer amplifier separates the source and the load.

= 10 m W

Again, find the load voltage, the load current, the source current, and the power delivered by the op amp to the

2. Consider the non-inverting circuit o f Figure P 4.2, in which v- = 4 V. (a)

If R^ = 2 kQ, find 7?2 so that the power delivered to = 100 Q is 4 W.

(b)

Now suppose i ?2 = 13 kO. Find R-^ so that the power delivered to

load.

+ Rs

= 2 kH

' l >r

is 450 mW. Then find the power con-

■6

sumed in R-^ and i?2(a)

(b)

Figure P4.4 SCRAMBLED ANSWERS: (a) 0.256, 3.2, 0.08 5. Figure P4.5 contains three circuits that explore loading and the elimination o f loading effects using either a dependent source or an equivalent buffering op amp circuit. (a) For the circuit o f Figure P4.5(a), com pute and in terms o f v^. Observe

Figure P4.2 Check: (b) 1500 < < 3000 3. For the circuit o f Figure P4.3, find the volt age gains, G,

andG , =

^

^ Vin in terms o f the literal resistor values.

that the 80-Q -240 Q resistor combina tion loads down the 320—0 resistor. (b)

For the circuit o f Figure P4.5(b), com pute and in terms o f v^. Notice that

is different from the answer

V,

180


computed in part (a) because the 80-

A N SW ER: (a) R, = 5 kQ; (b) 6 .76 m W

£2-240 Q resistor combination is iso lated from the 320-Q. resistor. (c)

For the circuit o f Figiire P4.5(c), again compute v-^ and

7. (a)

in terms o f v^. Your

answers should be the same as those in part (b). The buffering op amp circuit again isolates the 80-£l-240 Q. resistor comination from the 320 Q resistor.

<,v, 320 n<

Repeat part (a) for the circuit o f Figure P4.7(b).

(c)

If for Figure 4.7(b), 3 kn ,

12 kD,

= 4 kQ, Rj = 1 k£2,

= = 1.5

V, and v^ 2 = 2 V, find the power deliv ered to the load R^= 100 Q.

-o + V r + ^

as a function

and the R^. (b)

80 n

80 n

For the op amp circuit o f Figure P4.7(a), find

AN SW ER: (c) 0.04 watts

240

(a) 80 n

son

-o + 240 n 320 n<

(b)

80 0

son

-v \ ^ -

-o +

240 n 320 n

Figure P4.5 AN SW ER: (a)-V / 3

0 5V

6. In the circuit below, R^= 10 kQ. (a)

Find R^ and R^ so that

(b)

Given correct answers to part (a), sup

= -2t^^j 8. (a)

">â-

Find the power delivered to the load if

= 20. If

= “ 600 mV.

= 0.6, find the power deliv

ered to the 8 -Q load. (b)

Now suppose R^^'^ kQ, and find R^

so that (c)

out

=

20

Finally, suppose Rj =

and find

their common value so that F igu re P 4 .6

out

= 6 kQ, and find R^ so that

pose a 1 kO resistor is attached as a load.

= 200 mV and

For the circuit o f Figure P 4.8, suppose

out

=

20


181

A N SW ER: (a) 40 k£l; (b) /?, = 15 k£2 2kn -O-

11. Consider the circuit shown in Figure P 4 .l l.

-0 -, sn

and v^2 -

(a)

Find

(b)

If = 250 m V and v^ 2 - 500 mV, find the power delivered to the 1 kQ

in terms o f

load resistance. Figure P4.8 A N SW ERS (in random order): 14 kO, 8 kf2, 10 k n , 18

lOkn

lokn

lokn

5kn 2kQ

9. In the circuit below, Rr= 12 ki2 and ^3 > 1 k£2. (a)

Find

and ^2 so that

1 kn<

= -lOr^^j -

© '

20^,2(b)

Given correct answers to part (a), find the power delivered to the load if

=

- 2 0 0 m V and

Figure P 4.11 C H EC K : (b) Pj^ = 12.25 m W 12.

(a)

For

the

circuit

o f Figure

P4.12a, the input voltage

= 2 V and

the input voltage v^2 = 3 V. Find and the power delivered to the 1 kI2 load

1 kn

resistor. (b)

Repeat part (a), when

= 4 V and

the input voltage v^ 2 = 2 V. C H EC K : (b)

Figure P4.9 = 0.1 watt

(c)

Reconsider part (b). Find the mini mum value o f R so that the maximum amount o f power consumed in either R-ohm input resistors is 2 mW.


Find the value o f R = = R2 ’^ât the power delivered to R^= 1.25 kO is 0.5 watt when

(b)

3R

= 1 V.

Suppose 6R^ = i ?2 v^ = 2Y . Find and R2 so that the power delivered to

= 1.25 k il is 2 watts.

Figure P4.12 13. (a)

For the circuit o f Figure P4.13(a), the input voltage = 1 V, and the input voltage v^2 ~ 500 mV. Find /?, in terms

Figure P 4.10

o f R so that

= 10 V.

182

(b)


Repeat part (a) for the circuit o f Figure P 4.13(b), given that

= 2.5 V.

12R

r e

R, -V S/V

2R

5R

-O +

3R _ d

(a)

Figure P 4 .15 C H EC K : (a) 12R 2R

rO

5R

16. For the circuit o f Figure P 4.16, find -o +

terms o f

in

V, v^2

2.5R

(b)

Figure P4.13 AN SW ERS; (a) 6R-, (b) 3R 14. For the circuit in Figure P4.14, the input voltages are

= 2 V,

= '1-5 V, and

=2

V. (a)

Find

(b)

If /? = 10 kQ, find the power delivered

17. For the circuit o f Figure P 4.17, find

by each o f the operational amplifiers. 4R

2R

0.5 R

+ Av^2 -

1.5R

-o+

0.75R

0 - .

in terms o f R so that

and

^

Figure P4.14 C H EC K :

i = 0.9 mW,

= 0-20667

watts 15. For the circuit o f Figure P4.15, the input volt age = 5 V, and the input voltage (a) (b)

If = 8 R and Rj - Ry find If R^ = 8R and R^ = ARy find

(c)

If /?2 = that

find = lOV.

in terms o f R so

CH ECK:

Figure 4.17 = 8R

18. For the circuit o f Figure P 4.18, find R^ and i?2> and Rj in terms o f R so that

= 8f^j +

10v^2~2^s3- Hint: Consider Problem 17 first.


183

Time in s

AN SW ERS:

= AR, R^ =

(0

R, = O m

Figure P4.19

NON -IDEAL OP AM P-SATURATIO N EFFECTS 19. The op amp in Figure 4.19(a) has V,

= 4 ld2, and /?2 = 20 kQ. (a) Plot the versus for

20. Repeat Problem 19 for the op amp circuit of Figure P4.20 when

= 4 ld2, and R2 = 20 kfl.

=15 -O+ given in

Figure 4.19(b). for 0 < /■< 6 s for v(J) in

(b)

Plot

(c)

Verify your analysis in part (a) using

Figure 4.19(c). Figure P4.20

SPICE. Assume that the op amp is a type 741 whose model should be available within your SPIC E program.

21. For the circuit o f Figure P4.21, each amplifier■saturates at = 15 V. sat

(a)

Suppose the input voltage

= 5 Y

and the input voltage v^2 = “ 2.5 V.

R,

Find

-o (b)

and

If = -2 -5 V and = 15 V, find so that no amplifier saturates.

(a)

4R

1.5R

2R

-O

Figure P4.21 AN SW ER: (a) 15 V; (b) = 3.75 V 22. For the circuits o f Figure P 4.22, suppose R^ (b)

= 40 k n , and R^ = 120 Q. (a)

For the circuit o f Figure P4.22(a), compute the power delivered by the source, and power delivered to the load R^ in terms o f v-^.

184

(b)


For the circuit o f Figure P4.22(b), compute

(b)

the power delivered by

compute

the source, and power delivered to the load (c)

For the circuit o f Figure P4.24(b), the power delivered by

the source, and power delivered to the

in terms o f v-^^.

load

Discuss the differences in your solu

(c)

in terms o f

Discuss the differences in your solu

tions to (a) and (b). Specifically, dis

tions to (a) and (b). Specifically, dis

cuss the effect o f using a voltage fol

cuss the effect o f using a voltage fol

lower to isolate portions o f the circuit.

lower to isolate portions o f the circuit.

(a)

(a)

(b) Figure P4.24

25. For the circuits o f Figure P4.25, suppose /?, = 20 Q, and /?2 = 160 Q, R^ = 40 Q., and R^ = 120

a. (a)

For the circuit o f Figure P4.25(a), compute


the source, and power delivered to the Figure P4.22 23. In the circuits o f Figure P4.22, all resist ances are 100 Q, and = 1 V. (a)

(b)

the source current again when all resistances are 100 Q. I f a buffer amplifier separates the source and the load, as in Figure

For the circuit o f Figure P4.25(b), compute


the source, and power delivered to the

For the circuit o f Figure P4.22(a), find the load voltage, the load current, and

(b)

load Rj^ in terms o f v-^.

load R^ in terms o f v-^. (c)

Discuss the differences in your solu tions to (a) and (b). Specifically, dis cuss the effect o f using a voltage fol lower to isolate portions o f the circuit.

P4.22(b), find the source current, the load voltage, the load current, and the current supplied by the op amp. AN SW ERS: (a) 0.5V, 5 mA, 5 mA; (b) 0 A, 1 V, 10 mA, 10 mA 24. For the circuits o f Figure P4.24, suppose

= 40 Q , and Rj = Rl = 120 Q. (a)

For the circuit o f Figure P4.24(a), compute the power delivered by the source, and power delivered to the load Rj^ in terms o f v-^.

Figure P4.25


AN SW ERS: (a) 3.7037 x l O 'V .

185

= 0.003i^y„; (b) Pl =

26. Figure P 4.26 contains three circuits that explore loading and the elimination o f loading effects using either a dependent source or an equivalent buffering op amp circuit. (a)

27. Two non-ideal voltage sources are each rep resented by a connection o f a (grounded) inde pendent voltage source and a series resistor. Denote the parameters o f each connection by

{Vs^,

^st)' Design an op amp cir

cuit such that the output voltage with respect to ground is

For the circuit o f Figure P4.26(a), compute I'j and

Observe that the

8-Q -24 Q. resistor combination loads down the 3 2 -Q resistor. (b)

For the circuit o f Figure P4.26(b),

for all values o f and R^2 tie greater than or equal to 100 kS2 so that only small

compute

is

amounts o f current are drawn from the buffer

different from the answer computed

amplifiers. Note that the general difference

in part (a) because the 8-Q -24 Q resis

amplifier circuit o f the chapter will not work

tor combination is isolated from the

here because o f the presence o f the resistances

and

Notice that

3 2 -Q resistor. (c)

and R^2 - To achieve such a design, it is nec

For the circuit o f Figure P4.26(c),

essary to isolate the (practical) sources from the

again compute Pj and

Your

difference amplifier inputs using buffer ampli

answers should be the same as those in

fiers, as shown in Figure P 4.27. Explore your

part (b). The buffering op amp circuit

design for various values o f

again isolates the 8-Q -24 O, resistor

SPICE. Do the SPIC E simulations verify that

combination from the 32 Q resistor.

the output is independent o f the values o f

and R^2 using

Figure P4.27 8Q

80

-o +

circuit o f Figure P 4.28, suppose the op amp has 24 n

32 0 <

(c)

Figure P4.26 AN SW ERS: (a) 0.6665 V^, 0.5 V^; (b) and (c) 0.8 V^, 0.6

28. Following Example 4.9, for the comparator infinite gain and a saturation voltage =15 V. Find the versus relationship and plot as a function o f v-^. Verify your analysis using SPICE. Assume that the op amp is a type 741 whose model should be available within your SPIC E program.

186


Check your design using SPICE. Hint: How

80 kO

can the circuit o f Figure P4.29 be modified to achieve the correct polarities? /?] _

A N SW ER: 32. Find the comparator

Figure P4.28 29. (a)

Find the

1 1.5

versus v-^ relationship for the circuit

of

Figure

P 4.32.

Specifically, show that when

versus

relationship for

Vin >

, then

and when

the comparator circuit o f Figure P4.29. Specifically, show that when

Vin <

=

R,

and when v-^ > ??, then

then V = - V^sa f 'ôut (b)

în<-^^ref , then

Now suppose

^ref ~ characteristic if

= R2 = 100 kQ and ôut versus v-^ = 15 V. Verify

Figure P4.32

your analysis using SPICE. Assume that the op amp is a type 741 whose model should be available within your SP IC E program.

G EN ERA L SUM M IN G AM PLIFIER (ID EA L OP AM P M O D EL) 3 3 .(a)

Assuming the op amp in Figure P4.33 is ideal, derive the relationship

Figure P4.29 AN SW ERS: (b) = 15 V if

< 2 V, and

(b)

Rf

Rf

R,

Ro

Suppose so that

25 kO. Find , i?2> is the negative o f the

average o f

, V2 , and Vy

30. Using a 1.5 V battery, an op amp with = 10 V, and some resistors, design a comparator circuit such that and

= 10 V when

< 1 V,

= - 1 0 V when v-^ > 1 V. Check your

design using SPIC E. Use part (a) o f Problem 29 as a guide. 31. Using a 1.5-V battery, an op amp with = 10 V, and some resistors, design a comparator circuit such that = 10 V when v-^ > 1 V, and

= - 1 0 V when v-^ < I V.

V l-

Figure P4.33 AN SW ER: (b) R. = 75 kO


187

34. Using the topology o f Figure 4.13, design an op amp circuit to have the input-output


relationship

relationship +H i

- 3^.2 +

Two different designs are to be produced for

+H i Two different designs are to be produced for comparison and selection:

comparison and selection: (a) Design 1: Rjr= 100 kQ. (b)

1:

. = 50 kQ.

^

iqq

Design 2: Rj^= 50 kQ. Specify all final

values in terms o f Q.. C H ECK : = 25 kO, and R^2fiil ^ 33.33 k£2

^ 39. Generalizing the topology o f Figure 4.13, design an op amp circuit to have the input-output relationship


ôut ~â\ ~ '^'âl ~ I„ .he final circm, Rf - 40 k a

" ' “ “ “ '■‘P

ôut ~ ~^â\ ~ '^âl

‘^^h\

^'^bl

Two different designs are to be produced for comparison and selection:

40. Generalizing the topology o f Figure 4.13, design an op amp circuit to have the input-output relationship out

(a) (b)

Design 1: Rr= 100 k£l. Design 2: R^= 50 k tl. Specify all final values in terms o f Q.

,

^

,

= -4v^j + 2z;^j + n / r^ ^/=

VARIABLE GAIN AM PLIFIERS


4 1

x h e circuit o f Figure P4.41 is a modificanon o f the basic non-inverting amplifier. In the

relationship

modification, a potentiometer R^ is connected between the output terminal and Rq, with the

ôut ^

~ '^'âl ^

^'^bl

Your design must have Rj-= 10 kX2 in the final

sliding contact between points A and B, as shown. Show that as the sliding contact o f the potentiometer is moved between positions A

circuit, and all other resistors should be within

and B, the range o f voltage gain achievable is

the range 2 kQ to 20 kQ.

ôut

P

37. Using the topology o f Figure 4.13, design an op amp circuit to have the input-output relationship

ôut =

- ">âl + ^^b\ +

hi

Your design must have all resistors, including

Rjr, in the range 5 kQ to 25 kQ.

Figure P4.41 Variable gain non-inverting amplifier.

188


42. The circuit o f Figure P4.42 is a simple modification o f the basic inverting amplifier circuit in which a potentiometer is connected to the feedback resistor Rp as shown. Show that the range o f gains achievable by this circuit is

h .

SIM ULATION OF C O N TR O LLED SO URCES USING OP AM PS 44. Design an op amp circuit to simulate the grounded VCVS in Figure P 4.44 when p > 1. Hint: Consider the non-inverting amplifier o f Example 4.3.

Figure P4.44 Grounded VCVS. 45. Design an op amp circuit to simulate the grounded V CV S in Figure P4.44 for any p > 0. Hint: Try a voltage follower in cascade with two inverting op amp circuits.

Figure P4.42 43. T he circuit o f Figure P4.43 is another mod ification o f the basic inverting amplifier to obtain a variable gain amplifier. Show that as the sliding contact o f the potentiometer is moved between the two extreme positions, the range o f achievable voltage gain is R

f

U a = 1, H— where

:-a -

—H— -

Rq

Rf

46. Reconsider the design o f Problem 45 so that only two op amps are used. In this case, one still needs the voltage follower. Why? Hint: Consider using a voltage divider followed by a non-inverting amplifier circuit.

R

f

R, Rp R■0n // /Ii\f Rf

Hint: Apply KCL to the non-inverting input terminal, and make use o f the virtual ground property o f an ideal op amp.

47. Design an op amp circuit to simulate the grounded VCVS in Figure P 4.44 when p < 0. Hint: Consider an inverting amplifier configu ration in conjunction with a buffer amplifier. 48. For the circuit o f Figure P4.48, show that the load current

equals V/R^, which is inde

pendent o f the load resistance R^. Hence, this op amp circuit converts a grounded voltage source into a floating current source. (This is sometimes called a voltage-to-current con verter.)

■o

Figure P4.48 Op amp circuit simulating a floating current source.


49. In Problem 4.48, since 7^ depends on Vand

only, the load need not be a resistor. For

example, Rj^ may be replaced by an LED (lightemitting diode), as shown in Figure P4.49. Then by turning the knob o f the 10-kf2 poten tiometer, one can control the brightness o f the LED. The current through the load is supplied by the op amp. The potentiometer, which con trols the brightness o f the LED , uses a low-voltage part o f the circuit. Find the magnitude o f the LED current if the potentiometer is set at (a)

= 5 k£^ and (b) 7?, = 8 k tl.

A N SW ERS: 1.32 niA, 2.1 niA 10 kn

50. This problem is a variation o f Problem 4.49 in which the load current flows in the opposite direction. For the circuit o f Figure P 4.50, show that the load current

equals v^JR^, which is

independent o f the load resistance R^. Hence, this op amp circuit converts a grounded voltage source into a floating current source in which the current enters the op amp output terminal. (This is sometimes called a voltage-to-current converter.)

Figure P4.50 Op amp circuit simulating a float ing current source.

189

C

H

A

P

T

E

R

Linearity, Superposition, and Source Transformation H ISTO RICAL NOTE In the mid-nineteenth century, before the introduction o f the alternating current (ac), electricity was available mainly as direct current (dc). This time period saw the evolution o f basic laws for the analysis o f electrical circuits composed o f dc voltage sources and resistors: Ohm’s law, KVL, and KCL. Application o f these laws to the analysis o f circuits led to the development o f the mesh and nodal techniques requiring the solution o f simultaneous equations. Before the computer age, manual solution o f a (large) set o f equations was very difficult. To circumvent this difficulty, researchers developed a number o f network theorems that (i) simplified the aforementioned man ual analysis, (ii) reduced the need for repeated solution o f the same set o f equations, and (iii) pro vided insight into the behavior o f circuits. These network theorems remain useful even in the pres ent day o f high-powered computing.

CH APTER O BjEC TIV ES 1. 2. 3.

Introduce and apply the property o f linearity. State and explore the two consequences o f linearity called superposition and proportionality to simplify response computation. Use superposition and proportionality to simplify manual analysis and to gain better

4.

insight into circuit behavior. Introduce and apply the source transformation theorem to again simplify manual analysis.

SECTIO N HEADIN GS 1. 2. 3. 4. 5. 6.

Introduction Linearity Linearity Revisited: Superposition and Proportionality Source Transformations Equivalent Networks Summary

192

Chapter 5 * Linearity, Superposition, and Source Transformation

7. 8.

Terms and Concepts Problems

1. IN TRO D U CTIO N Chapter 3 covered nodal and loop/mesh analyses. Node voltage or loop current calculation pro ceeds by constructing a set o f simultaneous node or loop equations and solving them by hand, by MATLAB, or with some equivalent software package. Few o f us will attempt a paper-and-pencil solution o f four equations in four unknowns. Yet, MATLAB, Mathematica, or some other com putational software program, can easily and reliably crunch numbers, relieving us o f tedious hand calculations. Nevertheless, manual analysis in some form remains important for a deeper under standing or insight into a circuit’s behavior, as well as a way to check the validity o f a program output. Experience teaches us that manual analysis is ordinarily practical only for small circuits. Fortunately, the network theorems studied in this chapter and the next can often reduce seemingly complex circuits to simpler ones amenable to manual analysis. They also provide shortcuts for computing outputs and allow us to obtain deeper insights into a circuit’s behavior. This chapter talks about linearity and superposition, which are motivated by the following ques tions: What is the effect on the circuit output (voltage or current) o f a single independent voltage source, say

acting alone. “Acting alone” means that the independent source, Vj^, has a nonzero

value, while all other independent sources are set to zero. A deactivated voltage source acts as short circuit (see Chapter 2), and a deactivated current source acts as an open circuit (again, see Chapter 2). Is there a shortcut to computing the response if Vj^ is doubled in value? To answer the above questions and others, our discussion begins with the important property o f linearity. Linearity relates the values o f independent sources to a circuit output with a very com pact equation. This equation defines the effect o f any independent source on a circuit output. After studying linearity, we discuss two special consequences called superposition and propor tionality. Each o f these concepts helps reduce manual computation o f responses, and each provides insight into circuit behavior. Next, we state the source transformation theorem and show how this method can reduce a complex circuit to a more simple form. Finally, we set forth the notion o f an equivalent two-terminal network and then outline a proof o f the source transformation theorem.

193


2. LIN EARITY This section investigates the circuit property o f linearity, which we introduce with a motivating example. E X A M P L E 5 .1 . For the circuit o f Figure 5.1, find the outputs current /^j, and the source voltage V^2 -

A

3

and Vg in terms o f the source

will derive the relationships Vg = 40/^j + ^ 1^2 and

jgQ .sz I.

60 0

+ V„

120Q

FIG U RE 5.1. Resistive circuit driven by current and voltage sources.

So l u t io n Step 1. Find the voltage Vg. A node equation at the top o f the current source is Vb , V B - V s l ^ r 120 60 Solving for Vg yields = 4 0 7 ,1 + - V ,2

Here, Vg appears as a constant times /^j, plus another constant times K^2>^ so-called linear com bination. Step 2. Find the current I

From Step 1, we know Vg. The current

40 / , i + - V , 2 - V ,2

Similar to Step 1, the output current

linear combination.

3

satisfies

180 •'

is a constant times /^j plus another constant times V^2>^

194

Chapter 5 ' Linearity, Superposition, and Source Transformation

Exercise. 1. In Example 5.1, suppose the 60 Q resistor is changed to 120 Q. Find the outputs and Vg in terms o f the sources, /^j and V^2 AN SW ER: Vg = 60/ ,, + 0 .5 V ,2 and

= 0 .5 / ,, -

2. For the circuit o f Figure 5.2, find Vg in terms o f

and V^2-

--- ---------- ---- --------120 Q + 60 Q > 120Q

V

(j

V..

-

FIG U RE 5.2 Resistive circuit for Exercise 2. A N SW ER:

= 0.251/, + 0 .5 V^2

In the above example and exercises, the desired output voltage or current was a so-called linear combination o f the independent source values. This is, in fact, a quite general phenomena, as indi cated by the linearity theorem below.

LIN EA R ITY TH EO R EM For all practical linear resistive circuits, as per Figure 5.3, any output voltage,

or any cur

rent, ig, can be related linearly to the independent source values, as in the following equa tions:

= î^s\ + - +

+•

^nfsm

(5.1a)

= “ l Kl + - +

+ ■•• + ^mhm

(5.1b)

or

where the a - and

are properly dimensioned constants.

Chapter 5 • Linearity, Superposition, and Source Transformation

V

195

Linear Circuit containing no independent sources.

V

+

■

FIG U RE 5.3. A linear circuit driven by n independent voltage sources and m independent current sources with outputs of and A rigorous proof o f the linearity theorem entails solving a set o f modified nodal or loop equations using matrix algebra and is beyond the scope o f this text. EXAM PLE 5.2. For the circuit o f Figure 5.4, our objective in this example is to express ear combination o f /^j, Iq, and

lin

as per equation 5.1a. In doing this, we review nodal analysis. g .v .

)v .

FIG U RE 5.4

So l u t io n Step 1. Write nodal equation at A. For node A, (5 .2 )

196


Step 2. Write nodal equation at output node. At the output node,

or equivalently,

=

(5.3)

Step 3. Write equations 5.2 and 5.3 in matrix form. In matrix form, the nodal equations are

0

G ^ -g^ G

8m

2+

■ G

3

■

/.I

■

•

(5.4)

J s l + G ^ V ,,

Vou t,

Step 4 . Solve equation 5.4. Solving equation 5.4 for

and

yields

-1

-V

a

■

Vou t,

0

'G l- g m 8m

G 2

+

G

h i

3

1

G2 + G 3

{G :-gJ(G 2+ G ,)

-8 m

0

‘ s\

G, - I

fs2+G3V,3

It follows that

Vout--

8m (G i-g ,)(G 2 + G 3 )

hl+-

Gi-g. (G ,-g „,)(G

jGl-8m)G3

2+

G3)

(G i-g ,„)(G

2+

G

-V.s3

3)

(5 3 )

as set forth in equation 5.1(a).

Exercise. 1. In equation 5.5, suppose G j = 1 S, G2 = 2 S, G3 = 3 S, and^^ = 5 S. Find the numer ical expression for A N SW ER:

= 0.25/,, + 0.27,2 + 0-6 ^^,3

2. Suppose the dependent current source in Figure 5.4 is changed from^^V^ the new expression for

if G j = 1 S, G2 = 2 S, G3 = 3 S, and

Smôuf Compute

= 5 S.

AN SW ER: V ;„,= 0.17,2+ 0 .31/3 3. Suppose the dependent current source in Figure 5.4 is changed from Compute the new expression for AN SW ER:

if G j = 1 S, G2 = 2 S, G3 = 3 S, and

to

+ ^0 ^)-

= 5 S.

^ ^.'i + ^ h i + J ^v3

E X A M P L E 5 .3 . A linear resistive circuit has two inputs and z,2 with output as shown in Figure 5.5. Rows 1 and 2 o f Table 5.1 list the results o f two sets o f measurements taken in a lab oratory. The measurements are taken in a practical way by first setting the value o f the current

197

ch a p te r 5 " Linearity, Superposition, and Source Transformation

source to zero, i.e., i^ 2 = 0

exciting

with a dc power supply set to 5 V; then the voltage

source is removed and replaced by a short circuit using a jumper cable, i.e., rent source is excited by a power supply producing a constant current o f (a)

Derive the linear relationship

(b)

Find

when

= 0, and the cur = 0-2 A.

+ 50/^2 using the data in Table 5.1.

= 10 V and ^'.2 = 0.5 A, i.e., complete the third row o f Table 5.1.

FIG U RE 5.5. Linear resistive circuit driven by two sources. TABLE 5.1. Two Sets of Measurements o f a Linear Circuit in which One is Allowed to Set Each Source Value to 0

i^ 2 (amperes)

Vout (volts)

5

0

4

0

0.2

10

10

0.5

(volts)

So l u t io n From the linearity equation 5.1(a),

ôut =

+hhl

for appropriate ttj and ^2 - From the data in rows 1 and 2 o f Table 5.1,

4 = a , x 5 + P2><0 = 5a, ^

U j = 0.8

and 10 = a , X 0 + p2 X 0.2 = O.2 P 2 => P 2 = 50 in which case, + 5 0 i,2 So from row 3 o f Table 5.1, if i',] = 10 V and z'^2 = 0.5 A, we have that 0 .8 X 10 + 5 0 X 0 .5 = 33 V.

(5.7)

198


Exercise. 1. For Example 5.4, suppose

= 50 V and i^ 2 ~ 0-4 A. Find

AN SW ER: 60 V 2. Suppose the data in row 1, column 3, o f Table 5.1 is changed to 10 V. Find y^^^when V and

= 50

= 0.4 A.

AN SW ER: 120 V

Comparing the development o f equation 5.7 in Example 5.3 with equation 5.1 suggests that the coefficients Qj and P 2 can be defined as ratios:

«1 =

''out

and 132= — I,-,v2

Example 5.3 and these equations suggest the algorithm for finding the coefficients in equation 5.1 by setting all inputs to zero except the input associated with the desired coefficient. This approach is sometimes impractical. It is not always possible to set an independent source voltage or current source to zero: imagine turning off a generator for downtown Manhattan to obtain a coefficient. T he following example illustrates an alternate approach. E X A M P L E 5 .4 . Consider Figure 5.6, which has two inputs

and

output

Table 5.2

lists measurement data taken in a laboratory. Row 1 ofTable 5.2 lists the nominal operating con ditions o f the circuit. Rows 2 and 3 illustrate measurements in which one source has its value only slightly changed (although the change may be arbitrary) while keeping the other source value the same. From the linearity theorem, we know find to complete row 4 ofTable 5.2.

+ ^2^s2- Compute

FIG U RE 5.6. Linear resistive circuit driven by two sources.

and P j, and then


199

TABLE 5.2. Two sets o f measurements of a linear circuit.

i^ 2 (amperes)

hut (amps)

5

0.25

-1

5+0.1

0.25

- 1 .0 3

5

0.25+ 0.05

- 0 .9

15

0.5

????

(volts)

So l u t io n From rows 1 and 2 o f Table 5.2, - 1 = ttj X 5 + P2 X 0.25

(5.8a)

- 1 .0 3 = ttj x (5 + 0.1) + P 2 X 0.25

(5.8b)

and

Subtracting equation 5.8(a) from equation 5.8(b), we have - 0 .0 3 = ttj X 0.1 => ttj = - 0 .3 Similarly, from row 3, we have that - 0 .9 = ttj X 5 + P 2 X (0.25 + 0.05)

(5.8c)

Again, subtracting equation 5.8(a) from equation 5.8(c), we have 0.1 = p 2 x 0 .0 5 ^

Pa = 2

Equation 5.1 for the given data has the linear form (5.9) Hence, for row 4 o f Table 5.2, we have that 2,„, = - 0 . 3 x 1 5 + 2 x 0 . 5 = - 3 .5 V

200


Exercise. Find the unknown value in Table 5.3 using linearity. TABLE 5.3. Two Sets of Measurements of a Linear Circuit (volts)

i^2 (niA)

20

100

15

22

100

15.9

20

110

15.6

28

80

???

AN SW ER: 17.4 A

As a final comment on linearity, we note that by simply using the data o f rows 1 and 2 o f Table 5.2, one can solve for the coefficients by solving simultaneous equations. Specifically, using the data o f rows 1 and 2 ofTable 5.2, we have the following matrix equation ■5

0 .2 5 ' ■ «i’

5.1

0.25

.^1.

■ -1

■

-1 .0 3

whose solution yields the proper coefficients o f equation 5.9. ■5

■«r

5.1

.A .

0 .2 5 ' 0.25

-1

■ -1 - 1 .0 3

"

= -4 0

'0 .2 5 -5 .1

- 0 .2 5 ' ■ -1 5

■

■-0.3'

-1 .0 3

2

Exercise. Find the unknown entry in Table 5.4 after finding a , and (3j in the equation + Pl^.2TABLE 5.4. Two Sets of Measurements of a Linear Circuit (volts)

AN SW ER:

ia (itiA)

ôut

10

100

15

20

100

20

30

150

???

= 0.5z^,i + 100/^2

V

201


3. LIN EARITY REVISITED: SUPERPO SITION AND PRO PO RTIO N A LITY T he linearity principle o f equation 5.1 has the more simple form

y=

( 5 . 10)

+ ... + a„u„

Here, j denotes an output, whether it be current or voltage, and each

denotes a source input,

whether it be voltage or current. A special consequence o f the linearity principle is the superpo sition property. Equation 5.10 says that the total response jy is the sum o f the responses '"aju". Each “a-u” is the response o f the circuit to u- acting alone, i.e., when all other independent sources are set to zero. Although implied by linearity, this property is so important that we single it out.

THE SUPERPOSITION PROPERTY For almost all linear resistive circuits containing more than one independent source, any out put (voltage or current) in the circuit may be calculated by adding together the contributions due to each independent source acting alone with the remaining independent sources deacti vated, i.e., their source values are set to zero.

EXA M PLE 5 .5 . A linear resistive circuit has two inputs

and

with output

as shown in

Figure 5.7, where = 2 Q,, Rj = 2.5 £2, and R^ = 10 Q. Find by the principle o f superposi tion. Then, compute the power absorbed by the 10 resistor. We show that = 0.5 V^l + 0.4V^2’ where

is the contribution o f the source

acting alone for k = 1, 2.

FIG U RE 5.7 Linear resistive circuit driven by two voltage sources;

Rj = 2 Q , R 2 = 2.5 So l u t io n Step 1. Fini/ the contribution to

and R^ = 10 Q.

due only to V^j. Denote this contribution by

W ith V^2

= 0, the equivalent circuit is shown in Figure 5.8(a). Here, the 2.5 Q and 10 Q, resistors are in par allel, yielding an equivalent resistance o f 2 = 2.5 x 10/12.5 f i. By voltage division,

V

-

2 +2

202


+ Voût

FIG U RE 5.8 (a) Circuit equivalent to Figure 5.7 when (b) circuit equivalent to Figure 5.7 when Step 2. Find the contribution to

due to

= 0;

= 0.

Denote this contribution by

W ith

= 0,

the equivalent circuit is shown in Figure 5.8(b). Here, the 2 Q and 10 O resistors are in parallel, yielding an equivalent resistance o f 5/3 = 2 x 10/12 Q. By voltage division,

2.5 + 3 Step 3. Compute

by superposition. Using superposition, Vout = ]/’o u t ^+

^ out

= 0 5 K , ^+ 0 4K .,

Step 4 . Compute the power absorbed by the R^= 10 Q resistor.

Pr3 =

(youtT

= 0 . 1(0.5

+ 0.4V,2 f = 0 . 1(o .2 5 v /i + 0 .2 V ,iy ,2 + 0 . 16V / 2 )

Note that the total power, Pj^^, is not the sum o f the powers due to each source acting alone because o f the presence o f the cross product term. Hence, in general, superposition does not apply to the calculation o f power.

For dc circuit analysis, the principle o f superposition does NOT apply to power calculations.

Exercise. Reconsider Figure 5.7 in which principle o f superposition. AN SW ER;

0.5K^, + 0.25

2

= 2 Q , 7?2 = 4 Q, and R^ = A Q.. Find

by the

203


The next example adds a controlled source to the circuit o f Figure 5.7 and repeats the superposi tion analysis.

E X A M P L E 5 .6 . For the circuit o f Figure 5.9, suppose i?, = 2 Q, S. Using superposition, find in terms o f and v^2 show that

= 0 . 5 z^^j + 0 .4 t/^2’ where

=5 -^3 = 10 = 0-2 superposition theorem to

is the contribution from the source

acting alone for k = 1, 2 .

q V.A

FIG U RE 5.9 Circuit containing a dependent source for illustrating the principle o f superposition.

So l u t io n Step 1. Compute the contribution due only to

Setting v^ 2 ~ ^ leads to the circuit o f Figure 5.10,

where we note that

Applying KCL to the top node yields 0 - 5 ( - L - " . l ) + (0-2 + 0 . 1 + 0 .2 ) .i „ ,= 0 Therefore, d r = 0 - 5 ^ .i Step 2. Compute the contributions due only to o f Figure 5.11, where this time,

= ôut~

Setting

= 0 in Figure 5.9 leads to the circuit

204


As in Step 1, we apply KCL to the top node to obtain (0.5 + 0.1 + 0 .2 ).2 ^ ,- 0.2.^2 + 0 - 2 K i - -. 2) = 0 Therefore,

Step 3. Using superposition, add up the contributions due to each independent source acting alone.

ôut =

Exercise. Repeat Example 5.6 with AN SW ER:

+ 0-4 ^',2

= /?2 = ^3 = ^ ^

Sm ~

(5.11)

S.

= 0.25i',i + 0.5t^,2

The above examples used voltage division and superposition to compute an output voltage due to two independent voltage sources. E X A M PLE 5 .7 . This example illustrates the principle o f superposition for the three-input op amp circuit o f Figure 5. 12 . Show that is the contribution o f

acting alone for k = 1 ,2 , 3.

FIG U RE 5.12 Three-input op amp circuit.

+ 2 . 5 K^2 +


205

So l u t io n Step 1. Find the contribution to

due only to Denote this output by W ith V^ 2 = ^ 3 = 0, the circuit o f Figure 5.12 reduces to that o f Figure 5.13(a). The properties o f an ideal op amp

ensure that i^ = 0, making

= -0.5i?z^ = 0. Thus, v_ = v^=Q implies V’i r = - — K . = - 4 V ,

2R

(a)

FIG U RE 5.13 Step 2. Find the contribution to

due only to V^2 - W ith

shown in Figure 5.13(b) where we denote the output as

= 0, the equivalent circuit is From op amp properties and volt

age division, ^ -V ,2 = 0 .5 V ,2

R+R Hence, from Example 4.3,

Step 3. Find the contribution to

due only to

as that o f Figure 5.13(b) with V^ 2 replaced by

T he equivalent circuit in this case is the same Therefore, the output due to source

acting

alone is

Step 4 . Sum up contributions due to each source. By the principle o f superposition.

Vout =

out

out

Exercise. 1. For Example 5.7, suppose

out

= -^4 sK\ , + 2 .5 Ks2t + 2 .5 K. ^ si

“ ^3 “ ^

(5 . 1 2 )

ôuf

AN SW ER: 2 V 2. Now suppose A N SW ER: - 1 2 V

= 8 V,

= ^s3 =

°P

saturates at |

= 12 V; compute

206


The above examples have generated the linearity formula, equation 5.1, using superposition, i.e., the response o f a circuit is the sum o f the responses due to each source acting alone. The technique is equiv alent to that described in Example 5.3. However, superposition alone is not equivalent to linearity. Linearity is equivalent to the properties o f superposirion AND proportionality, which is now stated.

T H E P R O P O R TIO N A LITY PRO PERTY For almost all linear resistive circuits, when any one o f the independent sources is acting alone, say « j, with output 7 , then y =

for some constant a^. Proportionality says that if

is multiplied by a constant K, then the output is multiplied by K, i.e.,

However, for dc analysis, the proportionality property does N O T apply for power calculations.

T he proportionality property is easily illustrated by equation 5.12 o f Example 5.7:

- V'.., • V I , * V i , ■ . K,

If

and

* 2-5^2 ^ 2-5^3

. V„ . 0, then V ^ . - 4 (« V „ ) . « - 4 K „ ) -

Exercises. For certain nonlinear circuits, the principle o f superposition may be satisfied, but pro portionality not satisfied, or vice versa. This exercise explores these distinctions. 1.

sV ^^ôw that the principle o f

If a circuit has input-output relationship

superposition is satisfied, but proportionality is not satisfied. 2.

If a circuit has input-output relationship

= a,Wjj +

^ôw that the

principle o f proportionality is satisfied, but superposition is not satisfied.

A very interesting and significant application o f the proportionality property occurs in the analy sis o f a resistive ladder network. A resistive ladder netw ork is one having the patterned structure shown in Figure 5.14, where each box represents a resistor.

-H

v.Q

V

FIG U RE 5 .14 A ladder network. A typical analysis problem follows: Given and all resistances in Figure 5.14, find all node voltages. O ne can, o f course, solve the problem by writing and solving a set o f mesh equations or node equations. A simple trick using the proportionality property allows us to solve arbitrarily long ladder networks without simulta

207


neous equations, as follows: assume

= 1 V. We can sequentially compute currents and voltages

= \ V. Suppose we call this

in a backwards fashion to obtain the required source value to yield voltage

Define K =

to be the proportionality constant, where is the actual source

voltage. Then the correct output voltage is E X A M P L E 5 .8 . Find all the voltages V^-, i = 1, ..., 6 in the resistive ladder network o f Figure 5.15.

----------- —

— ^ L

"

R =100

'4 W R4 = 6 Q

R =5Q

-

—

+

r

1. tS

V =50V

-------- — '—

^ V.

V, R3 = 5 0

R j= 1 0 Q

t .

V,

R, = 1 0 Q

FIG U RE 5.15 A simple resistive ladder network.

So l u t io n Assume Vj = 1 V. Repeatedly apply Ohm’s law, KCL, and KVL as follows: (Q, V and A are used throughout): (Ohm’s law) /2 = / i = V2 = /?2 / 2 = V3 = /3

=

/2

1 0

(KCL)

0 .1

x

0 .1

+ V2 =

=

(Ohm’s law)

1

(KVL)

2

(Ohm’s law)

= ^ = 0.4 ^3

+ / 3 = 0 .1 + 0.4 = 0.5

V4 = R4 / 4 = Vg =

\/3

6

X 0.5 = 3

(Ohm’s law) (KVL)

+ y4 = 5

(Ohm’s law)

7, = "^ = 0 .5 ^5 ^6 = 7 4 + 7 5 = 0.5 + 0.5 = V6 = 7?67e=5 y, = y5 + y6 = io

(KCL)

1

(KCL) (Ohm’s law) (KVL)

We conclude that if Vj = 1 V, the source voltage must be = 10 V. But the actual source voltage is 50 50 V. Define K = = — = 5 . By the proportionality property, if = 50 V, then - I^x 1 = 5 V Similarly,

- 5 V,

= W V,

= 15 V, and V5 = 25 V.

208


In the solution given above, we have separated the expressions into calculation blocks to empha size the repetitive pattern. For example, the expressions in block #3 are simply obtained from block #2 by increasing all subscripts by 2. When the ladder network has more elements, the sequence o f expressions contains more blocks, each o f which entails two additions and two mul tiplications. This method then allows us to straightforwardly solve ladder networks o f any size without writing or solving simultaneous equations.

Exercise. In Example 5.7, change all resistances to 2 Q and find V^. Would it make any differ ence in the voltage Kj if all the resistors were changed to R ohms? 50 AN SW ER: Vi = — = 3.85 V, and no difference. ' 13

4. SO U RCE TRAN SFO RM ATIO N S The words “source transformation” refer to the conversion o f a voltage source in series with an Rohm resistor to a current source in parallel with an R-ohm resistor, and/or vice versa. This section explains the details o f such transformations and how they can simplify analysis. But first we must recall from Chapter 2 that voltage sources in series add together (such as batteries added to a flash light) and that current sources in parallel combine into an equivalent single current source. This is illustrated for multiple voltage sources in series in Figure 5.16. Similarly, Figure 5.17 shows how multiple current sources combine into a single source.

4.5 V

FIG U RE 5.16 (a) Three voltage sources in series; (b) equivalent single voltage source.

s,eq

FIG U RE 5.17 (a) Three independent current sources in parallel; (b) equivalent single source circuit.


209

SOURCE TRANSFORM ATION THEOREM FOR INDEPEN DEN T SOURCES A 2-terminal network consisting o f a series connection o f an independent voltage source and a nonzero finite resistance R is equivalent to a 2-terminaI network consisting o f an independent current source, /^ = VJR'm parallel with R. Conversely, a 2-terminal network consisting o f a parallel connection o f an independent cur rent source

and a nonzero finite resistance R, is equivalent to a 2-terminai network con

sisting o f an independent voltage source,

= RIy, in series with R. T he reference directions

for voltages and currents are as shown in Figure 5.18.

V =RL

FIG U RE 5.18 Illustration of source transformation theorem for independent sources. A justification for the source transformation theorem will be given in the next section. Practically speaking, it can save significant computational effort. For example, in the circuit o f Figure 5.19 in Example 5.9 below, a solution approach using mesh analysis requires writing and solving three simultaneous equations. Nodal analysis at A and B requires writing and solving two simultaneous equations. Applying the source transformation theorem is a third avenue that avoids all simultane ous equations. E X A M P L E 5 .9 . Find

in Figure 5.19 by repeated applications o f the source transformation

theorem. Then find the power absorbed by the 4 kD resistor.

5 kO

6kQ

50 V

FIG U RE 5.19 Circuit for Example 5.9.

210


So l u t io n Step 1. Substitute all series V^- R combinations by their parallel

- R equivalents, where in each

case, /^ = — . Applying the source transformation theorem four times results in Figure 5.20.

R

10 mA

10 mA

FIG U RE 5.20 Circuit equivalent to that o f Figure 5.19 by source transformation theorem. Step 2. Combine the parallel resistances and the parallel current sources. To the left o f point A are two independent current sources and two resistors, all in parallel. Similarly, to the right o f B are two current sources and two resistors in parallel. Combining current sources and resistors to the left o f A results in a single current source o f 5 mA directed upward and an equivalent resistance o f 4 kD. To the right o f B, the current sources can cel each other out, and the equivalent resistance is 2 kXl. T he resulting simplified circuit is shown in Figure 5.21.

5 mA

10 mA

FIG U RE 5.21 Simplification o f the circuit in Figure 5.20. Step 3. Apply the source transformation theorem a second time to each o f the allel I ; - R pairs become series V^- R pairs, as illustrated in Figure 5.22. 4kO

--A 20 V

4kQ

B 20 V

FIG U R E 5 .2 2 Further simplification of Figure 5.21.

- R pairs. These par


Step 4 . Find

and P^j^- From Ohm’s law, ÂB -

Thus,

211

20 + 20 4+4+2

= 4 mA

= 4 0 0 0 X (0 .0 0 4 )2 ^ <54

Exercises. 1. For the circuit o f Figure 5.23(a), /^j = 50 mA and R combination to a series V^- R combination, where

= 500 Q. Convert the parallel

= ? and

= ? in Figure 5.23(b).

AN SW ER: 25 V, 500 Q

f^series

A

-o Circuit

V.

Circuit

r + B ■O

B -O (b)

(a) FIG U RE 5.23

2. For the circuit o f Figure 5.24(a), /^j = 50 mA and R^ = 500 Q , while I ^ 2 = 1 5 0 mA and R^ = 300 Q . Convert the two parallel 1^—R combination to a single series V^—R combination, where

= ? and

= ? in Figure 5.24(b).

AN SW ER: - 2 0 V, 800 Q

A

A ho-

-o Circuit

Circuit

V B -O -

B -O (b)

(a) FIG U RE 5.24

3. Consider the circuit in Figure 5.25(a). Using a source transformation and resistance combina tions, determine the values o f

and Rp^^^ in Figure 5.25(b).


212

100Q

100Q

FIGURE 5.25 AN SW ER: 150 LX 50 niA

5. EQ U IVALEN T N ETW ORKS The source transformation theorem above is based on the notion o f equivalent networks, as is the material o f the next chapter. So we now explore a precise understanding o f equivalent 2-tenninal net works. Figure 5.26 illustrates four 2-terminal networks, all enclosed by dashed boxes, labeled N j, N 2, Ng, and N^. Their characteristic is that there are only two accessible nodes for connection to other circuits. Note however that any controlling voltage or current must be contained within the dashedline box. Such dashed-line boxes are often omitted to avoid cluttering in circuit diagrams.

FIGURE. 5.26 Examples of 2-terminal networks, i.e., networks in which only two terminals are available for connection to other networks. Observe that networks N j and N 2 in Figure 5.26(a) and (b) have the same terminal characteris tics: at the terminals o f N j, the v - i characteristic is

v = 2i+ 10


213

At the terminals o f N 2, the characteristic is v = 2( + 10

i= - - 5 2

The two equations are identical. We then say that a pair o f 2-terminal networks are equivalent if they have the same terminal characteristics. Therefore, and N 2 are equivalent. Now, observe that networks

and

are also equivalent to A^j and A^2- To see this, note that

for N y V

v-1 5

6/ = 3 v - 3 0

v = 2i + 10

And for N^, first observe that i = \0i^ from KCL, in which case 10 + 20/'a ^

= O.lz; further, from KVL,

v = 2 i + 10

as was to be shown. Because equivalent 2-terminal networks have the same terminal v —i characteristic, if one network is interchanged with its equivalent, all currents and voltages outside the box remain the same as illustrated in Figure 5.27; i.e., all voltages and currents in the “rest o f the circuit” are the same as before. A

■o N N.1

+

V J

Rest of Circuit

-O B FIG U RE 5.27 The networks denoted N^, i = 1,2, are equivalent when the v-i values at the terminals are identical; logically then, all voltages and currents inside the “Rest o f Circuit” remain the same. These examples allow us to justify the source transformation theorem as follows. Both 2-terminal networks in Figure 5.18 have the same v - i relationship: v = Ri + v = Ri + Rlj, = Ri +

and .

Therefore, the two networks o f Figure 5.18 are equivalent, and the source transformation is a valid analysis technique.

214


6. SUM M ARY This chapter covers the notions o f Unearity, superposition, proportionaHty, and source transfor mations. Linearity states that for any hnear resistive circuit, any output voltage or current, denot ed as y, is related to the independent sources by the formula j =

+ ... +

are the voltage and current values o f the independent sources, and ate constants. O nce values for the

where

through

through

are appropri

are known, one can compute the output for any (new) set o f

input values without having to resolve the circuit equations, a tremendous savings in time and effort. A special consequence o f linearity is the widely used principle o f superposition. Superposition means that in any linear resistive circuit containing more than one independent source, any output (voltage or current) can be calculated by adding together the contributions due to each independent source acting alone with the remaining independent source values set to zero. Practically speaking, this is the customary path to computing the coefficients, a^, in the linearity formula. Proportionality, another consequence o f linearity, means that if a single input is scaled by a constant, with the other inputs set to zero, then the output is scaled by the same constant. This property led to a clever technique for analyzing ladder networks without writing simultaneous equations. Since

power

is

proportional

to

the

square

o f a voltage

or

current, P = — =

R

for dc resistive circuits, the principle o f linearity and its consequences, superposition and propor tionality, D O N O T APPLY for power calculations. Using the notion o f an equivalent 2-terminal network, the chapter set forth the theorems on source transformations for source-resistor combinations: a 2-terminal network consisting o f a series connection o f an independent voltage source

and a nonzero finite resistance R is equiv

alent to a 2-terminal network consisting o f an independent current source,

= VÎR, in parallel

with R, as illustrated in Figure 5.18. These transformations, applied multiple times to a circuit, often simplify the analysis o f a circuit.


215

7. TERM S AND C O N C EPTS 2-term inal network: an interconnection o f circuit elements inside a box having only 2 accessible terminals for connection to other networks. T he concept is extendible to n-terminal net works. Equivalent 2-term inal networks: two 2-terminal networks having the same terminal voltage-cur rent relationship. I f two 2-terminal networks

and N 2 are equivalent, then one can be

substituted for the other without affecting the voltages and currents in any attached net work. and Uj, each acting alone, be y-^ and y 2 . U2 are applied simultaneously, the response is =

Linearity property: let the responses due to inputs W hen the scaled inputs

and

0 2

+ ®2^2- Linearity implies both superposition and proportionality, and vice versa. Linear resistive element: a 2-terminal circuit element whose terminal voltage and current rela tionships is described by Ohm’s law. Linear resistive circuit/network: a network consisting o f linear resistive elements, independent voltage and current sources, op amps, and controlled sources. Proportionality property: when an input to a linear resistive network is acting alone, multiply ing the input by a constant, K, implies that the response is multiplied by K. Source transformation: a 2-terminal network consisting o f an independent voltage source in series with a resistance is equivalent to another 2-terminal network consisting o f an inde pendent current source in parallel with a resistance o f the same value. Superposition property: when a number o f inputs are applied to a linear resistive network simul taneously, the response is the sum o f the responses due to each input acting alone.


216

4 . Consider the circuit shown in Figure P5.4.

Problems

(a)

Find the coefficients a , and (3j in the linear relationship

LIN EARITY

(b)

and |3j in the

linear relationship v^ 2 =

1. Consider the circuit o f Figure P5.1 in which = 5 £2 and = 20 Q. (a) Using linearity,

= a j V^j +

Find the coefficients

^l^sT

rnay be expressed Compute

ttj and (b)

If v^{t) = 10 cos(10?) V and I 2 = 2 A,

(c)

find v^Jt). Redo part (a), but this time express ttj and P 2 in terms o f the hterals

G] = — and G t = — .

Figure P5.4 5. Consider the circuit o f Figure P5.5. (a)

Find the linear relationship between

(b)

If

(c)

(Challenge) W hat is the effect o f dou

and the input sources V^j and l^j= 20 V and ^ = 0.5 A, find 1/

bling all resistance values on the coef ficients o f the linear relationship

’Q

'^(t)

found in part (a)?

Figure P5.1 A N SW ER: (b) 8 cos(lO^) - 8 V 2. For the circuit o f Figure P5-2, (a) (b)

find Vg in terms o f and Gy and

*^2’

find Ig in terms o f and Gy

V^2 > î> ^ 2’

CH ECK:

= 0 .2 5 ^ 1 +????/,2

6. For the circuit o f Figure P5.6, find the linear relationship between

and the independent

sources. Hint: Write a single loop equation.

Figure P5.2 3. For the circuit o f Figure P5.3,

Figure P5.6

(a)

find Ig in terms o f 7^,, /^2>^1 >-^2’ Ry and

(b)

find Vg in terms o f /^,, /^2> +

-

. ^2>

7. Consider the circuit shown in Figure P 5.7 in

^3-

which /?, = 80 Q. R2 = 20 Q, R^ = 80 Q. and r^ = 20. (a)

ôia the two independent sources. Hint: Write two loop equations.

-t" ’ (b) Figure P5.3


If = 20 V and 7^2 = 0.125 A, com pute the power delivered by the dependent source.


217

G,

— <+ i;

I .Q >R,

© '

Figure P5.7 C H EC K : (a)

Figure P5.9

????? I/, + 16/^2’ 39 m W >

10. A linear resistive circuit has two independent sources, as shown in Figure P5.10. If with v^2 ^t) = 10cos(2 ?) V, then

8. Consider the circuit o f Figure P5.8. in which

V. On the other hand, if

Ry = 18 Q. =9 and R^= 18 D.

with v^^{) = 0, then

(a)

^3 = 18

Ra = 36 Q,

it) and the four independent sources. out (b)

= 20 cos(2 i)

= 10cos(2^) mA = 2 cos(2z-) V. Find the

linear relationship between


and

= 0

and the inputs,

Now compute

when

= 20cos(2/) mA and v^2 (^) = 20 V.

(Challenge) If each o f the resistances is doubled, what is the new linear rela tionship. (Reason your way to the answer without having to resolve the circuit. Hints: Investigate the effect o f Linear resistive circuit with dependent sources

changing the resistance in Ohm’s law for fixed current. Investigate the effect o f equal changes in all resistances on a

V________________^

voltage divider formula.)

Figure P5.10 11. Again, consider the configuration o f Figure = 0 with v^2 ^t) = 10 V, then

P5.10. If

= 55 V. On the other hand, if z^j(r) = 4cos(2^) A with v^2 ^t) = 0, then

= -2cos(2?) V.

(a)

If z^j(?) = 2cos(2z-) A and v^2 ^t) = -

(b)

If /^j(z-) = -4cos(5^) A and v^2 ^t) =

10cos(2z-) V, find 20cos(5^) V, find C H EC K : (b) v^Jt) = 108cos(5^) V Figure P5.8 CH ECK:

27 - - Vi + ?? *4 - —

12. Consider again Figure P5.10. Suppose the + ?? V2

measured data are as follows: (i) when

9. For the circuit o f Figure P5.9, express a linear combination o f /^j, /^2> equation 5.1(a). Assume = 0.4 S, ^2 = G j = 0.05 S, and^^ = 0.1 S.

= 15 V

= 2 A and v^ 2 = 10 V, and (ii)

=10

V when = 3 A and v^ 2 = 5 V. (a) Determine the linear relationship (b)

Find

AN SW ER: 7.5 V

when z^j = 1 A and

=


218

13. Consider the linear network o f Figure

(a)

P 5.13, which contains, at most, resistors and linear dependent sources. Measurement data is given in Table P5.13.

Compute the coeiTicients o f a linear relationship among the output and three inputs.

(b)

(a)

Find the linear relationship

(b)

Find the power consumed by the 10 resistor when

= 20 V and

=

If 7,1 = - 1 A, 1/2 = 4 0 V, and I/3 = 10 V, find the power absorbed by R^. AN SW ER: 16 W

= 500 mA

Resistive Circuit witli Dependent Sources

+

V. . -

Figure P 5.I3 Figure P5.15

Table P5.13 I',, (volts)

z'^2 (amperes)

5

0.4

-1

16. Again consider Figure P5.15. Suppose the data measurements are given in Table P5.16.

10

1

2

Table P5.16

Km

14. Reconsider Figure P5.13 Two separate dc

/j, (mA) ^.2 (V)

measurements are taken. In the first experi ment, = 7 V and hi - 3 A, yielding =1 A. In the second experiment, = 9 V and z'^2 = 1 A, yielding = 3 A. (a)

Find the coefficients o f the linear rela

(b)

tionship + p 2i,2 Given the equation found in part (a),

compute when = 5 A. AN SW ER: (b) 90 watts

= 15 V and z^2

15 . The box in the circuit o f Figure P 5.15 con

Case 1

30

2

-1

11.5

Case 2

40

2

-1

13

Case 3

30

2.2

-1

11.6

Case 4

30

2

- 0 .9

11.9

Case 5

40

8

10

(a)

ments.

Case 1

50

-2

Case 2

0

Case 3

0

^^.3 (V) 1

(b)

Find the power consumed by Rj^ for

the data in Case 5. A N SW ER: 25 watts

Table P5.15 (mA) 1^.2 (V)

Find the coefficients in the linear rela tionship + a2^^2 + without any matrix inversions.

tains resistors and dependent sources. 7?^ = 100 Q. Table P5.15 contains various data measure

(V)

17. Again consider Figure P 5.15. Suppose the

5

-1 3

3

5

2

2

4

0

data measurements are given in Table P5.17.

219


Table P5.17

h\

Table P5.18

^ 2 (V)

^",3 (V)

Kut (V)

i,4 (mA)

-out (V)

Case 1

30

2

-1

11.5

1

6

Case 2

-2 0

4

2

27

2

10

Case 3

-1 0

-3

1

-1 4

5

?

Case 4

40

10

10

???

>

0

(a)

Find the coefficients in the linear rela tionship using a matrix inversion.

(b)

Find the power consumed by 7?^ for

the data in Case 4. A N SW ER: 102.01 watts

SUPERPO SITIO N AND PRO PO RTIO N ALITY 19. For the circuit o f Figure P5.19,

R2 = 50 Q, ‘'sX = 12 V, and (a)

=

1^2

using

Find

= 200 Q,

mA. superposition.

18. The linear resistive circuit o f Figure P5.18 has

Specifically, first find

due to

four independent sources. Three o f these sources

acting alone, and

to z,2 act-

have ftxed values. Only one,

is adjustable. In

a laboratory, the data set forth in rows 1 and 2 o f

out

ing alone. AN SW ER: 2.4 V, 2.4 V, 4.8 V

Table P5.18 were taken. Complete the last two rows o f Table P5.18 using linearity and the data

(b)

Find

and i^ 2

from the first two rows. For the data in row 3, find the power delivered by the current source z^. Hint: To solve this problem, recall from the lin

in terms o f the literals

compute the spe

cific numerical relationship. (c)

If

= 10

determine

earity equation 5.1

, R^-,

X

12 V and «j2 = ^ x 60 mA, using the proportional

ity theorem by first computing

ôut =

+ ^2^.2 +

\‘'s\

due to the modified and

'ih-i

acting alone,

due to the modified z'^2 act

ing alone. We have used the fact here that the term (a,v^ + ^2^j2 constant because the associat ed source values are constant. Thus, + ^2î2 ^^ some K. Hence, one can use the data from the first two rows o f Table P5.18 to solve for

and K.

Figure P5.19 20. Consider the circuit o f Figure P5.20 in which = 20 £2 , R2 = 60 Q, R^ = 20 Q. (a) Find the coefficients o f the linear rela

Linear resistive network with dependent sources

tionship + a 2 i, 2 + ^3^,3 by superposition. Specifically, first find —a

due to z/^j acting alone, due to z^2 acting alone, and ing alone.

Figure P5.18

g^f due to z^g act

220


(b) (c)

Repeat part (a), but express your answers

Using superposition, find acting alone, and then find

Find power delivered to

v^2 acting alone. What is

V, (d)

(a)

in terms o f the literals R-,i= 1, 2, 3.

= 2 A, and

and when

(b)

due to

^

Redo part (a) using the literals G/ = —

Ri

= 4 A.

Repeat part (c) when is tripled, and

and the = 100

due to

is doubled, i^2

is halved.

1 AN SW ER: (a)

5 2 7 = -''.v i- '' m » = - ‘',v2

AN SW ER: (c) 100 V, 500 watts

Figure P5.22 Figure P5.20

23. For the circuit o f Figure P5.23, suppose R^ =

21. In the circuit shown in Figure P5.21, R-^ = 180 Q, R^ = 360 a , T?3 = 90 Q, an R^ = 720 Q. (a)

Find the coefficients o f the linear rela tionship +

fi by superposition. Repeat part (a), but express your

0 -2 1

(b)

answers in terms o f the literals G- =

20 Cl, /?2 = 50 Q, (a)

= 100 Q and

= 0.02 S.

Using superposition, find

due to

acting alone, and then find due to v^ 2 acting alone. W hat is A N SW ER: = 0.5j',| + 0.9i^,, (b)

Redo part (a) using the literals G,- = —

l lR- ,i = 1 , 2 , 3 , 4. (c)

Find

and the power absorbed by

R^ when = 100 V and v^ 2 = 50 V. = 60 V; = 5 watts

AN SW ER: (d)

Repeat part (c) when

= 0.5 x 100 V

and v^ 2 = “ 10 x 2 V. Figure P5.21

Figure P5.23 24. For the circuit o f Figure P 5.24, find the contribution to

from each independent

source acting alone, and then compute '.,© .,.> U

- 0

power absorbed by the 900 Q resistor. AN SW ER: 38 V and 1.6 watts

22. For the circuit o f Figure P5.22, suppose R^ = 20 Q., /?2 = 50 Q and R^ = 100 Q.

by

the principle o f superposition. Finally, find the

221


i8on

225 n

0

20V

Figure P5.27

0.1 A

28. (a)

■^900 0

MATLAB program.

Figure P5.24 25. For the circuit shown in Figure P5.25, 160 V. Find

=

(b)

If it is known that

(c)

Find the equivalent resistance seen by

Then find the intermediate

node voltages. Hint: Assume

For the circuit shown in Figure P 5.28, If /j = 1 A, find /j by writing a = 200 mA, find /j.

the current source.

= 1V

and use proportionality, as per Example 5.11. C H E C K : Answer is an integer. +

Figure P5.28

LIN EARITY AND OP AM P CIRCU ITS 26. For the circuit shown in Figure P5.26, 64 mA. Find

Hint: Assume

=

29. Consider the circuit in Figure P5.29. (a)

= 1 A and

then use proportionality. 1n

1n

1n

1n III

©

2n

1n 2n

2n

2n

Find the contribution to

due only

to (b) Find the contribution to

due only

“ '^^2o>î (c) Find

by superposition.

r2n

Figure 5.26 2 7 .(a)

For the circuit shown in Figure P5.27, If Vj = 1 V, find V; by writing a MATLAB program to solve the prob lem, given that R-^ = 10 Q , = 10 = 5 Q, ^4 = 6 Q , = 10 Q ,

Figure P5.29 =

(b)

5 Q, = 20 Q, and = 5 Q.. If it is known that = 175 volts, find

(c)

^1Find the equivalent resistance seen by

(d)

the voltage source. Suppose is changed from 1 to 10 Q in steps o f 0.25 O.. Obtain a plot o f vs by modifying the MATLAB code o f (a). Assume Vj = 1 V.

30. Consider the circuit in Figure P5.30. (a) Find the contribution to due only (b)

to I/j. Find the contribution to

(c)

to Find

by superposition.

due only

222

Chapter 5 ®Linearity, Superposition, and Source Transformation

O


31. (a) For the circuit of Figure P 5.31, find

Vût^ the voltage due to each source acting alone in terms of the literal values, (b)

Find

(c)

Now suppose that = 2Rq, = 3i?o» = 4 i^ , Rr= URq, and = 100 Q,

33. (a)

in terms of the literals. (b)

If the input voltages are = - 2 .5 V and Vout.

Suppose each voltage source has value 2 V: (i) Find the power absorbed by the load to each source acting alone, and

For the circuit in Figure P 5.33, find the linear relationship between

(c)

= 5 V, V^2

= 2 V, determine

If the voltages are all halved, what is the new

(ii) the actual power delivered to the load when all sources are active.

r^

34. (a)

For the circuit in Figure

P 5.34, find the linear relationship

Figure P5.31

Kd-

between V^^^ând 32. Consider the circuit in Figure P5.32. (a)


due only

(b)

to Ki Find the contribution to

due only

to (c)

Find

(d)

If = ^3 = 0 . 5 ^ = 5 kQ and V;i = 2V;2 = 4 V, find the

(b)

If the input voltages are

V^2 = ” 0-5 V and Vour (c)

n

= 0.25 V,

= 2 V, determine

If the voltages are all halved, what is the new

by superposition.

power delivered to the 1

load.

n Figure P 5 .3 4

n


37. Consider the circuit in Figure P 5.37 in

35. Consider the circuit in Figure P5.35. (a)

Find the contribution to K^^^due only

(b)


(c)

to 1/2. Find

(d)

223

(a)

= Q.25R^= R2 = Ry Find the contribution to

due only

(b)

to Find the contribution to

due only

(c)

to Find

which

due only

by superposition.

by superposition.

Find the power delivered to Rj^ when is acting alone, i.e., ® then find the power delivered to R^

(e)

when V^2 acting alone when Rj = R and R^ = R. Find the total power delivered to R^ when R 2 = 4 R and R^ = R.

Figure P5.37 38. Consider the circuit in Figure P5.38. Find the contribution to Vout due only (a) to l/i(b) Figure P5.35 (c) 36. Consider the circuit in Figure P 5.36 in

(b)

= Q.25Rjr. “Find■ the ■ contribution to only to Find the contribution to

(c)

only to V^2 Find by superposition.

(d)

Find the power delivered to Rj^ when

which (a)

(d)

Find the contribution to Vout due only to K2Find the contribution to Vout due only to V^,3. Find by superposition.

due due

is acting alone, i.e., V^2 = and then find the power delivered to (e)

when V^ 2 is acting alone when R ^ ARy Find the total power delivered to

Figure P5.38

when Rjr=

SO U RCE TRAN SFO RM ATIO N S 39. Use a series o f source transformations to simphfy the circuit o f Figure P 5.39 into one consisting o f a single voltage source in series with a single resistance.


224

200 n

0 Figure P5.39 AN SW ER: 6 V source in series with 12 Q resis

40 0

80

40 n

20 n

son

Figure P5.41 AN SW ER: 13.5 and 9.1125 42. Use source transformations on the circuit o f

tor

Figure P5.42, to compute the value o f 40. Consider the circuit o f Figure P 5.40 in which /,i = 10 mA, 1^2 = 20 V, and I/3 = 80 V. (a)

need

ed to deliver a current o f I = 0.25 A.

Use a series o f source transformations to find a single voltage source in series with a resistance that is in series with the 9.6 kQ resistor.

(b)

Then find the power absorbed by the 9.6 kQ resistor.

Figure P5.42 A N SW ER: 28 V

Figure P5.40 AN SW ER: (a) 48 V in series with 3.2 kQ; (b) 135 mW

43. For Figure P5.43, use a series o f source transformations to find the value o f so that the power delivered to

is 16 watts.

40 A

41. In the circuit o f Figure 5.41, 1/j = 240 V and 1^2 = 0-25 A. (a) Use a series o f source transformations to reduce the circuit o f figure P5.41 to

©

sn 2A

< 50<

= 1n

a current source in parallel with a sin gle resistor in parallel with the 20 resistor across which

V ^ appears.

(b)

Find Find the power dissipated in the 20 Q resistor.

(c)

I f both sources have their values increased by a factor o f two, compute the new value o f Can you do this by inspection? Explain.

— is n Figure P5.43 44. Apply source transformations to the circuit shown in Figure P 5.44. Then write two nodal equations to find and V2 .


Figure P5.44 Source transformations simplify writing node equations. AN SW ER: 25 V, 20 V 45. Apply source transformations to the circuit o f Figure P5.45. Then write two nodal equa tions to find K| and Vj-

Figure P5.45 AN SW ER: 2.8 V, - 0 .4 V

225

C

H

A

P

T

E

R

Thevenin, Norton, and Maximum Power Transfer Theorems H ISTO RICAL NOTE In the early days o f electricity, engineers wanted to know how much voltage or current could be delivered to a load, such as a set o f street lamps, through a complex transmission network. Before the days o f computer-aided circuit simulation, simplification o f complex circuits allowed engi neers to analyze these very complex circuits manually. In 1883, a French telegraph engineer, M. L. Thevenin, first stated that a complex (passive) network could be replaced by an equivalent cir cuit consisting o f an independent voltage source in series with a resistor. Although stated only for passive networks, the idea o f a Thevenin equivalent evolved to include active networks. Its wide spread use has simplified the homework o f students for many years now and probably will con tinue to do so for many years to come. A more recent but quite similar idea is the Norton equivalent circuit consisting o f an independ ent current source in parallel with a resistance. At the time o f E. L. Norton (a scientist with Bell Laboratories), the invention o f vacuum tubes made independent current sources a realistic possi bility. Many electronic circuits were modeled with independent and dependent current sources. The appearance o f Norton’s equivalent circuit was a natural outcome o f advances in technology.

CHAPTER OBJECTIVES 1. 2. 3. 4.

Define and construct the Thevenin and Norton equivalent circuits for passive networks. Define and construct the Thevenin and Norton equivalent circuits for active networks containing dependent sources or op amps. Illustrate several different techniques for constructing the Thevenin and Norton equiva lent circuits. Investigate maximum power transfer to a load using Thevenin or Norton equivalents.

228

Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorem s

SECTIO N H EADIN GS 1. Introduction 2. Thevenin and N orton Equivalent Circuits 3. A General Approach to Finding Thevenin 4. Thevenin and N orton Equivalent Circuits 5. Thevenin and N orton Equivalent Circuits 6 . Thevenin and N orton Equivalent Circuits 7. M aximum Power Transfer Theorem 8 . Summary 9. Terms and Concepts 10. Problems

for Linear Passive Networks and N orton Equivalents for Active Networks for Op Amp Circuits from Measured Data

1. IN TRO D U CTIO N Practicing electrical engineers often want to know the power absorbed by one particular load. The load may be a large machine in a factory or a lighting network in the electrical engineering build ing. Simple resistances often represent such loads. Usually the load varies over time in which dif ferent resistances are used at different times to represent the load. W hat is the effect o f this load variation on the absorbed power and on the current drawn by the load? To simplify analysis, the rest o f the linear network (exclusive o f the load) is replaced by a simple equivalent circuit consist ing o f just one resistance and one independent source. For our purposes, a (resistive) load is a two-terminal network defined in Chapter 1, meaning that the current entering one o f the terminals equals the current leaving the other. More generally, a two-terminal network is any circuit for which there are only two terminals available for connec tion to other networks. (See Figure 6.1.) T he important question for our work in this chapter is:

How does one characterize a two-terminal networks As is shown in Figure 6.1(a), there is a voltage v{t) across the terminals and a current i(t) entering one terminal and leaving the other. T he rela tionship between the voltage v{t) and the current i{t) characterizes the two-terminal network. For example, if v{t) = Ri{t), we would recognize the terminal network as an equivalent resistance R. Or, if v{t) = Ri{t) + Vq, we might recognized this equation as that o f a resistance in series with a voltage source. In fact, this equation could be represented as graph, e.g. Figure 6.1(b). This leads to our next question: When are two 2-terminal networks equivalent'^ As developed in Chapter 5, two 2 -terminal networks are said to be equivalent when their terminal v-i characteris tics are the same. O f particular interest for this chapter is an equivalent network consisting o f a voltage source in series with a resistance, called the Thevenin equivalent network, and a current source in parallel with a resistance, called a Norton equivalent network. Figure 6.1c shows a Thevenin equivalent for a linear resistive circuit.

Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorems

N-

229

i(t)

2-terminal Linear

—o +

Resistive Network

V(t)

FIG U RE 6.1. (a) a 2-terminal linear network with terminal voltage v{t) and current i{t)\ (b) graphical representation of the equation v{t) = + Vq, (c) Thevenin equivalent network

having the same terminal v{t) and i{t) relationship as (b).

This chapter investigates the replacement o f a network N by its Thevenin equivalent or its Norton equivalent. The first section describes the Thevenin and Norton equivalent theorems for passive net works, those containing only independent sources and resistors. Following that, we generalize the statements to include active networks. However, because op amps have peculiar properties, Thevenin and Norton equivalents o f circuits with op amps are explored exclusively in Section 4. Following this, in Section 5, we describe how to obtain a Thevenin or Norton equivalent from measured data without having to know anything about the internal circuit structure. This is particularly useful when one has equipment such as a power supply but no schematic diagram o f the internal circuit ry. Unfortunately, not all linear devices have a well-defined Thevenin or Norton equivalent. The homework exercises illustrate a few cases. Section 6 explores the problem o f maximum power trans fer to a load in the context o f the Thevenin equivalent circuit, which ends the chapter.

2. TH EVEN IN AND NORTON EQ UIVALEN T CIRCU ITS FOR LINEAR PASSIVE N ETW ORKS Our first objective is to develop and illustrate the celebrated Thevenin theorem for passive net works. Then we will state and illustrate Norton’s theorem, dual to Thevenin’s theorem. To develop Thevenin’s theorem, consider Figure 6.2(a) consisting o f two 2-terminal networks, N and Nj^ , joined at A and B. Only resistors and independent sources make up N, while con


230

tains arbitrary even nonlinear elements. Suppose

undergoes various changes as part o f an

experiment, while N, complicated in its own right, remains unchanged. To simplify repeated cal culations, N is replaced by its Thevenin equivalent, as illustrated in Figure 6.2(b). The more sim ple Thevenin equivalent consists o f a single voltage source, in series with a single resistance.

Rthf

------ o-----+

Resistances and independent Sources

B

J

V

Arbitrary Networi<

r-\

J

V

(a)

i, = 0 -A

NResistances and independent Sources

-o -I-

-o

NReslstances with independent Sources

-o R..

Deactivated

B

(c)

(d)

FIG U RE 6.2 (a) Network TVattached to an arbitrary network load, N^, (b) N replaced by its so-called Thevenin equivalent, (c) circuit for computing

still attached to 7V^;

(d) circuit for computing

in which all independent sources inside N are deactivated. This brings us to a formal statement o f Thevenin s theorem for passive networks.


231

TH EVEN IN 'S TH EO REM FOR PASSIVE N ETW ORKS Given an arbitrary 2-terminal linear network, N, consisting o f resistances and independent sources, then, for almost all such N, there exists an equivalent 2-terminal network consisting o f a resistance,

in series with an independent voltage source,

Th e voltage,

called the open-circuit voltage, is what appears across the 2 terminals o f N. R^j^, called the Thevenin equivalent resistance, is the equivalent resistance o f N when all independent sources are deactivated. Figure 6.2(c) shows the appropriate polarity for v^J^t),

In the above theorem, “for almost all” means there are exceptions. For example, an independent current source does not have a Thevenin equivalent. More generally, any two-terminal network characterized by i{t) = constant does not have a Thevenin equivalent. This leads us to suggest that there ought to be an equivalent current source formulation o f an equivalent network. From Chapter 5, the source transformation theorem tells us that the Thevenin equivalent o f Figure 6.2(b) when

0 is equivalent to a current source in parallel with R^j^, as in Figure 6.3(b).

Figure 6.3 leads us to a formal statement o f the so-called Norton theorem. ■NResistances and Independent Sources

(c) FIG U RE 6.3 (a) Arbitrary 2-terminal linear network o f resistors and independent sources; (b) Norton equivalent circuit; (c) circuit for computing with computed, as per Figure 6.2(d).

N O RTO N 'S TH EO REM FOR PASSIVE NETW ORKS Given an arbitrary 2-terminal linear network, N, consisting o f resistances and independent sources, then for almost all such N, there exists an equivalent 2-terminal network consisting o f a resistance, R^^, in parallel with an independent current source, z'^^(r). Th e current, called the short circuit current, is what flows through a short circuit o f the 2 terminals o f N, as per Figure 6.3(c). R^^ as before, is the Thevenin equivalent resistance o f N computed when all independent sources are deactivated.

A single voltage source does not have a Norton equivalent, and— as mentioned— a single current source does not have a Thevenin equivalent. Both Thevenin and Norton equivalents exist for a 2terminal linear circuit when R^^

0 and is finite. W hen both the Thevenin and Norton equiva

lents exist for the same network, the source transformation theorem and Ohm’s law imply that thhc'kt)

(6.1a)

232

Chapter 6 • Thevenin, Norton, and M axim um Power Transfer Theorem s

4^ 0, then

and when

r,

th -

(6.1b)

■

This formula turns out to be useful in calculating

for a variety o f circuits, especially op amp

circuits. E X A M P L E 6 .1 . For the circuit o f Figure 6.4, using literals, find the open circuit voltage, short circuit current, 200 Q ,

and the Thevenin equivalent resistance,

the

Then, if R^ = 50 Q,

=

= 100 V, and i^ 2 = 2 A, construct the Thevenin and Norton equivalent circuits.

o

A

A

-O

-O

40 n

160V

40 0

4A

-o

-O

(c)

(b)

FIG U RE 6.4. (a) Resistive 2-terminal network; (b) Thevenin equivalent; (c) Norton equivalent. S

o l u t io n

Step 1. Find

Using superposition, we have by voltage division and Ohm’s law, /?2

^1^2

~ R\+R2

^1 + ^2

Substituting the given values into this formula yields = 0.8

X

100 + 4 0 x 2 = 160 V

Step 2. Find i^^. As per Figure 6.4, with terminals A and B shorted ^together, all the current from flows through the short circuit. From superposition, isc = 's2'^----- ■ Substituting numbers into this formula yields

i^^ = 2 + 0.02

X

100 = 4 A

233


Step 3. Find

Replacing

by a short circuit and z'^2 t>y an open circuit implies that ^

^ 40

q

R, +R2 Step 4 . Determine the Thevenin and Norton equivalent circuits. The Thevenin equivalent circuit fol lows from Steps 1 and 3 and is illustrated in Figure 6.4(b). The Norton equivalent circuit follows from Steps 2 and 3 and is illustrated in Figure 6.4(c). We also note that R ,, =

V =155 = he

40 Q

4

as expected. It is important to note that for many circuits, especially when the deactivated circuit is a seriesparallel connection o f resistances, one can obtain the Thevenin equivalent by a series o f source transformations.

Exercises. 1. Redo Example 6.1 using a series o f source transformations. 2. In Example 6.1, suppose = 100 Q, = 400 Q, v^^ = 100 V, and i^ 2 = 2 A. Find

v^^, and i^^.

AN SW ER; 80 Q, 240 V, 3 A

Among the three quantities, R^j^, v^^, and

if two have been calculated, then the remaining one

follows easily from Equation 6.1. In some cases, the choice o f which two to find first either increases or decreases the amount o f calculation. The following exercises illustrate this point.

Exercises. 1. For the circuit o f Figure 6.5, R^^= 200 Q, R2 = 50 O., R^= 10 Q, = 50 V. Find R^f^, i^^ and v^^ in this order.

= 100 V, and v^2

A N SW ERS: 8 Q, 1.5 A, 12 V 2. For the circuit o f Figure 6.5 with the same values as in Exercise 1, find v^^, i^^, and

this order.

AN SW ER: Same as in 1, but v^^^. is harder to find.

3. For the circuit o f Figure 6.5, find the Thevenin equivalent circuit using a series o f source trans formations. The next example illustrates the computation o f the Thevenin and Norton equivalent circuits using loop analysis.

Chapter 6 • Thevenin, Norton, and M axim um Power Transfer Theorem s

234

E X A M PLE 6 .2 . Find the Thevenin and Norton equivalent circuits seen at the terminals A-B for the circuit depicted in Figure 6.6, where = 100 V and = 3.2 A. We show that = 4 0 0 Q,

= 200 V, and

= 0.5 A. 1500

500 Q

A

FIGURE 6.6 Two-source circuit for Example 6.2 with loop currents shown;


= 100 V and i^ 2 = 3.2 A.

To compute R^^^, we set all source values to zero. Each voltage source becomes

a short, and each current source becomes an open. This leads to the circuit o f Figure 6.7. Here, we have a 500 Q in series with 100 Q, yielding 600 Q. Since this 600 Q resistance is in parallel with 400 £2, the resulting equivalent resistance is 240 Q. Hence, 500 0

= (150 + 240 + 10) = 400 Q..

1500

FIG U RE 6.7 The circuit of Figure 6.6 with all independent sources deactivated. Step 2. Compute an expression for A-B is N O T present. Hence,

Because we are computing

the short across the terminals

= 0 and no current flows through the 150 Q resistor. This means

its voltage drop is zero. (One ofi:en says that the 150 Q resistor is dangling.) Thus, from KVL we have

= 'ôozj + ioz;2

(6 .2)

Chapter 6 •Thevenin, Norton, and Maximum Power Transfer Theorems

23^

Step 3. Compute iy The only unknown in Equation 6.2 is /j, since

~ 3-2 A. Hence, around

loop 1,

^s\ = + 100/^2= 1000/jand

in which case,

/| = 0.0 0 ly^j + 0.1/^2 Thus, from Equation 6.2, = 400(0.001

+ 0.1/^2) + lO/^, = 0.4r/^, + 50/^2 =

+ 160 = 200 V

Step 4 . Construct the Thevenin and Norton equivalent ciraiits. Equation 6.3 with

(6.3)

= 400 Q yields

the Thevenin equivalent o f Figure 6.8(a). Further, from the source transformation theorem, (6.4)

Equation 6.4 leads to the Norton equivalent circuit o f Figure 6.8(b). 400 Q

A

FIGURE 6.8 (a) Thevenin equivalent o f circuit of Figure 6.6; (b) Norton equivalent o f Figure 6.6. Step 5. Compute i^^ directly so as to verify the above calailation. This step is merely given to illus trate the direct calculation o f i^^ and is unnecessary at this point to the solution o f the problem. Referring again to Figure 6.6 and assuming that the short acro.ss A-B is present, then /2 = i^^. Hence, around loop 1, ^S\

= 500/, + 400 (/j - / J

+ 100(/j - /^2)

in which case. v^\ + 100/^2 = 1000/j - 400/^^ = 420 V Around loop 2 we have 560/;^-400/, = 10/^2 = 32 V In matrix form, the pertinent equations are 1000

-4 0 0 '

-4 0 0

560

h ■ Jsc

■420' 32

2M^

Chapter 6 • Thevcnin, Norton, and Maximum Power Transfer Theorems

Thus,

' i\ ■

1000

-4 0 0 '

j.sc

-4 0 0

560

-1

■420‘

■0.62'

32

0.5

A

Consequently, / = 0.5 A as was found earlier using the easier method o f

=

Vo,. th

Exercises. 1. Suppose all source values in the circuit o f Figure 6.6 are doubled. What is the new v j Does

change?

A N S W E R : /•, =-)()() V. no

2. Suppose all resistances in the circuit o f Figure 6.6 are multiplied by 4 and the independent cur rent source is changed to 0.6 A. Find

and

Hint: For

in equation 6.3, the value “5 0 ”

is in ohms, so if the resistances are multiplied by four, what is the new value? AN SW ER: r

= 160 V. A', //» 4 x 400

1600 Q. and /SC . = 0.1 A

3. A 400 £L resistor is connected in series with terminal A o f the circuit o f Figure 6.6. Find the

V

“ ‘I V

ANSWLR:

im

V, A',,,

=»K) ti. ,nd

- «.2S A

4. A 400 Q resistor is connected across terminals A and B o f the circuit o f Figure 6.6. Find the V “" ‘I V ANSV('-UR:

O.SA, /(•, titJi - 2 0 0 U . a n d r oC- lOl) V

In the above two examples, deactivation o f all independent sources led to a series-parallel network. Calculation o f

wâs then straightfor\vard. In fact, we can state a corollary to Thevenin and

Nortons theorems.

CO RO LLA RY TO TH EVEN IN AND N ORTO N'S TH EO REM S FOR PASSIVE N ETW ORKS When a network contains no independent sources,

= 0 , and the Thevenin or

Norton equivalent consists o f a single resistance R^f^. For a series-parallel net%vork, R^f^ can be computed by straightforward resistance combinations.

3. A GEN ERAL APPROACH TO FINDING THEVEN IN AND NORTON EQUIVALENTS Consider Figure 6.10(a) where we have a network N connected to the remainder o f a larger cir cuit. Our goal is to replace the net\vork N by its Thevenin equivalent, as shown in Figure 6.10(b).

Chapter 6 •The\’enin, Norton, and Maximum Power Transfer Theorems

23'

The terminal v-i characteristics o f the network N and its Thevenin equivalent must be the same. Consider that the v-i characteristic at A-B o f the Thevenin equivalent o f N is (6.5) while the Norton equivalent o f N as per Figure 6.10(c) has the v-i relationship •

'a

- 1 ^th

(6 .6 )

- ^sc -

These relationships tell us that if we have a linear net\vork and assume there is a voltage

across

its terminals and a current /^j entering the network, as shown in Figure 6.10(a), then obtaining an equation o f the form (6.7) or o f the form

(6.8)

Vj n —

allows us to match the coefficients o f equations 6.7 and 6.5 to determine the coefficients o f equations 6.8 and 6.6 to determine ^th = — - and

and

or to march

This sometimes proves

^th an easier approach for non-simple circuits, as the next two examples illustrate. Remaining Network ^ A ------oLinear Network

A8

- o .......+ — B (a) Remaining

Remaining Network

(c) I'lG U R M 6.10 (a) Nervvork N attached to an unknown network; (b) theTheveinin equivalent of N attached to the unknown nervvork; (c) the Norton equivalent of N attached to the unknown network.

23.S

Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems

E X A M PLE 6 .3 . This example revisits Example 6.2 using the new approach. Again, we find the Thevenin and Norton equivalent circuits seen at the terminals A-B for the circuit depicted in Figure 6. I I , where = 100 V and = 3.2 A. Our goal is to find the v-i characteristic at the ter minals A-B.

FIGURE 6.11 Two-source circuit for Example 6.2 with loop currents shown; = 100 V and ip = 3.2 A.

So l u t io n Step 1, Consider i^ loop. Around the loop for i^, we have ÂB ~ 560/^ + 400/j + 10/^2 = 560/^ + 400/j + 32

(6.9)

Step 2. Consider loop 1. From Example 6.2, around loop 1 we have, = 100/^2 = 420 = 1000/, + 400;;^ V Thus, =

42 0 - 400/^

(6 . 10)

1000

Step 3. Substitute. Substituting equation 6.10 into 6.9 yields 4 2 0 - 400/4 + 32 = 400/.A + 200 V V. o = 560/.A + 4 0 0 ----------------^ AU jQQQ Step 4. Match coefficients. Matching coefficients o f equations 6.11 and 6.5 implies that

R,h = 4 0 0 Q,\-

= 200 V and /,, = ^

= 0.5 A.

th

(6.11)

Chapter 6 • Thcvenin, Norton, and Maximum Power Transfer Theorems

ly )

EXAM PLE 6.4. For the circuit o f Figure 6. ] 2, find theThevenin equivalent o f the 2-terminaJ Network N defined by the dashed line box. We show that = 9.6 V, = 4.4 Q., and = 2.1818 A.

FIGURE 6 . 12 A current source is is attached to N for computing

and

So l u t io n Our objective is to compute the relationship o f the form o f equation 6.7 using Nodal analysis and then match coefficients with equation 6.5 to obtain R^j^ and

Assume /^, = 2 A and

~^

Step 1. Write nodal equations. For writing the equations o f this circuit, the reader might first review Example 3.2. Alternately, using the inspection method, the matrix nodal equations are - 0 .2

■0.8

- 0 .4 ■ ■ ^’1 ■

- 0 .2

0.6

-0.1

- 0 .4

-0 .1

0.5

Step 2. Solve equation 6.12 fo r

h\

■ ' ’1 ■

=

\'2

\>2

( 6. 12)

0

js l + U .

^'ab

using Crammer's rule. First, we note that 0.8

det(A/) = det

- 0 .2

0.2

0.6

- 0 .4

-0 .1

-

- 0 .4 ' -

0.1

= 0.:

0.5

From Crammers rule, ■0.8

- 0 .2

det - 0 .2

0 .6

h\ 0

- 0 .4

-0 .1

‘s2 + U

^’ab -

V

d et(M )

which from the properties o f determinants becomes ■0.8

- 0 .2

O'

det - 0 .2

0.6

0

- 0 .4

-0 .1

1

d et(M )

=

4 . 4/4

■0.8

- 0 .2

r

■ 0.8

- 0 .2

O'

det - 0 .2

0.6

0

det - 0 .2

0.6

0

- 0 .4

-0 .1

0

- 0 .4

-0 .1

1

. /. -1------"

d et(M )

+ 2 . 6/^1 + 4 .4 /^ 2 = 4.4/,^ + 9 .6

• / . 4- -----

d et(M )

( 6 .1 3 )


240

Equation 6.13 shows that

is calculated finding four determinants numerically using MATLAB

or equivalent. Step 3. Match coefficients o f equations 6.13 and 6.5. Matching coefficients o f equation 6.13 with equation 6.5, we obtain V,

= 4.4 a and

2.1818 A

Exercises. 1. If the independent current sources in the circuit o f Figure 6.11 arc set to zero, find the Thevenin equivalent circuit. AN.SW'j-^R: The I hcvcnin ci|uiv.ilcnt «.onsist.s oi a single rc.sisn)r.

2. Find

= 4.4 12.

when i^^= 10 A and i^2 = 5 A.

A N S W I- R ; -48 \’

3. A 4.4 d resistor is connected in series with terminal A o f the circuit o f Figure 6.12. Find the V

V

•ANSW'l-.R:

. •).(, \’.

= 8.8 12. and

=1

A

4. A 4.4 Q resistor is connected across terminals A and B o f the circuit o f Figure 6.12. Find the ANS\V1;R:

- 2 .1SIS A.

= 2.2 f l and

M .S V

At this point, we end our development in this section with an example that shows how to com pute a Thevenin equivalent from measured, e.g., in a laboratory setting where there is a power sup ply with an adjustable voltage. EX A M PLE 6 .5 . Consider Figure 6.13, which show^s the Thevenin equivalent o f an unknow'n netw’ork N attached to a variable voltage,

power supply, which also shows the current delivered

to the unknown network N, i.e., Two measurements o f the unknown network N are taken, and the data is displayed in Table 6.1. Find the Thevenin equivalent o f N. N r R..

Variable Voltage Power Supply

oc

B MCURI-. 6.13 Thevenin equivalent of an unknown network N connected to a variable voltage power supply.


2/f

TABLE 6.1 /Î (mA) -

S

(V)

10

24

20

40

o l u t io n

Substituting the measured data in Table 6.1 into equation 6.5

yields

24 = 0.01«,;,+ V from row 1 o f Table 6.1, and 40 = 0 . 0 2 « ,^ * V from row 2 o f Table 6.1. In matrix form, 0.01

r

0.02

I

^th

•24‘ 40

Solving produces

^r/i 'V>(Hence, R,,, = 1600

■ 1 -1 ■■24‘ r -I ■24‘ = -100 -0.02 0.01 40 40 0.02 1 0.01

—

and

=

1600' 8

= 8 V.

Thus, one can use the technique o f Example 6.5 to determine Thevenin equivalent circuits in the laboratory.

4. TH EVEN IN AND NORTON EQ U IVALEN T CIRCU ITS FOR A CTIVE N ETW ORKS Constructing Thevenin and Norton equivalents for active networks, those containing dependent sources and op amps, presents us with some unique challenges. Except with one extra condition, Thevenin and Norton’s theorems and their corollary are valid for active networks. Because active networks contain dependent sources, the extra condition is that all controlling voltages or currents be within the 2-terminal network whose Thevenin/Norton equivalent are being sought.

Chapter 6 *Thevenin, Norton, and Maximum Power Transfer Theorems

242

TH EVEN IN AND N ORTO N'S TH EO REM S FOR A CTIVE N ETW O RKS For almost every 2-terminal linear network, N, as in Figure 6.14(a), consisting o f resistances, independent sources, and dependent sources whose controlling voltages and currents are con tained within N ‘, there is an equivalent 2-terminal network consisting o f either (i) a resist ance,

in series with an independent voltage source,

(Figure 6.14(b)), or (ii) a resistance,

called the Thevenin equivalent

in parallel with an independent current source, /y^(^),

called the N orton equivalent (Figure 6.14(c)). In most cases, both the Thevenin and Norton equivalent circuits exist. Computation o f y^îs characterized by Figure 6.14(a), computation of

by Figure 6.14(d), and computation o f b y Figure 6.14(e).

/^Neq

NResistances, independent and

+

R..

V

dependent sources

(a)

(b)

N-

(c)

NResistances,

Independent sources deactivated

independent and

(d)

(e)

-o—

dependent sources

M GURE 6.14 (a) Arbitrary linear nerwork N; (b) Thevenin equivalent of N; (c) Norton equivalent of N; (d) N with independent sources deactivated for calculating R^j', (e) N with short circuited terminals for calculating As in the previous section, a corollary to Thevenin and Nortons theorems is that if the nerwork N has no internal independent sources, then the Thevenin and Norton equivalent circuit consists o f a single resistance /?^yr. However, in contrast to passive networks, R^f^ can be negative. As a first example illustrating the above theorems, we consider an active nervvork containing no internal independent sources.


EXA M PLE 6.6 . Find the Thevenin equivalent circuit for the 2-terminal network (marked by dashed line box) in Figure 6.15(a) using the method o f Section 3. (The dependent source acts as a voltage amplifier.)

B O(b)

FIGURE 6.15. (a) circuit with terminal voltage i/^ând input current /^; (b)

/?„, = /?,//■

SO L u n o s Step 1. Since there are no independent internal sources, the Thevenin equivalent consists o f a single

resistance,

i.e., v^^ = i^^= 0.

Step 2. Write a nodal equation. Writing a single node equation we have

Step 3. Match coefficients with equation 6.6. Matching coefficients implies that

= ((7j + (p +

1)G^) in which case, ^2

R^R j A2

1

-

{ 1

J_ \

R2+(M + ^)R\

d 4.

R^

We recognize equation 6.14 as the parallel combination o f the resistance To illustrate a typical calculation, suppose p = 199, /?j = 100

(6.14)

'(A^ + 1)

and _.^2—

and Rj = 4 kf2.

Then

R.u = 500 / /20 = 19.23 a 20 =

iJ + \

Exercises. 1. For the above example, suppose p = 99, R^ = 500 Q, and /?2 = 1 A N SW ER: 10 Q

P*rid R^jy


244

2. For the circuit o f Figure 6.16, find the Thevenin equivalent resistance by obtaining of

in terms

V

'AB-

FIG U RF 6.16 A circuit having no independent sources, in which case

=

0 and

the Thevenin equivalent consists only of a single resistance,

îh -

('jth

E X A M PLE 6 .7 . Find the Thevenin and Norton equivalent circuits seen at the terminals A-B in Figure 6.17 when = 50 mA. Our computations will proceed using loop analysis to find the ter minal v-i characteristic A-B. 5001

MCiURF 6.17 Arbitrary network for finding Thevenin equivalent. S

o l u t io n

We first note that \.=

and

= 0.05 A.

Step 1. Write a set o f loop equations for the circuit o f Figure 6.17. For loop 1, vvc obtain 0 = 1000(/j + i^) + 1000(/, - i^) - m i ^ which simplifies to - 5 = 2000/, - 1500/;^ For loop A, we have

Chapter 6 * Thevcnin, Norton, and Maximum Power Transfer Theorems

Jn

Step 2. Write the loop equations in matrix form and solve. Writing the loop equations in matrix form yields ■ 2000

- 1 5 0 0 ' 'i\

-1 0 0 0

■-50'

■

1100

Solving for i^ using for example Crammers rule produces ■ 2000

del

IA= det

-5 0 ' det

-1000

[2 0 0 0 -1 0 0 0

-1 5 0 0

-= -5 0 -

■ 2000

1•

-1 0 0 0

0

det

AB

700x10-

■ 2000

()■

-1 0 0 0

1

700x10-

1100 50 2 ------- 1* AR-----700 700

(6.15)

or equivalently, (6.16)

‘'/IS = 350/;^ + 25

Step 3. Match coejficietits o f equation 6.15 with equation 6.6 or equation 6.16 with equation 6.5 to

obtain

50 1 1 = — = — A, G,i, = — = — S. R„, = 350 Q, and v,,,. = 25 V 700 14 700 350

Exercises. I. In Example 6.7, if /j= 5 n-L\, find I 140

A.

=

350

S.

R^f^, and

Hint: Use proportionalir)^

= 350 a a n d i„ ^ = 2.5 V

2. In Example 6.7, if /^= 5 mA and the 100 Q resistor is replaced by a short circuit, find and ANSWKR:

Hint: We have removed the dangling resistor in this case. = 2.S \'. K,i, = 250 Q.G,,, =

S, and

= - ^ = 0 .0 1 A

3. Find the Norton equivalent at the terminals A-B o f the circuit o f P'igure 6.18 when mA. AN SW ER: = 200 LI and = 0.1 25 A 8 0 0 I.

200 0 --- O

<— I < ^ 800 0

800 Q

----- ( B FIG U R H 6 .1 8 Modification o f the circuit o f Figure 6 .1 7 .


2-i6

5. TH EVEN IN AND NORTON EQ UIVALEN T CIRCU ITS FOR OP AM P CIRCU ITS Op am circuits arc active circuits. However, because the op amp is a device with special proper ties, such as the virtual short circuit in the ideal case and such as output saturation in the non ideal case, their discussion warrants special consideration. Our discussion begins with a Thevenin equivalent o f a non-inverting amplifier with a dangling resistor at the output terminal. EXA M PLE 6 ,8 . Find the Thevenin equivalent seen at the terminals A-B for the op amp circuit o f Figure 6.19.

S o lution Step 1. Find

2 0 -h 5 ' ’Cfi Step 2. Find v^g. By inspection.

Step 3. Match coefficients with equation 6.5. Matching coefficients we obsen'e that

R.i. = 16 Q. V’ r =

, and

= 16

Our next example illustrates how to construct a negative resistance using an ideal op amp.

Chapter 6 *Thevenin, Norton, and Maximum Power Transfer Theorems

EXA M PLE 6 .9 . Find the Thevenin equivalent seen at the terminals A-B for the (ideal) op amp circuit o f Figure 6.20. lO k O

S o lution

By V-division and the properties o f the op amp,

VcB -

5 ,

Thus, computing /^j we have .

^’AB - Vc

_

10x10^

lOxlO-"' ■■

Matching coefficients with equation 6.6 we have -4

S, R„, = - 2 .5 kQ, and VV;,. = 0

10x10^

Exercise. For the circuit o f Figure 6.21, find the Thevenin equivalent circuit at A-B.

AN SW ER:

= 0 and R^,, = -R

Chapter 6 • Thevenin, Norton, and Maximum Power Transfer 1 hcorems

:-i«

Our third example constructs a Thevenin equivalent of the standard inverting op amp configura tion with a terminal resistance. However, we will consider both the ideal and non-ideal cases. EXA M PLE 6 .1 0 . Find the Thevenin equivalent seen at the terminals A-B for the op amp circuit o f figure 6.22 when (a) w'hen the op amp is assumed ideal, and (b) when the op amp has a saturation voltage,

= 15 V.

S o lution

Step 1. Fiuci the Theveniti equivalent seen at the terminals A-B assuming an ideal op amp. I'h e prop erties of an ideal op amp imply that On the other hand,

with set to zero,

ed into node A. Hence,

x

= 0 for all possible currents, inject -5V^ implies = 0 and = -5K^. The

= 0. In flict,

Thevenin equivalent seen at the terminals A-B consists only o f a voltage source o f value

=“5

for the ideal op amp case. Step 2. Find the Thevenin equivaloit seen at the termitials A-B assuming an op amp ivith output sat

uration. When the non-ideal op amp operates in its linear region, the Thevenin equivalent by Step I is a voltage source having value v^^^ = - 5 V^. When, |-5 V} > = 15 V , or equivalently, when I > 3 V, then the op amp saturates at ± 15 V. Specifically, when K > 3 V, then v^^ = - \5V and when K, < - 3 V, then v^^ = 15 V. The Thevenin equivalent for an op amp with output saturation is summarized in Figure 6.23, where v^^. takes on three separate values depending on the region o f operation o f the amplifier.

{

1 5 V fo rV ^ < -3 V

- 5 V f o r |V j< 3 V 1 5 V fo rv / > 3 V

FIGURE 6.23 Thevenin equivalent at output terminals of an inverting amplifier (Figure 6.22) with non-ideal op amp.

Chapter 6 • Thcvcnin, Norton, and Maximum Power Transfer Theorems

249

This cnd-s our investigation of I hevenin equivalents o f op amp circuits. There arc many more interesting examples that are beyond the scope o f this text.

7. M AXIM UM POW ER TRANSFER THEOREM Figure 6.24 shows the Thevenin equivalent o f a network N connected to a variable load desig nated Rj. 1'he load voltage, Vj, the load current, ij, and the power,

delivered to the load arc

all functions o f Rj. The main objective o f this section is to show that for fixed R^j^, maximum power is transferred to the load when R^ = R^f^. We illustrate this assertion with an example that shows the power delivered to Rj as a function o f Rj. Throughout this section, it is assumed that all resistances are non-nesative.

FIG URE 6.24 Thcvcnin equivalent o f network N connectcd to a variable load, Rj. E XA M PLE 6 .1 1 . For the circuit o f Figure 6.24, suppose that R^j^ = 20 Q and

= 20 V, Plot the

power delivered to the load R^ as a function o f SO L U T IO N The power delivered to the load Rj is L

(6.17)

Plugging in the known values yields

I’L = —

^ (2 0 )'

To obtain the plot w'e use the following MA'FLAB code, resulting in the plot o f Figure 6.25. »voc = 20; Rth = 20; »RL = 0 :0.25:100; »PL = RL .* voc^2 ./ ((RL + Rth) .^2); >>plot(RL,PL) »grid

Chapter 6 •Thevenin, Norton, and Maximum Power Transfer 1 heorems

250

RL in Ohms

FIG URE 6.25 Plot of power delivered to the load in Figure 6.24 as a function of Rj.

= R^j^ = 20

From the curve, maximum power is transferred at

In a neighborhood o f R^ =

20 ^2, the curve remains fairly flat. At R^ = 40 Q. and Rj = \0 Q, the curve shows that about 88% o f maximum power is transferred.

This experimentally observed fact, that maximum power transfer occurs when R^ =

plays an

important role when matching speaker “resistances” to the output “resistance” o f a stereo ampli fier or when trying to get as much power as possible out o f an antenna and into a receiver.

M AXIM UM POW ER TRAN SFER THEO REM Let a two-terminal linear network, N, represented by its Thevenin equivalent, as in Figure 6.24, be connected to a variable load, R^. For fixed R^j^, maximum instantaneous power is transferred to the load when =«,/, and the maximum instantaneous power is given by

rL.max

. „

In the dc case, the instantaneous power is a constant for all t.

A verification o f the maximum power transfer theorem proceeds using differential calculus. From equation 6.17, the power absorbed by the load is

2S1


R Pl =

L____ VV? o c

[^L + ^th)‘ Following the standard procedure o f calculus for determining a maximum/minimum, we com pute the derivative o f w i t h respect to Rj, set to zero, and solve for R^.

dPL ^ d cJRl (IRl

Rr

2

-9 .

oc

Rl v I c

= y2 /

\3

from which R^ = R^^^ and R^ = ^ are the only possible solutions. But, if R^ = oc, then

= 0.

Hence, because equation 6.17 is positive for /?^ > 0, /?^ = R^j^ produces maximum power,/>^, deliv ered to the load. Further, substituting

= R^f^ into equation 6.1 7 yields

Rfll P L ,m a x

=

^n/(6.18)

This completes the verification o f the maximum pow'er transfer theorem. E X A M PLE 6 .1 2 . Consider the circuit o f Figure 6.26a. Find (i) the value o f 7?^ for maximum power transfer and (ii) the corresponding

■.......N ................................

600 O

<

V -/

300 0

Thevenin equivalent (b) FIGURE 6.26 (a) A network N connected to a load /?/-, (b) Thevenin equivalent of N connected to


2^2

SO L U T IO N

■urrent the independent voltage source becomes a short and the independent cur,

Step 1. To compute

= 200

source becomes an open. Finding the equivalent: resistance seen at the terminals produces Hence, maximum power is transferred when Step 2.

= 200

may be computed by any o f the methods discussed throi4ghoi4t this chapter. For example, by

repeated source transformations, the network N reduces to its Thevenin equivalent shown in Figure 6.26(b) with Plugging

16 V. In fact, this approach would have found R^j^ and 16 V and

at the same time.

= R^^ = 200 Q into equation 6.18 yields

PU

vi

(1 6 )“

4R,th

800

= 320 mW

Exercise. Suppose the 400 Q resistor in Figure 6.26(a) is changed to 100 Q. Find

R^j^, and pj

A N S W l'R S : 24 V. ISO 12, 0.96 wat t s

EXA M PLE 6 .1 3 . This example shows that the Thevenin equivalent cannot be used to calculate power consumption within the network N it represents. For this demonstration, consider the net work N given in Figure 6.27(a) with its Thevenin equivalent given in Figure 6.27(b). Compute the power loss within the actual network N and within its Thevenin equivalent. We show that these are different. •N. -O1n

1A

20 3V --Thevenin equivalent -(a)

(b)

FIGURE 6.27 (a) A network N; (b) Thevenin equivalent of (a). S O L U T IO N W ithin the network N, the power loss is ; watts PN, actual = 2 x 2 “ + 2 x 1 - = 10 w Within

the power loss is

Pm

- 1 X 1- -

1

Chapter 6 • Thevenin, Norton, and iMaximum Power Transfer Theorems

2^3

This means that theTIicvenin equivalent is not, in general, representative o f pov/er relationships within the network, i.e., the losses that are dissipated as heat, for example.

When a network N is a voltage source in series with a resistance R^, and hence is its own Thevenin equivalent, one may ask about maximum power transfer when fixed, assuming

is variable and the load Rf is

is also fixed. The following example is an experiment for investigating this sit

uation. E X A M P L E 6 .1 3 . For the circuit o f Figure 6.28, suppose /?^ = 20 Q and delivered to the load as a function o f R^ along with the power loss,

FIGURE 6.28 A network N in which R^ can be adjusted with

= 20 V. Plot the power in R^.

and /?^ fixed.

SO L U T IO N The power delivered to the load R^ is

R, (R

l

+ Rs )

0 •vr =

20 X 400 (20+ / ?,)

To obtain the plots, we use the following MATLAB code, resulting in the plot o f Figure 6.29. «vs = 20; RL = 20; ..Rs = 0:.25:50; >>PL = RL .* vsA2 ./ ((RL + Rs) .^2); »plot(Rs,PL) »grid »hold »Ploss = Rs .* vs'^2 ./ ((RL + Rs) .'^2); »plot(Rs,Ploss,’b’)

Chapter 6 • Thcvcnin, Norton, and Maximum Power 1 ransfer Theorems

25-1

FIGURE 6.29 Plot of power delivered to load as a function of

for circuit o f Figure 6.28.

According to Figure 6.29, the maximum power o f 20 warts is delivered when if R^< R^ (the usual case), then mi n i m i z i n g m a x i m i z e s Ploss

= 0. Observe that

However, if R^ > Rj, minimizing

niaximizep^.

The proof for the maximum power transfer theorem given earlier considers Rj as the independ ent variable and sets

dRi

to zero, standard practice in calculus. There is, however, an alternate

approach whose derivation is simpler mathematically, but is more meaningful for applications in the sense that the load can be a general 2-terminal linear network, N^, instead o f a single resistor. For this alternate derivation, refer to Figure 6.30. We ask the question. What v-i characteristic

should the load network

have so that maximum power is transferredfrom N to N J

FIGURE 6.30 The Thcvcnin equivalent of a network N connected to a loading ncrvvork N^.


To find the value o f v, we note that the power

= y i= y l^ - V R,th

p

To find the value o f v that maximizes

transferred from N lo Nj^ is

we differentiate

with respect to v and set the result to zero:

lIPL dv

0

ôc R,i,

Solving for v yields

V = 0.5v„

(6.19a)

at which value (6.19b)

ih

th

which are the conditions on i/and i for maximum power transfer to resistor

If

consists o f a single

it has the v-i characteristic o f equations 6.19. Then from O hm s law,

^ L - ~ - îh I At V= 0.5v , the corresponding maximum power is ( 6 .20 )

PLmax = V X /= -

4/?,th

EX A M PLE 6 .1 4 . In the circuit o f Figure 6 .31,

and R^f^ = 2 0.. Find the value o f that

maximizes power transfer to the network N^.

S o lution

According to equation 6.19(a), maximum power transfer occurs when v = equation 6.19(b), i =

= 0.5 A. Thus,

^th

V= \ = 0 .2 / +

= 0.\ + Vj

=o

Vf = 0 .9

V

= 1 V and from

2S 6

Chapter 6 • Thcvcniii, Norton, and Maximutn Power Transfer'I’hcorcnis

Exercises. 1 If the 0.2 £2 resistor is changed 4 Q, find the value o f

that maximizes power trans

fer to the network N^. ANSWHR; -1 V 2. If the 0.2 Q resistor is changed 2 Q, find the value o f Vj that maximizes power transfer to the network N^. AN SW ER: 0 V 3. IF the 0.2 Q. resistor is variable and Vj = 0.5 V, find the new' value of the 0.2

resistor that

maximizes power transfer to the Nj. AN SW ER: 1 Q

8. SUM M ARY This chapter has set forth a powerful strateg)' for analyzing complex networks by replacing portions o f the nerwork by their simpler Thevenin and Norton equivalents. The Thevenin and Norton theorems assure us that almost any 2-terminal linear nerwork, no matter the number o f internal elements, is equivalent to a simple nerwork consisting of an independent source either in series with or in parallel with a resistance. O f course, an independent current source does not have a Thevenin equivalent, and an independent voltage source does not have a Norton equivalent. \4ore generally, there are some circuits that have one but not the other. Further, some circuits have neither.

'

rhe chapter has illustrated various techniques for constructing the Thevenin and Norton equiv-

^

alents. For passive networks, the ordinary approach is to find

first by deactivating all internal

independent sources. If the resultant circuit is series-parallel, then

can be found by combin-

ing series and parallel resistances as learned in Chapter 2. If the resultant nerwork is not seriesparallel, then one should use the main technique set forth in this chapter, which is to find the vi characteristic o f the terminals. This technique is valid for all circuit t\'pes. With the ideas o f a Thevenin and Norton equivalent circuit, we then investigated the problem of transferring power to a load. When R^f^ is fixed, maximum power is transferred when R^ is adjust ed to be R^j^. If

is adjustable and R^ is fixed, then maximum power is transferred when R^j^ =

0. It is important to imderstand that a practical dc voltage source (such as a battery in an auto mobile) is designed to provide nearly constant output voltage for the intended load current. Accordingly, it has a rather small source resistance R^. Any attempt to transfer the maximum power from such a source continuously will overload the source and may cause damage to its internal structure. For example, in a lead acid battery, the plates may warp or the solution bub ble. Hence, maximum power transfer is not o f critical importance for power transmission net works, whereas for communication networks, maximum power transfer is important.

^

Chapter 6 •Thcvcnin. Norton, anil iMaximum Power Transfer Theorems

25'

9. TERM S AND C O N CEPTS 2-term inaI network: an interconnection o f circuit elements inside a box having only 2 accessible terminals for connection to other nervvorks. D eactivating an independent current source: replacing the source by an open circuit. D eactivating an independent voltage source: replacing the source by a short circuit. Equivalent n-term inal networks: t\vo n-terminal networks having the same terminal voltagecurrent relationships. Alternately, two n-terminal networks N j and N-, are equivalent when substituting one for the other in every possible network N; the voltages and cur rents in N are unaffected, the current through a short circuit placed across the output terminals of a 2-terminal network. M aximum Power Theorem : let an adjustable load resistor

be connected to the Thevenin

equivalent o f a 2-terminal linear network. Maximum power is absorbed by the resistor when Rj = Rî^. N orton’s equivalent circuit: any 2-terminal net\vork consisting o f independent sources and lin ear resistive elements is equivalent to an independent current source in parallel with a resistance. R(h (Thevenin’s equivalent resistance): the resistance that appears in the 7'hevenin equivalent circuit o f a 2-terminal linear network. It is also the equivalent resistance of the 2-terminal net\vork w'hen all internal independent sources are deactivated. Thevenin’s equivalent circuit: any 2-terminal network consisting o f independent sources and lin ear resistive elements is equivalent to an independent voltage source in series with a resist V

ance. : the open circuit voltage o f a 2-terminal network N when no load is connected.

' For a generalization o f this condition to the case where the controlling voltage or current is out side o f N, see the article by Peter Aronheim entitled “Frequenc)' Domain Methods” in The Circuits

and Filters Handbook, BocaRaton, FI.: C R C Press, 1995, pp. 682-691.


258


Prob ems

= 12 V,

=

= 60 Q, /?2 = 60 Q, and R^ = 40 Q.

0.4 A,

(a) Find the Thevenin and Norton equiva

TH EVEN IN /N O R TO N FOR PASSIVE CIRCU ITS

lents seen at the terminals A-B. (b) If a load resistor o f 90 Q is connected to AB, find the power absorbed by this resistor.

1. For the circuit o f Figure P6.1, find Rî^, and

in terms o f the literals. Hint: Consider

(c) Repeat (b) for a 30 Q resistor. Which resis tor, 30 Q. or 90

absorbs die most power?

using C/ = — . R.

Figure P 6.1 Figure P6.4

2. Find the Thevenin and Norton equivalent circuits seen at the terminals A-B for the circuit depicted in Figure P6.2.

5. In the circuit o f Figure P6.5, R^ = 2 kH, Rj

= 8 k n , Ri = 6 kQ,

= 60 V, i^ 2 - ^9 mA and

= 5 mA. Find the Thevenin and Norton equivalents o f the circuit in the dashed box. Then find ij and the power absorbed by R^.

5kQ -o-r

Figure P6.2

-O -v

3. For the circuit o f Figure P6.3, /?, = 3 kH, Rj = 6 kD,

= 30 V, and /p = 10 mA.

(a) Find the Norton and Thevenin equiva lents. (b) Suppose a variable load resistor attached

across

M ATLAB

A-B.

Plot

is using

or equivalent the power

Figure P6.5 C H E C K : 200 V, 10 kQ 6. For the circuit o f Figure P6.6,

R.

^«(l)

'■ 6

«

Figure P 6 .6

Figure P6.3 ANSWF.R: /?;/; =2 kil, isr = 20 mA, // . = 40 V

= 48

V. Find the Norton and Thevenin equivalents.

absorbed by R^ when \00 < R^< 4 kH.

- 6

= 18 kH,

/?2 = 9 kQ, ^3 = 3 kD, R^ = 6 kQ, and

CH ECK:

= 2 mA

Chapter 6 • Thcvcnin, Norton, and Maximum Power Transfer Theorems

2S9

7. Find the Thevcnin equivalent seen at A-B o f

(c) Find the value o f Rj for maximum

the circuit o f Figure P6.7, where R, = 18 kH, R-,

power transfer and the resultant power

= 9 kD, /?3 = 3 kD. /?4 = 6 kD.

= 3.6 kQ.

= 32 k n .

= 48 V, and s2 = 8 mA. Hint: Use the result o f Problem 6 to find the Thevenin equivalent of the network between C and D.

delivered to the load. (d) If the value o f

is doubled, what is the

power delivered to the load under the condition o f maximum power transfer?

R,

'■ 6

8R

6R

Figure P6.9 C H ECK S: (a) 900 Q, 30 V; (c) 250 mW, (d) 1 Figure P6.7 C H EC K : R.th = 10 kD

10. Find the Thevenin equivalent seen at A-B o f the circuit o f Figure P6.10. Hint: For this t)'pe o f problem, the more natural solution

8. (a) Find the Thevenin equivalents for the

technique is source transformations. Why?

circuit o f Figure P6.8 in terms o f the lit erals and i^2 (b) If the A-B is terminated in a 15 kl^ load, and = 30 V, and 1^2 = power delivered to the load. (c) What is the proper resistance across the terminals A-B for maximum power transfer and what is the resultant power

Figure P6.10

delivered to the load? ANSW'ER:

= 2R,

V()

= — +

V

11. Find the Thevenin equivalent o f the circuit o f Figure P 6 .ll enclosed in the dashed-line box. Then compute

and the power absorbed

by the 2 kQ resistor. Hint: W hat resistances are extraneous to the solution? Figure P6.8 A N SW FR: R^f, = 10 k ii,

= 204 V

9. Consider the circuit Figure P6.9, in which v^= 120 V and R = 300 Q. (a) Find the Thevenin equivalent circuit to the left o f the terminals A-B. (b) For Rj^ = 300 Q, 600 iX and 1200 Q., find die power absorbed by R^. Does the use o f a Thevenin equivalent reduce the effort needed to obtain these answers?

Figure P 6 .11 A N S W E R : 5 2 V. 2 4 k £ X 2 mA

Chapter 6 • Thev enin, Norton, and Maximum Power Transfer Theorems

260

12.(a) Find theThevenin equivalent circuit for the 2-terminal non-series-parallel net work shown in Figure P6.12. Use the

1 kn 2kO

general method. (b) If a load resistance

is connected to

terminals A-B, use MATLAB to calcu late and plot the power absorbed by the Figure P6.14

load for 1/ = 30 V, and 10 < /?^ < 200 ^2 in 5

steps. At what value o f

is max 15. (a)

imum power achieved?

= 3 kH for the circuit

Find a so that o f Figure P6.15.

(b) 100 o

100 Q

Repeat part (a) so that

R^^^ = -1 kQ.

Hint: Do problem 14 first, and then

100 Q

modify the Thevenin resistance appro

200 Q

priately. C H EC K : (b) a = 4000 Q i.

Figure P 6.12 ANSW FR: For (a),

------O

= 1(H) LI

1 kO

1 kO

A

2kn

TH EVEN IN /N O R TO N FOR A CTIVE CIRCU ITS 13.(a) Find the value o f

so that the

Figure P6.15

Thevenin equivalent resistance o f the 16.(a)

circuit shown in Figure P 6 .13 is 5 (b) Repeat part (a) for the case when

=

- 2 5 0 Q. C H EC K : (b) 6.25 mS

Find a so that R^^^ = 5 kQ for the cir cuit of Figure P6.16.

(b)

Repeat part (a) so that

200 n

1 kn

2k O

»800 n

A N SW FR:

= -1 kl^ .

Figure P 6.13 = 1000 uS

Figure P 6 .16 17. For the circuit shown in Figure P6.17, find the Norton equivalent circuit.

14.(a) Find a so that G,/, = — S for the circuit ^ ih o f Figure P6.14. (b) Repeat part (a) so that G,/, =

200 n

= -0 .0 0 1 A

7/l

18kQ

Figure P 6 .17 ANSWFR: - 6 0 0

/„ = 0

261


18, Use loop analysis to compute the Thevenin equivalent for the circuit shown in Figure

2 1 .(a) Find the Thevenin and Norton equiva lent circuits for the network shown in Figure P6.21, assuming that k = 0,025 S

P6.18, What is the Norton equivalent?

and

= 20 V,

(b) For what value o f k is the open circuit voltage zero. For this value o f k, deter

100 0

-----1----- -

loon

mine R^fj.

'300 0

' 0.01

0>

800 n

ANSW I-R;

Figure P 6,18 = 0, = 250 12

19.(a) Find the Norton and Thevenin equiva lents o f the circuit o f Figure P 6 ,19, (b) If a load resistor output

is attached across the

terminals,

plot

the

AN SW ER:

Figure P6.21 and

= 60

= 18 V

power

absorbed by the load for 1 <^Rj < 24 Q usingMATLAB or equivalent. For what

22. For the circuit shown in Figure P6.22, b = - 0 .0 2 S and a = 25 Find the Norton equiv

value o f

alent circuit.

does the load absorb maxi

mum power? Determine the power delivered to the load at maximum power transfer.

i.j 50 0

V.
'■© 50 0 bv

Figure P 6.19 A X 'S W I- .R :/'DC=6/<. / w' = U Mf, P,//

Figure P6.22 •W SW I'R; R^,, = 100 12. = 50/,

20. Find the Thevenin equivalent o f the circuit

23. Consider the circuit shown in Figure

in Figure P6.20 where

P6.23. (a) Find the Thevenin equivalent.

= 0.2 A.

(b) If a load resistor R^ is connected across 400 Q

—►

'• 0

1 kO

'400 0 200i

terminals A-B, determine R^^ for maxi mum power transfer and determine the maximum power delivered to R^. (c) If a resistor /?, were added in series with terminal A o f figure P6.23, what is the

Figure P6.20 AN SW FR: r^. = 60 \‘. A’,;, = SOO LI

Thevenin equivalent resistance o f the augmented network.

Chapter 6 •Thevenin, Norton, and Maximum Power Transfer Theorems

2(>:

ANSW ER: (a) R,i, = R^

R

'■ 6 Figure P6.23

27. Find the Thevenin and Norton equivalents of

2 4 .(a) Find the Thevenin equivalent for the network shown in Figure P6.24. (b) If the values of each source are cut in

the circuit o f Figure P6.27 when

= 20 mA.

AA

half, what is the new ANSWER: (a) 1.6 kLl - 260 V: (b) /• = - 130 V

400 n 2000i

io o v (^ ^

0

0.1 A

ANSWI-:R:

= 12.S V,

= 650 Q

Figure P6.24 25. Find the Norton equivalent for the circuit shown in Figure P6.25 when = 30 niA, = 0.04 S, /?j = 100 Q and

= 400 n .

28. Find the Thevenin equivalent seen at A-B for the circuit o f Figure P6.28 when /^= 10 mA, = 24 0 0 Q and /?2 = HOOO Q, R^ = 5600 Q, R^ = 1500 n , R^ = 1000 Q, and = 0.25 x 10-^ S.

C H EC K :

Figure P6.25 0.09 A

26. Consider the circuit o f Figure P6.26, where = 32 V, /?, = 80 Q, /?2 = 240 Q, and

(a) Replace the circuit to the left o f nodes A and B with its Thevenin equivalent. (h) Given your answer to (a), assume that Rf = 150 n , and find and the power con C H EC K : R^,^ = iQ Q.

Figure P6.28 = 8 mA

= 60 Q,

= 2.

sumed by Rj.

CH ECK:

OP AM P PROBLEM S 2 9 .(a) Find the Thevenin equivalent seen at the terminals A-B for the op amp circuit o f Figure P6.29.


(b) What is the value o f a load resistor /?^

263

C H EC K :

= 5000 Q

3 2 .(a) Find

the

attached across the terminals A-B for maximum power transfer. What is the power absorbed by this Rj}

Thevenin

and

Norton

Equivalent circuits o f the op amp con figuration o f Figure P6.32 seen at A-B. (b) Repeat (a) for the terminals C-B.

Figure P6.29 C H EC K : /?,/,=/?3 30. (a) Find the Thevenin and Norton equiva lents seen at the terminals A-B for the op amp circuit o f Figure P6.30.

Figure P6.32 AN SW ER: ( b ) = A*,, =0 3 3 .(a) Find

the

Thevenin

and

Norton

Equivalent circuits o f the op amp con figuration o f Figure P6.33 seen at A-B. (b) Determine the value o f a load resistor R^ connected across the terminals A-B for maximum power transfer. If

= 4 V

and ryp = 5 V, determine the maximum power transferred to this R^. C H E C K : (b) 1 ( /?! + /?2 rA

«i

\

20 kQ

= 0.9 watts 50 kO

‘I

3 1.(a) Find the Thevenin equivalent to the right o f the terminals A-B for the (ideal) op amp circuit o f Figure P6.31. (b) If the practical source indicated in the figure is attached to A-B, find the current

in terms o f 15 kn

34. Find the Thevenin equivalent seen at the terminals A-C for the op amp circuit o f Figure P 6.34 when the op amp has output saturation, >5 V.

Chapter 6 • Thcvcniii, Norton, and Maximiun Power Transfer Theorems

2 64

N Linear resistive network with dependent sources and fixed independent sources

Power Supply

Figure P6.36 Tabic P6.36 3 5 .(a) Find the Thevenin equivalent seen at the terminals A-C for the op amp circuit of

/^l (mA)

Figure P6.35 when the op amp has out put saturation,

1

6

4

12

= 12 V.

(b) Find the Thevenin equivalent seen to the left o f the terminals B-C and the maxi

C H E C K : ( b )P „ ,,,= 2 m W

mum power that will be absorbed by the 24 k n resistor for all variations in V..

37. Repeat Problem 36 with the data given in ■Rible P6.37. Table P6.37 r j (mA)


/W =

144 2800

10

54

40

66

C H E C K : (b)

.5625 W

= 6 mW

THEVEN IN AND NORTON EQUIVALENTS FROM M EASURED DATA 36. In a laboratory, the data set forth in rows I and 2 o f Table P6.36 were taken. (a) Compute the Thevenin and Norton equivalents o f N.

38. The data listed in Table P6.38 was taken for the network N o f Figure P6.38. (a) Fill in the values for the third column o f Table P6.38 and find the Thevenin and Norton equivalents o f the linear resistive nervvork N. (b) To what resistance should

be changed

to achieve maximum power transfer? What is P..

(b) After the power supply is removed, what resistance, Rj, should be connected

Table P6.38

across A-B for maximum power transfer?

(niA )

What i s />„„„? 2

4

10

10

j>

Chapter 6 •lhc\xnin, Norton, and Maximum Power Transfer Theorems

26S

41. This problem is the first of t%vo problems

N

that outline a laboratory measurement proce dure for finding the 'Fhevenin equivalent o f a

Linear resistive network with dependent sources and fixed independent sources

linear resistive 2-terminal nerwork. For this problem, consider Figure P6.41 in which the circuit under test contains no independent sources. I'he experimental apparatus includes a

C H EC K :

Figure P6.38 = 10.667 mW

resistance decade box, denoted R, a dc volt meter with internal resistance

and a signal

generator having known internal resistance, R_. 39. Repeat Problem 38 using the data in Tible P6.39.

To begin the procedure, one sets R = R^ = 0 and adjusts the dc level,

o f the signal generator

to obtain a reasonable meter reading, say Table P6.39

R, (£2)

=

£q, where the subscript “ 1” indicates our first (niA)

meter reading. (For an analog meter, the read ing should be almost full scale.) Leaving the sig

200

2

>

nal generator set at this value o f V^, increase R

1200

6

>

until the meter reading drops to Record this value o f R as R^.

^''o-

(a) Suppose R^^^ = x and R^ = 0. Show that CH ECK:

31.25 m\V

«,/, = «2(b) Now suppose R^^^ = x> and R^ ^ 0. Show

40. The data listed in Table P6.40 was taken for the network N o f Figure P6.40 with a volt meter (VM ) whose internal resistance is 10 M n . Fill in the values for the third column o f Table 6.3 6 and find theThevenin equivalent o f the linear resistive network N.

that/^,/^=y?2(c) F'inally, suppose R^^ând Râre nonzero and finite. Show that and then solve ^ th for R.th-

= R j - R^ ^m

Table P6.40 (pA)

R, (M Q) 2

0.4

>

10

1

>

N Linear resistive network with dependent sources and fixed independent sources

42. 'Fhis problem is the second of rwo problems that outline a laboratory measurement procedure for finding the The\'enin equivalent of a linear resistive 2-terminal net\vork. For this problem, consider the new configuration o f Figure P6.42 in which the circuit under test contains independent

Figure P6.40 =4 V

sources and has a non-zero

'Fhe experimental

apparatus includes a resistance decade box, denot ed R, and a dc voltmeter with internal resistance

Chapter 6 • Thcvenin, Norton, and Maximum Power Transfer Theorems

266

All devices are connected in parallel. Because ?= 0, a sign;il generator is not needed, as in Problem 41. l b begin the procedure, open circuit

AN SW ER: 4 WLl and 20 V 45. The linear circuit shown in Figure P6.45 is

= co,

found experimentally to have the voltage and

and set the scale on the voltmeter to obtain a rea

current relationship shown. Find its Thevenin Norton equivalent.

the decade box, or equivalently set /? = sonable meter reading, say

= £q, where the

subscript “ 1” indicates our first meter reading. (For an analog meter, the reading should be almost full scale.) Next, reconnect the decade and decrease R until the meter reading drops to V^p = 0.5

Record this value o f R as /?2. (a) Suppose R^^^ = co. Show that R^f^ = R-,. (b) Suppose R^^^ is nonzero and finite. Show R„,R that = /? 2 înd then solve for

R ,l,+ R ,

R^!^. Then show that

1+

Rm/

'0 ANSWT.R: 0.5 Q, 4 A 46. Repeat Problem 45 for the measurement curve shown in Figure P6,46, Then determine the value of a load resistor for maximum power transfer and compute .

Figure P6.42 43. The Thevenin equivalent o f a linear resistive network containing no independent sources is to be found experimentally using the method o f Problem 41. fhe voltmeter has an internal resist ance R^^^= 20 k n . The dc signal generator has an internal resistance, R^ = 2 kI2. The following measurements are taken: (i) with R = 0,

is

adjusted until the voltmeter reads 4 V; (ii) keeping

fixed, the decade box is adjusted until the

voltmeter reads 2 V. For this voltage, the decade box shows R = 6 kf2. Find R^/^. AN SW FR: 5 kLl

47. The i-v curve o f the network N in Figure P6.47a is measured in a laboratory, and is approximated by the straight-line segments shown in Figure P6.47b. The meter readings are shown in Table P6.47. Table P6.47

44. The Thevenin equivalent of a linear resistive network containing independent sources is to be found experimentally by the procedure o f Problem 42. The voltmeter has an input resistance

R^j = 1 M ti. The following me;isurements are taken: (i) when R is opcn-circuited, the voltmeter reads 4 V, and (ii) when R is decreased to 800 kl^, the voltmeter reads 2 V. Find R^jând

A

0.2 V

0.1 mA

B

0.7 V

10.1 mA

(a) Find the Thevenin equivalent for the range 0 < i < 0.1 mA. (b) Find the Thevenin equivalent for the range 0.1 < / < 10.1 mA.

Chapter 6 ♦ rhcvcnin, Norton, and Maximum Power Transfer Theorems

(c) If R = 500 Q, Vît) = 50 sintdOOO t) mV, 100 mV, find i{t). Hint: Use a

and

dated voltage

267

and the power delivered to

the load.

suitable Thevenin equivalent for N. (d) If R = 50 and

v^{t) = 200 sintdOOO t) mV,

= 500 mV, find /(/).

R scale)

'■ 6 Figure P6.50 (a) Figure P6.47 ANSWl-.R: (a)

= 0.

= 2 kl2; (h)

=

51 (a) For the circuits o f Figure P6.51, find the load resistance R^ needed for maximum

0.195V, R,/, = 5 o 'h ; (c) 0.04 + 0.02 sin (1000

power transfer, the associated voltage and the power delivered to the load.

/) mA: (d) .^.05 + 2 sindOOO t) mA

(b) If the load resistance is constrained as 5 k n < R^ < 10 kD, repeat part (a).

M AXIM UM POW ER TRANSFER

(c) If the load resistance is constrained as 15

48. For the circuit o f Figure P6.48, /?, = 160 Q.,

R-, = 480

and

k n < R^ < 20 k n , repeat part (a).

= 80 V. Find the value o f

for maximum power transfer and

e 2 mA

12 kO

8kn

6

24 V 6kfi

Figure P6.48 C H EC K : 7.5 watts Figure P 6.51 49. For the circuit o f Figure P6.49, R^ = 900 Q,

R j= 180

/?3 = 50 Q,

21 V. Find the value o f

= 60 mA, and v^ 2 = for maximum power

52. Consider the circuit o f Figure P6.52. (a) Find the value o f

for maximum

power transfer to the three-resistor load.

transfer and

(b) Find the power delivered to each load resistor, i.e., to R^, R J2 , and /?^/3.

40 V

Figure P6.49

lon 50. For the circuits o f Figure P6.50,

= 10 V

and v^ 2 - ^5 V. Find the load resistance R^^ needed for maximum power transfer, the asso-

Figure P 6 .5 2

Chapter 6 • I'hevcnin, Norton, and Maximum Power transfer Theorems

268

53. For the circuits o f Figure P6.53, /?j = 200 = 1000 Q, /?3 = 400 n ,

>^ v.,(V)

= 8 mS, and /;, =

0.4 A. Find the load resistance

80 -

i

needed for

<—

maximum power transfer, the associated voltage,

--- 0 + a

and the power delivered to the load.

— o b

40 -

/

0

(a)

> 1 0.2

1 1 0.4

i(A) ^ r

(b)

Figure P6.56 57. (a) Figure P6.53

For the circuit o f Figure P6.57 com pute, (i) the value o f R which leads to maximum power transfer to the load,

54. T he circuits o f Figure P6.54 have the load

(ii) the voltage across the load, and

resistor

(iii) the power absorbed by the load.

connected in different ways. For

each circuit, (i) compute the value o f

which

Hint: In MATLAB, the roots o f a

leads to maximum power transfer, (ii) the volt

quadratic, aQ x- + a, x + a2> are given

age across the load, and (iii) the power absorbed by the load. Which configuration absorbs more power?

by “roots([aO al a2])”. (b)

To verify the results o f (a) write a MATLAB program to calculate and plot the power absorbed by the load as

son

+

R varies from 0 Q to 400

15Q

30 V

15

in 2 Q

increments.

^ (b

Load

(a) R. 30 Q

(b

+ V

15Q

I

.s v Q

30 V

Figure P6.57 (b)

C H EC K : (a)

Figure P6.54 10 watts

55. Suppose the polarity o f the 15-V-source in Problem 54 is reversed. Repeat Problem 54 and determine which configuration transfers more power to the load.

58. The i-v relationship o f certain type o f LED (light emitting diode) in its operating range o f 1-7 V -3 V is represented by a 2 V voltage-source in series with a 50 H resistance. The load con sists o f a network o f n such diodes connected in parallel. The source network is represented by a 5 V voltage-source in series with a 50 ^2 resist ance. Assume that the power delivered to each diode is totally converted into light. Determine

56. T he linear resistive circuit o f Figure P6.56(a) is found experimentally to have the voltage-current relationship plotted in Figure P6.56(b). Find the maximum power that can be absorbed by placing a load resistor across ter minals a-b?

how' many LEDs should be connected in paral lel for maximum brightness. W hat is power dis sipated by each diode? ANSW ER: M - 5,

= 2S mW

C H A P Inductors and Caoacitors

CAPACITIVE SM O O TH IN G IN POW ER SUPPLIES Every non-portable personal computer contains a power supply that converts the sinusoidal volt age o f the ordinary household outlet to a regulated dc voltage. “Regulated” means that the output voltage stays within very tight limits o f its nominal value (e.g., 12 ± 0.1 V) over a wide range o f power requirements. Engineers design power supply circuits with regulators that produce voltages with a small oscillation because to generate a truly dc voltage is practically impossible.

REGULATION

RECTIFICATION

O O SMOOTHING

This process o f converting ac to dc has three stages: First, the ac waveform is rectified into its absolute value. Then a smoothing operation takes place that reduces the variation in the voltage to a reasonable but still unacceptable level. This first level o f smoothing is nccessar)’ becausc the voltage regulator is a precision subcircuit that requires a fairly constant voltage for its proper oper

270

Chapter 7 • Inductors and Capacitors

ation. The partially smoothed waveform is fed into a voltage regulator, which limits the voltage oscillation between critical levels even when the load drawn by any connected device (e.g., your computer) varies in the course o f its operation. As mentioned, the rectified sine wave is smoothed before entering the voltage regulator. A crude smoothing can be accomplished with a capacitor, a device studied in this chapter. Intuitively, capacitors resist voltage changes and are designed to steady the voltage at a constant level. In this chapter, we will study the capacitor and investigate a simplified smoothing operation for a power supply.

CH APTER O BJECTIVES 1.

Define the notion o f inductance and introduce the inductor, whose terminal voltage is proportional to the time derivative o f the current through it.

2. 3.

Investigate the ability o f an inductor to store energy and the computation o f the equiva lent inductance o f series-parallel connections. Define the notion o f capacitance and introduce the capacitor, whose current is propor tional to the time derivative o f its terminal voltage.

4.

Investigate the ability o f a capacitor to store energy and the computation o f the equiva lent capacitance o f series-parallel connections.

5.

Define and illustrate the principle o f conservation o f charge.

CHAPTER O U TLIN E 1. 2.

Introduction The Inductor

3.

The Capacitor

4. 5. 6.

Series and Parallel Inductors and Capacitors Smoothing Property o f a Capacitor in a Power Supply Summary

7. 8.


1. IN TRO D U CTIO N This chapter introduces two new circuit elements, the linear inductor and the linear capacitor, hereafter referred to as an inductor and a capacitor. The inductor, shown in Figure 7.3, is a device whose voltage is proportional to the time rate o f change o f its current with a constant o f propor tionality I , called the inductance o f the device, i.e.

as set forth in equation 7.1. The unit o f the inductance Z., is the henry, denoted by H. Macroscopically, inductance measures the magnitude o f the voltage induced by a change in the current through the inductor.


The capacitor, shown in Figure 7.15, is a device whose current is proportional to the time rate o f change o f its voltage, i.e.,

ic(0= C

dvcO) (it

as set forth in equation 7.5. Here, the constant o f proportionality, C, is the capacitance o f the device with unit farad, denoted by F. Capacitance measures the devices ability to produce a cur rent from changes in the voltage across it. By adding the inductor and the capacitor to the previously studied devices (the resistor, inde pendent and dependent sources, etc.), one discovers an entire panorama o f possible circuit responses, to be explored in the next four chapters. Together, these devices allow one to design radios, transmitters, televisions, stereos, tape decks, and other electronic equipment. In this chap ter, our goal is to understand the basic operation o f inductors and capacitors.

2. TH E IN DU CTO R

SomePhysics In Figure 7.1, a changing current flowing from point A to point B through an ideal conductor induces a voltage

between points A and B according to Faradays law. Joseph Henr}' inde

pendently observed the same phenomenon at about 1831. The induced voltage, to be proportional to the rate o f change o f current, i.e.,

was found

=— .

dt

FIG URE 7.1 A time-varying current flowing through an ideal conductor. The following experiment illustrates the idea. Suppose the conductor in Figure 7.1 is 6 feet o f #22 copper with resistance 16.5 Q/1,000 ft. The 6-foot length has a resistance o f about 0.1 U. Using a current generator, we apply a pair of ramp currents (shown in Figure 7.2a) to the conductor, as per Figure 7.2b. The measured responses are shown in Figure 7.2c and, as expected, satisfy Ohms law.

Chapter 7 * Inductors and Capacitors

(a) (b)

(c)

(d)

(e) FIGURK 7.2 (a) Ramp currcnt inputs to iincoilcd and coilcd wire, (b) Six feet of #22 wire attached to a current generator, (c) Voltage responses to ramp current inputs of uncoiled wire. (d) Six feet o f #22 wire coiled into 45 turns 1” long and 1” in diameter. (e) Voltage responses to ramp current inputs o f coiled wire.


273

Now suppose the wire is coiled into a qrlinder 1” in diameter and 1” long, as in Figure 7.2d. Apply the same ramp currents o f Figure 7.2a to the coiled wire. This time, the measured responses are as shown in Figure 7.2e. These responses have the same shape as those of Figure 7.2c, except for the offsets o f 30 mV and 60 mV, respectively. These offiet voltages are proportional to the derivatives of the input currents, i.e.,

Offset -

for k = 1 , 2 , where L is the proportionality constant, called the inductance o f the coil. Since the derivative o f i s lO'^ A/sec, and the derivative o f the coil can be computed as

^

3x10

Offset

0.03

= —j:— = — ^ 10^

ini^t)

is 2 x lO'^ A/sec, the inductance

0.06

L

of

, . hennes

2x10^*

dt As mentioned earlier, the heniy, equal to 1 volt-sec/amp and abbreviated H, is the unit of induc tance. Also, from the above experiment, one concludes that the inductance o f a cylindrical coil o f wire is much greater than the inductance of a straight piece of wire, which in the above experi ment was not measurable by our apparatus. The physics of the preceding interaction is governed by Maxwell’s equations, which describe the interaction between electric and magnetic fields. A time-varying current flow through a wire creates a time-varying magnetic field around the wire. The magnetic field in turn sets up a time-varying electric field, i.e., an electric potential or voltage. One can verify the presence o f this magnetic field by bringing a compass close to a wire carrying a current. The magnetic field surrounding the wire will cause the compass needle to deflect. Physically speaking, a changing current causes a change in the storage o f energy in the magnetic field surrounding the conductor. The energy trans ferred to the magnetic field requires work and, hence, power. Because power is the product of volt age and current, it follows that there is an induced voltage between the ends of the conductor. W hat is even more interesting is that if a second wire is immersed in the changing magnetic field of the first wire, a voltage will be induced between the ends of the second wire. A proper (mathematical) explanation of this phenomenon is left to a fields course. For our purposes, three fects are important: (1) energy storage occurs, (2) the induced voltage is proportional to the derivative o f the current, and (3) the constant of proportionality is called the inductance of the coil and is denoted by L. As mentioned, a straight wire has a very small inductance, whereas a cylindrical coil o f the same length o f wire has a much greater inductance. This inductance can be increased many times over, possibly several thousand times, simply by putting an iron bar in the center o f a cylindrical coil. Alas, the calculation o f inductance is the proper subject of more advanced texts, e.g., on field the ory or transmission line theory. Nevertheless, there are empirical formulas for estimating the inductance of a single-layer air-core coil as described in the homework exercises.


274

BASIC DEFIN ITIO N AND EXAM PLES D EFIN ITIO N OF TH E LINEAR IN D U CTO R The linear inductor, symbolized by a coiled wire as shown in Figure 7.3, is a two-terminal energy storage device whose voltage is proportional to the derivative o f the current passing through it. The constant o f proportionality, denoted by Z,, has the unit o f H enry (H), equal to 1 volt-sec/amp. L is said to be the inductance o f the coil. Th e specific voltage-current rela tionship o f the linear inductor is given by (7.1)

dt i,(t)

h/Y Y V + V jt) FIG U RE 7.3 The inductor and its differential voltage-current relationship as per the passive sign convention.

EXA M PLE 7.1 Compute Vi{t) for the inductor circuit o f Figure 7.4 when ij{t) = e'‘~.

0.5H

FIG URE 7.4 A 0.5 H inductor driven by a current source.

S o lution

From equation 7.1, direct differentiation o f the inductor current /^(/) leads to y^(t) = 0 . 5 - " '

dt

^ = 0 .5 (-2 f)e-''

V

Exercises. 1. In Example 7.1, suppose ijit) = 0.5sin(20r + 7t/3) A. Compute v^{t). AN SW FR: 5 cos(20t + tt/3) V. 2. In Example 7.1, suppose ii{t) = (1 AN SW ER: ///(/) = lOOe--”'*'V.

V for / > 0 and 0 otherwise. Find

t> 0 .

The differential equation 7.1 has a dual integral relationship. Safely supposing that at / = inductor had not yet been manufactured, one can take

= 0, in which case

the


275

(7.2)

L-fh The time

represents an initial time tiiat is o f interest or significance, e.g., the rime when a switch

is thrown or a source excitation is activated. The quantity

specifies the initial current flowing through the inductor at ^q. This quantity,

sums up the

entire past history o f the voltage excitation across the inductor. Because o f this, the inductor is said to have memory. EXA M PLE 7 .2 For the circuit o f Figure 7.5a, determine /^(O) and ij{t) for / > 0 when Vj{t) =

V as plotted in

Figure 7.5b. \(t)

ijt)

,(t,

L = 0.5H

(b)

(a)

Inductor Current (A)

FICJURE 7.5 (a) Simple inductor driven by a voltage source, (b) Source waveform Vj{t). (c) Resulting inductor current


1~ G

S olution

A direct application o f equation 7.2 leads to I

T,

= z j l ’.

1 i / o " " ' ' ’’ = z + i ’

It follows that

il^(0) = — = 2 A and //^(0 = — 2 - e ' -

The graph o f ij\t) for all t is given in Figure 7.5c.

Exercises. 1. In Example 7.2, compute an expression for i^{t) for / < 0. A N S V V K R : i j U ) - 2<-'' A tor t < 0.

2. Repeat Example 7.2 with L = — H and with v,{t) = cos(27ir) V for r > -0 .2 5 sec and zero I . 4jt otherwise. ANSWl'.R: //(()) = 2 A, /y(/) = 2 + .sinUni) A for r > 0.

E X A M PLE 7 .3 Consider the circuit o f Figure 7.6a with voltage excitation v^{t) shown in Figure 7.6b. Find the inductor current /^(r) for f > 0, assuming that /^(O) = 0.

i,(t)

,(t )

Q

L = 0.5H

(a) FIGURE 7.6 (a) Voltage source driving inductor, (h) Square wave excitation !»£(/). S o lution

It is necessary to apply equation 7.2 to each interval, [0, 1], [1, 2], ... , [;;, n + 1], .... For this we need to first specify the initial conditions for each interval. Step 1. Compute i^\). From equation 7.2,

Chapter 7 • Inductors atul Capacitors

Step 2. Cotnpute ii{2). // (2 ) = /^(0) + — f \'i^{T)dT = — X Net Area = 0

L

0

Step 3. Compute the iuitial condition for the interval [n,u + 1] for u even. Again from equation 7.2, with t = n and n even, we liave 1

Hence i^{n) = 0 for ail even values o f n. Step 4. Compute the initial condition for the interval [n,n + 1] for n odd. From equation 7.2, with

t = n and n odd, we have, utilizing steps 1 and 3, I

+“

/I-1

f

j

+j

/j

f ri-1

j

«

y i(r)d T = — J

v7 (tV /t = 2

n-l

since n - 1 is even. Step 5. Compute ij^t) over [n,n + 1] ivith n even. If n is even, then the value o f the inductor cur rent over the interval [;/,;/ + 1] is

ilSt) = iL{n) + — ^ cIt = i i{ n) + l { t - n) = 2{t - n) A

l^Jn

Observe that iît) = 2 t - In A is the equation of a straight line having slope +2 and^-intercept -In . Step 6. Compute i^(t) over [n,n + 1] with n odd. If n is odd, then for the inter\'al [n,n + 1], the inductor current is /^(/) = //(/2) + — ( dT = i i i n ) - 2{t - n ) = 2 - 2{t - n) A

[^J It

Here, i^{t) = 2 + 2n - 2t is the equation o f a straight line, with slope - 2 and )'-intercept 2+2«. Step 7. Piece segfnentsfivm steps 5 and 6 together. Thus the segments computed in steps 5 and 6 inter cept the /-axis at the same points. Figure 7.7 sketches the resulting triangular response for /> 0.

FIGURE 7.7 Triangular shape o f inductor current for the square wave voltage excitation of Figure 7.6b applied to the circuit of Figure 7.6a.


27H

Exercises. (All time is in seconds.) 1. Again consider the circuit o f Figure 7.6a. Compute iyr (?) for (i) 0 < r < 1, (ii) 1 < t < 3 , and (iii) 3
( < 4, and (iv) 4 < /, for the waveform o f Figure 7.8b, assuming i^(0) = 0.

F IG U R U 7.8 Voltage excitations for Exercises 1 and 2.

It is important to recognize that the square wave voltage input o f Figure 7.6b is discontinuous but the current waveform o f Figure 7.7 is continuous. Integration (computation o f “area”) is a smooth ing operation: it smoothes simple discontinuities. This means that the inductor current is a con tinuous function o f t, even for discontinuous inductor voltages, provided that the voltages are bounded. A voltage or current is bounded if the absolute value o f the excitation remains smaller than some fixed finite constant for all time. Thus, equation 7.2 leads to the continuity property o f the inductor: if the voltage Vf{t) across an inductor is bounded over the time interval /] < t <

tj, then the current through the inductor is continuous for

< t < tj. In particular, if

then /^(ô”) = ^ The notation and “+” on /q is used to dis tinguish the moments immediately before and after /q- For example, in Figure 7.9, t = 2 shows a discontinuity o f

The value o f *^^(2“) is 1 and the value o f

is - 1 . The value Vj{2*) can

be seen as the limiting value o f z^^(r) when approaching r -» 2 from the right, whereas Vf{2~) can be seen as the limiting value o f v^{t) when approaching t

2 from the left.


279 \ {2 )

/

c

(U Lu T3 C

rtJ 01 CT>

TO *-> o

> u ■O c

Time (seconds) F IG U R E 7.9 A possible discontinuous voltage v^{t) appearing across an inductor of 1 H, and the resulting continuous inductor current.

PotverandEnergy Rccall that the instantaneous power absorbed by a devicc is the product o f the voltage across and the current through the device assuming the passive sign convention. For an inductor,

dt

Plit)=\'L{t)ilU) =

where

w atts.

is in volts, i^{t) in amps, and L in henries.

Since energy (absorbed or delivered) is the integral o f the instantaneous power over a given time interval, it follows that the net energy

stored' over the inter\'al [/q, /■,] in the magnetic

field around the inductor is f'o V

(h

/ (7.3)

=

/£(/,)-//^(/q ) joules.

for L in henries and in amps. From equation 7.3, whenever the current waveform is bounded, the net energy stored in the inductor over the interval [/q, rj] depends only on the value o f the inductor current at times r, and /q, i.e., on //(^j) and //(/q)’ respectively. This means that the stored energy is independent o f the particular current waveform between and If the current waveform is periodic, i.e., if ij{t) =

+ T) for some constant 7'> 0, then over any

time interval o f length T, the net stored energ)' in the inductor is zero because

= /^(/-q + 7)

Cliapter 7 • Iiuluctors and Capacitors

280

forces equation 7.3 ro zero. To further illustrate this propert)', consider Figure 7.10a, which shows a 0.1 H inductor driven by a periodic current /^(^) = sin(27tr) V. This current signal has a funda mental period T = I, i.e., the smallest 7 'over which the signal repeats itselh From equation 7.3, VV^(OJ) =

pi iOdt = I L/7(l)

L/7(0) = 0

However, we can interpret this result in terms o f the waveform o f pi{t). First note that the volt age across the inductor in Figure 7.10a is Vjit) = 0.27Tcos(27if) V. Hence, the instantaneous power is pf{t) =

= 0.2jtcos(2Tt^)sin(27tr) watts, as plotted in Figure 7.10b. Observe the shaded

regions o f Figure 7.10b in which the area under the power curve has equal parts ot positive and negative area. This means that all the energ)' stored by the inductor over the part o f the cycle o f positive power is delivered back to the circuit over the portion of the cycle when the power is neg ative. Fhis is true for all periodic signals over any period. Because no energ)' is dissipated, and because energy is only stored and returned to the circuit, the (ideal) inductor is said to be a loss less device.

IlW

v jt)

sln(27T) A

0.1 H

= 0.2n cos(2n) V

(a) P lW

(b) FIGUllE 7.10 (a) Inductor excited by periodic current, (b) Plot of the power absorbed by the inductor. It is convenient to define the instantaneous stored energy in an inductor as

=

(7 .4 )


281

for all t. Equation 7.4 can be viewed as a special case o f equation 7.3 in which r,) = -oo and /^(-oo) = 0. Thus, equation 7.4 can be interpreted as the change in stored energ)' in the inductor over the inten'al (^x>, t]. E XA M PLE 7 .4 Find the instantaneous energy stored in each inductor o f the circuit o f Figure 7.11 a for the source waveform given in Figure 7.1 lb. In Figure 7.1 lb , note that ij^t) = 0 for r < 0.

FIGURE 7.11 (a) Series inductors excitcd by a source current, (b) Graph o f the source current.

S olution

From KCL, i^{t) =

for all t. Since i^{t) = 2r A for 0 < r < 1 and i^{t) = 2 A for /> 1, equa

tion 7.3 or 7.4 immediately yields the instantaneous stored energies (in J) as plotted in Figure 7.12:

r

0^/
4r

0:sr
1s

> t

> t (a)

(b)

FIGURE 7.12 (a) Encrg)' (in J) stored in inductor Z.,. (b) Energy (in J) stored in inductor Ly


282

Exercises. 1. For the circuit o f Figure 7.1 la, find analytic expressions for the instantaneous stored energ)' for the current excitation in Figure 7.13a for r > 0. 2. Repeat Exercise 1 for Figure 7.13b.

FIG U RF 7.13 Current excitation for Exercise 2. ANSWUPvS: 1.

0 .2 5 r

UV/) =

{) ^ / < 2

t-

()s/
0 .2 5 (9 - 6/+ /■ )

i s / < 3 and U '2(0 = 3s I

0

r

0 s /< 2

4

2s/ 4/“

0 s /< 1

(9-fv + r )

I s / <3

0

3
EXA M PLE 7.5 For the circuit o f Figure 7.14 in which v^{t) = cos(t) V for r > 0 and 0 otherwise, find the input current i^{t) for r > 0 and the energy stored in each o f the inductors for the intervals [0, t] for 0 < t < 1 and [0, t] for 1 < t. i.(t)

L^ = 1H

FIGURE 7.14 Parallel inductive circuit with switch in which v^{t) = cos(t) V for /> 0 and 0 otherwise.


283

S olution

Step 1. Since no voltage is applied to either inductor for / < 0, /,(0) = 0. Further, no voltage appears across the second inductor until r > 1. Hence, /^(l) = 0. Step 2. Equation 7.2 implies that, for 0 < r < 1, ( s ( 0 = /i(/) =

L|

v ,(T )r/ T =

-jL|

/ ^ c o s (t)Jt =

sin(/) A

Step 3. At ^ = 1, the switch closes. T he r\vo inductors are then in parallel, and the source voltage appears across each. Hence, by equation 7.2, co s (t

W t = sin( I ) + sin(/) - sin( I ) = sin(/) A

Also, equation 7.2 applied to L-, implies

i2{t)= /2(l)+ Jj^

cos(T)r/T = sin(/) - s i n ( l ) A

From the KCL, the input current ij^t) = /,(/) + ijit) = 2sin(/) - sin(I) A for / > 1. Step 4. Compute the energy stored in the inductors over the interval [0, t]. From equation 7.3, it fol lows that for 0 < r < 1,

t) - 0.5 sin^(r) joules, whereas

->((), t) = 0.

Step 5. Compute the energy stored in the inductors over the interval [0, f] for 1 < t. Again from equa tion 7.3, for 1 < t, sin^(l)] joules.

t) = 0.5 sin^(^) joules and

t) = 0.5[sin~(r) - 2 sin (l) sin(/) +

Exercise. Repeat the calculations o f Example 7.5 for = 2 sin(r) V for / > 0 and 0 otherwise. A N SW ERS: For 0 < / < 1. U'} ,(0, f) = [2 - 2 cos(r)]“ J, whereas W) ,(0, t) = 0; for 1 < U'^^,(0. /) = [2 - 2 cos(/)]- J and U'} ,(0, ;) = [ 1.0806 - 2 cos(/)]- J.


28-»

3. THE CAPACITO R

DefinitiojisandProperties D EFIN ITIO N OF T H E C A PA CITO R Like the inductor, the capacitor, denoted by Figure 7.15a, is an energy storage device. Physically, one can think o f a capacitor as two metal plates separated by some insulating mate rial (called a dielectric) such as air, as illustrated in Figure 7.15b. Placing a voltage across the plates o f the capacitor will cause positive charge to accumulate on the top plate and an equal amount o f negative charge on the bottom plate. This generates an electric field between the plates that stores energy. Hence, for a capacitor. (7.5)

clt

d\

where q{() is the accumulated charge on the top plate, which is proportional to the voltage

V({t) across the plates; thus q{i) = Cv^t), with proportionality constant C denoting capaci tance and having the unit o f Farad (F). One Farad equals 1 amp-sec/volt. The capacitance C is a measure o f the capacitor’s potential to store energ)' in an electric field.

ic(t) >r

^ V ,(t)

(a)

+ -h

,

-i- + 4-

4- + + -I--1- +

A- + + + + + +.

(b)

FIG URE 7.15 (a) The symbol for the capacitor with conventional voltage and current direc tions. (b) Illustration o f electric field between plates of a parallel-plate capacitor.

Modern-day capacitors take on all sorts o f shapes and sizes and materials. In keeping with craditio n , the parallel-platc concept remains the customar)' perspective. Calculating the capacitance o f

t^vo arbitrarily shaped conducting surfaces separated by a dielectric is, in general, ver)- difficult Fortunately, the ordinary capacitor o f a practical circuit is o f the parallel-plate variety, with the plates separated by a thin dielectric. The two plates are often rolled into a tubula, ruu„, .,„>1 complete structure is sealed. EXA M PLE 7.6 For the capacitor circuit o f Figure 7 . 16a, compute i^t) when v j t ) = r5""'sin(1000r) V for / > 0.

28S


ic(t) v Jt ) 2mF

(a)

Time in milli-seconds (b)

FIG URE 7.16 (a) A 2 mF capacitor connectcd to a voltage source, (b) Plots of capacitor voltage and current waveforms. S o lution

A direct application o f equation 7.5 yields

i^{t) =

sin(1000/) +

dt

Exercises. 1. In Figure 7.16, suppose =e (preferably in MATLAB) i^^t) for 0 < r < 0.5 sec.

cos(l()00/) A

V for r > 0. Compute

for r > 0. Sketch

ANS\V1-:R: - 0 . 0 5 , A. 2. Repeat Exercise 1 with = e~-^^ cos( 100/) V for f > 0 but plot over die time interval [0, 0.15 sec]. A.\’S\V1-:K: -.- -^ q 0 .0 5 COS. KJOr) r 0.2 sin(lOOr)] A.

The differential relationship o f equation 7.5 has the equivalent integral form ' t < '> =

^ f ° J c W d T + I j ; " ic W d T

=

(7.6) C •'^0

where

is in volts, /^r) is in amps, and C is in farads, and where we have taken

=0

because the capacitor was not manufactured at t = -oo. The time /q represents an initial time o f interest or significance, e.g., the time when the capacitor is first used in a circuit. The quantit)'

286


specifies the initial voltage across the capacitor at ^q. This initial voltage,

sums up the entire

past history o f the current excitation into the capacitor. Because o f this, the capacitor, like the inductor, is said to have memory. EXA M PLE 7 .7 Suppose a current source with sawtooth waveform

shown in Figure 7.17h, drives a relaxed

0.5 F capacitor (zero initial voltage) as in the circuit o f Figure 7.17a. Compute and plot the volt age across the capacitor. i.(t) A/v^(t)V

(a)

(b)

FIG U RE 7.17 (a) Current source driving a capacitor. (b) Sawtooth current waveform and voltage response of a 0.5 F capacitor. S o l u t io n

The input waveform is periodic in that it repeats itself every 2 sec. Therefore, the solution will pro ceed on a segment-by-segment basis. Step 1. Consider the interval 0 < f < 2. For this interval ij^t) = {It - 2) A. With

= 0, it fol

lows from equation 7.6 that v^(/) =

(2 t - 2)ilT = 2 ( r - 2/) V for 0 < t <2

Step 2. Consider the interval 2 < t
= 0; hence, the capacitor volt

age over the interval 2 < /“< 4 is simply a right-shifted version o f the voltage over the first inter val. Right-shifting is achieved by replacing t with t - 2 . In other words,

v^{t) = 2[{t - 2)2 - 2{t - 2)] V for 2 < r < 4 Step 3. Consider the general interval 2k < t < 2{k + 1). For interval 2k < t < 2{k + 1),

v^t) = 2{{t - 2k)^ - 2{t - 2k)], /^= 0 , 1, 2 ,

...

Lastly, obser\ê that the voltage across the capacitor, as illustrated in Figure 7.17b, is continuous despite the discontinuity o f the capacitor current. Again, this follows because the capacitor volt age is the integral (a smoothing operation) o f the capacitor current supplied by the source.


287

Exercise. Consider the capacitor circuit o f Figure 7.18. Suppose the current source is i^{t) = e~‘ A for /■> 0 and

= 1 V. Compute the capacitor voltage

the resistor voltage

and the

voltage vj^t) across the current source for r > 0. -H V„(t) -

+ I (t) = e-'u(t) 0

20

M O

;

.

0.5 F

v,(t)

FIGURE 7.18 Scries RC circuit driven by a currcnt source for accompanying exercise. AN SW ERS: v^^t) = 3 -

for t> 0,

= lc~' for t > 0. and, by K\'l., v^{t) = 3 V for t > 0.

It is important to emphasize that the sawtooth current input depicted in Figure 7.17b is a dis continuous function, but the associated voltage waveform is continuous because integration (equation 7.6) is a smoothing operation. This means that the capacitor voltage is a continuous function o f t even for discontinuous capacitor currents, provided they are bounded. This obser vation leads to the continuity property o f the capacitor: if the current i(^t) through a capacitor is bounded over the time interv'al

< ^ < ^2> then the voltage across the capacitor is continuous

for fj < r < tj. In particular, for bounded currents, if fj <

< tj, then

= V(- (tQ"^), even when

At the macroscopic level, there appear to be some exceptions to the continuit}' propert}' o f the capacitor voltage, e.g., when two charged capacitors or one charged and one uncharged capacitor are instantaneously connected in parallel. In such cases, KVL takes precedence and will force an “instantaneous” equality in the capacitor voltages, subject to the principle o f conservation o f charge, to be discussed shortly. Another example is w'hen capacitors and some independent volt age sources form a loop. When any o f the voltage sources has an instantaneous jump, so will the other capacitor voltages. Upon closer examination, however, we see that there is really no excep tion to the stated continuity rule: it can be shown that in all o f the cases where the capacitor volt age jumps instantaneously, an “impulse” current flows in the circuit. Physically, an impulse cur rent is one that is ver)' large (infinite from an ideal viewpoint) and o f very short duration. The cur rent is not bounded, and consequently, the capacitor voltage may jump instantaneously. This jump does not violate the rule, which presumes that the currcnt is bounded.

Relatio7ishipofChargetoCapacitorVoltageandCurrent We have defined the capacitance o f a two-terminal device strictly from its terminal voltage-current relationship— the differential equation 7.5 and the integral equation 7.6, which is now repeated: v 'c ( 0 = V c(fo) + ^ f ^ C •'M)

Physically speaking, the integral o f i(^t) over [/q, t\ represents the amount o f charge passing through the top wire in Figure 7.19 over [rQ, r].

288


ic(t)

/ ''+ + +q + + / /+ + + ^ / ++++++

A- + + + + + +y

-q FIGURE 7.19 Capacitor cxcitcd by a currciu. Bccausc o f the insulating' material (the dielectric), this charge cannot pass through to the other plate. Instead, a charge o f +q{t) is stored on the top plate, as shown in Figure 7.19. By KCL, if

i(it) flows into the top plate, then of

must flow into the bottom plate. This causes a charge

to be deposited on the bottom plate. The positive and negative charges on these two

plates, separated by the dielectric, produce a voltage drop

from the top plate to the bottom

plate. For a linear capacitor, the only t)'pe studied in this text, the value o f V(^t) is proportional to the charge

The proportionalit)' constant is the capacitance o f the device. Specifically,

qit) = C\U)

(7.7)

where q{t) is in coulombs, Cis in farads, and tî^) is in volts. Thus, equation 7.6 has the following phys ical interpretation: the first term,

is the capacitor voltage at /q; the integral in the second term,

ic(T )d r. represents the additional charge transferred to the capacitor during the interval [r,j, /]. Dividing this integral by Cgives the additional voltage attained by the capacitor during [^q, ^]. Therefore, the sum o f these rwo terms, i.e., equation 7.6, is the voltage o f the capacitor at r. Since q(/) = it follows direcdy that =

(7.8)

(It

cl!

ThePj'hicipleofConservationofCharge It is important in terms of modern trends in circuit applications to further investigate the rela tionship o f charge to capacitor voltages and currents. The principle o f conservation o f charge requires that the total charge tramferred into a junction {or out o f a junctiori) be zero.~ This is a direct consequence of KCL. To exemplify, consider the junction o f four capacitors shown in Figure 7.20.


289

v,(t)

+ V ,(t) -

i,(t)

i3(t)

- V3(t) + v,(t) + l4(t)

FIGURE 7.20 Junction of four capacitors. By KCL /,(/) + ijit) + i^{t) + i^{t) = 0 Since the integral o f current with respect to time is charge, the integral o f this equation over ( - 00, t] is

/ ^ (m

'2

+'3

M^

=^/1 ) +^/2(0 +qj, (/) +f/4(/) =0

(7.9)

where qj^{t) is the charge transferred to capacitor k. By equation 7.6, at ever}' instant of time,

qi{t) = C-v.{t)

(7.10)

which defines the relationship between transported charge, capacitance, and the voltage across the capacitor. Hence, from equations 7.9 and 7.10, at every instant of time,

C^vît) + C2 V2 U) +

+ C^v^{t) = 0

This simple equation relates voltages, capacitances, and charge transport. The following example provides an application o f these ideas. EXA M PLE 7 .8 This example shows that under idealized conditions, capacitor voltages can change instanta neously. Consider the circuit o f Figure 7.21 , in which £^q(0~) = 1 V and and V(^{t) for f > 0.

= 0 V. Find


290

t=0

C l = 1F C2

Cl

C2 = 1F FIGURE 7.21 Two parallel capacitors connectcd by a switch. S o lution

At t = 0“ , the charge stored on C, is C, ^ ^ (0 “ ) and that o f C j is requires that

For t > 0, KVL

Therefore, after the switch is closed at r = 0, some charge must be

transferred between the capacitors to equalize the voltages. According to the principle o f conser vation o f charge, the total charge before and after the transfer is the same. Thus, conservation o f charge requires that

F,quivalently, <7i(0'^) - <7i(0 ) + qîO*) - ^2(0 ) = 0- From equation 7.10, C,

=

Since ^ ^ (0 ) = 1 V, V(^{0 ) = 0, and from K \T C,

= V(y{0*), it follows that

v - c , ( 0 ^ ) - l l + C 2 [ i ’c i ( 0 ^ ) - 0

Hence, (Cj + C ,)y Q (0 ‘^) = I implies that

0

=

0

= Vq^{0*) = 0.5 V.

Exercises. 1. In Example 7.8, make C, = 0.75 F and C2 = 0.25 F, and compute ^/^(O'^). AN SW ER: /Y-,(0") = 0.75 V. 2. In Example 7.8, sufipose ^’q ( 0 " ) =10 V and Vqj,{Q~) = - 8 V. Also let C, = 0.75 F and C-, = 0.25 F. Compute ANSW ER: /.v.,(0^) = 5.5 V.

Example 7.8 is illustrative o f a charge transport that is germane to switched capacitor circuits, which are o f fundamental importance in the industrial world.

EnergyStorageinaCapacitor As with all devices, the energy stored or utilized in a capacitor is the integral o f the power absorbed by the capacitor. The net energ}' entering the capacitor over the interval [/q, /J is

Jff,

Pci'^)dT = f'' Vc(T)/c(T)i/T


2 ‘)1

ch’ciT )]

= c r '' (

(It

Vcih) Oo)

dT =

(7.11)

= -C

1

for C in farads,

in volts, and energy in joules (J). From equation 7. 11, the change in energy

stored in the capacitor over the inter\'al [rQ, r j depends only on the values o f the capacitor volt ages at times /q and

i.e., on

and v^t^. This means that the change in stored energy is

independent o f the particular voltage v/aveform between odic, i.e., if V(\t) =

and r,. If the voltage waveform is peri

+ T) for some r > 0, then over any time interval [t, / + 7], the change in

the stored energy in the capacitor is zero because

+ 7) =

forces equation 7.11

to zero. Analogous to the inductor, for all periodic voltages, the capacitor stores energy and then returns it to the circuit and is thus called a lossless device. As with the inductor, it is convenient to define the instantaneous stored energy in a capacitor as

Wc{t) = l^Cvc{t)

(7.12)

which is really the integral o f power over the interval (—x , /], assuming that all voltages and cur rents are zero at r = - x . E X A M PLE 7 .9 Consider the circuit o f Figure 7.22, in which

= 0. It is known that

for f > 0, the source current is i^{t) and the voltage across the capacitor is V for ^ > 0. Compute (i) the energ)', in joules, stored in the capacitor for / > 0, (ii)

and (iii)

FIGURE 7.22 Parallel RC circuit. S o lution

(i) Since

= 0, from equation 7.11 (or 7.12),

VV^^(0,/) = lc v J ( / ) = 8 C / ?

\-e

RC

(ii) To find the capacitor current, recall = C ^ ^ ^ = 4—

RC

A

) - 4/? \ - e


2 ‘)2

(iii) To find

we first compute

flowing fi-om top to bottom:

\ -e

A

Thus 1- e-

+ 4^

E X A M PLE 7 .1 0

= 4 A

/

\

ijt ) — ►

For the circuit o f Figure 7.23a, it is known that the voltage across the capacitor is

= 20sin(2f + rr/6) V for r > 0. Compute and

plot the instantaneous power absorbed by the capacitor and the energy stored by the capacitor during the time interval [0, f].

+ N 5mF

k -

T im e t in seconds (b) F IG U R E

7.23 (a) Capacitor with known voltage v^^t) connectcd to a network N. and the net cncrg)',

t), stored over the interval lO, ^].

(b) Plot of power,


S o lution

Step 1. Compute

From equation 7.5, for / > 0 ,-^(,) = C ^ ^

dt

= 0 .2 co s 2t + -

6}

Step 2. Computep(\t). By direct multiplication and a standard trig identit}', / k\ / k\ {A watts P c(t)= V(-(/)/(-(/)= 20 sin 2 / + - X 0. 2 cos 2/ + - = 2 sin 4t + 6j \ 3! l 6j Step 3. Compute

t). From equation 7.11 with

= 20 sin(7r/6) = 10 V, we obtain

\V^(0.0 = 0.5 C \’c {t) - 0.5 C v c (0 ) = sin-

2t + - - 0 . 2 5 J 6}

Plots o f pf4f) i^nd V\^^0, t) arc given in Figure 7.23b. Notice that WTÔ, /) can be negative, because W^—oo, 0) = 0.25 joules, meaning that at r = 0, there is an initial stored energy that can be returned to the circuit at a later time . Figure 7.23b substantiates this.

4. SERIES AND PARALLEL IN D UCTO RS AND CAPACITO RS

Sei'iesInductors Just as resistors in series combine to form an equivalent resistance, inductors in series combine to form an equivalent inductance. As it turns out, series inductances combine in the same way as series resistances. E X A M PLE 7 .1 1 . Compute the equivalent inductance o f the series connection o f three inductors illustrated in Figure 7.24. Then find the voltages

as a fraction o f the applied voltage

Leq

o

+ V ..

+ Leq

Vl2 + V .,

Q-

o+ êq o -

(a) FICIJRH

Leq

(b)

(a) Scries connection of three inductors, (b) Equivalent inductance.

29-4


S o lution

First we must answer the question o f what it means to be an equivalent inductance. Earlier, we defined the inductor in terms o f its terminal voltage-current relationship. Two 2-terminal induc tor circuits have the same inductance if each circuit has the same terminal voltage-current rela tionship as defined in equation 7.1. Step 1. The voltage labeled

il

appears across the series connection, and, by KCL, the current = ‘l y

flows through each o f the inductors, i.e.,

equivalent inductance,

is defined by the relationship =

in terms o f Z j, L-,, and Ly

O ur goal is to express Step 2. Find

(7.13)

dt

in terms ofi^^^. To obtain such an expression, observe that, by KVL, ^Leq = ^/.l + ^12 +

Since each inductor satisfies the v-i relationship

dt it follow's that

^'U’q - (^1 + ^2 + ^ 3 )

dt

Hence, the series inductors o f figure 7.24a can be replaced by a single inductor with inductance = -^-1 + ^2 +

Finally, since

= Lj

dt

= Lj —^

dt

and

^

_ dt

= { L, +L^ + ~ '

dt

, it follows that

Lj (L| 4- Zy-> -l- L-^)

^

which is analogous to the voltage divider formula for resistances.

Exercises. 1. If, in Example 7. 11, Z,, = 2 mH, AN SW ER: = 8 mH. 2. Find

in terms o f 3

ANSW ER: '■/.:=

S

= 5 mH, and “

= 1 mH, find

.

295


Extension o f the formulas in the above example to n inductors is fairly clear, and we state the results without rigorous proof: the formula for series inductances is (7.14a) and the formula for voltage division o f series inductances is

^'IJ =

(7.14b)

L\ + Lo + ... + Lfj

InductorsinParallel The same basic question as with inductors in series arises with a parallel connection o f inductors: what is the equivalent inductance? Rather than derive the general formula, let us consider the case o f three inductors in parallel, as illustrated in Figure 7.25a. E X A M PLE 7 .1 2 For this example our goal is to show that the equivalent inductance o f the circuit o f Figure 7.25a is given by the reciprocal o f the sum-of-reciprocals formula,

- ~\

i T — +— +•

U

(7.15)

Ly

We then show a formula for current division. Leq

Leq

O +

L3

L2,

êq

Leq

L,

L,

o-

o (a)

(b)

FIG URE 7.25 (a) Parallel connection of three inductors, (b) Equivalent inductance.

So l u t io n Once again, equation 7.13 defines the relationship for the equivalent inductance:

Û-q The goal is to construct

in terms o f Z ,,

êq and

in a way that satisfies equation 7.13. This

will produce equation 7.15. Step 1. Write KCL for the parallel connection shown in Figure 7.25a. Here, by KCL, ‘U q = 'Z.1 +

+ 'L 3


2%

DifFerentiating both sides with rcspect co time yields

(Hl\ ^ dt dt

_

ill

I

dt

di,

Step 2. Find — Ul in terms o f a n d L^,. From equation 7.1, For each inductor

dt dt

Li

~ î\ ~ ^L1 ~ ^L5

Substituting into the result o f step 1 and noting that

di, dt

( \ Ln

1 n — +— +— Li-q L] L t^)

L'S

This has the form o f equation 7.13, which implies equation 7.15, i.e.,

Le, =

1

1

1

— + ---- + — Z/j L~i L,'^ lb generate a currcnt division formula we first note that

= ^/1 =

1 ' 1 ' = — J y i J c i T ) d T = — J \ ' , ^ { T ) d T and

^ _-TT

' = / v^,^(tV/t

^ _nr.

Thus

(/)

Exercises. 1. If, in Flxample 7.12, Z.j = 2.5 mH, Z., = 5 mFl, and ANSWq-R:

= 1 mH, find

= 0.625 mH.

The above arguments easily generalize. Suppose there are u inductors, /,,, Z-,, ... , Z.,^, connected in parallel. Then the equivalent inductance is given by the reciprocal o f the sum-of-reciprocals formula.

(7.16a)

and the current division formula.

Chapter 7 * Inductors aiul Capacitors

U

ii.jn

~i-------- [

(7.16b)

Exercise. For two inductors Zj and L-, in parallel, show chat the equivalent inductancc satisfies the formula (7.17)

Series-ParallelCombinations This subsection examines series-parallel connections o f inductors. This allows us to use the for mulas developed above in an iterative way. EXA M PLE 7 .1 3 Find the equivalent inductance,

o f the circuit o f Figure 7.26.

S o l u t io n

Step 1. In the circuit o f Figure 7.26, several inductors are enclosed by an ellipse. Let

denote

the equivalent inductance o f this combination. Observe that the series inductance o f the 5/6 H and 0.5 H inductors equals 4/3 FI. This inductance is in parallel with a 1 H and a 4 Fi induc tance. Hence,

4

, =

I 1 I 3= 1 —!----- !---1

4

H

4

Step 2. The equivalent circuit at this point is given by Figure 7.27. This figure consists o f a series combination o f a 1.5 FI and a 0.5 FI inductor connectcd in parallel with a 6 H inductor. It fol lows that

29S


=

1

I

-------------+ 0 .5 + 1.5 6

= -= 1 .5

4

H

Exercise. In Example 7.13, suppose the 5/6 H and 0.5 H inductors are both changed to 0.4 H inductors. Find L o f the circuit. AN SW ER: 1.443 H.

CapacitorsinSeines Capacitors in series have capacitances that combine according to the same formula for combining resistances or inductances in parallel. Similarly, capacitances in parallel combine in the same way that resistances or inductances in series combine. This means that the equivalent capacitance o f a parallel combination o f capacitors is the sum o f the individual capacitances, and the equivalent capacitance o f a series combination o f capacitances satisfies the reciprocal o f the sum-of-reciprocals rule. These ideas are illustrated in the examples to follow. EXAM PLE 7 .1 4 Compute the equivalent capacitance,

o f the series connection o f capacitors in Figure 7.28a.

2‘)9


o-

Cl

'C2

o+ C3 eq

Q-

o

(a) (b) FIGURI^ 7.28 (a) Series combination of three capacitors, (b) Equivalent capacitance,

So l u t io n The equivalent capacitance denoted in Figure 7.28b is defined implicitly by the current-voltage terminal conditions according to equation 7.5, i.e.,

^

; - r

^

dt

Our goal is to express this same terminal v-i relationship in terms o f the capacitances, C j, C 2, and

Cy After this we set forth a formula for voltage division. Step 1. Set forth the i-v relationship fo r each capacitor. For each capacitor, k = 1, 2, 3,

‘Ck - Q But, by KCL, i^ =

dt

Hence,

dt

C,

Step 2. Apply K V L From KVL,

Differentiating this expression with respect to time and using the result o f step 1 yields dv/^

d v f^ ]

d v 'c '')

d v (--i

(

\

1

1

\ .

— ^ = —— + —— + —— = \— + — + — \ir dt dt dt dt l,C| Cj c J

3D0


Step 3. Compute

. From the result o f step 2, solve for Iq to obtain

\ iQ -

[C l

dt

H--------h --

Cl

dt

C^j

It follows that Q ,= C,

G

C3

To set forth a formula for voltage division, we first note that

= /q =

1 ' 1 ' ' ''CA-(0 = — J ick^T)dr = — J ic{T)dT and C,^ Vci 0 = f ic(r)dT ^ -rr-

^ —rri

Thus

''a - ( 0 = — f ic(r)dT = -—

v c(f) = -^------- j------- p v c (0 C,

C2

Exercises. 1. In Example 7.14, suppose Cj = 5 pF, Cj = 20 |.iF, and

C3

= 16 pF. Compute

ANSW ER: 3.2 uE 2. Find

in terms of

.ANSWER: /V;: = 0 .1 6 /y ;

Generalizing the result o f Example 7.14, we may say that capacitors in series satisfy the reciprocal o f the sum-of-reciprocals rule. Thus, for n capacitors C j, C2, ... , C^, connected in series, the equivalent capacitance is

C

1 - 1 -------------- r — + — + ... + — C, G C„

=■

(7.18a)

and the general voltage division formula is 1 '• c ( 0

------- i---------

— + --- +... C,

Co

C.,

( 7 .1 8 b )


3 01

Exercise. Show that if two capacitors Cj and C-y are connected in series, then

c

(7.19)

c ,+ c .

CapacitorsinParallel If rvvo capacitors are connected in parallel as in Figure 7.29a, there results an equivalent capaci tance

= Cj + C2 and a simple current division formula to be derived. I. a +

'c ‘Cl

o +

'C2

eq

a-

a(a)

(b)

FIGURE 7.29 (a) Parallel combination of two capacitors, (b) Equivalent capacitance, Since the voltage

appears across each capacitor, and since /^= /q +

by KCL it follows that

Hence,

êq -

^2

One surmises from the above example that, in general, capacitors in parallel have capacitances that add. And, indeed, this is the case: if there are n capacitors C j, C2, ... ,

in parallel, the equiva

lent capacitance is C

= C, + C2 + ... + C„

(7.20a)

Exercise. Show that in the above derivation Ck .

Cf,

Ceq

.

Q + Q

and that for n capacitors in parallel. 'CA- -

Q C] + C j +••• + C„

'c

( 7 .2 0 b )


302

Series-ParallelCombinations Wc round out our discussion o f capacitance by considering a simple series-parallel interconnection. EX A M PLE 7 .1 5 Consider the circuit o f Figure 7.30. Compute the equivalent capacitance,

0.45 mF

0.6 mF

FIG U RE 7.30 Series-parallel combination o f capacitors. S

o l u t io n

Step 1. Combine series capacitances. Observe that the rwo series capacitances o f 0.5 mF and 0.5 mF combine to make a 0.25 mF capacitance. Step 2. Combine parallel capacitances. First, as a result o f step 1, the three capacitances, 0.3 mF, 0.25 mF, and 0.45 mF, add to an equivalent capacitance o f 1 mF. Further, the two parallel capac itances, 0.3 mF and 0.6 mF, at the bottom o f the circuit, add to make a 0.9 mF capacitance. The new equivalent circuit is shown in Figure 7.31.

a-

II mF 1.125 mF

0.9 mF

a-

FIGURE 7.31 Circuit equivalent to that o f Figure 7.30. Step 3. Combine series capacitances. From equation 7.18,

^ eq-

\

1

1 1 -------- + - + ----1.125 1 0.9

1

Exercise. Suppose the two 0.5 mF capacitors in Figure 7.30 are changed to 2.5 mF capacitors. Find the new A N SW ER: 0.4 mF.


303

5. SM O O TH IN G PROPERTY OF A CAPACITO R IN A POW ER SUPPLY As mentioned in the chapter opener, a power supply converts a sinusoidal input voltage to an almost constant dc output voltage. Sucii devices are present in televisions, transistor radios, stere os, computers, and a host o f other household electronic gadgets. Producing a truly constant dc voltage from a sinusoidal source is virtually impossible, so engineers design special circuits called voltage regulators that generate a voltage with only a small variation between set limits for a given range o f variation in load. The voltage regulator is a precision device whose input must be fairly smooth for proper operation. A capacitor can provide a crude, inexpensive sm oothing function that is often sufficient for the task. This section explores the design o f a capacitive smoothing cir cuit. In practice, such a circuit is used only for low-power applications. i.(t)

FIGURE 7.32 Simple power supply with capacitive smoothing for low- power applications. Consider, for example, the circuit shown in Figure 7.32. The four (ideal) diodes are arranged in a configuration called a fidl-wave bridge rectifier circuit. An ideal diode allows current to pass only in the direction o f the arrow. The diode configuration ensures that i^{t) remains positive, regard less o f the sign o f the source current. Specifically, the diodes ensure that /j(f) =

Using the

integral relationship (equation 7.6) o f the capacitor voltage and current, it follows that

V cit)= yc(h)) + ^ f

ici-^)ch = VcUo) + ^ f

[M t)|-/o(t)]^/t

(7.21)

Because o f the difference |/j(r)| - iît) inside the integrand o f the integral, i^{t) tends to increase the capacitor voltage, whereas i^^{t) tends to decrease the capacitor voltage. Further, because the diodes are assumed ideal, it follows that

v^t) > I v^{t)

(7.22)

To see this, suppose the opposite were true; i.e., suppose One o f the diodes would then have a positive voltage across it in the direction o f the arrow. The diode is said to be forivarcl

biased. But this is impossible, because an ideal diode behaves like a short circuit when forward biased. The consequence is that V(4,t) will be 12 V whenever |?>'^(^)| is 12 V. This occurs every 1/120 o f a second. Thus, the rectifier output will recharge the capacitor every 1/120 o f a second. Between charging times, the current, i(){t), will tend to discharge the capacitor and diminish its voltage.

Chapter 7 • Inductors aiui Capacitors

304

The design problem for the capacitive smoothing circuit is to select a value for C that guarantees that v^t) is sufficiently smooth to ensure proper operation of the voltage regulator. Here, “suffi ciently smooth” means that the maximum and minimum voltages differ by less than a prescribed amount. To be specific, suppose that i>(\t) must remain between 8 V and 12 V. Recall that i^{t) tends to increase the capacitor voltage, while

tends to decrease it. The design requires select

ing a value for C to ensure that i^{t) can keep up with

so that the capacitor voltage remains

fairly constant. The value for /(,(^) is obtained from the specification sheet o f the voltage regulator. Suppose this value is a constant 1 A. It remains to select C so as to ensure that V({t) remains above 8 V between charging times. From equation 7.21, it is necessary that

CJio Now we need consider only values for t between 0 and 1/120, because the capacitor will recharge and the process will repeat itself every 1/120 o f a second. Thus, because i^{t) will only increase the capacitor voltage, to ensure that v^{t) remains above 8 V, it is sufficient to require that

With i^{t) = 1 and

= 12, it follows that 1 A X 120 sec = 2.083 mF 4 V

A 2,100 |.iF capacitor satisfies this requirement. A method for computing the capacitor voltage waveform is described in Chapter 22 o f o f 2"^ edition. However, using SPIC E or one o f the other available circuit simulation programs, one can generate a plot o f the time-varying capacitor volt age produced by this circuit, as shown in Figure 7.33. In the figure, it is seen that the capacitor voltage varies between 12 and 9.02 V, which is smaller than the allowed variation o f (12 - 8) V. Two factors contribute to this conservative design: (1) we used C = 2,100 uF instead o f the cal culated value, C = 2,083 uF, and (2) the increase in the capacitor voltage due the charging current is is not included in the calculation. volts

FIGURE 7.33 Time-varying capacitor voltage generated by the circuit in Figure 7.32 when C = 2,100 |.iK


The preceding brief introduction made several simplifying assumptions to clarify the basic use o f a capacitor as a smoothing or filtering device. Practical power supply design is a challenging field. A complete design would need to consider many other issues, some o f which are the nonzero resistance o f the source, the non-ideal nature o f the diodes, the current-handling abilit)' o f the components, protection o f the components from high-voltage transients, and heat-sinking o f the components.

6. SUM M ARY This chapter has introduced the notions o f a capacitor and an inductor, each o f which is a lossless energ}' storage device whose voltage and current satisfy a differential equation. The inductor has a voltage proportional to the derivative o f the current through it; the constant o f proportionalit}^ is the inductance L. T he capacitor has a current proportional to the derivative o f the voltage across it; the constant o f proportionalit}' is the capacitance C. It is interesting to observe that the roles of voltage and current in the capacitor are the reverse o f their roles in the inductor. Because o f this reversal, the capacitor and the inductor are said to be dual devices. That the (ideal) inductor and the (ideal) capacitor are lossless energy storage devices means that they can store energ)- and deliver it back to the circuit, but they can never dissipate energ)^ as does a resistor. The inductor stores energy in a surrounding magnetic field, whereas the capacitor stores energy in an electric field between its conducting surfaces. Unlike energ)' in a resistor, the energy stored in an inductor over an interval [r^, fj] is dependent only on the inductance L and the val ues o f the inductor current //(/^()) and //(/^j). Likewise, the energ)' stored in a capacitor over an interval [r^, /,] is dependent only on the capacitance C and the values o f the capacitor voltage and Both the inductor and the capacitor have memor)'. The inductor has memory because at a partic ular time Tq, the inductor current depends on the past histor}' o f the voltage across the inductor. The capacitor has a voltage at, say, time

that depends on the past current excitation to the capac

itor. The concept o f memory stems from the fact that the inductor current is proportional to the integral o f the voltage across the inductor and the capacitor voltage is proportional to the integral o f the current through the capacitor. This integral relationship gives rise to the important proper ties o f the continuity of the inductor current and the continuit)' o f the capacitor voltage under bounded excitations. rhe dual capacitor and inductor relationships are set forth in Table 7.1. Finally, we investigated the smoothing action o f a capacitor in a power supply.

306


lABLE 7.1. Summary of the Dual Relationships ot the Capacitor and Inductor

ic « )

icit) = C

d\'c ) dt

V[{t) = L

dilit) dt

/■/(0 =/■/(ô)+ t /^ V;(T)^/T

LJ k>

7. TERM S AND CO N CEPTS Bounded voltage or current; voltage or current signal whose absolute value remains below some fixed finite constant for all time. Capacitance o f a pair o f conductors: a propert)' o f conductors separated by a dielectric that per mits the storage o f electrically separated charge when a potential difference exists between the conductors. Capacitance is measured in stored charge per unit o f potential difference between the conductors. Capacitor (linear): a two-terminal device whose current is proportional to the time derivative o f the voltage across it. C oil: another name for an inductor. Conservation-of-charge principle: principle that the total charge transferred into a junction (or out of a junction) is /.ero. C ontinuity property o f the capacitor: property such that if the current i(\t) through a capaci tor is bounded over the time interval

< t < t-,, then the voltage across the capacitor is < tj, then t^(^tQ~) = when

continuous for /, < r < tj. In particular, if fj <

C ontinuity property o f the inductor: propert}' such that if the voltage is bounded over the time interval r, < r < continuous for r, < r <

In particular, if r, <

across an inductor

then the current through the inductor is then /^(/‘o~) =

when

Coulom b: quantit)' o f charge that, in 1 second, passes through any cross section o f a conductor maintaining a constant 1 A current flow. Dielectric: an insulating material often used between two conducting surfaces to form a capacitor. Farad: a me;Lsure o f capacitance in which a charge o f 1 coulomb produces a 1 V potential difference. Faradays law o f induction: law' asserting that, for a coil of wire sufficiently distant from any mag netic material, such as iron, the voltage induced across the coil by a time-varying current is proportional to the time derivative o f the current; the constant of proportionality,


307

denoted Z, is die inductance o f the coil. Faradays law is usually stated in terms of flux and flux linkages, which are discussed in physics texts. H enry: the unit o f inductance; equal to 1 V-sec/amp. Inductance: property of a conductor and its local environment (a coil with an air core or iron core) that relates the time derivative o f a current through the conductor to an induced voltage across the ends o f the conductor. Inductor (linear): a two-terminal device whose voltage is proportional to the time derivative of the current through it. Instantaneous powen p(t) = */(/)/(<), in watts when v{t) is in volts and i{t) in amps. Lossless device: device in which energy can only be stored and retrieved and never dissipated. Lossy device: a device, such as a resistor (with positive R), that dissipates energy as some form o f heat or as work. Maxwell’s equations: a set o f mathematical equations governing the properties o f electric and magnetic Beids and their interaction. M emory: property o f a device whose voltage or current at a particular time depends on the past operational history o f the device; e.g., the current through an inductor at time /q depends on the history o f the voltage excitation across the inductor for t< /q. Unbounded voltage or current: a voltage or current whose value approaches infinity as it nears some instant o f time, possibly r = oo. Voltage r^ u la to r: circuit that produces a voltage having only a small variation between set lim its for a given range o f load variation from a fairly smooth input signal.

^The word “stored” emphasizes that the energy in the inductor is not dissipated as heat and can be recovered by the circuit, whereas the word “absorbed” is used to mean that the energy cannot be returned to the circuit. In a resistor, energy absorbed is dissipated as heat. ^ More generally, conservation of charge says that the total charge transferred into a Gaussian sur&ce (or out of a Gaussian surface) is zero.

31)8

Chapter 7 • Induccors and Capacitors

Problems

4. (a) For i^{t) = 10sin(2000r) mA in Figure P7.4, calculate and sketch

for 0 <

t < \5 ms assuming both inductor cur

TH E IN D U CTO R AND ITS PROPERTIES

(b) What is the instantaneous power deliv

1. If the length o f a single-layer air coil is

(c) Compute and sketch the energy stored in

rents are zero at f = 0. ered by the dependent source?

greater than or equal to 0.4 times its diameter,

the 2 niH inductor for 0 < f < 15 msec.

then its inductance is approximately given by the formula L =

lOv ft)

o f tiirns)~

4 X

18 {dkimeter)+ 40 (lengths)

length o f the coil are in meters. A 2 cm diam

Figure P7.4

eter coil has 48 turns wound at 12 turns/cm. Compute the approximate value o f the induc tance. C H EC K : 18 pH < I < 2 0 pH.

2mH

0.2 mH

where L is in henries, and the diameter and

5. For Vsii) sketched in Figure P7.5a, compute and sketch for the circuit o f Figure P7.5b. What is the instantaneous power deliv ered by the dependent source?

2. (a) Find and plot for 0 < r < 5 sec the V (t) (mV)

inductor voltage Vj{t) for the circuit o f Figure P7.2a driven by the current

2•

waveform of Figure P7.2b. 1 1”

(b) Find and plot the instantaneous stored energ)^

t I

(c) Find and plot the stored energy U^(l,r)

1

-1

2

3

4

5

1 6

^

as a function o f time for 5 > / > 1. -1 -

L(A)

-2 L(t)

©

t(s)

(a) Figure P7.2

-2

(a)

+

, 0.8 mH

(b)

„(t)

3. Repeat Problem 2 for: (a) and (b) the waveform sketched in Figure P7.3.

6 ijt)

v,{t)

0.6 mH 0.75 mH

1.5 mH

i.JA ) (b)

Figure P7.5 6. (a) Find and plot for 0 < r < 6 sec the induc Figure P 7.3

tor current /^(f) for the circuit o f Figure

3 0 ')


(a) the voltage waveform sketched in Figure

P7.6a driven by the voltage waveform of

P7.8b, and

Figure P7 .6 b.

(b) the voltage waveform sketched in Figure

(b) Find and plot the instantaneous stored energy.

P7.8c.

(c) Find and plot the stored energy as a function o f time for 5 > ^ > 1. (d) Find and plot t/j (/)

i,(t) vJt)

(a)

(b) Figure P7.6 7. Repeat Problem 6 for (a)

u(t), and

(b) the voltage waveform in Figure P7 .7 .

Figure P7.8 9. Consider the circuit in Figure P7.9 in which Z,

= 0 .2 H , Z2 = 0 .5 H , and

=

100sin(0.257t/) mV for / > 0 and zero other wise. (a) Find the current /■^(/) for r > 0. (b) Compute the energy stored in each inductor over the intervals 0 < r < 2 sec and 2 < 8 . For the circuit in Figure P7.8a, suppose Z, = 0.8 H and L2 = 0.2 H. Compute and plot the waveforms / j (/) and /^J f ) for

0

310


+

v,(t)

-

% iQ ° ic(t) Figure P7.12 Figure P7.9

13. In Figure P7.13a, the capacitors C, = 4 mF

10. The circuit o f Figure P 7 .10 has two induc tors, Z., = 20 mH and Lj = 50 mH, in parallel. The input is v^{t) = 200cos(5007rr) mV for t >

and C j = 12 mF are driven by the voltage specified in Figure P7.13b. Plot /q(/), and iA i).

0 and zero otherwise. The switch between the two inductors moves down at r = 4 ms. Compute the currents

and

for 0 <

r < 4 ms and 4 ms < /. Also find the energy stored in each inductor as a function o f t for the same time intervals. (a)

Figure P 7.10

THE CAPACITOR AND ITS PROPERTIES 11.(a) Suppose that a 20 pF capacii charged to 100 V. Find the charge that

(b)

resides on each plate o f the capacitor.

Figure P7.13

(b) If the same charge (as in part (a)) resides on a 5 pF capacitor, what is the voltage

14. T he C = 2 pF capacitor o f Figure P 7 .l4 a

across the capacitor? (c) What is the voltage required to store 50 pC on a 2 pF capacitor? (d) Find the energy required to charge a 20

has current ;^ r) shown in Figure P 7 .l4 b . If = 4 V, compute at ^ = 1, 2, 3, 4 ms. Now compute the energy stored in the capaci tor over the intervals, [0, 2 ms], [2 ms, 3 ms],

pF capacitor to 100 V.

and [0, 4 ms]. 12.(a) The C = 2 pF capacitor o f Figure P7.12 has a terminal voltage o f

= 100[1 +

cos(lOOOTtr)] V. Find the current i(^t) through the capacitor. (b) Now suppose the voltage is v^^t) = 10sin(2000r) V and Iq = 10cos(2000r) mA. Find the capacitance C.

31

Chapter 7 • Induaors and Capacitors

15. Suppose

as specified for all time in

Figure P7.15a, excites the circuit of Figure P7.15b, in which Cj = 0.2 pF and C2 = 0.1 pF. (a) Plot

for 0 < ^< 8 msec.

and

(b) Compute and plot the energy stored in the 0.2 pF and 0.1 pF capacitors for 0 < r < 8 . Hint: use MATLAB to plot the answers. (c) Find

and

as t

00.

O '

w O '

0

Figure P7.16

w

(b) 0 Figure P7.15

0

16. For the circuit in Figure P7.16a, C = 0.25 mF. Compute and plot the waveforms of the

0

voltage,

given

as sketched in Figures

P7.16B and c.

17.(a) Consider the circuit sketched in Figure P 7.17 in which Cj = 20 pF and C2 = 0.1 mF. Suppose v^{t) = 5sin(2000f) V for t

> 0 and suppose

= 10 V. Find

for r > 0. Is the output voltage independent of the initial voltage on Q ? Why?

0 (b) W hat is the instantaneous power deliv ered by the dependent source?

w

(c) Find the energy stored in Cj over the interval [0 , t].

0

i jt ) (a)

0 0 0

Figure P 7 .1 7

312


18. Repeat Problem 17 when v^{t) = V for / > 0 and 0 otherwise. 19. Reconsider the circuit o f Figure P 7 .17. Suppose, however, that

is given by the plot

in Figure P 7 .19. (a) Find and sketch

Figure P7.21

for 0 < / < 6

msec. (b) What is the instantaneous power deliv

22. In the circuit o f Figure P7.22 v^{t) = 25 V and the

ered by the dependent source? (c) Compute and plot the energ)' stored in C,.

= 100 mF capacitor is uncharged at

f = 0 . Compute V({t) for 0 < r < 2 sec and 2 sec < t when C, = 4 0 0 mF. t = 2s

Figure P7.22

23 . Fhe circuit ol Figure P7.23 has two capac itors in parallel, C| = 30 mF, C-, = 50 mF. Fhe 20. Repeat Problem 17 for the waveform of

input current is iî) = 360f’~'®^^ mA for r > 0

Figure P7.20.

and 0 otherwise. Suppose each capacitor is imcharged at /= 0 . The switch between the t%vo capacitors opens at r = 2 msec. (a) Find the voltage,

for 0 < r < 2 ms

and 2 ms < t. (b) Compute the energy stored in each capacitor as a function o f t for the same time intervals. Figure P7.20

(c) Com pute the current through each capacitor over each time interval.

21. For the circuit o f Figure P7.21, suppose C, = 0.6 ml', Ct = 1.2 mF,

t = 2 ms

= 0.4 mF, Q = 1.6

mF, ijj) = 120sin(100r) niA for / > 0 and 0 for r < 0. (a) Find the equivalent capacitances the series combination and

for for the

parallel combination.

Figure P7.23

(b) Find and sketch (c) What is the instantaneous power deliv ered by the dependent source? (d) What is the instantaneous energ)' stored in C4?

M IX ED C A PA CITO R AN D IN D U C TO R PRO BLEM S 24. Consider the circuit o f Figure P7.24,

=

2.5 H, Cj = 1 mF, which is excited by the cur-

313


rent waveform

= 200fr“ ’^' mA for ^ > 0

27. For the circuit o f Figure P7.27, as a function o f i^{t) and

(a) Compute

and 0 otherwise. (a) Compute and sketch I'lit),

the capacitances Cj and C j.

and

in terms o f i^{t) and the

(b) Now find

circuit parameter values.

(b) Compute and sketch the energy stored in the inductor for r > 0. (c) Compute and sketch the energy stored

L.

in the capacitor for t> 0.

+

Figure P7.27 28. For the circuit o f Figure P7.28, compute and

Figure P7.24

as a function of

and the circuit parameters. 25. In the circuit o f P'igure P7.25, suppose Z., = 0.25 H, Cj = 2.5 mF, ij^t) = 20sin(400/‘) mA for r > 0 and 0 otherwise. Ail initial conditions are zero at r = 0. (a) Find Vjit).

+ “" ’0

c:

\|>

-5 L ,2

L,

(b) Find V(^t). (c) Find the instantaneous stored cnerg)' in the capacitor.

Figure P7.28

SERIES-PA RALLEL IN D U C TO R S 29. In the circuit o f Figure P7.29, all inductors are initially relaxed at /^= 0 and /.j = 6 mH, L-, Figure P7.25 26.(a) In the circuit o f Figure P7.26, (i = 10, C,

= 38.5 mH, = 22 mH. A voltage 200re~' mV is applied for r > 0. Find, Vjj(t), and Challenge: Find

= 20 pF, C , = 80 mF, Z., = Z , = 20 mH are

initially

uncharged.

If

vj^t) =

10/sin(20^) V for r > 0 and 0 otherwise, (a) Find i^{t) for r > 0. (b) Now find

for / > 0.

(c) Compute the energ)- stored in the 20 mH inductor for r > 0. i.(t)

C H E C K S: 20 mH and \AOte~' mV.

Figure P 7 .2 6

=

314


30. Consider the circuit o f Figure P7.30. Suppose Z., = 3 mH, L j= \2 mH,

(mA)

= 36

= 120cos(1000r) mA.

mH, and (a) Find L

and

(b) Find (c) Plot the instantaneous power deliv ered by the source for 0 < r < 14 msec. /

Y

Y

V

L.

-200

--

Figure P7.32

(!)

33. In Figure P7.33, Z-j = 5 mH, L-, = 20 mH, = 20 mH,

= 80 mH, and ij^t) =

lOsin(lOOOr) mA for r> 0 and 0 otherwise. (a) With the switch in position C, find the


12 mH, - 1 .4 4 sin(lOOOr) V, 90

equivalent inductance,

Vj^, and

(b) Repeat part (a) with the switch in posi

cos(lOOOr) mA.

tion D. 31. For the circuit o f Figure P7.31, Z., = 260 mH,

= 26 mH, L-^ = 39 mH, and

=

10^"^“ tiiA. (a) Find (b) Compute (c) Com pute stored in

and i[^2 and the

instantaneous

energy

as a function o f t.

C H EC K S: 52 mH. 0.2

0.8 i j t ) ,

=

-0.104^6’" '“ V, Vj2 i^) = 34. Consider the circuit o f Figure P7.34a with voltage source excitation given in Figure P7.34b. Let the inductor values be those given in Problem 33. Suppose the switch is in position C. Note that each inductor is relaxed at r = 0. (a) Find Z^^^. (b) Compute and sketch

for r > 0.

(c) Find the instantaneous (total) energ)' stored in the set o f four inductors as a function o f time. Figure P7.31 32. Repeat Problem 31 for the waveform of Figure P7.32.

(d) Compute and sketch //2(0CH ECK: L

= 20 mH; /.„(1) = /.„(3) = 0.8 A while /y,/2) = =0

31


ijt )

____rvY V v X

L.

L,

1-6

L,

i„(t)

f r r \ ____ T Y Y \ (a)

i-N/YYA___ TYYV

v„(t)(V) >k 16'

0

1

2

3

4

16-

L. (b)

rOA___ TYYV

Figure P7.34 35. Find

L,

(b)

for the circuit o f Figure P7.35, (a)

when the s\vitch is open, and (b) when the switch is closed. The unit o f L is henries.

Figure P7.36 SCRA M BLED ANSWERS: 0.1, 0.08, 0.6 (in 11) 36. Find


P 7.36, where Z., = 5 niH, = 20 mH, niH, I 4 = 150 mH, Z.5 = “50 mH, mH, L j = 120 mH,

= 40 = 180

= 35 mH. Notice that

the circuit o f (b) is a modification o f (a) and that o f (c) is a modification o f (b). Connections can create interesting behaviors.

37 . Three 60 mH inductors are available for interconnection. List all equivalent inductances obtainable over all possible interconnections o f these elements. C H EC K : There should be seven different val ues.

316


38. Find L P7.38.


/

4 mH

Y

Y

AN.SWER: C//I for . all / values. 40. Like Problem 39, this is a conceptual prob lem and requires no calculations for the answer.

Y

lOmH

1 mH

I

Consider circuits 1 and 2 o f Figure P7.40. All

5 mH'

wish to determine the relationship bet\veen

inductors are 1 H except the one labeled L. We 3mH'

36 mH

and L^^-, in the presence o f the finite posi

Bo(a)

tive inductor o f L henries between points a and b. W hich of the following statements is true?

/YYV 7mH /

Y

Y

V

/

2.4 mH

Y

Y

(h)

V

< /v,2-

1.2 mH

(d) There is no general relationship between

êq\

0.6 mH

relationship depends

on the value o f L Explain your reasoning.

(b)

Figure P7.38 ANSWHR: (a) 13 m il: (b) 2 in 11 39. This is a conceptual problem and requires no calculations for the answer. Consider circuits 1 and 2 o f Figure P7.39. All induc tors are 1 H except the one labeled

L. We wish to determine the rela tionship between

and

presence

o f the finite positive inductor o f L henries between points a and b. Which of the following statements is true?

êq\ ^ êql(b) < K ,l(d) There is no general relationship between and Any relationship depends on the value o f L. Explain your reasoning.

Circuit 1

Circuit 2 Figure P7.40

ANSW ER: /.

SERIES-PARALLEL CAPACITORS 41. (a) Find the indicated equivalent capacitance for the circuit o f Figure P 7.4la where C, = 4 pF, C 2 = 3 pF, C3 = 2 pF, Q = 4 pE

Then find vj^t) when ii^t) = lOcos(lO'^r) mA for r > 0 and 0 otherwise. (b) Repeat for Figure P 7 .4 lb in which C, = 60 pF, C , = 18 pF, C 3 = 18 pF, Q = 36 pF, and C 5 = 10.8 pF. Then find vj^t) when i^{t) =

1 Osin( 1O^f) mA for r > 0 and 0 otherwise.

31'

Chapter 7 • Induaors and Capacitors

-C,-

Vw/' C. c,

^ 1

Figure P7.43 44. For the circuit of Figure P 7.44, Cj = 8 mF, C2 = 6 mF, C3 = 12 mF, and

=

240sin(200r) mA for ^ > 0 and 0 otherwise. (a) Find C,^. (b) Find v- {i). Note: All capacitors are ini

(b)

tially uncharged. Why?

Figure P7.41 CHECKS: 6 pF, 66 ^F

If if if if c, c.

+

4 2 .(a) Find the indicated equivalent capaci tance for the circuit o f Figure P7.42a assuming Cj = 48 pF, Cj = 16 |jF,

c,

c

=

20 )jF, Q = 80 pF, and C5 = 8 pF. (b) Repeat for Figure P7.42b assuming Cj =

Figure P7.44

3 pF, C2 = 6 pF, C3 = 3.6 pF, Q = 6 pF, C5 = 4.5 pF, Q = 48 pF,

= 48 pF, Cg

= 24 pF, Cg = 24 pF.

4 5 . Three 12 pF capacitors are available for interconnection. List all equivalent capaci tances obtainable over all possible interconnec tions of these capacitors.

O ’

oc

C -L

c.

C,

(a)

4 6 . This is a conceptual problem and requires no calculations for the answer. Consider cir cuits 1 and 2 o f Figure P7.46. All capacitors are 1 F except the one that is labeled C. We wish to determine the relationship between S ' and

êql

presence o f the finite positive C F

capacitor between points a and b. Which of the following statements is true?

(b) Figure P7.42

^ êq2 ' (W

êq\ - êql43. Find

for the circuit of Figure P 7.43, (a)

when the switch is open, and (b) when the

(d) There is no general relationship between and

Any relationship depends

switch is closed, assuming that Cj = C4 = 12 pF,

on the value of C.

C2 = C5 = 40 pF, C3 = Cg = 2 0 pF. (c) Repeat parts (a) and (b) for Cj = 12 pF, Cj

Explain your reasoning.

= 40 pF, C3 = 20 pF, C4 = 4 0 pF, C5 = 20 pF, Cg = 100 pF.

318


ages at f = 0 “ are zero, find all three voltages,

I’Cjit), i = 1,2,3, for r > 0. Then find the instan taneous stored energy at / = 0.05 sec.

v jt)

c,

©

V,

Figure P7.48 C H EC K ;

= 6(1 -

V

49. In the circuit o f Figure P7.49, suppose Cj = 3 pF, C3 = 0.5 pF, C2 = 1.5 pF, and

=

10sin(400r) mV for f > 0 and 0 otherwise. (a) Find v^^{t) and v^ît) for ^ > 0. (b) Find an expression for the energy stored in C] and C , over the interval [0 , /]. Circuit 2

Figure P7.46 ANSWMR: C ;„ <

v jt)

©

47. In Figures P7.47a and b, the charge on Cp and C3 is Q = 72 X

C. hi Figure

P7.47a, the voltages on C ,, C , and

Figure P7.49

are 2 V,

3 V, and 4 V, respectively; and in Figure 7.47b

Vq = A \ while the charges on Cp C , and

50. In the circuit o f Figure P7.50, suppose C,

are Qj = 48 x 10“^ C, Q 2 = ^0 ^ ^0“^ C, and Q 3 = 72 X C, respectively.

= 5 mF, Cj = 20 mF,

(a) Find

for the circuit o f Figure P7.47a.

(b) Find

for the circuit o f Figure P7.47b.

= 4 mF, Q = 80 mF,

= 10 0 e~^^ V for f > 0 and 0 otherwise. (a) Find

and

for r > 0 .

(b) Compute the energy stored in the C-, over the interval [0 , t].

/

(b)

(a)

ANSWHK: (a)

Figure P7.47 = 8 mF

48. In the circuit o f Figure P7.48, C| = 6 mF, C2 = 12 mF, C3 = 36 mF, v J t ) = 20(1 V for t> 0 and 0 otherwise. If all capacitor volt-

v,„(t)

Figure P7.50

51 . In the circuit o f Figure P7.51, suppose Cj = 4 mF, C-) = 80 mF, C3 = 20 mF, and 1OO^*”^^ mA for t > 0 and 0 otherwise.

=

(a) Find /q(^) and for t > i). (c) Compute the energy stored in C-, over the interval [0 , t].

31')


the green left turn signal. The interesting variation is that the v-i inductor relationship is different for time-var\'ing inductances:

WYiLit)

v^(/) = -

at

The following highly simplified circuit illus

Figure P7.51

trates the principle o f operation, although the

M ISCELLAN EO U S 52. Find and sketch

configuration and values may not be what are for 0 < r < 4 sec for

actually used. Consider the circuit o f Figure

the circuit o f Figure P7.52a, assuming all

P7.54a consisting o f an inductor driven by a

capacitors are initially at rest for the excitation

current source. When the car with its steel

o f Figure P7.52b. Are any o f the capacitors

frame moves over the coil o f wire, the induc

redundant as far as

tance o f the coil changes from I , to some larg-

is concerned?

er vâlue 0.5 F

0.2

+ pkv,

y+\

__ ^ . 4 f^ . 2

-_ y f

3L^ as illustrated in Figure

P7.54b, where the time

depends on the

speed at which the car is slowing down and

i/y

0.4 F

1-^4

2 T f)

(a)

Plot vît) for t >0 assuming ij{t) =

V ,(t)

a con

stant value, and that the front edge o f the car begins to cross the first edge o f the coil at t = 0 . Explain how this voltage signal might be u.sed to control the traffic light.

6

Figure P7.52

0.5H

(a)

C H EC K : v,„,,(/) = :^ v ,(/ ) 53. Using the circuit given in Figure 7.32, select a capacitor value to filter the voltage for a regulator requiring 14 V < Use

< 20 V.

= 20 cos(2007tf) and /^(f) = 2 A.

C H E C K : C > 1.667 mR 54. When driving a car into a left-hand turn lane, one often sees a large circular or hexago nal cut in the concrete. Embedded in these cuts is a coil o f wire. When your car (contain ing a large percentage o f iron) passes over this coil, its inductance changes. This change o f

Time (in sec)

inductance can be used as a sensor to activate a

(b)

circuit that stops oncoming traffic and lights

Figure P7.54

L(t)

C

H

A

P

T

E

R

First Order RL and RC Circuits

When watching a manufocturing process, a visitor might see a pair of robotic arms assemble an engine or machine a block o f metal with perfectly timed maneuvers. Timing is a critical aspect of a manufac turing process. In T V transmitters there is a signal called the raster, which is critical to the generation of the screen image. In an oscilloscope a timing signal called a horizontal sweep acts as a time base, which allows one to view measured input signals as a func tion o f time. All these applications utilize a signal hav ing sawtooth shape and called a linear voltage sweep. The linear voltage sweep is nicknamed the sawtooth, rhis sawtooth is pictured here together with an approximating exponential curv'e for comparison. Linear Sweep or Sawtooth Waveform

Exponential Approximation Ideally, the sawtooth voltage increases linearly with time until reaching a threshold where it imme diately drops to zero, which reinitiates the process. The threshold voltage corresponds to a fixed unit o f time. The linear voltage increase then acts as an electronic second hand, ticking o ff the

322

Chapter 8 • First Order RL and RC Circuits

smaller units o f time. In practice, the linear increase in voltage is approximated by the “linear” part o f an exponential response o f an RC circuit. ^X1ên the voltage across the capacitor reaches a cer tain threshold, an electronic switch changes the equivalent circuit seen by the capacitor, allowing the capacitor to discharge ver)^ quickly, i.e., the capacitor voltage drops to zero almost instanta neously. Once the voltage nears zero, the electronic switch reinstates the earlier circuit structure, causing the capacitor to charge up again. The process repeats itself indefinitely.

CHAPTER O BJECTIVES 1.

Explore the use o f a constant-coefficient first-order linear differential equation as a model

2.

for first-order RL and RC circuits. Derive from the differennal equation model, the exponential response form (voltage or

4.

current) o f first-order RL and RC circuits without sources and with constant excitations. Interpret the solution form o f the differential equation model in terms o f the circuit time constant and the initial and final values o f the capacitor voltage or inductor current. Develop techniques to handle s\vitching and piecewise constant excitations within first-

5.

order RL and RC circuits. Investigate waveform generation and RC op amp circuits.

3.

SECTIO N HEADIN GS 1. 2.

Introduction Some Mathematical Preliminaries

3.

Source-Free or 2^ro-Input Response

4. D C or Step Response o f First-Order Circuits 5. Superposition and Linearity 6 Response Classifications 7. Further Points o f Analysis and Theory 8 . First-Order RC Op Amp Circuits 9. Summary 10. Terms and Concepts 11.

Problems

1. IN TRO D U CTIO N Our study prior to Chapter 7 focused exclusively on resistive circuits. Recall that all nodal equa tions and loop equations for resistive circuits lead to (algebraic) matrix equations whose solution yields node voltages and loop currents, respectively. Chapter 7 then introduced the capacitor and the inductor. Interconnections o f sources, resistors, capacitors, and inductors lead to new and fas cinating circuit behaviors. How? Inductors and capacitors have differential or integral voltage-cur rent relationships. Interconnecting resistors and capacitors or resistors and inductors leads to cir cuits that must satisfy both algebraic (KVL, KCL, and Ohm’s law) and differential or integral rela tionships for L and C values. When only one inductor or one capacitor is present along with resis tors and sources, these relationships lead to first-order RL and RC circuits. When the sources are


dc, such circuits have vohages and currents o f the form A + Be~^ for constants A, B, and X. The main purpose o f this chapter is to develop techniques for computing the exponential responses o f first-order RC and RL circuits driven by dc sources. A simple example serves to explain some of these points. In the series RC circuit o f Figure 8.1, suppose an initial voltage

is present on the capacitor,

where 0~ designates the instant immediately before zero. Often vve distinguish among 0“, 0, and O'*' when switching occurs or when discontinuities o f excitation functions occur at r = 0.

R

v(t)

+

©

v,(t)

FIG URE 8.1 Series/?Ccircuit. A loop equation for the series RC circuit leads to

vp) = Ri(ir) + Since iciO = C —

(8.1)

equation 8.1 becomes v,(/) =

at

+Vc(/)

Dividing through by RC yields the constant-coefficient first-order linear diflferential equation ( 8 .2 )

dt

RC

RC

subject to the initial condition Vf^Qr). This equation says that the derivative o f the capacitor volt age plus MRC times the capacitor voltage equals MRC times the source voltage. The equation enforces constraints on the capacitor voltage, its derivative, and the source voltage, and is differ ent from the algebraic node or loop equations studied earlier. The terminology first-order differ ential equation applies because only the first derivative appears. Equation 8.2 is linear because it comes from a linear circuit. Our goal is to find capacitor voltage waveforms that satisfy the con straints imposed by the differential equation 8.2.

Exercise. For the circuit o f Figure 8.1, show that the capacitor current i({t) satisfies a differential equation o f the form

cti.it) dt

1 .

RC

1 dv,{t)

R

dt

32 4


Our scope in this ciiaptcr is limited to circuits containing one inductor or one capacitor— equiv alently, first-order RL or RC circuits. W ithin this category we further constrain our investigation to circuits with no sources but nonzero initial conditions, circuits driven by constant (dc) sources, circuits driven by piecewise constant sources, and circuits containing switches. First-order circuits driven by arbitrary source excitations are covered in later chapters using the Laplace transform method.

2. SOM E M ATHEM ATICAL PRELIM INARIES Ver)' often our interest is in source excitations such as v^{t) = 2e~^’ V for /> 0 and 0 otherwise. To conveniently represent such time-restricted waveforms, we define a signal called the unit step function, denoted by u{t), as «(/) =

1 0

/ < ()

The unit step function is a universally used function and will appear many times in the remain der o f this text. MATLAB code for specifying the step function is function f = ustep(t) t = t + le-12; f = (sign(t)+l)*0.5; With the unit step so defined, v^{t) - 2e~"'u(t) V, and both relations are plotted in Figure 8.2.

F IG U R E 8 .2 Unit step function and v^{t) = 2 e

V.

32S

Chapter 8 * First Order RI. and liC Circuits

Further, if v<^t) = l e

for t>t^ and 0 for r < /q, then v^{t) = lOf’

r^) would be the prop

er representation because the shifted unit step function, //(/- /q), means

Plots o f v^{t) = 2e~^‘u{t - ^q) and u{t - t^) are given in Figure 8.3 for

= 0.5.

FIGURH 8 . 3 . Plots of u{t - 0.5) and v^{t) = 2e ^'u{t - 0.5).

Exercise. Plot //(—/) and «(/q — t). Hint: For what values o f t are the functions zero and for what values are they 1? A working model ot a physical system underlies an engineer’s ability to methodically anai)'ze, design, or modify its behavior. Linear circuits are physical systems that have differential equation models. The RL and RC circuits investigated in this chapter have differential equation models o f the form

dx{t) dt

.

= > ..V (0 + /(/ ).

-Y(/(,) = .Vo

(8.3a)

or, equivalently.

dxU) dt

(8.3b)

valid for t> /q, where a-(/q) = is the initial condition on the differential equations 8.3. T he term J{t) denotes a forcing function. Usually, y(r) is a linear function o f the input excitations to the cir cuit.

326

Chapter 8 • First Order RI. and RC Circuits

Before proceeding, it is appropriate to explore the intuitive nature o f a differential equation. Equations 8.3 are first-order constant-coefficient linear differential equations. They are first order because o f the presence o f only the first derivative o f some unknown function x{t). For example, in equation 8.3a the derivative o f x{t) equals a constant X times x{t) plus a known forcing func tion y(r), w h e r e in c o r p o r a te s the effect o f all the circuit excitations. Rigorously speaking, “lin ear” means that under the assumption o f zero initial conditions, if the pairs o f voltage waveforms (/j(f), A'j(r)) and (fjit), X2 it)) each satisf)' equations 8.3, then for any scalars a^ and a^, the pair

{a/ît) + ajfjU),

+ ajXjU))

also satisfies equations 8.3. The parameter X, denotes a riaturalfrequency o f the circuit. Natural frequencies are natural modes o f oscillation such as, for example, in the ringing o f a bell. For physical objects natural frequen cies are called natural modes o f vibration. All physical objects have a vibrational motion even though it may be imperceptible. Knowledge o f these modes is important for the safety and reliabilit)' o f large buildings and bridges. For example, the Tacoma Narrows Bridge had natural modes o f vibration that the wind excited. Undulations in the wind intensit}' resonated with the natural vibrations of the bridge, causing a swaying motion to increase without bound until the bridge col lapsed. In circuits, the natural modes o f oscillation are reflected in the shapes o f the voltage and current w'aveforms the circuit produces. A more thorough and mathematical discussion o f the notion o f natural frequency will take place in the next chapter, when we study second-order {RLQ circuits. Let us return to the goal of finding a solution to the differential equations 8.3. The solution to equations 8.3 (a derivation will appear shortly) for t >tQ has the form

( c ^ ' - " /( x U k Jk)

(8.4)

This means that the expression on the right-hand side o f the equal sign (1) satisfies the differen tial equations 8.3 [its derivative equals K times itself plus/r)], and (2) it satisfies the correct ini tial condition, xit^) = x^^. A simple example illustrates this point.

E XA M PLE 8.1. Compute and verify the solution o f equation 8.3a using equation 8.4. SO L U T IO N Suppose in equation 8.3a., J{t) = u{t - 1), a shifted unit step function, X = - 1 , 10, in which case cit

From equation 8.4, for ^ > 1, ,v(/) =

\)(k=

+ I

= 1, and at(1) =

32'

Chapter 8 • First Order Rl. and RC Circuits

To verify that [9^

+ 1] does indeed satisfy the differential equation, observe that for ? > 1,

dx{t) _ d dt

+ I = -,v(/) + 1

= -9 e

~~dt

1] = 10, which is the mandatory initial condition. Thus, x{t) = 9^

Further, at ^ = 1, [9f’

+ 1 is a valid solution for r > 1. Example 8.1 spells out the application o f the solution (equation 8.4) to the differential equation 8.3a. It also verifies that the computed solution satisfies the differential equation and the proper initial condition. Although not shown, equation 8.4 also satisfies equation 8.3b. A formal deriva tion o f the solution o f equation 8.4 requires the use o f the integrating factor method, the sub ject o f a differential equations course. Briefly, the first step o f this method entails multiplying both sides o f equation 8.3a or 8.3b by a so-called integrating factor e~^. For equation 8.3b, this results in

e

dx(t) _>j — ------ h e x i l ) = e dt

f/,\ f{ t )

(8.5)

By the product rule for differentiation, the sum on the left equals

d_ dt

e""x{t)

in which case equation 8.5 becomes ( 8 .6)

dt One can integrate both sides o f equation

f -dr J'o

e

8 .6

from

Tq

to t as follows:

dr = e (8.7)

JI q Bringing the term e'^‘Ox{tQ) to the right-hand side o f equation 8.7 and multiplying through by results in the solution to the differential equation 8.3a or 8.3b, given by equation 8.4. This completes the derivation o f the very powerful formula o f equation 8.4. There are four points to remember about the preceding discussion: (1) circuits have behaviors modeled by differential equations such as equations 8.3; (2) the solution to a first-order differen tial equation is a waveform (also called a signal or response) satisfying equation 8.4; (3) the for mula o f equation 8.4 works for all continuous and piecewise continuous time functionsy(/); and (4) a solution to a differential equation means that the waveform satisfies the given differential equation with the proper initial condition.

328

Chapter 8 • Hirst Order RL and RC Circuits

Exercise. Show that the hinction .v(/) = (1 — 0. ôr r > 0, is a solution to the difFerential equation ^

dt

= -.v(/) + //(/) with initial condition a-(0) = 0 by showing that x{t) satisfies the difFerential

equation and has the proper initial condition at r= 0.

3. SOURCE-FREE OR ZERO-IN PUT RESPONSE Figure 8.4 depicts the most basic (undriven or source-free) RL or RC circuit: a parallel connection of a resistor with an inductor or a capacitor without a source. In these circuits, one assumes the pres ence of an initial inductor current or initial capacitor voltage. The complication introduced by a volt age or current source is taken up later. Once the source-free or zero-input behavior is understood, one can understand more easily the responses resulting from constant source excitations. i,(t) >r

+ ^

yf

+

v^(t)

R ^

ijt )

\ fi)

+

^R(t) < -

L

S v jt )

*

■

(a)

(b) FIGURK 8.4

Our first goal is to derive differential equation models for the RL and RC circuits o f Figures 8.4a and 8.4b, respectively. We do this in parallel. (1) At the top node o f Figure 8.4a, KCL implies

(1) Similarly, for Figure 8.4b, KVL implies

= v^t)

ifiU) = -iiit) (2) However,

(2) However,

Vi it)

.d\'c{t) n it ) — R icit) ——RCdt

L d ii U)

(3) Making the obvious substitution and

(3) Making the obvious substitution and

multiplying by R/L yields the differential

dividing by RC yields the differential

equation model

equation model

dii it)

,

with //(/()) a given initial condition.

dV( {t) _

(8.8a)

dr with

1

RC

v'c(/)

a given initial condition.

(8-81’ )


Both differential equation models have the same general form,

^ dt

= hc{t) = - - m

(8.9)

X

i.e., the derivative of x(t) is a constant, X = -1/x, times itself. Applying equation 8.4 to equation 8.9 implies that both equations 8 .8a and 8 .8 b have solutions given by

where x is a special constant called the time constant o f the circuit. Equation 8.10 means that the responses for W

W

(q o f the undriven JiL and /?C circuits are, respectively, given by

1^(0 =e

-^ '-'0 )

^

vc(0 = e

- ■ ^ ' - ‘0)

(8.11) vc(/q)

L

where the time constant o f the RL circuit is T = — and the time constant o f the RC circuit is x

= RC.

^

The time constant of the circuit is the time it takes for the source-free circuit response to drop to

e~^ = 0.368 o f its initial value. Roughly speaking, the response value must drop to a little over onethird of its initial value. This is a good rule of thumb for approximate calculations involving decaying exponentials. The mathematics that underlie the solution to the differential equation 8.9 given in equation 8.10 is nothing more than elementary calculus. To see this, consider the exponential solution form

(8. 12) where K 'ls an arbitrary constant. The fiinaion

has the property that its derivative is----- e~' ^

This is precisely what equation 8.9 requires. Therefore equation 8.12 satisfies the differ-

^

ential equation 8.9 and is said to be a solution. To completely specify x{t) it only remains to iden tify the proper value of K from the initial condition. Evaluating x(r) at ^

yields

Mt„) = /Cf-Vr in which case

Substituting this value o f i n t o equation 8.12 produces the solution given in equation 8.10, which is adapted to specific RL and R C circuits in equations 8 . 11. Figure 8.5 plots equation 8.12 for arbitrary K and x > 0 . This plot proves instructive for understanding how the response decays as a function o f the time constant.

330

Chapter 8 • First O rder RL and RC Circuits

Time

FIGURI: 8.5 Plot o f equation 8.12. For f = x, one time constant,

decays to 0.368

of its maximum value. In summary, the circuits o f Figure 8.4 motivate the development o f the rudimentary machinery for constructing solutions to undriven RL and RC circuits. For more general circuits, those con taining multiple resistors and dependent sources, it is necessary to use the Thevcnin equivalent resistance seen by the inductor or capacitor in placc o f the R in equation 8.11. Figure 8.6 illus trates this idea.

f

Linear

\ >f

Resistive Circuit

i,(t)

f

\ Linear Resistive Circuit

+

L p

No Sources

-N V jt) No Sources

C

i,(t) + v,(t)

F I G U R I : 8 .6 R ep la c em en t o f “resistive” part o f c irc u it by its T h e v e n in eq u iv alen t.


331

These facts imply that the general formulas for computing the responses o f undriven RL and RC circuits have the structures

(8.13)

VrU) = e

The difference between equations 8.11 and 8.13 is that in equations 8.13 R^f^ is the Thevenin equivalent resistance seen by the inductor or capacitor.

EXA M PLE 8.2 For the circuit o f Figure 8.7, find i^{t) and v^(t) for r > 0 given that S closes at r = 0.4 sec. Then compute the energy dissipated in the 5 val [0.4, co).

= 10 A and the switch resistor over the time inter

t = 0.4sec S

20Q 5Q

+ v,(t)

8H

FIGURE 8.7 Parallel RL circuit containing a switch. S

o l u t io n

Step 1. With switch S open, compute the response for 0 < /< 0.4 sec. From the continuity property o f the inductor current, /^(O'*’) = ^^(0“) = 10 A. Using equation 8.13,

'1 ( 0 = c

A

We note that //(0.4) = 3.679 A. Step 2. With switch S closed, compute the response for t > 0.4 sec. For this time interval the Thevenin equivalent resistance seen by the inductor is

= 20||5 = 4 Q, i.e., the equivalent resistance o f a par

allel 20 Q and 5 ^ combination. According to equation 8.13, the response for t>tQ = 0.4 sec is

,- ,( 0 = /

t

Step 3. Write the complete response as a single expression using step fitnctions:

i^{t) = 10^>-2-5^[«(^) - «(/- 0.4)] +

0.4)A

(8.14)

Chapter 8 * First Order RL and R C Circuits

332

Step 4. Plot the complete response. To plot this using MATLAB, we use the following m-file along with the code given earlier for the unit step function: »t = 0:0.005:1.4; »iL= 10*exp(-2.5*t) .* (ustep(t).* ustep(t - 0.4)) + 3.679*exp(-0.5*(t-0.4)) .* ustep(t - 0.4); »plot(t,iL) »grid Using this code, Figure 8.8 illustrates the complete response, showing the two different time con stants. The 0.4 sec time constant has a much faster rate o f decay than the lengthy 2 sec time con stant.

a E <

Time (seconds) FIGURE 8.8 Sketch of response i^{t) for Example 8.2.

Step 5. Compute i^^(r). It is a simple matter now to compute v^^t) since vi^t) = In particular, t'^(0'*') = -200 V. Hence for 0 < t< 0.4

.IL l

For t> 0.4, however, the circuit structure changes and = 14.716 V. Thus,

yi(f) = e

(0 .4 "') =

4

in which case i'^(0.4'^) = 4 x 3 .679

-

1

4

.

7

1

V


Step 6. Compute the energy dissipated in the 5 O. resistor over the interval [0.4, oo). The power absorbed by the 5

resistor for 0.4 < / is v’i ( 0 5

5

= 43.31

The energ)' dissipated over [0.4, oo) is given by IVjn ( 0 .4 ,cc) =

PfaU/)A/ = 4 3 .3 12 j;^ ^

= 4 3 .3 12 J

Exercises. I. Plot vît) using the above m-file, ustep, and the appropriate .code. 2. Repeat the calculations o f Example 8.2 with the 8 H inductor changed to 8 mH and a switch closing time o f 0.4 ms. A N SW ER: ij{t) = l{)r-“^'^"'//(f)//(0.4 x lO"-^ - r) +

- 0.4 x Ur-^) A

E X A M PLE 8.3 Find V(^t) for r > 0 for the circuit o f Figure 8.9 given that y^^Ô) = 9 V.

So l u t io n Because there is a switch that changes position at r = 1 sec, there are two time intervals to consider. Step 1. Compute the response forO < t < 1. Over this time interv\il, the equivalent circuit is a par allel /?C circuit, as shown in Figure 8.10a.


334

0.1 F

0.1 F 80

3Q

V jt)

+ V jt)

(a)

(b)

FIG URE 8.10 Equivalent circuits for Figure 8.9: (a) 0 :s r < 1 and (b) 1 s t.

) = 9 V. Therefore from equation 8.11,

By the continuity o f the capacitor voltage, 1

------- 1 v c(t)= e Vc(O^) =9^’- ’ “5^V Step 2. Compute the response for r > 1. Figure 8.10b depicts the pertinent equivalent circuit. Observe that

= 2.58 V and

=3

Again by equation 8.11, for r > r-I

-St-to) V c (0 = f

= 1,

v .c (^ ) = 2 .5 8 e

V

Step 3. Use step functions to specify the complete response. By using the shifted unit step function, the two expressions obtained previously can be combined into a single expression: t-\

V cir)=9e~^-'^'[u{t)-u{f-\)] + 2.5Se

- \) V

Step 4. Obtain a plot o f the response. Using MATLAB and code similar to that used in Example 8.2, the plot in Figure 8 . 1 1 w'as obtained. Here the part o f the response with the 0.3 sec time con stant shows a greater rate o f decay than the longer 0.8 sec time constant.

Time (seconds) F IG U R E 8 .1 1 Response,

for the circuit o f Figure 8.9 .

Chapter 8 • First Order RI, and RC Circuits

Exercises. 1. Show that ti{t) - u{t - 1) =

- t)-

2. Suppose that in Example 8.3 the switch moves to the 4.5 ^ resistor at r = 0.5 sec instead oF 1 sec. Compute the vakie V(\t) at f = 1.2 sec. ANSW ER: 0.4671 V

For all o f these examples x > 0 and the response is a decaying exponential. Intuitively, the response decays because the resistor dissipates as heat the energy initially stored in the inductor or capaci tor. One o f the homework exercises will ask the student to show that the total energ)' dissipated in the resistor from

to oo equals the decrease in energ)^ initially stored in the inductor or capac

itor at ^Q. When controlled sources are present,

may be negative, in which case x < 0. Here the

negative resistance supplies energ)' to the circuit and the source-free response will grow exponen tially. This is illustrated in the next example. E X A M PLE 8 .4 Find Vf^t) for the circuit o f Figure 8.12, assuming that^^^^ = 0.75 S and i^(;(0“) = 10 V.

—

------ o -----q V

0.25F

N

:

^

4Q

+ v,(t)

<

0 -----0.25F

-2 Q

v,(t)

-o-

-o-

b

b

FIGURE 8.12 Parallel /?Ccircuit with dependent currcnt sourcc.

So l u t io n It is straightforward to show that theThevenin equivalent seen by the capacitor is a negative resist ance,

- -2 Q, as shown in Figure 8.12b. Again, by equation 8.11, v cit) = e

vc{0^)=\0e^'u{t) V

Because o f the negative resistance, this response grows exponentially, as shown in Figure 8.13. A circuit having a response that increases without bound is said to be unstable. Practically speaking, an unstable circuit will destroy itself or exhibit a nonlinear phenomenon that clamps the voltage at a finite value, as in the case o f saturation in an op amp.


336

Tim e (seconds) FIGURE 8.13 Plot of unbounded voltage response due to presence of negative resistance. Circuits with such responses arc said to be unstable.

Exercises. 1. For Example 8.3 show that icU ) = e 2. in Example 8.3, f > 0. ANSWI-.l^S: 8 £2.

ic{0-^) = 5e-'ii{t)A

= 0.125 S. Find the equivalent resistance seen by rhc capacitor and V

3. Show that in general, for / >

the form o f the capacitor current is similar to the voltage form.

Hint: Apply the capacitor v-i relationship to equation 8.11.

4. DC OR STEP RESPONSE OF FIRST-ORDER CIRCU ITS The circuits o f the previous section had no source excitations. This section takes up the calcula tion o f voltage and current responses when constant-voltage or constant-current sources are pres ent. It is instructive to start with the basic series RL and RC circuits as shown in Figure 8.14.


O

Linear Resistive Circuit with

i^Ct)

Linear Resistive Circuit with

Constant Sources

Constant Sources

O

(a)

(b)

+

\{t)

FIGURE 8.14 (a) Driven first-order RL circuit, (b) Driven first-order RC circuit, (c) Thevenin equivalent representation oF (a), (d) Thevenin equivalent representation of (b). Given these basic circuit representations and initial conditions at Tq, what is the structure o f a dif ferential equation mode! that governs their voltage and current behavior for t>tQ' The first objec tive is to derive the “differential equation” models characterizing each circuits voltage and current responses. It is convenient to use ij{t) as the desired response for constructing the differential equation for the series RL circuit (Figure 8 .l4 c ), whereas for the series RC circuit (Figure 8 .l4 d ), is the more convenient variable. (1) The circuit mode! for the inductor is vl O )= L

(i) The circuit mode! for the capacitor is

dv(^(t)

dilit) ic(t)= C

dt

dt

(ii) By KCL and Ohms law,

(ii) By KVL and O hm s law,

ic (t) -

(iii) Substituting for ;^(^) leads to the differential equation model

d iijt) dt

^

R.

(iii) Substituting for /^r) leads to the differentia! equation model

dvc(r) +

(8-> 5a)

^’oc R,ill

dt

_ -^ ^ 'C (0 + -

—

v o c (8 .1 5 b )


338

initial condition V(\t^ ) =

with initial condition //(ô") = is constant (not impulsive).

since

is constant (not impulsive).

Exercise. Construct differential equation models for the parallel RL and RC circuits o f Figure 8 . 15 . Note that these circuits are Norton equivalents o f those in Figure 8.14a and Figure 8.14b. Again choose /^(r) as the response for the RL circuit and v^^t) as the response for the RC circuit.

(constant)

(constant)

FIG URE 8.15 Driven RL and RC parallel circuits.

dll (/) .\NSWI-RS: — V di

V/( I.

R

//(/)+ — L

and ^ -■ ■ = ------^— ''r ( n + — d! R,i,C ^ C

A simple application o f basic circuit principles has led to the two differential equation models of equations 8.15. The next important question is: What do these t%vo differential equation models tell us about the behavior o f each circuit? Equivalently, how do we find a solution to the equa tions? Observe that both differential equations 8.15 have the same structure:'

dx(t)

1

dt

X

(8.16a)

for RL circuits and x = R^j^C for RC circuits, and F= v J L for RL

where the time constant T =

circuits and F = vJiRî^j for the RC case. This equation is valid for t>

Equation 8.4, rewritten

here with/^y) = F, presents the general formula for solving the differential equation 8.16a: ) + ( e^^^'-‘f^Fdcj

A-(/)= where A = —

(8.16b)

•'h)

j

is a natural frequenc)' o f the circuit, and where we have emphasized the use o f

T

the initial condition at

Note that as long as x{t) is a capacitor voltage or inductor current, the

initial condition is continuous, i.e., x{t^p = x(fQ+), because F h a . constant (non-impulsive) forc ing function. A straightforward evaluation o f the integral o f equation 8.16b yields t

x(/) = e

^ \v(fo)+ F e

( I-In

f e Jif)

dq = e

^ 'x (t^ )+ F T

Some rearranging o f terms in equation 8.16c produces the desired formula

- e

(8.16c)


339

(t-tn x(

0=

f t

+ U 4 )-

ft

”

(8.17)

= -r

which is valid for t > /q. After some interpretation, this formula will serve as a basis for comput ing the response to RL and RC circuits driven by constant sources. A homework exercise will ask for a different and direct derivation of this formula. At this point it is helpful to interpret the quantity i r in equation 8.17. For RL circuits, when x(r) = /*£(/), equation 8.15a implies that

v J L , x = UR^f^ and hence F l = vJR^f^ =

For RC cir

cuits when ;c(/) = V({t), equation 8.15b implies that F = vJR ^ C , x = Rf^,C, and hence Fz = This interpretation is valid for both positive and negative values of x. If x > 0, then

t-tQ y jr(oo) = lim x(t) = lim

/—*00

t~ * 0 0

Fx + \ x {t^ )-F x y

T

isc =

= Ft =

for RL case

Rih

(8.18) for R C case

This means that for the RL case, /^(oo) =

= vJR^f^ and for the RC case, V({) =

O f course,

is computed by replacing the inductor with a short circuit, and is computed by replacing the capacitor with an open circuit. See Chapter 6 for details. Mathematically, any constant, such as x(/) = constant, that satisfies a differential equation is called an equilibrium state of that dif o

ferential equation. Since the constant x(/) = F i satisfies the differential equation 8.16, / r is an equilibrium state of the differential equation 8.16. Whenever X > 0, equation 8.18 implies that the formula (equation 8.17) for the solution o f equa tion 8.1 6 given constant or dc excitation becomes .

a: ( 0

_ ^-'o (8.19a)

= j :(«>) + U ( ^ ^ ) - J c( oo) e

and when x(^) = /£(/), (t-to )

iL (0 = / z .(“ ) + k ( 4 ) - ' L ( “ ) o

and when x(/) = V({t)y

(8.19b) t-t

vc(0 = vc(00) + Vc(ô )-^ c (°°)

K/.C (8.19c)

Note that x > 0 is true whenever R^j^ > 0, C > 0, and Z, > 0, i.e., the circuit is said to be passive. This allows us to state a nice physical interpretation of equation 8.19a: elapsed time

w w

's . ;

x { t )= [ F in a l value^ + iÎnitial v a lu e ]-[F in a l value\)e ttê constant Graphically, equation 8.19a is depicted in Figure 8 .1 6 for xipo) >

x (/ q ).


340

Elapsed Time

I'iG U RE 8.16 Graphical interpretation o f equation 8.19a for the case xico) > x{tQ).

Exercise. Redo the curve o f Figure 8.16 for the case x{
computed from initial conditions and possibly the value o f the source

excitation, or it can be computed from past excitations up to trate the use o f equation 8.19.

Several examples will now illus

EXA M PLE 8.5 For the circuit o f Figure 8.17, suppose a 10 V unit step excitation is applied at r = 1 when it is found that the inductor current is = 1 A. The 10 V excitation is represented mathematically as = 1 0 « (/ - 1) V for r> 1. Find ij{t) for r > I. R = 5Q

v^(t)

1^(1-)=1 A

F IG U R L 8 .1 7 Driven series R L circuit for Example 8.5 with /^(1“) = 1 A.


.Vtl

S o l u t io n

Step 1. Determine the circuit’s differential equation model. Since the circuit o f Figure 8.17 is a driv en series RL circuit, equation 8 .1 5a implies that the differential equation model o f the circuit valid for r > 1 is

cl'iAt)

R

1

1

10

w here the tim e c o n sta n t t = 0,4 sec.

Step 2. Determine the form o f the response. Since /^(1“ ) =

equation 8.19b implies that

/^(/)=/^(oo)+^/^(r)

^

Here the presence o f u{t - 1) emphasizes that the response is valid only for / > 1. Step 3. Compute iî^X)) and set forth the fin al expression for /^(r). Since x = 0.4 > 0, we replace the inductor in the circuit o f Figure 8.17 with a short circuit to compute i^^. = //(oo) = 2 A. It follows that [2 + (1 -

- 1) = (2 -

- 1) A

Step 4 . Plot iît). One cannot presume that the response is zero for r < 1. Hence, using MATLAB or the equivalent, one can construct the graph o f i^{t) for / > 1 as given in Figure 8.18.

c 0;

3

u O tj 3 ■o c.w

Time (sec) FIGURE 8.18 Plot of /^(r) for Example 8.5. Step 5. Compute vît). Given the expression for the inductor current in step 3, it follows that for

t> 1, cliLit) Vi{t) = Ldt

/-I

u{t -\^ ) = 5 e

* ^ / (r -l'* ')V


342

Exercises. 1. Verify that in Example 8.5 vît) can be obtained without differentiation by

=

V -V iW 2. In Example 8.5, suppose R is changed to 4 Q. Find iît) at r = 2 sec. ANSW ER: 1.8647 A

Note that we have used the differential equation 8.16 (or equations 8.15) to obtain the solution form o f equation 8.19. However, when using equations 8.19, it is not necessary to reconstruct the differential equation of the RL or RC circuit. Specifically, we need only compute xit^), x{cc), and the time constant x = LIRî^ or The method described for computing final values can also be used to find the initial values o f and i^ at f =

if dc excitations have been applied to the circuit for a long time before t = ?q. The

next example illustrates this technique and extends the preceding discussion. EXA M PLE 8.6 The source in the circuit o f Figure 8.19 furnishes a 12 V excitation for / < 0 and a 24 V excita tion for r < 0, denoted by yy^(r) = [\2u{-t) + lAu{t)] V. The switch in the circuit closes at / = 10 sec. First determine the value o f the capacitor voltage at r = 0", which by continuity equals Next determine

for all ^ > 0. R = 6 kn

+ v,(t)

FIGURE 8.19 Switched driven

circuit for Example 8.6.

So l u t io n Step 1. Compute initial capacitor voltage. For r < 0, the 12 V excitation has been applied for a long time. Therefore, at r = 0“ , the capacitor has reached its final value and looks like an open circuit to the source. Hence the entire source voltage o f 12 V appears across the capacitor at r = 0~, i.e.,

^(40~) =

= 12 V by the continuity property o f the capacitor voltage.

Step 2. Use equation 8.19c to obtain vît) fb r Q < t< 10 sec. Equation 8 .1 9c requires only that we know V(iQ*) (step 1), x, and Vf^co). For 0 < t< 10 sec, r = R^C= 3 sec. It is important to realize here that for 0 < r < 10 sec the circuit behaves as if the switch were not present. Hence, the com putation o f v^(co) proceeds as if no switching would take place at r = 10 sec. Here v^^ = v^ôo) = 24 V. Hence, for 0 < ^ < 10, equation 8.19c implies


.vi3

v'cW = ''c(= o ) + [wc(0^)

= 2 4 + (1 2 -2 4 )c^ - ^ = 2 4 - 1 2 ^

Step 3. Compute the initial condition fo r the interval 10 <

V

(8.20)

i.e., t'^^lO'^). Plugging into equation

8.20 and using the continuity property o f the capacitor voltage yields

= 2 4 - 12f>-io/3 = 2 3 .57 V Step 4. Find V(^t) for / > 10. For r > 10, the resistive part o f the circuit can be replaced by its Thevenin equivalent, which yields Figure 8.20. Rtn = 2 k O

+ V(t)

FIGURE 8.20 Circuit equivalent to that of Figure 8.19 for / > 10. Here, equation 8.19c applies again. The value for y^^co), however, is now 8 V and the new time constant is

= 1 sec. Hence, for ; > 10, /-lo

v c (/ )= V c (“ ) + [v c (IO *)-V (-(= c )]e

= 8 + ( 2 3 . 5 7 - 8 ) t '- * '- '" ’ = 8 + I5 .5 7 e ‘ * ' '" ” V

Step 5. Set forth the complete response using step functions. Using step functions, the response V(\() for f > 0 is

v ^ {t)= 2 A -\ le

+ S + \5.51e^'

Step 6. Plot V(it). Plotting i^(^t) yields the graph of Figure 8.21.


3-m

o OJ

O)

Q.

u

Time (seconds)

FIGUllK 8.21 Capacitor voltage

for r > 0 .

Exercise. Suppose the switch in Example 8.6 opens again at r = 20 sec. Find vît) at r = 25 sec. ANSW ER: 20.98 \’

EXA M PLE 8 .7 The circuit o f Figure 8.22a has a capacitor voltage given by the cur\'e in Figure 8.22b. We note that ^fÔ.l) = 7.057 V. Find, Kq, y^Ô), the time constant r = RC, the exact value of j/(j(Q.25), and the value o f C i f /? = 100^2.

------ O-

v.(t)= V „u (t)

v,(t)

-o(a)

345

Chapter 8 • First Order lU. and RC Circuits

01 IB

I Q.

fO U

T im e (s)

(b)

FIGURE 8.22 (a) Scries /?Ccircuit, (b) Capacitor voltage, So

lu t io n

. A simple inspection o f the graph indicates that

V(^{Q) = 2 V. One recalls that /

) - \’(-(0C) e ^

Hencc as / -> oo, v^{t) -> v^co) = 10 V. Since the capacitor looks like an open circuit at / = oo, Vq =

= 10 V. From the given problem data, y ^ 0 .1 ) = 7 .0 5 7 = 10

Simplifying yields -

T =

0.1

/ 1 0 - 7 .0 5 7

= 0.

8 Therefore C = 0.01 F and M 0 .2 5 ) = 9..34.33 V.

When switching occurs frequently, or the excitation changes its constant level frequently, then hand analysis, as in lixample 8.6, becomes very tedious. For such problems a SPIC E simulation (or the equivalent) proves useful and saves time. The next example u.ses SPIC E to compute the waveform o f a simple RC circuit whose input excitation is a square wave. Like the previous exam ple, the solution is broken down into time intervals such that during each time interval inputs are constant. Because no switching occurs, the time constants for all time intervals are the same. In applying equation 8.19c, the quantities that vary from one time interval to the next are the initial values and final values.

346


EXA M PLE 8.8 The first-order RC circuit o f Figure 8.23a is excited by the 50 Hz square wave input voltage o f Figure 8.23b given that the capacitor is initially relaxed. (a) (b)

Plot

for 0 < r < 60 ms, using SPIC E or equivalent software.

Find the initial value and the final value in equation 8.19 when t is very large, for exam ple, at the beginning and end o f the interval 1 < t< 1.01 sec. Plot the v^t) wave for this interval using MATLAB or the equivalent. V (t) (V)

■> t (msec)

10

20

30

(b) FIG U RE 8.23 (a) Series RC circuit excited by the 50 Hz square wave o f (b). So

lu t io n

Part (a) Doing a SPICE or equivalent simulation gives rise to the response curve shown in Figure 8.24, over which the square wave input is superimposed. Observe that the response V(\t) has an approximate tri angular shape. What is happening is that from zero to 10 msec, the circuit sees a step and hence the capacitor voltage rises toward one volt. At 10 n:isec, the square goes to zero for 10 msec. The capacitor then discharges its stored energ)' through the resistor, causing a decrease in its voltage value. The decrease does not go to zero, however. So when the square wave again is at 1 volt the capacitor voltage begins to rise again and achieves a slighdy higher value at f = 30 msec compared to f = 10 msec. In fact, one notices in Figure 8.24 that the peak and minimum values are increasing slighdy as time increases. Eventually the peak and minimum values will reach their respective fixed viilues, c;illed steady-state values. To find these values, a simulation program could require a very lengdiy simulation interv'al, which often proves impraaical. The steady-state values can be computed analytically as in part (b).

10

20

30

40

50

t(m se c)

FIG URE 8.24 Response of circuit of Figure 8.23a calculated using SPICE. For reference, the input square wave excitation is superimposed on the plot.


_____________________________________________________

Part (b) Let Tq = mT, where 7'= 20 msec is the period o f the square wave and m is some large integer. Then, 0.57-

+ 0.571 = 1 + {v^t^) - 1) / Further, in steady state, V(it^ + T) =

= 1+

- 1]^’- ’

(8.21)

which implies that

0.5T + 71 =

+ 0 .5 7 ) ^ ' KC .

+ 0.571^'-' =

(8.22a)

Equivalently, equation 8.22a implies that + 0 .5 7 ) = V(itQ)e^

(8.22b)

Substituting equation 8.22b into equation 8.21 yields

= 1 + [i/c(ro)the solution o f which is

^ "

1+e

= 0-2689 V

It follows that r/J/o + 0 . 5 7 ) = i^(3 = 0 .7 3 1 1 V

An examination o f the response in Figure 8.24 shows that the minimum and peak values are approaching the steady-state values o f 0.26 8 9 V and 0.7311 V, respectively.

Exercise. Based on the response in Figure 8.24, roughly sketch the capacitor current,

At

what time instants is the capacitor current discontinuous?

5. SUPERPO SITION AND LINEARITY Superposition, a special case o f linearity, helps simplify the analysis o f resistive circuits, as discussed in Chapter 5. Recall that linear resistive circuits are interconnections o f resistors and sources, both dependent and independent. Does superposition still apply when capacitors and inductors are added to the circuit? The answer is yes, provided one properly accounts for initial conditions. In order to justify the use of superposition for RC and RL circuits, consider that resistors satisfy O hm s law’, a linear algebraic equation. Capacitors satisfy the differential relationship ; - r -----ir ^ dt

r\ 348


which is also a linear equation. To see linearity in this i-v relationship, suppose voltages

and

V(^ individually excite a relaxed capacitor producing the respective currents :

,■

Let

_^ dvc2

, >c2 - c —

‘a - c

be the current induced by a voltage equal to the sum o f

and

i.e.,

‘C3 = (^-^(^Cl+^C2) However, the linearity o f the derivative implies the property of superposition; .

ic=C-

_ dvf ^ + C — — = ic i + t 2 dt dt

By the same arguments, the current due to the input excitation t/Q - ^\^c\ + On the other hand, suppose two separate currents / q and

^2^Cl

^C3 “ ^l^Cl

individually excite a relaxed capac

itor C Each produces a voltage given by the integral relationship

^Ci (0 =

(”<:) d t , vc2 (0 =

ic2 (T^) ^

By the distributive property of integrals, the combined effect of the input,

+
be a voltage,

vc3 (0 = ^ f_ Ja \ ic \ (T ) + a2ic2('^)] d r

= ai

\ 7 S ' - J c 2 W ‘‘^

Thus linearity and, hence, superposition hold. Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship, and thus superposition is valid, whether the inductor is excited by currents or by voltages. The interconnection of linear capacitors and linear inductors with linear resistors and sources sat isfying KVL and KCL produces linear circuits because KVL and KCL are linear algebraic con straints on the linear element equations. Hence, the property of linearity is maintained, and as a consequence superposition holds for the interconnected circuit. To cap off this discussion we must account for the presence of initial conditions on the capacitors and induaors o f the circuit. For first-order RC and RL circuits, this need is clearly indicated by the first term o f equation 8.4. For a general linear circuit, one can view each initial condition as being set up by an input that shuts off the moment the initial condition is established. Hence the

r-


efFect o f the initial condition can be viewed as the effect o f some input that turns o ff at the time the initial condition is specified. This means that when using superpositioti on a circuit, one first looks at the effect o f each independent source on a circuit having no initial conditions. Then one sets all independent sources to zero and computes the response due to each initial condition with all other initial conditions set to zero. The sum o f all the responses to each o f the independent sources plus the individual initial condition responses yields the complete circuit response, by the principle o f superposition. A rigorous justification o f this principle is given in a later chapter using the Laplace transform method. The following example illustrates the application o f these ideas. E XA M PLE 8.9 The linear circuit o f Figure 8.25 has two source excitations applied at r = 0, as indicated by the presence o f the step functions. The initial condition on the inductor current is

= -1 A.

Compute the response /^(r) for r > 0 using superposition.

So

lu t io n

Because the circuit is linear, having a linear differential equation, superposition is but one o f sev eral methods for obtaining the solution. An alternative approach is to find the Thevenin equiva lent circuit seen by the inductor. As we will see, the superposition approach sometimes has an advantage over the Thevenin approach. Superposition must be carefully applied, however. First one computes the response due only to the initial condition with the sources set to zero. Second, one computes the response due to Vj with all initial conditions and all other sources set to zero. Third, one computes the response due to /j with all initial conditions and all other sources set to zero. Finally, one adds these three responses together to obtain the complete circuit response. Step 1. Compute the part o f the circuit response due only to the initial condition, with all independ

ent sources set to zero. With both sources set to zero, there results the equivalent circuit given by Figure 8.26. The Thevenin equivalent resistance is = 4 Q, resulting from the parallel combi nation o f and R->. Figure 8.26 depicts the equivalent undriven RL circuit having response

352

Oliapccr 8 • First Order RI. and RC Circuits

An approach based on the Thevcnin equivalent circuit seen by the inductor would allow one to quickly compute the complete response, but not in a way that identifies the contributions due to each o f the individual sources. Answêrs to the preceding three questions would have required repeated solutions to the circuit equations. However, if one keeps the source values in literal form, then the Thevenin equivalent approach would be as efficient.

6. RESPONSE CLA SSIFICA TIO N S Having gained some understanding of the form o f the behavior of RL and RC circuits, it is instruc tive to classify the responses into categories. The zero-input response o f a circuit is the response to the initial conditions when all the inputs are set to zero. The zero-state response o f a circuit is the response to a specified input signal or set of input signals given that the initial conditions are all set to zero. By linearit)', the sum o f the zero-input and zero-state responses is the com plete response o f the circuit. This categorization is the convention in advanced linear systems and lin ear control texts. I'Vequently circuits texts include two other notions o f response, the natural response and the forced response. However, decomposition o f the complete response into the sum of a natural and a forced response applies only when the input excitation is (1) dc, (2) real exponential, (3) sinu soidal, or (4) exponentially modulated sinusoidal. Further, the exponent of the input excitation, for example, a'm j{t) =

must be different from that appearing in the zero-input response.

Under these conditions it is possible to define the natural and forced responses as follows: (1) the natural response is that portion of the complete response that has the same exponents as the zeroinput response, and (2) the forced response is that portion of the complete response that has the same “exponent” as the input excitation provided the input excitation has exponents different from that of the zero-input response. This decomposition is important for rwo reasons. First, it agrees with the classical method o f solv ing linear ordinary differential equations with constant coefficients where the natural response cor responds to the com plem entary function and the forced response correspontls to the particular integral. Students fresh from a course in linear differential equations will feel quite at home with these concepts. The second reason is that the forced response is easily calculated for dc inputs. For general systems this type o f decomposition is not used.

7. FURTHER POINTS OF ANALYSIS AND TH EO RY In deriving equations 8.17 and 8 .1 9a, the quantin' x{t) was thought o f as a capacitor voltage or an inductor current. It turns out that any voltage or current in an RC or RL first-order linear circuit with constant input has the form

{I- to) x{l) = X^,+

~ r~

(8.23)

For T negative or positive equation 8.23 is identical to equation 8.17 with Pi = X^. Further, T is the circuit time constant and X is that voltage or current of interest computed under the condi-

Chapter 8 • First Order Rl. ami RC Circuits

3 S3

tion that the inductor is replaced by a short circuit for the RL case or the capacitor is replaced by an open circuit for the RC case. How do we justify the form o f equation 8.23 for all variables? We invoke the linearity theorem o f Chapter 5 and the source substitution theorem o f Chapter 6 o f 2"^ edition. Suppose in a firstorder RC circuit we have found V(\t). By the source substitution theorem, the capacitor can then be replaced by a voltage source whose voltage is the computed V(it). This new circuit consists o f constant independent sources, one independent source o f value

and resistors and depend

ent sources. By linearity, any voltage or current in the circuit has the form x(r) = for appropriate ATj and Kj- By equation 8.17,

which implies that

Mr) = K , + ^6^ for appropriate

"

This is the same form as equation 8.23 for proper choices o f

Exercise. Show that, for t > 0,

and

-X^.

^rid

Note that equation 8.23 requires that the initial value be evaluated at ^

instead o f r =

This

is because only the inductor currents and the capacitor voltages are guaranteed to be continuous from one instant to the next for constant input excitations. The capacitor current and the induc tor voltage as well as other circuit voltages and currents may not behave continuously. E XA M PLE 8 .1 0 This example illustrates the application o f equation 8.23. For the circuit o f Figure 8.29, -18//(-r) +

V. Find i- it) for f > 0.

6kQ

3kQ

2kQ v jt ) 0.5 mP

FIGUFIK 8 .2 9 R C circuit with /■ (/) as the desired response.

=


354

SO L U T IO N Step 1. Compute /y„(0^). To obtain /y„(0*), we first compute V(\Q~) =

Since for r < 0, - 1 8

V has excited the circuit for a long time, the capacitor looks like an open circuit. By voltage divi sion, j = —( - 1 8 ) = - 6 V Thus at r = 0*, the equivalent circuit is as shown in Figure 8.30.

rigurc K.30 Circuit equivalent to that of Figure 8.29 at t = 0"^. Application o f superposition to Figure 8.30 shows that Kj = - 1 .5 V. Flence, 9 - ( - 1 .5 ) 6x10-^

= 1.75 mA

Step 2. Find the circuit time constant and the equilibrium value o f ijj,t). From Figure 8.29, the equivalent resistance seen by the capacitor is

= 4 kl^. Hence, the time constant is t = 2 sec.

Further, since t>
=

9 x 1 0 -'

= 1 mA

Step 3. Apply equation 8.23. Using equation 8.23, we have, for r > 0,

= 1 + 0.75^'

-0.5/

mA

Exercise. In Flxample 8.10, find /Ô ), /(Ô"^), and i^^t) for r > 0 using equation 8.23 directly. AN SW ERS: 0, 2.25 mA. 2.25^’-** "^ mA

Note that in Example 8 .1 0 we used

instead o f /,„(0~) to obtain the correct answer. Some

straightforward arithmetic shows that /y„(0“) = - 2 mA. Since /y„(0'^) = 1.75 niA, the input current is discontinuous at f = 0, unlike the capacitor voltage, which is continuous at f = 0. This empha sizes the need to use x{t^) in equation 8 .23.


3S5

In several o f the examples o f sections 3 and 4, the circuits contain switches that operate at pre scribed time instants. In some electronic circuits, the switch is a semiconductor device whose on/off state is determined by the value o f a controlling voltage somewhere else in the circuit. If the controlling voltage is below a certain level, the electronic switch is off; if the voltage moves above a fixed level, the electronic switch is on. The time it takes for a controlling voltage to rise (or fall) from one level to another is very important because timing is as critical in electronic circuits as is scheduling for large organizations. For first-order linear networks with constant excitations, cal culation o f the time for a voltage or current to rise (or fall) from one level to another is straight forward because all waveforms are exponential functions, as per equation 8.19. The situation is illustrated in Figure 8.31.

c 0) k_ I— D u
4->

o >

FIGURE 8.31 First-order response showing a rise from the voltage or current level A'j to the voltage or current level A'2, for which the elapsed time is h - ty In equation 8.19a, let .v(rj) = and .v(^^) = X , be the two levels o f interest. A straightforward manipulation o f equation 8.19a leads to the elapsed tim e formula for first-order circuits.

h -ri

[^1 --V(oo)

(8.24)

^2 - -v(x)

E X A M PLE 8.11 This example uses the elapsed time formula o f equation 8,24 for the circuit o f Figure 8.32. The switch in this circuit is used to produce r^vo different “final” capacitor voltages. When the switch is open, the final capacitor voltage is 12 V. When the switch closes, the final capacitor voltage, by V-division, changes to 4 V. Thus the switch causes the capacitor to charge and discharge repeat edly. For our purposes we show that the choice o f resistances produces an approximate triangular waveform. For the purposes o f this example, suppose the switch in Figure 8.32 is controlled electronically so that it closes when SNvitchings.

rises to 9 V and opens when V(^ falls to 5 V. Find and plot v^^t) for several


FIGURE 8.32 Switched driven RC circuit used to generate an approximate triangular waveform. So

lu t io n

subsequently opening at ^ = tj^ and closing again at t^tc,

Suppose the switch first closes at f = and so on. For 0 < r <

the time constant T = 3 sec, V(\Q) = 0, and

= 12 V. From equa

tion 8 .1 9c,

t VcU) = 12 1 - e 3

V

From the elapsed time formula o f equation 8.24,

- 0 = 3//? Now, for

19-12^

= 3 X 1.386 = 4 .1 5 9 s

t < tj^, there is a new' time constant x= 1 sec,

= 9, and

= 4. .^gain using

equation 8.19c,

From the elapsed time formula, equation 8.24, //9-4\

- t „ =/n Finally, for the time interval

f

- /n{5) = 1.61 s

\5-4/

= 5, and t'^co) = 12. Using equation 8 .1 9c,

, x= 3 sec,

JJz Itl v c(/) = 1 2 - 7 e

^

and from the elapsed time formula, ^ ///5- 12\

- th =3/n rh e w aveform o f

v^t)

for 0 < / <

9-12/

^ / ; (1\ \3

= 2.54 s

is plotted in Figure 8 .3 3 .

Chapter 8 • 1-irst Order RL and RC Circuits

Capacitor Voltage (V)

From the preceding solurion

= 4.16 sec,

= 4.1 6 + 1.61 = 5.77 sec, and t^ = 5.77 + 2.54 = 8.31

sec. If we proceed to calculate the waveform for t >

the waveform begins to repeat itself, as is

evident Irom Figure 8.33. Practically speaking, the first c)'cie o f a periodic, approximately trian gular waveform occurs in the time intervâl [t^, r j , and the period is

= 8.31 - 4.1 6 = 4.15

sec. Note that the triangular waveform has a frequency I / =

period

1

1

2.5 4 + 1.61

4. 15

= 0.241 Hz

The waveform in figure 8.33 is approximately triangular. This is due the fact that two time con stants, 1 and 3 s, have the same order o f magnitude. If we select the resistances so that the charg ing time constant is much larger than the discharging time constant, then the capacitance voltage waveform will look more like a sawtooth waveform. Sawtooth waveforms arc used to drive the horizontal sweep of the electronic beam in an oscilloscope or a T V picture rube.

8. FIRST-ORDER RC OP AM P CIRCU ITS RC op amp circuits have some singular characteristics that set them apart from standard passive RC and RL t)^pes o f circuits. Specifically, because ol the nature o f the operational amplifier, the time constant o f the circuit will often depend only on some o f the resistances. We present four important examples to illustrate the behavior o f RC op amp circuits.

3^H


EXA M PLE 8 .1 2 Compute the response

for the ideal op amp circuit o f Figure 8.34.

1. FIGURE 8.34 Differentiating op amp circuit. So

lu t io n

Observe that Chapter 4. Also,

by the virtual short-circuit propert)' o f the ideal op amp, as set forth in Hence, from these equalities and the definition o f a

capacitor, = « ,,( ,) = -« / c (0 = - R C ^

=- R C ^

(8.25)

Since the output is a negative constant (user chosen) times the derivative o f the input, the circuit is called a differentiator.

Exercise. Suppose y/„(^) = cos(250^). Find R for the circuit o f Figure 8.34 so that

= sin(250/)

V a n d C = 1 pF. AN SW FR: 4 kH

EXA M PLE 8 .1 3 Compute the response

for the ideal op amp circuit o f Figure 8.35 assuming V(^0~) =

= 0.

1.

FIG U R K 8 .3 5 Integrating op amp circuit.


So

lu t io n

= v-^{t)IR by the virtual short circuit property o f the ideal op amp. Also, i^^t)

Observe rhat

=

3S9

Hence, from these equalities and the integral i>-i relationship o f a capacitor,

(8.26)

Since the output is a negative constant (user chosen) times the integral o f the input, the circuit is called an integrator.

Exercise. Suppose = cos(250r) V. Find R for the circuit o f Figure 8.35 so that sin(250r) V and C = 1 pF.

=

A N SW ER: 4 kD

EXA M PLE 8 .1 4 This example considers the so-called leaky integrator circuit o f Figure 8.36, which contains an ideal op amp. The input for all time is v^{t) =

V. Rj represents the leakage resistance o f the

capacitor. Given C and Rj, the resistance /?| is chosen to achieve a dc gain o f 10. The objective is to compute the response

assuming

= 0 and compare it to a pure integrator having a

gain o f 1. This problem is reconsidered in Chapter 13. +

v jt )

1 FIG U RE 8.36 Leaky integrator op amp circuit in which v^{t) = -5u{t) V. So

lu t io n

Because there is only one capacitor, the circuit o f Figure 8.36 is a first-order linear circuit. Because the inverting terminal o f the op amp is at virtual ground,

and the capacitor sees

an equivalent resistance Rî^ = /?2- Hence x = /?,C = 10 sec > 0, Equation 8.19 implies that


360

(8.27) Because the voltage source v^{t) = 0 for r < 0,

) = -V(i^ ) = 0. For f > 0, v^{t) = - 5 V. Since

the source voltage is constant, the capacitor looks like an open circuit at r = co having final value ..„ „ ( co) = - A ( . 5 ) = 5() V Entering numbers into equation 8.27 yields = 50 + (0

= 5 0 (l

ii{t) V

A plot o f the op amp output voltage appears in Figure 8.37 along with that o f an ideal integrator. O ne observes that the more realistic leaky integrator circuit approximates an ideal integrator only for 0 < r < 0 .1 5 t before the error induced by the feedback resistor R-^ becomes noticeable. Such integrators need to be reinitialized periodically by resetting the capacitor voltage to zero.

OJ Ol ra > 4-' D CL 4-1 D

o

Q. E

< Q. O

Time (s)

FIGURE 8.37 Outpiu voltage of leaky integrator that approximates an ideal integrator. So h r we have assumed an ideal op amp. In practice, the output voltage will saturate at a level determined by the power supply voltage and the specs o f the particular amplifier used. Further, practical op amps have complex models. To evaluate the preceding analysis, Figure 8.38 shows a SPIC E simulation using the standard 741 op amp.

361

Chapter 8 * First Order RL and RC Circuits

Leaky lntegrator-Transient-0

Time (s)

Observe rhar the response approximaces the ideal up to about 0.1 5t = 1.5 sec, which corroborates our analysis using the ideal op amp. Note, however, that the simulation accounts for saturation present in practical op amps but absent from the ideal.

EXA M PLE 8.1 5 In trying to build an inverting amplifier in a laboratory, a student inadvertently reverses the con nection o f the two input terminals, which results in the circuit of Figure 8.39a. Assume a practi cal op amp model with output o f

= 15 V and a finite gain of/I = 10“^. Instead o f seeing the expected

= - 4 V, the student observes a 15 V output. Explain how this 15 V output could

possibly exist. 4kO

Capacitance (a)

(b)

F IG U R E 8.39 (a) Incorrectly connectcd Inverting amplifier, (b) Circuit model, including a stray

capacitancc C = 1 pF and a finite gain A = lO"^.

362


So l u t io n C'hapter 4 op amp models contain only resistors and controlled sources. One way to explain the situation described in this example is to postulate a small stray capacitance, C = 1 pF, across the input terminals. In fact, this is a more accurate circuit model and is shown in Figure 8.38b. This means that the response will be o f the form o f equation 8.23. The first quantit)' to compute is the circuit time constant t =

C. The equivalent resistance

looking to the right of C, is obtained from the circuit o f Figure 8.40.

FIG URE 8.40 Circuit for computing

note the artificial 1 V excitation.

From Figure 8.40 and our knowledge o f constructing Thevenin equivalems,

'

4000

Hence,

R

= 1 = /, 1 -1 0 ^

Observe that in the actual circuit,

Q

is in parallel with 1 k il. Hence, the Thevenin equivalent

resistance seen by C is

R^,, = 1000

= lOOOll ( - 2.5) = - 2 .5 0 6 Q.

The time constant o f the first-order circuit is T = R ^^C =

- 2 .5 0 6

X

10"'^ sec, or -2 .5 0 6 picosecond (psec)

The negative time constant spells instabilit)'. The complete response may be written directly with the use o f equation 8.23, where x{t) =

Suppose a very small noise voltage,

To use equation 8.23, we need aÔ"^) =

=E V, appears across C Then

= lO-le V

and


363

To compute the equilibrium output voltage

Kg =

we open-circuit the capacitor and compute

with C open-circuited / -2 5 \ j = 1 X ------ — 10'’ = -2 5 .0 6 3 V V 1000-2.5/

From equation 8.23, the complete response is

= -2 5 .0 6 3 + (lO'Ê + 2 5 .0 6 3 )e

0.339xl0‘^r

(8.28)

For any small positive initial capacitor voltage E, equation 8.28 implies that the output would increase exponentially, had the op amp been ideal. Because this particular real op amp saturates at 15 V, the output more or less instantaneously saturates at 15 V, the phenomenon observed by the student. Had the initial capacitance voltage been sufficiently negative, -2 5 .0 6 3 £ < ------- j ---- V, 10-^ equation 8.28 implies that v

would saturate at - 1 5 V.

9. SUMMARY This chapter has explored the behavior o f first-order RL and RC circuits (1) without sources for given ICs, (2) for constant excitations (dc), (3) for piccewise constant excitations, and (4) with switching under constant excitations. In general, first-order RL and RC circuits have only one capacitor or one inductor present, although there are special conditions when more than one inductor or capacitor can be present. Our discussion has presumed only one capacitor or one inductor is present in the circuit. Using a first-order constant-coefficient linear differential equation model o f the circuit, the chap ter sets forth rwo t}'pes o f exponential responses, the source-free response and the response when constant independent sources are present. The source-free responses for the RL and RC circuits have the exponential forms

ii(0= e where

^

iiito)

vcit)=e

is the initial condition for the inductor and

itor. For an RC circuit, the time constant x=

the initial condition on the capac

where R^^^ is theThevenin equivalent resistance

seen by the capacitor. For an RL circuit, the time constant T= L/Rî^, where R^^^now is theThevenin equivalent resistance seen by the inductor. When independent sources are present in the circuit, the response o f a first-order RC or RL cir cuit has the general form

I-10 x{t) = ,v(3c) + [,y(/o ) -

lie-

36-i


forT > 0. Stated in words, this Formula is elapsed lime

x{t) =[Final value] +{[Initial value]- [Final value])e provided the time constant

t

> 0. When the time constant

t

<^<>nsvdn{

< 0, then it is necessary to modify

the interpretation as discussed in section 7, with equation 8.23 identifying the form o f any volt age or current in the circuit:

The time constants of- a circuit can be changed by switching within the circuit. By changing time constants in a circuit, one can generate different t)'pes o f waveforms such as the triangular wave form o f Figure 8.32. As mentioned at the beginning o f the chapter, wave shaping is an important application o f circuit design. When inductors, resistors, and capacitors are present in the same cir cuit, many other wave shapes can be generated. RLC circuits arc the topic o f the next chapter and allow even greater freedom in waveform construction. As a final application o f the concepts o f this chapter, we looked at the leaky integrator op amp cir cuit. Integrators are present in a host o f signal processing and control applications. Unfortunately, ideal integrators do not exist in practice. The leaky integrator circuit o f Figure 8.35 provides a rea sonable model o f an ideal integrator.

10. TERM S AND CO N CEPTS Com plete response: sum o f zero-input and zero-state responses. D ifferential equation o f a circuit: equation in which a weighted sum o f derivatives o f an impor tant circuit variable (e.g., a voltage or current) is equated to a weighted sum o f derivatives o f the source excitations to the circuit. D ifferentiator circuit: op amp circuit whose output is a constant times the derivative o f the input. Equilibrium state o f a differential equation: c o n stan t, say x{t) = X^, th at satisfies the differential eq u atio n in the variable A.*(r).

First-order differential equation o f a circuit: difterential equation whose highest derivative is first order. Forced response: that portion o f a complete response that has the same “exponent” as the input excitation, provided the input excitation has exponents diflêrent from that o f the zeroinput response, under the condition that the input excitation is either (1) dc, (2) real exponential, (3) sinusoidal, or (4) exponentially modulated sinusoidal. Integrating factor method: mathematical technique for finding the solution o f a differential equation in which multiplication by the integrating factor e~^’‘ on both sides o f the dif ferential equation leads to a new equation that can be explicitly integrated for a solution. Integrator circuit: op amp circuit whose output is a constant times the integral o f the input. Leaky integrator circuit: op amp circuit having a response approximating an ideal integrator, as described in Example 8.13.


365

Natural iirequency of a circuit: natural mode of “oscillation” of the circuit. For a first-order circuit having a response proportional to it is the coefficient X in the exponent. Natural response: that portion of the complete response that has the same exponents as the zeroinput response. Passive RLC circuit: circuit consisting of resistors, inductors, and capacitors that can only store and/or dissipate energy. Sawtooth waveform: triangular waveform resembling the teeth on a saw blade and typically used to drive the horizontal sweep of the electronic beam in an oscilloscope or a T V picture tube. Source-free response: response of a circuit in which sources are either absent or set to zero. Step response: response, for ^> 0, of a relaxed single-input circuit to a unit step, i.e., a constant excitation of unit amplitude. Stray capacitance: small capacitance always present between a conductor and ground. It usually can be ignored, but as Example 8.14 shows, it can critically affect the response of a cir cuit. Superposition: in linear RC and RL circuits, the complete response is the superposition of the relaxed circuit responses due to each source with all other sources set to zero, plus the responses to each initial condition when all other initial conditions are set to zero and all independent sources are set to zero. Time constant: in a source-free first-order circuit, the time it takes for the circuit response to drop to e~^ = 0.368 of its initial value. Roughly speaking, the response value must drop to a litde over one-third of its initial value or rise to within one-third of its final value. For RL circuits x = LIR^f^ and for RC circuits x =R^/jC. Unit step (unction: function denoted «(/) whose value is 1 for f > 0 and 0 for ^< 0. Unstable response: response whose magnitude increases without bound as t increases. The time constant for first-order circuits is negative for an unstable response. Zero-input response: response in which all sources are set to zero. Zero-state response: response to a specified input signal or set of input signals given that the initial conditions are all set to zero.

w w ^ It happens that all variables in a first-order RL or R C circuit satisfy a differential equation o f the same form. The interpretation o f the solution is somewhat different fi-om what follows. A detailed explanation o f the general solu tion is presented in section 7.


366

Problems

i,(t)

UNDRIVEN RESPONSE WITH GIVEN INITIAL C O N D ITIO N S 1. For the RC circuit o f Figure P8.1, R = 2.5 kiQ and C = 50 |.iF. (a) If V(^Q) = 10 V, find

Plot your

answer for 0 < / < 5x, where x is the cir cuit time constant. (b) If V(\Q) = 10 V, find

with

out differentiating your answer to part (a). Plot your answer for 0 < f < 5x, where x is the circuit time constant. At what time is the energy stored in the capacitor about 1% o f its initial value? (c) Compute ;^(r) for «^^0) = 5 V and y<;^0) = 20 V without doing any further calcu lations, i.e., by using the principle o f lin earity.

Figure P8.2 ♦ 3 . In Figure P8.1, suppose R = 25 kH and

vÔ) = 20 V. (a) Find C so that vÔ.25) = 2.7 0 6 7 V. (b) Given your answer to part (a), find C H EC K ; C = 5

pF

♦ 4 . In Figure P8.2, suppose R = 2.5 kl^ and /^(O) = 20 mA. (a) Find L so that at /, = 1 msec,

=

2.7 0 6 7 mA. (b) Given your answer to part (a), find C H E C K ; Z.= 1.25 H 5. The response o f an undriven parallel RC cir cuit is plotted in Figure P8.5. Find the time

ic(t)

constant of the circuit, at least approximately. If

+

C= 0.25 mF, find R.

. v,(t)

Figure P8.1 2. Consider the RL circuit o f Figure P8.2 in which R = 50 Q and Z, = 0.1 mH. (a) If the energy stored in the inductor at t = 0 is 2 [ij, find /^(O) and

Plot your

answer for 0 < r < 5x, where x is the cir cuit time constant. (b) If the energy stored in the inductor at t = 0 is 2 pj, find Vjit) without differenti

Figure P8.5

ating your answer to part (a). Plot your answer for

0

< t < 5x, where x is the cir

cuit time constant. (c) Repeat part (b) for /j^(0 ) = 50 mA and /^(O) = 250 mA. Hint; What principle makes this a straightforward calculation given your answer to part (a)?

6. In the circuit o f Figure P8.6, suppose /?j = 50

Q., Rj = 200 Q., L = 2 H, /^(O) = 100 mA, and the switch opens at t = 50 msec. (a) Find ij {t) for r ^ 0. Plot your answer in MATLAB for 0 s / ^ 0.1 sec. (b) Find y^(0) and t > 0. Plot your answer in MATLAB for 0 < / < 0.1 sec.


36"

where T is the circuit time constant. (b) Let a = - 1 1 . Compute the equivalent resistance seen by the capacitor. Find y^ f) for

> 0. Plot

for 0 < f < 2 t

where T is the circuit time constant. (c) Find the range o f

(X

for which the time

constant is positive.

Figure P8.6

R. 7. In the circuit o f Figure P8.7, suppose /?j = 5

= 20 V,

kQ, /?2 = 20 k n , C = 50 pF H, and the switch opens at r = 0.4 sec.

R,

(a) Find v^^t) for f > 0 . Plot your answer in

0.25 mF

MATLAB for 0 < f < 4 sec. (b) Find

Figure P8.9

and /^(/), t > 0. Plot your

answer in MATLAB for 0 < r < 4 sec. t=0.4s

C H E C K : (c) a > - 3 10. In the circuit o f Figure P8.10, /?, = 100 Q, = 20 P = 2 0 0 ,1 = 0.5 H, and /'^(O) = 250

i„{t)

+

mA. The switch opens at / = 0.03 sec. Find the

,Vc(t)

Thevenin equivalent resistance seen by the inductor before the switch opens, and then compute /^(r) and

for r > 0.

Figure P8.7 Consider the circuit o f Figure P8.8. (a) Find the value o f

and the initial con

dition /^(O) so that /^(0.05 msec) =

R,

9.197 niA and /^(0.15 msec) = 1.2447 mA.

ijt)

1

(b) Given your answer to parr (a), compute and plot /^(r) for 0 < r < 5x, where I is the circuit time constant.

-


= -2 5 n

ijt )

11. Consider the circuit o f Figure P 8 .ll in

1 kf)

which /?j = 25 ^2, “ 50 Q, and L = 2.5 H. Suppose /^(O) = 2 A.

80 mH

(a) With

CX

= 0.1, compute the Thevenin

equivalent resistance seen by the induc

Figure P8.8

tor; then compute /^(f) for f > 0. Plot in

AN SW ER: (a) 600 Q, 25 mA

M A T I^ B for 0 < r < 5 t, where T is the 9. Consider the circuit of Figure P8.9, in which = 100 Q and /?2 = 50 Q. Let (a) Let a

= 500 mV.

= 7. Compute the equivalent

resistance seen by the capacitor. Find i/^r) for f > 0. Plot

for 0 < ? < 5 t

time constant o f the circuit. (b) W ith a = 0.1, compute (c) Repeat part (a) for a = 0.02. Determine the time, say r,, when the inductor has lost 99% o f its initial stored energy.

Chapter 8 • First Order RL and R C Circuits

368

-

+

t=4RC

A

i,(t)

R.

R

—

1=0 .4R

\4R

R,

Figure P 8 .1 1

Figure P8.14

12. In Figure P8.12, the current source

has

been applied for a long time before the switch opens at r = 0. Find /^(O^) and /^(r) for f > 0 in terms o f /^, R, and L, where is in A, R in and L in H. Sketch /^(r) for 0 < ^ < 4x where x is the circuit time constant for f > 0.

15. In Figure P8.15, the current excitation is

= Vîi{-t) V. Find and V(^t) for ^ > 0 in terms o f R, and C, where R is in Q and C in F. Sketch v^^t) for 0 < f < 3x,

given by

where x is the circuit time constant for f > 0. t=RC

.4R

Figure P8.15

Figure P8.12 13. In Figure P8.13, the current excitation is given by

A. Find /^(O^) and iît)

for r > 0 in terms o f /^, R, and L, where R is in and L in H. Sketch /^(r) for 0 < r < 4x, where X

4R

is the circuit time constant for r > 0.

16. Repeat Problem 14, except find i(^t) for /■> 0.

and

17. Repeat Problem 15, except find

and

i(^t) for r > 0.

RESPONSE OF DRIVEN CIRCUITS 18. Consider the RC circuit o f Figure P8.18 in which R = 10 k n and C = 0.4 niF.

Figure P8.13

(a) If

14. In Figure PS. 14 the voltage source

=

has been applied for a long time before the switch opens at r = 0. Find

and t»^r) for

r > 0 in terms o f V^, R, and C, where

is in V,

R in Q, and C in F. Sketch V(^t) for 0 < r < 3x, where x is the circuit time constant for / > 0.

= 0 and

V, find

U(^t). Plot your answer for 0 < r < 4x, where x is the circuit time constant. (b) II /^(^-(O) = 10 V and = 0, find Plot your answer for 0 < / < 4x, where x is the circuit time constant. Now making use o f linearit)' and its associated properties, compute the indi cated responses without any further cir cuit analysis. (c) If V(^Q) = 10 V and find V(^t).

= 2i)ti(t) V,


(d) If «c<0) = - 2 0 V and i/,.„(») = -I0 « (» ) V, find V(it).

369

20. In Figure P 8.20, = 50 Q , 7?2 = 200 Q, C = 2.5 mF, and the voltage excitation is given by

(e) If i/Ô) = 10 V and v-J^t) = 20«(r) V. find i({t) without differentiating your answer to part (c). Plot your answer for 0 < r < 4 t , where x is the circuit time constant.

where

= -1 0 V

and V^2 = 20 V. (a) Find ^(;^0'*^) and V(it) for / > 0. (b) Sketch v^^t) for 0 < f < 5x, where x is the circuit time constant for r > 0. (c) Identify the zero-input response (f > 0) and the zero-state response (r > 0) for the

yoj

answer computed in part (a).

'.w O

(d) Now compute

for f > 0 assuming

the switch opens at r = 0.2 5 sec. Plot your result ForO < f < 0.5 sec.

Figure P8.18 19. Consider the RL circuit o f Figure P 8.19. Suppose /? = 100 Q, Z = 0.2 H. (a) If /^(O) = 0 and

= 20«(^) V, find

i.

I

'j» (D

Plot your answer for 0 < / < 5t , where x is the circuit time constant. Figure P8.20

(b) If ij{G) = - 5 0 mA and v-J^t) = 0, find /^(r). Plot your answer for 0 < ^ < 5x, where x is the circuit time constant.

21. In Figure P 8.21, R^ = 50 Q , i?2 = 200 Q, £

Now making use o f linearity and its

= 2 H , and the voltage excitation is given by

Kl = -1 0 V

associated properties, compute the indi cated responses without any further cir cuit analysis. (c) If/^(O) = - 5 0 mA and find

= 20«(r) V,

Plot your answer for 0 < r <

5x, where x is the circuit time constant. (d) If/^(O) = 25 mA and y.„(r) = -1 0 « (f) V, find

Plot your answer for 0 < ^ <

5x, where x is the circuit time constant. (e) If /^(O) = - 5 0 mA and v-^{t) = 20«(/) V, find Vj{t) without differentiating your

and 1^2 = 20 V. (a) Find /^(O^) and ijit) for t> 0. (b) Sketch i^{t) for 0 < r < 4x, where x is the circuit time constant for t> 0. (c) Identify the zero-input response (t > 0) and the zero-state response (^ > 0) for the answer computed in part (a). (d) Now compute Vj{t) for ? > 0 assuming the switch opens at ^ = 0,0 4 sec. Plot your result for 0 < f < 0 .2 sec.

answer to part (c). Plot your answer for 0 < ^ < 5x, where x is the circuit time constant.

I' '»
Figure P8.21 Figure PS. 19

22. Consider the RC circuit o f Figure P8.22a in which = V’q «((), where Kq = 100 V. (a) Find

'Vw>

the time constant o f the cir-


370

cuit, Rp and /?2 circuit response v^^t) is given by Figure P8.22b.

n

Assume C = 0.25 mE and C so that the circuit

(b) Find

response v^t) is given by Figure P8.22b. Assume i?2 = 10 n

+ ;vc(t)

'.( s O

r\ (b)

Figure P8.23

(a)

24. In the circuit o f Figure P 8.24, V(^Qr) = 2$ V,

= 50u{t) mA, and

= 25«(/) mA

(a) Find the zero-input response, i.e., the response due only to the initial condi tion. (b) Find

for r > 0 due only to

(c) Find V(^t) for ? > 0 due only to (d) Find the zero-state response. (e) Find the complete response

for t >

r>

0.

Time in seconds

(f) Suppose the initial condition is doubled

(b)

and each independent source is cut in

Figure P8.22

half Find the new complete response 23. Consider the RL circuit o f Figure P 8.23a in which

using linearity.

= Vqu(/), where Vq = 100 V.

(a) If Z = 2 H , find

the circuit time

constant, R^, and /?2 so that the circuit

1 kn

response i^(t) is given by Figure P8.23b. (b) If /?2 = 2 k n , find ij^(0*), R-^, and L so that the circuit response ij{t) is given by Figure P8.23b.

V„u(t)

r>

i,(t)

d)

Figure P8.24 25. In Figure P 8.25 Ri = 2 0 0 Q, R2 = 6 0 0 Q, (a)

T?3 = 650Q , Z = 20 H,

= -1 0 0 « (-r ) +

50u{t) V, and = 5 0 « ( ? - 0.5) V. Compute iît) for ^> 0. Plot your answer using MATLAB or its equivalent for 0 < r < 8x.


371

directly without differentiating your answer to part (a). (c) W hat are the new responses if the value o f each source is doubled?

Figure P8.25 26. Repeat Problem 25, except compute VjjJ) for t>Q. 27. In Figure P 8.27 = 2 0 0 Q, /?2 = 6 0 0 Q, = 8 5 0 a . C = 2.5 mF, v^^{t) = - 5 0 « M + 100«W V, and

Figure P8.30

= - 5 0 « ( f - 5 ) V. Compute

31. The switch in the circuit o f Figure P8.31

V(^t) for t > 0. Plot your answer using

has been open for a long time before it is closed

MATLAB or its equivalent for 0 < ? < 6x.

at t = 0 . Suppose = 6R, R2 = 30i?, R^ = 20R, and = Vqu{T - /). In terms of Vq, R, C, and T = 6RC, (a) Find V(^0~) and t/(;;(0'^). (b) Find the Thevenin equivalent seen by the capacitance for 0 < ^ < T. (c) Using the Thevenin equivalent found in pan (b), find an expression for Vf^t)

Figure P8.27

valid over 0 < t < 7’. 28. Repeat Problem 27, except compute the

(d) Find the expressions for V(iT~) and

capacitor current i^t) for t > 0 . 29. For the circuit o f Figure P 8.29,

=

-2Qu{-t) + 20«(^) V. Find V(iO~) and v^^t) for r > 0. Plot V({t) for 0 < ? < 40 msec.

(e) Find the time constant valid for t> T. (f) Find V(^t) for t> T. (g) Plot

for 0 < r < 4 7 using MATLAB.

20 msec

2kO '. w ( D

8kO

=0 1.6 kn

+ Vc(t). S jjF

o

20 V

+ .Vc(t)

vjt) Figure P8.29 Figure P8.31 30. Consider the circuit o f Figure P 8.30 in

3 2 . For the circuit of Figure P 8.32, -V^ = - 1 0

which /?j = 300j^, /?2 = 800 Cl, R^ - 600 Q, L = 4 H, - -24u{-t) + 24«(/) V, and =

V, Kj = 20 V, R^ = 6 0 0 Q, /?2 = 2 0 0 Cl, and C

24u{t) V.

= 12.5 hF(a) Find yÔ"^).

(a) Compute the response ii{t) for ? > 0. Plot for 0 < ^ < 4x where x is the circuit time constant. (b) Find the inductor voltage v^{t) for r > 0

(b) Using the initial condition computed in part (a), find

< t< 160 msec.

Plot the result for 0


372

34. The voltage waveform

of Figure

P8.34a drives the circuit o f Figure P8.34b. The voltage-controlled switch Si closes when the capacitor voltage goes positive and opens when the capacitor voltage v^^t) goes negative. Compute the voltage Vf^t) across the capacitor. Assume that

has been at - 1 0 V for ?< 0

for a very long time. Hint: Use the elapsed time formula as needed. > 20V

o

v„,(t) 5

t(Msec)

2.5 (b)

--10V

Figure P8.32 (a) Pulse driving RC circuit of part (b).

----------(a)

IMegO

33. For the circuit of Figure P 8 .3 3 ,, -V^ = - 1 0 V, Kj = 20 V, /?! = 80 Q.7?2 = 20 Q, andZ = 4 H. (a) Find (b) Using the initial condition computed in part (a), find

Plot the result for 0

< r < 160 msec.

(b)

t

Figure P8.34 (a) Pulse waveform exciting RC circuit in part (b).

V

35. Repeat Problem 34, except find i(^t) for t> 80

t(msec)

40

36. Consider the circuit o f Figure P8.36. Suppose

--V.0 (a)

V ({0 ) = 0

and find V(î) for

0

as follows:

(a) Find the Thevenin equivalent circuit seen by the capacitor. (b) Find the complete response V(^t) for t > 0. W hat is y(j(oo)? - v,(t) +

400

(b)

6

5kO C = 0.1 F

IV

V,

Figure P 8 .3 3 Figure P 8 .3 6

lOlv.

373


OP AMP CIRCUITS

41. Figure P 8 .4 la shows an op amp integrator with positive gain, and Figure P 8 .4 lb shows a = K


sin(cor)«(r) V and all capacitor voltages are zero at f = 0. Find and of/e, C,/C and (0.

differentiator with positive gain for the con stant K> (a) For each o f the circuits find a literal

for / > 0 in terms

expression for

in terms of

(b) For /?j = 10 k n and (7 = 0.1 mF, find

Vo,„{t) when

= 100sin(20r)//(r) mV

assuming that

= 0 in each case.

v.(t) R. O v/W + Vjt)

A .

Figure P8.37

KR

38. In the circuit o f Figure PS.38, /? = 10 klT2, C = 10 jiF,

= 10 sin(50f)«(r) mV, and all

capacitor voltages are zero at / = 0. Find and plot it for 0 < r < 6 sec.

Figure P8.41

M ISCELLAN EOUS

Figure P8.38

42. Although most o f the first-order circuits 39. Repeat Problem 38 for

= \QOe~~‘ii{t)

mV, But plot from 0 to 6 seconds.

considered in the text have only one capacitor or one inductor, it is possible to have a firstorder circuit containing more than one energ}'

40. In the circuit o f Figure P8.40, /?j = 10 kH,

storage element. Consider the situation depict

Rf = 40 kQ, C = 12.5 |.iF, and v^{t) = -lOOw(f)

ed in Figures P8.42a and b. Here

mV.

Vcii^*) are given. The networks A^, and N j are

(a) If V({0~) = 0, compute

for r > 0.

and

equivalent under two conditions:

(b) Repeat part (a) for V(jS^~) = 50 mV. - v,(t) +

r C, + Co Prove this equivalence using the integral rela tionship o f a capacitor to show that the i-v ter minal conditions are the same for both

N,.

Figure P 8 .4 0

and

374


C .d - v ,

N1 (a)

(b)

Figure P8.42 43. As mentioned in Problem 42, although most of the first-order circuits considered in the text have only one inductor or one capacitor, it is possible to have a first-order circuit contain ing more than one energy storage element. Consider the situation depicted in Figures P8.43a and b. Here and given. The networks and A/'j are equivalent under two conditions: êq -

~

Prove this equivalence using the int^ral relation ship of an inductor to show that the i-v terminal conditions are the same for both A^j and A^2 -

(a)

45. In the circuit of Figure P8.45, Cj = C2 = 2 F and the switch closes at time ^= 0. The initial conditions on the two capacitors are t/^(0 ) = 4 V and vc2(0-) = 0 V. (a) For = 0.5 ft, find an expression for the current /^(f) for ^> 0. (b) For /? = 0.5 ft, find for ? > 0 and for f > 0. Note that ^ 0 for all r > 0. (c) Compute the energy stored in each capacitor at r = 0^. Also compute the energy stored in each capacitor at r = c». Finally, compute the decrease in total energy stored in the capacitors fi-om t = O'^ to ^= 00. (d) Compute the energy dissipated in the 0.5 ft resistor fi-om r = 0"^ to r = 00. Verify that the energy dissipated in the resistor equals the decrease in total energy stored in the capacitors fi-om r = 0* to ^= 00. (e) Does the dissipated energy depend upon the value of P?. What does R affect? Verify that conservation of energy holds for the circuit.

(b)

‘r

Figure P8.43

Figure P8.45

44. In the circuit of Figure P8.44, suppose /?j = 50 Q, /?2 = 200 a , q = 0.06 F, Cj = 0.3 F, yQ(0“) = 15 V, and i'qCO") = 5 V. Let = 40tt(r) V. Use the equivalence set forth in Problem 42 to compute for r > 0. C. - ±

C, - t

V,

46. Repeat Problem 45 for the circuit of Figure P8.45 when q = 1 F, Cj = 0.25 F, i/îCO-) = 3 V, and t/f^(0~) = 8 V. 47. In the circuit of Figure P8.47, = 1 lOtt(-f) + 220u(t) mV, Zj = 110 mH, ^2 = 11 mH, and /? = 10 ft. Compute and plot the waveforms for inU) and /^(/). Hint: Adapt the results of Problem 43 to ê case of two paral lel inductors with initial currents.

Figure P 8 .4 4

r>

375


and K 2 . Find

for appropriate

'JO

and

K2 in terms of K^.

iA n i'“

R

(b) Suppose the input v^{t) = -\2u{-t) +

Q

24«(/). Find y(0“ ) by inspection of the

L

circuit, and v(0+) by the principle of conservation of charge.

Figure P8.47

(c) Use equation 8.1 7 to write down direct 48. Repeat Problem 4 7 for the case where v-J^t)

ly the answer for v{t), f > 0. Had y(0“ )

= 220«W mV,

been used, would the answer still be cor

= 4 4 mA, and

=

rect?

11 mA.

51. The solution to the basic RL or RC differ

49. Consider the circuit of Figure P 8.49 in which Cj = 1 F, Cj = 4 F, v-^ = 10 V, and R = 2

ential equation in this chapter, equation 8.3,

a.

builds on the integral solution o f equation 8.4, (a) Compute v^Qr) and Vj^{Q*). Hint: How

which is valid for arbitrary y(/). This powerful

does the charge distribute over the two

formula will be studied in a course on differen

capacitors at ^ = 0^?

tial equation theory. Wheny(r) = F, a constant,

(b) Compute Vjfi) for r > 0.

it is possible to develop an alternative deriva tion o f equation 8 .1 7 using no more than some basic knowledge of calculus. Since the solution to the source-free case is the exponential, it is reasonable to expect (or to try) a solution for the constant input case o f the form

'> ^ '

( 1)

Figure P8.49

and K2 are two constants to be deter

where

mined. The constant K 2 arises from the con 50. The circuit o f Figure P 8.50 contains two

stant input, suggesting that the response would

capacitors.

intuitively contain a constant term. (a) Substitute equation 1 into equation 8 .16. You should obtain the result

=

Pi. (b) With K2 determined, evaluate x(/) at f =

tQ* to obtain an expression for

. Your

result should be Figure P8.50 (a) Suppose vjî) = Kû{t) V. Show that for t W

> 0, the voltage v{i) satisfies the firstorder differential equation

dv It

= -K

iv + K 2

x Uq

) - F t

(2)

(c) Finally, substitute K2 = Fx and equation 2 into equation 1.


3 "6

(ii) S is at position B when v is greater than

APPLICATIONS 52. An approximate sawtooth waveform can be

60 V and decreasing, and it moves to

produced by charging and discharging a capac

position A when i drops to 1 mA and v

itor with widely different time constants. The

drops to 60 V.

circuit o f Figure P8.52 ilkistrates the idea. Vj„{t) = 20 kQ,

= 10 V,

Assume that at r = 0, switch S is at A,

= 1 kQ, and C= 10 pF.

The switch S is operated as follows: S has been

and z^„,^;(0) = 60 V. Find

at position B for a long time, and S is moved to

cycle o f operation (i.e., charging and dis

for one

position A at /■ = 0 to charge the capacitor.

charging the capacitor), and roughly

When V(~increases to 9 V, switch S is moved to

sketch the waveform. Whau is the fre

position B to discharge the capacitor. When

quency o f the sawtooth waveform?

decreases to 1 V, switch S is moved to position A to charge the capacitor again. The process repeats indefinitely. (a) Compute the waveform o f

for four

switchings. (b) Plot

Is the name “sawtooth wave

form” appropriate? What is the frequen cy in hertz o f the sawtooth waveform? i (mA)

idealized i-v curve ofa neon lamp • (not to scale)

/:

A

/

slope

= 1m U /

Figure P8.52

V

slope

53. The sawtooth waveform is used in T V sets

negative resistance region

i

:

= 0.5 mU

60

(b)

:

N fc i

v(V)

----------- ► 90

(0

Figure P8.53

and oscilloscopes to control the horizontal motion of the electron beam that sweeps across

m is based on a hypothetical

the screen. One method o f generating such a

energy storage system using an inductor and a

waveform is to repeatedly charge a capacitor

solar cell. Consider the circuit o f Figure P S.54.

with a large time constant and then repeatedly

During the day, the solar cell stores energy by

discharge it with a very small time constant.

increasing the current in the inductor. During

The circuit in the shaded box o f Figure P8.53a

the night, the stored energy is used to power

is a crude functional model for the neon bulb

lights and appliances. Energy from the solar cell

in Figure P8.53b (type 5AB, costing about 75

is scored in the inductor during 0 < r < 7’j. At /

cents), whose i-v characteristic is shown in

= 0, the beginning o f the storage interval, /^(O”)

Figure PS.53c. The switch S in the model oper

= 0. At r =

ates as follows;

storing energy in the solar cell via the source

(i) S is at position A when v is less than 90

the device

is switched from

V and increasing, and it moves to B when

Vsolar <^0 powering a light denoted by Ry Note that diere is some overlap in the switching

reaches 90 V (the breakdown volt

movement; this is to ensure continuity o f the

V

age).

inductor current. At r = T^, the TV, represent ed by Rj, is also turned on.


377

Remark: All answers to parts (a) to (f) should be in terms o f /?,. and R^.

photo tim er used for timing the light in pho

(a) Draw a simplified equivalent circuit

tographic enlarger and printing boxes. Briefly,

5 5 . The circuit o f Figure P8.55 is a transistor

with three circuit elements for 0 < ? < T j,

the circuit operates as follows. When the relay

indicating all device values.

contact closes, the lamp is lit. When the contact

(b) Construct an expression for

0 < ^<

opens, the lamp is turned off. The relay has a 4 0 0 0 Q dc resistance and a negligible induc

(c) Draw a simplified equivalent circuit

tance. The pickup current is 2 mA, and the

with two circuit elements for T-^
dropout current is 0.5 mA; i.e., the contact

indicating all device values.

closes when the relay current increases from

(d) Construct an expression for /^(/), T^^
< Tj. You will need a value or expression for

zero to 2 mA, and it opens when the current drops below 0.5 mA.

After obtaining your expres

sion, for simplicity, let

denote

To use the timer, switch S2 is closed first. Switch Sj is normally in the B position. When

(e) Draw a simplified equivalent circuit

it is thrown momentarily to position A, the bat

with two circuit elements for T2 < t,

tery charges the 1000 pF electrolytic capacitor C to 1.5 V. When Sj is then thrown back to

indicating all device values. (f) Construct an expression for

Tj < t.

position B at /^= 0, the capacitor discharges and

You will need an expression for

produces a current iy, which, after amplifica

After obtaining your expression, for sim

tion by the transistor, actuates the relay and

plicity, let ^Tl- denote /^(jT2 )• Remark: For the remaining parts, all

amplified current drops below a point for the

answers are to be given in terms of

relay to open and the lamp is turned off.

K kf tori’ ^1> ^ 2> “ d This is to prevent the substitution of

Compute

possibly incorrect answers from prior

the middle of its full range (i.e., only 5 kQ is

parts for

used in the circuit).

(g) For each o f the four devices

^ston>

^store down an expres sion for the power absorbed at time t. Call the results and PRstore’

(h) For time t give an expression for the energy Wj(J) stored in the inductor if W i( r = 0 ) = 0. ^soljr -N -------------- ^

Figure P 8 .5 4

turns on the lamp. At some later instant

the

if the 10 kH potentiometer is set at

378


1.5V

lO kn E Potentionmeter

crude transistor circuit model

Figure P8.55 56. The circuit o f Figure P 8.56 suggests a way of generating a sustained sinusoidal oscillation. All op amps are assumed to be ideal. Capacitors, C = 0.1 fiF are uncharged at f = 0. The first two op amps are differentiators and the last is an inverting amplifier. (a) With switch S at position A and v^{t) = sin(lOOOr) V, find vj^t),

and

for r > 0.

(b) If at a later instant switch S is quickly moved to position B, what would you expect to be?

Figure P8.56

C

H

A

P

T

E

R

Second Order Linear Circuits

Warming up snacks in a microwave oven is a common activity in student dormitories. It works much faster than a conventional

s B s m

oven: heating a sandwich takes about 30 seconds. How does the microwave oven do this? While a precise explanation is beyond the scope o f this text, the basic principle can be understood through the properties o f a simple LC circuit. Recall that two conducting plates separated by a dielectric (insulating material) form a capacitor. Suppose some food were placed between the plates in place o f an ordinary dielectric. The food itself would act as a dielectric. Ordinar)' food contains a great number o f water molecules. Each water molecule has a positively charged end and a negatively charge end, with their orientations totally random for uncharged plates, as illustrated in part (a) o f the figure below. Applying a sufficiently high dc voltage to the plates sets up an electric field produced by the charge deposited on the plates. This causes the water molecules to align themselves with the field as illustrated in parr (b). If the polarity o f the dc voltage is reversed, the molecules will realign in the opposite direction as illustrated in part (c). If the polarity o f the applied voltage is reversed repeatedly, then the water molecules will repeatedly flip their orientations. In doing so, the water molecules encounter considerable friction, resulting in a buildup o f heat, which cooks the food. Microwave cooking is therefore very different from conventional cooking. Instead of heat coming from the outside, the heat is generated inside the food itself.

380

Chapter 9 • Scconcl Order Linear Circuits

food

food

conducting plate

conducting plate

(b)

(a)

food

0 Reversal of the polarity o f the applied voltage at a low frequency can be easily achieved with the circuit elements studied in earlier chapters: the resistor, the capacitor, and the inductor. However, the friction-induced heat production is inefficient at low frequency, l b produce a useful amount o f heat for cooking purposes, ver)' high frequencies must be used. The t}'pical frequency used in a microwave oven is 2.45 gigahertz, i.e., the water molecules reverse their orientations 2 x 2.45 x 10^ times per second. At such a high frequency, capacitors and inductors are quite different in their behavior from their conventional forms. For example, the LC circuit becomes a “resonant cavit}” and the connecting wire becomes a “waveguide.” These microwave components will be studied in a future field theory course. T he theory studied in this chapter will enable us to under stand the low-frequenc}' version o f the phenomenon, i.e., how a connected inductor and capaci tor can produce oscillator)' voltage and current waveforms.

CHAPTER OBJECTIVES 1.

Investigate the voltage-current interactions that occur when an ideal inductor is con nected to an ideal capacitor with initial stored charge.

2.

Use a second-order differential equation for modeling the series RLC and parallel RLC circuits.

3.

Learn to solve a second-order differential equation circuit model by first finding the nat ural frequencies o f the circuit, then looking up the general solution form, and finally determining the associated arbitrary constants.

Chapter 9 • Sccond Order Linear Circuits

3
4.

Define and understand the concepts of underdamped, overdamped, and critically

5.

damped responses. Investigate and understand the underlying principles o f various oscillator circuits.

SECTION HEADINGS 1.

Introduction

2.

Discharging a Capacitor through an Inductor

3.

Source-Free Second-Order Linear Networks

4.

Second-Order Linear Networks with Constant Inputs

5.

Oscillator Application

6.

Summary

7.

Terms and Concepts

8.

Problems

1. IN TRO DUCTION The previous chapter developed techniques for computing the responses of first-order linear net works, either without sources or with dc (constant) sources, having first-order linear differential equation models. Recall that the source-free response contains only real exponential terms. This chapter focuses on second-order linear networks having second-order linear differential equa tion models. Usually, but not always, a second-order net\vork contains t^vo energ)^ storage ele ments, either {L, Q , (C Q , or {L, L). Second-order circuits have a wide variety o f response wave forms: exponentials

sinusoids (/l,cos(coy) + y4-,sin(coy)), exponentially damped

sinusoids, and exponentially growing sinusoids, among others. Tables 9.1 and 9.2 catalogue the various response types. W ith no sources or with constant-value sources, some straightforward extensions o f the solution methods o f Chapter 8 are sufficient to compute the various responses. The behavior of second-order circuits is a microcosm o f the behavior o f higher-order circuits and systems. Many higher-order systems can be broken down into cascades o f second-order systems or sums o f second-order systems. This suggests that our exploration o f second-order circuits can build a core knowledge base for understanding the behavior o f higher-order, more complex phys ical systems. Many introductory texts discuss only parallel and series RLC circuits, stating separate formulas for the responses o f each. Our approach seeks a unified treatment. To this end, we formulate a basic second-order differential equation circuit model. The associated solution techniques become applicable to any second-order linear nersvork and, for that matter, to second-order mechanical systems. An oscillator circuit (section 5) motivates our study o f second-order linear networks. The chapter contains several other practical examples illustrative o f the wide variety o f second-order circuit

382

Chapter 9 • Second Order Linear Circuits

applications. Some advanced applications pertinent to higher-level courses include low-pass, highpass, and bandpass filtering (covered later in the text); dc motor analysis; position control; and many others. Most important, the concepts presented in this chapter are common to a host o f engineering problems and disciplines. Hence, the techniques and concepts described here will prove useful time and time again.

2. DISCHARGIN G A CAPACITOR TH RO UGH AN IN DUCTO R Chapter 8 showed that the voltage o f an initially charged capacitor in parallel with a resistor decreases exponentially to zero: the capacitor discharges its stored energy through the resistor. When an inductor replaces the resistor, as in Figure 9.1a, very different voltage waveforms emerge for V(-{t) and

In order to construct these new waveforms, we first develop a differential equa

tion model o f the LC circuit.

EX A M PLE 9 .1 . The goal o f this example is to develop a differential equation model o f the cir cuit in Figure 9.1b. In Figure 9.1a, with the switch S in position A, the voltage source charges the capacitor to

volts. At / = 0, the switch moves to position B, resulting in the new circuit o f

Figure 9.1 b, valid for f > 0. A

B

+

+

(a)

(b)

FICiURF 9.1 (a) A voltage source charges a capacitor, (b) An LC second-order linear network in which the cnerg)’ stored in the capacitor in part (a) is passed back and forth to the inductor. S

o l u t io n

.

Step 1. Write down the terminal i-v relationship for the capacitor and itiductor; then apply KCL and

KVL, respectively. Using the i-v relationships for L and C (see Chapter 7) in conjunction with KCL and KVL, it follow's that

ih C

_ >c _

dt and

±L_

c

V J V C definition KCL

di

(9.1a)

(9 .1 b )

V

L definition

c

KVL


383

Step 2. Obtain a dijferential equation in the capacitor voltage, V(j^t). For this, first differentiate equation 9.1a to obtain 1 dii^

dt

C dt

Substituting equation 9.1b into this equation yields ^

= - — VC

dt

LC ^

(9.2)

Equation 9.2 is a second-order linear differential equation circuit model o f Figure 9.1b in terms o f the unknown capacitor voltage, v^t). Equation 9.2 stipulates that the second-order derivative

j

o f the unknown function, vît), must equal the function itself multiplied by a negative constant,-----

LC

Step 3. Obtain a differential equation in the current, ij{t). An alternative circuit model in i^ is obtained by first differentiating equation 9.1b and substituting equation 9.1a into the result to produce

Equation 9.3 has precisely the same form as 9.2: the second-order derivative o f the unknown function, ij{t) , equals the function itself multiplied by a negative constant, -----. This similarity suggests a similarity o f solutions, which we shall pursue further.

Exercise. Fill in the details o f the derivation o f equation 9.3 from 9.1.

Our next goal is to construct the waveforms V(^t) and ij{t), which are the solutions o f the differ ential equations 9.2 and 9.3. Although differential equations are not usually part o f the common background o f students in a beginning course on circuits, the solutions o f equations 9.2 and 9.3 do not demand this background. Some elementary knowledge o f differential calculus is sufficient. Specifically, recall the differential properties o f the sine and cosine functions:

d

d

dt

dt

— sin(co/+ 0 ) = (o co s(o )/-t-0 ) and — c o s ( o ) / + 0 ) = - a ) s i n ( c o / + 0 )

Differentiating a second time yields

d2 ^y2 — 7 sin((or-f-0) = - 0) “ sin(co/+ 0 ) and — 7 cos(o)/+ 0 ) = - c a “ cos(co/+ 0 ) dt-

dt-

In both cases, the second derivative equals the function itself multiplied by a negative constant. This is precisely the propert}' required by equations 9.2 and 9.3. Thus one reasonably assumes that the solutions o f equations 9.2 and 9.3 have the general forms

V(^t) and

=

K cos(ojr

+ 0)

(9 .4 a )

38^

Chapter 9 • Sccoiul Order Linear Circuits

/•^(r) = /^cos(tor+0)

(9.4b)

7'hese forms are general because the cosine function can be replaced by the sine function with a proper change in the phase angle. Specifically, we note that A'sin((i)/ + (j)) = A'cos(co/ + (j) - 0.5ti) = K cos((or + 0) with 0 = ([) - ().5ti. Computing values for (o, A', and 0 specifies the solutions to equations 9.2 and 9.3.

E X A M PLE 9.2. Find A"and B for the capacitor voltage in equation 9.4a. So

lu t io n

.

Step 1. Differentiate equation 9.4a to obtain

dvrit)

—

—

(It

^

= -A c o sin (O )/+ 0 )

/'o S'!

Step 2. Dijferentiate n second time. Differentiating equation 9.5 (the second derivative o f 9.4) yields

d~Vr

^

«

">

-----^ = -AÔ)“ cos(O)/+0) = -(O “ i ’<7

dt~

___

(9.6)

Step 3. Match the coefficients o f equation 9.6 with those o f 9.2 to specify O). Under this matching, ' o1r CO = (0 ^ = -----

LC

7)

— j= =

fo

fiC

^ ^

^

Equation 9.7 specifies co, the angular frequency o f oscillation, in rad/sec, o f the capacitor voltage. Step 4. Compute K an d f) in equation 9.4a. These two constants depend on the initial conditions as follows: when the switch is at position A, the capacitor is charged up to I^q volts and the induc tor current is zero; immediately after the switch moves to position B, i.e., at r = 0+, the continu-

^

it>' properry o f the capacitor voltage ensures that y^Ô*) = Vj, and^the continuity property o f the inductor current ensures that //(O^) = 0. The initial value, ^ , ’ is now calculated from equation 9.1a as ^/r(--(0 ) i( ^ ( 0 ) ------------ = ----------- = ------------- = ()

dt

C

^

C

Evaluating equations 9.4a and 9.5 at / = 0^ , we have

= K c o m = v;,

(9.8a)

and
(9 g y

From equation 9.8b, 0 = 0. Consequently from 9.8a, K = V^^. Hence the capacitor voltage, i.e., the solution o f the second-order differential equation 9.2, is

Vc(f) = VqCOS

[J

lc

)

(9.9)

^


38S

As per equation 9.1a, one can obtain ij{t) directly by differentiating equation 9.9 and multiply ing by —C. However, one could aLso solve equation 9.3 by repeating the above steps to arrive at the same answer.

Exercise. Assuming that /^(r) = K coslcor + 6), solve for (o, K, and 0 in terms o f the initial condi tions, and show that

Several very interesting and significant facts about this parallel ZCcircuit and the solution method are apparent: (1)

For the source-free LC circuit o f Figure 9.1, the voltage and current responses are sinusoidal waveforms with an angular frequency equal to —_L _ . Since the amplitude o f sinusoidal oscillations remains constant (i.e., does not

VZc

damp out), the circuit is said to be undamped. (2)

rhe frequenq^ (o, depends on the values o f L and C only, while the amplitude K and the phase angle 0 depend on L, C, and the initial values o f the capacitor voltage and inductor current.

(3)

Although the instantaneous energ}" stored in the capacitor, \V^{t), and the instantaneous energy stored in the inductor,

both vary with time, their sum is constant. (This is

investigated in a homework exercise.) Physically there is a continuous exchange o f the energ)' stored in the magnetic field o f the inductor and that stored in the electric field of the capacitor, with no net energy loss. This is analogous to a frictionless hanging mass-spring system: because o f the absence o f friction, the up-and-down motion of the mass never stops; in such a mechanical system there is a continuous interchange between potential and kinetic energy. Figure 9.1 shows what is, in theory, the simplest circuit that generates sinusoidal waveforms. Such an electronic circuit is an (idealized) oscillator circuit. Oscillator circuits play an important role in many communication and instrumentation systems.

3. SOURCE-FREE SECON D-ORDER LINEAR N ETW ORKS Unlike their ideal counterparts, real capacitors and inductors have resistances. A better under standing o f a realistic oscillator entails the analysis o f an RLC circuit. This section investigates source-free RLC circuits having two energ)' storage elements. Our investigation begins with the development o f the differential equation models o f the series and parallel /^//.'circuits. Both mod els are special cases o f an undriven general second-order linear diflerential equation. Hence we will discuss the solution o f a general second-order linear differential equation and adapt the solution to the series and parallel RLC circuits. We will also illustrate the theory with a second-order cir cuit that is not a parallel or series RLC.

386


Developmeut o f Dijferential Equation Models for Series/Parallel RLC Ciraiit The first goal o f this section is to develop differential equation models for series and parallel RLC circuits as detailed in the following example.

EXA M PLE 9 .3 . For the series and parallel RLC circuits shown in Figure 9.2, develop two sec ond-order differential equation models (one in

and one in

for each circuit.

+

+ V

''c = V''l

= V

''r

(b)

FIGURE 9.2 (a) Series RLC circuit, (b) Parallel RLC circuit. Passive sign convention is assumed as usual. S

o l u t io n

Wc do this in “parallel” rather than in “series.” Step 1. Apply KVL to the series RLC.

Step 1. Apply KCL to the parallel RLC.

Step 2. Choose i^âs circuit variable and express v^, and Vq in terms

Step 2. Choose

and express i^,

as circuit variable and iQ in terms o f

^ f‘L(ii 1 ‘ Ril + L — + — / ^ ( t V t = 0 dt C— •oo '

\’f^ I f , „ dv/" — +— vrix)dx + C — ^ = 0 R L ^ ^ dt

Differentiate, rearrange terms,

Differentiate, rearrange terms, and

and divide by L to obtain

divide by C to obtain

dîl Rdi^ 1 . ^ — ^ + ------ ^ + ----- 1, = 0 dr L dt LC ^

d^Vf dt-

d\>r RC dt

1 H-------= 0

LC

Step 3. Choose Vq as the circuit

Step 3. Choose ij as the circuit

variable and express Vjând terms o f Vq

variable and express terms o f i^.

in

Again using the KVL o f step 1,

d ir Rin -f" L ------- 1= 0 ^ dt ^

arid i^ in

Again using the KCL o f step 1, — -H/, + C — ^ = 0

R

^

dt


38‘

= C dv^^dt,

Hence, substituting for

L dij/dt,

Hence, substituting for

rearranging, and dividing through by

rearranging, and dividing through by

LC yields

LC yields R dvr L dt

d t‘

llL dr

LC

RC dt

+—

LC

Each circuit has two second-order differential equation models, one each for

= 0

and

as the

unknown quantity.

Exercise. Show that

and Vj^ satisfy second-order differential equations similar to those

derived in Example 9.3.

Solution o f the General Second-Order Dijferential Equation Model The final differential equations o f Example 9.3 force the current

or the voltage

to satisfy cer

tain differential constraints. All four (differential) equations have the general form

d^x dx — ~ + h ----- h (.'.V = 0 dtdt

(9 . 10)

for appropriate constants b and c, where .v is either ij or

Equation 9.10 stipulates that the sec

ond derivative o f the function x{t) plus h times the first derivative o f x{t) plus c times x(t) itself adds to zero at all times, t. Unlike the example o f section 2, where a sinusoidal solution was easi ly predicted, the present differential equation requires a more careful mathematical analysis. Recall from elementary calculus that the derivative o f an exponential is an exponential. Thus the first and second derivatives o f an exponential are proportional to the original exponential. This sug gests postulating a solution o f the form at(^) = Ke^‘ where we make no a priori assumptions about

s. If it is truly a solution, it must satisfy equation 9.10. Under what conditions will x{t) = Ke^‘ sat isfy equation 9.10?

E X A M PLE 9.4. Determine conditions under which the postulated solution x{t) = Ke^‘ satisfies equation 9.10. S

o l u t io n

.

Step 1. Substituting Ke^‘ for ,v(/) in equation 9.10 produces

K

rfV '

dr

+ bK —

+ cK e” = Ke" (,v^ + hs + c) = 0 ^

I

(9.11)

Step 2. Interpret equation 9.11. For nontrivial solutions, K\s nonzero. The function e^‘ is always different from zero. Hence the quadratic in s on the right side o f equation 9.11 must be zero. This necessarily constrains $ to be a root of

r + bs + c =0

(9 .1 2 )

3
Chapter 9 * Sccond Order Linear Circuits

Step 3. Solve equation 9.12. From rhe quadratic formula, rhe roots o f equation 9.12 are

.V|, .v. =

-b ±

- 4c

(9.13)

C O N C LU SIO N : x(/) = Ke^' satisfies equation 9.10 provided s tatces on values given by equation 9.13. Equation 9.10 does not constrain K\ however, the initial conditions will.

Equations such as 9.12 whose solution is given by equation 9.13 are a common characteristic o f second-order networks. Hence, equation 9.12 is called the characteristic equation o f the secondorder linear circuit. The associated roots, equation 9.13, are called the natural frequencies o f the circuit. These are the “natural” or intrinsic frequencies o f the circuit response and are akin to the natural frequencies o f oscillations o f a pendulum (for small swings) or o f a bouncing ball. From elementary algebra, a quadratic equation (the above characteristic equation) can have dis tinct roots or equal roots. Distinct roots can be real or complex. Thus

and Sj can be two dis

tinct real roots, two distinct conjugate complex roots, or two repeated (equal) roots, depending on whether the discriminant, Ip- - 4r, is greater than, less than, or equal to zero. This trifold grouping separates the solution o f equation 9.10 into three categories, listed below as cases 1, 2, and 3:

Case 1. Real and distinct roots, i.e., b~ -A c> 0. If the roots are real and distinct, then for arbitrar}' and K2 , both

constants

.v( o = .vi (/) =

a:,6^"''

and

xif) = X2it)=K2e-'^-'

^

satisfy the second-order linear differential equation 9.10, i.e., are solutions to the differential equation. Since equation 9.10 is a

differential equation, by superposition the sum x(r) = x,(^) +

X2 (t) is also a solution, a fact easily verified by direct substitution. Therefore, whenever

s-y, the

most general form o f the solution to equation 9.10 is

xit)=Kê^^' + K2e^^-‘ The constants

(9.14)

and K-, depend on the initial conditions o f the differential equation, which

depend on the initial capacitor voltages and inductor currents. For example, if a.*(0'^) and a''(0'*') are known, then from equation 9.14, .v (0^ )=

A y '*'-h

and

.v-(O^) = — ^

(It

= .s'l 1= 0 ^

+ stK')

^

Chapter 9 * Second Order Linear Circuits

389

These are simultaneous equations solvable for If

and

and $2 are negative, the response given by equation 9 .1 4 decays to zero for large f and the cir

cuit is said to be overdamped.

Case 2. The roots,

and $2 , o f the characteristic equation are distinct hut complex, i.e., iP- -A c
Since Jj ^

general form of the solution to equation 9.1 0 is again given by equation

9.14, i.e.,

= with complex

and

$2

given by

+K2e"‘^^ _______

-b . y l^ c -b ^ ^2 = Y ± J ------ ------- =

, ± J^ d

(9.15)

y l^ c-b ^

where a = bH and co^ = ------ ^ S i n c e s^ and S2 are conjugates, so are dîând e^2în equation 9.14. For x(t) to be real, the constants /Cj and A'j in equation 9.1 4 must also be complex conjugates, i.e.,

Using Euler’s formula, giy = cos

+ j sin y

the two terms in equation 9.14 combine to yield a real time function:

= e

co s(co jf) + jKisinioi^t)] + e

ATj cos((0^/) - yATj sin(cojO

(A"! + A"! )cos(to^f) + (yî + jK i )sin((Ojf)

Thus the solution to equation 9.1 4 with

and S2 complex is given by the (damped) sinusoidal

response x(r) = e~^^ [A cos((o/) + B sin((0/ )] where ^4 =

+^

= 2 Re[^j] =

+ A^2 and 5

(9.16)

- y Af] = - 2 ImATJ = /A j - yATj are real con

stants and where Re[ ] denotes the real part and Im[ ] denotes the imaginary part. The solution expressed in equation 9.16 is completed by specifying ^4 and B. As before, A and B depend on the initial conditions, x(0‘^) and xXO"*^) as follows: jc(0'*‘) =

[Acos(tOjf) + Bsin((0^r)])^_^^ = A

390


o

jr'(0'*’) =

-a e ’‘^^^y4cos (o)^0 + 5sin (coô) + e

co^y4sin (coÔ + (O^Bcos (coô)

= - o A + (o^B and which are easily solved for A and B. Making use of a standard trigonometric identity, the general solution of equation 9.16 has the equivalent form

x{t) =

[A cos(coj) + B sin(o)^)] = Ke~^^ cos(o)^ + 0)

(9.17a)

where

K = ylA^ + B^ ,Q = tan"‘

f-B \ (9.17b)

KA)

and the quadrant of 0 is determined by the signs o f - B and A. In MATLAB, one uses the command “atan2(-BvA)” to obtain the angle in the proper quadrant. Note that the response waveforms have oscillations with angular frequency O)^ These oscillations are bounded by the envelope ±Ke~^*. If Re[jj] = - a < 0, the amplitude of the oscillations decays to zero and the response is said to be underdamped. If Re[jj] = - a > 0, the amplitude of the oscillations grows to infinity.

C ased. The roots are real and eqml, i.e.,lP' - 4 c = 0 . When the two roots of the characteristic equa tion are equal, equation 9.1 4 does not represent the general solution form because if two terms collapse into a single term. However, the general solution for

the

is

x(t) = {K^+K2t)e^^^ (This is investigated in a homework exercise.) Calculation of

and

in equation 9.18 is

straightforward:

x{0-) = and i(0+) = Substituting the value o f

into x'(O^) yields a simple calculation for Kj.

If^l = i2 is negative, the response decays to zero and is said to be critically damped. “Critically damped” defines the boundary between overdamped and underdamped. This means that with a slight change in circuit parameters, the response would almost always change to either over damped or underdamped. T he discussion o f the three cases is summarized in Table 9.1.


391

TABLE 9.1. General Solutions for Source-Free Second-Order Networks

General solution o f the homogeneous differential equation

d~x

dx

dr

dt

— ^ + h — + cx - 0 having characteristic equation

+ r = (j - 5 j)(j -

= 0, where

- h ± yjb" —4 c

^2=

-----------

Case 1. Real and distinct roots, i.e., ^ - 4 c > 0: x {t)=

+ Kie^-

where 40^) =

and .v (0") = s^K^ + s^K^,

Case 2. The roots, i| = - a + p i^ an d ^2 = ~j^d> tion are distinct but complex, i.e., tr - Ac < Q:

characteristic equa

x{t) = e~^‘ [A cos(o)/) + B sin(o)/)] = Ke~^' cos(to/ + 0) where x(0^) = A . x'(0+) = - 0 .4 + CO/ and +

, e = tan"^

Case 3. The roots are real and equal, i.e.,

(-B \ A

= Sj and Ir - Ac

xU) = (K^ + K2t)e^'’ where

x{0*) =

and a-'(O^) = s^K^ + Kj

Figure 9.3 displays the various response forms described above for the case where Re[xj] and Re[j-,] are negative or zero. Because o f their similarity, it is not possible to distinguish between the over damped and the critically damped responses by merely looking at the waveforms. Both types o f response may have at most one zero-crossing.

392

Chapter 9 • Second Order I.incar Circuits

> t

t

FIGURE 9.3 (-'tncric waveforms corresponding to the four cases o f damping: (a) undamped (sinu soidal) response, (b) undcrdamped (exponentially decaying oscillatory) response, (c) overdamped (exponentially decaying) response, and (d) critically damped (exponentially decaying) response. The terms “undamped,” “underdamped,” “overdamped,” and “critically damped” stem from an intuitive notion o f “damping.” The sourcc-frce response o f an undamped second-order linear sys tem, whether electrical or mechanical, has an oscillatory response (waveform) o f constant ampli tude. Damping, due to system elements that consume cnerg\', means a monotonic decrease in the amplitude o f oscillation. In electrical circuits, resistances produce the damping effect. In mechan ical systems, friction causes damping. When the amount o f damping is just enough to prevent oscillation, the system is critically damped. Less damping corresponds to the underdamped case, where oscillation is present but eventually dies out. A greater amount o f damping corresponds to the overdamped case, where the waveform is non-oscillatory, and a ver\’ small perturbation o f any circuit parameter will not cause oscillations to occur. In summary, once the roots o f the characteristic equation are found and the expression for the general solution selected from the above cases or Table 9.1, it remains to find the constants A', and


3^)3

K j (or A and 5 ) from the initial conditions on the circuit. In the above development, and Kj (or A and B) are given in terms of a:(0^) and x'(O^). Since jf(/) represents either a capacitor voltage or an inductor current, its value at f = 0"^ is usually given, or can be determined from the past his tory o f the circuit. (See Example 9.5.) The value o f x'(O^), on the other hand, is often unknown and must be calculated. If >:(/) =

then the capacitor v-i relationship implies that

xX0*) = v'c{0*) = ^ £ ^ .

^ If x{t) =

then the v-i relationship of an induaor implies that

The problem then reduces to finding an unknown capacitor current,

, or an unknown

inductor voltage, y^(O^). To find

or y^(O^), we construct an auxiliary resistive circuit valid at r = 0^. Since the initial

values,

and /^(O^), are known, we replace (each) capacitor in the original circuit by an inde

pendent voltage source o f value

and (each) inductor in the original circuit by an inde

pendent current source o f value /£(0'*^). Here the current /(;^0^) retains its original direction and the voltage ^^(0^) retains its original polarity. After the replacements, the (new) circuit is resistive. Values for

and y^(O^) follow by applying any of the standard methods of resistive circuit

analysis learned earlier. This allows us to specify x(0^) and x'(O^) in terms of the initial conditions on the circuit. Two equations in the two unknowns A'j and K2 (or A and B) result. Example 9.5 and, in particular. Figure 9.4c illustrate this procedure.

Response Calculation o f Source-Free Parallel and Series RLC Circuits Before any additional circuit examples, let us summarize the solution procedure.

Procedurefor Solving Second-Order RLC Circuits Step 1. Determine the differential equation model of the circuit. Step 2 . From the differential equation model, construct the characteristic equation and find its roots using the quadratic root formula. Step 3. From the nature of the roots (real distinct, real equal, or complex), determine the general form of the solution from Table 9.1; the solution form will contain two unknown parameters. Step 4. Find the two unknown parameters using the initial conditions on the circuit. The following example illustrates these calculations for the three cases described in Table 9.1.

W

39-'i


E XA M PLE 9 .5 . In the circuit o f Figure 9.4a, the 1 pF capacitor is assumed to be ideal, and the inductor is modeled by a 10 mH ideal inductor in series with a 20 Q resistor to account for the resistance o f the coiled wire. Suppose the switch S in Figure 9.4a has been in position A for a long time. The capacitor becomes charged to 10 V. Then the switch moves to position B at f = 0. Find and plot V(\t) for r > 0 for the following three cases; (1)

= 405

(2) /?2 = 0, (3)

180 ti.

Each o f these cases produces a different response type.

:L=10mH

lO O +

10V C=1

R=200

Practical Inductor

mF

(a)

i,(0*) = 0

L=10mH + C=1

mF

(b) FKJURH 9.4 (a) Discharge of a capacitor through a practical inductor in series with a resistance /?,. (b) Kquivalent circuit for t > 0. (c) Equivalent circuit at / = 0^ for calculating

in which the

inductor has been replaced by an independent current source of value //(O^) and the capacitor by an independent voltage source of value

.

SoL u rioN

From the problem statements,

= 10 V. When the switch moves to position B,

=

- 10 V by continuit)' o f the capacitor voltage; the circuit now becomes a series RLC, for t > 0, with / ?= / ?!+ R-y as shown in Figure 9.4b. The first step in the calculation o f the circuit response is to find a second-order differential equation in the unknown

From Example 9.3,

for the series RLC,

-

R dvf'

*- + -------^ +

dt~

L dt

\’r = 0 LC ^

(9.19)

Since L and C are known, the series RLC characteristic equation is o R 1 s~ H— .y H------ = .v“ + ( 2 0 + /?-,)10^y-r-1 0 ^ = 0 L LC With this framework, we can separately investigate each o f the three cases.

( 9 .2 0 )


Case I: /?2 = 405 Q or /? = 425 O. Step 1. Find the characteristic equation and the generalform o f the response using Table 9.1. If R-, = 405

the characteristic equation o f 9.20 is

+ 42,500x + 10^ = 0. Solving for the roots by the

quadratic formula yields 2 = - 2 1 ,2 5 0 ± 18,750 = -2 5 0 0 , - 4 0 ,0 0 0 sec’ ^ Real distinct roots imply an overdamped response o f the form

v c(0 =

+ K2e^~ =

(9.21)

valid for r > 0. Step 2. Fijid

and K-,. Evaluating at f = 0"^ implies z;^0") = 10 = /Tj + /f2 (9.22a)

Differentiating equation 9.21 implies

vL(0+) = -^^^^ = -2.5xlO^/r, - 4 0 x \ 0 ^ K 2 C

From the circuit o f Figure 9.4c, v’ ^ (0^ ) =

C

^

^

C

C

(9.22b)

^ = 0 , where //^(O'*') = i[{0~)

by the continuity o f the inductor current. Solving equations 9.22a and b after substituting the above values yields AT, = 10.667 and K-, = - 0 .6 6 7 Step 3. Set forth the solution for V(^t). For / > 0,

v^t) = 10.667^-2’500^-0.667e’-^0’®®°'V This function is plotted in Figure 9.5.

Exercise. You may verify' the above answer with the Student Edition o f MATLAB (version 4.0 or later) by typing the command: y = dsolve(‘D 2y+42500*D y + leS^y = 0,y(0) = 10, Dy(0) = O’).

Case 2: /?, = 0 or /? = 20 Q Step 1. If /?2 = 0> then from equation 9.20, the characteristic equation is + 2,000^ + 10^ = 0. Since tr - Ac = -3 9 6 ,0 0 0 ,0 0 0 < 0, the roots are complex. From the quadratic root formula, = - 1 0 0 0 + y9950 = - a + yo)^ and Sj = - 1 0 0 0 -y '9950 = - a From Table 9.1, the underdamped response form is

v^t) =

[A cos(w/) + B sin(co/)] = ^’"1000/

cos(9950f) + B sin(9950f)]

(9.23)

396


Step 2. Fiiid A and B. Ir remains to determine A and B in equation 9.23. From equation 9.23 and its derivative, i / ^ 0 ^ ) = 1 0 = /I

(9 .2 4 a )

and

= -oA + ( .) / = -1000/1 + 9950/y

(9.24b)

^ = — — - = 0. Substituting into equations 9.24 and

As in case 1, V(- '(O"^) = — — ^ solving yields C

C

C

A = 10 and

= — = 1.005 CO,/

Step 3. Set forth the solution for v^{t). For t > 0,

v^t) = ^-'OOOrjio cos(9950?) + 1.005 sin(9950^)] = 10.05^'-^^®^' cos(9950^ + 5.7°) V This waveform is also plotted in Figure 9.5.

Exercise. You may verify the above answer with the Student Edition oF MATLAB (version 4.0 or later) by typing the command: y = dsolve(‘D 2y+2000*D y + le8*y = 0,y(0) = 10, Dy(0) = O’).

Case 3: ^2 = 180 Step 1. If

=180

or R= 200 Q then the characteristic equation from 9.20 is

+ 2 0 ,0 0 0 j + 10^ = 0, whose

roots are = -lO '*, implying a critically damped response. From Table 9.1, the general criti cally damped response form is, for r > 0, v^(/) = (/r, + K2t)e^^' = (A", + K2t)e~^^'^

(9-25)

Step 2. Find /Tj and Kj. From equation 9.25, its derivative, and the known initial conditions from cases 1 and 2,

vîO^) = 10 = A",

(9.26a)

and ^yO ") =

+ K, = -10^ A', + A'2 = 0

(9.26b)

Solving equations 9.26 yields A', = 10 and Kj = --^lA", = 10^ Step 3. Set forth the sohition for vî). For /^> 0, v'c(n = ( l0 + 1 0 ^ ) e " '^ ‘’ ' The waveforms o f V(fJ) for the three cases (underdamped, critically damped, and overdamped) are plotted in Figure 9.5.

39'


FIGURU 9.5 Waveforms of vj,t) in Example 9.5 for three different degrees of damping. Critical damping represents the boundary between the overdamped condition and the oscillatory behavior of underdamping.

Exercise. Verify the answer calculated in Example 9.5 using the Student Edition o f MATLAB and the “desolve” command.

On a practical note, commercially available resistors come in standard values each with an associ ated tolerance. Tolerances vary from ±1% (precision resistor) to as much as ±20% . Further, because o f heating action over a long period o f time, resistance values change. Given the above example, in which the type o f response depends on the resistance, one can imagine the care need ed in the design o f such circuits: without consideration o f precision and long-term heating effects, a desired critically damped response could easily become oscillator)'. Not all second-order circuits arc RLC. Some are only RC but with two capacitors and some are

RL with two inductors. Passive RC or RL circuits cannot have an oscillatory response. The proof o f this assertion can be found in texts on passive network synthesis. However, with controlled sources a second-order RC or RL circuit can have an oscillatory response that is not characeristic o f a first-order circuit, but o f a second- or higher-order circuit. The example below illustrates the analysis o f a second-order RC circuit containing controlled sources that has an oscillator)^ response.

398


EX A M PLE 9 .6 . This example illustrates the analysis o f the second-order RC circuit shown in Figure 9.6. The objective is to find = 10 V and = 0.

'Cl

1 kn I

for / > 0 given the initial conditions V(^{0)

and

r

mF

FIGURE 9.6 Sccond-order RC circuit with controlled sources that has an oscillatory response. S o l u t io n

Step 1. Write a dijferential equation in V(^^{t). From the properties o f a capacitor and KCL at the left node,

di Multiplying through by 10*^ yields

d\

■^^ — 10

— 10^ v^2 ~ 1 0 'v’f'i

dt

(9.27a)

We expect a second-order differential equation, so differentiating a second time yields _ 1()5

, Q3

dt

dt~ To obtain a differential equation in

(9.27b)

dt

equation 9.27b must be eliminated. This

requires another relationship between t»Qand

. At the right node,

10“6 — C l _

_ -0 .

dt or equivalently,

dt

= -1 0 ^ Vc,

(9.28)

Substituting this expression into equation 9.27b produces

dt

*2

After rearranging terms, we obtain a second-order differential equation in

'2 dt^

dt

(9.29)

Step 2. Determine the characteristic equation, its roots, and the form o f the response. The differential equation 9.29 has characteristic equation


+ 1 0 ’° = 0 From the quadratic formula, the complex roots are

= - 5 0 0 + 799,998.75 = - a ± y W From Table 9.1, complex roots imply an underdamped response o f the form

[A cos(co/) + B sin(co/)] = ^’" 500' [A cos(99,998.75r) + B sin(99,998.75r)] (9.30) Step 3. Find A and B. \x. t = 0, z^qCO) = \() = A. Also, from equation 9.27a and the initial con ditions, ^^^'Cl(Q) ^ i0 - \ ;^ .,(0 ) - iq 3 v ^ ,(0 ) = - 10^

dt

(9.31)

Differentiating equation 9.30, evaluating at r = 0, and equating the result with equation 9.31 produces

dl

"

in which case B = 5.0001 x 10“^. Step 4. Set forth the final form o f V(-^{t). The final form o f the response is

v^{t) = ^'-500^ [10 cos(99,998.75^) + 5.0001 x 10"2 sin(99,998.75r)] V = 1 0 ^ 5 0 0 'c o s (9 9 ,9 9 8 .7 5 r- 0.2865°) V Step 5. Plot the response. A plot o f the (underdamped) response is given in Figure 9.7.


Exercise. Construct a parallel RLC circuit to have the same second-order difFerential equation model as 9.29. Note that there is no unique solution. C H EC K : /?C= 10-^ and ZC = \0~^^

It is important to observe here that the design o f Example 9.6 achieves a second-order RLC response without the use of an inductor, which is important for integrated circuit technolog)'.

4. SECON D-ORDER LINEAR NETW ORKS WITH CON STANT INPUTS The preceding section studied source-free second-order linear networks. When independent sources are present, such as in the circuit o f Example 9.7 below, the network (differential) equa tions are similar to the source-free case except for an additional term that accounts for the effect o f the input: ^

clf-

+ / ; ^ + c-.v = /(/)

dt

(9.32)

wherey{r) is a scaled sum o f the inputs and/or their first-order derivatives. Ordinarily one might expecty(f) to be the value o f the input. A homework problem illustrates that j{t) can depend not only on the source input, but also on the derivatives o f the input. For general circuits, those not reducible to parallel or series RLC circuits, constructing equation 9.32 can be a challenge. Further, the solution o f 9.32 for arbitrary inputs and initial conditions is no less challenging but is best obtained via the Laplace transform method, which is a topic studied in a second circuits course. However, when the input excitations are constant,/r) = F, the solution to 9.32 is a straightfor ward modification o f the source-free solution, as explained in the remainder o f this section. Since the expressions o f Table 9.1 satisfy the homogeneous diflferential equation 9.10, the gen eral solution to equation 9.32 follows by adding a constant

to each o f the solution forms given

in Table 9.1. Specifically, the general solution o f the driven differential equation

tl~x dx - + h — ^ cx = F

(9 . 33 )

xit) = .V,//) + Xj:

(9.34)

dr

dt

where-v^j(f) is the solution to the homogeneous equation 9.10 (equivalently, equation 9.33 with F = 0). Recall that the form o f x j t ) is determined by the roots of the characteristic equation r + bs + r = (i - ^])(j - ^2) = 0, given by the quadratic formula

h yjh~ - 4 c i |-) = — ± -------------9 9

*0 1

Chapter 9 * Scconcl Order Linear Circuits

To verify that the structure o f equation 9.34 is a solution to 9.33 and to compute the value o^Xp sub stitute the structure given by equation 9.34 into 9.33. Since

satisfies the homogeneous equation

9.10, it contributes zero to the left-hand side. What remains is cXp= F. Therefore,

= — which

is independent o f the roots o f the characteristic equation. However, if Re[j|] and Re[^2] < 0, then tends to zero for large t. Hence x{t) tends to Xp for large t. Consequently Xp is termed the final value o f the response. Because o f the trifold structure o f

as summarized in Tible 9.1, the solution form o f equation

9.34 once again breaks dow'n into three distinct cases. We summarize this trifold structure for the constant-input case in Table 9.2. TABLE 9.2 General Solutions for Constant-Source Second-Order Networks

General solution o f the driven differential equation

-z- + h — + cx = F

dr

dt

having characteristic equation p- + bs + c = (j - Jj)(^ - ^2) = 0 with roots

h yjb^ - 4 c -7 = ---- ± --------------“ 2 2 Case 1. Real and distinct roots, i.e.,lP' —4 r > 0: x {t)=

+ Xp

F with Xp = —. Further, c

x(0'^) =

+ K2 + A'yând x'(0'^) = s^K^ + s^K^

Case 2. The roots, = - a + y'co^y and S2 = - o o f the characteristic equation are distinct but complex, i.e., - 4 c
F

where again Xp = —, with

c

x{0^) = A + X p x(0^) = - g A +

K V

)

Case 3. The roots are real and equal, i.e., s, = s, and Ir -A c= Q . The solution form is .v(/) = (/ r ,+ ^ 2 0 ^ ' where Xp = —, and

c

x{Q^) = K^+Xp

and .v W "-) = .v ,^ i-h AT2

402


The interpretation Xp = F/c. is a mathematical one. When the differential equation describes a lin ear circuit with constant inputs, there is a physical interpretation o f Xp and a circuit theoretic = Xp

method for computing its value, even without writing the differential equation. Since

= a constant” satisfies the differential equation 9.33, it is also a constant solution to the circuit. Hence Xp\s either a constant capacitor voltage or a constant inductor current. If a capacitor volt age is constant, its current is zero; this is interpreted as an open circuit. Similarly, if an inductor current is constant, its voltage is zero; this is interpreted as a short circuit. Therefore, Xp is an appropriate (capacitor) voltage or (inductor) current obtained when the capacitor (or capacitors) are open-circuited and the inductor (or inductors) arc short-circuited. The value o f Xp can be obtained by analyzing the resistive network resulting when all capacitors are open-circuited and all inductors are short-circuited. Recall that if Re[^j] and Re[^2] < 0, then x{t) tends to the constant value Xp Physically speaking, then,

equals either

or /^(o))when Re[;,] and Re[^2] < 0.

Once the proper general solution structure is ascertained from Table 9.2 and the constant Xp is found, the parameters

(or A and B) are computed by the same methods used in the

and

source-free case. The following example illustrates the procedure for a parallel RLC circuit.

= u{t) A, excites the parallel /^ZCcircuit o f Figure 9.8,

E XA M PLE 9.7 . A step current input,

whose initial conditions satisfy /)(0) = 0 and y^Ô) = 0. This simply means that the current source turns on with a value of 1 amp at r = 0 and maintains this constant current excitation for all time. The objective is to find the inductor current,

for / > 0, for three values o f R: (i) R = 500 Q,

(ii) R = 25 n , and (iii) R = 20 Q..

I'IGURE 9.8 Parallel RLC circuit cxcited by a step current input.

So l u t io n Because the circuit is a parallel RLC, the characteristic equation is

. r + - L , + - L = ,2 + 1 0 l,v + 4 x l 0 ' " = 0

RC

LC

(9.35)

R

For all positive values o f R, the roots of the circuits characteristic equation have negative real parts. Thus for large t or ideally at “t = oo,” the inductor looks like a short circuit and the capacitor like an open circuit. Hence, for all cases o f this example, Xp = i[{^) = 1 A; note that y^co) = 0 because the inductor looks like a short at t = .

Case I. For R = 500 Q, the characteristic equation 9.35 reduces to r + 2 0 ,0 0 0 j + 4 From the quadratic formula, the roots are

2 = - 1 .0

X

lO'^ ± 7 I .9975

X

105 = - a ±yco^

x

10*^^ = 0.

'103


which indicates an underdamped response o f the form (Table 9.2)

[A cos(coy) + B sin(o)^)] + Xp= Ke~^‘ cos(to^ + 0) +

/^(r) =

= A + Xp = A + 1, then A = -\. Further,

Since 0 =

elicit)

0=

dt ^

\

= -0/4 + (HjB 1 = 0^ From derivative o f expression for ij^t)

From physical circuit

This implies that fi = — = - 5 .0 0 6 3 x 1 0 " “ (0,y Hence for / > 0, /•^(r) = ^.-10.000^ [ cos(1.9975 x IQ5f) + 5.0063 x 10"2 sin(1.9975 x lO^r)] + 1 = 1 .0 0 13^>-*^’‘^^®'cos( 1.9975 x 105^+ 2.866'’) + 1 A

Case 2. For R = 25 O., the characteristic equation 9.35 reduces to From the quadratic formula, the roots are

+ 4 x 10^5 + 4 x lO'® = 0.

j, 2 = - 2 . 0 x 105 indicating a critically damped response o f the form (Table 9.2)

iL^t) = (K ^ + K 2t)e‘ '' + Xp Since 0 = /^(O^) = A^, + Xp = Q

+ 1, then

= - 1 . Further,

_ vc(0'*') _ vz.(Q~^) _ d ilit )

L

L

dt

= SiKi + K2 = 2xW ^ -hK2 1=0*

This implies that K y - - I y. 10^ and for r > 0, -2xio-\

Case 3 . For R = 20 Q, the characteristic equation 9.35 is quadratic formula, the roots are = - 1.0

X

10^ and S2 ~ ~

specifying an overdamped response o f the form (Table 9.2)

+ 1A

+ 5 x 1O^j + 4 x 10^® = 0. From the

^

40 4


+ Xp. Furtiier,

Evaluating this response at f = 0+ yields

L

L

dt

t=(f 1

Equivalently, -K^ ~ Solving these two equations yields K\ = ---- and K-, = —. Therefore the actual response for r > 0 is ^ ^

Figure 9.9 displays a graph o f the response for each o f the three cases.

FIGURE 9.9 Underdamped, critically damped, and overdamped response curves for the parallel RLC circuitof Example 9.7.

Exercises. 1. Show that for t > 0 , the differential equation for the circuit o f Example 9.7 with R = 500 Q is

2

^

^

clr

+ 2 x l 0 ‘' ^

dt

+ 4xl0'»,',.
2. Use MATLAB s “dsolve” command to verify the solution obtained for case 1 in Example 9.7.


In a linear circuit or system, the response to a step input often indicates the quality o f the system performance. The problem o f measuring a batter}' voltage using a voltmeter is illustrative o f this indicator. Here the battery dc voltage is the input and the output is the meter pointer position. Connecting the meter probes to the batter)'^ terminals amounts to applying a step input to the voltmeter circuit that drives a second-order mechanical system consisting o f a spring and mass with friction. Naturally, one would like the pointer to settle on the proper voltage reading quick ly. If the mechanical system is underdamped, then the pointer oscillates (undesirably) for a short time before resting at its final position. On the other hand, if the mechanical system is over damped, the pointer will not oscillate but may take a long time to reach its final resting point. This also is undesirable. A near critically damped response is the most desirable one: the pointer will come to rest at the proper voltage as quickly as possible without being oscillatory, and small changes in the mechanical system will not make it oscillatory. In the next example, we reverse the process o f analysis and ask what the original circuit parame ters are given a plot o f the response that might have been taken in a laboratory.

E X A M PLE 9 .8 . Consider the circuit o f Figure 9.10, which shows the response,

of a

(relaxed) series RLC circuit to the voltage input i/y^(r) = 10«(r) V. In laboratory, you have meas ured the capacitor voltage values (approximately). If the response has the form vît) = Ke~^^ cos(ojy + 0) + vYp find

a , 0, K, and the values o f R and C i f it is known that L = 0.5

H. Your lab instructor has told you that toând a are integers.

m

(a)

\

___________,

406


(b)

F IG U R t 9.10 (a) Series RLC\ (b) Response to

= 10 u{t) V.

TABLE 9.3 Tim e (sec)

0.316

0 .5 2 3 6

0.839

1.5708

vcKt) (V)

10

13.509

10

10.432

First crossing

First

Second crossing

Second

o f 10 V

peak

o f 10 V

peak

S olution Step 1. FindXp. By inspection, the curve is settling out at X p - 10 V.


Step 2. Find

40

Now obscr\'e that the first two crossings o f v^^t) = 10 occur at r = 0 .3 1 6 sec,

0.839 sec (Table 9.3). This means that a full k radians is traversed by tlie cosine over [0.316, 0.839] , which is a half c>de or half period. So the period o f the cosine is 7 = 2(0 .8 3 9 - 0.316)

2 ti

= 1.046 sec, making CO^/ =

= 6.007 = 6 rad/sec.

Step 3. Find O . From Table 9.3, we know that two successive “peaks” occur at

= 0.523 sec

and t-y = 1.5708 sec. This means that for ^ = 1 ,2 , cos(COj//. + 0 ) + X/r

(9.36)

After some manipulation, equation 9.36 implies

COS(Ci),y/l + 0 )

C0.s((0^/r2 + 0 )

Thus cos((0 ,//| +Q) V c i t j ) - ^ ^,-a(/2-/,)

Xf,-

cos(co^y/'> + 0 )

(9.37)

Equation 9.37 simplifies because two adjacent positive peaks must be 2 tu radians apart, i.e., (co/2

+ ^) =

which means cos(to^^^ + H) = cos((i)y, + 0). It follows that

X’c d o - ^ F

3 .509

(9.38)

Solving leads to a = 2. Step 4. Find 0 and K. At the first crossing o f 10 V, we have 0 =

0.316

0 . 3 1 6 + B)

Thus 6 X 0.316+ 0 must equal 0.5?! or 1.5tt radians. We also know that since

=0 =K

cos(0) + 10, we must have K cos(0) = - 1 0 . Since A"> 0 (always by convention), the value o f cos(0) must be negative. This means 0.57t < 0 < 1.5Tt. So therefore, it must be that at the first cross ing o f 10 V D 6 X 0 .3 1 6 + 0 = —

or 0 = 2.1864 rad.

Therefore K = — ^— = 10.553. cos(0) Step 5. Find R and C. We know that the characteristic equation o f the series RLC circuit must be

Therefore

f? 1 02 ^ ^ s~ H—- . V.V +H-------= 2Rss +H—— = (5' + 2)*' + 6 " = s~ + 4.V + 40 ------- = .v“ + IR L LC C /? = 2 n and C = 0 .0 5 F

-H)K

Chapter 9 • Sccontl Order Linear Circuits

In the previous examples one obsen'e that the characteristic equations are independent o f the source values. I'his is a general property of linear circuits with constant parameters. Hence when constructing the characteristic equation we may without loss of generality set independent source values to zero; i.e., independent voltage sources become short circuits and independent current sources become open circuits. With this operation, some circuits that appear to be non-series/parallel, become series/parallel. This allows us to easily compute the characteristic equation and then use Table 9.2 and physical reasoning to obtain the solution without having to construct the dif ferential equation explicitly. The following example illustrates this procedure for a pseudo-parallel/.series RLC. The example will also illustrate the computation o f initial conditions due to past excitations and the computation o f the complete response w'hen the input changes its dc level.

EXA M PLE 9.9 . The circuit of Figure 9. l i b is driven by the input o f Figure 9.11 a, i.e., vj^t) -60//(-r) + Gû{t) + 60//(/‘ - 1) V. Our goal is to find the response

for / > 0.

FIG URE 9.11 (a) Input cxcitation whose dc level changes at r = 0 and t = 1 second, (b) A pseudoparallel RLC circuit; i.e.,when the voltage source is replaced by a short, the circuit reduces to a paral lel RLC whose characteristic equation is p- + — RC

s+—

= 0.

LC

So l u t io n Step 1. Analysis at 0“. Here the circuit has been excited by a constant - 6 0 V level for a long time. Therefore at ^ = 0", the capacitor looks like an open circuit and the inductor a short cir cuit. Because the inductor looks like a short, the entire - 6 0 V appears across the 6 i l resistor, making

= - 60 V and /^(O") = - 60/6 = — 10 A.

Step 2. Analysis at 0^. By the continuity o f the capacitor voltage and the inductor current, the equivalent circuit at 0"^ is given in Figure 9.12.


409

= -6 0 V

^ (0 *)= i,(0 ) = -10A

FICJURE 9.12 Equivalent circuit for analysis at 0^; the capacitor is replaced by a voltage source o f value

and the inductor by a current source o f value /^(O*) = //;(0“).

From the circuit diagram o f Figure 9.12, v^{0*) = 60 - ( - 60) = 120 V,

= - 60/6 = - 1 0 A, and

iff-) = if^{0*)l5 = 40 A. It follows that Step 3. Find the characteristic equation and the form ofthe response using Table 9.2. To find the char acteristic equation, we set the independent voltage source to zero. I'he resulting circuit is a paral lel RLC with characteristic equation -) 1 I T -) .v“ + ----- .V+ ------= .v“ + 4.V + 4 = (.V + 2 )“ = 0

RC

LC

where R = 2 LI is the parallel combination o f 6 Q and 3 i i . The characteristic roots are sî = - 2 , which correspond to a critically damped response o f the form (Table 9.2)

V(4,t) = (/f,

+

Kjt)

exp(j,

t) + Xf:

Step 4. Find constants in the response form for 0 < /■< 1. The input is constant for 0 < ^ < 1, but changes its value to 120 V at r = 1 sec. However, the circuit does not know the input is going to change, and so its response behaves as if the input were to remain at 60 V for all time: the circuit cannot anticipate the future, and thus its response over 0 < r < 1 behaves as if no further switch ing were going to occur. If no further switching were to occur and if the input remained at 60 V, then in Figure 9.10b for large t the capacitor is an open circuit and the inductor is a short circuit; hence Xp= 60 V. Under these same conditions we find

and K-y. To find A^j, observe that from = - 6 0 V. Evaluating the response form o f step 3 yields + Xp. Equating these rwo expressions produces - 6 0 = ^'(;(0■^) = + X^ which implies thar = - 1 2 0 . To calcu late Kj, observe that from step 2, /(-(O^) = 40 A. Since C = 0.125, it follows that

step 2,

c!t H en ce,

K-,

C

/={)■'

= 320 =

+ A'2 = 240 + ^2

= 8 0 . T h u s, the response o f the circu it for 0 < r < 1 is

V({t)

= (-1 2 0 +

m)e~^‘ +

60 V


410

Similarly, one can compute, for 0 < r < 1,

il{t) = (-2 0 +

+ 10 A

Step 5. Analysis at t = 1“. Although the circuit does not know the input will change at r = 1 sec, we do and we must prepare for the analysis for f > 1. To do this we must evaluate the initial con ditions at / = 1“ and then use the continuity o f the capacitor voltage and inductor current to obtain the initial conditions at / =

At r = 1“, using step 4 we have V(^\~) =

= 54.59 V

and/^d") = /^ (r) = 10 A. Step 6. Analysis att= I

This step mimics step 2 for r = P . The capacitor is replaced again by an inde

pendent voltage source and the itiductor by an independent current source as shown in Figure 9.13. Here v^{V) = 1 2 0 - 54.59 = 65.41 Vand

=-in\ +

+ //(O'*') =

= - 9 .0 9 8 + 21.8 + 10 = 22.71 A.

’

^ 120 - 54.59 ^ I

6

3

>

Step 7. Computation o f the responsefor t> 1. Because the characteristic equation is independent o f the input excitation, the form o f the response is almost the same as in step 3, except for the replacement o f t by (/ - 1); this substitution follows by the time invariance (constant parameter values) o f the circuit. Thus, for r > 1,

v^t) = \K^ + K\{t- 1)1 e x p U ,(r- 1)] Since the source excitation for r > 1 is 120 volts, by inspection o f Figure 9.1 lb V. To find /f], 54.59 = ^'c^l"^) = ^p+ A',. This implies

f^ôo) = 120

= -6 5 .4 1 . Finally, to find K-, consider

that = dt

w hich makes

1 8 1 .7 = .ViA:, + K2 = 1 3 0 .8 + K 2

/=r

= 5 0 .9 . T h u s, for / > 1,

v^t) = [ - 6 5 .4 1

+ 5 0 . 9 ( r - l)]^>-“( ' -

+ 120 V

(9 .3 9 )

Chapter 9 • Scconcl Order Linear Circuits

41

Time in seconds FIGURE 9.14 Complete analytical response of the capacitor voltage for 0 < r < 3 sec.

Exercise. Fill in the details for the computation o f iît) = (-2 0 + 20/)^^ + 10 A for 0 < r < I and then compute /y(/) for 2.5 > ^ > 1. Also, compute /^(/) for t> 2.5.

Despite the idea illustrated in Example 9.9, many second-order RLC circuits are not reducible to series or parallel RLC circuits when the independent sources are set to zero. Furthermore, when a dependent sourcc is present, the circuit is generally not reducible to a series or parallel RLC. In such cases one ordinarily uses a systematic methodolog)' to compute the circuit’s differential equa tion and, subsequently, the characteristic equation. This systematic procedure is described in more advanced texts and in the second edition o f this text. Nevertheless, for some situations one can use the earlier method s integro-differential equations, which must be differentiated again to eliminate the integral. This is illustrated in the example o f the next section.

•il2


5. OSCILLATOR APPLICATION An imporrant difference between first-order and second-order linear networks is the possibilit)' o f oscillatory responses in the latter. In some applications sinusoidal oscillations are intended responses, while in other applications oscillations arc undesirable. This section presents an exam ple o f a Wien bridge oscillator circuit. The goal is to build a circuit that generates a pure sinusoidal voltage waveform at a specified fre quency. In theory, as per section 2 of this chapter, this is achievable by discharging a capacitor through an inductor. In practice, both capacitor and inductor have losses. Losses cause the oscil lation amplitude to decay eventually to zero. For sustained sinusoidal oscillations, some “active” element such as a controlled source or op amp must replenish the lost energy. Note that these active elements require a dc power supply for their operation. Ultimately the dc power supply replenishes the power losses due to various resistances in the circuit.

E XA M PLE 9 .1 0 . Figure 9.15 shows a Wien bridge oscillator constructed with an op amp as the active clement. Find the condition on the circuit parameters R^,

and C for sustained sinusoidal

oscillation, and the frequency' o f oscillation. r

R.

(b) FIGURE 9.15 (a) Wien bridge oscillator, (b) Equivalent circuit. S o l u t io n

From the principles described in Chapter 4, the non-inverting amplifier enclosed in the dashed box of Figure 9.15a is equivalent to a voltage-controlled voltage source with a gain equal to {2Rj- + RJ}IRr= 3. (See Chapter 4.) Replacing the dashed lx)x with this equivalent yields the simplified circuit o f Figure 9.15b. Using the simplified circuit, the first task is to derive die differential equation model of the circuit. Step 1. Write a single-loop equation. (9 .4 0 a )


413

To eliminate rhe integral, we differentiate again to obtain

dt

(9.40b)

7?1 dt

/?,C

Step 2. Express /^’j in terms o f V2 - By inspection o f Figure 9.15b we obser\'e that ,

V’2

(9.41a)

^c\ ~ ^ ----- '---dt

/?,

dt^

l<2

and thus, differentiating again, (9.41b)

dt

Step 3. Substitute equations 9.41 into equation 9.40b. Substituting as indicated yields

dt~

/?2 dt

R^C ^

dt

Rj y

- ± ^ =0 /?, dt

(9.42a)

Grouping terms and dividing by C produces

d Vf

1 dv-> 1 dv-y - + ----------- ^ + ---------- ^ + dt~ Ro_C dt /?,C dt R^R^C-

R^C dt

which simplifies to

dvo

d-V2 dt~

RjC

Vo

=

0

(9.42b)

R\C}

Step 4 . Compute the characteristic equation and determine the conditions for sustained oscillations. The resulting characteristic equation is

s~ + hs + c = s~ +

1

1

\ R ,C

= 0

R ,C )

(9.43)

For sustained sinusoidal oscillations to occur, the roots must be purely imaginary. Thus the coef ficient o f s must be zero, i.e.,

b=

R2C

/?,c

-

^

/e|/?2C

=0

(9.44)

Thus the condition for sustained sinusoidal oscillations reduces to /?, = R-,. Step 6. Find the frequency o f oscillation. Under the condition R^ = Rj, the roots o f the character istic equation are /?,C We conclude that the frequenq^ o f oscillation (in rad/sec) is (Oo =

(9 .4 5 )

/?,C

A14


> R^, then ^ > 0 and the unforced response is an exponentially decreasing sinusoid. On the other hand, if R^ < Rj, then b <0, and the unforced An examination o f equation 9.44 shows that if

response is an exponentially growing sinusoid. For the oscillations to start, the value of/?| should be designed to be slightly smaller than Rj- Then the value for b in equation 9.58 will be negative, producing an exponentially growing sinusoidal response. If all circuit parameters are truly con stant, the amplitude o f oscillation would theoretically grow to infinit)'. In real oscillator circuits, such growth is limited to a finite amplitude by saturation effects or nonlinearities that clamp the response when the voltage swing grows large. The resulting waveform then only approximates a pure sine wave. The analysis o f this nonlinear effect is beyond the scope o f this book. However, the next example illustrates the growing oscillation when

< /?2 and also shows the effect o f sat

uration to produce an approximate sinusoidal oscillation.

EXA M PLE 9 .1 1 . The circuit o f Figure 9.16a is a B2 Spice schematic for the Wein bridge oscilla tor o f Figure 9.15. The op amp is a 741 with = 0. Observe that

= 10

= 15 V. Suppose that

(0) = 10 m V and

= 9.5 kQ. According to the analysis o f Example 9.10, the out

put voltage labeled IVout should be a growing sinusoid. The output response o f Figure 9.16b shows this growth and the saturation effects induced by the op amp. The waveform is not a pure sinusoid due to these saturation effects. Also note that the frequency o f oscillation is approximately 16 Hz, which is consistent with equation 9.45, i.e.. =

2k

C2

R1

(a)

2 k ^R^R2 C

16.3 Hz

15


Example 9.11 Oscillator-Transient-4

Time(s)

(b)

FIGURE 9.16 (a) Schematic diagram of Wein bridge oscillator, (b) Voltage response showing grow ing oscillation clamped at ±15 V due to saturation effects of op amp.

An alternative approach to initiating oscillations and simultaneously limiting amplitude is to use a temperature-sensitive resistor, R^, with a positive temperature coefFicient. Any incandescent lamp is an example o f a temperature-sensitive resistor. For small voltages the temperature o f an incandescent lamp is lower than for larger voltages because the dissipated power is lower. Hence the lamp temperature (and thus its resistance) increases with increasing voltage. In the case o f our oscillator, we have a desired output voltage swing. The nominal value o f /?, is designed to be slightly less than R-y when the output voltage swing is bclow' a pre-specified voltage less than This causes a growing oscillation. As the voltage swing increases, the temperature o f R^ and thus its resistance increase. When the resistance o f /?, reaches Rj, the amplitude will settle (stabilize) at the pre-specified voltage swing, at least theoretically. If /?, happens to increase beyond R^, a decay ing sinusoid wîll result, decreasing the temperature and hence the resistance o f R^. Should the amplitude o f oscillation decrease for any reason, /?j will decrease, causing a growing sinusoid. Although the resistance o f R^ may dither about /?2>

amplitude o f oscillation will nevertheless

restore itself to the equilibrium level. In practice this equilibrium level only approximates the spec ified value due to imperfections in the circuit parameter values. The resulting waveform is almost a pure sinusoid.

4 16


6. SUMMARY 'I'his chapter has explored the differential equation modeling and response computation of^ sec ond-order linear circuits having either no input or constant input excitation. Such second-order circuits contain at least two dynamic elements, either an LC, CC, or LL combination. Secondorder circuits may also contain active elements such as op amps. In contrast to first-order circuits, second-order linear circuits allow for the possibilit)^ of damped and undamped sinusoidal oscilla tions. Analysis o f second-order linear circuits has two phases. Pha.se 1 entails the formulation o f the sec ond-order differential equation circuit model. For simple I C parallel RLC, or series RLC, the cir cuit model can be found by inspection. Phase 2 o f the development centers on the solution o f the second-order differential equation model o f the circuit. I'he first step here is to compute the (quadratic) characteristic equation and then solve for the two roots. The roots o f the characteristic equation determine the t}'pe o f response. The three t)'pes o f roots for a quadratic— real distinct, real identical, and complex— specify the three response types of overdamped, critically damped, and underdamped, respective ly. These three types o f responses characterize all second-order linear differential equation models, be they o f electrical circuits, mechanical systems, or electro-mechanical systems. Since sinusoidal waveforms are germane to many electrical systems, this chapter presented an oscillator circuit that generates a sinusoidal waveform. O f the many types of oscillator circuits, we chose one containing an RC circuit built around an op amp, avoiding the use o f an inductor.


41

7. TERM S AND CO N CEPTS Characteristic equation: for a linear circuit described by a second-order difTerentia! equation

= j{t), the algebraic equation

,v"(f) +

+ r = 0 is called its characteris

tic equation. Characteristic roots: roots o f the characteristic equation, also called the natural frequencies o f the linear circuit. Critically damped circuit: a second-order linear circuit having characteristic roots that are real and identical. The source-free response o f such a circuit has a non-oscillatory waveform, but is on the verge o f becoming oscillator)'. Damped oscillation frequency: in an underdamped second-order linear circuit, the source-free respon.se has the form [Kc^^' cos(o)y + ()}. The angular frequency

is the damped o.scil-

lation frequenc)', which is the magnitude o f the imaginary part o f the characteristic roots. Homogeneous differential equation: a differential equation in which there are no forcing terms. For example, x"{t) + bx\t) + cx{t) = 0. Natural frequencies: the characteristic roots. Oscillator circuit: an electronic circuit designed to produce sinusoidal voltage or current wave forms. Overdamped circuit: a second-order linear circuit having a characteristic equation whose charac teristic roots are real and distinct. Scaled sum o f waveforms: let

...

be a set of waveforms. A scaled sum of these wave

forms is an expression o f the form /r) =

+ ... +

for real (possibly complex)

scalars ^/j, ... , Second-order linear circuit: a circuit whose input-output relationship may be expressed by a second-order differential equation o f the form .v"(/') +

+ cx{t) = /(r).

Source fi-ee: there are no independent sources, or all independent sources have zero values. Step function: a function equal to zero for r < 0 and equal to 1 for r > 0. Step response: the response o f a circuit to a step function input when all capacitor voltages and inductor currents are initially zero. Undamped circuit: a second-order linear circuit where the characteristic roots are purely imagi nary and the unforced response is purely sinusoidal. Underdamped circuit: a second-order linear circuit whose characteristic roots are complex with nonzero real part. ' T h e notations

and KO arc used interchangeably in the literature to den ote the first derivative ol v{t).


418

PROBLEMS

where all coefficients are real (but not necessar ily positive). (a) Prove that x(r) = 0 at some r = T, 0 <

TH EO RY RELATED

r < 00, only if

1. In section 2, the solution to the undriven LC circuit is given by v^^t) = K cos(cof + 0) V, Observe that

and /f, have oppo

site signs. (b)

Prove that the x(^) vs. t curve has at most one zero crossing for r > 0.

(c) cos(tt)0 = - W “ cos(0)/)

State the necessary and sufficient condi tions on the coefficients

>

and Sj for the presence o f one zero and

d

— 7 sin((0 / ) = - 0 )

2

crossing. sin(coO

dt^

5. The voltage or current in a second-order

imply that A cos(tijr) and B sin(tof) are both solutions to the differential equation

eral form

x(,) = (A:, + K 2 t)e-'>^'

d^\>C dr

source-free critically damped circuit has the gen

where all parameters are real, but not necessar

LC

ily positive. Prove that x(/) = 0 at some t = T <

By superposition, then, v^t) = A cos(ojr) + B sin((or) V. Show that for a given A and B, there exist K and 0 such that Vf\t) = A cos(tof) + B

6. (a)

underdamped

v^t) =

and

satisfies the differential equation

x"{t) + 2}^'{t) + h~x{t) = 0 and AT, are arbitrar)^ constants,

4. The voltage or current in a second-order source-free overdamped circuit has the general

[10 cos(9950^) + 1.005

the voltage waveform before the peak

W^{t) is constant.

x{t) = (A'l + K^j)e-^‘

the

+ 5.7°) How many cycles o f “ringing” occur in

energy stored in C and L for the circuit o f

3. By direct substitution, show that

and

sin(9950/)] = 10.05^’" ‘®‘^®' cos(9950f

2. Find the expressions o f the instantaneous

form

is

response is given by

the two solution forms are equivalent.

where

Consider case 2 o f Example 9.5. The circuit

sin(cof) = K cos(tor + 0). One concludes that

Figure 9.1b. Show that the sum o f

and K j have opposite signs.

COif and only if

value drops from its largest value o f 10.05 to 10.05/f = 0.3 6 8 X 10.05? (b)

Suppose the characteristic polynomial is written as response

+ 2as +

form

x{t)

+ to^ with =

cos(w y + 0). Prove that for the under damped case, the circuit will ring for

N = iy ijiln o) cycles before the ampli tude decreases to Me o i its initial value. 7. When a dc voltage o f

volts is applied to a

series LC circuit with no initial stored energy, the voltage across the capacitor reaches a peak value o f twice the source voltage. To investigate

x{t) =

419


this phenomenon, consider the circuit o f Figure

UNDRIVEN RLC PROBLEMS

P9.7 where switch S is closcd at r = 0. Assume

9. The switch S in the circuit o f Figure P9.9 has

the inductor current and the capacitor voltage

been closed for a long time and is opened at / =

are zero at / = 0.

0. Express vj,t) and /^(r) for r > 0 in terms o f the literals R, L, C, and /q. Also compute the

t =0

initial stored energy in the inductor and capac S ^

itor.

L

Figure P9.7 Show that for f > 0

Figure P9.9 1

.V l

] c

',

and

10. In the circuit o f Figure P9.10, suppose

Vi„{t) = 10 V, /? = 10 n , C = 0.4 mF, Z = 0.25 H, and the switch opens at /^= 0.

yc(t) = Vo 1 -

COS

f

(a)

Compute

, //(0~), and

(b)

Compute the energ)^ stored in the

(c)

Using only energy considerations,

1

■Jlc\

inductor and the capacitor at / = 0.

8. The circuit in Figure P9.8 is a dual o f the

compute the maximum value o f

previous problem.

f > 0. (d)

Find the analytical expression for and verify the maximum value of

Figure P9.8

computed in part (c).

R

i,(t) ,v (t)

v„(t) I

The switch S is opened at r = 0. Both the induc tor current and the capacitor voltage are zero at r = 0. Show that for / > 0

Figure P9.10 11.

vU)

Consider the circuit o f Figure P 9 .ll in

which i r jf ) = -2 0 u (-r ) V, R = 10 Q, C = 0.4 mF, and L = 0.25 H.

and ' l ( 0 = / o 1 - cos

f

1

(a)

Compute

/^(O"), and

(b)

;,( 0 -) . Compute the energ}' stored in the inductor and the capacitor at r= 0.

(c)

Find the analytical expression for Plot using MATLAB for 0 < / < 200 msec.


420

(d)

Find the analytical expression for Plot using MATLAB for 0 < r < 200 msec.

i,(t) v„(t)

(a) (t>0)

Figure P 9.11 12. Reconsider the circuit o f Figure P 9 .ll = 50«(-/) V, R = 25 Q, C = 0.8 mF, and Z, = 1 H. Repeat Problem under the conditions 11.

13. Reconsider the circuit o f Figure P 9 .11 under the conditions v-^{t) = -5 0 « (-r) V, R = 25 O., C

Figure P9.15

= 0.8 mF, and Z, = 2 H. Repeat Problem 11. 16. Figure P9.16 shows an overdamped source14. For the circuit o f Figure P9.14, suppose C = 0.8 mF and determine L so that the fre quency o f the sinusoidal response, for r > 0, is 2500 rad/sec. Now find y(;;(0~),

free circuit in which R = 0.4 Q., L = 0.5 H, and C = 0.5 F. (a)

/(^0“),

If VfÔ) = - 2 V and /^(O) = 2.5 A, find

V(it) for r > 0. Use MATLAB or the

/Ô"^), and Vf^t) for r > 0.

equivalent to plot the v^^t) waveform and verify that there is no zero cross

^

t=0/\ L

25 0

_L

100 0

©

ing. 25 mV

(b)

If vÔ) = 2 V and /^(O) = 2.5 A, find

v^t) for t> 0 . Use MATLAB or equiv alent to plot the V(^t) waveform and verify that there is no zero crossing.

Figure P9.14 15. Consider the circuit o f Figure P9.15, in which = 2 QV, R^ = 2 Cl, Rq = %Q., R = 2 Q, Z. = 0.5 H, and C = 62.5 mF. (a)

Figure P9.15 a shows a source-free parallel RLC circuit whose past history

Figure P9.16

is depicted by Figure P9.15b where the switch S has been at position A for

17. In Figure P9.17, the switch S has been at

a long time before moving to position

position A for a long time and is moved to posi

B at r = 0. Find

y^Ô'*^), /^0~),

/Ô"^), and V(^t) for r > 0. Plot V(^t) (b)

tion B at t = 0. Suppose 0.5 a

using MATLAB for 0 < f < 1.25 sec. Find i^{t) for / > 0. Plot V(^t) using

(a)

MATLAB forO < r< 1.25 sec.

(b)

= 100 mV, R =

/. = 1 H, and C = 0.01 F. Find

and

for f > 0.

Plot for 0 < / < 50 sec. Find i^{t) for r > 0. Plot for 0 < / < 50 sec.

421


20. The voltage across the capacitor for the cir cuit of Figure P9.20a is given by Figure P9.20b. Suppose R = 25 k£2. v J t)

(a)

Using the plot, estimate the values of

L and C (b)

Figure P9.17

O '

18. For the circuit of Figure P 9.18,

current.

=

Clearly show and explain all steps in your cal

0,5u{-t) A. (a)

culations. Hint: You might assume a general

I f /? = 20 Q , Z = 1 H, and C = 8 mF, find and plot V(^t) and

(b)

Now estimate the value o f the initial capacitor voltage and initial inductor

response form v^^t) = Ae~^^ cos(co^ + 0) V and

for f > 0.

then use the plot to estimate a and O)^; what is

Repeat part (a) for R = 22.5

the relation of a and Ci)^ in the characteristic polynomial of a parallel RLQ

0

ijt)|

Figure P 9.18 19. In Figure P 9.19. ^^<0") = 25 V, /^(O-) = 50 mA, /? = 2 kQ, Z = 0.1 H, and C = 0.1 ;/E The switch closes at f = 0.

3

(a)

Compute y(;;(0^), ^^(0"^), and /^(O^).

(b)

Compute /^(/),

(c)

Plot i^(t) and Vjit) forO < t< 1 msec. |

(d)

and V({t).

| |

Find the energy stored in the circuit ^ over the interval [0, 0.2 msec], i.e., in the capacitor and the inductor over this interval. Is this energy positive or negative? Also compute the energy dissipated in the resistor over this same (b)

interval. The sum o f the energy dissi pated in the resistors, the energy

Figure P9.20

stored in the capacitor, and that stored in the inductor should equal zero.

21. In the circuit of Figure P 9.21, R = AO., C

Why?

= 6.2 5 mF, and i,(t)

/YYV< L

for f > 0. (a) (b)

t=0

Figure P9.19

Compute the value of L. Find the value of the initial condi tions, ^^(0*) and

(c) 'n^

= (20 - \200t)e~^^* mV

Find ijit) for t> 0.

422


i jt )

/ Y W


22. For the circuit o f Figure P9.22, L = 0.04 H, C = 2.5 mF, and /^, = 10 LI. (a)

Find the value o f R (in ohms) that

25. Figure P9.25 shows a critically damped

makes the circuit o f Figure P9.22 crit

source-free circuit.

ically damped. (b)

0.1 H

Given this value o f R, suppose V(^Q) = 160

mV

and

/^(O)

= -6 0

mA.

Compute V(\t). (c)

40 0 0.25 mF

Determine the first time at which the

+ sVc(t)

capacitor voltage is zero. Plot your result using MA'FLAB or the equiva

F'igure P9.25

lent to veri5' your calculation. (a)

If vÔ) = - 5 V and /^(O) = 1 A, find ;^(f) for r > 0. Determine the differen

,_ r Y Y \ =

tial equation for the circuit. Let^ = ij and use “y = dsolve(‘D2y + 400*D y + 40e3*y = 0,y(0) = l,D y(0) = -3 5 0 ’)” in MATLAB to verify your answer. Figure P9.22

(b)

If i;JO ) = 5 V and /^(O) = 1 A. find

V(^t) for t> 0. Plot the V(^t) waveform 23. Reconsider the circuit o f Figure P9.22 with

and verify that there is one zero cross

/?///?, = 0.8

ing. Again use the dsolve command in

(a)

and A = 0.04 H.

MATLAB to verif}' your calculations.

Find the value o f C so that the circuit is underdamped with to^= 100 rad/sec.

(b) (c)

Suppose = 160 mV and /^(O) = - 3 0 mA. Compute Determine the first time at which the

26. The capacitance voltage o f a source-free parallel RLC circuit, with R = 2.4 LX has the form

V(^t) =

capacitor voltage is zero. Plot your result using MATLAB or its equiva

(a)

lent to verify your calculation.

(b)

cos(8r + B)

Find the values o f L and C. If y^Ô) = 10 and /^(O) = 0, determine

A and 0. 24. In the circuit o f Figure P9.24, /? = 20 Q and

(c)

If the values o f L and C remain unchanged, find the value o f R for the

//:.(0 = 500f?"'^' sin (l0 V 3 /)

circuit to be critically damped, and the

mA for r > 0. Find the proper values o f L and

general

C to produce this response. Now find t'^Ô'^),

response under this condition. Then determine the source-free response when V(^0) = 10 and /^(O) =

y'cÔ"^), and vît) for r > 0.

0.

form

of

the

source-free

423


27. Almost 75% of fliilures in circuits, i.e., situa tions where a circuit dramatioilly fails to perform as designed, are due to opens and shorts o f indi vidual circuit elements. Heating, c\'cling a circuit on and off, etc., cause degradation in the circuit parameters, resistances, capacitances, inductances, etc. that often precipitates the short or open situa tion. For example, the material inside a resistor might become brittle over a period of time and finally crumble, leaving a break in the circuit. On the other hand, the material might congeal or become dense, decreasing the resistance. In the problems below you are to determine the length of time it takes for a circuit to move from an over damped behavior to an imderdamped behavior due to changes in the resistor characteristic as a fiinction of time. (a) For the parallel RLC circuit in Figure P9.27a, suppose R = Rq + exp(/ - 5) H where f > 0 constitutes time in years. Determine the time f’ for which the cir cuit changes its behavior from over damped to underdamped. (b) For the series RLC circuit of Figure P9.27b, the resistor satisfies R = + exp(r - 5)] n , where again t is time in years. Here it is presumed that the circuit is pan o f a larger piece o f electronic appa ratus, such as a TV, which is used exten sively over a period o f years. The tiine t' then is not connected with the response time o f the circuit. Determine the time f ' for which the circuit changes its behav ior from overdamped to underdamped.

DRIVEN SERIES AND PARALLEL /?/.C CIRCUITS 28. (Initial condition calculation) For the cir

//(O'*'),

cuit shown in Figure P9.28, find ;^ 0 ^ ), and

in two steps:

Sff/> I. Find V(^0~) and /^(O") by open-circuit ing C and short-circuiting L. Step 2. Construct a resistive circuit valid at t = O'*’ and from this find

29.

and

Consider the circuit o f Figure P9.29 in

which

= -10w(-/) + 20«(f) V, 7? = 20 Q,

C = 0.1 niF, and L = 0.25 H. (a)

Compute

/^(0“), and

(b)

Compute the energ)'^ stored in the inductor and the capacitor at r = 0.

(c)

Find the analytical expressions for the zero-input, zero-state, and complete responses for

Identif}' the tran

sient and steady-state responses. Plot

V(^t) using MATLAB over [0, 40 msec]. (d)

Find the analytical expressions for the zero-input, zero-state, and complete

R=

R„ = 0.8 Q

responses for i/{t) . Plot ij{t) using MATLAB over [0, 40 msec].’ i,(t)

ijO

L= 1H

R„ = i s n C = 1/36

(b)

Figure P9.27 Parallel and series RLC circuits subjccc to resistor degradation over time.

30. Reconsider the circuit o f Figure P9.29 under the conditions

+ 25u{t) V, /? = 25

Q, C= 0.8 mF, and Z, = 2 H. Repeat Problem 29 but construct plots over [0, 400 ms].


424

31. Reconsider the circuit of Figure P9.29 under tiie conditions = -50«(-r) +25u{t) V, 7? = 25 ii, C = 0.8 mF, and L = 0.2 mH. Repeat Problem 29 but construct plots over [0, 40 ms]. 32. In Figure P9.32 v-J^t) =-250«(-r) +750«(f) mV, ^ =0.5 a , I = 1 H, and C= 0.01 F. (a) Find /^(O^), and the zeroinput, zero-state, and complete responses of vît) for r > 0. Identify the steady-state and transient parts of the complete response. Plot in MATLAB for 0 < ^< 50 msec. (b) Find the zero-input, zero-state, and complete responses of i^{t) for t > 0. Plot in M ATLAB for 0 < r < 50 msec.

ijt)

fY Y \ h ^

36. In Figure P9.36,

Figure P9.32

T he switch closes at f =0.

(a) (b)

(c)

ComputeyÔ*), y^(O^), and /^(O^). Compute the zero-input, zero-state, and complete responses of and v^t). Identify the transient and steady-state parts of the complete response. Plot iît) and Vjit) for 0 < r < 1 msec. Find the energy stored in the circuit over the interval [0, 0.2 msec], i.e., in the capacitor and the inductor over this interval. Is this energy positive or negative? Also compute the energy dissipated in the resistor over this same interval.

o

n

i,(t)

33. Repeat Problem 32 for /? =40 Q and v-JJ) = -0.5tt(-/) + 2«(/) V. Plots in M ATLAB should be for 0 < f < 800 msec. 34. Repeat Problem 32 for /? = 50 Q and and v.^{t) =-0.5«(-/) - 2u{t) V. Plots in M ATLAB should be for 0 < ? < I sec.

=-1 0 « (-^ ) + 40«(/)

mA, /? = 4 k n , Z = 0.1 H, and C = 0.1 pF .

(d)

vJt)

o

Figure P9.35

fY Y \ k L

+ ,Vc(t)

(t)

Figure P9.36 35. For the circuit of Figure P9.35, = -0.5«M + 2u{t) A. 37. For the circuit of Figure P9.37, R^= 5 0.^ (a) I f /? = 2 Q, Z = 1 H, and C = 8 mF, = 2 0 a , C = 2.5 mF, L = 0.25 H, and v j t ) = find and plot the zero-input, zero- 20«(^) - 20u(t-7) V, where T = 0.25 sec. state, and complete responses of v^{t) (a) Find the zero-input, zero-state, and and ij\t) for / > 0. Identify the tran complete responses of v^t) for f > 0. sient and steady-state parts of the Plot the complete response for 0 < r < complete response. 0.25 sec. (b) Repeat part (a) for R= 22.5 Q.

n

o

o

n


(b)

■li')

Find the zero-input, zero-state, and

42. Repeat Problem 40 for

complete responses o f

200 n .

for t > 0.

= 50

and Rj =

Plot the complete response for 0 < r < 4 3 . Consider the RLC circuit in Figure P 9.43

0.25 sec.

where /?^, = 60 Q, R^2 = = 5m F (a)

Q, I = 4 H, and C

= 100«(-f) mA and r^2(^) = 2 0u (-t) V. Find the response,

(b)

for

t> 0. Plot for 0 < f < 1 sec. /y,(r) = 100u(-t) + 500u(t) mA and 1/^2 ^) = 20u(-t) V. Find the response, for ^ > 0. Plot for 0 < / < 1 sec.

= 50 Q, /?j = 200 Q, C = 0.05 mF, L = 0.5 H , and v j t ) = - 5 0 « (- r ) + 50k(?) - 50«(r - 7) V, where T = 38. Repeat Problem 37 for

0.08 sec. However, only plot the complete

I V ,(t)

responses for 0 < f < 200 msec. 39. Repeat Problem 37 for R^= 100 Q ,

=

100 Q , C = 0.25 mF, L = 2.5 H , and v-J^t) =

Figure P9.43

-5 0 tt(-f) + 50«(f) - 50«(^ - 7) V, where T = 100 msec. However, only plot the complete

44. Repeat Problem 4 3 , except find y^(r), ^> 0.

responses for 0 < ^ < 300 msec. 45. Consider the RLC circuit in Figure P9.43 40. Consider the RLC circuit o f Figure P 9.40 in

where Rî = 20 Q, R^2 = 20 Q, L = 0.4 H, C =

which R^ = 100 Q, /?! = 4 0 0 Q , C = 0.125 mF,

4 mF.

L = 0.2 H, and v.„{t) = 50u{t) - 50u U - 7) V, where T = 0.025 sec, vîQr) - - 2 5 V, and

(a)

/^(0“ ) = 10 mA. (a)

(b)

(r) = -u (-t) A and Find the response,

= 40«(-^) V. for t> 0 . Plot

for 0 < f < 0.8 sec.

Find the zero-state, zero-input, and complete responses of V(it) for r > 0.

(b)

(f) =

+ 2u{f) - 2uU - 0.4) A

and v^2 (^) = 4 0 « ( -/) V. Find the

Plot for 0 < r < 60 msec.

response,

Find the zero-state, zero-input, and

r < 0.8 sec.

for ^ > 0. Plot for 0 <

complete responses of vît) for f > 0. Plot for 0 < r < 60 msec.

46. Repeat Problem 4 5 , except find

t> 0 .

47. Consider the RLC circuit in Figure P9.43

w

where /?j| = 16

R^2 = 32 Q, I = 0.4

H, and C = 4 mF. w

(a)

'n- ^

f > 0. Plot for 0 < r < 0.3 sec. (b)

Vw>

i M = - 0 .5 « ( - / ) A and v,y{t) = 24«(-^) V. Find the response, for

41. Repeat Problem 40 for R^= 140 = 360 a

and Rj

/■^,(r) = -0.5u{-t) + 0.5«(f) A and v^2 ^t) = 24u(-t) V. Find the response, for f > 0. Plot for 0 < f < 0.3 sec.

426

Chapter 9 * Sccond Order Linear Circuits

r> 0.

48. Repeat Problem 47, except find

49. riic current source with /^|(^) = 5//(-/‘) mA and the voltage source

= 10 V in Figure

P9.49 drive the circuit in which /? = 1 k^2, C = 0.5 //F, and Z. = 0.184 H. (a)

Find

(b)

.,(0 ^ ). Compute

t'(-(0‘"),

/^(O"^),

/^(0‘^),

and

for r > 0. Plot for 0 < r

PSEUDO SERIES AND PARALLEL /?/.C CIRCUITS 52. Consider the circuit o f Figure P9.52 in which /?! = 4 a = 4 a /. = 5/12 H, C = 25 mF (a) Find the roots o f the characteristic equation. C H EC K : - 8 , - 1 2 (b)

/j^(0"), î(0*), and

< 5 msec. (c)

Compute Compute

for / > 0. Plot for 0 < f

ages and currents, draw the equivalent circuit valid for t > 0. Check your

for r > 0. Plot for 0 < r

answer for

< 5 msec. (e)

Compute

for r > 0 . Plot for

0

for t >

0. Hint: After finding the initial volt

< 5 msec. (d)

If v.^,{t) = -2Qu{-t) V, find 1/(^0-),

using the “dsolve”

command in MATLAB. Plot

for

0 < r < 2 sec using MATLAB or the

< 5 msec.

equivalent. Be sure you properly label your plot.

(0

If

(d)

If y,./r) = -2 0 u (-t) + 20u(r) -2 0 u (t- 1)

= - 2 0 « (- f) + 20u(r) V, find for / > 0.

V, find and plot v^^t) for 0 < r < 2 sec. Figure P9.49 50. Repeat Problem 49 for the new source cur rent /jj(r) = 0 < r <_10 ms.

-5 ti(r-

0.005) mA. Plot for

51. The switch in the circuit o f Figure P9.51 is in position A for a long time and moves to posi tion B at / = 0. Find the voltage

for t > 0

when L equals (a) 0.625 H, (b) 0.4 H, and (c) 0.2 H.

53. Repeat Problem 52, but find i^{t) and v^{t) without differentiating 54. Repeat Problem 52 for

= 4 Cl, Rj = 4 Q,

1 = 0.2 H, a n d C = 0 .2 F . 10V

55. Reconsider the circuit o f Figure P9.52 in which /?, = 600 Q, /?2 = \ 2 0 n ,L = 2 H, C = 1 Figure P9.51

mF and v j t ) = -72u{-r) + 72u(t) -72u{t - 1). Find the response

for r > 0 as follows:

(a)

Find

(b)

Find z^f;(0'*') and i^CO"*").

(c)

and

Draw the equivalent circuit valid at O'*" and find ^'/(O'^) and Z(;;;^0‘'').

^1/


(d) (e)

Find die characteristic equation and

60. Reconsider the circuit of Figure P9.57.

natural frequencies o f the circuit.

Suppose /?j = 80 Q, /?2 = 4 0 ^ = 2 H, and C = 0.625 mF with = -1 5 0 tt(-/) + 150«(/) mA.

Determine the general form o f the response, ij{t)y valid for 0 < ? < 1. Determine all coefficients in the gen

(a)

Find V(iO~) and V(^0*).

eral form o f the response.

(b)

Find ii(0~) and /^(O'*’).

(g)

Determine the form o f the response,

(c)

Find t//^(0+) and /’c(0+).

for f > 1. Find the response.

(d)

(h)

Plot the response

(f)

for / > 0 using

MATLAB or the equivalent.

Find the characteristic equation and natural frequencies o f the circuit.

(e)

Find the response,

(f)

Find the response,

t> 0. t> 0.

56. Repeat Problem 55, except find Vf4,t). 6 1 . Consider the circuit of Figure P9.61 57. Consider the circuit o f Figure P 9.57 in which

= 2 £2, /?2 = 2 Q, Z = 0 .4 H, and C =

0.1 E (a) (b)

Suppose /?j = 80 Q, R2 = 40 Q., L = 2 H, and

C = 0.6 2 5 mF with

= 300u(f) mA, Vq =

50 V, and /?3 = 20 Q. Find V(iO~) and y^Ô'*’).

Find the roots of the characteristic

(a)

equation.

(b)

Find i^(0-) and z^(0+).

(c)

Find v^{0*) and /’c (0 ‘*‘).

(d)

Find the characteristic equation and

= -2u{-t) A. find t;c<0-),

If

^> 0, Hint: After finding the initial volt

natural frequencies o f the circuit for t >0.

ages and currents, draw the equivalent circuit valid for ^ > 0. Check your

(e)

Find the response, v^t), t> 0.

answer for ij{t) using the “dsolve”

(f)

Find the response, ijXt), t> 0.

command in MATLAB. Plot

for

0 < ^ < 1 sec using MATLAB or equiv alent. Be sure you properly label your plot. (c)

If

= -2u{-t) + 2«(?) A, find

for ^ > 0.

(d)

If

= -2u{-t) + lu{t) - l u { t - 1) A,

find and plot

for ^ > 0.

62. The switch in the RLC circuit of Figure

'S-> ijt)

d)

P 9.62 opens at ^ = 0 after having been closed + > F L

for a long time. The purpose o f this problem is to find the complete response o f the capacitor voltage, ? > 0. Suppose

Figure P9.57 58. Repeat Problem 57, except find V^

^

(b)

(c) 59. Repeat Problem 57 with W

(t) = 1 A, v^2 (^) = 20

V, C = 4 mF, and I = 0.625 H. (a) Using a dc analysis, find the initial

Using a dc analysis, find the final value o f the capacitor voltage, Vfôo).

= 80 £2, R2

2 0 a , L = 10 H , and C = 1/240 E

conditions /^(O") and V(^0~). Find and

(d)

Find the characteristic equation and

■\2H

(e)


compute its roots. Given tiie roots,

tion with ij{t) as the unknown. Observe that

write down the general form o f the

the derivative o f

response

hand side.

is present on the right-

Solve for the unknown coefficients in the response form o f part (d) and write

ANSW'F.R: /"(/) + i\t) + i{t) =

^

down the exact expression for v^^t) valid for ^ > 0.

40 fi 40 0

Figure P9.62


GENERAL SECOND-ORDER CIRCUITS

P9.65. (a)

roots o f the characteristic equation.

63. In the operational amplifier circuit o f Figure P9.63 is a second-order circuit. Suppose

Write a second-order differential equa tion with the unknown. Find the

(b)

If

V, find V(^t) for r > 0.

= R^ = 50 k n . (a )‘

Determine the values o f C, and Cj that produce a characteristic equation having natural frequencies at - 5 and -

(b)

10 .

Adjust the value o f /?j so that for a step function input voltage, the value o f the output voltage for large t is for all practical purposes is 5.

(c)


when v^{t) = 2u{t)

P9.66.

and all capacitor voltages are zero at t

(a)

Compute =

0.

Write a second-order differential equa tion with

the unknown. Find the

roots o f the characteristic equation. Then find V(^t) when (b)

Repeat part (a), for when unknown.

Figure P9.63 Cascade of leaky integrator circuits having a second-ordcr response. 64. Consider the circuit shown in Figure P9.64. Write a second-order differential equa-

= u{t) A. is the


429

67. The second-order circuit shown in Figure

nearly 1 V in about 5 nsec, it behaves approxi

P 9.67 contains two capacitors.

mately as an open circuit. The 0.1 pF capacitor

(a)

(b)

Find the second-order differential

is then charged up with a time constant of

equation with V(^ as the unknown.

about 0.1 msec. As the larger capacitor is

Give the values o f s.

charged up, the output across the smaller one

If vci (0) = 2 V, and ^ ^ (0 ) = 4 V, find V(^{t) for t> 0.

decreases toward zero.

0.5 0 2F

0.5 n 0.5 0

2F

Figure P9.67 Figure P9.69 68. In the circuit of Figure P 9.68, the voltagecontrolled voltage source has a gain A > 0 . Find the ranges of damped,

(b)

for the circuit to be (a) over underdamped,

(c)

critically

damped, and (d) undamped.

70. Find the value o f the negative resistance

-R„ for the circuit shown in Figure P 9.70 required to generate sinusoidal oscillations with constant amplitude.

-R .

V,F 120

Figure P9.70

69. The second-order circuit shown in Figure 's-/' O '

P 9.69 is of the overdamped type. Find the step response, i.e., the expression for Vg{t) for ^ > 0, when the input is vi{t) = u{t) V, and the capac itors are initially uncharged. Roughly sketch the waveform of Vpit). Verify your sketch by doing a SPICE simulation o f the circuit.

so

Remark: The waveform vît) consists of a very fest rise toward 1 V, and then a relatively slow

o

exponential decrease toward 0 V. This can be explained using the first-order RC circuit prop

o

erties studied in Chapter 8. During the first few microseconds, the 0.1 jiF capacitor behaves

o

almost as a short circuit, and the 1 nF capacitor is charged with a time constant of about 10 nsec. After the smaller capacitor is charged to

' o

71. Refer to the Wien bridge oscillator of Example 9.11. Suppose the op amp has a satu ration voltage 15 V. C = 1 jiF and /?2 = 500 Cl. is a temperature-sensitive resistor whose resistance is a fixnction of the amplitude of the sinusoidal current passing through

. The fol

lowing relationships are given: ?l(f) =

sin(C0f + 0)

/?, = 500 + 1 0 0 ( 7 ^ -0 .0 1 )

(a)

Find the frequency of oscillation (iHn-

(b)

Find the amplitude of voltage at the op amp output terminal (with respect to ground).

(c)

Suppose /?j is a fixed resistance of 4 9 0 t^ci(O) = 100 mV, and V(^{0) = 0 in the Wein bridge oscillator circuit. Perform a SPICE simulation. Does the circuit behave as expected?

430


72. In the Wein bridge oscillator example o f this chapter, let

= R-, = \ kH, Rr= 10

and

C=0.1pF.

(a)

Determine the frequency o f oscillation in Hz.

(b)

If

(0) = 5 V and v^{0) = 0 V, find

v^{t) for r > 0. (c)

Use any circuit

simulation

(e.g.,

SPICE) to verify the waveform o f v^{t) in part (b). 73. In the Wein bridge oscillator o f Figure 9.15a, /?2 = 1 k n . ^/=10 and C = 0.1 pR The lamp resistance R^ is a function o f the peak value o f the sinusoidal current as shown Figure P9.73.

(a)

Determine the frequency o f oscilla tion.

(b )

Determine the amplitude o f the sinu soidal waveform v^{t).

C

H

A

P

T

E

R

Sinusoidal Steady State Analysis by Phasor Methods A HIGH-ACCURACY PRESSURE SENSOR APPLICATION The control o f high-performance jet engines requires highly accurate pressure measurements, with errors less than one-tenth o f 1% o f a full-range measurement, over a wide range o f temperatures, - 6 5 ° to 200° F. The pressure range may be as low as 20 psia or as high as 650 psia. In jet (turbine) engine applications, knowing pressure and temperature allows one to compute the mass (volume) air flow, a critical aspect o f an engines performance. A pressure sensor is also a critical component in the regulation o f aircraft cabin pressure. Such a sensor is depicted here along with a functional block diagram o f its operation. A diaphragm consisting o f t\\'o fused quartz plates separated by a vaciumi has a capacitance that changes as a function o f pressure and temperature. This quart/ capacitive diaphragm is an element in a bridge circuit. It is this bridge circuit, in conjunction with detailed knowledge o f the characteristics o f a pair o f quartz capacitors over the required operating range o f pressure and temperature, that enables accurate pressure measurements.

Functional block diagram courtcsy o f AllicdSignal Aerospace Com pany.

432

Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods

Because o f the small capacitances, on the order o f picofarads, associated with the quartz (diaphragm) capacitor, the bridge circuit is driven by an ac source and is called an ac bridge. Driving the bridge by an ac source moves its analysis outside the realm of the dc and step response techniques studied in earlier chapters. New methods o f analysis, such as phasor analysis, are nec essary. Phasor methods, the primary focus o f this chapter, allow us to analrze capacitive and induc tive circuits excited by sinusoidal (ac) inputs. In particular, phasor techniques permit us to anal)'ze an ac bridge circuit. Although the analysis ot the pressure sensor shown here is beyond the scope o f this text, the chapter will end with a simplified pressure sensor circuit based on the one shown.


Review and elaborate on the basic arithmetic and essential properties o f complex num bers pertinent to sinusoidal steady-state analysis o f circuits.

2. 3. 4.

Develop two complementary techniques for computing the response o f simple RL, RC, and RLC circuits excited by sinusoidal inputs and modeled by differential equations. Define the notion o f a (complex) phasor for representing sinusoidal currents and voltages in a circuit. Using the notion ol phasor, introduce the notions of impedance, admittance, and a gen eralized Ohm’s law lor two-terminal circuit elements having phasor currents and voltages.

5.

Utilizing the methods o f nodal and loop analysis and the nerwork theorems o f Chapters 5 and 6, analyze passive and op amp circuits by the phasor method.

6.

Introduce the notion o f frequency response for linear circuits, i.e., investigate the behav ior o f a circuit driven by a sinusoid as its frequency ranges over a given band.

SECTION HEADINGS 1. 2. 3.

Introduction Brief Review o f Complex Numbers Naive Technique for Computing the Sinusoidal Steady State

4. 5.

Complex Exponential Forcing Functions in Sinusoidal Steady-State Computation Phasor Representations of Sinusoidal Signals

6. 7. 8. 9. 10. 11. 12.

Elementary Impedance Concepts: Phasor Relationships for Resistors, Inductors, and Capacitors Phasor Impedance and Admittance Steady-State Circuit Analysis Using Phasors Introduction to the Notion o f Frequency Response Nodal Analysis o f a Pressure-Sensing Device Summary Terms and Concepts

13.

Problems

Chapter 10 • vSinusoicial Steady State Analysis by Phaser Methods

-133

1. INTRO DUCTION Perhaps you have experienced the bouncing motion of a car with broken shock absorbers or watched the (mechanical) oscillations o f a sw'inging pendulum. These motions reflect the sinu soidal and damped sinusoidal oscillations in

circuits with conjugate poles o f the characteris

tic equation, as detailed in Chapter 9. In this chapter we allow sources with sinusoidal forcing functions (such as such as

cos(ov + 0) or perhaps

sin(d)/ + f))), which almost always result in

sinusoidal responses regardless o f the root locations o f the characteristic equation. A sinusoidal voltage source models the voltage from the ubiquitous wall outlet. If one hooks up an oscilloscope to measure a voltage in a linear circuit driven by sources with sinu soidal values, the voltage may not look sinusoidal at first. However, if the circuit is stable, after a sufficiently long period o f time the screen o f the scope will trace our a sinusoidal waveform. (Here “stable” means that any zero-input response consists o f decaying exponentials or exponentially decreasing sinusoids.) The eventual sinusoidal behavior is not immediately apparent because at startup, stable circuits exhibit a transient response. “Transient” means that the circuit response is transitioning— for example, from an initial voltage or current value to another constant value. Flickering lights during a thunderstorm illustrate the phenomenon o f transient behavior: light ning may have struck a transmission line or pole, causing the power system to waver briefly from its nominal behavior. Because sinusoidal excitations and sinusoidal responses are so common, their study falls under the heading o f sinusoidal steady-state (SSS) analysis. Here “sinusoidal” means that source excita tions have the form we take + 0) =

cos((or + 0) or K sin((0/ + 0). For consistency with traditional approaches,

cos(tor + 0) as the general input excitation, as shown in Figure 10.1, because

sin(to/

cos(tof + 0 - 7t/2). Steady state mean that all transient behavior o f the stable circuit has

died out, i.e., decayed to zero. Observe that every sinusoidal waveform is periodic with angular argument (to/ + 0). In terms o f angle, each cycle o f the waveform traverses 2 n radians. In terms of time, each cycle covers a time interval o f T = 27r/o) seconds, called the period o f the waveform. The number o f cycles contained in 1 second is called the frequency o f the sinusoidal waveform and is denoted by / T h e unit for/is the herrz (Hz), meaning “cycles per second.” The quantity OJ, which specifies the variation o f the angular argument (tor + 0) in 1 second, is called the angu lar frequency o f the wâveform. The unit o f (O is radians per second (rad/sec). From these defini tio n s,/ = 1/7'= ti)/27i and o) = 2 k / Stable circuits driven by sinusoidal excitations produce sinusoidal voltages and currents, as illus trated in Figure 10.1. The output excitation in Figure 10.1 has the general form to distinguish it from the input excitation,

cos((0/ + (j))

cos((0/ + 0). Because o f linearity, the circuit can

change only the magnitude o f the input sinusoid is changed to K^^) and the phase angle o f the input sinusoid (0 is changed to (j)) while ensuring that the angular frequenq^ to remains the same. For nonlinear circuits, to can and usually does change.

Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods

434

Input Excitation: Vjcos(cot + 0)

CO

FlGURl'l 10.1 Graphical illustration ot steady-state sinusoidal linear circuit behavior. (Kând could just as well be

and

or any combination thereof.) Note that 0 and (}) are often different and

that (0 is the same for both input and output excitations. In Figure 10.1 the steady-state (voltage) response is been a current response, +

8

cos(a)r + ({)). Alternatively this could have

cos(a)r + (j)). Such waveforms have the equivalent structure A cos(wr)

sin(oj/), deducible from trigonometric identities, cos({or + (}))=

cos((j)) cos(tor) -

= A cos(a)f) + B sin(tor) where A = obtains

cos(({)) and B = —

sin(i|)) sin(cor)

( 10 . 1)

sin((})). Conversely, by summing the squares oi'A and B, one


V„, =

4 3 “^

n ------ ^ + B-

(10.2a)

By taking the inverse tangent o f the ratio o f - B and A, one obtains

(10.2b) In using equation 10.2b it is important to adjust the resulting angle for the proper quadrant o f the complex plane. Equations 10.1 and 10.2 turn out to be useful in developing a conceptually simple, although naive, technique for computing the steady-state response using a differential equation model o f the circuit, as explained in section 3. Sinusoidal steady-state (SSS) analysis o f circuits draws its importance from several areas. The analysis o f power systems normally occurs in the steady state where voltages and currents are sinu soidal. Music is a rhythmic blend o f different notes. Mathematically, a musical (voltage) signal can be decomposed into a sum o f sinusoidal voltages o f different frequencies. The analysis o f a sound system typically builds around the steady-state behavior o f the microphone, the amplifier, and the loudspeakers driven by sinusoidal excitations whose frequency varies from around 40 Hz to 20 kHz. Indeed, almost any form o f speech or music transmission requires an understanding o f steady-state circuit behavior. There are many other areas o f applicability. This chapter will introduce three techniques for computing the SSS response. The first two, some what naive, approaches map out a natural motivation and path to the third, ver)' powerful tech nique o f phasor analysis. Phasor analysis builds on the arithmetic o f complex numbers and the basic circuit principles studied thus far. To set the stage for phasor analysis, section 2 reviews the necessary basics o f complex number arithmetic. O f course, the student is assumed to have stud ied complex numbers in high school and in prerequisite calculus courses.

2. BRIEF REVIEW OF C O M P LEX NUMBERS Let

= a + jb be an arbitrary complex number, where ^

. The real number a is the real

part o f 2 j, denoted hy a = Re[z,]. The real number b is the im aginary part o f z^, denoted by b = Im[ 2 j]. It is simple to verify that « = Re ^1 ^ i L ± i L 2 and

b = Im -1

where

_

2y

a - jb is the complex conjugate* o f Zy The magnitude or modulus o f 2 ,, denoted by

|2 ,|, satisfies |z|P =

+ Ip-


•»36

The number Z| =

+ jb is said to be represented in rectangular coordinates. Another represen

tation o f 2 p called polar form or polar coordinates, follows from the simple geometry illustrated in Figure 10.2.

FIG UIIE 10.2 Diagram showing relationship between polar and rectangular coordinates of a complcx niunber. In Figure 10.2, the number Zj can be thought o f as a vector o f length p = >/«“ + b" = |zj|, which makes an angle 0 = tan"*(^/
= p cos(0) + yp sin(0) = p[cos(0) + j sin(0)] = peJ^ = pZ.0

where p Z. 0 is a shorthand notation for

and

t’/ ’ = cos(0) + ;s in ( 0 )

(10.3)

is the famous Euler identity. The Euler identity can be demonstrated by writing the Taylor series for

and recognizing it as the sum of the Taylor series for cos(0) added to j times the Taylor

series for sin(0). Note that the symbol Z. has two meanings, depending on the context o f its use: (1) L z means angle o f the complex number z, and (2)pZ. 0 means the complex number whose magnitude is p and whose angle is 0. The properties o f the exponential immediately imply that (10.4)

Exercises. 1. Compute the polar coordinates o f AN SW ERS: 2. Let z = 6

/T

= -1 - j and Zj = \ + j-

r ; /4S'’ where ti/6 has units o f radians; for example, n rad equals 180°. Find the real

and imaginary parts o f z. ANSW ER; 2 = S. 1962 + /3


43*

3. Show by direct computation that since eJ^\ = cos(0j) + j sin(0j) and eJî = cos( 02) + j sin(f)2), then = cos(0,+ 0-,) + j sin(0,+ (),) =

With these simple definitions, the product oF t^vo complex numbers z-^ = a +jb =

and z-y

can be found using rectangular coordinates as

z^zf = {a +jb) {c + jd) = a c - bd +j{bc + ad) or using polar coordinates as

= p ,p 2 cos(G,+ 62 ) + yp ,p 2 sin( 0 ,+ dj) which in shorthand notation is

~ P 1P 2

^2^

EXA M PLE 10.1. Suppose z^ = 3 -jA = 5 Z .-5 3 .1 3 ° and ^2 = 8 + ;6 = 1 0 ^ 3 6 .8 7 °. Then £,Z 2 = (24 + 24) + y(18 - 32) = 48 - ; 1 4 Equivalently, z ,z 2 = 5 ^ - 53.130 X 1 0 ^ 3 6 .8 7 ° = 50 Z ( - 5 3 .1 3 ° + 3 6 .8 7 °) =

+36.87«)

= 50 cos(16.26°) - ; 5 0 sin (l6.26 0 ) = 48 - ; 1 4

Exercise. Let Zj = 2 + j l and z^= - 2 + j 6 . (a)

Compute the polar form o f 2 , and z ,.

(b)

Compute ZjZ-, in rectangular coordinates.

(c) Compute ZjZ2 in polar coordinates. A N SW ERS: 2.8284^'/''^", 6.3246^>-/>"''^- *-^‘’, - 1 6 + ;8 , 17.8885^->''^-^-'''-'‘’

Similarly, in rectangular coordinates the arithmetic for the division o f two complex numbers is Cl _ a + jb _ {a + jb ){c - jd ) _ {a + jb ){c - jd )

Z2

c + jd

ic + j d ) { c - j d )

{ac + bd) + j ( b c - a d ) ~

2

^2

c- + d

c -+ d ~

-i38


In polar coordinates the calculation is more straightfor\vard;

il- =

=-

B

i

Z2

-

) = -Hi-cos(6, - 0 2 ) + 7 — sin(6| - G j) P2 P2

In our shorthand notation,

Exercise. Let Zj = 2 + j l and Zj = +76 . (a) Compute z^tzj in rectangular coordinates. (b) Compute z'^lzj in polar coordinates. A N SW ERS:

(J.i

-

/ 0 . 4 , 0 . 4 4 7 2 ^ ' / ’ ’ "*^^'’

O f particular concern in this chapter are equations involving mixed representations o f complex numbers. For example, suppose an unknown complex number z =

satisfies the equation

{a + jb) = f + j d Then dividing through by

+ jb yields c + jd

a + jh Since Vis the magnitude o f the complex number on the right-hand side o f the equal sign, it fol lows that +

_ V c-+ t/ ~

+

V «^+/r

Here we have used the fact that a complex number that is the ratio o f two other complex num bers has a magnitude equal to the ratio o f the magnitudes. To determine the angle 0, one uses the property that 0 equals the angle o f the complex number in the numerator minus the angle o f the complex number in the denominator, -1 0 = Z (c + jd ) - Z {a + jb )= tan' tan

Exercise. Let z, = 2 - 2j and ANSW ER: 0.471 - 6 8 .4 7 "

— -ta n

= 5.5 + jlA . Find V and 0 when

-I

—

= 2 ,/z,.


EXA M PLE 10.2. Suppose v(t) =

^ a n d

-5VeJ^ + j6VeJ^ Find V, (j), and v{t). S o l u t io n

Factoring Ve-^ out to the left and dividing by (-8 + j6) yields

- 8 + j6 Hence. K= 2, (}) = - 9 8 .1 3 ” and

v{t) =

‘j’)] = 2 cos(to/- 98.13°) V

Sometimes a function v{t) is a complex number for each t, such as if{t) =

and v{t) will

satisfy some specific algebraic or differential equation. W hen this is the case, it is possible to use the differential equation to find values for V and (j). The next two examples illustrate this strategy.

E X A M PLE 10.3. Suppose the function

v{t) satisfies the differential equation

^

dr

+ 2 - + 2v = 10e-'<“ ” “ °'

dl

Find the values o f A and (}) if O) is known to be 2 rad/sec. S o l u t io n

Since the function v{t) must satisfy the differential equation, the first step is to substitute into the differential equation. Substituting

‘1’^ into the differential equation and taking appropriate

derivatives yields

The

term, which is always nonzero, cancels on both sides, leaving

AeJ^ [2 - co2 + ;2co] = Since cu = 2, one can equate magnitudes and angles to obtain

2 - ( 0 " + y2 (0

-2 + y4

440


Exercise. Repeat the above example if (o = 3 and ANSWERS: A = 2. (|) = 13‘J.4 "

is changed to

' HI.2”)

The techniques of circuit analysis in this chapter will often require complex number arithmetic. The voltages and currents o f practical interest are always real. T he complex arithmetic is a short cut to computing “real” voltages and currents. The real quantities are obtained by taking the real part o f the complex number or complex function. The various manipulations depend on some general properties related to the real part o f complex numbers. Property 10.1. Re[Z| + Zj\ = Re[z,] + Rel22j. This property has a particularly nice application to summing trigonometric waveforms. Let v^{t)

= cos(oj/ + 55°) and r-y(t) = 10 sin(d)/ - 3 0 °) = 10 cosCd)/ - 120°). Note that a - 9 0 ° shift con verts the sine to a cosine. Hence, ?'l(r) +

= cos(o)r + 55°) + 10 co s(to f- 120°) = Reld’-^^^'^^

+ Ret 10^-^^^'^^“

= Rek>("*'" = Relt’-''^^^

by the Euler identity, equation 10.3 by Property 10.1

+ lOf*"/’ ’ *' )] by equation 10.4 and then factoring

to the left

= Re{f-^‘'^'1(0.5736 +/).8192) + (-5 - 78 .66 )]} after conversion to reaangular form = Rel^>’'(-4 .4 2 6 - ; 7 . 8 4 l ) ] = Re[9<'>(‘"^ “

after simplification and con

version back to polar form = 9 cos(o)/ - 119.4°) after taking the real part. This sequence o f manipulations shows that the magnitude and phase o f two cosines at the same frequency O) can be represented by distinct complex numbers. One can then add the complex numbers and determine the magnitude and angle o f a third cosine equal to the sum o f the origi nal two cosines. This presents a shortcut for adding two cosines together. Property 10.2 {proportioiuilityproperty). R elazj] = aRefZ]] for all real scalars a . Properties I and 2 taken together imply R e [a , 2 , +

= a jR e [ 2 j] + a,Re[z-,]

which is a linearity property for complex numbers with multiplication by real scalars. 'I'he next property, which underpins the techniques of this chapter, defines how differentiation can be inter changed with the operation Re[-].


Property 10.3 {differentiation property). Lex. A = ê^^. Then = Re - ( / t e - '" ' )

= Re

dt

Exercise. Find Re

(10 + cit

ANSXXTR: 1118 cosdOO ; + 116.57

Our fourth property tells us the conditions for the equality o f two complex-valued time functions. Property 10.4. For all possibly comple.x numbers A and B, Re[/lf’-^^'^T = only

for all t if and

A = B.

Taken together, the preceding properties imply a fifth, very important propert)'. Here note that a complex exponential is sometimes referred to as a complex sinusoid. Property 10.5. I he sum of any number o f (1) complex exponentials, say AjeJ^'^‘, or (2) derivatives o f any order o f complex exponentials o f the same frequency co, or (3) indefinite integrals of any order o f a complex exponential o f the same frequency O), is a complex exponential o f the same fre quency (1). This property is another foundation stone on which the phasor analysis o f this chapter builds. Table 10.1 summari?,es the properties o f complex numbers. TABLE 10.1 Summar}' of Properties o f Complex Numbers cos((jL)r + 0) = A cos(cor) +

B sin(co^)

= V-4- +

, ({) = tan"'

A

Euler identit)'

eJ^ = cos(0) + j sin(0)

Real part o f sum

Re[z, + z-y] = Re[zj] + Re[z2l

Proportional it)' property

RelfXZ]] = a Re[2 i] for all real scalars a RelcXjZ, + a ,Z 2 ] = ct]

Linearity property Differentiation propert)' Equality propert)'

dt

L

J

dt^

+ «2

}

= Re

Re[y4f’-^^^T = Re[5^^'^T for all t if and only i f A = B Sum o f complex exponentials AêJ^^ or deriva

Single-frequency propert)'

tives, or their indefinite integrals o f any order is a complex exponential o f the same frequency co

442


3. NAIVE TECH N IQ U E FOR C O M PU TIN G THE SINUSOIDAL STEADY STATE Property 10.5 o f the previous section suggests a technique for computing the SSS response o f a circuit. The technique uses a differential equation model of an RL, RQ or RLC circuit as devel oped in Chapters 8 and 9. In contrast to the dc sources in those chapters, suppose the source exci tations have the form

cos(cor+0). In addition we assume that the zero-input response consists

o f (eventually) decaying exponentials or (eventually) exponentially decaying sinusoids to ensure that there is a valid sinusoidal steady state. Thus the form o f a first-order circuit differential equa tion model with a sinusoidal excitation is ^/■v(/)

cit

+ ax{t) = K^cos(co/+ 0 )

(10.5a)

or in the second-order case,

df

(10.5b)

dt

where x{t) is a desired voltage or current, such as

V(\t).

Property 10.5 guarantees that the sum o f any number o f cosines or derivatives o f any order o f cosines o f the same frequenc)' OJ is a cosine o f the same frequency O). Hence, the circuit response

x{t) in equations 10.5 has a steady-state cosine form o f frequency O). Further, the scaled sum ofx(f) and its derivatives on the left-hand side o f each differential equation 10.5 must equal V^cos(o)/+0), the input excitation. This also implies that the steady-state circuit response, x{t), is a cosine o f the same frequency as the input, but not necessarily the same magnitude or phase. We conclude that •’^■(^) = cos(co/‘+(j)) = A cos(tor) + B sin(to/). The SSS response is then specified upon finding A and B. The following example illustrates this calculation.

E X A M PLE 10.4. Let the source excitation to the circuit o f Figure 10.3 be Compute the SSS response /jr(r).

______________

i^lt)

i,» ©

F I G U R E 10.3 Parallel

RL circuit

for Exam ple 10.4.

= /^cos(o)^).


So

443

lu tio n

Step 1. Determine the differential equation model o f the circuit. From KCL applied to the top node o f the circuit, i jt ) = i^(t) + //_(/)• Sincc the resistor and inductor voltages coincide, the t/-i rela tionship o f the inductor implies that the inductor current satisfies the difTerential equation

( 10.6) which has the form o f equation 10.5a. Step 2. Determine the form o f the response. Since the input is a cosine wave, the SSS response will have the sinusoidal form

ij{t) = A cos(cor) + B sin(oj/)

(10.7)

Step 3. Substitute the form o f the response {equation 10.7) into the differential equation 10.6. Inserting equation 10.7 into the differential equation 10.6 and evaluating the derivatives yields —

L '

cos(co/) = — f/\cos(o)/)+ Z?.sin(o)r)l + — /4cos(co/)+ fisin(( 0 /)

(It

L

= - 0) /\sin(coO + to

RA

RB

I-t

L

cos((or) + — cos(co/) + — sin(co/)

Step 4. Group like terms and solve for A and B. Grouping like terms leads to r

R

R

]

r

R

Bco H— A ----- 1^ cos(o)r) + — B -A ii) L L ', .L

] sin(coO = 0

(10.8)

To determine the coefficients A and B, we evaluate equation 10.8 at two distinct time instants. Since equation 10.8 must hold at every instant o f time, it must hold at / = 0; i.e., at /= 0,

=0

or, equivalendy,

(10.9a) In addition, equation 10.8 must hold for t = 7r/(2to), in which case

-COA + - 5 = 0

L

( 1 0 .9 b )

444


Solving equations 10.9 simultaneously for A and B yields /?-/, = — ------f — ,

„

CO/?L/

B=

R~ + L^(0-'

/?- + L“co“

Step 5. Determine the steady-state response. Since A and B are known, .

,

,

Il (0 = — ------ ^

R~ + L~0)~

cos(o3 0 + —

R~ +

In the more common alternative form o f i^{t) =

.

O^RLL ^

SI n(co /)

cos((or + ({)) as per equation 10.2a,

where = - tan * (10.10b) is adjusted to reflect the proper quadrant o f the complex plane.

This example has illustrated a procedure for finding the SSS response o f a circuit. Step 1 is to sub stitute an assumed sinusoidal response form, such asy4cos(cor) + Bs\n{a)t), having unspecified con stants A and B, into the differential equation and evaluate all derivatives. Step 2 is to group like terms, and step 3 is to compute the constants A and B. After finding A and B, one computes the magnitude,

and phase (j) o f the cosine

cos(oj^ + (|)) via equations 10.2.

The next section offers an alternative approach. Using complex excitation signals o f the form yêJ^Mt +0)^ computes and (j) by a more direct route.

4. C O M P LEX EXPON EN TIAL FORCING FUNCTIONS IN SIN USOIDAL STEADY-STATE COM PUTATION Complex exponential forcing functions are simply complex exponential input excitations o f the properties o f complex numbers in section 2, we can form or ^;(o)/ + 0) PfQji-j replace the input excitation

cos{wt + 0) and the assumed circuit response

cos(to^) + j5sin(co^) with their complex counterparts

and

cos(cor + cj)) = /I respectively, with

out any penalt)'. To recover the actual real-valued responses, we simply take the real parts o f the complex quantities. Again this is justified by properties 10.1 through 10.5. This process o f sub stitution and subsequent taking o f real parts actually simplifies the calculations developed in sec tion 3, because of the simple differential and multiplicative properties o f the exponential function. The following example illustrates a more efficient calculation o f the steady-state response using complex exponentials.

■vn


EX A M P L E 10.5. For the series RC circuir o f Figure 10.4 let vît) = K^cos(cof). Compute the steady-state response

FIG URE 10.4. Series /?Ccircuit for Example 10.5. Step 1. Construct the dijferential equation o f the circuit. Writing a loop equation and substituting for i(^t) yields

( 10. 11)

at

Step 2. Substitute complex forms o f the input and response into the differential equation. If v^{t) were to be equal to the complex exponential if vj^t) =

then the response would be

= V}cos(cor) (as is the case), then V(^t) =

o f complex numbers. Hence, for the moment, let us set vj^t) =

^

However,

from the properties and agree that

=

appropriate real parts. Substituting the complex expressions into the circuit differential equation 10.11 yields

After canceling the

terms, factoring V„,e^^ out to the left, and dividing through by {jixiRC +

1), we obtain

1 + jOdRC

Step 3. Determine the magtiitude

( 10 . 12)

and the angle (}). Equating magnitudes on both sides o f equa

tion 10.12 yields

Vl + ( o V c '

(10.13a)

and equating angles yields (j) = - tan '(to/?Q

(10.13b)

Step 4. Determine the steady-state response. Using equations 10.13 the desired response is comput ed by taking real parts:


4 i6

\'cU) = Re

(10.14)

Ks

cos[co/ -tan ha^RC)]

Vl + (0“/e-C^ In deriving the relationship 10.12 from the ciifterential equation 10.11, we utilized a complex exponen tial ftinction as the circuit input. A complex exponentid input is not a signal that cm be generated in the laborator)'. Nc-vertheless, it is often used in advanced circuit theor)' to simplift' the derivation of many important results, as was done in the preceding example. If one does not mind a more lengthy derivation, then the s;imc result (equations 10.12 through 10.14) am be obtained without the flaitious complex exjxjnentid excitation. For example, let the voltage source in Figure 10.4 represent a reiil signal source

v^{t) = V^cosim) = R e[K / > T Then the steady-state response has the form

v^t) =

cosiMt + (j)) =

Substituting these expressions into the differential equation 10.11 yields

/ec— ( rc v...e

+ Re

= Re

Making use of properties 10.2 and 10.3, move the position o f the operator Re[] outside the first term to obtain

Re

+ Re

= Re

Evaluating the derivative and using property 10.2 (linearity) produces

(10.15) By property 10.4, equation 10.15 holds if and only if

^

+ ) =

This is precisely the equation following equation 10.11 that leads to equations 10.12, 10.13, and finally 10.14. As we can see, the use o f complex exponentials does indeed lead to a more direct calculation of the SSS response. However, this method and the method o f section 3 require a difFerential equa-


44 /

tion model o f the circuit. For circuits with multiple sources, dependent sources, and many inter connections o f circuit elements, finding the differential equation model is ohen a nontrivial task. In the next section we eliminate the need to find a differential equation model o f the circuit by introducing the phasor concept.

5. PHASOR REPRESENTATIONS OF SIN USOIDAL SIGNALS Recall

is shorthand iov

= /lcos(({)) +yy4sin((})). If the frequency co is known, then the

complex number AL^ completely determines the complex exponential known, then AL^ completely specifies A cos(ojr +({)) =

In turn, if to is

This means that the com

plex number A/L^ can represent a sinu.soidal function A cos(ojf + (})), whenever (O is known. Complex number representations that denote sinusoidal signals at a fixed frequency are called phasors. A phasor vo\iz^c or current will be denoted by a boldface capital letter. A typical voltage phasor is V =

and a ty'pical current phasor is I =

Por example, the current i{t) =

25cos((Of + 4 5 °) has the phasor representation I = 25Z-45°. The voltage v{t) = - 1 5 sin(tor + 3 0 °) = 15 cos(w^ + 120°) has the phasor representation V = 15/-120°. As all voltages and currents satisfy KVL and KCL, respectively, one might expect phasor voltages and currents to do likewise. This is not patendy clear. The following simple example demonstrates why this is true for KCL. Consider the circuit node drawn in Figure 10.5.

FIGURE 10.5 Single node having four incident branches. From KCL it follows that ^4(^) =

- ijit) + = 10 cos(tor) - 5.043 cos(cor + 7 .5 2 °) + 8 cos(cof- 9 0 °)

Using trigonometric identities or property^ 10.1 to combine terms on the right-hand side leads to

iît) =

1 0 c o s (tO f- 6 0 °)

4
Chapter 10 • Sinusoidal Steady State Analysis by Pliasor Methods

For the corresponding phasors to satisR' KCL, it must follow that

10Z-60O =

= I, - I 2 + I3

The right-hand side o f this equation requires that

I, - I2 + I3 = lOZQO - 5.043Z.7.52O + 8Z.-9()0 = 10 - (5 + >0 .66 ) + ( - ; 8 ) = 5 - 78.66 = 10/1-60° = I 4 Thus the phasors (which have both a real and an imaginar)' part) satisfy KCL. KCL is satisfied because i^{t) = /j(^) - i-,{t) + /^(r) implies

= Re[10^>n - Re[ 5. 043^>>( ' ^' ’^] + Re(8f>i<“ ^ -^0-’)] = Re[( 10 - 5.043
(10.16)

for all t. By property 10.4, equation 10.16 holds if and only if

IOZ- 6OO = 10 - 5.043e’^‘7-52° + In phasor notation this stipulates that

It is the properties o f complex numbers and the fact that an equation is true for all t that guaran tee that phasors satisfy KCL. Although not general, the argument is sufficient for our present ped agogical purpose. A similar argument implies that phasor voltages satisf}' KVL, as illustrated by the following example.

EXA M PLE 10.6. Determine the voltage across the resistor in the circuit of Figure 10.6 using the phasor concept. Vj(t) = 19.68 sin(a)t

152.8°)

V3(t) = 4 .2 1 5 cos(cot + 71.61 °)

+ . --------------

•+

v,(t) = 20co s(cot + 53.13°)

F K i U R E 10.6 Resistive circuit w ith three sourccs.

^


S

44^)

o l u t io n

Firsr note that 19.68 sinltDr+l 52.8”) = 19.68 cos(co/+152.8“—90") = 19.68 cos(cor+62.8°). From KVL,

Vfît) = v^{t) - Vjit) + v^{t) Since voltage phasors must satisfy KVL, V/^ = V, - V 2 + ¥3 = 20 z i 5 3 . 1 3 0 - 19.68 z^62.80 + 4.215 ^ 7 1 . 6 ° Changing to rectangular coordinates and adding yields = 12 + y i6 - (9 + y i7 .5 ) + 1.33 + j4 = 4.3 3 + jl.5 Equivalently,

= 5Z-30® V, and = Rc[5r>(‘" ' " ^0")] = 5 co${o)t + 300) V

Exercise. In Figure 10.6, suppose yj(r) = 10 cos(cor) V, v-^ (/) = 10 co s((o r- 0.5tc) V, and V3(/)= I 0 V 2 cos(cOf - 0.25ti ) . Find the phasorVând then ANSWKR:

= 20 - /20.

vrU) = 2 ()V : cos(O)/ - 4 5 " ) \’

Given that phasor voltages and currents satisfy K\^L and KCL, respectively, it is possible to devel op phasor O hm s law-like relationships for resistors, capacitors, and inductors operating in the SSS. This would allow us to do SSS circuit analysis with techniques similar to resistive dc analy sis. The next section takes up this thread by introducing the notion of (phasor) impedance.

6. ELEMENTARY IM PEDANCE CO N CEPTS: PHASOR RELATIONSHIPS FOR RESISTORS, INDUCTORS, AND CAPACITORS Ohm’s law-like relationships do exist for resistors, capacitors, and inductors operating in the SSS. The constraint, operating in the SSS, suggests that any Ohm s law-like relationship should be dependent on the sinusoidal frequency. The first objective o f this section is to derive three Ohm’s law-like relationships, one each for the resistor, the capacitor, and the inductor. The relationships each take the form V = Z(/to)I, where V is a phasor voltage, I is a pha.sor current, and Z(/to) is called the impedance o f the device: Zy^(;tij) for a resistor,

for a capacitor, and Z/(/co) for an inductor. The fact that the phasor voltage V is a

function Z(/co) times a phasor current I indicates a clear kinship with Ohm’s law for resistors. Indeed the unit o f impedance is the ohm because it is the ratio of phasor voltage to phasor current. The impedance Z(/cij) explicitly shows that the relationship is potentially frequency dependent.

450


The derivation o f these elementary impedance concepts will build on the assumption that all volt ages and currents are complex sinusoids o f the same frequency represented by complex phasors. This is permissible because real sinusoids can be recovered from complex sinusoids simply by tak ing the real part. To this end consider the resistive circuit o f Figure 10.7a.

V „(t) =

i,(t) = j

0j(u)t + 6)

(b)

(a)

FIGURE 10.7 (a) Resistive circuit driven by complcx current, (b) Equivalent phasor representation of the circuit in (a). From Ohms law,

vjiit) = In terms o f the phasors

= RIj^

+ 0)

ej^\ this relationship reduces to

= 1^-^^^ and = /?

= Zpfjia) \j^

(10.17)

and Z^(yco) = R\s the impedance o f the resistor defined by equa

where

= Rlj^ Z.0. If

tion 10.17. Ideally the resistor impedance is independent o f frequency. Thus =

cos(u)r + 0) = Kq[

I

^

then Vjît) = Rlj^ cos(co/ + 0) =

This phasor

relationship restates Ohm’s law for complex excitations. The distinctiveness o f phasors comes with their application to inductors and capacitors. Now consider the inductor circuits o f Figure 10.8. Assume the circuit o f Figure 10.8a is in the steady state. i jt ) =

A

Remainder

Remalnder

of circuit

of circuit

(a)

L

-I-

V = jcoL 1^

(b)

FIGURE 10.8 (a) Inductor having complex exponential voltage and current, (b) Phasor relationship of (a).

/is:


The complex current and voltage associated with the inductor are, respectively, /^(f) = and vît) =

Substituting these expressions into the defining equation for an inductor

yields j = jcoz.

Canceling out on both sides yields the relationship is

In terms o f the phasors

and

( 10. 18)

in which case the inductor Impedance is derived as Z^(/co) = p iL . The inductor impedance clear ly depends on the value ol the radian frequency CO. Specifically, if U) = 0, then the impedance o f the inductor is 0, i.e., in SSS the inductor looks like a short circuit to dc excitations. If co = oo, the impedance is infinite, i.e., in the steady state the inductor looks like an open circuit to signals o f very high frequency. Equation 10.18 exhibits a frequency-dependent Ohm’s law relationship for the inductor. From the properties o f the product o f two complex numbers, the polar form o f the voltage phasor is

\ l = (yco£)I^ = (03/./,) ^ ( 0 + 9 0 °) Hence if /^(/) =

^

= /, cos(tor + 0) A

then

p^(t) = Re[/wZ/^f’>(‘'^' ^ 0)] = RelcoL/ê^^^"

" ‘-^0“)] = coL/^ cos(tof + 0 + 9 0 °) V

From this relationship one sees that the voltage phase leads the current phase bv 9 0 ” . Equivalently, one can say that the current lags the voltage by 9 0°. This leading and lagging takes on a more con crete meaning when one views phasors as vectors in the complex plane, as per Figure 10.9, which shows that the voltage phasor o f the inductor always leads the current phasor by 9 0 ” .

452


The capacitor has a similar impedance relationship, derived as follows. Assume the circuit o f Figure 10.1 Oa is in the steady state.

C'

+

Remainder of

►

Remainder

circuit

of

C

L

J

V

circuit

-

(a)

J (b)

FIGURF. 10.10 (a) Capacitor having complex exponential voltage and current. (b) Phasor relationship o f (a). The complex current and voltage associated with the capacitor are, respectively, i^^t) = and V(^t) =

Substituting these expressions into the defining equation for a capacitor

yields

Vce 7(o)r+)■

Canceling out

on both sides yields

In terms o f the phasors

= I(-e^ and V (-=

this relationship becomes

I^=yo)C V ^ or, equivalently.

- — — i c - ^ cO ^ ^ )Ic ycoC

(10.19)

Equation 10.19 defines the capacitor impedance as Z(^j(a) = l/(/a)Q. if co = 0, the impedance o f the capacitor is infinite in magnitude. This means that in SSS the capacitor looks like an open circuit to dc signals. On the other hand, if OJ = co, then the capacitor has zero impedance and looks like a short circuit to large frequencies.

4 S3

Chapter 10 * Sinusoidal Steady State Analysis by Phaser Methods

Looking again at equation 10.19, observe that

Vr =

70) C I c

=-

t:

(oC

- 90°)

(10.20)

Equation 10.20 has a vector interpretation in the complex plane, as shown in Figure 10.11

Imaginary axis

F ' l C l J R l i 10, 1 1

Diagram of capacitor voltage and current phasors where the voltage phasor lags the current phasor by 90°.

The diagram o f Figure 1 0 .1 1 indicates that the capacitor voltage lags the capacitor current phasor by 9 0 ° or that the capacitor current leads the capacitor voltage by 9 0 °, which is the opposite o f the case for the inductor.

Exercises. 1. For the circuit o f Figure 10.12a, show that

^

ycoL ^

and that

V/ i r ( t ) = — ^COS(Cl)/ + 0 - 90°) (OL


2. For the circuit o f Figure 10,12b, show that I^ = ya)C V ^ and that

i(^t) =

+ ^ + 90”)

iiCt)

ijt )

V ,(t) =

V ,(t) =

\J

\J

0 j((i)t + 0)

g j((» t + 0)

(a)

(b)

I-ICURE 10.12 (a) inductor driven by vohagc source, (b) Capacitor driven by voltage source.

Recall that resistance has a reciprocal counterpart, conductance. Likewise, impedance has a recip rocal counterpart, admittance. Admittance has units o f siemens, S, as does conductance, fh e admittance, denoted by l^yto), associated with an impedance, Z(/co), is defined by the inverse rela tionship

K(./co) =

( 10.21)

ZOCO)

provided Z{j(M) is not equal to zero ever)^vhere. What this means is that the phasor i-v relation ship o f a resistor, capacitor, and inductor satisfies an equation o f the form I = K(y(o)V. Hence, the admittances o f the resistor, inductor, and capacitor are respectively given by

>"/eO'w) = - ,

R

r^(./co) = — ./(oL

Kc-(yco) = ycoC

( 10.22)

The impedance and admittance relationships o f the resistor, capacitor, and inductor are summa rized in table 10.2. TABLE 10.2 Summary of Impedance and Admittance Relationships for Resistor, Capacitor, and Inductor

Impedance

Admittance

2/?(y(0) - R

K^(;o)) =

i

yV-(./co) = ycoC

j(oC

JY Y V

Z Lijoi) = j(i)L

j(i)L


In the next section the notion o f impedance is applied to an arbitrar)' two-terminal network. This generalization will allow us to consider the impedance and admittance o f interconnections o f capacitors, inductors, resistors, and dependent sources.

7. PHASOR IM PEDANCE AND ADM ITTAN CE For the resistor, the inductor, and the capacitor, the impedance equals the ratio o f the respective phasor voltage to the phasor current. Analogously, the impedance o f any t\vo-terminal circuit, as illustrated in Figure 10.13, is the ratio o f the phasor voltage to the phasor current, i.e..

Z „ S j^ ) = ^

= R + jX

(10.23a)

1 in

O+

Two Terminal Circuit

Z (j(o)orY,„(j(o) FIGURE 10.13 Two-terminal device with phasor voltage \ p h a s o r current 1^^^, and input impcdance Zy^^(/co). Because impedance is the ratio o f phasor voltage to phasor current, its unit is the ohm. Inverting the relationship o f equation 10.23a defines the adm ittance o f a two-terminal device as the ratio o f phasor current to phasor voltage, i.e..

I; (10.23b)

Provided Z(yto) ^ 0 for all

cd,

in contrast to a short circuit, then

As an example, the impedance o f an inductor is jwL and its admittance is \/(J(.oL). Historically, impedance and admittance were first defined as per equation 10.23. However, with the wide spread use and utility o f the Laplace transform (Chapter 12) in the past several decades, imped

‘0 6


ance and admittance have become understood as much broader and more useful concepts than the steady-state presumptions o f equation 10.23, as set forth in Chapter 13. In general, admittances and impedances are rational functions with real coefficients o f the com plex variable Ju). At each d) the impedance and the admittance are generally complex numbers. Since a complex number has a real part and an imaginary part, we can further classify the real and imaginary parts o f an impedance or an admittance. For an impedance Z(yw) the expression lm[Z(/co)] = X is called the reactance o f the two-terminal element, while Re[Z(/(o)] = R refers to its resistance. 1‘urther, for an admittance Vijto), Im[K(/(o)j = B is called the susceptance of the two-terminal device whereas Re[)1[yoj)] = G is referred to as the con d u aan ce. These definitions are summari/,cd in Table 10.3. I'ABIJ-. 10.3 Summary Definitions of Various Terms

Admittance

Impedance V/„

Y( jo) ) = ^

Z{J(}^) = — = R + jX I in

V/«

= G + jB

Resistance

Reactance

Conductance

Susceptance

R = Re[Z(yw)]

X = Im[Z(;to)l

6’ = R e [n / o )]

im [K(>j)]

Using equations 10.23, one can compute the equivalent impedance series, as in Figure 10.14a. Flere

= ^] +

o f two devices in

% Ohm’s law for impedances, V , = Z|(/to)I, and

V 2 = Zol/w) I ,. But I, = I 2 = I/„- Mence,

I.e.,

în (

)

= —^ = Z, ( ./O)) + Z 2 (./(O)

*iit

(10.24)

This simple derivation has another consequence: given Zy^j(yco) = Z,(/to) + Z 2 (/co) and the fact that

i = 1, 2, a simple substitution yields the voltage division formula,

Z/O ) '

Z,(./(D) + Z2(7(0)

V;., (10.25)

Kquations 10.24 and 10.25 are consistent with our early development o f series and parallel resist ance.

4V


Y,(jco)

V.

V.

V,

oYJjco) (b) FIGURE 10.14 (a) Two impcclanccs in series, (b) Two admittances in parallel.

Exercises. 1. Duplicate the derivation ol" equation 10.23 for three impedances in .series. 2. Derive a formula for voltage division when there are three impedances in series.

The admittance o f two devices in parallel, as sketched in Figure 10.14b, satisfies

v,„

v,„

V,

V,

since

^1 (./“ ) = “ f

V|

Y2 ijc o ) = ^

Vo

we conclude that (10.26)

Exercises. 1. Duplicate the derivation o f equation 10.26 for three admittances in parallel, i.e., show that )^y„(/co) = Kj(/o)) + 2. Show that the equivalent impedance o f two devices, Z,(/co) and Z t(/co), in parallel is given by

(10.27)

3. Show that the equivalent admittance o f two devices, Kjlyco) and Y-,{ji.o), in series is given by

458


>U/co)>'2(yw) ■

K|aco) + K ,0 (o )

4. Show that the admittance o f two capacitors, Cj and C^, in series is y’CO

(10.28)

C 1C 2

C\ + C-)

5. Show that the impedance o f two inductors, Z.j and Z.^, in parallel is /O)

Lt + Lo

Now the derivation o f equation 10.26 leads to a current division formula as follows. Since

= Kj(/co) +

and since

for /= 1 ,2 , one immediately obtains the current divi

sion formula,

I; = '

n-(7(0) K ,(7co)+ r2(y(o)

I,

(10.29)

Since devices represented by impedances or admittances must satisfy KVL and KCL in terms of their phasor voltages and currents, and since each device so represented satisfies a generalized Ohms law, i.e., V = Z(;o))I

or

I =

it follows that impedances can be 7nauipiilated in the same manner as resistances, and admittances in the same manner as conductances. The voltage division formula o f equation 10.25 and the current divi sion formula o f equation 10.27 illustrate this fact. Example 10.7 further clarifies these statements.

Exercises. 1. Derive a current division formula for three admittances in parallel. 2. Find the admittance and then the impedance of each parallel connection in Figure 10.15. AN SW FRS: Admittances are 3. Compute the equivalent inductance for Figure 10.15a and the equivalent capacitance for the circuit ol Figure 10.15b. ANSWI-RS:

,

L|

L.

4. Find ^ in terms o f

Cj + Ct+C^

L, for each circuit in Figure 10.15.

I O-

I

+ V

(a)

(b)

FIGURE 10 . 15 . (a) Set of three parallel inductors, (b) Set of three parallel capacitors.

4S9


EXA M PLE 10.7. For the circuit o f Figure 10.16, compute the input impedance ^/„(/Co) when co = 500 rad/sec.

FIGURE 10.16 Series-parallel interconnection of different impedances. S

o l u t io n

As shown in Figure 10.16, Z^-^^(/500) can be seen as the sum o f three impedances,

+ ^2 + Zy

Our approach is to first calculate Zy for each /. Step 1. Compute Z y Since this is an I C series combination,

A = j 5 0 0 x 0 .0 0 5 -

1

5 00

X

0 .0 0 0 4 )

= - j 2 .5 Q

Step 2. Compute Z-) = MYj. From the propert)' that parallel admittances add and series imped ances add.

Y2 = y'500 X 0.0 0 0 2 -h

1

:—

10 + ( 1 0 -H7 IO)

= JO. 1 -f 0 .0 4 - jO .02 = 0 .0 4

H en ce,Z 2= 1/^2 = 5 - ; 1 0 a Step 3. Compute Z 3 = 1/

=

Here

= j --------------- = o.i-H 70 . 1 - 70.2 = 0 . 1 - 70.1 5 0 0 x 0 .0 1

Hence, Z 3 = 5 + J5 Step 4. Compute Z-^^. Adding the three impedances together yields Z.„ = Z j + Z 2 + Z 3 = - ;2 .5 + 5 - 7 IO + 5 +75 = 10 - 77.5 Q = 12.5 Z - 3 6 .8 7 ° Q

y0.08

160

Chapter 10 * Sinusoidal Steady State Analysis by Phasor Methods

Calculations performed in this example are most easily done with an advanced calculator or in M ATLAB.

For example,

in

M ATLAB

the command

for computing

is “Z3

=

l/(sqrt(0.02)*exp(i*pi/4) - j/(50()*0.01)).” EXA M PLE 10.8. Compute the input impedance Zy^^(/a)) o f the ideal op amp circuit o f Figure 10.17. I,

V,

FIGURE 10.17 Op amp circuit callcd an impedance converter. S o l u t io n

The trick to solving this problem entails full use o f the ideal op amp properties discussed in Chapter 4. Step 1. From the properties o f an ideal op amp, from KVL, and from Ohms law, V 2 - V „ , - M 3 = V,„

(10.30)

This follows because the voltage across the input terminals of each ideal op amp is zero and no current enters the + or - terminal o f each ideal op amp. This implies that Step 2. Using the phasor voltage division formula o f equation 10.25, it follows that

R =

T " '

R+ j(oC or, equivalently,

v,=

1+ -

1

Jc^RC)

Here, o f course, because o f the idealized properties of the op amp, the voltage the resistor R in the leftmost op amp.

(10.31)

appears across


Step 3. Writing a node equation at the inverting terminal o f the rightmost op amp yields

again by the properties o f an ideal op amp. Simplifying this equation yields 2V,„ = Vi + V ,

(10.32)

Step 4. Substituting equations 10.30 and 10.31 into equation 10.32 yields V.

=\. + — + jCdRC

-/?!■

Equivalently,

Z i„ u ^ )= ^ = m ~ c I//J

(10.33)

Equation 10.33 suggests that the op amp circuit o f Figure 10.17 can replace a grounded inductor whose impedance is jii)L with proper choice of R and C, i.e., L = R^C. In integrated circuit tech nology it is not possible to build a wire-wound inductor. Instead, inductors are “simulated” by cir cuits such as that o f Figure 10.17. The next section continues to develop our skill with and deepen our understanding o f the phasor technique by computing the steady-state responses o f various circuits.

8. STEADY-STATE CIRCU IT ANALYSIS USING PHASORS This section presents a series o f examples that illustrate various aspects of’ the phasor technique. Our purpose is not only to demonstrate how to compute the SSS, but also to illustrate the pha sor counterparts o f Thevenin equivalents, nodal analysis, and mesh analysis. Our first example reconsiders the parallel RL circuit o f Example 10.4, together with the series RC circuit o f Example 10.6. We will demonstrate the superiority o f the phasor technique over the methods presented in sections 3 and 4.

i6:


E X A M PLE 10.9. Compute the steady-state voltage V(^t) for the circuit o f Figure 10.18 when i^{t) = cos(lOOr) A.

R=10Q

21,(t)

i ( t ) 0

C=1mF

R = 100

L = 0.1H

FIG U RF 10.18 7?ACcircuit for Example 10.9. S

o l u t io n

Step 1. Determine I^. Since the phasor

R

= 1Z.0° A, by current division, 1

1

1 Z -4 5 "

A

(10.34)

R Step 2. Use equation 10.34 and voltage division on the RC part o f the cirniit to compute \ q Using voltage division and equation 10.34, the capacitor voltage phasor is

ycoC

j(oC Step 3. Determine V(^t). Converting the phasor

o f equation 10.35 to its corresponding time

function yields

v^t) = co s(1 0 0 ^ - 9 0 °) = sin(100/) V The next example illustrates voltage division with phasors as well as the basic impedance relationships. E X A M PLE 10 .1 0 . Consider the circuit in Figure 10.19 where = 2 H, and vi^t) = 10 cos(2r) V. Find Vfj^t) and ij{t) in steady state.

= 5

C = 0.1 F, /?2 =

^


463

SO LU TIO N Step 1. FindZj^^{jl). Y,^cU'^) = “ +

= 0.2 + 70.2 = 0.2> / 2Z 45" . Hence

R

ZrcU2) =

Step l.F in dZ j^ iijl).

= 2 .5 V 2 Z - 4 5 " = 2.5 - J2.S

=-+ — R

= 0 . 2 5 - ;0 .2 5 = 0 .2 5 V 2 Z - 4 5 " . Hence

j(d L

Zr,.U2) = -

4

-

= 2 v /2 Z 4 5 " = 2 + j2

Step 3. F in d Y a t id Vq (t). From volrage division V

^ -------- Z r c (J2) -------- Y ^ ------- 2 .5 -j2 .:> ------ ^ 2 r c 0 '2 ) + Z r i,0 '2 ) 2 . 5 - j 2 .5 + 2 + j2

= 7 g 0 9 ^ _ 3 8 .6 6 “

4.5-J0.5

It follows that V(^{t) = 7.809 cos(2^ - 38.66°) V. Step 4. FindYj^, 1^, a n

d From step 3,

= 10 - 7.809 Z - 38.66° = 1 0 - (6.098 - ;4 .8 7 8 ) = 3 .9 0 2 4 + ;4 .8 7 8 Hence

l,= ^

=

1 ^ 2 3 ilM iZ ! =

,5 6 ,7 Z - 3 8 .6 6 "

Thus /^(f) = 1 .5 6 l7 c o s (2 f- 38.66«) A.

The next example illustrates the computation o f a Thevenin equivalent circuit with the aid o f nodal analysis. Because impedances may be manipulated in the same manner as resistances and admittances in the same manner as conductances, the Thevenin theorem, the source transforma tion theorem (Chapter 5), and node and mesh analysis (Chapter 3) carr)' over directly.

46^


EXA M PLE 10.11 (a) Find cheThevenin equivalent o f the circuit o f Figure 10.20 if to = 4 rad/sec. (b) Determine the voltage

when a 1.2

load resistor is connected across terminals a and b.

FIGURE 10.20 Z.Ccircuit for Example 10.11. S o l u t io n

Find theThevenin equivalent circuit, and then using theThevenin equivalent, find

Vj{t).

Step 1. Establish nodal equation. A nodal equation at the left node o f Figure 10.20 in terms o f phasors is given by I.V = — r ^ L + J^CWoc = -J^ L + y ^ o c ,/(oL Step 2. Determine the relationship between

and

The relationship between

(10.36) and

as

determined by the dependent source is V z - V , , = 0 .2 5 [ ;2 V J Equivalently, 10.37)

V , = (1 Step 3. Substitute equation 10.37 into equatio7i 10.36. Substituting yields

h = m -j)y o c Solving for

with

= 1Z.0° yields Vôc =

: I , = (0 .4 - y0.8)I^ = 0 .8 9 4 Z - 6 3 .4 3 ° V 0.5 + j

Step 4 . Compute the Thevenin equivalent impedance

(10.38)

Consider the circuit o f Figure 10.21,

which is the phasor version o f Figure 10.20 with the output terminals short-circuited. Hence, the short-circuit current phasor is 1=

A

4(n


FIGURE 10,21 Phasor version o f Figure 10.20 with short-circuited terminals. Therefore, from equation 10.38, Z,/,0'4) = ^ = ( 0 . 4 - y O . S ) Q .VC

Step 5. Interpret

to generate the Thevenin equivalent circuit. To physically interpret theThevenin

equivalent impedance, consider that

= (0-4 -yO .8 ) = {R^,, + MjAQ Q. Thus,

= 0.4 Q and C = 0.3125 F. Hence, the desired Thevenin equivalent circuit (valid at O)

= 4 rad/sec) has the form sketched in Figure 10.22. 0 .3 1 25F

-OFIGURE 10.22 Thevenin equivalent of Figure 10.20. Step 6. Compute

by voltage division. Using voltage division on the circuit o f Figure 10.22, =

1.2 l.2 + ( 0 .4 - y 0 . 8 )

= (0.6 + y 0 .3 )(0 .8 9 4 Z 6 3 .4 3 °)

= 0 . 6 Z - 36.87° V Converting the load voltage phasor to its corresponding time-domain sinusoid yields y^^(t) = 0 . 6 cos(4/^ - 3 6 . 8 7 ° ) V

166


EX A M PLE 10.12. Determine the phasor voltage

and the corresponding time function vj,t)

for the circuit o f Figure 10.23 if co = 100 rad/sec.

j60Q

I'lG U R E 10.23 Phasor domain circuit for Example 10.12. Ail clement values indicate phasor impedances at 100 rad/sec. S o l u t io n

To solve this problem, it is convenient to execute a source transformation on the independent cur rent source and to combine the impedances o f the parallel combination o f the capacitor and inductor on the right-hand side o f the circuit. After executing these rwo manipulations, one obtains the new circuit of Figure 10.24.

FIGURE 10.24 Phasor domain equivalent circuit to that of Figure 10.23. All element values indi cate phasor impedances at 100 rad/sec. I denotes a phasor loop current. For the circuit o f Figure 10.24, the indicated loop equation is 250Z.-90O = (50 - ; 2 5 ) I - 0.4(501) - ; 1 5 I = (30 - ; 4 0 ) I Solving for I yields I = 4 - J 3 = 5 ^ - 3 6 .8 7 ° A Consequently,

= 501 = 2 5 0 ^ - 3 6 .8 7 ° V and vU) = 250 c o s (1 0 0 r- 3 6 .8 7 °) V.

467


9. IN TRO DUCTION TO THE NOTION OF FREQUEN CY RESPONSE The frequency response o f a circuit is the graph o f the ratio o f the phasor output to the phasor input as a function o f frequency, i.e., as the frequency varies over some specified range. Since the phasor input and the phasor output are complex numbers, the frequency response consists o f twô plots: (1) a graph o f the magnitude o f the phasor ratio and (2) a graph o f the angle o f the phasor ratio. Such graphs indicate the magnitude change and the angle change imposed on a sinusoidal input to produce a steady-state output sinusoid. In steady state, the magnitude o f the output sinu soid is the product o f the magnitude o f the input sinusoid and the magnitude o f the frequency response at the frequency o f the input. Similarly, the phase o f the output sinusoid in steady state is the sum o f the input phase and the frequency response phase at the input frequency. This prop erty takes on greater importance once one learns that arbitrary input signals can be decomposed into infinite sums o f sinusoids o f different frequencies, i.e., each signal has a frequency content. This notion is made precise in a signals and systems course, where one studies Fourier series and Fourier transforms. The frequency response o f a circuit describes the circuit behavior at each fre quency component o f the input signal. This permits one to isolate, enhance, or reject certain fre quency components o f an input signal and thereby isolate, enhance, or reject certain kinds of information.

EX A M PLE 1 0 .1 3 . Plot the frequency response o f the RC circuit o f Figure 10.25.

-o-

-O +

0.01 F

10 -o FIG URE 10.25 RC circuit passing high-frequency content of an input signal. S o l u t io n

to the input phasor voltage N^

Using voltage division, the ratio o f the output phasor voltage is given by ___ 1 out __________________yO.Ola) Vi„ " i + ----- ! _

= m p ))

“ l + iO .O lc o

/).01(0

where we have designated this ratio as //(/co). The two universally important frequencies are

O) =

0 and co=

oo.

At these frequencies, H{jO)

0Z .90° and //(;“ ) = 1^-0°. Asymptotically then, the magnitude |//(/ca)|

1 as

CD

^

oo

=

and

468

|//(/to)|


0 as to

0. W ith regard to angle, Z.//(/co)

0 as to

oo and Z.//(/to)

9 0 ° as oj

0. Also, a close scrutiny o f //(/w) indicates that to = 100 rad/sec is also an important frequency. Here H{j\00) = 0.707^^45°. These values give us a prett)' good idea what the magnitude and phase plots look like. Using a computer program. Figure 10.26a and Figure 10.26b show the exact magnitude and phase plots. These plots are consistent with our earlier asymptotic analysis.

Frequency (rads/sec) (a)

Frequency (rads/sec) (b)

FIG URE 10.26 (a) Magnitude plot of frequenc)' response for Example 10.13. (b) Phase plot of frequency response.

Do these frequency responses make sense? They should. Going back to the circuit, observe that at to = 0, the capacitor impedance is infinite. Physically, then, in steady state, the capacitor looks like an open circuit for dc, i.e., at zero frequenc)'. The magnitude plot bears this out. For frequencies


close

CO

zero, the capacitor approximates an open circuit and, hence, the magnitude remains small.

O n the other hand, for large frequencies, the capacitor has a very small impedance. This means that most o f the source voltage appears across the output resistor. The gain then approximates 1, as indicated by the magnitude plot. The frequency response o f the circuit is such that the highfrequency content o f the input signal is passed while the low-frequency content o f the input sig nal is attenuated. Such circuits are commonly called high-pass circuits.

EXA M PLE 10.14. Investigate the frequency response o f the parallel /^Z,C circuit o f Figure 10.27.

R=10

L=0.04H

C=0.25F

FIGURE 10.27 A parallel RLC circuit having a hand-pass frequency response. S o l u t io n

The input admittance o f the circuit o f Figure 10.27 is given by

LC R

RC . CO

jOiL

C Inverting to obtain the input impedance yields

J

w C

J_ _ ^ 2 ^ ._ o L LC RC

Clearly,

y'4co

I00-co-+y4co

Hence the ratio o f the output phasor to the input phasor is simply

Zy^^(yoj). Once again, co = 0 and co = oo are the first two frequencies to look at. Here Zy^,(0) = 0Z .90° and -2^,„(oo) = 0Z.-90^’. Also at co = 10, the impedance is real, i.e.,

0) = I . These three

points provide a rough idea o f the magnitude and phase response. Two more points are necessary for a real .sense o f the frequency response. At what frequency or frequencies does the magnitude drop to 0.707 o f its maximum value or when does the phase angle equal ±45°? This will occur when 1100 - co^| = |4co|. This is a quadratic equation. Flowever, because o f the absolute values, there are rvvo implicit quadratics, co" - 4co - 1 0 0 = 0 and co^ + 4co - 100 = 0. Solving using the quadratic formula yields co = ±8.2, ±12.2. Since the magnitude plot is symmetric with respect to the vertical axis (co = 0 axis), we consider only the positive values o f co. This information provides a good idea o f the magnitude and phase plots. A computer program was used to generate the fre-

470


quenc)' response plots in Figure 10.28a (magnitude) and Figure 10.28b (phase). The magnitude plot shows that frequencies satisfying 8,2 < O) < 12.2 are passed with little attenuation. Frequencies outside this region are attenuated significantly. Such a characteristic is said to be o f the band-pass type, and the corresponding circuit is a band-pass circuit.


Frequency (rads/sec) (b) F IG U R t 10.28 (a) Magnitude plot o f frequency response for band-pass circuit o f Figure 10.27. (b) Phase plot of frequency response.


•n

EX A M PLE 10.1 5 . As a final example vve consider the so-called band-reject circuit o f Figure 10.29. A band-reject circuit is the opposite o f a band-pass circuit. A band-reject circuit has a band o f frequencies that are significantly attenuated while it passes with little to no attenuation those frequencies outside the band. In this example our goal is to compute the magnitude and phase o f the frequenc)' response o f the band-reject circuit o f Figure 10.29. R =10

FIGURE 10.29 Band-reject circuit for Example 10.15. S o l u t io n

Once again using voltage division, we obtain the phasor ratio

'

LC

t)Ul

-(0 ‘

1 0 0 - CD"

_ L _ c o 2 + /o)-5

LC

■

1 0 0 - 0 ) 2 +^25co

L

H{jLo) = 1Z.0®. Hence, as}^mptotically, |//(/to)| approaches 1 as OJ approach es 0 and CO. Also at (o = 10“, //(/co) = 0 Z .-9 0 ” while at to = 10+, Hijo)) = 0 Z .-2 7 0 ° = 0Z.90". For

At to = 0 and

CO = o o ,

this example, to find the frequencies where |//(/‘to)| drops to l/ V I o f its maximum value o f 1, it is necessary to equate the magnitudes o f the real and imaginary parts o f the denominator. This produces two quadratics whose positive roots are to = 3 .5 0 7 8 and OJ = 28 .5 0 7 8 . At these frequen cies the angles o f //(/w) are —45® and 45°, respectively. Fhe computer-generated plots o f Figures 10.30a and 10.30b are, o f course, consistent with these quickly computed values.

472



Frequency (rads/sec) (b) F I G U R t 10.30 (a) Magnitude plot of frcquenc)^ response for band-reject circuit of Figure 10.29. (b) Phase plot of frequency response.

As wc can see, a wealth o f different kinds of frequency response are obtainable by different inter connections o f resistors, inductors, and capacitors. Historically, phasor techniques were the essen tial tool for the analysis and design of such circuits. Nowadays, engineers ordinarily use either M A T L A B or SPICE to obtain frequency response plots. Two examples follow where we use MATLA B , SPICE, or both to obtain the frequenc)’ response.

4 '’3


E XA M PLE 10.16. Compute the frequency response o f the circuit o f Figure 10.31 using MATLAB and SPICE. R=10Q

FIG URE 10.31 /^/.Ccircuit for Example 10.16. S oluti on

This circuit was originally analpêd in Example 10.9. You might want to refer to that example before proceeding.

SPICE Part. A SPIC E simulation produces the result shown in Figure 10.32. EX1016 FreqRsp-Small Signal AC-0 +20.000

+40.000

+60.000

+80.000

Frequency (Hz)

+100.000 +120.000 +140.000

FIGURE 10.32 SPICE plot of capacitor voltage for the circuit of Figure 10.31.

MATLAB Part. Although the analysis appears in Example 10.9, we can use MATLAB to more easily obtain the frequency response. First define Zj(/co) = ytoA and Z-,{p)) = l/y'coC Then from current division, = -7 “ " r 7 —

R+

Z|(yco)

1 ,0 )

and from voltage division, 2-. (/CO)

R+

Z 2 (y co )

I

R

K2( 7(0) +

•21^( 700)

474


Assuming a frequency range o f 0 < co < 1000 rad/sec, the following MATLAB code will result in a suitable magnitude frequency response plot, as shown in Figure 10.33. »L = 0.1;R = 10;C = 0.001; »w = 0:1:1000; »Z1 = j*w*L; »Y2 = j“w*C; >>IL= R./(R+Z1); »VC = 2 ’ 1L./(R *Y 2+1); »plot(w/(2*pi),abs(VC),’b’) »grid »xlabel(‘Frequency in Hz’) »ylabel(‘Capacitor voltage (V )’)

FIGURE 10.33 Magnitude plot of frequenc)’^response o f capacitor voltage in the circuit of Figure 10.31. The response is of the low-pass t)'pe. Now suppose the inductor in the circuit o f Figure 10.31 is replaced by a capacitor C, = 1 mF with the controlling current changed to /q (0- The frequency response is easily computed with a sin gle change to the MATLAB code, namely, “Z , = 1. ./(j*w *0.001).” The resulting plot shows a band-pass characteristic, as illustrated in Figure 10.34.


47S

F IG U R E 10.34 Magnitude plot of frequency response of capacitor voltage in circuit of Figure 10.31 when inductor is replaced by a 1 mF capacitor. The response is of the band-pass type.

E X A M PLE 10.17. In Chapter 9 we investigated the Wien bridge op amp oscillator circuit, redrawn in B2 Spice in Figure 10.35.

4" 6

Chapcer 10 • Sinusoidal Steady State Analysis by Phaser Methods

Two difTerences are notable: ( !) there is a current sourcc present across the

combination,

and (2) is now 10 kQ, as opposed to 9.5 kH in Example 9.14, forcing /?, = /?2- This means that the characteristic equation for the circuit is 1 ■V*' + / ? i ' + ( • = .y~ +

{RlC

R,C)

which indicates a purely sinusoidal oscillation at the frequenc)'

f o = - ^ = -----!— = 15.92 Hz "

2n

2 k R,C

for any initial condition on C ,. In fact one might recall that R^ < Rj causes a growing oscillation that is limited by the saturation effects o f the op amp. The current source, set at 1 A, is present in Figure 10.35 so that we can obtain the frequency response cur\'e shown in Figure 10.36. In Figure 10.36 observ'e that the magnitude response peaks at/q, as expected from the theoretical analysis. In an actual circuit, the current source would not be present. Nevertheless, a sustained sinusoidal oscillation will occur because o f the presence o f noise. W ithout going into the analysis, noise contains an infinite number o f frequency compo nents, each o f which has a minute magnitude. In particular, noise contains frequency components around / q that drive the circuit into oscillation. This is precisely what the peak in the frequency response means: a very small (noise) voltage on Cj will cause a very large-magnitude sinusoid out put voltage at /q. However, the presence o f nonlinearities such as saturation keep the magnitude at an acceptable level.

Exi0.17-Small Signal AC-13 +12.000

+13.000

+14.000

+15.000

+16.000

+17.000

Frequency (Hz) +18.000

F I G U R E 10.36 Frequeno,- response plot o f W ie n bridge oscillator.

+19.000

+20.000


10. NODAL ANALYSIS OF A PRESSURE-SENSING DEVICE The bridge circuit presented in Figure 10.37, or some variation of this bridge circuit, has been and continues to be a widely used approach to accurate measurement technology. In this section we will analyze the ac bridge circuit o f Figure 10.37 as a pressure measurement device. The capaci tance C t is a diaphragm capacitor consisting o f a hollow cylinder capped on either side by fused quartz wafers. Bersveen the wafers is a vacuum. The capacitance o f the diaphragm changes with temperature and pressure. For our analysis we will assume that the temperature is constant and that the pressure is constant for a time period greater than five times the longest time constant o f the circuit. This will allow the voltages and currents in the circuit to reach steady state and thus allow us to use phasor analysis to compute their values. R =100 0

15cos(20,000nt)V

F1GUR1-' 10.37 Bridge circuit diagram of pressure-sensing device. The capacitance function of pressure, which causes the voltage

changes as a

to changc as a function of pressure. This is

registered on the attached voltage meter, which has a 1 MQ internal impedance. As a rule o f thumb, the capacitance C-, » Q.llAKAId. This means that the capacitance is inverse ly proportional to the distance d between the plates and proportional to the area A o f the plates and to the dielectric constant K o f the material between the plates. Increasing the pressure on the diaphragm decreases the distance rf'between the wafers, increasing the capacitance. Conversely, a decrease in pressure will increase the distance between the wafers, thereby decreasing the capaci tance. As the capacitance changes, the magnitude o f the ac voltage appearing across the voltage meter will vary accordingly. Hence, two relationships are necessary: (1) the relationship berween the capacitance C , and the magnitude o f the voltage

- V e and (2) the relationship berween

the pressure applied to the diaphragm and the associated capacitance. Our first task will be to specify the relationship between the pressure applied to the diaphragm and the resulting capaci tance. Following this, we will use nodal phasor analysis to determine the magnitude of and finally, the relationship between pressure and the magnitude V ^ - V e Pressure is measured in various units. Millimeters o f mercury (mm Hg) is a common standard; 1 mm Hg = 1 torr, and 760 torr = 1 atmosphere (atm), where 1 atm is the pressure o f the earth’s atmosphere at sea level, which supports 76 0 mm o f mercur\' in a special measuring tube. Suppose

4 ■’8


it has been found experimentally that the capacitance C j (in pF) varies as a function o f pressure according to the formuhi

C 2 ( A P ) = Q ) + ^ lo g 10

= 2 6 .5 + 68 log 10

^0

(10.39)

(16i)+A P^ 760

A plot o f C2 as a function o f AP is given in Figure 10.38.

QJ

(T3

Q. U

Change in Pressure FIG URE 10.3H Plot of capacitancc versus pressure. Our next task is to develop the relationship between the capacitance of^ the bridge circuit and the magnitude o f the phasor voltage and

In our analysis, G, = (/?,)“ ', Gj =

= (^ 3)"^

= 10~^ S is the conductance o f the meter , According to Figure 10.37, Cj = 20 pF. We

will let Cj range as 0 < C 2 < 40 pF. Finally, co = 2h x 10"^ rad/sec. The following phasor analysis will be done symbolically so as not to obscure the methodology. Summing the phasor currents leaving node A leads to the phasor voltage relationship (G j + G 2 + yujCj)v^ — G-jV^ —ycoCj

= Gj 15

Similarly, summing the currents leaving node B leads to the relationship - G 2 V ^ .( G 2 .G 3 .G J V ^ - G ,,V

c

=0

479


Finally, summing the currents leaving node C produces - > C ,V ^ .;( o ( C , . q

t

G JV c-

=0

Writing these three equations in matrix form yields

G] + G 2 +

-/■(oC,

-G i G 2 + G 3 + G ,„

-G j -yco C ,

- G ,„

■ l5 G ,'

V / =

~^m G „i + ^ ( C | + C 2 )

Vc.

0

(10.40)

0

The matrix on the left is said to be a nodal admittance matrix. Its entries can be real or complex, as indicated. It is nor advisable to solve such a set o f equations by hand over the range o f possible

C2 values. However, using MATLAB one can solve this matrix equation over the range 0 pF < C, < 40 pF to produce the plot o f Figure 10.39.

c CO

QJ

01 nj

*->

-o o

cu ■o D 'E

cn fO 10

15

20

25

30

35

40

C, in pF

FIGURE 10.39 Plot of the magnitude o f the phasor voltage

V ^ as a function of capacitance.

O f course, one could measure the voltage appearing across the meter, from Figure 10.39 deter mine the associated value o f Cj, refer to Figure 10.38 for AP, and then determine P = 760 + AP. This is a long route. To complete our analysis, then, we need to develop the relationship between pressure and bridge voltage. As we have the relationship between C j and APand the relationship between C2 and | |, it is a matter o f using equation 10.39 to derive the value o f in equation 10.40. This is best done with a simple MATLAB routine, which yields the plot given in Figure 10.40.

480

Cliaptcr 10 • Sinusoidal Steady State Analysis by Phasor Methods

> m

> 'o 01 T3 D 'c

ro

300

400

500

600

700

900

800

1000

1100

1200

Pressure in mm Hg

I'lG U RE 10.40 Relationship herween magnitude of bridge output voltage and pressure applied to diaphragm capacitor C-,. An actual pressure sensor would, oFcourse, be more complex. For example, there would probably be a difterential amplifier such as the one shown in Figure 10.41 across the terminals oFthe bridge circuit, and this would probably drive a peak (ac) detector to determine the maximum value oF the ac signal appearing at the output oF the dlFFerential amplifier. Fiu ther, the peak value would probably be read by a digital voltmeter. Nevertheless, our analysis illustrates the basic principles involved in such a measurement. OF course, one could just as easily use loop analysis to solve the problem. This is leFt as an exercise in the problems. kR.

-o V,

F IG U R E 10.41 Difierential amplifier having output voltage

= k { i >2 - /î).

481


Exercise. Prove that

= k{v-, -

for the differential amplifier o f Figure 10.41.

11. SUM M ARY The two primar)' goals o f this chapter were (1) to de\'elop the phasor technique for the analysis of circuits having a sinusoidal steady state and (2) to illustrate how this technique leads to die idea of a circuit frequenc)^ response, which characterizes the circuits behavior in response to the frequency content of an input excitation. In the development, sinusoids were first represented ;is the real part of a complex sinusoid. As a motivation for the delineation of die phasor method, we showed how the complex sinusoids could be urilized to compute the sinusoidal steady-state response using difierential equation circuit models. We then pointed out that a complex (voltage or current) sinusoid is specified by a complex number or pha.sor rep resenting its magnitude aind phase. After introducing the notions of impedance and admittance for the capacitor, the inductor, the resistor, and a general two-terminal circuit element, we showed how the pha sor voltage and phasor current for each such element satisf}" a frequency-dependent Ohms law. I'his iillowed us to adapt the ;inalysis techniques and network dieorems of Chapters 1 through 6 to the steadystate analj'sis of circuits excited by sinusoidal inputs. For example, diere are voltage division formulas, cur rent division formulas, source transformations, and The\'enin and Norton theorems all valid for phasor representations. This permits us to effectively analv/e circuits diat have a src*ady-state response. The phasor technique opens a door to seeing how circuits behave in response to sinusoids. Given that input excitations are composed o f different frequenc)' sinusoids, such as a music signal, phasor analy sis shows why a circuit will behave differently toward the different frequencies present in the input sig nal. This fact prompts the notion o f a circuits frequency response, which is defined as the ratio o f the phasor output to the phasor input excitation as a function o f (u in the single-input, single-output case. The frequency response consists of two plots. The magnitude plot shows the gain magnitude o f the circuits response to sinusoids o f different frequencies, and the phase plot shows the phase shift the cir cuit introduces to sinusoids o f different frequencies. The notion o f frequency response will be gener alized in Chapter 14 afrer we introduce the notion of the Laplace transform.

12. TERM S AND C O N CEPTS Admittance: o f a two-terminal device, the ratio o f the phasor current into the device to the 1;in phasor voltage across the device, y< ( /CO) = in Band-pass circuit: circuit in which frequencies within a specified band are passed while frequen cies outside the band are attenuated. Band-reject circuit: circuit in which one band o f frequencies is significantly attenuated while those frequencies outside the band are passedes with little to no attenuation. Com plex exponential forcing function: function o f the form v{t) = ^ = a +yco are complex numbers. A special case (a = 0), f(t) = out the chapter as a shortcut for sinusoidal steady-state analysis.

, where V =

and

is used through

Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods ^

Conductance: real part of a possibly complex admittance. Current division: in a parallel connection of admittances driven by a current source, the current through a particular branch is proportional to the ratio of the admittance of the branch to the total parallel admittance. Euler identitjr: = cos(0) + j sin(0). Frequency: in a sinusoidal function A cos((j)t + d) or B sin((Of + 0), the quantity co is the angular frequency in radians per second (rad/sec). Equivalendy, A cos(o)f + 6) = A cos(2K/t + 0), where/is the frequency in hertz (Hz or cycles per second). Note that O) = 2n f. Frequency response: (of a circuit) graph of the ratio of the phasor output to the phasor input as a function of frequency. It consists of two parts: (1) a graph of the magnitude of the pha sor ratio and (2) a graph of the angle of the phasor ratio. High-pass circuit: circuit with a frequency response such that the high-frequency content of the input signal is passed while the low-frequency content of the input signal is attenuated. Imaginary part: the imaginary part of a complex number z = a + Jh for real numbers a and b, denoted by Im[z], is b. Impedance: ordinarily complex frequency-dependent Ohms law-like relationship of a two-ter minal device, defined as Z(/a)) = V/I, where V is the phasor voltage across the device and I is the phasor current through the device. For the resistor, = /?; for the capacitor, = l/(/a)Q; and for the inductor, = ycoZ. Magnitude (modulus): the magnitude of a complex number z = a + jb , denoted by |z|, is

Phason complex number representation denoting sinusoidal signals at a fixed frequency. Bold&ce capital letters denote phasor voltages or currents; a typical voltage phasor is V = and a typical current phasor is I = Polar coordinates: representation of a complex number z as p?-^, where p > 0 is the magnimde of z and 0 is the angle z makes with respea to the positive horizontal (real) axis of the complex plane. Reactance: imaginary part of an impedance. Real part: real part of a complex number z = a + jb for real numbers a and b, denoted by Re[«], is a. Rectangular coordinates: representation of a complex number z as coordinates in the complex plane, i.e., zs a + jb for real numbers a and b. Resistance: real part of a possibly complex impedance. Sinusoidal steady-state response: response of a circuit to a sinusoidal excitation after all transient behavior has died out. This definition presumes that the zero-input response of the cir cuit contains only terms that have an exponential decay. Stable circuit: circuit such that any zero-input response consists of decaying exponentials or expo nentially decaying sinusoids. Susceptance: imaginary part of an admittance. Voltage division: in a series conneaion of impedances driven by a voltage source, the voltage appearing across any one of the impedances is proportional to the ratio of the particular impedance to the total impedance of the connection. Zero-input response: response of the circuit when all source excitations are set to zero. ’ In the literature, both z and z* are used to denote the conjugate o f a complex number z. However, in matrix arithmetic, Z* usually means the conjugate transpose o f the matrix Z. We will sometimes interchange the usage. In MATLAB, * means multiplication and conj(Z) means conjugated. So there is some ambiguity in the usage.

^ ^

^


483

Problems SO LUTIO N OF DIFFEREN TIAL EQUATIONS W ITH CO M PLEX EXPO N EN TIALS

O

'

Figure P I0.3 A N SW ER S: 0 .8 sin (2500r -

3 6 .8 6 ‘>). 80

1. Construct the difFerential equation model o f

s in (2 5 0 0 r-3 6 .8 6 » )

the series RL circuit o f Figure PI 0.1 in which L = 0.25 H and R= 100 Q. Then use the method

4. Construct the differential equation model of

o f section 4 to compute the steady-state response when v’,„ (/) = 20V 2 cos(400/) V.

the parallel /?Z.Ccircuit o f Figure PI 0.4 for R = 100

C = 1 pF, and Z. = 40 mH. Then use

the method o f section 4 to find the steady-state response when / (f) = 20 cos(2500f) mA. vJt)

R < v .J t )

Figure P I0.1 AN SW TR: 20 cos(400/ - jr/4) V

Figure P I0.4 AN SW ER: 1.6 cos(2500; + 36.8~») V

2. Find the differential equation model o f the series RC circuit o f Figure P I0.2 in terms o f and V(it) assuming that C = 5 pF and R = 800 Q. Write

as a function o f

and V(^t). Then use the method o f section 4 to determine the steady-state response when iV„(/) = 20>/2sin(250/) V. + v,(t) -

KCL AND KVL W ITH PHASORS 5. Find the phasor current I and /(/) for each circuit o f Figure Pi 0.5 when (O = IOOti rad/sec.

(2+j4)A

+ v„(t)

R

V „(t)

Figure P i0.2 ANS^XTR: 20 sin(250/ + 0.25k) V

Figure P i0.5 AN SW TRS: (a) 10 cos( 1OOTif - 0 .9 :" )

A. fb)

S.6626 cos( 1OOtt/- 1.798) A 3. Construct the differential equation model o f the series RLC circuit o f Figure Pi 0.3 in terms o f /^(r) and assuming L = 1 0 mH, C = 4 pF, and R = 100 Q. Then use the method o f section 4 to find the steady-state response /^(t) when

= 100 sin(2500/) V. Next compute

6. T he circuit o f Figure P i 0.6 operates in the sinusoidal steady state with the indicated pha sor currents when i^{r) = 10 cos( 1OOOf) A. Find the value of the phasor currents and and the associated /,(r) and

484


j20 (26+j12)V

0

20

(a)

Figure P i0.6

e

C H E C K : /,(/) = 25 cos(100r + 0.9273) A

j20

(26+ jl2)V

L 20

7. Suppose that in Figure P I 0.7, v^{t) = 4 cos(o)f) V and ''2^^ ^~ 4-s/2 cos(o)/ —0.25ti ) V. Find V[{t) = A"cos(ior+ (J)). +

v^(t)

(b)

-

e (26 + jl2)V

Vj(t)

v,(t)

0

j20

20

(0

Figure P I0.7 ANSW ER; vjU) = 4 cos((.»/ - ‘)0") \'

Figure P I0.9 8. Use KVL to determine the phasor voltage 10. For the circuit o f Figure P i 0.10, use KCL

in the circuit o f Figure PI 0.8. 8jV

4jV

and KVL to find the phasor voltage V^. and the phasor current I^,. If the frequency co = 2 0 0 0 ;: rad/sec, find the associated voltage and current time functions. -

(4 + ]4)V

I

Figure P10.8

(2-j10)A

-

(2+j4)A V _y (2+ j

/ Ti

(t)

AN SW ER: - 4 V

V

(4 + j6) A

£)

(4+j6)V

1

(5

9. For the circuits of Figures PI0.9a, b, and c,

Figure PIO.IO

compute the indicated phasor currents assuming

ANSW'ERS: V^.= 18.98^17156« V. I. = 2 0 .4 ^

R = 2 0 ., L = A mH, and C = 1 mF. If oj = 500

-1 0 1 .3 '* A

rad/sec, determine the associated time functions.

BASIC IM PED AN CE AND A D M ITTAN CE CO N CEPTS 11. (a)

A capacitor has an admittance

= y’8

mS at OJ = 400 0 rad/sec. Find the

48S


value o f C in pF. Compute the value of the capacitor’s impedance at (0 = 500 rad/sec. An inductor has an impedance Zj =

(b)

j20 Q at (0 = 4000 rad/sec. Find tlie value o f L in mH. Compute the value

Figure P I0.14

of the inductors admittance at 10,000

A N SW ERS:

rad/sec.

3 0 ‘> A. I,

= 0 .0 2 Z 6 0 ‘> A, 1/ = 0 .0 4 Z = 0 .()2 Z ]5 0 " A. i j t ) = 0.02S3

cosi 11)00/ + 1S‘>) A 12. In the circuit o f Figure P I0.12, cos(lOf) V, C = 0.2

(f) = 10

F, Z: = 0.1 H, R= 2.5 Q.

Determine the phasors I j, I 2, and

Find

15 . Consider the circuit o f Figure P i 0.15 where R = 200 H, A = 80 mH, C = 1 pF and

i- p ) =100 sin(2500r) niA. Find the voltage

".a/')-

phasors r

n

I'' *

' r

t

V^,

and

and then compute

" R

JY Y V v„ -

l

+ V. -

Figure P i0.12 13. The circuit o f Figure PI 0.13 is operating in the sinusoidal steady state with = 20

Figure P i0.15

C = 1 mF.

C H E C K : v ;p ) = 28.28 co s (2 5 0 0 r- 135°) V or

(a)

v .p ) = 28.28 s in (2 5 0 0 f- 45") V

Suppose /^|(r) = 10 cos(100r + 30'*) mA and v^Û) = 200 cos(lOOr) mV.

(b)

Find the phasor I^. and then the cur

16. In the circuit o f Figure P IO .I 6 , C = 0.03 S,

rent ip ) .

I = 0.1 H, C = 0.4 mF, /j(r) = 1.2 cos(200r) A

Now let /j,(r) = 10 cos(50/ + 3 0 ”) mA

and

and v^2 ^t) = 200 cos(lOOr) mV. Find

(a)

the current

(b)

How does this part

= 40 sin(200^) V. Find the phasors Ij and Find the phasor 1^ and the associated time function ijit).

differ from part (a)? fi.(t)

o

v,(t)

Figure P I0.13 14. Find the phasor currents

Figure P I0.16 AN SW ER: (b) 1.2Z-9()'\ 1.2 sin(200r) A and

and then determine i j p ) for the circuit o f Figure P I 0.14 in which /? = 1 k li, C = 1 pF, and

= 0.5 H,

= 20 cos(1000r + 60°) V.

17. In the circuit o f Figure P 10.17, /? = 6 f i, Z. = 80 mH, C = 0.5 mF, Vp{t) = 8 cos(200r) V, and I, = 0.5Z 90® A. Find the source voltage, which operates at the same frequency o f 200 rad/sec.

■m


V jt)

Z Jj(o )

Figure P I0.17 ANSWHR: S.831 1 cos(200r +30.96<’) V

Figure P 10.20 Parallel Z.C circuit. ANSWl-.RS: (a) -/1.25 12, (b) 1.6 ml-

18. The circuit o f Figure PI 0.18 operates in the sinusoidal steady state at a frequency o f cOq =

21. Consider the circuit o f Figure P I0.21.

2000 rad/sec, /?, = 7?^ = 10 Q, V-„ = 50 V, and

(a)

Find the impedance at (o = 100 rad/sec.

= 2Z. - 53.13® A. Compute the phasor volt

(b)

What happens to the impedance as to

age across Rj and then find the impedance Now construct a simple series circuit

gets large? (c)

Ifv/„(/) = IOV 2 cosdOOOV, find /.(/).

that represents this impedance at cOq. ijt )

1 mF v jt)

I

6

lOO

0.1 H

Figure P I0.18 Figure P I0.21

ANSWHR: Z = 2.5 +ylO Q

A N SW FR: (a) S - p U 19. (a)

Find the steady-state response o f the circuit o f Problem 3 using the phasor

22. For the circuit o f Figure PI 0.22, suppose R

method. Discuss the relative advan

= 100 Q, ^ = 0.5 H, C = 5 îF. (a) If = 0.1 cos(500/) A, find vît). (b) Find (O ^ 0 in rad/sec so that the input

tages o f the phasor method. (b)

Find the steady-state response o f the

admittance is real.

circuit o f Problem 14 using the phasor method. Discuss the relative advan tages o f the phasor method.

+ V ,( t)

SERIES-PARALLEL IM PED A N CE AND A D M ITTA N CE CA LCU LA TIO N S 20 . Consider the circuit o f Figure P I 0.20. (a) (b) (c)

Figure P i0.22

If C = 0.01 F, findZ.„(/100). If Zy,^(/100) = 25j O., find the appro

23. Consider the circuit o f Figure P I0.23 in

priate value o f C.

which /?j = 20 /?2 = 10 Z, = 20 mH. (a) I f C = 0.3 mF, find K.„(;-500).

Using the impedance o f part (b), if

ij„U) = 100 cos(100r + 45®) mA, find v^t).

(b)

Find the value o f C that makes the input admittance real at OJ = 500 rad/sec.


(c)

If C = 0.3

mF and /.„(f) = 100

487

zero. Determine the minimum value

cos(500r) niA, find /^](^) and /^-(r)

o flK „(/ 0 ))l.

using current division. /Y Y V L

(a)

Figure P I0.23

(b)

A N SW ERS: (a) 0.1 + yO.l S, (b) 0.1 mF

Figure P I0.25 24. Consider the circuit o f Figure P I 0.24 in which C = 1 F. At C0= 2 rad/sec,

= 4 + jl

n.

26.

For a particular two-terminal device, = 0.002 + y0.002 at O) = 500 rad/sec.

(a) (b) (c)

Find the appropriate values o f R and

Construct a parallel RC circuit having this

L

admittance at (0 = 500 rad/sec. If the circuit is

For 0 < Z < 0.2 H, specify the range o f

excited by a currcnt source with ij^t) = 10

possible reactance values for

cos(500f) niA, find the voltage appearing across the current source.

If

= 10 cos(2/) V, find Vjî).

A N SW ER : C = 4 uF, A* = 500 Q .. 5 \'^(/) = —;= cos 500/ - 4 5 " ) V V2 27. For a particular two-terminal device,

R

'jt ) Q

Z .„(;1000) = 2000 + ;2 0 0 0 Q.. Construct a series RL circuit having this impedance at CO = 1000 rad/sec. If the circuit is excited by a volt

Z,„(jw)

age source with v^{t) = 10 cos(1000/) V, find the current through the resistor.

Figure P I0.24

AN SW ER: Z. = 2 H, /^ = 2 kQ., 5 /(/) = - = cos 1000/ - 4 5 " ) mA

AN SW ERS: 4 Q, 0.2 H, 0 to x 25. Consider the circuits o f Figure P I 0.25 in which R = 5 O., L = 32 mH, and C = 5 ^F. (a)

Find

28. The circuit o f Figure P i 0.28 operates in the

Zy^^(/(0) as a function o f (0. Then com

sinusoidal steady state at the frequency' (0 =

pute the frequency at which Zy^^(/CO) is purely real, i.e., the reactance is zero.

5000 rad/sec with R =4 Q., L = 0.4 mH, and C = 0.1 mF. Find Z-^^{j5000) and ^/„(y5000).

Determine the minimum value o f

Construct a simple series circuit that is equiva

For the circuit o f Figure P 10.25b, find

construct a simple parallel circuit that is equiv

Ky^;(/0)) as a function o f (0. Then com

alent to this circuit at 0) = 5000 rad/sec. In both

pute the frequency at which

cases specily the element values.

Consider

Figure

P 10.25a.

lent to this circuit at 0) = 5000 rad/sec. Finallv, (b)

is

purely real, i.e., the susceptance is

488

Chapter 10 * Sinusoidal Stoatly State Analysis by Phasor Mcthoils

(b) R

Find the value o f to in terms o f R and C at which the phase angle difference bet\veen \ a n d

(c)

is 4 5 “.

At the w computed in part (b), deter-

Figure IM0.28 -H f

C H EC K : Z.„(/‘5000) =1 + ;2 ^

■6

29. For the circuit o f Figure P i 0.29, let Z.=4 mH, C = 10 |iF, and v^{t) = K^cos(cor) V. Compute

Find the frequency co (in

Figure P I0.31

rad/sec) at which tlie steady-.state current /^(/) = 0. At this frequenc}', what is the vohage across

32. Consider the circuit o f Figure PI 0.32.

the LC parallel combination?

(a)

If vj^t) = V^^(zos{tlRQ, find the sinusoidal steady-state response Vf{t) in terms of V;,, k and C.

(b)

lf/?= 1 0 Hand

10 n/2 cos(10/)

V, find the value o f C so that V(^t) = 10 cos(10r + 0) V. (c)

For the value o f C found in part (b), compute the corresponding value ol 0.

Figure P I0.29 AN SW FR: (I) = 5000 rad/sec

R

SERIES/PARALLEL IMPEDANCES WITH V/l DIVISION

^,(t)

30. Consider the circuit o f Figure P I0.30 in which

/? =

20

Q,

Z.

=

4

H,

( 0 = 10>/2 cos(5/) niA. (a)

Find the input impedance ■^,„(/w) and the input admittance

(b)

Figure P I0.32

and

At OJ = 5 rad/sec determine the steady-

ANSW ER: (a) r(^(/) = ^ c o s v2

RC

-4 5 ‘

33. Consider the circuit o f Figure P I 0.33 in which y? = 8 (a)

state current ;’^(r).

Z, = 8 mFi, and C = 0.125 mF.

Determine the values o f the phasors I^, and V(^ when

i,(t)

= 2 A and (d =

1000 rad/sec. Specify the correspon ding time functions.

©

(b)

Repeat part (a) for o) = 500 rad/sec.

Figure P I0.30 ANSW'F'.KS: (a) .b; I (I 31. (a)

y20co 5+

/(I)

0.25 . 0.05 - /■ (I)

5; - k !4) niA For the circuit o f Figure P I0.31, find the ratio in terms of/^, C, and CO.

Express the answer in polar form.

Figure P i0.33 Parallel y?/.Ccircuit. 34. In Figure P I0.33, suppose R = 500 Q, I. = mH, C = 0 .1 2 5 i:,Ar) = loV2co.s(co/+ 60^) A.

0 .5

mF,

and

•hS9


(a)

Compute the values o f the phasors I^

Using the phasor method, find

I^, and

v j t ) = 50 cos(4000^) mV.

when

= 2 A and OJ =

4000 rad/sec. Specify the correspon

He

ding time functions. (b)

Repeat part (a) for (O = 8000 rad/sec.

when

100v„,(t) v„.(t)

35. In the circuit o f Figure P i 0.35, suppose R

= 2 0 n , L = 0.5 H, and C = 0.625 mF. (a) If = 16 cos(40r) V, find V(^t) and vît) using phasor voltage division. (b)

If

Figure P i 0.38 Iw o coupled /^Ccircuits.

39. Consider the circuit o f Figure P i 0.39 in which /?j = 200 n , Z = 0.2 H, R, = 200 Q, and

= -32sin(40r) V, find

v^t), and vît) using phasor voltage division. Hint: avoid repeating the cal

C = 0.05 pF. Use the phasor method to find

i^t) when

= 10 cos(lO'^r) mA.

culations o f part (a); this can be done by inspection.

R

6

Figure P i0.39 Two coupled circuits. 40. Consider the circuit o f Figure P i 0.40 in

Figure P 10.35 Series /?/,C circuit. 36. Reconsider Figure 10.35 lor /? = 10

Z. =

Find

0.08 H, and C = 0.02 R (a)

If

which /?! = 500 Q, I = 0.125 H, R ,= m Q, and C = 5 pF. Suppose Vj^^) = 120 cos(400/) V. and

= 10 cos{25t) V, find

and vît) using phasor voltage division. (b)

If za//) = 16 sin(50f) V, find and

V[ {t)

V({t),

using phasor voltage division.

37. In the circuit o f Figure PI 0.37, /?, = 20 Q., i = 2 H ,a n d / ? j = 1 0 a . If|V„„/V,„| = 0.2 at co= 40 rad/sec, find the necessary value(s) o f C (in mF). Figure P i0.40 41. Consider the circuit o f Figure P I 0.41 in which 7^1 = 500 QX = 0.125 H, R, = 100 iX and C = 5 pF. Suppose

''" 6

mA. Find

Figure P I0.37 AN SW TR: 0.625 mF or 0.2083 mF 38. In the circuit o f Figure PI 0.38, /?, = 50 Q, q

= 1 uF, Rj = 300 Q, and q

= 0.625 pR

and

= 120 cos(400r)


490

AN SW ERS: 2 Z 9 0 ‘>, - 2 sin(400r) A 44. For the circuit o f Figure P i 0.44, find the and the phasor voltage Y q If

phasor current

the circuit is known to operate at a frequency o f CO = 1000 rad/sec, find /^(f) and

and the

values o f L and C 2Q > j2 Q

-j2 0

NETWORK THEOREMS IN CONJUNCTION WITH V/l DIVISION.

AN SW ERS; t'fit) = 2 cos(lOOOr) V, i/(/) = cos

(You should consider applying one or more

(1000/-O.S3T) A

Figure P I0.44

network theorems to simplify the solution to the problems in this section.)

45. In the circuit of Figure Pi 0.45, R=20Q., L = 20 mH, C= 100 pF, and v^.(0 = 2 0V 2 cos(IOOO/)V.

42. In the circuit o f Figure P 10.42, R^ = 60

Compute the value of

in steady state.

/?2 = 40 n , and C = 0.1 mF. Find the phasor Ij /Y Y V L

and the corresponding steady-state current /j(/) when /^.(/) = 5>/2cos(100/) mA. This prob lem can be solved by direct current division or

6

by source transformation and impedance con cepts. Which method is easier?

Figure P i0.45 A N SW ER: 20 co s(1 0 ()0 r- 135°) V

i(t)

0

46.

Consider the linear circuit o f Figure

PI 0.46, which operates at 50 Hz and for which V, = + 1,1^2(a) Find the values o f a and b. (b)

Figure P i0.42

If v^^{t) = 10 cos(lOOTCf) V and 200 sin(lOO)/^) mA, find vît).

43. In the circuit o f Figure P I 0.43, CO = 400 rad/sec,

= Rj = 2

L = 5 mH, and C =

625 |.iF. Find I^ând the corresponding steady state. vp)= 12 cos(400^) V

in j5on

■ -© J200Q + V. -

Figure P i0.46 AN SW ERS (in random order); Fiaure P I 0.43

- 0 .8 . ./40 Q. - 1 6 cos(IO().-t O V

=


491

47. The linear circuit o f Figure P I 0.47 is such that in the steady state, if

can do this with some straightforward

= 10 cos(2007U/)

reasoning without writing any equa

A with v^2 ^t) = 0, then

\t) = 20 cos(2007ir + 45®) V. On the other hand if i^^{t) = 0 with /;2 (/) = 10 cos(2007if + 45°) V, then v^{t) = 5

tions.

cos(200)r + 900) V (a)

Find a linear relationship between

v^2 >and (b)

,

.

If"/^,(/) = 5 cos(2007tr - 45«) A and

= 20 cos(2007rr) V, then in the steady state find v^{t). (c)

Find

and

Zj

Z - ,.

Develop simple cir

Figure P I0.48 AC Wheatstone bridge circuit.

cuit realizations o f these impedances valid at OJ = 20071 rad/sec.

THEVENIN AND NORTON EQUIVALENTS 49. Find theThevenin equivalent for the circuit o f Figure PI 0.49 when R = AQ. L = 20 niH, C = 1.25 mF, = 2Z.45® A and (O = 200 rad/sec. Be sure to express the open-circuit voltage as a

Figure P I0.47 ANSWHRS: (b) r,(/) = 18.46 cos(200nr -

time function.

22.5") V; (c) Z, = 0.763 + y2.6()5 H Z, = 4 Q

OA

48. The circuit o f Figure P I 0.48 is a general Wheatstone bridge circuit (the dc version o f which is described in Problem 35 o f Chapter 2). Here the circuit is used to measure the value o f the unknown inductance L. (a)

Suppose

= 0. Show that the steady-

state voltage i'„,^f(t) = 0 when

=

UC. Note: In general the condition for a null voltage, v{t) = 0, in the steady state is that the products o f the cross impedances be equal. (b)

Again suppose that

= 0. You are given

= 2 sec and that the voltage

source v-^j^t) is a sinusoid with a fre quency

of

5

rad/sec.

W ith

the

unknown inductance L inserted in the circuit as shown, you adjust R[ until you reach a sinusoidal steady-state voltage null, ing value for R^ is 3 (c)

= 0 V. The result Find the value

ofZ,. Now suppose R^ ^ 0. Show that the condition o f part (a) is still valid. You

Figure P10.49 Parallel /?ZC circuit. A N SW ERS: Z^;, = 4 i l , /-^.(/) = 8 cos(200/ + 45*’) V

50 . For the circuit o f Figure P I0.50, let 1 = 1 0 mH, /? mA. (a)

=20 a, C = 20

îF, and

l-„ = 1 0 0 ^ 0 ”

Find the Thevenin equivalent at the terminals a and b if to = 2000 rad/sec.

(b)

If the circuit is terminated with a load consisting o f a series connection o f a 20 Q resistor and a 20 mH inductor, find the sinusoidal steady-state voltage across the load.


492

OA

Figure P i0.53 Two coupled circuits. 54. The circuit o f Figure P 10.54 operates in the SSS at cOq. (a) 20 krad/sec and

= 2Z.0". Find the Thevenin

Find the Thevenin equivalent imped ance

51. The circuit o f Figure PI 0.51 operates at to = (b)

Find

if

10 a , ^ = 5 a

equivalent circuit (in the phasor domain) at ter

= 1^^^cos(w^/), R = and

20 Q.

minals A and B. Use this Thevenin equivalent to find the magnitude o f the phasor

when the

10 mH and 1 kH series combination load is con nected to A and B.

i jt )

-jx

Figure P I0.54 55. In the circuit o f Figure PI 0.55, assume co = 100 rad/sec, I = 40 mH, C = 5 mF, R = S Q., and a = A Q.. Find the Thevenin impedance

Figure P I0.51

seen at terminals A and B. If CH ECK : 190 < |V J < 205 V. |V^^| . 0.5| V J 52. For the circuit o f Figure PI 0.52,

Q, Cj = 0.2 pF,

= 20

cos(lOOf) V, find

= 1000

= 500 Q, and C , = 1 \i¥.

Find the Norton equivalent circuit when v- (t)

= 50 cos(4000f) V.

e

v.(t) -OA

R, Figure P I0.55 ■OB

GENERAL SSS ANALYSIS (NODE OR LOOP ANALYSIS)

Figure P I0.52 Two coupled R C circuits. 53. For the circuit o f Figure P I 0.53 R = 2.5 k n . Find the Thevenin equivalent when

56. (a)

Find the phasors

in the cir

cuit o f Figure Pi 0.56 when

=

= 2 0 V 2 Z 4 5 " V, R = AQ., L = A mH, and oj = 1000 rad/sec. Specify the corresponding time functions.

10 cos(4000r) mA.

(b)

Determine the value o f oj for which the magnitude o f the output voltage


phasor is 20% o f rhe magnitude o f rhe

(0

input voltage.

493

At to = 100 rad/sec, determine the Thevenin equivalent circuit in which Z^y^(/'100) is a series combination o f two circuit elements seen at terminals

/Y Y V L

A and B.

'“6

-OA 2R 2R

;v,(t)

Figure P i0.56 C H E C K S: 5 A, 2o’ v

-O B

Y.(j«) 57. In the circuit o f Figure P I0.57,

= y'30 Q, Figure P I0.59

= - ; 4 0 Q, V^, = 28 V, V^2 = ^ sinusoidal sources have been operating for a

60. Consider the circuit o f Figure P i 0.60 in

long time, and Z = 50 - j 4 0 Q. Find V^.

which v(,t) = 20 cos(lOOO^) V,

= 40 Q., Rj =

20 Q.,L = 20 m H .an d C = 7 5 \i¥. (a)

Write and solve a nodal equation at the top node for

6 ''“

Then write the

corresponding time-domain expres sion for v^t). (b)

Figure P I0.57

z^ = y io a

z^^ = - y i o a

and then write the corre

sponding time-domain expression for

58. For the network o f Figure P i 0.58, a = 20

a

Calculate

^ = lo a

and

= 20 V. Find the phasor current I^.. Z,

the inductor current (c)

Find the Flievenin equivalent circuit at the source frequenc)' relative to terminals A and B. Draw the Thevenin equivalent circuit showing the Thevenin impedance as a series circuit of two elements.

Figure P I0.58 AN SW FR: O.r. + 0.2/ 59. Consider the circuit o f Figure P i 0.59 for which C = 0.8 mF, r.„(;100) = 0.01 + ;0 .0 4 S, and v-^,(r) = 80 cos(lOOf) V. Find R. (a) (b) (c) (d) (e) -

Find L. Find Find At CO = 100 rad/sec, determine the Thevenin equivalent circuit phasors

Woe and Z^;^(ylOO) at terminals A and B.

Figure P I0.60 ANSWKRS: V^. = S - /S = -/ 0 .:5 . Z./. = 1 0 - ; 10 12 61. The circuit o f Figure P i 0.61 operates at (o = 2 krad/sec a n d =1. 5 mS with = 10 0 Z 0 ° V. Find the Thevenin and Norton equivalent circuits seen at terminals A and B.


494

determine the asymptotic behavior for large (0 / 'A

J \

■9

l.s f

and for at least one other frequency without a

0-251

computer or calculator. List these properties in

.

writing along with your reasoning.

Figure P I0.61 = ;1 0 0 0 Q,

C H EC K :

'■©

= 25 + ;2 5 V,

= 25 + y25 mA 62. Consider the circuit o f Figure PI 0.62. If i^{t)

Figure P I0.64

= 40 cos(lOOO^) mA, C, = 0.25 1-iF. = 10 mS, /?j = 1 kQ, and R-, = 5 kH, find the Thevenin

65. Compute the magnitude and phase o f the

equivalent circuit parameters

frequency response o f the circuit o f Figure

and

P I 0.65 where I = 25 mH and /? = 50 Q. Plot your response in MATLAB (0 < 0) < 8000 rad/sec) and determine the frequency at which '• « '©

the magnitude is I/ V 2 o f its maximum value.

R,

Before sketching the responses, determine the asymptotic behavior for large (0 and for at least

Figure P I0.62 C H EC K : = 128 - ;3 3 7 .8 8 Cl, 1 4 0 .2 5 Z 9 5 .2 8 “ V

=

one other frequency without a computer or cal culator. List these properties in writing along with your reasoning.

63. This problem tests whether you can synthe size ideas from two different parts o f the text.

L

In the circuit o f Figure P i 0.63, R = 20 Q, L =

(b

1 H, vît) = 50 cos(100r)«(r) V (notice the step function), and /^(O'*') = 1 A. If the response for

t > 0 has the form i^{t) = A cos(l OOf + (j)) + then determine the constants A, (j), X, and B. ijt)

R v,(t)

6

Figure P i0.65 66. Inside the black box o f Figure P I0 .6 6 a there is a two-element circuit composed o f a resistor o f 10 ^2, capacitors, inductors, or some combination o f these elements. A variable-fre-

Figure P i0.63

FREQUENCY RESPONSE 64. Compute the magnitude and phase func tions o f the frequency response o f the circuit o f Figure P I 0.64 in which L = 4 mH and C = 0.25 mF. Plot your response in MATLAB (0 < CO < 5000 rad/sec). Before sketching the responses,

quency voltage vp) = lOcos(O)r) V is applied to the box and the voltage v(t) =

cos(cor + 0) is

observed. A plot o f the magnitude o f v{t) with respect to CO is given in Figure P 10.66b. (a)

Draw the circuit contained inside the

(b)

box. (There are two solutions.) Specify the element values.


v,(t) = 10cos(cot)V

+

Black Box

v(t)

(a)

(b)

69. Reconsider the pressure-sensing example of

Figure P i0.66

section 10. Specify a set o f mesh currents and 67. Compute the frequency response o f the cir

write a set o f mesh equations that describe the

cuit o f Figure PI 0.67, where R= 100^2, L = 10

circuit. Solve the equations for 1 pF < € 2 ^ pF using MATLAB or some other, equivalent

mH, C = 0.1 mF, and

is the output. Use

MATLAB or its equivalent to generate the magnitude and phase (in degrees)

plots.

Consider 0 < (O < 3000 rad/sec.

software program. Plot the magnitude o f V ^ V âs a function o f Cj- Now construct a plot o f the magnitude o f V ^ - V ^ a s a function o f pres sure in mm Hg.

OP AMP CIRCUITS 70. (a) Figure P I0.67

a single inductor. Let v^{f) be the input excita tion and ip ) the circuit response. The magni tude frequency response is given by Figure P I0 .6 8 b . Draw^ the circuit inside the box and assign component values if it is known that L = 40 mH.

when

= sin(200r)

mV for the circuit of Figure P10.70a. (b)

For the circuit o f Figure PI 0.70b, find C so that when

68. The box labeled V{joi) in Figure P I0 .6 8 a contains a single resistor, a single capacitor, and

Compute

(c)

= cos(400r) mV,

= sin(400r) mV. Find the phasor transfer function, //(yco), and plot the magnitude o f the frequency response (using iMATL^B or the equivalent) as a function o f 03 =

Inf, where/is in Hz and to in rad/sec.


496

r> 100 kfi

20 kO

80 kO

o n (a)

(b)

Figure P I0.70 Op amp differentiation circuits. (a)

n

Compute when = sin(400/) V for the circuit of Figure P10.71a. For the circuit of Figure P I 0.7 lb, find C such that when =sin(500/) V, = 5 cos(500/) V. This represents an integration of the input with gain. Find the phasor transfer function, and plot the magnitude of the frequency response (using M ATLAB or its equivalent) as a function of O) = In fi where/is in Hz and O) in rad/sec.

71. (a)

(b)

(c)

n o

(b) Figure P I0.72 Leaky integrator circuits.

n

r s

10|j F

H e-

50 kO

+

He-

200 kn +

o— +

+

vjt) JL . (a)

(b)

Figure P I0.71 Op amp integrators.

73. (a) At (0 = 2 X 10^ rad/sec, find the phasor voltage gain ôuPîn °P circuit of Figure P 10.73. (b) Find the phasor transfer function, //(/to), and plot the magnitude of the frequency response as a function of O) = In f, where / is in Hz and (O in rad/sec using M ATLAB or equivalent software.

1kn 72. (a) If an 800 Hz sine wave of unit ampli tude excites the leaky integrator circuit of Figure PI0.72a, determine the steady-state output voltage. (b) For the circuit of Figure P 10.72a, find the phasor transfer function, //(/(o), and plot the magnitude of the fi*equency response (using M ATLAB or Figure P I0.73 its equivalent) as a function of (O = 271/where/is in Hz and O) in rad/sec. 74. For the circuit of Figure P i0.74, find the (c) If the input to the circuit of Figure expression for the phasor transfer function P I 0.72b is = cos(2000)r) V, //(/ cd) = Assume an ideal operational determine the values of R and C so amplifier. Plot the magnitude of the transfer that v i t ) = 5cos(2000ti^ +135°) V.

n r^ o

n


O '

^ function as a function o f COusing M ATLAB or

w

•^19'

the equivalent, assuming /? = 10 k fl and C =

0.01 mE

frequency increases to infinity? What happens as the frequency decreases to zeroi^

VO o o

Figure P I0.74

Figure P I0.76 Ideal op amp circuit.

75. In the circuit of Figure PlO.75 assume the operational amplifier is ideal and that = 25 ki2, Cj = 1 ^F, /?2 = 5 k£2, and C2 = 0.2 pF. Compute the gain of the circuit as a function of (0. Then use M ATLAB or the equivalent to plot the magnitude and phase of the fi-equency response as the logarithm of the frequency for 1 < C0< lO'^ rad/sec.

77. Consider the circuit of Figure P I0.77 in which = 200 £2, Cj = 0.05 |iF, /?2 = 28 IdQ, and C2 = 0.05 pF. Use nodal analysis to compute the ratio =I ^Hz. Now use physical reasoning to obtain the approximate the ratio at/= 1 Hz and /= 100 kHz.

o

o o o o o o o o o o o o o o o

w

Figure P I0.75 76. For the operational amplifier circuit of Figure P I0.76, /?j = 5 kQ, C, = 0.02 |iF, /?2 = 5 kfi, and C2 = 0.08 pF. (a) Write two node equations and solve to find a relationship between the output phasor and the input phasor at the frequency/= 1000 Hz. Note that the voltage from the minus termi nal of the op amp to ground is which equals the voltage from the plus terminal to ground, assuming the op amp is ideal. (b) Repeat the calculation at/= 100 Hz and/= 3000 Hz. What happens as the

Figure P I0.77 Op amp circuit having a band pass type of response. Note that is an intermediary variable useful in the nodal analysis of the circuit.

C

H

A

P

T

E

R

Sinusoidal State State Power Calculations The AM or FM receiver that is often part o f a home stereo system receives signals from radio sta tions through an attached antenna. The intensity o f these signals or radio waves depends on the power radiated into the atmosphere by the broadcasting station, the distance between the receiv ing and transmitting antennas, and the design o f the receiving antenna. The intensity or magni tude o f the signals picked up by the receiving antenna is very small. The power available from the antenna and deliverable to the receiver is typically in the microwatt range. Again, this is ver)^ small. Hence, it is important to have maximum power transfer from the antenna to the receiver input so that the music signals received can be properly amplified and enjoyed. Since the signals in the antenna are sinusoidal at very high frequencies, the antenna is represented by a phasor Thevenin equivalent circuit as is the input circuit o f the receiver. Hence we must understand maximum power transfer in the context o f sinusoidal steady-state analysis to describe and analyze this prob lem. An example at the end o f the chapter illustrates some impedance matching techniques to achieve maximum power transfer from an antenna to a receiver.

CHAPTER OBJECTIVES 1. 2.

Define and investigate the notion o f average power. Define the notion o f the effective (rms) value o f a periodic voltage or current and its rela

3.

tionship to the average power absorbed by a resistor. Define the notion o f complex power and its components— average, reactive, and appar ent power— and investigate the significance o f each and their relationships.

4.

Introduce the notion o f power factor associated with a load and describe reasons and a

5.

method for improving the power factor. Prove the maximum power transfer theorem for the sinusoidal steady-state case, and illus trate its significance for the input stage o f a radio receiver.

500

Chapter 11 • Sinusoidal State State Power Calculations

SECTION HEADINGS 1. 2. 3. 4.

Introduction Instantaneous and Average Powers Effective Value o f a Signal and Average Power Com plex Power and Its Com ponents: Average, Reactive, and Apparent Powers

5.

Conservation o f Com plex Power in the Sinusoidal Steady State

6.

Power Factor and Power Factor C orrection

7.

M aximum Power Transfer in the Sinusoidal Steady State

8. Summar}^ 9. Terms and Concepts 10. Problems

1. INTRODUCTION Chapter 1 defined the concept o f power. The following chapters were primarily devoted to the cal culation o f voltages and currents. This does not mean that the consideration of power is o f sec ondary importance. The very opposite is true. A homeowner pays for the energ}' used, not for volt age and current. The integral o f power over, say, a 30-day period determines the household ener gy consumed in a month. Hidden in the homeowners cost is an adjustment to cover the power losses incurred in transmitting energy from the generating station to the home. Thus power con siderations have a significant impact on everyday life. A second reason for understanding ac power usage is safety. Each appliance, and its cord that plugs into the wall outlet, has a maximum safe power-handling capacit)'. Misunderstanding such infor mation and/or misusing an appliance can lead to equipment breakdown, fire, or some other lifethreatening accident. Even for electronic equipment in which power consumption is low, such as laptops and handheld PDAs, power consumption and, thus, battery life are important design factors. Power drainage direct ly determines the PDAs operating time before the battery needs recharging. In fact, optimizing power management in laptops and hybrid electric vehicles is an important research area in todays world. In this chapter we will investigate different notions o f power in ac circuits and discuss their sig nificance and application. The term “ac circuits” has a narrow meaning here. It refers to linear cir cuits having all sinusoidal sources at the same frequency and consideration o f responses only in steady state. The basic analysis tool is the phasor method o f Chapter 10.

2. INSTANTANEOUS AND AVERAGE POWERS Figure 11.1 shows an arbitrary two-terminal circuit element isolated from a larger circuit. With the voltage (in V) and current (in A) having indicated reference directions, the instantaneous power (in watts) absorbed by the element is given by equation 11.1: p{t) = v{t)i{t) (11.1)


501

FIGURE 11.1 Instantaneous power delivered to an arbitrary two-terminal element. Evaluating battery life or the length o f operation o f your cell phone involves consideration o f a quantit)' called average power, P (or

for emphasis), defined as the average value o f the instan

taneous power over an interval [T*!, T^. The idea is based on the average value o f a function, say

j{t) , which is defined as Le =

T-,

1

T h - TM r,J

Using this idea we define the average power consumed by a rwo-terminal element as shown in Figure 11.1 over the interval [T j, Tj] as

1

Pave(TiJ2) =

T, piOclt

r .-r ,

( 11. 2)

When the signal is periodic with period T, we speak o f the average power consumed by an ele ment over the period T as T

T (11.3)

It is not necessary that T b e the fundamental period; the evaluation o f the integral is the same for any integer multiple o f the fundamental period. EXA M PLE 11.1. Compute the average power absorbed by the resistor R connected to an inde pendent voltage source as shown in Figure 1 1.2b with the excitation shown in Figure 1 1.2a.

v.(t) f +

F IG U R E 1 1 .2 Triangular voltage waveform

driving resistor R .


S{)2

So

lu t io n

Step 1. Compute the instantaneous power for 0 < f < 7q. Here

IV J P {t) =

0 < / < 0 .5 ^ o .5 r „ < / < 7 ;

Step 2. Compute

Using equation 11.3 and observing that the Fundamental period is Tq, we

have 0.57n

/ ‘ >ave

p {t)d t = —

rj.

TnR

6R

Exercises. 1. Suppose the sawtooth in Figure 11.2a does not drop to zero at r = 0.5 T'q , but rather continues to increase until reaching ^ = T’q when it drops to zero and repeats. Find the average power consumed by R. AN SW ER: 3/^ . K? 2. Show that the average power absorbed by an R Q resistor in parallel with a Vq V dc source is— over any time interval [T'j, 7'^]. ^

O f particular importance is the average power consumed by devices in the SSS assuming all exci tations are at the same frequency, to. Consequently, all voltages and currents are sinusoids at the

sa7nefrequency. To compute the average power absorbed by a circuit element as depicted in Figure 11.1 (assuming a linear circuit), suppose v{t) = cos(u)t + 9^^ and i(t) = cos(a)r + 0^) . The associated instantaneous power is p ( f ) = v(f ) i { t ) = V,„ cos(co/ + e^,) X /,„ cos(cor + 0 ,-)

(11.4) =

cosO , -

e,-) +

cos( 2 (o/+ e,, + e,.)

Equation 11.4 follows from the trigonometric identit)' cos(x) cos(y) = 0.5 c o s (x - y) + 0.5 cos(a' + y). Observe that the instantaneous power o f equation 11.4 consists o f a constant term plus anoth er component varying with time at tivice the input frequency. Figure 11.3 shows typical plots o f />(/), v{t), and i{t).

Chapter 11 • Sinusoidal State State Power (lalculations

503

0.01

0 .005

0 .0 1 5

0.02

t in secs FIGURE 11.3 Plots o f tit) = 10 cos(377f) A, v{t) = 2 cos(377/ + 45°) V, and p{t). Using equation 11.3 witii T = 23t/ol), and observing that the integral o f a sinusoid over any peri od is zero, we obtain the following formula for average power in SSS:

T _ *Pnvp = —

co s(0 , - 0 ; )dt + ^ j c o s ( 2 c o r + 0 , + 0,-)d! =

c o s(0 , - 0 , ) (11 -5)

If the two-terminal element is a resistance R, then v{t) = Ri{t) and 0^^ - 0y = 0 . It follows from equation 11.5 that for a resistor

p _ Vm‘I m _ RI^m _ m âve,R 2 2 2R If the two-terminal element is an inductance L, then

âveL ~ ^

( 11.6)

= (/(.oZ,)!^ and 0^, - 0^ = 9 0 ° . Hence,

cos(±90°) = 0. Similarly, if the two-terminal element is a capacitance C, then

c~ ^ ^neans that the average power consumed by or delivered by a capacitor or au inductor is zero. Even though an ideal capacitor and an ide;il induc tor neither consume nor generate average power, each may absorb or deliver a large amount ot = (/coQV^ and 0^ - 0y = - 9 0 ° . Hence,

instantaneous power during some particular time inten-al. Before closing this section, we need to investigate the question o f superposition o f average powder. Is there a principle o f superposition o f average power? If so, when is it valid? When is it not valid? The following example provides the answers.


504

EXA M PLE 11.2. Consider the circuit o f Figure 11.4, which consists o f a series connection o f two (sinusoidal) voltage sourccs in parallel with a 1 Q resistor. For this investigation v^{t) = cos(ojjr + 0|) (having fundamental period /'] = 2:t/tO|) and v-,(f) =

cos(t02^+ B2) (having fun

damental period T-, = 271/(0,). For simplicit)^ we assume that v^{t) and v-,{t) have a common peri od o f 7 'seconds, i.e., there exist Integers ni and ;/ such that T = uT^ = niT^.

v,(t) Q 1n

< v_(t)

FIGURE 1 1.4 Circuit for investigating superposition o f average power. So

lu t io n

Compute the average power consumed by the 1 Q resistor. First observe that the power consumed by the 1 Q resistor with source 1 acting alone, i.e., v-j{t) = 0, is 1

1 0

0

Also note that the power consumed by the 1 H resistor with source 2 acting alone, i.e., u^{t) = 0, is I T , r ^ 0

0

= v^{t) +

W ith both sources active, linearit)^ (or KVL) implies that

1

1

(^ ’ 1 (/)

Pave = - \ V R {t)i,i{t)d t = - \ 0

. By equation 11.3,

+ v'2 (r ))“ dr

0

vf(r)f/^ + - J v ? ( r V r + - Jv| (t)v2{t)dr 0

0

0

T ôve.l

â\r,2 '

^ Vj (/)V2(/)f/f 0

âve,\

= Pave.\ + f'ave.l + '

ôve.l

T

COS((Oi /+0| )COS(C02/ +02)<^/^

j [ c o s ( ( ( 0 | + C O 2 )/ + (0 i + ©2 ) ) + C 0 s ((0 3 , - ( O 2 ) / + (0 i - 0 2 ) ) ] ^ / ^ 0


S()S

When the integral term in this last equation is zero, then

, indicating that

superposition o f average power holds. When this integral term is nonzero, superposition o f aver age power does not hold. The next question is, under what circumstances is the integral zero and nonzero.^ There are three cases to consider. Case 1 is when cuj

co^ , which will result in a zero

value o f the integral. In this case, the integral consists o f rwo sinusoids integrated over a common period T. The integral of a sinusoid over any period is zero. Thus, the integral is zero and super position o f power holds when cOj ^ O)-,. Case 2 is when C0 | = co-, but with (Bj - O2) = ±knl2, k an odd integer. In this case, the integral is again 0 . This follows because the first term o f the integrand is a sinusoid whose integral is zero over the period T. The second term o f the integrand is a constant, cos(0j - (),) = cos{±kKl2) = 0 , also resulting in a zero integral. Hence for case 2, superposition o f power holds. Finally, we have case 3, for which tOj = co-, but with (6 j - 62 ) ^ ±kKll, k an odd integer; here superposition ot power does not hold. The second term of the integrand is a constant, cos(0j - 62 ) 0 , resulting in a nonzero integral over the period T. So

P^^^^ j +

For case 3, it is

desirable to use the phasor method o f Chapter 10 to compute the desired voltage and then use equation 11.5 to compute average power.

Exercises. 1. In Example 11.2, suppose t/j(r) = 3cos(107rr) V and V2 {t) = 4 cos(15n:r + 0.25ti) V. Compute T, a common period for the two sinusoids, and then compute the average power con sumed by the 1 Q resistor. C H EC K ; T = 0.4 sec w-ill work, and 2. In Example 11.2, suppose age power consumed by the 1

= 12.5 watts

= 3cos(107Tf) V and

= 4 sin(lOJur) V. Compute the aver

resistor.

C H EC K ; P^..= ave 12.5 watts 3. Now suppose v^{t) = 3cos(10Tcr) V and

=4

c o s

(1 5 7 T /

+ 0.257i) V. Compute the average

power consumed by the 1 H resistor. C H EC K ; P^^^ = 20.99 watts

Equation 11.6 resembles equation 1.18b for the dc power absorbed by a resistor connected to a dc source. However, in equation 11.6 the factor 1/2 is present. With the introduction o f a new concept called the efifective value o f a periodic waveform, the formulas for the average power absorbed by a resistor can be made the same for dc, sinusoidal, or any other periodic input wave forms.


S06

3. EFFECTIVE VALUE OF A SIGNAL AND AVERAGE POWER GENERAL CONSIDERATIONS From section 2, a resistor o f R ohms excited by a periodic voltage or current absorbs an average power,

. T he effective value o f any periodic current, i{t), denoted by

is a positive con

stant such that a dc current o f value /^-exciting the resistor causes the same amount o f average power to be absorbed, i.e.,

The same holds for a resistor excited by a periodic volt

age v{t). Mathematically,

(11.7a) or (11.7b)

R Equation 11.7a suggests that 1

Iq+T'

I f " )

T Hence, the mathematical definition o f the effective value of a periodic current i{t) is /m+7 U>ff.R

T

(11.8a)

and, similarly, the effective value o f a periodic voltage u{t) is h\+T

ef/M

(11.8b)

In general, the effective value o f any periodic signal y(r) is In + T

F e jf-

(11.8c)

Observe that the expressions under the radical sign in equations 11.8 constitute the average value ot the square o f the signal. Hence, the expressions give rise to the alternative name for the effec tive value, the root-m ean-square (abbreviated rms) value oiJ{t), since mean value o f the square j{t) over one period.

the square rooto'i the


SO'

Exercises. 1. Show that the average power absorbed by an R Q resistor carrying a periodic current

2. Suppose i{t) = 3cos(2n:/) + 4 cos(47i:f) A flows through a I Cl resistor. Find ANSWlvR:

= 1 2 .5 watts, and l^,g- =

5

7?

= 2.5\/2

and

A

EXA M PLE 11.3. Compute the efFective value o f the periodic voltage waveform sketched in Figure 11.5.

So

lu t io n

From equation 18.b, = T

Therefore,

1 1

.2

4J

4J

3

4J

.

3

= 2.3094 V,

Exercise. Repeat the calculation o f Example 11.3 for the case where the values on the vertical axis o f Figure 11.5 are doubled. A N SW l'R :

V

For a sinusoidal signal/r) =

cos(tor + 0) , the effective value can be calculated using the identit}' cos^ (.v) = 0.5 + 0.5 cos(2.v)

as follows:

f~{t)

Fm -

= F “jC o s “( o j/+ 0 ) = —

F~m

+ —

c o s (2 o j/+ 2 0 )

S08


Since by assumption to 7^ 0, the average value o f the cosine term is zero. The average value o f the first (constant) term is itself Hence, by equation 11.8c,

F~ I itt

(11.9)

Eml 72

Thus, for a sinusoidal waveform, the effective or rms value is always 0 .707 times the maximum value or, equivalently, the ma.ximum value divided by >/2— a basic fact well worth remembering. The ac voltage and current ratings o f all electrical equipment, as given on the identification plate, are rms values unless explicitly stated otherwise. For example, the household ac voltage is 110 V, with a maximum voltage o f 1 lOx-s/^ = 1 5 6 V. A typical appliance such as a coffee maker will have a 110 V rating, ac, at say, 900 watts. The effective values o f a few other periodic waveforms are listed in Figure 11.6, with their derivations assigned as exercises.

Feff «=F

r

dc

sinusoidal

triangular

square F.»= F .

F IG U R E 1 1 . 6 Efleccive values o f some com m on periodic waveforms.

S09


Exercises. 1. Derive the formula for riie cfFectivc value o f a triangular waveform shown in Figure 11.6. 2. Compute the effeccive value of the waveform shown in Figure 11.2a. AN SW FR:

SINGLE-FREQUENCY ANALYSIS WITH EFFECTIVE VALUES We recurn now to the case of single-frequency SSS analysis. The average power as per equation 11.5 absorbed by an arbitrary rwo-terminal element mav now be rewritten in terms o f effective values:

^

cos(0,, - 0y) = - ^

2

V2 v 2

COS(0^. - 0,-) = yeffêjf COS(0,, - 0, )

For the remainder o f the chapter, all voltage and current phasors will be taken as being effective values unless the subscript m or

appears, indicating the maximum value. The subscript ^ w ill

be added sometimes for emphasis, however. This practice is widely accepted in the power engi neering literature. Omitting the subscript eff\v\ equation 11.10 yields K / c o s (0 ^ -0 .)^ V7cos(0p where 0^ = 0,^ - 0 y , V= 0.707

( 11.11)

, and / = 0.7071^^ . The angle 0^ is the angle o f the impedance

Z (/co) o f the two-terminal element and is also interpreted as the angle by which the voltage phasor leads the current phasor.

EXA M PLE 11.4. Figure 11.7 shows two t)'pes o f household loads connected in parallel to a 110 V, 60 Hz source, = 1 1 0 ^ 2 cos(120:tr) V. Lainp 1 and lamp 2 have effective hot resistances o f 202 (a) (b)

and 121

respectively. The impedance o f the fluorescent light is Zjfjwi) = 60 + j70

Find the average power consumed by each light. Find the average power delivered by the source.

Lamp 1

Lamp 2

FIGURK 1 1.7 An example of load current calculation. So

lu t io n

(a) For lamp 1, Z|(/to) = 202Z.0® Cl. Hence, Ij = Vy^Zj = 0.5446Z .0° A. From equation 11.11,

^\ave ^

z\^ = 110

X 0 .5 4 4 6 cos(0°) = 59.9 watts

5 10

Cliapier 11 • Sinusoidal State State Power Calculations

This means that lamp 1 is a 60 watt bulb. Similarly, for lamp 2, Z^C/co) = 121Z.0°

Hence, 1, = ^ iJZ j = 0.909 IZ.O^’ A. From equation

11. 11, = 110 X 0.9091 cos(0°) = 100 watts

= 56 + y'66 = 86.56Z.49.7® O.. Hence,

Finally, for the fluorescent light,

^ jJZ jj =

1.2 7 Z .-4 9 .7 ° A. From equation 11.11,

^Jlave =

^

(b) For this part we first compute

1.27cos(49.7°) = 90.4 watts

and then apply equation 11.11 to compute the average power

delivered by the source. Here by KCL, = I, + I 2 + I 3 = 0 .5 4 4 6 Z 0 " + 0 .9 0 9 1 ^ 0 ° + 1 .2 7 ^ -4 9 .7 ° = 2.2759 -y 0 .9 6 9 0 = 2 .4 7 3 6 -^ -2 3 .0 6 ° A By equation 11.11, the average power delivered by the source is

Pave = |V/,,l|lm|co.s(0, - 0 , ) =

cos(6 , - 0 ,- ) = 110 X 2 .4 7 3 6 cos(23.06^ ) = 250.35 watts

Observe that the sum o f the individual average powers is 250.3 watts, which equals the power delivered by the source within the accurac)' o f our calculations, where we have rounded our answers.

4. COMPLEX POWER AND ITS COMPONENTS: AVERAGE, REACTIVE, AND APPARENT POWERS Recall the notion of a phasor. When all source excitations are sinusoidal at the same frequency, voltages and currents in the SSS can be represented by phasors. Our question here is, can the pha sor method aid the computation o f power consumption in a circuit? The answer is yes. However, the formulation will bring out several other concepts o f power associated with the sinusoidal steady state. In dc power calculations, the average power consumed by a two-terminal device is the product o f the voltage and current, assuming the passive sign convention. In SSS, the complex power absorbed by a two-terminal device, as shown in Figure 11.8, is a complex number defined by the formula (1 1 .1 2 ) where I^^ is the com p lex conjugate o f

51 1


eff o +

Two Terminal

Veff

Device

o-

FIGURE 11.8 Two-terminal device with phasor voltage and current consistent witli passive sign convention.

The first useful result o f this definition is that suppose v(r) =

i(t) =

= Re S = Re

. To see this result,

cos((Of-I-0J.) , which is represented by the phasor

y/l cos((or + 6,-) , which is represented by the phasor

Also suppose

The average power consumed by

a rwo-terminal device excited by this voltage-current pair (Figure 11.8) is given by equation 11.10 as

Now obseiAê that

S = \ //C /f =

j =

h r

in which case

= Re

R e[S]= R e

The curious reader may ask why a conjugate o f the current is used in the definition o f complex power.

Suppose

Re

one

did

not

have the conjugate o f the current. Then = VeffI(,ffCOs{Q^,+Qj)Î\,y^, i.e., the resulting product would have no physical

meaning. Now because S is a complex number, it has an imaginary part, a magnitude, and an angle. T he imaginary part o f S defines a quantity called the reactive power absorbed by the twoterminal device in Figure 1.18; i.e., reactivepoioer \s defined as

Q = Im [S]=

J=

sin(0 ,. - 0,) = reactive power

^

14)

The unit o f reactive power, Q, is VAR, which stands for volt-amp-reactive. It follows immediately that

where P =

. Also, the magnitude o f S is defined as the apparent power absorbed by the two-

terminal device o f Figure 1.18, i.e.,

________

\^\ = Kffêff = \P ^ JQ\ = 4 p ~ ^Q~ = apparent poNver


^12

The Linir o f apparenr power is VA, short For volt-arnp. The iinerrclationship o f these different pow ers is illustrated by the right triangle diagram in Figure 11.9, which is often helpful in solving problems. Observe that the apparent power is always greater than or equal to the average poiuer, with equality applying to the case o f a purely resistive load.

FIGURE 11.9 Relationships among complex, average, reactive, and apparent powers. I'h e distinction among these various powers is best understood by computing the powers for some basic circuit elements. For simplicit)', except when needed or for emphasis, from this point on we will often drop the subscripts <^’and rfw'as given in equations 1 1 .1 3 -1 1 .1 6 . EXA M PLE 11.5. This example explores the computation o f the various powers for a simple inductor. Given that 1^0 ) = v2/sin(CO/) in the circuit of Figure 11.10, compute

V^, S^,

Ql, the instantaneous absorbed power />/(/), and the instantaneous stored energ)' Wjit) in terms o f L, CO, /^, and Vj. After this show that (i)

max

Ql

(ii)

Wiit) max

Ql (0

Remainder of Circuit

FIG U R E. 1 1 . 1 0 Isolation o f an inductor for investigating the concept o f com plex power.


So

“SI 3

lu t io n

= -jl^, = /oZ.1^ = cdZ./^ = V^. jVJi^ = I\ + yQ^. This implies that = 0 and (X = V Ji- Further, the instantaneous absorbed power is Pi ( f ) = v = COLy/lIi^cosUor)x J l l i sin(o)/) = V^// sin(2(0/), which By inspection, and noting that wc again presume effective values, = VÎ

,„a\ ~

is consistent with equation 11.4. It follows immediately that

~ Ql •Further,

W' l it ) = i ) . 5 L i l u ) = L I I sin -((o ;) = 0.5 L/; [l - cos(2w/)] 9

[l-c o s (2 c o / )] [l-c o s (2 o )/ )] -------------- ^l ‘ l -------- z-----------= 2co 2(0

[ l - c o s ( 2 (0 /)] ----------2(0

(f)

Since the bracketed quantity varies between 0 and 2,

\Qi\ 0)

, as was to be shown.

EXA M PLE 11.6. This example, like the previous one, investigates the concept o f reactive power, but in the case o f a capacitor. The calculations will all be dual to those o f Example 11.5. Hence, given that ^ ^ ( 0 = >/2V(-sin((0/) in the circuit o f Figure 11.11, compute instantaneous absorbed power

S^^

and the instantaneous stored energ)^

'r»

Qq the

terms o f C,

(O ,

Vq and Ir-. After this show that (i)

P c (0

(ii)

lV c (0

Qc

and

Qc (0

Remainder of Circuit

FIGURE 11.11 Isolation of a capacitor for investigating the conccpt of complex power. So

lu t io n

^C~

By inspection, and noting that we again presume efteaive values, implies that P r= 0 and Q_(^ = —

S^= absorbed power is /;^(/) = which is consistent with

~

Further, the instantaneous

)/(-(/) = (oC>/2\^(7 cos((0/) X V 2 V^7 sin((0/) = sin(2(0/), equation 11.4. It follows immediately that

^

= | 0 c L Funher,

\V^(i) = 0.5Cv^(/) = CVc sin“ (03/) = 0.5CVf^ [l - cos(2(0/)]

^ ,,2 [l-c o s (2 (o / )]

=

(oCVr

------------------ 2w

,, ,

= Vr^c

[l-c o s (2 (o / )] -----------------2(0

^

= Qc

[ l - c o s ( 2 (0 /)] -----------------2(0


^\A

Qc

Since the bracketed quantity varies between 0 and 2, W(-(/)

These quantities,

’ I

p

r

(0

o

v

e

useful for identify

ing energy storage values in inductors and capacitors in systems where energy is to be recovered and stored, and for modifying the power factor (to be discussed shortly) in networks with motors. Energy storage in systems and power management are important research topics in todays world. In Examples 11.5 and 11.6, one observes that the inductor absorbs reactive power while the capac itor absorbs negative reactive power or, equivalently, delivers reactive power. This follows from the definition o f complex power (equation 11.13, i.e., S =

T he structure o f equation 11.13

derives from the convention that whenever the phasor current lags the phasor voltage (as with the inductor), the device is considered to absorb reactive power, whereas if the current phasor leads the voltage phasor (as with the capacitor), the devicc is considered to deliver reactive power. Indeed, the overwhelming majority o f loads (toasters, ovens, hair dryers, motors, transformers, TV s, etc.) have lagging currents. When a t%vo-terminal element absorbs an average power

, there is a transformation o f elec

trical energ)' into other forms o f energ}'— for example, heat or kinetic energy. In contrast, when a two-terminal element absorbs reactive power Q, no energy is expended. T he energy transferred into the two-terminal element is merely stored and later returned to the surrounding network. To distinguish it from real (expended) power, we use VAR (volt-ampere-reactive) instead o f watt as the unit for the reactive power Q. EXA M PLE 11.7. This example investigates the computation o f the various powers defined above for an /?Ccircuit. Here, consider the circuit o f Figure 11.12, where v,„(/) = 100 V 2 c o s (200071:/) V. Find the complex, average, reactive, and apparent powers absorbed by the load.

o

100 Q lO kO 16nF i

FIGURE 11.12 Simple RC circuit for investigating aspects of complex power. S o lu tio n S te p 1 .

Compute

Z,„ry20007Cj = 100 j -I- 720007T X 16x10 10

-9


Step 2. Compute

Converting

SI^

to a phasor, we have

I,v, = ^

= 10 + j9 .8 5

= 100 V. By O hm s law,

mA

în Step 3. Cornpute the complex power absorbed by the load. By equation 11.12,

S =

= 100(10 - y9.85)10"-'‘ = 1 - jO .985 VA

Step 4. Given the complex power, the average power is

The reactive power is Q = I m [ S ] = - 0 .9 8 5 VAR and the apparent power is |S| = 1.404 VA

Before doing a more complex example, we will discuss the particulars o f the principle o f conser vation o f power in the sinusoidal steady state.

5. CONSERVATION OF COMPLEX POWER IN THE SINUSOIDAL STEADY STATE Basics and Examples The basic principle o f power conservation is that instantaneous power is conserved.

GENERAL PRINCIPLE OF CONSERVATION OF POWER In all circuits, linear or not, instantaneous power is conserved; i.e., the sum o f the absorbed powers o f all the elements in a circuit is zero. If one thinks o f sources as generating power and other elements as absorbing power, then we can rephrase this statement as “the sum o f gen erated powers equals the sum o f absorbed powers.”

The validity o f this principle follows from KVL and KCL. This principle leads to the particular fact that complex power is conserved in ac circuits operating in the SSS.

PRINCIPLE OF CONSERVATION OF COMPLEX POWER IN AC CIRCUITS In ac circuits operating in the SSS, complex power is conserved; i.e., the sum o f the absorbed complex powers o f all the elements (operating in the steady state) in a circuit is zero. Consequently, average power is conserved and reactive power is conserved.

Note however, that the conservation principle does not hold for apparent power, i.e., for the mag nitude o f the complex power. The following example illustrates a basic use o f the conservation law.


S I6

EXA M PLE 1 1.8. This example illustrates the application o f the principle o f conservation o f com plex power in determining power delivered by a source and the input current to a circuit. We also show that conser\'ation o f apparent power does not hold. Consider the circuit o f Figure 11.13. Find the power delivered by the source and the phasor input current,

given that S j = 360 +

;1 6 0 VA, S 2 = 360 - ; 1 2 0 VA, S 3 = 420 + y540 VA, S 4 = 130 + >80 VA, S 5 = 40 - ; 1 0 0 VA.

/ / lOOV

V- \ S. \ A

\

FIGURE 11.13 Bridge circuit where S- represents the complex power absorbed by the element. S o lu tio n

By the principle o f conservation of power in ac circuits, = S , + S , + S 3 + S 4 + S 5 = 1310 + j5 6 0 VA This means that the circuit absorbs 1310 w'atts o f average power; the reactive power is 560 VAR, and the apparent power is 1425 VA. Notice that the large component o f reactive power makes the apparent (consumed) power larger than the actual consumed power, P

.

To compute I , recall that ^so u rce

Hence,

= 'OOI/,, = 1 3 1 0 + p 6 QVA

= 13.1 - J5.6 A.

Exercise. Repeat the above example calculations for S j = 300 + y'400 VA, S 2 = 300 - y 4 0 0 VA, S 3 = 600 + j\ 000 VA, S^ = 60 + y’80 VA, S^ = 120 -> 1 6 0 VA. What are the average and reactive pow ers delivered by the source? AN SW FR: S^

= 1380 + /920 VA :uid

= 13.8 + / ).2 A. 1380

w ;itts

and 020 VAR

The next example illustrates the computation o f various powers through basic definitions and application o f the principle o f conservation o f power. EXAM PLE 11.9. Consider the circuit of Figure 11. 14, which depicts a motor connected to a commer cial pow'er source. The motor absorbs 50 kW of average power and 37.5 kVAR o f reactive power, and has a terminal voltage

= 230 V. Find IIj, the complex power delivered by the source, S^, and IVÎ.


R,line = 0 .5 0

FICfURE 11.14 Motor absorbing 50 kW and 37.5 kVAR at a terminal voltage o f 230 V; the value of

^line

exaggerated for pedagogical purposes; electrical code requires that the size of the connecting

wire be large enough that the voltage drop is only a small percentage of the source voltage. S o l u t io n

Step 1. FiJid the apparent power, |S^^J, absorbed by the motor. Since S,„ = ^™V-Q™ = 50t/-37.5kV A it follows that |S,^J = 62.5 kVA. Step 2. Find |/J. Here, |S,J =

= 230| lj. Hence, [Ij = 2 7 1 .7 4 A.

Step 3. Compute the line loss. 0-5 X 271.72 = 36 .9 2 kW Step 4. Compute the complex power delivered by the source. From conservation o f power, S,

= S,„ . s,,.„ = S,„ *

= 5 0 ;3 7 .5 . 3 6 .92 = 8 6 .9 2 *y 3 7 .5 kVA

Step 5. Compute |Vj. s. V. =- *

h In the above example we choose

—

s.

= 3 4 8.4 V

I.

large to illustrate the calculations. In practice a line loss of

36.92 kW for a 50 kW motor operation would not be permitted.

6. POWER FACTOR AND POWER FACTOR CORRECTION In a resistor, average power is dissipated as heat. In a motor, most ol the average consumed power is converted to mechanical power, say, to run a fan or a pump, with a much smaller portion dis sipated as heat due to winding resistance and friction. The ratio o f the average power to the appar ent power is called the power factor, denoted by pf, i.e.. pf=

Average Power

P^ye

Apparent Power

|S|

= cos(0,-0,)

(11.25)


SIS

The right-hand portion o f equation 11.25 follows directly from equation 11.13. Equation 11.25 specifies the power factor as cos(6^, — Oy) , i.e., the cosine o f the difference between the angles o f the voltage phasor V and the current phasor I. Clearly, 0 < pf < 1 . The angle (11.26)

- 0y) = power factor angle (pfa)

Since cos(x) = cos(-a:), the sign o f (0^^- 0^) is lost when only the pf is given. In order to carry the relative phase angle information along, the common terminology is p f lagging ov p f leading. A l/igging power factor occurs when the current phasor lags the voltage phasor, i.e., 0 < (0^, - 0y) < 180® . A leaditjg power factor occurs when the current phasor leads the voltage phasor, i.e., 0 < (0y - 0^j < 180® . Practically all t)^pes o f electrical apparatus have lagging power factors. Some typical power factor values are listed in Table 11.1. TABLK 11.1. Power Factors for Common Electrical Apparatus T

ype o f

L oad

P o w e r F a c t o r ( L a g g in g )

Incandescent lighting

1.0

Fluorescent lighting

0 .5 -0 .9 5

Single-phase induction motor, up to 1 hp

0 .5 5 -0 .7 5 , at rated load

Large three-phase induction motor

0 .9 -0 .9 6 , at rated load

To illustrate the idea o f leading and lagging pf, consider the circuits o f Figure 11.16. Suppose the circuits operate at a frequency o f 400 Flz or co = 2 5 13.3 rad/sec. For the circuit o f Figure 1 1.6a, I = (1 -jO.995) 10“^ V = 1.41 10“^/.—44.85° V. Hence, the current phasor lags the voltage pha sor, i.e., ( 0 J ,- 0y) = 44.8 5 ° and the pf is cos(44.85°) = 0.709 lagging. On the other hand, for the circuit o f Figure 11.16b, I = (1 + ;2 .5 ) 10-^ V = 2.7 10“3 Z 6 8 .3 ° V. Hence, the current phasor leads the voltage phasor by 68.3", i.e., (0 — 0^^ = 6 8 .3 ° and the pf is cos(68.3°) = 0.688 leading.

o-

O-

i0 .4 H

IH F

1 kO

1 kQ O-

(a)

(b)

FIGURE 11.16 (a) A parallel RL circuit illustrating a lagging pf. (b) A parallel RC circuit illustrating a leading pf A load with a required average power demand, operating at a fixed voltage with a low pf, say 0.6, has a relatively high reactive power component. This results in a relatively high apparent power.


SI 9

Since the operating voltage is fixed, the line current needed to drive the load is higher than if the load operated at a higher pf, say 0.95. Relatively speaking, a higher pf has a lower reactive power component with correspondingly lower apparent power. Figure 11.9 helps to visualize the rela tionships. For fixed line voltage, lower apparent power (higher pO means lower line current and hence lower power loss in the connecting transmission line. In todays world o f energy conserva tion, it is important to be energy efficient. The following example illustrates how improved pf on a load can reduce line losses and thus decrease cost o f operation. EXA M PLE 11.10. This example reconsiders Example 11.9, involving a motor connected to a commercial power source as illustrated in Figure 11.17. The solution process will emphasize the basic definition o f pf and the use o f voltage and current phasors. Suppose the motor absorbs 50 kW (about 67 hp) o f average power at a pf o f 0.8 lagging. The terminal voltage,

is 230 V. The

frequency o f operation is 60 Hz or co = 120 j1 . For the first part o f the example the capacitor in Figure 11.17 is not connected to the motor. In part (c), the capacitor is connected to the motor to improve the p f This will reduce the magnitude o f the current supplied by the source and hence reduce the line losses. (a) (b)

Find the complex power delivered to the motor. Find I., V^, and the power delivered by the source, which might represent the power delivered by the local electric company.

(c)

Correct the power factor o f the combined motor-capacitor load to 0.95 lagging by choos ing a proper value for C.

(d)

Compute the new power delivered by the source to the combined motor-capacitor load.

R,line =0.5 0

FIG U RE 11.17 Motor absorbing 50 kW and 37.5 VAR at a terminal voltage of 230 V. Again, the value of

is exaggerated for pedagogical purposes; clectrical code requires that the size of the con

necting wire be large enough that the voltage drop is only a small percentage of the source voltage. S o l u t io n

(a) Find the complex power delivered to the motor. Step 1. Use the p f o f 0.8 lagging and the given average power to fin d the apparent power. From the definition o f pf,

= 50 kW = R e[S] = |S„,|cos(e -

6

) = |S,„| x p f = |S„,| x 0.8

As such, the apparent pow-er is 50

!s„,

0. 8

= 6 2 .5 k V A = \/„,/,„

( 1 1 .2 7 )

Chapter 1 1 • Sinusoidal State State Power Calculations

S20

where V„, = |V„J = 230 V and /„, = Step 2. Compute

. Ligging means that current phase lags behind voltage phase, i.e., 0 ^ -

= 0 - 0^ > 0 . Consider the diagram in Figure 11.18, which shows that the current phasor I^ la g s the voltage phasor, i.e., the current phasor makes an angle o f - 3 6 .8 7 “ = cos” ' (0.8) from the volt age phasor. Hence, = 3 6 .8 7 “ > 0

S m= V IDP m

FIGIJRI- 11.18 Phasor relationship ofV^^, Step 3. Compute the complex power, S

and

. By definition,

s,,, = | S ,„ k S m =

= 6 2 .2 Z 3 6 .8 7 ° kVA

= 5 0 + ;3 7 .5 k V A = /’„„^t>Q (b) Fi7ici

, V., and the power delivered by the source.

Step 1. FindX.. From equation 11.27 and the fact that 62.5 X 10'^ V„, And from Figure 11.18, again since

230

for this part.

= 2 7 1 .7 4 A

, Z.1^ = - 3 6 .8 7 ”. Hence

= 271.74 Z - 3 6 .8 7 " = 271.74 Z .-0 .6 4 3 5 rad = 217.4 - y l 6 3 A Step 2. FiudY^. From KVL and Ohms law, = 0.5 [217.4 - j\ 6 3 ]+ 230 = 3 3 8.7 - y 8 1.5 = 3 8 4 .4 Z - 0.2362 rad = 3 8 4 .4 Z - 13.533'" V


s :i

Step 3. Compute the complex power, , delivered by the source. S^. = V ,I* = 3 4 8 .4 Z - 13.533" x 2 7 1 .7 4 Z 3 6 .8 7 " VA = 94.6 6 4 Z 2 3 .3 3 7 ° kVA = 9 4 .664 LQA07 rad kVA = (86.918 + ;3 7 .5 ) kVA Norc that it takes 86.918 kW to run a 50 kW motor. The difference is the loss in the power line. If we have a way o f reducing the magnitude o f

this line loss will be reduced. In fact, we do,

and this strateg)^ is the goal o f the next part o f the example. (c) Correct the power factor o f the combined motor-capacitor load to 0.95 biggifig- Since motors are inductive, a properly chosen capacitor can improve the pf to 0.95 lagging. The new motor con figuration is that o f Figure 11.17, with the capacitor connected across the motor. The proper value o f C must be found. Step 1. What does a p f o f 0.95 lagging require in terms ofcomplex power absorbed by the motor-capac-

itor combination'^ = — Z c o s " l( 0 .9 5 ) = 5 2 .6 3 Z 1 8 .1 9 5 ‘’ = (50 + 7 1 6.4342) kVA 0.95

(11.28)

Recall that = (50 + > 37.5) kVA Step 2. Find a capacitor value to reduce the reactive power. For this step consult Figure 11.19.

FIGURU 11.19 Relationships bcrwecn new and old complex powers. In Figure 11.19, one observes that is the same for both the new and old complex powers since that is what the motor requires for its operation. The reactive powers are different. The new com plex power with the 0.95 lagging pf has a smaller reactive power component. The capacitor must be chosen to reduce the old reactive power to this new level. Hence,

jQoU -j^ncw = ;2 1 -0 7 kV.AR = -/reactive power o f capacitor) = -yQ^-

522


Therefore,

jQ c = - j 2 1.07 kVAR = V „,Ic = V,„ [K c(JO ))V ,„]' = -./coC| V,„p It follows that C = ^ = ^ H lZ iii4 = ,o 5 7 x lO - ’ F co|V,„ 12071(230)“ (d) Compute the new power delivered by the source. Step 1. Compute the complex power, denoted

absorbed by the motor-capacitor load. The com-

plex power absorbed by the load is the sum o f the complex power consumed by the motor and the reactive power of the capacitor, as illustrated in Figure 11.9, i.e., S « f'= (50 + y i6 .4 3 ) kVA Step 2. Compute the new

denoted

Since S ”^“'is the complex power o f the combined motor-

capacitor load, ♦ ^

V

m

^ 5 ( W ! M 3 ,0^ = (217 + ; 7 : .43) A 230 J !

Step 3. Compute . From KVL and Ohm’s law, V ;^ “’ = 0 . 5 l f + V,„ = (3 3 8 .5 - y 3 5 .7 2 ) V Step 4. Compute the new complex power delivered by the source. By definition,

S’r ' = v f ' ( i ; '" ’’ )* = (7 6 - ; i 6 .4 8 ) kVA Hence, the new average power delivered by the source is 76 kW with pf correction as opposed to 86.9 kW without pf correction. With this pf correction, there is a reduction o f 86.9 - 76 = 10.9 kW o f power loss in the line connecting the source to the load.

Example 11.10 illustrates how adding a parallel capacitor can improve the pf o f a load. The main motivation for improving the pf was to reduce the power loss in

However, even if

is

negligible, another strong reason exists for improving the load pf. Example 11.11 illustrates how an improved power factor allows a single generator to run more motors. Example 11.11 will fully utilize the principle of conservation o f complex power and the two consequences o f equation 11.25. From equation 11.25 and the fact that S = /^ + jQ, we can express pf directly in terms o f P an d Q as follows:

.— ^

523


with a lagging pf for Q > 0 and a leading pf for Q < 0. Solving for Q from equation 11.29, we obtain

e=

± P .H r - i P f‘

(11.30)

with Q > 0 if pf is lagging and Q < 0 if pf is leading. W ith these formulas we can simplify the process o f power factor correction. E X A M P L E 11 .1 1 . An industrial plant has a 100 kVA, 230 V generator that supplies power to one large motor and several identical smaller motors. The resistance o f the connecting line is assumed negligible in the approximate analysis below'. The large motor, labeled t^-^pe A, draws 50 kW at a pf o f 0.8 lagging. Each smaller motor, o f type B, draws 5 kW at a p f o f 0.7 lagging. The configu ration is illustrated in Figure 11.20.

Generator 230 V ^ 60 Hz lOOkVA Type B Motor

Type A Motor

Type B Motor

FIGURH I 1.20 A generator supplying power to one large motor and several smaller motors. (a) (b) (c)

(d)

Can the generator safely supply power to one large motor and three small motors? What are the generator current (magnitude) and the power factor o f the combined loads? Compute the number o f small motors (besides the one large motor) that can be run simultaneously without exceeding If the power factor for all motors, ing appropriate parallel capacitors (besides the one large motor) can

the generators rating. large or small, is corrected to 0.9 lagging by connect (as done in Example 11.10), how many small motors be run simultaneously without exceeding the genera

tors rating? Compute the capacitances required in part (c) for the large and the small motors.

S o lu tio n

(a) Compute the reactive power for each motor type. Using equation 11.30, the reactive power for each type o f motor is given as

- 1

1

=50

-1

= 37.5 kVA

0.8^ and

Q b = P ,i, / A - >

y

pfg

= \

/

V

-^ 2 -

0.7

i

= .‘i . l 0 1 k V A

‘’ ■2 I


By the principle ot conservation o f power, the complex power (in kVA) supplied by the generator is V

^ 3(7^^ + ;Q ^ ) = (50 . 15) + ;(3 7 .5 -h 15.303)

=

= 65 + ;5 2 .8 = 8 3 .7 4 ^ 3 9 .

kVA

By inspection, the apparent power is 83.74 kVA, which is below the generator capacity o f 100 kVA, meaning that the generator can safely operate the large motor and three smaller motors. The magnitude o f the generator current is 83,740/230 = 364 A. From equation 11.29, the pf o f the combined loads is

pf =

.

^ P- +

= 0 .7 7 6 2 \/65“ + 5 2 . 8 “

(b) Compute the number o f stnall motors {besides the one large motor) that can be run simultaneous ly. When one large type A motor and n smaller t)^pe B motors are connected in parallel, the com plex power delivered by the generator is

^gen

7 ^

= ^^0 + « X 5) + 7(37.5 + ;/ X 5.101) kVA

rhe apparent power is

^ (50 + n x 5 ) ~ + (3 7 .5 + /; x 5 .1 0 1 )" kVA Since the generator has a capacity ot 100 kVA, then c

‘ = ( 50+//x 5 ) “ + (37.5+/; X 5.101)- <

100“ =

10^

(11.31)

Replacing the inequalit)' sign in equation 11.31 by an equalit)^ results in the quadratic equation 51.020/r + 882.5750/7 - 6 ,0 9 3 .8 = 0 The resulting zeros are //, = 5.288 and //2 = - 22.58. The largest positive integer satisfying the inequality 11.31 is // = 5. Thus, at most, five small motors can be run simultaneously with the large motor without exceeding the generators capacit)' (c) I f all power factors are corrected to 0.9 lagging, fin d the number o f small motors {besides the one large motor) that can be run simultaneously. We essentially repeat the calculations o f part (b) with the new given power factor o f 0.9 lagging:

and

e :r = / '.. A V PfA

i

=5o

y - 1 = 2 4 . 2 1 6 kVA

V 0 .9 ^


The complex power (in kVA) supplied by the generator is

= Pa + j Q T + n(Pfl + j Q T ) = (50 + « X 5 ) + j(24.216 + /, x 2.416) The apparent power (in kVA) is = s j (50 + /2 X 5) “ + (24.2 \6 + n x 2.4216)“

s '" "

ficn

As before, n satisfies the inequaiit)' (50 + « X 5)2 + (24.216 + « x 2.4216)2 < 10^ To find n, we compute the largest positive root o f the quadratic equation 30.8641;/2 + 6 l7 .2 8 2 9 n - 6 ,9 1 3 .6 = 0 The roots o f this quadratic are

= 8 and «2 = —28. The largest positive integer that satisfies the

above inequalit}' is n = 8. Thus, eight small motors, as opposed to five in the earlier case, can be run simultaneously with the large motor without exceeding the generators capacity. (d) For the large motor, the capacitor must absorb a negative reactive power equal to

Q^.

Equivalently, the capacitor must supply a reactive power equal to Qw - Q T =

- 24.216) X 1000 = 13284 VAR

(10.32)

From Example 11.6, the reactive power supplied by a capacitor is

I Q-CA I = ^CA^CA =

6 0 Q X 2302 VAR

(10.33)

Equating equations 10.32 and 10.33, we have In x 60C^|2302 = 13284. Solving produces

=

66 6 .1 6 X 10-<^ E Similarly, for the smaller motors,

Q-B - Q"b ' = (5.101 - 2.4216) X 1000 = 2679.4 VAR Also, we have

Qcb\ ~Vcb^Cb\ -

= In

X60Q

X 2302 VAR

Equating these two quantities and solving for C^, w'e obtain Cg = 134.35 x 10“^ F. We note that in the power industry, such capacitors are usually specified only by their kVAR rating, with no mention o f their actual capacitive value in F.

In the above example the generator capacit}^ was given in terms o f VA, the unit o f apparent power. The example points out the importance o f reducing reactive power to more fully utilize the power

526


capacity o f the generator. Use o f VA for generator, motor, and transformer capacit)' arises out o f safety considerations. Most ac machiner)' operates at a specified voltage depending on the insula tion strength. The size o f the wire and other heat transfer factors determine the maximum allow able current o f a machine or transformer. Also, the cost and physical size o f most ac equipment are more closely aligned to the VA rating than to other measures. Hence, the VA rating better reflects the safe operating capacity o f ac equipment. Another motivation for improving the power factor is economical. A power company charges a consumer only for the actual electrical energy used. A meter measures this energy usage in units o f kWh (kilowatt-hour). As mentioned earlier, most clectrical loads have lagging currents. As shown in Examples 11.10 and 11.11, for a given required average power, a higher pf means lower transmission line losses. Also, loads that operate at low pf force power companies to pursue high er kVA ratings o f the generator equipment. Thus utilities companies encourage consumers to operate their equipment and appliances at high pfs. Since power companies can supply more power with the same equipment if the pf is high, they adjust their rates so that energy costs are less with a high pf and are greater with a low pf.

7. MAXIMUM POWER TRANSFER IN THE SINUSOIDAL STEADY STATE Chapter 6 outlined the basics o f maximum power transfer for linear resistive networks. Having introduced energ)' storage elements L and C, and having studied methods for sinusoidal steady analysis, it is time to extend the results on maximum power transfer to general linear networks in the sinusoidal steady state.

MAXIMUM POWER TRANSFER THEOREM FOR AC CIRCUITS Let a practical ac source be represented by an independent voltage source V^ {voltage phasor in rnis value) in series with an irnpedance + jX^. An adjustable load impedance = jX^j with R l > 0, is connected to the source {Figure 11.21). In steady state, for fixed Z^, Vp and O), the average power delivered to the load is maximum when Z[^ is the complex conjugate ofZ^, i.e., (11.33a)

and (11.33b)

and the maximum average power is given by p

^
max

(11.33c)

>2.


To derive the conditions o f the maximum power transfer theorem, observe that the current phasor, I, is (11.34)

1=

{R, + RO + j(X , + XO Thus the average power delivered to the load is

(11.35)

Here Pave

a hmction o f two real variables

and Xj . To find the conditions for maximum

the partial derivadves ------ and ------ to zero and solve for

set

and X^. Differendating equation 11.35

with respect to R^ yields V f [ ( R , + R i f + {X, + X 0 - - 2R l (R, + R , , )

dRi

{R^ + Rl)~ +

+

=

-)12

0

(1 1 .36a)

and differentiating with respect to Xj produces

dP ax,

v "-[-1R l {X,

+ Xl )

= 0

From equation 1 1.36b, the only physically meaningful solution is

(11.36b)

528


which is equation 11.33b. Substituting this result into the numerator of equation 11.36a yields

v3 “ ^

(R s + Rl T The only physically meaningful solution here is

^ 0 and

^

R ,-R , which produces equation 11.33a. (Note that this is the condition for maximum power transfer in purely resistive circuits.) Substituting these results into equation 11.35 produces equation 11.33c,

Ir

_

4/?,

which verifies the theorem. The theorem can be established less formally as follows. W ith any existing connected to the source, if the total reactance +X^) is not zero, we can always increase the magnitude of the current, and hence the power delivered to the load, by “tuning out” the reactance, i.e., by adjust ing to be -X ^. This implies condition 11.33b. Under such a condition, the circuit becomes resistive, and the maximum power transfer theorem of Chapter 6 may be applied to obtain equations 11.33a and c. The maximum power obtainable with a passive load, given by equation 11.33c, is called the available power of the fixed source. The conditions for maximum power transfer, as given by equation 11.33, are valid when both and Xj^ are adjustable. If X^ is fixed and only is adjustable, then the condition for maximum power transfer is =

+

(11.37)

which is obtained by solving equation 11.36a for fixed X^ and X^ .

n

r\

If the source is a general two-terminal linear network, then its Thevenin equivalent must be found before application of the maximum power transfer theorem. If the source is represented by a Norton equivalent circuit, we can use a source transformation to obtain the Thevenin form and then apply equations 11.33.

'

As pointed out in Chapter 6, maximum power transfer is not the objective in electric power systems, as the sources usually have very low impedances. On the other hand, it is a very important factor to be considered in the design of many communication circuits, as illustrated in the following example. o

S29


EXA M PLE 11 .1 2 . The radio receiver shown in Figure 11.22a is connccted ro an antenna. The antenna intercepts the electromagnetic waves from a broadcast station operating at 1 MHz. For circuit analysis purposes, the antenna is represented by the Thevenin equivalent circuit shown in Figure 11.22b. (a)

Find the input impedance

o f the receiver if maximum power is to be transferred

(b)

from the antenna to the receiver. Under the condition o f part (a), find the magnitude of the voltage across the receiver ter minals, and the average power delivered to the receiver.

receiver input

antenna equivalent circuit

equivalent circuit

(b)

(a)

FIGURE 11.22 Example of maximum power transfer. S o l u t io n

= 1\ i l and

(a)

From the maximum power transfer theorem, the answers are

(b)

Since the reactances in the circuit have been “tuned out,” the input current to the receiv er is simply 14.6/(21 + 21) = 0.348 mA. The input impedance has a magnitude 7.

= 1070.

+ 1070- = 10 7 0 .2 n

Therefore the magnitude o f the voltage across the receiver terminals is 0.348 x 1070.2 = 3 7 2.4 mV (rms). The power transferred from the antenna to the receiver is 0.348^ x 21 = 2.54 uW.

In the preceding discussions of maximum power transfer, we have assumed that the load is adjustable. In practice the load is often fixed, as for example, in the case o f a loudspeaker having a4

voice coil. In such cases, one designs coupling networks consisting o f lossless passive com

ponents. These coupling networks transform the fixed load impedance into one whose conjugate matches the fixed source impedance. This permits maximum power transfer to the load. The fol lowing example illustrates the principle. A design procedure for some simple coupling networks will be discussed in the second volume o f this text.

530


EXA M PLE 11.13. A fixed load resistance /?^ = 100

representing the input resistance o f an amplifier, is connect

ed to the source o f Example 11.12 through a passive coupling network, i.e., a network that does not generate average power, as shown in Figure 11.23. (a)

Show that the maximum voltage that can be developed across

(b)

Show that the coupling network shown in Figure 11.23 achieves this maximum voltage across

is 0 .504 V.

.

FIGURF, 11.23. Maximum power transfer through a coupling network. S o l u t io n

(a)

From equation 11.33c, as used in Example 11.12, the available power from the source is 2.54 j.iW. If all o f the power is delivered to R^ , then the voltage,

must be

= ylPmaxf^L = ^ 2 .5 4 X 10“^ X 100000 = 0 .5 0 4 V (b)

The input impedance

o f the coupling network with load must be the conjugate o f

the source impedance. Specifically

în = yw ^ + ------------- r~ j(dC + R, Substituting the values OJ = 10^’, L = 400.9 x 10“^, C = 109.8 x 10

and /? = 100 x 10^ into the

above expression yields = 21 + ;1 0 7 0 Q which is indeed the conjugate o f the source impedance. Since

is conjugate-matched to the

source impedance, the maximum power o f 2.54 |aW is transferred to the coupling network. Since the coupling network consists o f L and C, neither o f which consumes average power, the 2.54 f,iW power must be transferred out o f the coupling network and into the load resistance. The voltage across the load resistor, Vj, is given by = 4 ^

= V 2 .5 4 X 10“^ X 100000 = 0 .5 0 4 V

This verifies that the coupling network o f Figure 11.23 enables the largest voltage to appear across the load resistor.


531

8. SUMMARY Fundamental to the material in this chapter is the definition o f the effective value (rms value) o f a periodic voltage or current waveform. For a sine wave, the effective value is the maximum value divided by -v/2 • For a general periodic voltage or current, the effective value is the value o f a dc waveform that will produce the same amount o f heat as the periodic waveform when applied to the same resistance. Using the definition o f the effective value o f a waveform, formulas for the average power absorbed by a linear two-terminal network in ac steady state were set forth and derived. Recall that for a two-termlnal element with sinusoidal voltage v{t) = ^/2 V'^ cos(co/- 0y) an current i{t) =

0/), the absorbed average power \s P = ^^^^cos(Oj; - G/), assuming

the passive sign convention. Next w'e presented the definition o f complex power and its compo nent parts, which include its real part or average power, its imaginary part or reactive power, and its magnitude or apparent pow’er. Various examples illustrating the calculation o f these powers were given. Again, for a two-terminal element w'ith sinusoidal voltage v{t) and current i{t) as above, the reactive power absorbed is defined to be Q = ^ ^ ^ s i n ( 0 y - 0/) VAR (volt-amperereactive). After introducing these different types o f power, we proved the principle o f conserva tion o f complex power, which implies the consenâtion o f real power and the conservation o f reac tive power. This was followed by the definition o f power factor, pf, the ratio o f average power to apparent power, which takes on values between 0 and 1. The need for improving a low power fac tor and a method for achieving an improved power factor were illustrated with two examples. The maximum power transfer theorem, first studied in Chapter 6 for the resistive nervvork case, was taken up again in this chapter for the sinusoidal steady-state case. Here, maximum power transfer to the load requires that the load impedance be the conjugate o f theThevenin impedance seen by the load. As pointed out earlier, the theorem has no application in electrical power sys tems. However, for communication circuits the maximum power transfer theorem is o f extreme importance. The power that can be extracted from the antenna o f a radio receiver is usually in the microwatt range, a very small value. It is therefore necessary to get as much power as possible from the antenna system. Example 11.13 illustrates this principle.

9. TERMS AND CONCEPTS Apparent power: the apparent power absorbed by a two-terminal element is

assuming the

use o f a passive sign convention . The unit is VA (volt-ampere). Average power: the average value o f the instantaneous power. For a two-terminal element with sinusoidal voltage u{t) = yfz V^^cos{o)t + Qi,) and current i{t) = ^

V^jj-cos{o)t + 0/), the

absorbed average power is P = ^ ^ ^ ^ c o s (0 y - 0/), assuming a passive sign convention. Com plex power: for a two-terminal element absorbing average power P and reactive power Q, the complex power is defined to be S = P + jQ. The unit o f measuremcnr is VA (voltampere). The magnitude of S is the apparent power. Conservation o f powers: for any network, the sum o f the instantaneous powers absorbed by all elements is zero. For any linear network in sinusoidal steady state, the sum o f the aver age powers, reactive powers, or complex powers absorbed by all elements Is zero. This property is a consequence of KCL and KVL.

532

Chapter 11 * Sinusoidal State State Power Calculations

Efifiecdve value (nns value): for a sine wave, the effective value is the maximum value divided by V 2 •For a general periodic voltage or current, the effective value is the value of a dc wave form that will produce the same amount of heat as the periodic waveform when applied to the same resistance. Instantaneous power: the power associated with a circuit element as a function of time. The instantaneous power absorbed by a two-terminal element is p{t) = y(/)/(/), assuming that a passive sign convention is used. Maximum power transfer theorem: if a variable load 7.^^ = -^jî is connected to a fixed source Vy having a source impedance =7?^+/A^, then the largest average power is transferred to the load when is the complex conjugate of Z j , i.e., = R^ and AT^ =-X^. Power factor: the ratio of average power to apparent power. The pf value lies between 0 and 1. For a passive load, the power factor is said to be lagging when 90° >0j, - 6,->0, and lead ing when 90° > 0^- 0^, > 0. Real powen in ac circuits, real power means average power. It is the real pan of the complex power.

'

Reactive power: for a two-terminal element with sinusoidal voltage v{t) = -y/2 V^cosisat + 0^) and current i{t) = ^ /^ co s(co f + 0^, the reactive power absorbed, denoted by Q, is defined to be Q = ^ ^ ^ sin (0 y - 0/), assuming that a passive sign convention is used. The unit of measurement is VAR (volt-ampere-reactive).

o

n o o n n o


^33

Problems

AV(t)(V) 20

INSTANTANEOUS AND AVERAGE POWERS

t(sec)

1. For the source current waveform o f Figure PI 1.1a, which drives the circuit o f Figure

-

10- -

(a)

PI 1.1b, find the average power consumed by the 2 Q resistor.

(a) Figure Pi 1.3 (a) Rectangular waveform, (b) Triangular waveform.

EFFECTIVE VALUE OF NONSINUSOIDAL SIGNALS (b)

Figure P l l .l AN SW FR. 0.758 watrs

4. (a)

Compute the effective value o f each o f the periodic signals in Figures Pi 1.4a and b.

2. Compute the average power delivered to a 1

(b)

kQ resistor by a current o f the form (a)

1 0co s(10t)m A

(b)

10 |cos(10^)| niA

(c)

10 cos“( 1Of) mA

(d)

Plot each o f the instantaneous powers for 0 < r < 1 sec using MATLAB or its equivalent.

3. (a)

Compute the average power absorbed by a 10 resistor whose voltage is given by each o f the waveforms in

(b)

Using MATLAB, plot the instanta

Figure Pi 1.3. neous power associated with each waveform for 0 < f < 3 sec.

For the circuit o f Figure P i 1.4c, find the power absorbed by Rj if the volt age source v{t) is given by the wave form o f Figure PI 1.4a.

(c)

Repeat part (b) for the waveform o f Figure PI 1.4b.

534


i(t)0

60 O <

= 30 Q

(c)

Figure PI 1.5 6. (a)

Find the effective value o f the source current plotted in Figure PI 1.1a.

2Q v(t)

(b)

6

R = 80

part (a).

Figure P11.4

7. Compute the effective value o f (a)

Compute the effective value o f each o f

(b)

the periodic signals in Figures P11.5a

(c)

and b. (b)

resistor in Figure PI 1.1b

using the effective value computed in

(0

5. (a)

Find the average power absorbed by the 2

v^{t) = 10 + 2 cos(20r) = 10 cos(2r) + 5 cos(4/) = 10cos(2r) + 5 cos(4r) + 5 c o s ( 4 f - 4 5 ° ) V.

For the circuit o f Figure PI 1.5c, find the power absorbed by

if the cur

rent source i{t) is given by the wave form o f Figure Pi 1.5a. (c)

Repeat part (b) for the waveform o f Figure PI 1.5b. Ai(t)

AVERAGE POWER CALCULATIONS IN SINUSOIDAL STEADY STATE 8. Using equation

44

11.10, find the average

power absorbed by the resistor in the circuit

3

shown in Figure PI 1.8, where sin(5r) V . /? = 25 Q, C = 8 mF. 12

15

R

(a)

A

t(sec)

-4 -■

(b)

Figure PI 1.8 ANSWl-.R: 25 watts

= 50 x


9. In the circuit shown in Figure PI 1.9, i-{t) =

(b)

5 cos(30/) A, /? = 5 Q, and C = 5 niF. (a) Find vît). (b)

Compute the average power delivered by each source.

(c)

Repeat parts (a) and (b) for R = 50 Q,

Zi = p O Q, and a = 49.

Using equation 11.10, find the instan taneous and average power absorbed by the load.

/Y Y V — ► Load

' ■6

aV

Figure 1’ 11.1 I ANSW'FRS: (b) in random order: - 9 0 . 100 watts; (c) 4. - 3 .9 watts Figure P l l .9 AN SW ER: (b) 40 warts

12.

10.

= 120 R = 32 W, Z^ = ;2 0 0 Q. Z q = y’80 ri, and ^ = 4.

Consider the circuit o f Figure PI 1.12,

where

In the circuit o f Figure PI 1,10,

5 0 z i-9 0 °

/? = 6 Q,

= ;1 2 Q, and

=

(a)

=

- ;4 Q. (a)

Find the phasor current

(b)

and deter

Find the average power delivered by each source.

mine its magnitude. (b)

Find the average power (in watts) absorbed by the resistor.

Using equation 11.10, find the aver /Y Y V z. -

age power (in watts) delivered by the source. (c)

O nly

R absorbs

average

power.

' •6

III is

Therefore, once known, the average power consumed by R is = y?| (d)

bl Figure PI 1.12

Check your answer to part (b)

using this formula.

ANSWF.RS: (b) in random order: S7.6, 230.4

Repeat parts (a) and (b) when 7? = 30 Q, Z^ = y'50 Q, and Z q = -ylO Q..

watts

13. A coil is modeled by a series connection o f /Y Y V

L and R. When connected to a 110 V 60 Flz

R

source, the coil absorbs 300 watts o f average pow'er. II a 10 Q resistor is connected in series with the coil and the combination is connected to a 220 V 60 Hz source, the coil also absorbs 300 watts o f average power. Find L and R. AN SW FR: R = 0.9901 LI and L = 16.6 mH.

Figure Pi 1.10 A N SW FR; (b) ISO wans 11. For the circuit in Figure P i 1.11,

=

100.10° V ^ , /? = 5 n , Z^ = ;5 0 a , and ^ = 9. (a) Compute the current phasor I^.

14.

Consider

Figure

Pi 1.14,

where

V,•„(/) = 2 2 0 >/2cos( 120jir) V and R^^-^ =3

The inductance L is adjusted so

that IV 150 V„,„ a n d watts. Find the magnitude o f

250 the


^36

reactance o f the coil, and the values o f L and R.

v.(t)

Figure P 11.17 C H EC K : Complex power is 7+y’6 VA Figure P 11.14

18. In the circuit shown in Figure Pi 1.18, to =

ANSW ER: /. = i2.9 niH, R - 14.SS3 Q

= 4 £2,

15. Repeat Problem 14 for 400 watts, and

64 rad/sec, =

nns with all other values the

= 120/ i60"

= 20 £2, /?,

= 4 12, /., = 0.375 H, and L , = (^-125 H. Find the complex and average powers absorbed by the load.

same. Load

C H EC K : L = 7.958 mH, R = 17.794 £2 —TY"YA--- Q R.

L.

+

COMPLEX POWER CALCULATIONS 16. For the circuit o f Figure PI 1.16, R = 5 = 3 .jA a , = -yiO and v,„(r) = 100v 2 cos(1207i:/) V. (a)

Find the complex power absorbed by

(b)

Compute the apparent power in VA, the

Figure P 11.18 C H E C K : Complex power is 36 + p 2 VA

the load. average power (in watts), and the reactive

19. In the circuit shown in Figure Pi 1.19, Z| =

power in VAR delivered to the load.

1

Zc {-

+7a, z, =4 +ji2 a, z,^ 2+j i a, v, =

(104 +y50)~V, and (a)

= (106 + y48) V at 60 Hz.

Find the voltage V , in polar and rec tangular forms. C H EC K : ¥ , = 1 0 0 + ?

'-«’0

(b)

Find the complex power absorbed by each of the three impedances and then the

Figure P 11.16

power delivered

C H EC K : (b) 300 watts, 400 VAR

sources.

17. In the circuit o f Figure PI 1.17, v ,(r ) = IOOV2 cos(500/ + 30^) V, /?, = 100 Q,

Z 2 is 100 + ;5 5 0 VA

by the

two

C H EC K : complex power absorbed by

R, = 700 £2, and Z. = 1.2 H. Find /^(r), the complex power, average power, and apparent power absorbed by the load.

(c)

Using the results o f part (b), verify conservation o f complex power.


537

(a) r-

Z.

Z.

power, and the average power deliv ered by the source.

O

v .Q

the complex power, the apparent

-I

(b)

Find the source current

in rectangu

lar and polar form.

Figure Pi 1.19 C H EC K : complex power delivered by source h

Z,

is 10 + 260; VA. v .Q

CONSERVATION OF POWER 20. This problem should be done without any phasor voltage or current computations. In the circuit o f Figure P i 1.20,

= 2300

Figure PI 1.21 C H EC K S: P,

.= 1600 watts, |Ij = 8 .6957 A

at 60

Hz and the following powers (in kVA) are con sumed by various impedances and resistances: S , = 20 + y8 , S , = 20 + 7 I 8 , S 3 = 5+ 76 , and = 3 + j4. (a) Find

(rectangular form) and the

complex

power delivered

by

the

POWER FACTOR AND POWER FACTOR CORRECTION 22 . (a)

load that absorbs 2 kW o f average

source. (b) (c)

Determine

in polar form.

Find the complex power delivered to

(b)

(d) (e)

Find

in rectangular and polar form.

power with pf = 0.90 lagging. Find the complex power delivered to a load that absorbs 4 k\V o f average

the group o f impedances Z j, Z-,, and ^4Find V , in rectangular and polar form.

Find the complex power delivered to a

power with pf = 0.90 leading. 23. For the circuit o f Figure PI 1.23, Z j absorbs

1600 watts at pf = 1, while the apparent power absorbed by Z^ is 1000 VA at pf = 0.8 lagging, and V;„(/)= l20V 2cos(1207cr) V. Find (a)

the phasor current

(b)

the phasor voltage V|

(c)

the phasor voltage

in

Figure PI 1.20 C H EC K S: (a) Complex power delivered by source: 48 + y'36 kVA; (b) |lj = 26.087 A

v,(t)

21. In the circuit o f Figure PI 1.21, V^. = 230 V.

at 60 Hz. The complex powers (in VA)

absorbed by the five impedances are S| = 100 +

yioo, S , = 200 +7 IOO, S 3 = 100 + p o , S^ = 400 +J250, and S^ = 800 + J700. Find

C H EC K : - 1 8 .8 2 3 5 ;

Figure P i 1.23 = 20 - 5; A„„^ and V , = 75.2941

538

Chapter 1 1 * Sinusoidal State State Power Calculations

24. The circuit shown in Figure P I 1,24 is in the sinusoidal steady state. Suppose that absorbs 3000 W of average power at pf = 0.7905 lagging, Ri-^^ = 0.1 £2, and = 120 V rms’ (a) Find the average power absorbed in the transmission line resistance (in

26. The circuit in Figure P I 1.26 operates at CO = 500 rad/sec and = 100Z0° V ^ . The com plex power drawn by the load without the capacitor attached is S^; = 100Z30® = (86,6 + y‘50) VA. This constitutes a pf of 0.866 lagging. (a) Find the values of R and L (b) Find the value of C in fiF that pro duces a pf of 0.95 lagging.

W). (b)

Find =A cos(1207t^ + 0) V and the complex power delivered by the source.

Load

0 -"0 -

!'4<

Transmission line resistance

— 'N/S/'------ OR .,„

+

I

L

o - -oFigurePll.26

CHECK; 3 pF < C< 5 pF

Figure P I 1.24 CH ECK: (a) 100 watts 25. As shown in Figure P I 1.25b, a capacitor is put in parallel with a motor using average power = 40 kW operating at a power fector of 0.7 laggmg to boost it to a power factor of 0.9 lading. The voltage across the parallel motor-capacitor combination is 230Z0° V ^ . The power relationships are shown in Figure P I 1,25a. If the frequency of operation is CO = 120tc, compute the proper value of the capaci tance, C (in mF).

27. The circuit shown in Figure P I 1.27 is operat ing in the SSS with v^(0 = 120>/2cos(1207cO V. Device 1 absorbs 360 W with pf =0.9. Device 2 absorbs 1440 W with a pf of 0.866 lading. Find the value of the capacitor C such that the magnitude of the source current equals 16 A ^ . What is the pf of the two-device-plus-capacitor combination?

Device 1

Device

2

Figure P I 1.27 CH ECK: 0.108 mF

< C<2 mF

(b)

28. A group of induction motors is drawing 7 kW from a 240 V power line at a power faaor of 0,65 lagging. Assume CO = 12071 rad/sec. (a) What is the equivalent capacitance of a capacitor bank needed to raise the power factor to 0.85 lagging? (b) What is the kVA radng of the capacitor bank of part (a); i.e., what is the reactive power supplied by the capacitor bank?

Chapter 1 1 • Sinusoidal State State Power Calculations

(c)

539

Determine the annual savings from

By what amount in kVAR must

installing the capacitor bank if a

the reactive power be reduced to

demand charge (in addition to the

produce a pf o f 0.94 lagging?

charge for the kilowatt-hours used by

(iii) Compute the needed capacitor

the induction motors) is applied at S20.00 per kVA per month.

(iv) Compute Z^j(£)) as the ratio o f

AN SW ERS: (a) 0.1771

current

niF; (b) Minimum

the capacitor phasor voltage to

kVA rating: 3.8457 kVA; (c) S608.14

the capacitor phasor current, at the indicated frequency. (v) Compute the proper value o f C in

29. Consider a source that drives an electric

mF.

motor that consumes an average power o f 94 kW (about 125 hp) at a pf o f 0.65 lagging, as show'n in Figure PI 1.29, where

= 0.07 Cl.

(i)

Compute the new I"^ .

(j)

Compute V 'f":

(k)

Compute the complex power deliv ered by the source and the new effi ciency.

— oM o to r)

—

Later Addition of Capacitor for p.f. Correction

MAXIMUM POWER TRANSFER

Figure P11.29 30. Consider the circuit shown in Figure The phasor voltage across the motor is êff~ 2 3 0 Z 0 V. The sinusoidal frequency is 60 Hz. (a)

Find the apparent power delivered to Find the complex power,

(a)

(c)

Find the value o f the load impedance that will absorb maximum power at

delivered (b)

to the motor.

CO =

delivered to the motor. Compute I

(e)

(f)

Compute V^. Compute the complex power deliv

(g)

ered by the source. Determine the efficiency o f the con

= 100

find the average power

absorbed by the load. u—

f

R. v .Q o—

figuration, i.e., the ratio o f average power delivered to the motor to the

100 rad/sec.

Given the conditions o f part (a) and

Determine the reactive power in VAR

(d)

C=

1 mF.

the motor in kVA. (b)

= 100 Q, R-> = 25

PI 1.30 in which

Figure PI 1.30 C H EC K : 20 + ;1 0 Q

average power delivered by the source as a percentage. (h)

Add a capacitor across the motor to

31. In the circuit o f Figure PI 1.31, v^(r) = 100V 2cos(1000/) V, /?, = 80 Q, R, =

improve the power factor to 0.98 lag

20 Q, Z. = 5 mH. Find the value o f the resist

ging. Then (i) Compute

ance R^ (in Q) and the capacitance C (in mF) Compare with

sou (ii) Recall that the role of the capaci tor is to reduce the reactive power.

such that maximum average power is absorbed by the load.


540

Thevenin equivalent, i.e., a volt

Load

age source in series with either a

/Y Y V l L

:

series RC or a series RL as appro

R,

priate. (b)

Compute

the

load

impedance,

Z^(/‘150), necessary for maximum power transfer. Show this load as

Figure P i 1.31

either a series RL circuit or a scries RC

A N S W H R : R, = 16 12 and C= 0.2 mF

circuit. Should it be the opposite of

32. The circuit o f Figure P l l .3 2 operates in the sinusoidal steady state with W = 1000 rad/sec, R = 1 k n , C = 1 ;
= 3 and

the case in (a)(iii)? Why? (c)

Compute the average power con sumed by the load at maximum power

=

transfer.

2 ^ 0 ’A ^ (a) Find the value o f the load imped ance

for maximum average

power transfer. (b)

8Q

.

-L

Find the average power absorbed

A

3.334 mF - J 1.667 mF

16Q

by the load under the conditions o f part (a). Figure Pi 1.33

34. Consider the circuit o f Figure P l l . 3 4 , w'hich operates at

CO

=10 rad/sec. Suppose R =

l O a , L = 2 H , and /^(O = I 0 V 2 cos(lOr) A. (a) AN SW ER:

Q to deliver maximum average power

= 1 + \.5/ k li, 4000 warts

33. The purpose of this problem is to compute for maximum powder transfer by following a spe

to the load. What is this maximum average power? (b)

(a)

Compute the Thevenin equivalent of the circuit at terminals A and B: (i)

Use nodal analysis to compute Note that there is a floating

If Rj = 30 Q, determine Q for maxi mum power transfer to the load. What

cific procedure. Consider the circuit o f Figure PI 1.33 in which /^.(r) = 50V 2 cos(150/) A.

Choose the proper values o f Rj^ and

is this maximum average power? (c)

If Q = 8 mF, then determine R^ for maximum power transfer to the load. What is this maximum average power? Load

dependent voltage source. C H EC K : V^^.= 2 8 .4 7 -791.1 V (ii) Find the Thevenin equivalent impedance, Z^y^(/150), seen to the left o f terminals A and B. C H EC K : Re[Z,//yi50)] = 8.2846Q (iii) Show' the phasor form o f the Thevenin

equivalent

circuit.

Then show the circuit form o f the

Figure Pi 1.34 A N SW ERS: (a) 10 12, 5 ml-, 1230 wans; (b) 5 mF, 937.5 wans


35. In rhe circuit o f Figure PI 1.35,

541

= 100

^rms adjusted to achieve different goals. Assume R = GO Q., Z q = - ^ 8 0 Q. (a)

Find the value o f

that maximizes

(C H EC K : 24 Q .) What is the value o f (b)

Find the value o f

that maximizes

Figure PI 1.37

V /.,. W hat is the value o f IV,I L nutx

(b)

Now suppose y ? = 4 Q ,C = Im F , and /. = 0.1 mH. Choose R^ for maximum

—

R

— — • -------------

power transfer and find 38. Consider the circuit o f Figure PI 1.38

C

where

— • -------------

= 0.1

at

CO =

10^ rad/sec. Suppose

/?, = 100 a , /?, = 10-1 k n , Zî = -71000. (a) Find the impedance so that P j is

Figure P l l .35 ANSWHR: (a) 48 ih . r .0 3 7 watts

(b)

maximized. Find the values o f L and C , to achieve

36. The circuit shown in Figure PI 1.36 has

the impedance Zy computed in part

v>v,(/) = 10>/2 cos(60/) V, /?^ = 4

(a). Find the impedance Z^ such that Pj

(c)

I = 1/120 H, and Q = 1/30 F. What value o f

and IV-,1 are maximized.

R should be chosen so that maximum power is delivered to the load? Note: It is the source resis tor here, not the load resistance, that is being varied. What is this maximum average power consumed by the load?

Load /

Y

Y

V

L

R

zrc, Figure P 11.38

'" ‘" Q

A N SW FRS: (a) Z^ = 100 f ylOOO Q: (b) 99 piand 0.2 niH

39. The series RLC circuit o f Figure PI 1.39 has Figure Pi 1.36 ANS\\'|-:RS: 0. 25 waits

reached steady state and v^{t) = 110sin(l 207tr) V. /

37. The circuit o f Figure PI 1.37 operates in the sinusoidal steady state with v^(O = 5 0 V 2 co s(2 0 0 0 r) V. (a)

Y

Y

V

6Q v .(t )Q

-jisn

Choose R and C such that the maximum average power is absorbed in the load resistor R^^ = 5 when Z. = 0.1 mH. What is this maximum average power?

Figure P I 1.39


542

(a)

Find the instancaneous stored energy and

Lîve

at the moment when the

2(0

terminal voltage o f the source is zero. (b)

Find the instantaneous energ)^ the moment when

(c)

at

42. (a)

For the circuit o f Figure P H .4 2 a , show that the powers absorbed by the

= 0.

impedance Z = R + jX are

Find the instantaneous energ)' \Vf- at

the moment when = 0. ANSWHRS: in random order (J), 1.14, 1.65,

and Q = (b)

= /?|lp

where I is in

For the circuit o f Figure P H .4 2 b , show that the powers absorbed by the

4.115, 1.27

admittance Y= G +jB are

= (j|Vp

and Q = 5|Vp where V is in V

THEORETICAL PROBLEMS L 40.

Let the voltage across a capacitance C b e v{t) = sin(cor) V. (a)

Find p{t), the instantaneous power delivered to the capacitance and show that p{t) has a peak value o f 0 .5 c o C (V J2 watts and an average value o f 0.

(b)

Find

the instantaneous energy

store in C (or, rather, in the electric field) and show that W^t) has a peak value o f 0.5C(1/^^)2 joules and an aver age value o f 0 (c)

. 2

5

j oules.

Let Q^- be the reactive power absorbed by C. Show that

WC,ave _

Qc 2(0

41. Let the current flowing through an induc tance L be i{t) = sin(co^) A. (a)

Find p{t), the instantaneous power delivered to the inductance and show that

p{t)

has

a

peak

value

of

0 .5 co£(/,„)2 watts and an average value 0, (b)

Find the instantaneous energy store in L (or, rather, in the magnetic field) and show that W^{t) has a peak value ot

joules and an aver

age value o f 0.25I(/^,)^ joules. (c)

Let be the reactive power absorbed bv L. Show that

Z = R + jX

o --------(a)

a

1

+ V

.

Y = G + jB

a (b)

Figure Pi 1.42. Two-terminal elements modeled via impedance (a) and admittance (b).

C

H

A

P

T

E

R

Laplace Transform Analysis I: Basics HISTORICAL NOTE The Laplace transform converts a time function into a new function o f a complex variable via an integration process. The name Laplace transform comes from the name o f a French mathemati cian, Pierre Simon Laplace (1 7 4 9 -1 8 2 7 ). Pierre Laplace adapted the idea from Joseph Louis Lagrange ( 1 7 3 6 -1 8 1 3 ), who in turn had borrowed the notion from Leonhard Euler (1 7 0 7 -1 7 8 3 ). These early mathematicians set the stage for converting complicated diff-erential equation models o f physical processes into simpler algebraic equations. The Laplace transform technique allows engineers to analyze circuits and to calculate responses quickly and efficiently. In turn engineers became better able to design circuits for radio communication and the tele phone, not to mention other, earlier electronic conveniences. This chapter introduces the notion o f the Laplace transform, a mathematical tool that is ubiquitous in its application to an army o f engineering problems.

CHAPTER OBJECTIVES 1. 2. 3.

Explain and illustrate the benefits o f using the Laplace transform tool for solving circuits. Develop a basic understanding o f the Laplace transform tool and its mathematical prop erties. Develop some skill in applying the Laplace transform to differential equations and cir cuits modeled by differential equations.

SECTION HEADINGS 1. 2 3.

Introduction Review and Summary o f Deficiencies o f “Second-Order” Time Domain Methods Overview o f Laplace Transform Analysis

4. 5.

Basic Signals The One-Sided Laplace Transform

6. 7.

The Inverse Laplace Transform More Transform Properties and Examples

Chapter 12 • Laplacc Transform Analysis 1; Basics

8.

Solution o f Integrodififerential Equations by the Laplace Transform

9. 10.

Summary Terms and Concepts

11.

Problems

1. INTRODUCTION I'his chaprer introduces a powerful mathematical tool for circuit analysis and design named the Laplace transform. Later, more advanced courses will describe the design aspects. Use o f the Laplace transform is commonplace in engineering, especially electrical engineering. A student might ask why such a potent tool is necessary for the analysis o f basic circuits, especially since many texts use an alternative technique called complex frequency analysis. Complex frequency analysis does not permit general transient analysis; rather, it restricts source excitations to sinu soids, exponentials, damped sinusoids, and dc signals. This class of signals is small and does not begin to encompass the broad range o f excitations necessary for general circuit analysis and the related area o f signal processing. The Laplace transform framework, on the other hand, permits both steady-state and transient analysis of circuits in a single setting. Additionally, it affords gen eral, rigorous definitions of impedance, transfer fimctio}!, and various response classifications perti nent to more advanced courses on system analysis and signal processing. Introducing the Laplace transform early allows students an entire semester to practice using the tool and learn about its many advantages. Section 2 describes some of the difficulties associated with the methods o f circuit analysis intro duced in earlier chapters when applied to circuits o f order 3 or higher. Following this, we present an overview o f Laplace transform analysis in section 3, define important basic signals in section 4, and introduce the formal definition o f the one-sided Laplace transform in section 5. The inverse Liplace transform and important properties o f the transform process arc introduced in sections 6 and 7, with numerous illustrative examples. Section 8 applies the technique to circuits modeled by differential equations. Such models were developed in Chapters 8 and 9.

2. REVIEW AND SUMMARY OF DEFICIENCIES OF "SECONDORDER" TIME DOMAIN METHODS Recall that the output or response o f a circuit depends on the independent source excitations, on the initial capacitor voltages, and on the initial inductor currents. Calculation o f the output often begins with the writing o f an algebraic or a differential equation model o f the circuit for the out put variable in terms o f the source excitations or inputs and element values. For first- and secondorder circuits with simple .source excitations, such as dc or purely sinusoidal, the solution o f the differential equation circuit model has a known general form containing arbitrary constants. See, for example, Tables 9.1 and 9.2. The arbitrary constants depend on the initial conditions and the magnitude o f the dc excitation or on the magnitude and phase o f the sinusoidal excitation. Specifically, the steps in finding the response o f a second-order circuit to a constant input are as follows;

Chapter 12 • Laplace Traiisforni Analysis 1: Basics

Step 1. Generate a differential equation model o f the circuit. Step 2. Compute the characteristic equation o f the circuit/differoitial equation and then compute its

roots {say Aj and X-y) using, for example, the quadratic fonnula in the second-order case. Step 3. From the location ofthe roots o f the characteristic equation, determine the form o f the solution:

or if A, = X-),

Step 4. Compute the constant D by shorting itiductors, open-circuiting capacitors, and analyzing the

restdting resistive circuit. Step 5. Compute the constants A and B using the initial conditions on the circuit. For circuits beyond second order, the approach in the above algorithm tends to break down. Example 12.1 demonstrates how the approach breaks down with a simple third-order circuit. As mentioned earlier, the foregoing technique, although quite useful for simple circuits, has seri ous drawbacks for circuits with more than two capacitors or inductors. This is because higherorder derivatives o f circuit output variables generally have little or no physical meaning. Such derivatives are complicated linear combinations o f initial capacitor voltages and initial inductor currents. The following example illuminates the difficulties. E X A M P L E 12.1. Figure 12.1 shows three

circuits coupled through the use o f dependent volt

age sources. The goal o f this example is to construct a differential equation model, determine the solution form in terms o f arbitrary constants, and demonstrate the difficulties w'ith the simple recipe o f the above algorithm by attempting to relate the arbitrary constants to the initial condi tions. 1Q

1Q

1O -I-

FIG URE 12.1 A cascade of three RC circuits coupled by means of dependent voltage sources. The differential equation model of the circuit is third order. S o l u t io n

Step 1. Construct the differential equation o f the circuit. For this task, first write a dilTerential equa tion relating to Then write one relating to and finally, write one relating Some straightfor^vard algebra leads to the following three differential equations:

to

Chapter 12 • Laplacc Transform Analysis 1: Basics

^

at

+ v c i ( 0 = 0.5iv„

( 1 2 .!)

0 . 5 - ^ + it2(/ ) = 0 .5 i'c i

(12.2)

0 . 2 5 ^ . v „ „ , ( O = 0 ,5 v „

^ , ^ 3 ^

^

Successively substituring equation 12.1 into equation 12.2 and equation 12,3 into the result pro duces the input-output differential equation model,

^’ont

^

-7 ^~ôut , 1 A

dt^

dr

, o

(12-4)

d,

Step 2. Compute the characteristic equation and its roots. The characteristic equation for differen tial equation 12.4 is

^ + 7s^ + \As + ^ = {s - a) {s - b) {s - d) = {)

^

which has roots a = - \, b = - 2 , and d = - 4 . Step 3. Determine the form o f the solution. If v-^j^t) = form

Vou, it) =

+ Be^‘

then the complete solution has the

+ E = Ae~^ + Be~~' + De~^‘ + £

(12.5)

for r > 0. Step 4 . Compute A, B, D, and E in equation 12.5. Using the rule o f thumb mentioned earlier, a simple calculation yields E = 0.125

Calculation o f A, B, and D specifies the solution in equa

tion 12.5. Applying the recipe described earlier, we take derivatives o f equation 12.5. set t = 0, and relate them to the circuit initial conditions:

ôut (0) = ^ 'ôut (0) =

+ dD

Again, one dot over a variable means a first-order time derivative, and two dots denotes a second-order

time derivative. A, B, and D are computed by solving this set o f equations. The difficulty is in specifying

and

First,

is simply the initial capacitor voltage on the third capaci

tor. However, v^^^ît) is proportional to the current through it, which depends on all the initial capacitor voltages. Further, what is the physical interpretation o f

And how do v^^^{0)

and v^^fiO) relate to the initial capacitor voltages? The relationship is complex and lacks any mean ingful physical interpretation. Finally, even for this simple example, computation and solution o f the differential equation 12.4 proves tedious.

Chapter 12 • Liplacc Transform Analysis I: Basics

Exercise. For Example 12.1, compute expressions for and (0). A N SW ERS: = [2iv..(0) - 4 (0) = [\6r^J0) - 1 2 ^ ,.(0 ) + 2/v^(0)]

One o f the advantages o f Laplace transform analysis is that it does not destroy the physical mean ing o f the circuit variables in the analysis process. Chapter 13 addresses how the Laplace transform approach explicitly accounts for initial capacitor voltages and initial inductor currents.

3. OVERVIEW OF LAPLACE TRANSFORM ANALYSIS Laplace transform analysis is a technique that transforms the time domain analysis o f a circuit, sys tem, or differential equation to the so-called frequency domain. In the frequency domain, solu tion o f the equations is generally much easier. Hence, obtaining the output responses o f a circuit to known inputs proceeds more smoothly. To apply the technique, one takes the Laplace transform o f the time-dependent input signal or signals to produce new signals dependent only on a new' frequency variable s. In an intuitive sense, and as precisely derived later, one also takes the Laplace transform o f the circuit. Assuming zero initial conditions, one multiplies these two transforms together to produce the Laplace transform o f the output signal. Taking the inverse Laplace transform o f the output signal by means o f known algebraic and table look-up formulas yields the desired response o f the circuit. The effect o f initial conditions is easily incorporated. Figure 12.2 is a pictorial rendition o f the method. As just mentioned, one transforms the input signal, transforms the circuit to obtain an equivalent circuit in the Liplace transform world, and computes the Laplace transform o f the output by “multiplying” the two transforms together. Inverting this (output) transform with the aid o f a lookup table or MATLAB produces the desired output signal.

■> Output Signal

Input Signal \

T

Laplace Transfornn of Input Signal

y

'' r

Laplace Transform of C I R C U I T ^ ,

Laplace Transform of Output Signal

F I G U R E 12.2 D iagram show ing flow o f Laplace transform circuit analysis.

S4iS

Chapter 12 • Liplacc Transform Analysis I; Basics

In a mathematical context, one executes the same type o f procedure on a difFerential equation model o f a circuit and, indeed, difFerential equations in general. Figure 12.3 illustrates the idea.

Input Signal

DIFFERENTIAL' EQUATION

■> Output Signal

J

Laplace Transform of Input Signal”

Laplace Transform of DIFFERENTIAL EQUATION

Laplace Transform of Output Signal

FIG URE 12.3 Diagram showing flow of Laplace transform analysis for solution o f differential equations. The benefit o f this t)'pe o f analysis lies in its numerous uses. Some o f these uses include steadystate and transient analysis o f circuits driven by complicated as well as the usual basic signals, a straightforward lookup table approach for computing solutions, and explicit incorporation of capacitor and inductor initial conditions in the analysis. The forthcoming sections will flesh out these applications.

4. BASIC SIGNALS Several basic signals are fundamental to circuit analysis, as well as to future courses in systems analysis. Perhaps the most common signal is the unit step function.

«(r) =

1,

r>0

0,

r<0

( 12.6)

defined in Chapter 8. The bold line in Figure 12.4, resembling a step on a staircase, represents the graph o f u{i). u(t)

FKiURE 12.4 Graph of the unit step function. It often represents a constant voltage or current le\'el. The unit step function has many practical uses, including the mathematical representation o f dc voltage levels. Any t)'pe o f sustained, constant physical phenomenon, such as constant pressure, constant heat, or the constant thrust o f a jet engine, has a step-like behavior. In the case o f jet engine thrust, a pilot sends a command signal through the control panel to the engine requesting a given amount of thrust. The step function models this command signal.

Chapter 12 * Ijp lacc Transform Analysis 1: Basics

The shifted step, shown in Figure 12.5, models a rime-delayed unit step signal. u(t-T)

A

1 --

FIGURE 12.5 Graph of a unit step shifted T units to right. This function is often used to represent a delayed startup. Shifted steps, u (t- 7), often represent voltages that turn on after a prescribed time period T. The flipped step fioiction, u{T - t), o f Figure 12.6 depicts yet another variation on the unit step. Here the step takes on the value o f unity for time t ^ T. Often it provides an idealized model o f signals that have excited the circuit for a long time and turn o ff at time T. The key to knowing the val ues o f these various step functions is to test whether the argument is non-negative or negative. Whenever the argument is non-negative, the value is 1; when the argument is negative, the value IS zero. u(T-t)

•1 ■

FIGURE 12.6 Graph of flipped and shifted unit step. This function is often used to model signals that have been on for a long time and turn ofTat time ’/'.

Exercise. Represent each o f the following functions as sums o f step functions;

(0 / ( 0 =

1,

0
0,

otherwise

I, -3
(///■) m =

0.

otherwise

1,

/ < -l

0,

-l< r < l

1,

; > 1

ANSVC^R.S: in random t)rder: //(-I - r) +

u {t

- 1), wu! - u{i - 2),

ii{ t

+ 3) - i d t - {>)

The pube function, p-î), o f Figure 12.7 is the product o f a step and a flipped step or, equivalent ly, the difference o f a step and a shifted step. Specifically, a pulse o f height A and width 7" is =

Au{i)u{T -

1) =

Au{t) - A u { t - T)

(12.7)

Chapter 12 • Laplace Transform Analysis 1: Basics

S50

Pr(t)

>

t

T FIGURH 12.7 Pulse of width T and height A. This function is often used to model signals o f fixed magnitude and short duration. A signal sharing a close kinship with the unit step is the ramp function r{t) depicted in Figure 12. 8. r(t)

FIGURE 12.8 Graph of the ramp function, r{t) = Ramp functions conveniently model signals having a constant rate of increase. The ramp function is the integral o f the unit step, i.e.,

/•(0= w h e re t

J —oo

ii(T)dx =ln{t)

( 12.8)

is s i m p l y a d u m m y v a r i a b l e o f i n t e g r a t i o n .

Exercise. Plot r { - t) and r{t - 2). ANSW ER: i\-t) is the relleciion of ;■(;) about the vertical axis while >\t - 2) is siniplv the shift of lit) by two units to the rinht.

EXA M PLE 12.2 Express Figures 12.9a and b in terms o f steps and ramps. f,(t)

y t)

F I G U R E 12,9 T w o signals to be represented by steps and ramps.


S

5S1

o l u t io n

For the signal/j (/), observe that the signal begins with a ramp with a slope o f 2. Thus we have

f\ U) = 2r(f) + ? . At ; = 7, the signal/j(^) levels off. Since the 2r{t) part o f the signal continues to increase, the increase must be canceled by another ramp o f slope 2, but shifted to the right by T units. Thus,/j(r) = 2r{t) - 2r(t - T). The signal j 2(^) replicates/j(^) up to 3T . After 37', the signal drops to zero. Hence we must sub tract a shifted step o f height 2 7 ’ from/j(/‘), Thus fjit) =/j(f) - 2Tu{t - 3 7 ) = 2r{t) - 2r{t - T) -

2T u {t-5 T ).

Exercises. 1. Figure 12.10 depicts a sawtooth waveform denoted hy J{t). Sequences o f sawtooth waveforms are used as timing signals in televisions and other electronic devices. f(t) A

FIGURE 12.10 A sawtooth waveform. A N SW T .R : / / ) = fit) -;•(/- 1) - u{r - 1)

2. For/r) o f Figure 12.10, p lo t/ l - t) and represent the ftmction in terms o f steps and ramps. AN SW ER: /(I - /) = ;-(l - r) - >i-t) - u{-t)

EXA M PLE 12.3. Express Figure 12.11 in terms o f steps and ramps.

f(t)

> t FIG URE 12.1 I Triangular waveform to be represented by steps and ramps. S

o l u t io n

Observe that the signaly(f) begins with a ramp with a slope o f 1 at r = -T . Thus /(f) = r{t ■¥ 7) + ?. The signal falls o ff with a linearly decreasing ramp for 0 < r < T. If we subtract r{t) the signal would become flat for r > 0. Thus we subtract 2r{t) to obtain the linear decrease. Hence, y(r) =

Chapter 12 • Liplacc Transform Analysis 1: Basics

^52

r{t + 7) - 2>it) + ?. For r > T, this signal, r{t + T) - 2Kr), continues to linearly decrease. Hence for the signal to be zero for t > T, we cancel this decrease with an additional ramp. T h u s/ r) = r{t + 7) - lr{t) + r { t - T).

Newtonian physics provides a good motivation for defining the ramp signal. Applying a constant force to an object causes a constant acceleration having the functional form Ku{t). 'J'he integral o f acceleration is velocity, which has the form

a ramp function.

A very common and conceptually useful signal is the (Dirac) delta function, or unit impulse Function, implicitly defined by its relationship to the unit step as

h{cj)dq

(12.9)

The relationship o f equation 12.9 prompts a natural inclination to define s/ .

, V

I-

u{t)~ ii{t - h) h

0 {{) = — //(/)= hm----------------(h

//->{)

( 12. 10)

Strictly speaking, the derivative o f u{t) does not exist at t = 0, due to the discontinuit)' at that point. Without delving into the mathematics, one typically interprets equations 12.9 and 12,10 as follows: define a set o f continuous differentiable functions The derivatives,

as illustrated in Figure 12.12a.

5^(/) = -y//^(/) >of these functions are depicted in Figure 12.12b. 5A(t)

-►t (a)

(b)

F IG U R l'. 12.12 (a) Continuous differentiable approximation to the unit step. (b) Derivative of

the integral of 6^(/) produces

Clearly, 6^(/-) has a well-def'med area o f 1, has height I/A, and is zero outside the interval 0 s t s A. In addition, u[t) = lim //^(r), and lim 6^ = b{t) as A -♦ 0.

Hence, although the definition

o f equation 12.10 is not mathematically rigorous, one can interpret the delta function as the limit o f a set o f well-behaved functions. In fiict, the delta function can be viewed as the limit o f a variet)' o f different sets o f functions. A problem at the end o f the chapter explores this phenomenon. Despite the preceding mathematics, the delta function is not a function at all, but a distribution,^ and its rigorous definition (in terms o f so-called testing functions) is left to more advanced math ematics courses. Nevertheless, we shall still refer to it as the delta or impulse function. The stan dard graphical illustration o f the delta function appears in Figure 12.13, which shows a pulse of

Chapter 12 • Laplace Transform Analysis I: Basics

infinite iieight, zero width, and a well-defined area o f unit}', as identified by the “ 1” next to the spike. Visualization of the delta function by means o f the spike in the figure will aid our under standing, explanations, and calculations that follow.

FIGURE 12.13 Standard graphical illustration o f a unit impulse lunction having a wcll-dcfincd area of 1. The function typically represents an energ)' transfer, large force, or large impact over a very short time duration, as might occur when a bat hits a baseball. The unit area property follow's from equation 12.9, i.e.,

bU)(lr=

5(rV/r = / / ( 0 ^ ) - / / ( 0 " ) = l

where 0“ is infinitesimally to the left and 0"^ infinitesimally to the right o f zero. I f the area is dif ferent from unity, a number Kalongside the spike will designate the area; i.e., the spike will be a sig nal Kb{t). One motivation for defining the delta function is its abilit)' to “ideally” represent phenomena in nature involving relative immediate energ\' transfer (i.e., the elapsed time over which energy trans fer takes place is very small compared to the macroscopic behavior o f the physical process). An exploding shell inside a gun chamber causing a bullet to change its given initial velocity from zero to some nonzero value “instantaneously” is an example. Another is a barter who hits a pitched ball, “instantaneously” transferring the energy o f the s\vung bat to the ball. Also, the delta function pro vides a mathematical setting for representing the sampling of a continuous signal. Suppose, for example, that a continuous signal v{t) is to be sampled at discrete time instants t^ r,*

••• •

v{t) is to be physically measured at these time instants. The mathematical representation o f this measuring process is given by the sifting property of the delta function.

v(//)=

(12.11

v(l)5(X

In other words, the value o f the integral is the non-impulsive part o f the (continuous) integrand, replaced by that value o f r which makes the argument o f the impulse zero, in this case r = tj. Verify'ing equation 12.11 depends on an application o f the definition given in equation 12.9. Specifically, if v{t) is continuous at f = t-, then \'(T )6 (T

-

1: )dX =

=

v (T

)5(T 5(1 -

f j )dx

rj)(lx = v(tj)


SS4

by equation 12.9, where

are infinitesimally to the right and left o f f,.

Exercises. 1. Compute the derivatives o f the signals in Figures 12.9a, 12.9b, and 12.11. A N SW E R S;/,(/) = 2|//(/)-/ / (/ -T ) ] . / .(z) = 2[//(/) - / / ( / - 7 ')| - 27’6 (/ - 3 7 '). ) = //(/+

T)-2iiii)+

oo

2. Compute the integral ANSWHR; sin(l.5.T)

//(/-

T)

sin(27T/+ 0.57t)5(/ —0.5)^//.

-oo

5. THE ONE-SIDED LAPLACE TRANSFORM Intuitively, a transform is like a prism that breaks white light apart into its colored spectral com ponents. T he one-sided or tmilaternl Laplace transform is an integral mapping, somewhat like a prism, between time-dependent signals y(r) and functions F{s) that are dependent on a complex variable s, called complex frequency.

LAPLACE TRANSFORM Mathematically, the one-sided Laplace transform

J{t) is

( 12. 12) where s = (5 + ./CO(y = yf-A) is a complex variable ordinarily called a complex frequency im the signals and systems literature. As the equation makes plain, the Laplace transform integrates out time to obtain a new func tion, F{s), displaying the frequenc)' content o f the original time function/r). In the vernacu lar, F{s) is the frequency domain counterpart of/ r). Analysis using Laplace transforms is often called frequency domain analysis.

Exercises. 1. Find the Laplace transform o f a scaled Dirac delta function, Kb{t). ANSWF.R; K 2. Find £[sin(2;rr + 0 .5 jr ) 6 ( r - 0.5)]. ANSWHR: sin(l.5.T)<'-'^ ‘‘^

Chapter 12 •Laplacc Transform Analysis I: Basics

A number o f questions about the Laplace transfbrm promptly arise:

Question 1: Why is it called one-sided or unilateral? Answer: It is called unilateral because the lower limit o f integration is 0“ as opposed to -oo. If the lower limit of integration were -oo, equation 12.12 would be called the two-sided Laplace trans form , which is not covered in this text. Q uestion 2 : Why use 0“ instead ofQ * or 0 as the lower lim it o f integration? Answer: Our future circuit analysis must account for the effect o f “instantaneous energy transfer” and, hence, impulses at / = 0. The use o f 0"^ would exclude such direct analysis, since the Laplace transform of the impulse function would be zero. Using ^ = 0 is simply ambiguous.

Question 3 : What aboutJunctions that are nonzero fo r t < 0.^ Answer: Because the lower limit o f integration in equation 12.12 is 0“, the Laplace transform does not distinguish between functions that are different for f < 0 but equal for ^ > 0 (e.g., «(/) and

u{t + 1) would have the same unilateral Laplace transform). However, since ^ = 0 designates the universal starting time o f a circuit or system, the class o f signals dealt with will usually be zero for

t < 0 and thus will have a unique (one-sided) Laplace transform. Conversely, each Laplace transform F{s) will determine a unique time fiinaion J{t) with the property that f^t) = 0 for ^ s 0. Because o f this dual uniqueness, the one-sided Laplace transform is said to be bi-unique for signals/^) yfirh J{t) = 0 for ^ < 0.

Question 4 : Does every signalj{t) such th a tfj) = 0 fo r f < 0 have a Laplace transform? Answer: No. For example, the function f i ) = ^ «(/) does not have a Laplace transform because the integral o f equation 12.12 does not exist for this function. To see why, one must study the Laplace transform integral closely, i.e..

Observe that e~j^* = cosicot) -jsm{(Ot) is a complex sinusoid. As f approaches infinity, the real and imaginary parts o f the integrand in equation 12.13 must blow up, due to the

term. Hence,

the area underneath the curve e^~^ grows to infinity, and the integral does not exist for any value of a. W h enever//) is piecewise continuous, a sufficient condition for the existence o f the Laplace trans form is that I

f.

\f{t)\
for some constants

and

(12.14)

This bound restricts the growth of a function; i.e., the fimction can

not rise more rapidly than an exponential. Such a fiinaion is said to be exponentially bounded. The

Chapter 12 * Uplacc Transform Analysis 1: Basics

condition, however, is not necessary for existence. Specifically, the transform exists whenever the integral exists, even if the function/f) is unbounded. W ithout belaboring the mathematical rigor underlying the Laplace transform, we will presume throughout the book that the functions we are dealing with are Laplace transformable.

Question 5: Why does the existence o f the Laplace transform integral depend on the value o f a , men tioned in the answer to question 4? Answer: If the condition in equation 12.14 is satisfied, then there is a range o f a s (recall that s = a + yoo) over which the Laplace transform integral is convergent. This is explained in the follow ing example. EXA M PLE 12.4. Find the Laplace transform o f the unit step. By equation 12.12,

£{u(i)\ = U(s)= \Z

O'

(12.15)

1 a +

o

provided that o > 0. Notice that if a > 0, then

+ 7 (0

.V

-* 0 as t

This keeps the area under

neath the curve finite. For a < 0, the Laplace transform integral will not exist for the unit step. The smallest number Oq such that for all a > Oq the Laplace transform integral exists is called the abscissa o f (absolute) convergence. In the case o f the unit step, the integral exists for all a > 0; hence, Oq = 0 is the abscissa o f convergence. The region a > 0 is said to be the region o f conver gence (RO C) of the Laplace transform o f the unit step. Figure 12.14 illustrates the R O C for the unit step.

j (o-axis

-f->a-axis

FIGURE 12.14 Region of convergence, a > 0, o f the Laplace transform of the unit step function (i.e., the Laplace transform integral will exist for all a > 0).

Question 6: Is the unilateral Laplace transform valid only in its region o f convergence? Answer:

the answer is no. There is a method in the theory o f complex variables called “ana

lytic continuation” which, although beyond the scope o f this text, permits us to uniquely and anal)aically extend the transform to the entire complex plane.- Analytically means smoothly and also that the extension is valid ever)^vhere except at the poles (to be discussed later) o f the transform. Thus, the region o f convergence goes unmentioned in the standard mathematical tables o f one sided Laplace transform pairs.

Chapter 12 • Liplace Transform Analysis I: Basics

EXA M PLE 12 . 5 . Find F{s) iorJ{t) = Ke So

u{t), where K and <{ are scalars.

lu t io n

Applying equation 12.12 yields

K

poo /

Ke~^“ e~^‘dt = K

s + ci

(12.16)

The integral exists if Re[j + ^] > 0. If^7 is real, then the R O C is a > -a. As mentioned in the answer to question 6, by analytic continuation, F{s) = M{s + a) is valid and analytic in the entire com plex plane, except at the point s = -a . The point s ■=-a\s a pole o f the rational function M{s + a) because as s approaches -a , the value o f the function becomes infinitely large. The preceding discussion and examples set up the mathematical framework o f the Laplace trans form method. Our eventual focus rests on its application to circuit theor)\ which builds on two fundamental laws: Kirchhoffs voltage law (KVT) and KirchhofT’s current law (KCL). KVL requires that the voltage drops around any closed loop sum to zero, and KCL requires that the sum o f all the currents entering a node be zero. For the Laplace transform technique to be useful, it must distribute over such sums o f voltages and currents. Fortunately, it does. Linearity property: The Laplace transform operation is linear. Suppose j{t) = Then

L [ m = L [ a jl{t) +

= ^,£[/i(r)] + a.L\f,{t)]

= a^F^{ s ) â, f , St )

(12.17)

This property is easy to verify since integration is linear; l« l/ | (0 + «

= «l

O'

/]{t)e~^'dt + fl2 [q- fi{t)e~^'dt

This is precisely what equation 12.17 states. Hence, our curiosit)^ satisfied, we may rest peaceful ly in the knowledge that the Laplace transform technique conforms to the basic laws o f KVL and KCL. E X A M PLE 12.6. Find F{s) wheny(r) = K^ti{t) + So

for real scalars A', ,

and a.

lu t io n

The Laplace transform o f u{t) is \h by equation 12.15 and that of

is ]/{s + a) by equa

tion 12.16. By the linearity property (equation 12.17), F(.v) = ^

s

+- ^ ,

s +a

with region o f convergence (a > 0} H {a > - a }, where H denotes intersection.

By analytic con

tinuation, the transform is valid in the entire complex plane except at the poles, s = 0 and s = -a. (Henceforth we will not mention the RO C in our calculations.)


Exercise. Find the Laplace transforms o f {i) J{t) = + 2 r t ( r ) + 2u(t), (/■/)

J{t) = -2u(t) +

(iil)

J{t) = 5u(t) - 4e~^‘u{t) + 2e~"^‘u{t) ?>s - a

AN SW ERS: (/) - . ,v~ - (I~

- 2e~^û{t), and 2

.V/- A

2

...

3

.v“ - (/“

4

-

2 , + -----, 4

The transform integral o f equation 12.12 has various properties. These properties provide short cuts in the transform computation o f complicated as well as simple signals. For example, the Laplace transform o f a right shift o f the s i g n a l a l w a y s has the form e~^^F{s), T > 0. Shifts are important for two reasons: 1. 2.

Many signals can be expressed as the sum o f simple signals and shifts o f simple signals. Excitations o f circuits are often delayed from t = 0.

Hence, provisions for shifts must be built into analysis techniques. Tim e shift property: If £[/{i)u{t)] = F{s), then, for T > 0,

£ [ f { , - T l u i i - T ) ] = r ‘'rF{s)

(12.18)

Verification o f this property comes from a direct calculation o f the Laplace transform for the shift ed function, i.e.,

£ [ f ( t -T )u O - 7)1 = J " / ( r - T )u (t- D e'^ ’di = J " / (; -T )e~ ^ d t Let cj = t -T ,W k WJ q = dt. Noting tliat the lower limit o f integration becomes 0“ with respect to q,

L\S(t -T )u U - D 1 = r

fUl)e~'ê~^'^dq =

f ” /(qye-^ 'dq = O'

F^s)

Observe that if T < 0, the property fails to make sense, since J { t - T )ii{t- 7) would then shift left. Since the transform ignores information to the left o f 0“ one cannot, strictly speaking, recover J{t) from the resulting transform.

Exercises. I. Find L\J{t - T)] w hen/f) is (i) Ad(t), (ii) Au(r), and (iii) Ae~'’'u(t). 2. pyU) = /iu(t) - Aii{t - T). A N SW ERS: In random order:

-----

.\----- ^

s+ a

s

/

s


E X A M PLE 12.7. Using the tim e shift property, find F{s) for the signal 12.15.

sketched in Figure

f(t) 3 2 1

1

2

-1

-2

FIGURE 12.15 Signal for Example 12.7. S o l u t io n

Using step functions and shifted step functions, we obtain p )

= 5 u {t)-5 u {t-\ )^ 2 u {t-2 )

3 5e~^ 2e~ Direct application o f linearity and the time shift propert)’ yields F {s) = —------------f- - —

s

s

s

Exercises. 1. Find the Laplace transform o f the pulse signal o f equation 12.7. 2. Find F{s) when/^) = Aû[t - T^) + A2 ^(t - T-^ + Aû{t - T^). 1

AN SW ERS:

~sl

1

— v7i

.4

P j(s) = A— ----- . F ( s ) = ^ -------^ +

4

.—

---------------------

One more property allows us to revisit the signals discussed in section 3 and take their Laplace transforms. The new propert)^ is multiplication of//) by t. This always results in a Laplace trans form that is the negative o f the derivative o f F{s). M ultiplication-by-f property: Let F{s) =

Then

£[tf{t)\ = - — F(5) ds

(12.19)

Verification o f this property follows by a direct application o f the Laplace transform integral to with the observation that te~^^ = ------ • In particular,

ds


S60

-St

■oo

n o

j()

d

dt =

.d s

as

ds J

Table 12.1 lists this transform pair, as well as numerous other such pairs, without mention o f the underlying region o f convergence. As mentioned earlier, we shall dispense with any mention of the ROC, assuming that all functions are zero for r < 0. EXA M PLE 12.8. Find the Laplace transform o f the ramp function, r(r) = tu(t). S o l u t io n

Using equation 12.19,

M ds \ s ,

R {s)= L\r{t)\ =

E X A M PLE 12.9. Supposey{r) = te

( 12.20)

ds

where a is real. Find F{s).

S o l u t io n

The quickest way to obtain the answer is to apply equation 12.19. Specifically, since

^ r —nt , , T

1

s+a L

ds s + a

ds

1

{s + a ) -1

( 12.21)

An alternative, more tedious approach is to use integration by parts as follows:

F {s)= L where v = t and dti = e

=:

te

oo

fOO

0“

ydii = uv

•oo _

0"

O'

udv

dt. Thus, •oo

le

+

.V+ a

■dt

The RO C is a > -a , in which case the first term on the right-hand side is zero. Thus, in this RO C , evaluation of the integral term implies that

F {s)•'0“

s+a

dt =

1

(s + aY

Equation 12.21 is a special case ol the more general formula

£\t''e-^^‘u(t)] =

nl

(s +a)71+ 1

(1 2 .2 2 )

Chapter 12 • Liplace Transform Analysis 1: Basics

561

Exercise. Find the Laplace transform oij{t) = p-e AN SW ER: F{s) = (\ + a r EXA M PLE 12.10. Find F(s) for the signal depicted in Figure 12.16.

f(t)

FIGURE 1 2 . 1 6 A signal to be represented by steps and ramps. S o l u t io n

From Example 12.2,/ r) = 2r(^) - 2r(t - T) - 2 T u { t - 3 7 ). Hence by linearity, the time shift prop erty', and equations 12.15 and 12.20,

F {s) = L \ 2rU )- 2r{t - T ) - 2Tu(t - 3 7 )

2 - 2 £ ” '^

Exercises. 1. Note that the sawtooth o f Figure 12.17 is/ f) = t[u{t) - u {t- 1)]. Suppose

= z/(r)

- u {t- 1). Compute F (5 ) = - —-G ( i')-

d.s

f(t)

2. Use equation 12.22 to compute the Laplace transforms o f/ ;) = tr{t) for the ramp function r{t) and forjit) = p-r{t).

2

6

.s'

s

AN SW TRS: — . —


562

EX A M PLE 12 .1 1 . The circuit o f Figure 12.18a has two source excitations, /j(/) and /2W> shown in Figures 12.18b and c. Compute

V o Jt)

(a)

(c) FIGURE 12.18 (a) Resistive circuit driven by two current sources. (b) Triangular signal, /,(r), in A. (c) Pulse signal, ijit), in A. S o l u t io n

Step 1. Find the form ofV^^^{s). By superposition and Ohms law, = 10/, U) + 10/2(r) From the linearity o f the Laplace transform, = 10/, W + lOAW Step 2. Compute /,(j) and

Some reflective thought yields /,(?) = 2ti{t) - 2r{t) + 2r{t - 1) A

and ijit) = \.5u{t) - \ 5 u {t- 1) A. From linearity, the time shift propertys and the previously com puted transforms. 2

h{s) = - ~ S Step 3. Find V..Js). Since V

l + 2e

s'

and l 2 {s) =

(s) = 10/As) + lOAU), ir follows that

your(s) = - - ^

+ e~^

20

l5^

Chapter 12 * Laplace Transform Analysis I: Basics

S63

Step 4. As an introduction to the next section, by inspection we can compute the time function o f the output voltage: = 35«(/) - 20r{t) - \5 u (t- 1) + 2 0 r{t- 1)

Exercises. 1. Find the Laplace transform o f (i)^(^) and (iii)^ (r) =

■>r e

(ii)

=e

+e

+ te

+ te

2) + «(t - 3).

i\NSV('T,RS: in random order: — !----- 1-----!-----j------- !------- 1------- ?----- . — !------- y — 5— .v + c/ .v + /> (.v + f/)(.v + /?)~ (.v + f/) (.v-c/)

-— --------h .v+ ^/

2. Recall that cos(cor) = ------------------ . Show that the Laplace transform ofJ{t) = cos{cot)u{t) is . r + 0) 3. Recall that sin(CO/) = ------------------ •Show that the Laplace transform o f/ r) = sin(ct)f)z^(f) is (0

F{s) = - ------, 4 4 4. Find the time functions associated with Fi ( j) = — , F) (s) = ----------^ ^ (s + 2)-

AN SW ERS: /;(/) = Au{t)

= Atr"‘H{() , /^(/) =

~

4 . - “’ = ---------. -v+ 4

- 4)

We end this section with Table 12.1, which lists a number o f Laplace transform pairs. Some o f these will be developed later in the chapter and some in the homework exercises. We will refer to this table in the next section when computing inverse transforms.

.V


KV4

T A B L E 12.1 Laplace Transform Pairs

Item Number m t)

K

Ku{t) or K

KIs

m t)

KlP+1

\I {s+ a) ]/{s+ dp-

(OqUP- + OJ^) cos{ci)Qt)u{t) 10

e

s!{P- + co^) (Oo s2

s in {(O Q t)u {t)

~>

(i + a) + coq

is + a)

11

{s + Cl)~ +

(Oq

2(0o^

12

(.v^ +toi5)“ t C O s{0 )Q t)u {t)

13

is +0)o) sin((W()/ +

14

.vsin((|)) + coo cos(<)))

(j>)tiit)

s~ +(0n .vcos((t))-a)o sin(({)) 0 1 + (Oq

15

16

te

smiO)Qt)uit)

17

te

cosi(OQt)uit)

s +a

2(0

is + a)~-(OQ ((5 + rt)^ +C0n)^ 2(0o

18

[(5+C/)^ +(OoJ“ 19

20

21

Cl cos(coor) +

2Î a ^ + B ~

Cj ~ C\Ci

sin(o)o/) n it)

(.? + « )“ +0)^ A + jS

cos (Oq/- tan -1

2Â~ + B- te~^' cos cogr - tan *

C|5' + C 2

' B^

A + jB

.A)j

[s + a + ;cOo)“

A —j B

^

A -jB

{s + a - ycoo)"


6. THE INVERSE LAPLACE TRANSFORM For the Laplace transform tool to effectively anal)'ze circuits, one must be able to uniquely recon struct time functions /(r) from their frequenc}'" domain partners F{s). Theoretically, this is attained through the inverse l-aplace transform integral.

INVERSE LAPLACE TRANSFORM Intuitively, if £[/{()] = F{s), then J{t) = X “ '[F(s)]. Rigorously speaking, the inverse Laplace transform integral is a complex line integral defined as

/ ( ,) = r V u ) , = ^ J ^ f ( . ) e V ,

over a particular path V in the complex plane. T he path F is typically taken to be the vertical line Oj + jio where OJ ranges from -oo to +00 and Oj is any real number greater than Oq, the abscissa o f absolute convergence. This integral uniquely reconstructs the time structure o f F{s) to obia.\n J{t) in whichy(r) is zero for ^ < 0. Conceptually, the process resembles the reverse action o f a prism, to produce white light from its spectral components. An appreciation for the power o f this integral requires a solid background in complex variables and would not aid our purpose, the analysis o f circuits. In fact, the evaluation o f the integral is carried out using the famous residue theorem o f com plex variables. Further discussion is beyond the scope of this text.

Just as the Laplace transform is linear, so, too, is the inverse Laplace transform, as its integral structure suggests, i.e.,

Also, the unilateral transform pair

[fit), f(j)} is uftique, where by unique we mean the following: let F^{s) = L\f^{t)] and Fjis) = Z[^(r)] coincide in any small open region o f the complex plane. Then F^{s) = Fjii) over their com mon regions o f convergence, and/j(f) = f-y{t) for almost all r > 0, “Almost all” means except for a small or thin set o f isolated points that are o f no engineering significance. Hence, there is a oneto-one correspondence between time functionsy(f) for whichy(f) = 0 for f < 0 and their one-sided Laplace transforms. Linearity and this uniqueness make the Liplace transform technique a pro ductive tool for circuit analysis. Virtually all the transforms o f interest to us have a rational function structure; i.e., F{s) is the ratio o f two polynomials. Rational functions may be decomposed into sums o f simple rational func tions. These simple rational functions are called partial fractions and their sums are known as par tial fraction expansions. Two o f the more common “simple” terms in partial fraction expansions have the form K b and K!{s + a). Such simple rational functions correspond to the transforms o f steps, exponentials, and the like. Table 12.1 lists these known inverse transforms. With the table, direct evaluation o f the line integral in equation 12.23 becomes unnecessary. Our goal is to describe techniques to compute the simple rational functions in a partial fraction o f F{s). Once these are found, the transform dictionar)- in Table 12.1, in conjunction with some well-known properties o f the Laplace transform, will allow us quickly to compute the time function y(r).


PARTIAL FRACTION EXPANSIONS: DISTINCT POLES Our focus will center on proper^ rational functions, i.e., + --- + ^^l-y + ao _

^ (^ )_

+--- + biS + bo

5" + where m s « and p^, ... ,

+

is - Pi )(.v - 7^2

+

- Pn )

are the zeros o f the denominator polynomial,

+ ... +

+ l?Q, and are called the finite poles o f F{s). For the most part, rational functions are sufficient for the study o f basic circuits. There are three cases o f partial fraction expansions to consider: (1)

the case o f distinct poles, i.e., p - p - for all i

j;

(2)

the case o f repeated poles, i.e., pj = pj for at least one i

(3)

the case o f complex poles. Although case (3) is a subcategory o f case (1) or (2) or both,

j\ and

its attributes warrant special recognition. If F{s) is a proper rational function with distinct (equivalently, sitnple) poles />j, ... ,

F(s)=K +

(S-Pi)

{S-P2 )

+■■■+

(s-p„)

then (12.24)

where

K = lim F(s)

(12.25a)

5-400

The numbers Aj in equation 12.24 are called the residues o f the pole p- and can be computed according to the formula A =

[(s - Pi)f^(s)] = [(^ - Pi)Fis)\,^p,

(12.25b)

The rightmost equality o f equation 12.25b is valid only when the numerator factor {s - p ) has been canceled with the factor [s - p ) in the denominator of F{s)\ othenvise, one will obtain zero divided by zero which, in general, is undefined. As intimated earlier, this partial fraction expan sion should enable a straightforward reconstruction o f/ r). Indeed, from Table 12.1, we immedi ately conclude

f i t )= Kh{t) + Aê^’^'uU) +

) + •••-!- A„e^’‘‘u(t)

(12.26)

EXA M PLE 12.12. Findy(r) when

jr(^+ a ) S o l u t io n

The solution proceeds by executing a partial fraction expansion (equation 12.24) on F(s) to pro duce the Laplace transform o f two elementar>- signals, a step and an exponential. Specifically,

F{s) = ----------- = - - h

5(5 + a )

s

s +a


S6 '

where Ah is the Laplace transform o f a weighted step, Au{t), and B!{s + a) is that o f a weighted exponential,

To find A, multiply both sides by^, cancel common numerator and denom

inator factors, and evaluate the result at j = 0, to produce A = Ma. Similarly, to find B, multiply both sides by i

cancel common numerator and denominator factors, and evaluate the result

at j = -ay to obtain B = -Ma. Recall that, by iinearit)', X “ ’ [aF{s)] = aL~^

Hence,

/ (r) = - H(0 - - ( - “'uir) = - (I - e - “' )» (r)

a

a

a

I ^Cl

Exercises. 1. Findy(f) when F{s) = --------- ^ AN SW ER: J{t) = 2u(t) 'j

2. Find/r) when F(.s’) = ^ + 3 6 '+ 6 A N SW ER; {.s + \)(s+ 2){s + 3)

Jit) = 3. Find a partial fraction expansion o f /r^y) =

Â +

^

s

s+a

5(i' + a) AN SW ER: K = c . A = - , B =

a

(I

4. Find/r) for f{s) from Exercise 3. AN SW ER: / ( ,) = c 5 ( 0 + ^ K / ) - —

a

a

E X A M PLE 12.13. Suppose V',„(5)= 1 0 “ ^ +^>y + 2

circuit o f Figure 12.19. Find

and v„U) . Assume standard units. 2Q

V ..

'>

80>v^,

FIG URE 12.19 Series resistive circuit. S O L U T IO N Step 1. Detemihie

By voltage division,

= 0.8t'y„(r), in which case 2 5 “ + 35 + 2

C'haptcr 12 • Liplace Transform Analysis 1: Basics

56H

Step 2. Construct a partial fraction expansion ofV^^^^{s). Since the numerator and denominator are both o f degree 2, 16.v" + 2 4 i-+ 1 6 .v(.v + 2)

= K + — + ------.v + 2

(12.27)

The value o f K in equation 12.27 is determined by the behavior o f F(s) at infinity (equation 12.25a), i.e..

K - lim

^ 1 6 i“ + 24.v+ 16

i —ÔO

\

i(A- + 2)

To I'lnd A in equation 12.27, we use equation 12.25b: 165 “ + 24.V + 16 (.v + 2)

Ks + /4 + ,v=()

Bs .sT 2

.v=0

Similar!)’, 16.v- + 2 4 a + 16

Ii =

= -1 6 i= -2

Step 3. Find v^^^ît). Using Table 12.1,

v v ,„ ,( o = r '

r

1 8 161 ' 8 ' 16 16 + -------------- = r ' [ Li 6 Ji + r ‘ - r ‘ .V .v+ 2. s ..v + 2.

= 166(r) + 8u{t) - ]6e--'u{t) V

Exercises. 1. Repeat Example 12.13 for \A^^(.y)= = lOuU) \'

AN SW ER:

*v(.v + « )

2. Given the circuit o f Figure 12.20, find a partial fraction expansion o f /. (.v) = ^ '

+ {a + b + c ).v + b

assuming standard units.

.v(.v+l)

A N SW l'R: /■-,(/) =

+ ■\hn{t) + -\ce~‘u(t) A

ijt)|

1Q 40

F I G U R E 12.20 Parallel resistive circuit.


E X A M PLE 12.14. Compute the inverse transform oFthe function

F{s) = So

-e

lu t io n

From Example 12.12, 1

r '

This result and the shift theorem yield

.v(.v+l) By the linearit}' o f the inverse Laplace transform, /rt = ( l - e - O u W - ( l - r < ' - ' ) » ( ; - ! ) A sketch

appears in Figure 12.21..

FIG URL 12.21 Sketch ofy{/) = [ 1

^]«(r) - [1 -

!)•

PARTIAL FRACTION EXPANSIONS: REPEATED POLES Proper rational functions with repeated roots have a more intricate partial fraction expansion, and calculation o f the residues often proves cumbersome. For example, suppose

F{s) =

fijs) is-a)^d(s)

Chapter 12 • Liplace Transform Analysis 1: Basics

S70

- a)^ specifies a repeated root o f order k, d{s) is the remaining fac

where the denominator factor

tor in the denominator o f the rational function F{s), and n{s) is the numerator o f F{s). The struc ture o f a partial fraction expansion with repeated roots is

Ak

FCv) = - ^ +

where A^, ... ,

(s-af

(12.28)

are unknown constants associated with s - a, ... ,(s - a)^, respectively, and î^(s)

and <^(s) are whatever remains in the partial fraction expansion o f f(s). The formulas for comput ing the y4y o f equation 12.28 are

A k=(s-arF(s)

n(s) -*i-a

(12.29a)

f/(5)

n{s) [d(s)

Â-l = y ( ( . v - « / 'F ( ^ ) )

(12.29b)

and, in general, 1 /!

ds '

njs)''

(s-afF(s)

(12.29c)

O f these expressions, only the first looks like the case with distinct roots; the others require deriv atives o f (.f - a)^ F{s). Computation o f high-order derivatives borders on the tedious and is prone to error. The above formulas, equation 12.29c in particular, are included for completeness. Computer implementation circumvents these difficulties. An example that illustrates the use o f the preceding formulas, as well as a usefiil trick, comes next. EXA M PLE 12.15. The goal here is to illustrate the computation

F(s) = —

when

s +2

.v“ (.v + l)“

.V

.s-

.v + 1

(.v+l)'^

(12.30)

The two easiest constants to find are A2 and Bj, as their calculation requires no differentiation. From equation 12.29a, A, =

s-Fis)

-I

5-1-2

-5=0

=2 i=0

and

B 2 = ( s + \r F (s )

j'-i-2 .v=-l

_ s~

.9 = - !

Finding A^ and is more difficult, since formula 12.29b requires some differentiation. According to equation 12.29b,

as

Chapter 12 • Laplacc Transform Analysis I: Basics

571

To implement this formula multiply both sides o f equation 12.30 by resulting expression with respect to s, and evaluate at i = 0: d

’

ds

5+2

_ d

(5+l)-_

y4|5 +

ds

5=0

take the derivative o f the

^ B|iA-y H----------- h (5+1)“

S+\

i= 0

Observe that, on the right-hand side, it is not necessary to differentiate the terms that contain A 2 , and ^ 2» since these terms disappear at / = 0, as the formula for d ds

'

5+2

’

(,v+l)2

.v=0

1

.

(5+1)"

^+2

'

requires. Consequently,

= -3

“ (5+l)-\

i=()

Similarly,

Bx=ds

0 ( 5 + l ) ^ F ( 5 )

”

_ d

s= - 1

ds .

5

+2 5“

= . A--1

■|

^s + 2'

.5 * '

S'

= 3

. ,v=-l

Aj, and B2 were known, a simple trick allows a more direct computation o f Bji merely evaluate equation 12.29 at j = 1 (in fact, any value o f s, excluding the poles, will do),

Note that since to obtain

0.75 = - 3 + 2 + 0.25 + 0 .5 5 j As expected, solving yields Bj = 3. Hence -3 2 3 = — + ^ + ------ + i “ (.v + l)5 “ s + \ (5 + 1 )2

F(s) = —

.v+ 2

The above result can also be found with the MATLAB command “residue” as follows. Let F{s) =

7i{s)ld{s). It follows that n{s) = s + 1 and d{s) =

+ r . In MATLAB,

»num = [1 2]; >>den = [ 1 2 1 0 0 ] ; »[r, p, k] = residue(num, den) The answers from MATLAB are: r = -3 2 3 1 (the residues associated with the poles) p = 0 0 -1 -1

and constant k = 0.

Exercises. \. Find the partial fraction expansion o f F{s) =

2 5 ^ + 2 r S 3 ^ + 35 4-2 .v^(.9+n^

AN SW ERS (residues in random order): 2, 2. 2, - 1 . -1 2. The partial fraction expansion o f a rational function is given by

F{s) = -

3s^ + \0s + 9 + 45* + 55 + 2

A 5+ 2

B

+ ------ + 5+1

C


Compute A, B, and C ANSWHRS: In random order: 2. 1. 2 3. Use MATLAB to find a partial l^raction expansion o f F{s) =

(Iis)

where n{s) = G{s + 2)^(^ - 2)“

and d{s) = s{s + 1)“(^ + 4)“. Hint: Use num = 6*poly([-2 -2 -2 2 2]) and den = poly([0 -1 -1 -4 -4]). ANSWHR: [r,p,kj = residue(num,den) \'iclds r = - 4 4 4 8 - 1 6 - 6 12 p = _4 .4 _| _ ] 0 ,

n .r

-^8

1 his results m rlie r r h : r(.s) = ■

-1 6

-6

v+l

(v + l)~

12

^

-{------------ - h----- H--------------- -----------h 6

_______________________________ •'>+ 4

(.v + 4 )

^

________________________

The derivative formulas o f equations 12.29 are often difficult to evaluate for complicated ration al functions, such as

s 55^ + 955-^ + 692^^ + 2369.V- + 3715.9 + 2076

F{s) = ---------------------------------------------------- r--------------

(.v+l)(:? + 2)(.s- + 3)(.v+5r A

B

s+\

s +2

_ C ^ D\ i + 3

.y + 5

For these functions, it is very efficient to find A, B, C, and

1)2

D3

(.v + 5 ) -

(.^•+ 5 )'*'

directly. Then one evaluates F{s) at

two values o f j, e.g., j = 0 and s = 1, to obtain two equations in the unknowns D j and Dj. Typically, solving the resulting two equations simultaneously is much easier than evaluating Z), and D j directly by equations 12.29. Alternatively, one can use a software program such as MAT LAB to compute the answers. In particular, in MATLAB: n =[5 95 692 2369 3715 2076] d = [l 21 176 746 1665 1825 750] »[r,p,k]=residue(n,d) r= -l.OOOOe+00 -l.OOOOe+00 -l.OOOOe+00 3.0000e+00

2 .0000 e+00 l.OOOOe+00 P= -5.0000e+00 - 5 .0000 e +00 - 5 .0000 e +00 - 3 .0000 e +00 - 2 . 0000 e +00 -l.OOOOe+00 k=


573

PARTIAL FRACTION EXPANSIONS: DISTINCT COMPLEX POLES Distinct complex roots present challenges different from those for the repeated root case. Since the roots are distinct but not real, the methods o f equations 12.25 and 12.29 apply. Unfortunately, the resulting partial fraction expansion has complex residues, and the resulting inverse transform has complex exponentials multiplied by complex constants. Such imaginar}^ time functions lack mean ing in the real world unless their imaginar}- parts cancel to yield real-time functions. When they do, our goal is to find a direct route for computing the associated real-time signals. To do this, consider a rational function having a pair o f distinct complex poles as in the following equation:

F(s) =

n{s)

n{s)

|(.v + a ) - + ( 0 -](/(. 5)

(s + a + ju>)(s + a - j w ) d ( s )

(12.31)

Since the poles - a - j c o and - a + jco are distinct, the partial fraction expansion o f equation 12.24 is valid. Since the poles are complex conjugates o f each other, the residues o f each pole are com plex conjugates. Therefore, it is possible to write the partial fraction expansion of f{s) as

r(s)-

.v + « + yto

I

5 + a -y c o

(12.32)

d{s)

and d{s). As per equation 12.25b, the first residue in equation

for appropriate polynomials 12.32 is

(12.33)

jco

With A and B known, executing a little algebra on equation 12.32 to eliminate complex numbers results in an expression more amenable to inversion, i.e..

C\S + C-> nîs) /?|(.v) F{s) = ------ \2, ~ 2 + ^ 7 T = ^0(■'■) + (I{s) d(s) {s + fl) + to

(12.34)

C, =2/1

(12.35a)

C-, = 2aA + 2 ojB

(12.35b)

where and

with A and B specified in equation 12.33. W ith Cj and Cj given by equations 12.35, it is straight forward to show that

F'ois) =

C \S + C

2

{s + a)~ +oj^

=C

( C2 -C â \ 1,

x2

(.9 + a) + 0)

2

to

to

{s + « )"

From Table 12.1, item 19, or a combination o f items 10 and 11,

M O = e'

C| cos(tor) +

(C2-Cici) sin (to/) Hit) [ to )

+03"

(12.36)

574


Exercise. Suppose F { s ) =

. Compute/(r). a- 2 + 4

AN SW ER: C, cos(2r)//(f) + 0 . 5 ^ sin(2f)«(r)

The following example illustrates the algebra for computing C, and Cj without using complex arithmetic. EX A M PLE 12.16. Find/^) when 3 .r + 5 + 3 D A + jB A - j B D C^s + C^ F{s) = ------------^------- = -------- + ------ =;- + ------^ = ------ + - ^ ------- (5 + 1 )(5 “ + 4 ) ^+1 s + j2 s-jl i +1 s +A

, ^ (12.37)

Step 1. Compute the coefficients D, C ,, and C2 in the partial fraction expansion o f equation 12.37. First we find D by the usual techniques: 3 .r

+ 5

+ 3

.v^ + 4

= 1

s=-\

Given that D = 1, to find C , we evaluate F{s) at j = 0, in which case 0.75 = 1 + O.2 5 C2, or Cj = - 1 . With D = 1 and C2 = - 1 , we evaluate F{s) at j = 1 to obtain 0.7 = 0.5 + 0.2(C j - 1) or, equiv alently, Cj = 2. Thus,

■V+ 1

s^ + 4

+ 4

Step 2. Compute j{t). Using Table 12.1, items 8 and 9, to compute the inverse transform yields

J{t) = [e~‘ + 2 cos(2r) - 0.5 sin(2f) ]u{t) Alternative Step 1. Compute A and B in equation 12.37 by hand or with MATLAB. In MATLAB, »num = [3,1 3]; »den = conv([l 1],[1 0 4]) den = [1 1 4 4] »[r, p, k] = residue(num.den) r= 1.0000 + 0.25001 1.0000 - 0.2500i 1.0000 + O.OOOOi P= -0.0000 + 2.0000i -0.0000 - 2.0000i 1.0000

-

k =0


This implies that

.y+1

^+ . A -JB s + j2 s-jl

1

, l- y '0 .2 5 , l + yO.25

5+ 1

s+ jl

s-jl

(12.39)

Alternative Step 2. One must exercise caution here and note the difference between the MATLAB output and the form o f the partial fraction expansion. From equation 12.39, w = +2, A = 1, and B = - 0 .2 5 . Again using MATLAB to obtain the form needed in item 20 o f Table 12.1, »K = 2*sqrt(A^2 + B^2) K = 2.0616 »theta = atan2(B,A)* 180/pi theta = -1 4 .0 3 6 2 Thus

Example 12.16 illustrates not only the computation o f an inverse transform having complex poles, but also the computation o f Cj and C , without resorting to complex arithmetic, as was needed in equation 12.32. The trick again was to evaluate F{s) at two distinct ^-values different from the poles o f F{s). This yields two equations that can be solved for the unknowns Cj and C ,.

5 ^ 8

4

Exercises. 1. Find Kt) when F{s) = ------- z----------. s(s-+4) AN SW ER://) = [1 + 4 cos(2r) - 4 sin(2r)]//(/) , 5s" - 2 ^ + 5 2. Find/(r) when F ( s ) - — ^ . . v ( r + 2 5 + 5) A N SW ER://) = u(t) + 4 r ‘ [cos(2/l - sin(2/)l/<(/)

7. MORE TRANSFORM PROPERTIES AND EXAMPLES Another handy propert)' o f the L'lplace transform is the frequenc)' shift property, which permits one to readily compute the transform o f functions multiplied by an exponential. With knowledge o f the transforms o f u{t) , sin(o)/), and other functions, computation o f e~^‘u{t) and f’“"^sin(to/)«(^) becomes quite easy. Frequency shift property: Let F{s) =

Then

L[ e~^'p)] = F{s * a)

(1 2 .4 0 )

Chapter 12 • Laplacc'Iransform Analysis I: Basics

ThisS property can be verified by a direct calculation, = F (s + c)

Xlc’- “7 ( ' ) l =

where we have viewed the sum s + a m tlie integral as a new variable p, which leads to F{p) with

p replaced hy s + a. EXA M PLE 12.17. Let//) = sin(wr)//(r). D efine^/) = e~‘" p ) = e~‘*‘ s\n{iot)u{t). Suppose it is known that

Compute G{s). So

lu t io n

By the frequency shift property, G{s) = F{s + a), or CO

)| = L\e-"\m \ = F (s + « ) =

G( j ) = £|

CO

5“ + (0 “

Exercise. Lcijit) = cos(cor)u(i) for which r ( s ) = ___ - ___ D efin e^r) = e Compute G{s).

J{t) = e

cos{LOt)u{t).

+C0“ .V+ a

AN’SW I-R: i.s +

a )-

+ (0 "

Another property o f particularly widespread applicability is the time differentiation formula. Its utility resides not onl)' in obtaining shortcuts to transforms o f signals, but also in the solution o f differential equations. Differential equations provide a ubiquitous setting for modeling a large variety o f physical systems— mechanical, electrical, chemical, etc. In terms o f signal computation, recall that the velocity of a particle is the derivative o f its position as a function o f time. The accel eration is the derivative o f the velocity. After computing the Laplace transform o f the position as a function o f time, one finds that a differentiation formula allows direct computation o f the trans forms of the velocity and acceleration. Also, as discussed at the very beginning o f this chapter, cir cuits have differential equation models. For example, weighted sums o f derivatives o f the response of- the circuit are equated to weighted sums o f derivatives o f the input signal. Therefore, a differ entiation formula is an essential ingredient in the analysis o f circuits. First-order tim e difiFerentiation formula: Let L\j{t)] = F{s). Then

L jfU )

(It

= sF{s)-f{Q-)

(12.41)


The difFerenriarion property is validarcd using integration by parrs as follows:

4/(0V' ' d t = f U ) e

y / (/ )

(It

JO'

The following examples explore some clever uses o f the first-order time difTerentiation formula

E XA M PLE 12.18.^^ Recall that 6f/) = — u ( t ) . Using the sifting propert)', a direct calculation yields £ [(5(/)] = 1. Is this consistent with the differentiation propert}^? Interpreting the delta function as the derivative o f the step function and applying the differentiation formula yields I d £ 6 { l ) = £ - H i t ) = s £ u{r) - u { 0 ) = s dt s)

(12.42)

which demonstrates the expected consistency.

Exercises. I. The Liplace transform o f a signal/r) is F{s) = f o r m o f e - 2 '4 / W

— . What is the Laplace trans

?

dt'

2s + 4 AN SW I-R: -— — ^-----

(.Y+ 2 r + 4 2. £[sm{wt)6it)] = ? AN SW ER: u

EXA M PLE 12 .1 9 . Suppose^r) = sin(wr)w(/) and we know (for example, from Table 12.1) that Compute £ [cos(wr)//(r)] using the time differentiation formula.

F { s ) = £ sin((0/);/(/) .V" + (0 So

lu t io n

j

Since cos(coO/^0 = ------- sin{a)/)//(/)and sin(wr)«(/)]j ^ o = 0 , the differentiation property imme diately implies that ^ ^^ £ cos(coO//(0

(0 (0

■)

+C0‘

.S - + W

Exercises. 1. Express//) = s\n{o)t)u{t) in terms o f the derivative o f ^ r) = cos{(Ot)u{t). Note the presence o f the delta function. AN SW ER: - ' (1)

^ di

(ij

= sin (w ;)//(r '

2. Now suppose it is known that £ cos(co/)/KM

-. Use the result of Exercise 1 and the + 0)' differentiation propert)' to compute the Liplace transform of/r) = sin(oj/)«(/) noting that^Ô ) = 0 .

-t

S~

ANSW ER; £ sin((0/);/(/)

1 (0

(!)

_

1

1

(I)

(0

(0

V +(0'

s' - (O'

Chapter 12 • Laplacc Transform AnaK'sis I: Basics

578

EXA M PLE 12.20. Lety(r) and its derivative iiave the shapes shown in Figure 12.22. Th e goal o f this example is to explore the relationship between the Laplace transforms o f/ r) and f'{t) in light o f the differentiation property.

FIGURE 12.22 A pulse and its derivative for Example 12.20. Observe how the derivative o f the pulse leads to a pair o f delta functions. Using linearity and the shift theorem on j{t) yields

X|/(01 = £ | » ( 0 - - 1)] = £ [« (f)]- X | h(/ -1)1 = - (I Applying the linearity o f the Laplace transform x.of\t) yields £ l / ’(r)] = i : [ 5 ( 0 - 5 ( f - I ) ] = l - ^ - ^ From the differentiation formula, it must follow that L\f{t)] = sL\J{t)]. Thus, £|/'(/)l = sL U (t)] = j i d - « - * ) = I demonstrating consistency.

As might be expected, the formula for the first derivative is a special case o f the more general dif ferentiation rule:

L cit”

m

= s"F(s) -

)

(12.43)

This rule proves useKil in the solution o f general «th-order difTerential equations. O f particular use is the second-order formula: (1 2 .4 4 )


579

The inverse o f differentiation is int^ration. The following property proves useful for quantities related by integrals. Integradon p r o p e r ^ Let F{s) = -£[/(/)]. Then for ? > 0,

■ c Jo-/(9 )d? =—

(12.45a)

and

F (s)

(12.45b)

As with many o f the justifications o f the properties, integration by parts plays a key role. By direct computation (using equation 12.16),

To use integration by parts, let

fiq ) d q and dv = e

u= Then

-too

rr JO'

For the appropriate region of convergence, the first term to the right o f the equal sign reduces to fO' et

f(q ) d q

f( q ) d q JO*

Since the second term to the right of the equal sign is F{s)/s, as per equation 12.45a, the proper ty is verified.


S80

E XA M PLE 12 .2 1 . Find the Laplace transform o f the signal/f) sketched in Figure 12.23a using the integration property. f(t)

FIGUIIE 12.23 (a) A triangular signaly(/) for Example 12.21. (b) The derivative S o l u t io n

Observe that the triangular waveform y{r) o f Figure 12.23a is the integral o f the square wave^^). Since ^t) is easily represented in terms o f steps and shifted steps as

^t) = u{t) - 2u{t - 7) + u{t - 2 7) its Laplace transform follows from an application o f linearity and the time shift property:

x u (o i= The integration property implies that

g{q)dq .S -

EXA M PLE 12.22. This example explores the voltage-current {v-i) relationship o f a capacitor in the frequency domain by way o f the integration property. Recall the integral form o f the voltagecurrent dynamics o f a capacitor: 1 r'

Taking the Laplace transform o f both sides and applying the integration property produces

£[vcit)] = £

J

C->

I

1 fO-

ic(T)dx = 7T^c(-^) + — Cs Cs

But this expression depends on the initial condition

), because

S8]

Chapter 12 * Laplacc Transform Analysis 1: Basics

Therefore,

Cs

(12.46)

s

Equation 12.46 says that the voltage V(^s) is the sum of rvvo terms: a term dependent on the fre quency domain current I^^s) and a term that looks like a step voltage source and depends on the constant initial condition V(^0~). The quantity Z^^s) = MCs looks like a generalized resistance— “generalized” because it depends on the frequency variable s and a “resistance” because it satisfies an Ohm’s law-like relationship, V^^s) = Z^^s) I({s). These analogies prompt a series-circuit inter pretation o f equation 12.46 as depicted in Figure 12.24. An application o f this equivalent frequenc)' domain circuit to general network analysis appears in the next chapter.

\,{S) o — >+

ic(t) O— >■ 4-

£ [ •] c ----- > VJs)

V,(t)

Cs

O 'f O------FIG U RE 12.24 Equivalent circuit interpretarion of a capacitor in the frequency domain. This equiv alent is arrived at by applying the integration propert)' of the Laplacc transform to the capacitor volt age, seen as the integral of the capacitor current.

A second example interpreting the v-i characteristics o f the capacitor in the frequenc)' domain ensues from the differentiation rule. Instead o f winding up with a series circuit, one obtains a par allel circuit. The interpretation is thus said to be dual to the one just described. E X A M PLE 12.23. This example has two goals: (i) Verify that equation 12.46 is consistent with the differentiation formula interpretation o f the capacitor; (ii) Build a dual frequenc)' domain interpretation o f the v-i characteristic o f a capacitor analogous to that o f Example 12.22. As a first step, recall equation 12.46: Vc(-

Cs

s

which, after some algebra, becomes

I(is) = CsVfis) - CvfiO-)

(12.47)

Notice that equation 12.47 is consistent with the application o f the derivative formula to i(^t) =

C[civ(Jdt]. This consistency offers some reassurance in the accuracy o f our development. The interpretation o f equation 12.47, however, is quite different from that o f equation 12.46. In the latter equation, the current /^j) equals the sum o f two currents, CsV^s) and -C t/^ 0“). This sug


582

gests a nodal interpretation, resulting in an equivalent circuit having two parallel branches. One branch contains a capacitor with voltage V^s). The other, parallel branch contains a current source with amperage Cv(\Qr). The current through the capacitive branch is

where

“G ” now acts like a generalized conductance because it multiplies a voltage, similar to Ohm’s law. “Q ” is generalized because it depends on s. Figure 12.25 presents the equivalent circuit o f the capacitor in the frequency domain and is dual to the circuit o f Figure 12.24. Chapter 13 covers in detail the role o f these equivalent circuits in analysis. Ic(s)

FIGURE 12.25 Equivalent circuit to a capacitor in the frequency domain using the differentiation formula.

The last elementar)' property o f the Laplace transform that we consider in this chapter is the timescaling property, also called the frequency-scaling property. Its importance is fundamental to net work synthesis. Here, numerical problems, such as roundoff, prevent engineers from directly designing a circuit to meet a given set o f specifications. Instead, the design engineer will normal ize the specifications through a frequency-scaling technique. Once the normalized circuit is designed, frequency-scaling techniques arc reapplied in an inverse fashion to obtain a circuit meet ing the original specifications. Time-/Frequency-scaling property: Let ^ > 0 and L\J{t)] = F{s). Then

L[J\at)\ = - F a \ci)

(12.48)

or, equivalently, F{sln) = aL[f{nt)]. Since the proof o f this property is straightforward, it is left as an exercise at the end o f the chap ter. E XA M PLE 12.2 4 . Figures 12.26a and b show impulse trains that model sampling in signal-pro cessing applications. The impulse train o f Figure 12.26b is the time-scaled counterpart to that o f Figure 12.26a.

^83


f(2t)

f(t)

A 2 --

A 2 --

i< 1

1 --

- -

> t

> t 1

1

FIG URE 12.26 (a) Unit impulse train, (b) Time-scaled imit impulse train. Unit impulse trains such as these model sampling in signal-processing applications. The time-scaled impulse train in Figure 12.26b increases the frequency at which the impulses occur (twice as often as in the original signal). This is reflected in the Liplace transforms o f the two signals:

'Z&U-k) = I k=0 A-=0

\-e'

(12.49)

By the time-scaling property, £ [/ (2/ )l = 0.5

l - e -0 .5 5

(12.50)

Notice that what occurs at, say, Sq in equation 12.49 now occurs at 2s^^ in equation 12.50. Hence, time scaling by numbers greater than 1 concentrates more o f the frequenc)^ contcnt o f the signal in the higher frequency bands.

Exercise. Verify, by direct calculation, that L\J{2t)\ is given by the right side o f equation 12.50.

Several more properties o f the Laplace transform are germane to our purpose. However, these properties have a systems flavor and are postponed until Chapter 13. We close this section by pre senting Table 12.2, which lists the Laplace transform properties and the associated transform pairs.


lA Bl.E 12.2 Liplace Transform I’ropertics Property

Transform Pair

Linearity

L \ j{t-

Time shift

£\t/lt)n(t)]=-— F'{s) as

Multiplication by t

n cr r j s )

Multiplication by t“

ds"

= Rs + //)

Frequency shift

Tim e differentiation

£ jfO ) (it

d - f{ t )

Second-order differentiation

£

= sF(s)-J{0-)

= r F c v ) - 5 / ( ( r ) - / '\ ( D

dr

wth-order differentiation

T> 0

7)1 =

d''fU) dt’'

(i)X

fUl)dq

m dq

Fis)

Time integration

(ii) £

Time/Frequency scaling

.-fU l)d q

Fis)

£[f{at)\ = MaFista)

S8S


8. SOLUTION OF INTECRO-DIFFERENTIAL EQUATIONS USING THE LAPLACE TRANSFORM Differential equarions provide a cross-disciplinary mathematical modeling framework. Although difTerentia! equation models may represent only the dominant behavioral facets of a circuit or physical process, their widespread utility and importance to circuits and control systems warrant special discussion. To begin, recall the time differentiation formulas o f equations 12.41 and 12.43 and the integration formulas o f equations 12.45a and 12.45b. Also, recall that a differential equa tion relates a sum o f derivatives o f an output signal to a sum o f derivatives o f an input signal. For example, if the input and output signals are voltages, then the relation

d'\-

+ a..V.

dl for constants

dt'

+b

and bj might model the behavior o f a linear circuit. We may use the following steps

to solve this differential equation for

using the Laplace transform procedure:

1.

Take the Laplace transform o f both sides o f the equation, using the appropriate deriva

2.

Algebraically solve the resulting expression for

3.

Compute a partial fraction expansion o f the expression for

4.

Inverse-transform the partial fraction expansion to obtain the time function

tive formulas, equations 12.41 and 12.43.

If the equation is an integro-differential equation, i.e., a mixture o f both derivatives and integrals of the input and output signals, then we simply apply the same algorithm, except we use the inte gral formula where appropriate. Some examples ser\-e to illuminate the procedure. EXA M PLE 12.25. Consider the pulse current excitation o f Figure 12.27a) to the RC circuit o f Figure 12.27b. The goals o f this example are (i) to use and illustrate Laplace transform techniques to solve a difTerential equation derived from a simple RC circuit and (11) to find the response volt r > 0, when V(^0~) = 1 V.

age

F IG U R E 12.27 Excitation currcnt (a) fora simple /?Ccircuit (b) for Example 12.25. S o l u t io n

Step 1. Find L[i{t)]. Since i{t) = 0.5//(r) - 0.5//(^- 1),

X [/ (0 ] = 0 .5

\ -e


Step 2. Find the circuit’s dijfereutial equation model that links the excitation current i{() to the response

voltage, V(it). Since ijît) = 0.5v(\t) and

= ^.‘b d v jd t, summing the currents into the top node

of the circuit yields

d\’c {t) dt

After multiplying through by 2, the desired differential equation circuit model is

dv({f) + v c ( f ) = 2 i(t) dt

Step 3. Take the Laplace transform o f both sides, apply the differentiation rule to the left side, and solve for V(i{s). Applying the Laplace transform to both sides yields

sVcis) - v^Q-) + V^s) = ll{s) Solving for V^s) produces 2

v r(0 ” )

.v+1

i’ + l

Vcis) = ----- :/(.v) + - ^ ^

\-e ~ ^

= -----------+ ^(i-i-1) .v + 1

Some straightforward calculations show that 1

1 _

s

5(.v+ l)

1 (.v+1)

Thus, with the aid o f the shift propert)' and the transform pairs o f Table 12.1, we obtain

vcU) = r'[V c(s)\ = r '

.v(i + l)

V

s+\

/

= / / ( 0 - ( l- c '" ^ '“ '^)/K ^-l). Figure 12.28 presents the graph o f this response. Because o f the initial condition and the magni tude o f the pulse input, the capacitor voltage is constant for 0 < r < 1 second. At r = 1 second, the pulse magnitude drops to zero, making the circuit equivalent to a source-free RC circuit in which the capacitor voltage decays to zero as shown in the figure. V,(t)

F I G U R E 12.28 T h e response voltage

v^^t)

for Exam ple 12.25.

S8‘

Chapter 12 • Iâplace Transform Analysis 1: Basics

EXA M PLE 12.26. The goal o f this example is to compute the response, denoted here by the given the scries RLC circuit o f

input current /,„(^), to the input voltage cxcitation

Figure 12.29. Suppose the initial conditions are /^(O-) = 1 A and V(^0~) = -2 V.

— TYYY 40

1H

FIGURE 12.29 Series RLC circuit for Example 12.26. Here the current //„(^) = So

lu t io n

Step 1. Compute the Laplace transform o f the input. From the tables or by inspection, X[6(r)] = 1. Step 2. Compute the integro-differential equation o f the circuit o f Figure 12.29. The first task is to sum the voltages around the loop to obtain

Substituting for each o f the element voltages using the mesh current, ij„{t), yields the desired integro-difFerential equation,

diin (12.51) Step 3. Take the Laplace transform o f both sides, substitutefor R, L, C,

solve for

Vf^Qr), and

and

W ith the aid o f the differentiation and integration formulas, taking the Laplace

transform o f both sides o f equation 12.51 produces

/?/,„(.V) +Uli„ (,v) - L/i(0-) +^Cs

(s) +

= V;„(s)

s

This has the form i - + S.s + 7li

L ------ 1----- ^ h „ U )

= K „ (S ) + i , / t ( 0 - ) -

V>-(0‘ )

S

S

Plugging in the required quantities and solving for

produces

I •') ~

5 + 4 5 + 4

5 +2

~

5+ 2

~

( 5 + 2 )-

Step 4 . F in d iinit). Taking the inverse Laplace transform yields the desired result:

i d t ) = (2 - 2,)e-^-‘u(t) A

588


A plot o f this response appears in Figure 12.30.

Exercises. 1. An integro-differential equation for an LC circuit is given by

cli,C dl

C —oo

= 0, and V(^0~) = - 1 0 V. Compute

with C = 1 F,

L.(s) r . ( ( ) “ ) ANS>X^R: .v/,-.(.v) + -^----- + -^-------- = 0 ^

.S

= > / ..(/ )= 10sui(/)//{/) A

.V

2. If two signals x{^) and^(^) are related by the equations

dxU) dt

+ 2v(/) = 45(/) and 2.v(/)-

y{z)dz = 2ii(t)

where x(0 ) = 2 , u{t) is the unit step function, and 8{t) is the Dirac delta function, then findATi). AN SW ER: X{s) = •V

EXA M PLE 12.27. The final example o f this chapter looks at the leaky integrator circuit o f Figure 12.31, which contains an ideal operational amplifier (op amp). resistance o f the capacitor. Given C and

represents the leakage

/?, is chosen to achieve an overall gain constant, in

this case, 1. The objective is to compute the response

assuming that t'(j(0~) = 0, and com

pare it with that o f a pure integrator having a gain constant o f —1.

SH‘)

Chapter 12 • Liplacc Transform Analysis 1: Basics

R, = IMegO

V (t) = 5 u (t )

''ou.W

FIGURE 12.31 Leaky integrator op amp circuit. So

lu t io n

First, note that since the op amp is ideal, -V(^t) = equation that relates

to z^^-and solve for

The goal, then, is to write a difFerential

using the Laplace transform method.

Step 1. Determine the dijferential equation. Since the op amp is ideal, it follows that ijr= -i^. From O hm s law, i^ = vJR^^. On the other hand,

^

/?2

dt

This leads to the difFerential equation model o f the op amp circuit,

dt

where, as indicated before,

I<2C

~

R^C

= -v^^t). Note that if /?,C= 1 and R2 is infinite, then the cir

cuit works as a simple integrator. The circuit is called a leaky integrator because /?-,C is large but finite. Since /?, C = 1, one expects the gain constant to be 1 as well. Step 2. Substitute values, take the Laplace transform o f both sides, a7jd solve for Vg,,f{s). Taking the Laplace transform o f both sides, one obtains 'w ( 0 " ) + Since

=- -

= 0, it follows that -5 0

5( 5 + 0 . 1 ) Step 3. Invert

to obtain

.V

+

50 .9 + 0 .1

Solving for v^^^^{t) produces

^92


'Problems

5. Find the Laplace transform o f each o f the following time functions.

f\{t) = Ke-‘^‘u {t- 7), r > 0

(a)

BASIC SIGNALS, SIGNAL REPRESENTATION, AND LAPLACE TRANSFORMS

=

(b )

(d) (e)

r > 0

f^{t) = Kte-‘^‘u {t- 7), T > 0

(c)

r>0

=

f^{t) =

T>0

1. Find the Laplace transform o f each o f the following signals assuming T- > 0.

6. Compute the Laplace transform for each of the following signals.

(a) (h)

/ 2 W = / / - 7 „ ) 6 U - 7-,)

(c)

h (D = e-^' cos(0.5rt/ + f )5(2/ -

(d)

/^W = A-,6W + 7 - „ )

=

(c) (f)

(a)

/\{t) = Kt^[uU)-N{t- T)l T>()

(b )

f,U)

)

sin(2yTr - 2jr)u{t - 1) (c )

K^6(t-T„)

/sW = c o s h (2 / )a (t-2 )^

= s in ( 2 . T / ) « ( r ) -

(d )

f^{t) = r [ s i n ( 2 . T / ) w ( r ) s i n ( 2 ; r / ‘ - 2jt)u{t - 1)] f^{t) = 2 s i n ( 4 ; r r ) / / ( r ) « ( 2

-

r)

sinh(/ - 2 r , ) 6 ( f - r , ) (g)

7. Represent each o f the following signals using

fj{t) = sm{2m - Ji)d{2t - 4)

(sums of) steps, ramps, shifts o f basic signals, 2. Find the Laplace transform o f each o f the

etc. Then find the Laplace transform.

following signals. Use Tables 12.1 and 12.2 as

f,{t)

needed. (a)

f^{t) = lu{t) + u{t - 1) + u{t - 2)

2K -

-Au{t-A) (b)

f^{t) = 2 r { t ) - l r { t -

l ) - r ( r - 3 )

+

At - 5) ->

(c) (d)

t(s)

j\{t) = cosiS)5m)u{t) + c o s (0 .5 > -r(/ -2 ))« a -2 )

(a)

(e)

f,(t)

3. Sketch the indicated waveforms and find the

2K -

Laplace transform. Use Tables 12.1 and 12.2 as needed. (a) (b) (c) (d) (e)

f^{t) fJ,t) f^{t) f^[t) f^{t)

= ti{t) - l i t - \ ) . = u{t)-r{t-\)-^r{t-2) = 2u{t) u{2 - t). = Ar{t) u{\ - t) = 2rit)u{\ - t) + u { t - \)r{2 - t)

■>

2 (b)

A K

4. Find the Laplace transform o f each o f the following time functions. (a)

f^{t) = Ktu{t-\)

(b)

/,(/) = AT/- ])«(/)

(c) (d)

f^{t) = Ktr\t-\) f^{t) ^ K{t - \)r{t)

r

ir (c)

t(s)

S93


f,(t) 2K

K > t ■>

t

2T

T (d)

i k f^ft) ’ -2

f,(t)

1

K ■> t

■> t

1

-T.

2

-K (e)

f«(t)

(d)

9. Represent each of the following signals using

3 •

(sums oO steps, ramps, shifts o f basic signals,

2 ■

etc. Then find the Laplace transform.

1

^-------1-------1------ r— 1

2

3

4

■>

f,{t)

t

5 4K -

(f)

8. Represent each of the following signals using (sums of) steps, ramps, shifts o f basic signals, etc. Then find the Laplace transform.

2K -I---------- '----------1--------- 1— ► t(s) 1 2 3 4 (a)

(a)

(b)


5 94

r>

-► t r^

(0 11. For the circuit o f Figure P I 2 .11, suppose /?, = 6 0 0 Q, Rj = 1000 Q, and

= 1500 Q.

Use the Laplace transform tables to compute

\ltu{t) + 3e~^û{t) + 18^’“ ^sin(2jK)«W A. ■> t | U t) i,(t)

(!) Figure P I2,11

12. For the circuit of Figure P I 2 .12b, R^ = 600 Q, /?2 = 1000 Q, and R^ = 2 4 0 0 Q. Use the Laplace transform tables to compute Vg^t^s) =

£

for the input given in Figure P I2 .12a.

10. Represent each of the following signals using (sums of) steps, ramps, shifts of basic sig nals, etc. Then find the Laplace transform.

vJt)

f,(t) 2 ■> 1

(a) 2

■> t

o Figure P I2.12

(a)

13. For the circuit o f Figure P i 2 .13, suppose /?! = 6 0 0 Q, /?2 = 2 0 0 0 Q, and R^ = 3 0 0 0 Q. Find the Laplace transform of the voltage when yjj(f) = 2Ae~^û{i) V and v^2 ^t) = 30^-^Mf) V. r> >

t o r^


(e)

Using the formulas o f part (c) and the frequency shift property, compute (i) L [Ke-^^ cos(wr)] (ii) L [K e -‘^^s\n{o)t)]

17. Find a simple expression for each o f the waveforms shown in Figure P i 2.17 in terms of sines, cosines, shifted sines, and shifted cosines. Then find the associated Laplace transforms. Half-cycle of sin (nt)

LAPLACE TRANSFORMS VIA TABLE 12.1 AND PROPERTIES VIA TABLE 12.2 14. Find the Laplace transform o f J{t) =

Ke-‘"u{t)u{T- t), T> 0, as follows. (a) Express Ku{t)u( T - t) zs a difference o f step functions. (b) (c)

Find £ [Kn{t)u{T- t)]. Apply the frequency shift property on

Quarter cycle of 2cos(0.25m)

your answer to part (b) to compute

£ W )l 15. Prove the time-/frequenc)'-scaling property by direct calculation o f the Laplace transform integral. 16.(a)

Using the famous Euler formula,^'^‘ = cos((Ot) + j sin(wf), find an expression

(b)

for cos(a>/) and sin(wr) in terms o f the

Figure P I2.17

complex exponentials (b)

Determine the Laplace transform o f

(c)

Using the formulas developed in parts

18.(a)

Represent sin(7cO

0
0

otherwise

(a) and (b), show that

as the difference o f a sine and a shifted

(i) £[co s((o r)]=

sine. Find G{s). (ii) Xfsin(co/)] = (d)

(b)

■> ~> S +(D“

Using the formulas o f part (c) and the multiplication-by-/ propert)', compute (i)£[/irrcos(w r)] (ii) L[Kts\n[cot)]

Relate j^r) in Figure P 12.18 to ^ r) o f part (a). Then find F{s) from G{s).

S‘)6


Figure P I2.18 19.(a)

22. (a)

Using the formulas cosh(/7r) = and sinh(^/r) =

(i)

+

- e~“^, find

Idt

,

using the derivative property.

f(X )dx

(iii) Com putc£[//(0] = £

multiplication-by-r property, compute (1 )X [/ T r c o s h (/ 7 r )]

(c)

as a sum o f appropri

(ii) Compute £ \g {t ) \=£ — /(/)

Using the formulas o f part (a) and the

(2)

Express

ate step functions. Compute F{s)

{\) £ [cosh(/7r)] (2) L [sinh(^r)] (b)

Considery(/) in Figure P i 2.22.

using the integral property.

£ [Kt sinHat)]

Again using the formulas o f part (a)

(b)

=J{t + 4).

Repeat part (a) for

and the multiplication-by-r property,

f(t)

compute

{\) £ [K r cosh(/7r)] (2) £ [Kr

3 -

s m h ia t)]

2 -

20. Suppose/r) = 0 for r < 0 and F(.v) =

2.V + 4

s +1

■> t

Find the Laplace transforms o f the functions

1

below, identifying each o f the properties used

2

3

4

5

Figure P I2.22

to compute the answer. Solutions obtained by finding/f) are not permitted. (a) (b) (c) (d)

23. The Laplace transform

g^{t) = 5 J { t - T), T> 0 g,{t) = 2r^^t) gît) = 2e-‘^ % t - D , T > 0 g^{t) = 5 i f i t - r i , T > 0

F{s) =

2 1. Use Liplace transform properties to find G/is) 4^ + 20 as eiven below when F{s) = --------- :r .

and

(.9+1)-

State each property that you used. Assume that

J{t) = 0 for ^ < 0 . Solutions obtained by finding J{t) are not permitted. (a) g^{t) = 0.5^r) (b)

=

W W)

^^W = 2 ^ , ( 2 f - 4 )

(c)

^5W = 2 ( r - 2 ) / ; - 2 )

(f)

\-e

is given as

-(s-a)

s-a

(a)

Find the Laplace transform o f e “%t).

(b)

Find the Laplace transform of tj{2t).

AN SW LR: (a) .s' 24. The Laplace transform

F{s) = (a)

-V

J{t) is given as

l-e "_ ^

Find the Laplace transform of withy(0“) = 3.

cm cit

(b)

Now find the Laplace transform o f

(c)

Finally, find the Laplace transform o f

dt


25. Supposey(/) = S(t) —d{t — T), T > 0. (a)

Find L\J{2t)\ by direct calculation of

597

transform o f / r ) and then, using the relation ship, find the Laplace transform o f^ f).

the Laplace transform integral. (b)

Find L\J{2t)] by computing Hs) and then using the scaling property.

26. Use only Laplace transform propenies to answer the following question. Suppose that for w Vw/ O

-► t

t < 0 , / / ) = eû{-t), and the one-sided Laplace 5 “f" 1 transform o f/^ ) is X [ / ( f ) ] = F {s) = — ^ .

s

L et^ r) =J{t)u(t). Find the Laplace transform of v{t) when (a) (b)

(c)

v(t) = 2 g " {t)-g \ t) = t/ W = / a )+ J_ ^ q )d q

(d) -► t It is not necessary to have the answer be a rational function. 27. (a) W O ' O

Find the Laplace transform of the function f j )

sketched

in

Figure

P12.27a. (b)

Identify a relationship between and the function

Figure P i2,28

sketched in

Figure P I2.27b . Use your answer to

29. Develop a relationship betweenj^f) and g(t)

part (a) and the appropriate property

in Figure P I2 .29. Find the Laplace transform

from Table 12.2 to compute the

ofy(/) and then

Laplace transform o f ^ /).

tionship between the two functions. Assume

by making use of the rela

that 0 < A < B < C. Also, determine D and E in terms o f A, B, and C

O ->

o Figure P I2.27

28. In Figure P I 2.28, what is the relationship between j{t) and g^t) ? First find the Laplace

598


33. Find (i) the partial fraction expansion and (ii) the inverse Laplace transform for each o f the following functions by hand. Show all work. (No details, no credit.)

(a) F,Cv) = s{s

a){sb)

C H E C K : One residue is a. ^

- 7.v^ + 4.V + 2

( b ) F ^ s ) = ---------------------------- ^

Figure PI 2.29 30. Let j{t) and ^t) be as sketched in Figure P I 2.30. Find G{s) in terms o f F{s). f(t)

20 10

C H EC K : One residue is - 2 . . X s 2 / + 18i-^ + 4 6 5 ^ + 4 4 5 + 12 (c) F2,{s ) = --------------------r---------- z----------(5 + l ) ^ 5 + 2 r C H E C K : Two residues are at 2 and - 2 .

Remark: Check answers using MATLAB. Use the help command to make sure you

-

understand the terms used. For example ■

for part (b), ■>

n = [2 -7 4 2]; d = conv([l -1],[1 -4 4]):

Figure P12.30

INVERSE LAPLACE TRANSFORMS BY PARTIAL FRACTION EXPANSION 31. Using partial fraction expansions and your knowledge o f the Laplace transform o f simple signals, find j{t) when F{s) equals (a )

2 5 ^ + 1 3 r + 305 + 32

[r,p,k] =residue(n,d) 34. Find (i) the simplified partial fraction expansion and (ii) the inverse Laplace trans form for each o f the following functions by hand. Show all work.

(a) Fi(.v) =

{a + b ) s + l a b (5 + fl)(5 + b )

Check: One residue is a. (b ) F2{ s ) =

( a + b + c ) s ‘' + ( b e + 2 a b + a c ) s + a b c s { s + a ) { s + b)

s{s^ + 6^ + 8) C H E C K : One residue is a. (b)

- s-6

Cv + 2 ) ( . v - - l ) (c)

cs~ + { a + 2 a c ) s +

2.v'^ + 12.v“ +22.y + 8

( r +25 + 1)( j + 2) (d ) F4(s ) =

(d)

il + c ) a ^ - a

(c) F3(.v) =

/ +185^+ 9 8 5 -+ 2085+ 144 (5 + 2 ) “ (5 + 4 ) “

/ + 1 2 A - ^ - 2 4 r - 3 2 .y + 16

(e)

C H EC K : Two residues are at 2 and - 2 . (e ) F ^ ( s ) =

55^ + 1 4 4 5 + 2 0 4 (5+1)

32. Inadvertently left out by the authors.

(5+ 2)“ + 6 4

599


'w '

Vg„f{s) and then find

35. F in d /r) when F{s) equals

(a) (b)

(c)

(d)

for the

input current

25 + 16 +16

r

“ 45 + 9

/X

lin ( s ) = 2 0 -------------- 2— 2---------

245-72 5^ + 4 5 + 40

Check: One residue is at 20.

25^ + 885

(5 + 4)(5^ +64) 2 5 ^ + 2 5 ^ -2 5 -6 v.(t)

0

36. Find (i) the partial fraction expansion and

+ 20

80

24 0

(ii) the inverse Laplace transform for each of (a)

the following functions by hand. Show all work. , X r ., X

25^ + 125^ + 235 + 17 (5 + l)(5 + 2)(5 + 4 )

C H EC K : Residue at j = - 2 is - 1 .5 . (b) F2(5) =

25^ + 9 5 ^ + 1 6 5 + 1 1 (5 + i)(5 + 2 y

C H EC K : Two o f the residues are 2 and 1.

5^ + 4 5 ^ - 2 5 ^ - 9 5 - 3

(C) ^3 ( 5 ) = -------------=-------- T-----

( 5 - 1 ) 2 ( ^ + 2)2 CHECK: Two residues are at I and two are at - 1 . 4 5 ^ - 1 2 5 ^ + 325 + 16 (d) /=4(5)=r

(5 + ir+ 1

38. Suppose F {s)

it follows that (5 + 2)^ + 9

fj) =

cos{(ji)i)u{i) + Ar2^“'“sin(cflf)«(r). Find tf, A^j, Kj, and CO. Now express J{t) = cos{(Ot + 9)u{i) by finding K y a, (O and d.

(5 + 2)2 + 16

37. Find the partial fraction expansion and the inverse Laplace transform for each of the indi

39 .

The

Laplace

cated output voltages or currents. All answers

[A^jf“'^'cos(fi)?) +

transform

of f j )

=

sin(fi)r) + Kê~^û{t) is

must be in terms o f real functions with real coefficients or symbols. Show a ll work, (a)

- 5 2 5 + 228

F {s)

For the circuit o f Figure P 12.37a, find the partial fraction expansion o f and then find for die

(5 + 4 ) ( 5 + 0 ^ + 100 Find a, b, K^,

K y and 0).

input voltage 40. Consider the resistive circuit in Figure P 12.40. Use Table 12.1 and the shift property 5(5 + 4 ) (b)

For the circuit o f Figure P 12.37b, find the partial fraction expansion

for each Vj^(s) below. Sketch tôuM) by hand or with the help o f MATLAB.

to find of

600


(a)

Kv,(.v) =

IOf'"'' +

(b)

\-e'

Vi„{s)=\0

- 5e' ..-4 5

-2 0 -

(b)

Use the Laplace transform method to compute the inductor current, for t > 0.

(c)

If the input is changed to

=

1 0 « ( r - 7) V, where 7"= 10 msec, find /^(/), for t > 0. Hint: Your answer should be a shift o f the answer com

8kQ

puted in part (b).

-n/ V ^ 8kQ

0

(d)

for r > 0.

1 8 kQ / 9kQ,

(e)

for / > 0. Hint: Can you

use superposition?

(0

If ij{0~) = 100 mA and for t > 0.

tion expansions o f the rational functions listed below. Then use Table 12.1 and MATLAB to

R

obtain the associated time function. i,(t)

(c) F^{s) = (d) 74(‘' ) = (e) F^{s) =

= 10//(r

- 7) V, where 7 = 1 0 msec, find i/it),

41. Use MATLAB to compute the partial frac

(b) F2(s ) =

= 10//(r)

If //(0~) = 100 mA and V, find

Figure P i2.40

(a) F,(5) =

= 0, find

If /^(0~) = 100 mA and

3.v'^ + 3 0 5 - + 86.9 + 6 4

'■">6

.s'* + 8 i - +20.V+ 16

-O + v,(t)

- 4 6 .2 5 s - 6 9 2 . 8 125 .s-'* + 14.5.V- + 169.5625s + 510.25 -2.s-^ + 23.V- - 68.V - 3265

Figure P i2.42

.V - S 3 .55 " + 134.V + 797.5 10.5.r'^ + 47.875.s- + 151.875 - 108.5938 + 6.5

+ 36.5625.v- + 101,5625.v + 2 07.0312

- 1 .5.s-'^ - 25.75.v‘^ - 127.5.y'* - 2 9 1 .5.y~ - 330^ - 143.75 .s-^’ + 10.5.v-'^ + 50.v-^ + 141.V-'' + 250.V- + 262.5.V + 1 2 5

CIRCUIT RESPONSES VIA LAPLACE TRANSFORM APPLIED TO DIFFERENTIAL EQUATIONS

4 3 . For the circuit o f Figure Pi 2.43, suppose R = 10 £2 and C = 0 .0 1 F. (a) Show that the differential equation lor the circuit is

(h'rU) 1 1 ^ + ----- VcU) --------,ll RC RC

42. T he input to the circuit o f Figure PI 2.42 is = 10//(r) V, valid for r > 0, and has an ini tial inductor current of /^(0“ ) = 0. Suppose R =

(b)

30 Q. and L = 0.2 H. (a) Show that the differential equation for

for r > 0 when (c)

R. , . 1 , . , ({) = - Vi„U) L L

= 10«(r) V, valid

for t > 0.

the circuit is

di l it ) (If

Use the Laplace transform method to compute the capacitor voltage,

Now suppose the initial capacitor volt age ^'Ô") = —10 V and

= 0.

601


(a)

Find V(^t) , for r > 0. (d)

Find

the capacitor voltage,

for r > 0, wlien (b)

= 10/^r) V.

- 1 0 and

Construct a differential equation in Solve the differential equation by the Liplace transform method; i.e., show that V(it) =

R v jt)

/> 0 and for appropriate constants A'j,

K-,, and to, which are to be found in terms o f /y, L and C.

vJt)

6

sin(ojf) + KjCosUot) for

-O i,(t)

ijt) \r

Figure P I 2.43

;r

v^(t)

44. The circuit o f Figure Pi 2.44 has rwo source excitations, v^^{t) = 10//(f) V and

= lu{t)

A, both applied at / = 0. Suppose /?j = 5 Q, R-, = 20 Q, and Z = 2 H. The initial condition on the inductor current is /^(0“ ) = —I A. (a)

ANSWl-R; (a) —

the circuit, assuming the response is and

(c)

Find the response due only to

(e)

+

- \V' = 0 . ( b ) (I) = 7 =

LC ^

J lC

46. The circuit o f Figure P 12.46 is a series (loss less) LC circuit driven by a voltage source. Suppo.se (a)

= 0 and /^(0“ ) = 0. Construct the differential equation o f

assuming /^(0“ ) = 0.

the circuit in terms o f the capacitor

Find the response due only to ip{t)

voltage,

assuming /^(0“ ) = 0. (d)

dr

Construct a differential equation for /^(r). Leave the inputs in terms o f

(b)

Fieurc P I2.45

(b)

Solve the differential equation using

Find the response due only to the ini

the Laplace transform method, and

tial condition ;^(0~) = -1 A, assuming

show that

both inputs are zero.

for t > 0.

10 -

10

c o s(0

.5 jia )

for t

Find the complete response,

l = ih

> 0 by superposition.

^ '„ « ) Q

=

R.

0

' . ,(t)

4C=-,F

vJt)

Figure P I2.46 47. A pair o f (coupled) differential equations

Figure P i2.44

that represent a circuit are given as M t )

45. Consider the Z,C circuit ol^ Figure P i 2.45, for w'hich /^(O") = 7q and

= V^. Since

and

dt

there is no resistance present in the circuit, there is no damping; hence, one expects a pure ly sinusoidal response. Such circuits are called

lossless.

dt

-t-^/|.v(/) = oiyit)

602


with initial conditions ;c(0~) = 1 andyO “ ) = 2. Suppose , = 1 , = 1. ^3 = 1 . ^3 = 1. and/r) = 2u{t). Find j/(^) and A-(r).

51. The op amp in the circuit o f Figure Pi 2.51 is assumed to be ideal. /?, = 20 R-, = 40 ki2, and C = 10 |iF. (a)

= 2,

48. Reconsider Problem 47 with

Use nodal analysis to construct a first-

= 2,

order differential equation describing

= 4, b^ = 3, and/r) = 2u(t).

the input-output relationship o f the voltages.

49. The inductor current i{t) in a second-order

If u jt ) = 2uU) V, and j/c(0) = -1 V,

(b)

RLC circuit satisfies the following integro-dif-

and

ferential equation for / > 0.

then

=

Sketch the response in MATLAB.

(a)

If v-it) = 2e~^-">‘tt{t) V and ^CÔ) = 0,

(c)

v'c(() ) + 8 j^ _ //(X )f/r

find

If /(0-) = 8 A and vM~) = - 4 V, find

kis). (b)

Use your answer in part (a) to find /,(/).

50. Consider the circuit o f Figure PI 2.50. (a)

Use KVL and KCL to show that the differential equation relating the input Figure P I2.51

voltage to the capacitor voltage is

ci\\

d\ 'c (It-

1

RC clt

LC

''C

52. Reconsider the RC active circuit shown

LC

Figure

12.1

o f Example

12.1, where we

encountered difficult)' using the single third(b)

Take the Laplace transform o f both

order difTerential approach. Now we will solve

sides o f this equation to show that

the problem with Laplace transforms applied to three first-order differential

_ Vc(A) =

RC (c)

( .v + 5 ) v c ( 0 " ) -h v c (0 " ) I 1 s~ + ----- 5 +

1 •s + -----

RC

LC

Assuming that vc(0 ) =

) = 0,

equations

LC

(a)

Three node equations in the time domain have been given in equations

^ = 0.8 Q, Z. = 1 H, C = 0.25 F, and

12.1, 12.2, and 12.3. Take the Laplace

= 5(r), show that

transform o f each o f these three node v’c(/)

=

equations, accounting for initial con

~

(b) L=1H —

► /

Y

Y

v„(t)

6

+

;± R=0.8Q I

= lOu(t) V and the initial

capacitor voltages are ^^(O) = 12 V, = 6 V, and t/^(0) = 3 V, find

Y

ijt)

ditions. If

v,(t) (d)

Now do a partial fraction expansion of

Vgut^^^ and determine Figure P I 2.50

for /> 0.

C

H

A

P

T

E

R

Laplace Transform Analysis II: Circuit Applications A FLUORESCENT LIGHT APPLICATION Fluorescence

is a process for converting one

type o f ligiit into another. In a fluorescent light, an electric current heats up elcctrodes at each end o f a tube. T he hot clcctrodes emit free electrons, which, for a sufficiently high voltage between the electrodes, initiate an arc, causing mercury contained in the tube to vaporize. The energized mercury vapor emits invisible ultraviolet light that strikes a phosphorus coating on the inside o f the tube. The phosphorus absorbs this invisible short-wavelength energy and emits light in the visible spectrum. A starter circuit must quickly generate a sufficient quantity of free electrons and crcate a suffi ciently high voltage to initiate the arc that vaporizes the mercury inside the tube. One t)'pe o f starter circuit contains a special heat-sensitive switch in series with an inductor. We will model this special switch by an ideal heat-sensitive (bimetal) switch in parallel with a capacitor. The concepts developed in this chapter will allow us to analyze the operation o f such a starter circuit as set forth in Example 13.1 1 .


In terms o f the Laplace transform variable

s,

Z (j), and the notion o f admittance, denoted

define the notion o f impedance, denoted

y\s).

Impedances and admittances will sat

isfy a type o f O hm ’s law. These ideas are generalizations o f the phasor-based notions o f impedance and admittance introduced in Chapter 10.

2.

Learn the arithmetic o f impedances and admittances in the Laplace transform domain, which is analogous to the arithmetic o f resistances and conductances in the time domain.

Chapter 13 • Laplacc Transform Analysis 11: Circuit Applications

3.

Apply the new concepts o f impedance and admittance to redevelop the notions o f volt age/current division, source transformations, linearity, and Thevenin and Norton equiv alent circuits in the /-dom ain.

4.

Define /-domain-equivalent circuits o f initialized capacitors and inductors for the pur pose o f transient circuit analysis.

5.

Introduce the notion of a transfer function.

6.

Define rvvo special types

7.

Redevelop nodal and loop analyses in terms o f impedances and admittances.

8.

of

responses: the impulse and step responses.

Utilize the l-aplace transform technique, especially the /-domain-equivalent circuits o f initialized capacitors and inductors, for the solution of switched /?ZC circuits.

9.

Introduce the notion o f a switched capacitor circuit, which has an important place in real-world filtering applications.

10.

Set forth a technique for designing general summing integrator circuits.

SECTION HEADINGS 1.

Introduction

2.

Notions o f Impedance and Admittance

3.

Manipulation o f Impedance and Admittance

4.

Equivalent Circuits for Initialized Inductors and Capacitors

5.

Notion o f Transfer Function

6.

Impulse and Step Responses

7.

Nodal and Loop Analysis in the j-Dom ain

8.

Switching in RLC Circuits

9.

Switched Capacitor Circuits and Conservation of Charge

10.

The Design of General Summing Integrators

11.

Summary

12.

Terms and Concepts

13.

Problems

1. INTRODUCTION Chapter 12 cultivated the Laplace transform as a mathematical tool particularly useful for circuits modeled by differential equations. This chapter adapts the Laplace transform tool to the peculiar needs and attributes o f circuit analysis. W ith the Laplace transform methods described in this chapter, the intermediate step o f constructing a circuit’s differential equation, as was done in Chapter 12, can be eliminated. Available for the analysis o f resistive circuits is a wide assortment of techniques: O hm s law, volt age and cu rren t division, nodal and loop analysis, linearit)', etc. For the sinusoidal steady-state analysis o f RLC circuits, phasors serve as a natural generalization o f the techniques o f resistive cir cuit analysis. The Laplace transform tool permits us to extend the sinusoidal steady-state phasor analysis methods to a much wider setting where transient and steady-state analysis are both pos sible for a broad range o f input excitations not amenable to phasor analysis. Recall that transient an;ilysis is not possible with phasors.

Chapter 13 * Laplacc Transform Analysis II: Circuit Applications

T he keys to this generalization are the i-domain notions o f impedance and its inverse, admit tance. Instead o f defining impedance in terms ofyoj, as in phasor analysis, we will define it in terms o f the Laplace transform variable

s. This

definition allows the evolution of a frequency- or

j-dependent O hm s law, j-dependent voltage and current division formulas, and ^-dependent nodal and loop analysis; in short, all o f the basic circuit analysis techniques have analogous

s-

dependent formulations. W hat is most important, however, is that with the ^-dependent formu lation, it will be possible to define .^-dependent equivalents for circuits containing initialized capacitors, inductors, and other linear circuit elements. These equivalent circuits make transient analysis natural in the i-domain. In the final section o f the chapter, we introduce the notion o f a switched capacitor circuit. Switched capacitor circuits contain switches and capacitors, and possibly some op amps, but no resistors or inductors. Present-day integrated circuit technolog)' allows us to build switches, capacitors, and op amps on chips easily and inexpensively. This has fostered an important trend in circuit design toward switched capacitor circuits. A thorough investigation o f switched capac itor circuits is beyond the scope o f this text. Nevertheless, it is important to introduce the basic ideas and thereby lay the foundation for more advanced courses on the topic.

2. NOTIONS OF IMPEDANCE AND ADMITTANCE Chapter 10 introduced an intermediate definition o f (phasor) impedance as the ratio o f phasor voltage to phasor current, and admittance as the ratio o f phasor current to phasor voltage. In the Laplace transform context, impedances and admittances are j-dependent generalizations o f these phasor notions. Such generalizations do not exist in the time domain. To crystallize this idea, we Laplace-transform the standard differential

v-i relationship

o f an inductor,

at to obtain

Vîs) ^ Lsliis), assuming /^(0“ )

(1 3 .1 )

Ls multiplies an ^-domain current, /^(^), to yield Vjis), in a manner similar to O hm s law for resistor voltages and currents. Ls are ohms. The quantit)' Ls depends on the frequency variable s and gen

= 0. Here, the quantit)' Z^(s) =

an j-domain voltage, T he units o f Z^(j) =

eralizes the concept o f a fixed resistance, and it is universally called an im pedance. This complexfrequency or ^-domain concept has no time-domain counterpart. Although the inductor served to motivate ^-domain impedance, in general an impedance can be defined for any two-terminal device whose input-output behavior is linear and whose parameters do not change with time. A device whose characteristics or parameters do not change with time is called

time invariant.

606

Chapter 13 • Laplacc'Iransforni Analysis II: Circuit Applications

IMPEDANCE T he

impedance, denoted Z{s), o f a linear time-invariant

rwo-terminal device, as illustrated in

Figure 1 3 .1 , relates the Laplace transform o f the current, /(s), to the Laplace transform o f the voltage, V^j), assuming that all independent sources inside the device are set to zero and that there is no internal stored energy at ^ = 0 “. Under these conditions.

V{s) = Z{s)I{s)

(13.2a)

and, where defined,

Z(s) =

V(s) I(s)

(13.2b)

in units o f ohms.

l(s)

Device

vis)

Z(s) orY(s) FIG U R E 13.1 A two-terminal device having impedance

Exercise.

For an unknown linear circuit,

ANSW'HR;

Vj,j{s) = —------

and

5“ + 4

Z{s) or admittance

I- (s)

^(^).

= 2. Com pute

4 — .y" + 4

T he inverse o f resistance is conductance, and the inverse o f impcdance is admittance. For exam ple, if we divide both sides o f equation 13.1 by

Ls,

we obtain

Ls This suggests that

\/Ls acts

is defined as follows.

as

a. generalized conductance universally

(1 3 .3 ) called an adm ittance, which

607

Chapter 13 • Laplacc Transform Analysis II: Circuit Applications

ADMITTANCE T he

admittance,

denoted y(^), o f a two-terminal linear time-invariant device, as illustrated

in Figure 1 3 .1 , relates the Laplace transform o f the voltage, V(j), across the device to the Laplace transform o f the current, /{^), through the device, assuming that all internal inde pendent sources are set to zero and there is no internal stored energy at f = 0~. Under these conditions,

i{s) = ns)v{s)

(13.4a)

and, where defined,

Ijs)

(13.4b )

V{s) in units o f S.

From equations 13.2 and 13.4 , impedance and admittance satisfy the Inverse relationship

Y(s) =

Exercise.

1 Zis)

(1 3 .5 ) 16

=

For an unknown linear circuit, '

C om pute Ky,//).

(.S-

----

arid

+ 2)

A/l(-^)="

(>v +

2 )(i + 4 )

2 ANSWER:

^+ 4

As a first step in deepening our understanding o f these notions, we compute the impedances and admittances o f the basic circuit elements shown in Figure 13.2.

i,(t)

ic(t)

O-

O— +

-I-

v,(t)

V ,(t)

v«(t)

o — >■

o-

(a) FIGURi-l

(b) 1 3 .2

From O hm ’s law, the resistor o f Figure 13.2a satisfies sides yields the obvious, K^(j) =

(c)

(a) Resistor, (b) Capacitor, (c) Inductor.

Rljîs).

= Rijît).

Laplace-transforming both

From equations 13.1 and 13.2, the im pedance o f the

resistor is

Zfîs) = R

608

Chapter 13 • l-aplacc Transform Analysis II: Circuit Applications

and, from equation 13.5, rhe adm ittan ce o f the resistor is

Here the kinship of impedance/admittance with resistance/conductance is clear. rh e capacitor o f Figure 13.2b has the usual current-voltage relationship,

(!v(^{ t ) dt

/c ( 0 = C-

Assuming no initial conditions, the Laplace transform relationship is

I^s) = CsV^s) From equation 13.4 , the ad m ittan ce o f the cap acito r is

Y(is) = a and from equation 13.5, the im pedance o f the cap acito r is

Z eis) =

1 Cs

Repeating this process for the inductor o f Figure 13.2c ,

)=

L — ——

the im pedance and

ad m ittan ce o f the in d u cto r are

Z,,U) = U . YLU) = - jLs

Exercises. 1. Given the integral form o f the v-i capacitor relationship, assume no initial stored energy and take the Laplace transform o f both sides to derive the impedance o f the capacitor. This provides an alternative, more basic means of deriving the impedance characterization. 2. Given the integral form o f the

v-i

inductor relationship, assume no initial stored energy and

take the Laplace transform o f both sides to derive rhe admittance o f the inductor.

Throughout the rest o f the text, whenever we refer to an impedance the unit o f Ohm is assumed, and similarly, admittance is assumed to have the unit of siemens (S). The units for KW and

I{s)

are usually not shown, although strictly speaking they are volt-second and ampere-second, respec tively.

609

Chapter 13 • Laplace* Transform Analysis II: Circuit Applications

3. MANIPULATION OF IMPEDANCE AND ADMITTANCE Recall that the Laplace transform is a linear operation with respect to sums of signals, possibly multiplied by constants. KVL and KCL are conservation laws stating, respectively, that sums of voltages around a loop must add to zero and sums o f all currents entering (or leaving) a node must add to zero. Since the Laplace transform is linear, it distributes over these sums, so the sum of the Liplace transforms o f the voltages around a loop must be zero and the sum o f the Laplace trans forms o f all the currents entering a node must be zero. In other words, complex-frequency domain voltages satisfy' KVL and complex-frequency domain currents satisfy' KCL. Because of this, and because impedances and admittances generalize the notions o f resistance and conduc tance, one intuitively expects their manipulation properties to be similar. In fact, this is the case. M a n ip u la t io n

r u le .

Because impedances map j-domain currents,

I{s),

to ^-domain voltages,

K(j), and because all /-domain currents must satisfy' KCL and all /-dom ain voltages must satisf)’ KVL: 1.

Impedances,

Z{s), can

be manipulated just like resistances and, like resistances, have units

o f ohms. 2.

Admittances, K(/), can be manipulated just like conductances and, like conductances, have units o f S.

This manipulation rule suggests, for example, that admittances in parallel add. T he following example verifies this property for the case o f two admittances in parallel.

EXA M PLE 13.1. C om pute the equivalent admittance, general admittances, K,(/),

V^{s), and Y:^{s) in

and impedance,

o f three

parallel, as shown in Figure 1.3.3. Then develop the

current division formula.

Yji-s] K,(.v)+Ko(.v)+r,(.v)

(1.3.6)

An(-v)

Z Js) FIGURE 13.3 Three general admittances, Vjis), in parallel, having an equivalent admittance K- (/) or impcdance

Z- (s).

(>10

Chapter 13 * Liplace Transform Analysis II: Circuit Applications

S o lution W c seek the relationship

vvhich implicitly defines

o f the admittance o f a two-terminal device, /^(j) =

= AW

for

^3(5) =

From the definition

= 1 ,2 , 3. From KCL,

>3(5))

This relationship implicitly defines the equivalent admittance as

= y i(s)+ y 2 (s)+ y 3 (s)

y i„(s)=

affirming that admittances in parallel add. From the inverse relationship

Z;„ (5 ) = -------------- !-------------K,Cv)+K20v) + y3(.v) Returning to the relationship

/j^(s) =

we now note that

/;,(,C) = Y / : ( S ) V , „ ( S ) = y , ( s ) Z , „ ( s ) / , ; , ( s ) =

Y^{s )+Y 2( s )+Y2{ s )

Equation 13 .6 has the obvious generalization to any number o f parallel elements.

Exercises.

1. Show that for t%vo impedances,

Z^{s)

and

Zjis),

in parallel, Z ,„(^) =

2. Show that the equivalent impedance o f two capacitors in parallel is

Z (5 )=

and that the equivalent capacitance is

‘

‘

C]^+C25

(C j+ C 2 )5

2 i ( ‘^) +

= Cj + C 2 .

3. Derive the following formula for the impedance o f two inductors in parallel:

L^ + Lo 4. A 2 |.iF and a 0 .5 uF capacitor are in parallel. Find the equivalent capacitance. A N SW ER : 2.5

til

5. A 2 mH inductor is connected in parallel with a 0 .5 mF capacitor. Find the equivalent imped ance.

ANSWER: 2.^

'

.s- + i ( r "

61 1

Chapter 13 • Laplace Transform Analysis II: Circuit Applications

6. In the circuit o f Figure 13.3, suppose Kj(j) =

MR,

K,(j) =

M{L$),

and

=

Cs,

a resistance,

an inductance, and a capacitance. Derive the relationship

1 fin (s)

L C s ^ + -s + R

the equivalent admittance,

and find

terms o f I;„{s). in' ■ Ai\S\V1-:R: y j s ) =

I^{s)

s

1 ^ ^

7. In the circuit o f Figure 13.3, suppose Kj(j) = 0 .5 ,

25+1

in

Find

^ l and /^(.v) =

EXAMPLE 13.2. Compute the input impedance o f the parallel RLC circuit sketched in Figure 13.4.

o— +

VJs)

Z Js) =

Y Js)

FIG U R E 1 3 .4 Parallel

RLC circuit

for Example 13.2.

S o l u t io n

For parallel circuits, it is convenient to work with admittances, since

parallel admittances add.

Thus, for the circuit o f Figure 13.4, ^

1

r +—

RC

.v+'

1

LC

Since impedance is the inverse o f admittance, 1

1

C , 2 ^ _ 1L , +

RC w hich is the equivalent input im pedance o f a parallel

1

LC

RLC circuit.

(1 3 .7 )

612

Exercises.


1. C om pute the equivalent impedance o f a parallel connection o f three inductors hav

ing values 4 m H , 5 m H , and 2 0 m H . A N SW ER : 2 x 1()--S2. Com pute the equivalent impedance o f a parallel connection o f six elements: rsvo resistors, o f 6 kQ and 3 kQ; two inductors, o f 3 mH and 6 m H ; and two capacitors, o f 0 .2 |.iF and 0 .0 5 |.iF. A N SW E R ; 4 X 1()^V(r + 2 x 1

i)-^s + 2 x 1 0 * ’)

1 he dual o f the parallel circuit o f Figure 13.3 is a series connection o f three impedances as shown in Figure 13.5. T he following example verifies that impedances in series add, and simultaneous ly develops a voltage division formula. E X A M P L E 1 3 .3 . Com pute the equivalent impedance, general impedances,

Z^(s), Zîs),

and

Z^{s)

and admittance,

o f three

in series, as shown in Figure 13.5. Then develop the

voltage division formula,

Zj{s) =

Z,(.v) + Z2(.v) + Z3(i-)

+ V,(s) -

V^n(.v)

(1 3 .8 )

+ V^(s)

r i d l J R E 13.5 Series impcdance circuit illustrating voltage division. S

o l u t io n

O hm s law tell us that (1 3 .9 ) for / = 1 , 2 , 3,. From KVL (1 3 .1 0 ) Using equation 1 3 .1 0 and the definition o f input impedance, it follows that

Z,„ (s) = -^^4^ = Z| (s) + Z , ( i) + Z3(,s)

(1 3 .1 1 )

Chapter 13 * Laplacc Transform Analysis II; Circuit Applications

613

T he voltage division formula o f equation 13.8 follows from a modified form o f equation 1 3 .1 0 , and equation 13.9, to yield

Z i (5) + Z2(5) + Z3(5) T he voltage division formula is easily extended to the case o f w devices in series:

Z (s ) Z ,( 5 ) + Z 2 ( s ) ...+ Z„(5)

Exercises.

1. C om pute the equivalent impedance o f two capacitors, C, and C j, in series. 1

AN SW I-R;

1

c

C’|CS

V

Cj + C . 2. Show that the equivalent admittance o f rwo capacitors, Cj and C-,, in series is 3. Suppose

Zj(s)

= 10 Q,

Z-,(s)

=

2s,

and

Z^(s) = 6^ in

Figure 13.5. Find

Y{s) = -------- — s. C\ + Cl

V^-)W, and

.V A N SW ER S:

Z,,(s)

= 10 + 8^. \ s ( .0 =

■

4.V + 5

. /-.(/) =

'

2

Zj(s) = 10 Q, Z-,(s) = 2s, V^Js).

4. Suppose terms o f

and

-7 + 10.v + 2 A N SW ER S: Z,„(.v) = ---------------------, ■V

Z t^(s )

=-

=— ^

in Figure 13.5. Find

Z^-^^(s)

and K^(j) in

1

-------------- V;„(.v) ,v‘' + 5 .v + !

5. Verify that the equivalent inductance o f two inductors in series is

= -î +

O f course, there are series-parallel connections of circuit elements that combine the concepts illus trated in Examples 13.1 through 13.3, as set forth next.

EXA M PLE 13.4. Com pute the input impedance Z-J,s) o f a series connection o f t\vo pairs o f par allel elements, as shown in Figure 13.6, in which Then compute

in terms o f

If

=

= 10 Ci, C = 0.1

u{t),

find

Vjit).

¥,

= 5 O.,

and Z. = 1 H.

Chapter 13 • l^place Transform Analysis II: Circuit Applications

614

FIG U R E 13.6 Series-parallel connection of S

RC elements

for Example 13.4.

o l u t io n

Conceptually, view the circuit as shown in Figure 13.7.

V,(s)

FIG U R E 13.7 Conceptual series structure of the circuit in Figure 13.6. Here

Z ,U ) =

1

10

O.l+O.l.v

.y + 1

and

Z2{s) =

Is

7+2

in which case 10(.y + 2 )+ 2 .s;(.v + 1) _ 2^ “ + 1Is + 20 (5 + 1 )(5 +

(s + \)(s + 2)

2)

It fo llo w s th at

1//■ X

,

(.v + l )(i + 2 )

2.V

.v(.v+l)

2.v“ + 12.v + 20

-V+ 2

.v“ + 6 .y + I 0

Chapter 13 * Laplace Transform Analysis 11; Circuit Applications

Finally, if

61 S

= -,

s

(^■+1) (-^ + 0 ^2(^') = -3 ---------------= ---------- ^ .9“ + 6 . 9 + 10 ( 5 + 3)" + From Table 12.1, item 19,

Exercise.

Repeat Example 13.4 with the following changes: C = 0.01 F and /?, = 10 Q.

A N SW ER S:

Z Js )

= 10

VS(.v) =

\+10

Another basic and useful circuit analysis technique is the source transform ation property, exhibited now in terms o f impedances and admittances. The first case we will examine is the voltage-to-current source transformation, illustrated in Figure 13.8,

(b)

(a) FIG U R E

1 3 .8

Z^[s), as shown in Z^(s), as shown in part (b).

Illustration of the transformation of a voltage source in series with

part (a), to an equivalent current with a current source in parallel with

Often, voltage-to-current source transformations provide an altered circuit topology that is more convenient for hand or calculator analysis. Mathematically, the goal is to change the structure o f a voltage source in series with an impedance to a current source in parallel with an admittance while keeping both

Vjis)

and / 2 W fixed. To justify this, one starts with Figure 13. 8a, in which

voltage division implies

V,(.v) = ^ ^ V^îs)=Z.is)l2is) Z ,(5 )+ Z 2 (5 ) H en ce,

ZAs) ^ 0 ,

V ;(.v) =

Z2(.V)Z, is )

1

(V iu (s)\

Z,(.v)+Z2(.v) I 2 ,(5 )J ■r,(5)+K2(.v) U i„(.v )j

(1 3 .1 2 )

Chapter 13 • U p lace Transform Analysis 11: Circuit Applications

616

where

Yjis)

=

[Zj{s)]

^ This equation identifies the parallel structure of Figure 1 3 ./b ; i.e., Figure

13. 8b is a circuit equivalent o f equation 13.8. Reversing these arguments leads to the current-to-voltage source transformation, illustrated in Figure 13.9.

(a)

(b)

FIG U R E 13.9 Illustration of (a) current source to (b) equivalent voltage source transformation. Clearly, the manipulation of impedances and admittances parallels that o f resistances and con ductances, as suggested earlier. Indeed, for a rigorous statement o f the soiuce transformation tech nique developed above, refer to the source transformation theorem in Chapter 5 and replace by

Z{s),

by

and

by

Ijp).

R

Indeed, all such values in Chapters 5 and 6 have i-domain

counterparts. This section ends with a demonstration of finding a Thevenin equivalent in the /-dom ain. E X A M P L E 1 3 .5 . C om pute the Thevenin equivalent circuit o f Figure 13.10.

VJ s )

V Js )

v„(s)

(b) FIG U R E 1 3.10

SOLUTIO N From the material in Chapter 6, our new concepts o f admittance and impedance, and Figure 13.10b,

(13.13a) o r equivalently,

o s-B b )

Chapter 13 * L^placcTransform Analysis II: Circuit Applications

61

Now from Figure 13.1 Oa,

(d + 1)/(^(.V)+ GV'j-^C.v)

/(;’ (.v)+ CV^„(5) =

= {ii + i).vc[v;„(.9) - \/^(^)] + gv;„(.9) = I {a + 1).vC + C ]

(13.14)

( 5 ) - (« + 1).vCV/^(.v)

Rewriting equation 13.14 in the form o f equation 13.13a, we have

1

(a + \)sC + G

(13.15) '

(W +D.9C + G

Com paring equations 13.15 and 13.13a, we identify

(a + 1).vC + G

Exercises. 1.

In Example 13.5, what is the Norton short circuit current,

A N SW ER : /.^.(.v) = U + 2. Find

(fl + 1).?C + G

and

\)s(:Vjj,s)

VgJ
the circuit in Figure 13.11. 2-2t;\

..

2/ / , (. s)

2 + 2.V +

FIG U RE 13.11

3. For the circuit o f Figure 13.12, use source transformations to find I^p) and Y^jj^s) for the indi cated terminals.

.

A N SW ER S: I. is) = 0 . 2 .a ' (>■) and

= 0.2.^ + - + 0.4

0.2 F

(V Js )

2.5 Q 1 H F IG U R E 1 3 .1 2

618

Chapter 13 * Laplace Transform Analysis 11: Circuit Applications

4. EQUIVALENT CIRCUITS FOR INITIALIZED INDUCTORS AND CAPACITORS T he notions o f impedance, admittance, and transfer function do not account for the presence of initial capacitor voltages and initial inductor currents. Hoiv can one incorporate initial conditions into various analysis schemes? For an answer vve look at the transform o f an initialized capacitor and inductor and interpret the resulting equation as an equivalent circuit in the complex-frequency domain. For the capacitor and the inductor, rvvo equivalent circuits result for each; a series circuit containing a relaxed (no initial condition) capacitor/inductor in series with a source, and a parallel circuit with a relaxed capacitor/inductor in parallel with a source. Example 1 2 .2 3 pre viewed this notion. T he capacitor has the standard voltage-current relationship

c

^

(it

= /c (0

Taking the Laplace transform and allowing for a nonzero initial condition

Cs

yields

- Cv(J,Q~) = I (is)

(1 3 .1 6 )

T he left side of equation 1 3 .1 6 is the difference o f two currents, one given by the product o f the capacitor admittance and the capacitor voltage

{CsVf^s))

and the other by Cy^Ô"). Thus the cir

cuit interpretation o f equation 1 3 .1 6 consists o f a relaxed capacitor in parallel with a current source, as illustrated in Figure 13.13. In the time domain the current source o f Figure 1 3 .1 3 cor responds to an impulse that would immediately set up the required initial condition.

1^(5) ^

.....................................................

V,(s) Cs

FIGllRK 13.13 Parallel form of an equivalent circuit for an initialized capacitor. Here, the capacitor within the dotted box is relaxed while the current source

Cv(4S)~) accounts

for the initial condition.

Rearranging equation 1 3 .1 6 yields

Cs

s

(1 3 .1 7 )

Example 12.22 previewed this equation by taking the transform o f the integral relationship o f the capacitor. We observe that the right-hand side o f equation 1 3 .1 7 is the sum o f two voltages, one o f which is the product o f the capacitor impedance and the capacitor, current and the other z/^(0“)A'. Thus, the interpretation is a series circuit, as sketched in Figure 13.14.


619

V Js)

FIG URE 1 3.14 The series form of an equivalent circuit for an initialized capacitor. Here the capacitor in the dotted box is relaxed, and the voltage source accounts for the effect of the initial condition. Initialized inductors have similar j-domain equivalent circuits analogous to those o f the capaci tor. W ith the voltage and current directions satisfying the passive sign convention, the differen tial inductor current-voltage relationship is

Transforming both sides yields

( 13 . 18 ) Again, this equation consists o f a sum o f voltages,

Lslîs)

and - Z /^ ( 0 ). Thus equation 1 3.18 can

be interpreted as a series circuit, as depicted in Figure 1 3 .1 5 .

FIG U R E 13.15 Series form of equivalent circuit for an initialized inductor. Here the inductor with in the dotted box is relaxed; notice the polarity orientation of the voltage source. To construct a parallel equivalent circuit, divide equation 1 3 .1 8

Ls

s

by Ls and

rearrange to obtain (1 3 .1 9 )

T he right side o f Equation 13.19 is a sum o f currents that determines a parallel equivalent circuit, as sketched in Figure 13 .1 6 .

Chapter 13 * Laplace Transform Analysis II: Circuit Applications

620

F'lGURli 13.16 Parallel form of equivalent circuit for an initialized inductor. Again, the inductor inside the dotted box is relaxed. Two examples illustrate the use o f these four equivalent circuits for initialized capacitors and inductors. E X A M P L E 1 3 .6 . This example illustrates an ^-domain application o f superposition. In the /?Z,Ccir

= Au{t) V.

cuit o f Figure 13.17, suppose ^(--(0“ ) = 1 V, /^(0“ ) = 2 A, and

Find Vj{t) for f > 0.

+ vjt) --------- (-

F K iU R E 13.17 Circuit for Example 13.6. S

o l u t io n

In this example, it is convenient to replace the capacitor by its (series) 5-domain voltage source equivalent circuit, because the capacitor is in series with the input voltage source. On the other hand, it is convenient to replace the inductor by its (parallel) /-dom ain current source equivalent circuit, because the desired output is the inductor voltage. This results in a three-source or multi input circuit. O nce the equivalent circuits are in place, one can apply superposition to obtain the answer, although there arc many other ways to solve the problem.

Using the voltage source ynodelfor the capacitor and the current source modelfor the induc tor, draw the equivalent s-domain circuit. Using the equivalent circuits o f Figures 1 3 .1 4 and 1 3 .1 6 ,

Step 1.

we obtain the circuit of Figure 13.18. Here we note that

V;„(.v) = - .

s

---- = - ,and

s

s

-----= - .

s

s


+

621

>

—O +

1 .5 0

V,(s) 0.5s

FIG U R E 1 3 .1 8 j-domain equivalent accounting for initial conditions of the circuit of Figure 13.17. Step 2 . Fmd the contribution to

from

From voltage division,

S- + 3S + 2

I.5 + - + 0.55

s Step 3. Find the contribution to

from

^ = - . Again, from voltage division,

s

s

V l(s) = —

1

-.y

^

s^ + 3s + 2

X —=

.5 + - + 0.55

s

Step 4 . Find the contribution to Vj{s) from Z,;'^(0 ) = 1. Using O hm ’s law in the 5-domain,

0.5^ 1.5 + -

s/

V t{s) = -

1.5 + - + 0.55 5

2

-3 s - 2

^

^“ + 3^ + 2

--------------

X — =

Step 5. Su?n the three contributions and take the inverse transform.

V i(s) = v l u ) + v l { s ) + vl(.s) = -

-2 - + 35 + 2

2

2

-V+ 2

.V+ 1

in which case Vj{t) =

Exercise. Find Ij{s) and

2e

^hi{t)

- le

û{t) V

ij{t) for the circuit o f Figure 1 3 .1 7 using the equivalent circuits o f Figures

1 3 .1 4 and 1 3 .1 5 . Hint: Write one loop equation. ANSW FR:

/ ,(,)=

./•/(/) = M ’- ' i a i ) - 2( " “ '/M/) ( .V

I )(.V + 2 )


622

E X A M P L E 1 3 .7 . This example illustrates a single-node application o f nodal analysis. In the circuit o f Figure 13.1 9 , suppose

V(^Qr)

= 1 V, /^(0“ ) = 2 A, and

=

n{t) V.

Find

V(^t)

RLC t>

for

0. 0.5 H /Y Y \

-o

1.5 Q

+ v,(t)

v Jt) 1F

MGURH 1 3.19 Circuit for Example 13.7. S

o l u t io n

In this example, it is convenient to replace the inductor by its (series) coniplex-frequency domain voltage source equivalent circuit, because the inductor is in series with the input voltage source. On the other hand, it is convenient to replace the capacitor by its (parallel) complex-frequency domain current source equivalent circuit, because the desired output is the capacitor voltage. This results in a three-source, or multi-input, circuit. O nce the equivalent circuits are in place, one can combine the voltage sources and write a single node equation to find

V(As).

Using the voltage source model for the inductor and the current source ynodelfor the capaci tor, draw the equivalent complex-frequency domain circuit. Using the voltage source equivalent for

Step 1.

the initialized inductor and the current source equivalent for the capacitor produces the circuit o f Figure 13.20a. Combining the voltage sources and the series impedance into single terms results in the circuit shown in Figure 13.20b.

0.5 i^(O-) = 1 / Y Y V 1.50

_ Q

0.5 s CvJO-) = l

(a)

FIG U R E 1 3.20 (a) Complex-frequenc)’ domain equivalent accounting for initial conditions o f the circuit of Figure 13.19. (b) Circuit equivalent to part (a) with voltage sources combined.


Step 2.

Write a single node equation for V(^s).

623

Summing the currents leaving the top node o f

yields

1 V c ( s ) ---------------- l + 5V ’c ( 5 ) = 0

s

1.5 + 0.5.y Grouping terms produces

1

V c is )= -:

----- r+ 1 5 (0 .55'+ 1.5 )

+ 5

U . 5 + 0.5^

V(\s)

Solving for

leads to

s~ + 5s + 2

Vcis) = 5

( 5 + 0 ( 5 + 2)

Exemte a partialfraction expansion on V(^s), and take the inverse transform to obtain V(^t).

Step 3 .

Using the result o f step 2,

5^ + 55 + 2 1 2 -2 ^(7(5) = -------------------- = — I----------- f-

s{s +

1) ( 5 + 2 )

S

.V +

1

5

+ 2

Inverting this transform yields the desired time response,

v^t)

Exercises.

le-'-2e-^~û{i)V

1. In Example 13.7, change the resistance from 1.5 H to 2 .2 5

A N SW ER : 2. Find

= [1 +

v^p)

I^{s)

and

Find

V(\t)

for r > 0.

= [1 » 0.57l4i>-'^-^'- 0 .5 7 l4 f -^ q « (/) V

iît)

for the circuit o f Example 13.7, using the equivalent circuits o f Figures

13.14 and 13.15. Hint: W rite one loop equation. TV A N SW ER :

1,

EXA M PLE

1 3

( 5 ) = -------- ----------- . ( 5 + I ) ( 5 + 2)

i,{t) ^

=-

le -‘ii{t)

+

Ae--'u{t)

. 8 . T he chapter opened with a discussion o f the operation o f a fluorescent light

with classical starter, com m on in residential usage. For a fluorescent light to begin operating, there must be a sufficient supply o f free electrons in the tube and a sufficiently high voltage between the electrodes to allow arcing to occur. During arcing, mercury particles in the tube vaporize and give off ultraviolet light. The ultraviolet light excites a coating o f phosphorus on the inside o f the tube that emits light in the visible range. For a simplified analysis, assume that all resistances are negligible and refer to Figure 1 3 .2 1 . The source

VjJ^t)

is 120 V, 6 0 Hz, i.e., ordinary house voltage, which is too low to cause arcing inside

the fluorescent tube. Prior to arcing the gas inside the fluorescent tube acts like a very large resist ance betvN'een the rwo electrodes. W hen the switch is turned on, the starter, a neon bulb with a bimetallic switch inside, lights up and heats the bimetallic strip. This causes the metal to curl and


624

close the contacc. The bulb then looks like a short circuit, and a large current, limited by the inductive ballast, flows through the heating electrodes o f the fluorescent tube, making them bet ter able to emit electrons. During this time the neon bulb is shorted out and the bimetallic strip cools and opens the circuit after a few seconds. At this point in time, which we will call ^ = 0 , the inductor has an initial current

Because o f the Z.Ccombination, a very high voltage will then

appear across the electrodes of the lamp, resulting in ignition or arcing. After the lamp ignites, the voltage between the electrodes becomes “small” and is insufficient to relight the neon starter lamp. Hence, the ac current flows between the two electrodes inside the fluorescent tube. The ballast again serves to limit the current.

Heating

Direction of curl when

FIGIIRK 13.21 Wiring diagram of simple fluorescent light circuit, including an inductive ballast, a capacitor, and a starter within which is a neon bulb containing a bimctallic switch. Suppose

L = 0 .8

H , C = 1 nF, and

= 0.1 A. For r > 0, we find the com ponent o f

due

to the initial inductor current, i.e., the zero-input response. The other com ponent, the zero-state response, is not as important for ignition purposes. O ur strategy will be to use the ^-domain equiv alent circuit for

L,

as illustrated in Figure 13 .2 2 .

0 .8 s

Voltage due to

I,(S)

Li,(0) = 0.08

intial inductor current '....................................................

•

u- u •* High resistance prior to arcing

FIGUKI! 13.22 Equivalent complex-frequenc)' domain circuit immediately prior to arcing and normal lamp operation in fluorescent lighting.


625

Since we are assuming that all resistances are negligible and that the internal resistance (between electrodes) o f the fluorescent lamp prior to arcing approximates infinity, voltage division in terms o f impedances yields

— + Ls Cs

^

r + —

J l .25x10^

r

+ 1 .2 5 x 1 0

LC

J l . 2 5 x lo ' 2 ,8 2 8 ^ o . r + 1.25 X 10^ Hence, immediately prior to arcing, the capacitor voltage approximates = - 2 ,8 2 8 sin (3 5 ,3 5 5 /) V which is sufficiently high to induce arcing and cause the fluorescent lamp to operate.

See the homework exercises for an extension o f this analysis to the case where the ballast model includes a resistance o f 100

Q.

5. NOTION OF TRANSFER FUNCTION Besides impedances and admittances, other quantities such as voltage gains and current gains are critically im portant in amplifiers and other circuits. T he term transfer fu nction is a catchall phrase for the different ratios that might be o f interest in circuit analysis. Impedances and admit tances are special cases o f the transfer function concept.

TRANSFER FUNCTION Suppose a circuit has only one active independent source and only one designated response signal. Suppose fiirther that there is no internal stored energy at f = 0~. T he transfer func tion o f such a circuit or system is

H{ s) =Thus if the input

X fdesignated response signal

--------- r------ . ■ V

(13.20)

£ [designated input signal

and the response is^(f), then y(^) =

which is a handy for

mula for computing responses. Notice that if the input is the delta function, then and

Y{s)

=

H{s).

F{s)

= 1

This means that the transfer function is the Laplace transform o f the so-

called im pulse response o f the circuit, i.e., the response due to an impulse applied at the circuit input source when there are no initial conditions present. The idea is easily extend ed to multiple inputs and multiple outputs to form a transfer function matrix. This exten sion, however, is beyond the scope o f this text.


626

Exercise.

A transfer function o f a particular circuit is

H(5 ) =

response. Hint: Review Table 12.1.

—. Find the impulse (•5' + ^^)

A N SW ER : ^'-‘"[cos(Z^/) + S sin(Z^r)]/Hf)

A transfer function, as defined by equation 1 3 .2 0 , has broad applicability to electrical and electro-mechanical systems. For example, the designated output may be a torque while the input

netiuork driving point impedance, where the the current source; (ii) driving point

might be voltage. However, in the context o f circuits, a transfer function is often called a

function.

T he literature distinguishes four special cases: (i)

input is a current source and the output is the voltage across

admittance,

where the input is a voltage source and the output is the current leaving the voltage

source; (iii)

transfer impedance,

where the input is a current source and the voltage is across a des

ignated pair o f terminals; and (iv)

transfer admittance,

where the input is a voltage source and the

output is the current through another branch in the circuit. In cases (i) and (iii), the voltage polar ity must be consistent with the conventional labeling o f sources as set forth in Chapter 2. In gen eral, however, we will adopt the ordinary language o f transfer function. EXA M PLE

1 3

. 9 . T he circuit o f Figure 1 3 .2 3 has elements with zero initial conditions at f =

0 “ Find ôut

V :Js)

S

o l u t io n

There are many ways to solve this problem. O ur approach is to execute a source transformation on the

R-L

impedance in series with the voltage source. After the source transformation, we use

current division to obtain the necessary transfer function. Step 1.

Execute a source transformation to obtain three parallel branches as per Figure 13.24.

627


FIG U R E 1 3 .2 4 Circuit equivalent to Figure 13.23 after a source transformation. This circuit has the parallel structure o f Figure 13 .2 5 . ,(S)

V Js ) r A

©

Y,

sT T

FIG U R E 1 3 .2 5 Parallel admittance form of Figure 13.24. Step 2 . Use current divisiofi. Since the output current,

^ current through one o f three

parallel branches, the current division formula (equation 13.9) applies, producing

>3(^)

M l y,(A-)+r2(.s') + >3(-^'V 5 + 1 Hence,

H(s) =

Yîs)

(1 3 .2 1 )

U W + i 2 ( > ^ ) + W / s+ 1

Vinis)

Step 3 . Compute K, (^), Y2 {s), and Y^îs). Because impedances in series add, and admittance is the inverse o f impedance (equation 13 .7 ), some straightforward algebra yields

K,(.v) = ----- K2 (.v) = ------------. = ----- ^3(5) = S+\ 5+1 5

2.5s 0 .4 5 +

, 2^1

0.4

Step 4 . Substitute into equation 13.21 to obtain H{s):

2.5 s H(s) =

1 5+1

2.5s

1

5^ +1

2.55 + - —I— X---- \ 5 + 1/ 5

5+1

5

“ + 1 + 5 ( 5 “ + 1) +

s +l

2.5s

2.5s (5 + 1 ) ( 5 - +

2.5s +

1)

(s +

1)(^ + 0 . 5 ) ( 5 + 2 )

2

.5 5 ( 5 + 1)

(i2«

Exercise.


For

A N S W FR :

H{s)

as computed in Example 13.9, find the so-called impulse response

//{/) =

-

--C

-0 .5 /

-2 /

h{t)

=

u(t

3

E X A M P L E 1 3 .1 0 . C onstruct the transfer function o f the ideal operational amplifier circuit o f Figure 13 .2 6 , where

Zp)

and

Ip)

denote a feedback impedance and feedback current, respec

tively.

V,„(s)

FIG U R E 1 3 .2 6 Simple ideal operational amplifier circuit for Example 13.10. S

o l u t io n

Since no current enters the inputs o f an ideal op amp, I - p ) = - I p ) . Further, the voltage at the negative op amp terminal is driven to virtual ground; hence,

V-p)

=

Z -p )I-p ),

and

Vg^,f{s)

=

Z p ) I p ) . Combining these relationships with I - p ) = - I p ) yields

Vi„{s)

Zj„{s)

Yj{s)

(1 3 .2 2 )

Equation 1 3 .2 2 is a verv' handy formula for computing the transfer functions and responses o f many op amp circuits.

Exercises. Find R so

1. In the circuit o f Figure 1 3 .2 6 , suppose that the transfer function is

A N SW E R :

R=

H{s)

=

-Ms,

Z p)

is the impedance o f a 0.1 mF capacitor.

i.e., an inverting integrator.

1{) kQ

2. In the circuit o f Figure 1 3 .2 6 , now suppose a 0 .2 m F capacitor, and the transfer function,

Zis)

consists o f a 10 kQ resistor in parallel with

consists o f a 4 0 k ti resistor in parallel with a 0 .4 m F capacitor. Find

the dc gain, and the gain as x

A N SW E R : -{s + 0.5)/(2y + 0 .1 2S).

-O.S

oo.


629

3. Find the value o f C for which the transfer function o f the op amp circuit in Figure 1 3 .2 7 is

H {s) = ----------- ---------- . (s + 2)is + 4) A N SW ER : C = 0 .5 F

0.2 5 Q

FICJURE 1 3 .2 7 Op amp circuit.

E X A M P L E 1 3 .1 1 . T he ideal op amp circuit o f Figure 1 3 .2 8 is called a leaky ifjtegrator. If the input to the leaky in tegrator circu it is v-^^) = e~û{t), find the values o f /?,, an output response

Rj,

and Cleading to

= -lte~ ‘u{t), assuming that I'fjCO") = 0.

R,

FIG U R E 1 3 .2 8 Ideal operational amplifier circuit known as the leak)' integrator. S

o l u t io n

Step 1.

From the given data, compute the actual transferfunction o f the circuit.

By definition o f the

transfer function,

2

H{s) =

L[resp(mse\

Voui{s)

(s + l)“

2

£\input\

Vi„(s)

1 .y+1

5+1

( 1 3 .2 3 )

(i3 0


Step 2 . Using Figure 13.28, fin d the transferfiinction o f the circuit in terms o f R^, Rj, and C. Here, obsenê that Figure 1 3 .2 8 has the same topolog)^ as Figure 13 .2 6 , where

Ky.(,9)=— !— =

c j+

Z /(.v )

-^

w

'

R2

' «i

From equation 13.22 o f Example 13 .1 0 ,

Cs + Step 3 . Match coejfcients in equatiotis 13.23 and 13.24a to obtain the desired values ofR^, Rj, and

C. Equating the coefficients yields _1_

Cv+—

Rl O ne possible solution is

R^ =

0 ,5 Q, /?, = 1 Q, and C = 1 E If we rewrite equation 1 3 .24a as

H (s)= -

(1 3 .2 4 b )

CRj

other solutions are also possible. For example, for any =

> 0,

^2new ~

^new

represents a valid (theoretical) solution. In Chapter 14 we encounter a concept called

magnitude scaling.

is called a magnitude scale factor, which leaves this transfer function

unchanged but produces more realistic values for the circuit elements.

Exercises. 1. In equation 13.24b, it is required that C = 10 uE Find appropriate values o f /?j and R-^. AN SW ER:

R^ =

50

kLl R,

= 100

kLl

2. Given equation 13.24b , compute h{t) = L [//(^)].

A N SW ER :

_JL

t H,C

tl(l)


631

6. IMPULSE AND STEP RESPONSES H{s), with ^-domain input Y{s) = H W F{s). Assuming that

Suppose a circuit or system lias a transfer function representation denoted by

F{s)

and j-domain output given by y(y) in which case

all initial conditions arc zero, if /(f) =

^{t),

then the resulting^(r) is the system im pulse response.

Some simple calculations verify that the transform o f the impulse response is the transfer function, 1.e., X[y(r)] =

H {s)m t)]

=

His)

Hence, the impulse response o f the circuit/system, denoted /;(/“), is the inverse transform o f the transfer function

Ht) ‘ £-'[M(s)]

(13.25a)

H{s) = £[h{t)]

(1 3 .2 5 b )

and conversely

These equivalences represent another use o f the transfer function concept.

Exercises.

l.T h e transfer function o f a certain linear network is

H{s) = (s + 5)l[{s + 1)(j + 2)].

Find

the impulse response o f the network. A N SW E R :

[lc^‘ -

2. If the impulse response o f a circuit is a pulse ^(f) =

u{t) - u {t- T), T > 0,

compute the transfer

function. A N SW ER : (I 3. Suppose

- e^'^)/s

t/{t) = 2b{r-

ing an impulse response ANSXXHER: y(r) =

2hU -

1) - 3 6 (/‘- 3) is the input to a relaxed (zero initial conditions) circuit hav

h{t)

=

2u(t) - 2u{t—5).

Find the output ^(f).

1) - 3A(r - 3)

Why is the impulse response important? hs

we will see, it is because every linear circuit having con

stant parameter values for its elements can be represented in the time domain by its impulse response. This is shown in Chapter 15, where we define a mathematical operation called

tion

convolu

and show that the convolution o f the input function with the impulse response function

yields the zero-state circuit response. In addition to this significant theoretical result, the impulse response is im portant for identification o f linear circuits or systems having unknown constant parameters. Sometimes a transfer function is unavailable or a circuit diagram is lost. In such a predicament, measuring the impulse response on an oscilloscope as the derivative o f the step response is quite practical.

What is the step response o f a circuit? T he step response is merely the zero-state response o f the cir f^t) to the circuit is « (/), then F{s) = 1/^ and K(j) = H[s) (1/^). By the integration propert}’ o f the Laplace transform, it follows that the step response

cuit to a step function. Observe that if the input


032

is the integral o f the impulse response. Conversely, the derivative o f the step response is the impulse response. In lab, many scopes can display the derivative of a trace and hence can display the derivative o f the step response, which is the impulse response. Alternatively, a homework prob lem will suggest a means o f directly generating an approximate impulse response.

Exercises.

1. If the transfer function o f a circuit is

H{s)

= 1/j, what arc the impulse and step

responses? 2. If the Laplace transform o f the step response o f a circuit is given by

Y{s)

= I/fi'U + I)], what is

the impulse response? 3. If the step response o f a circuit

\sy{t)

= [1 - 0.5^’“ “^-

cos(2r)]//(r), what is the impulse

response?

u(t),

A N SW ER S: in random order: H/).

cus(2/ + 2 6 .5 7 ‘')//(r).

E X A M P L E 1 3 .1 2 . Figure 13.29a shows the impulse response o f a hypothetical circuit. If an input

= b{r)

+

b {r -

1), com pute the response,^(f).

y(t) A

h(t) 3

A 2

-

2

1

--

1 H-----

1

2

1

(a)

2

3

(b)

FIG URE 13.29 (a) Impulse response of hypothetical circuit, (b) Response to 6(r) +

b {t-

1).

S o lu t io n

Since X [6(r)

+ 6(/- 1)]

= I +

e~\ the

response,

is simply the sum

o f /}{[) and h (t- \)u {t-

1).

Doing the addition graphically yields the waveform o f Figure 13.25b.

E X A M P L E 1 3 .1 3 . The response o f a relaxed circuit to a scaled ra m p ,/(/) = = ( - 6 + 4r + 8 e~' S

Com pute the impulse response,

is given by;/(^)

h{t).

o l u t io n

T he relationship between/( /) and

b{t)

identifies the strateg)' o f the solution. If the step function

is the integral o f the delta function and the ramp the integral o f the step, then the delta function equals the second derivative o f the ramp. Hence, the impulse response

h{t)

y\t)

b{t)

= 0 .1 2 5 /" (r ). By the linearity o f the circuit,

= 0.1 2 5 7 "(^ ), and some straightforward calculations produce

= [4 -

+ [ - 6 + 4 / + 8 ^ * -'-

(t)

Chapter 13 • Laplace Transform Analysis 11: Circuit Applications

y"{t)

But the right-hand term is zero. (Why?) Hence,

633

= [Sf*

and

hit) = [e-^-

To see the utility o f this approach, try the alternative method o f computing

V(s)/f(s).

F(s), V(s),

and

H(s) =

The algebra is straightforward, but tedious and prone to error.

As a final example, we compute a circuits step response and verify that its derivative is the impulse response. E X A M P L E 1 3 .1 4 . C om pute the step response o f the

RLC circuit

o f Figure 13 .3 0 .

/m R = 40 v

. »

+

L=1H

Q

C = 0 .2 F

-o FIG U R E 13.30 S

circuit for Example 13.14.

o l u t io n

From voltage division.

1

5

s~ -\ --s+ — L LC

f.v + 2 ) “ + l

1

Cs /^ + L .V + —

Cs

(1 3 .2 6 )

From equation 1 3 .2 6 , the Laplace transform o f the step response is

His)

1

5

= - +—

.y (.y + 2 ) “ + l l

-.V - 4

(.v + 2)^ + l

Rearranging terms yields

.V

(5 + 2 ) - + 1

{s + 2)-+\

Taking the inverse transform produces the desired step response: = [1 -

cos(f) -

2e-~‘ sin(f)];K r)

(1 3 .2 7 )

fi3^

Chapter 13 • Laplacc IVanstorm Analysis II: Circuit Applicarions

As a check, obser\'e that the derivative o f equation 1 3 .2 7 is

—

, , ( / ) = 2 e ""^ [co s(/) + 2 s i n ( 0 k ( 0 -

(It

= 5e

+ 2 c o s(/)]« /(/) + (l - 1)(5(/)

sin{t)u{t)

Thus 5

£

Cv + 2 ) - + 1 in which case

h{{) =

5^’" “^sin(r)«(f)

as expected.

6. NODAL AND LOOP ANALYSIS IN THE S-DOMAIN This section develops ^-domain formulations of node and loop analysis. Nodal analysis o f circuits builds around KCL, whereas mesh/loop analysis utilizes KVL. In Chapter 3 and, indeed, in most beginning courses on circuits, loop and nodal analysis are taught first in the context o f resistanc es and conductances and then (in Chapter 10 here) in the phasor context. Recall that KCL requires that the sum o f the currents leaving any circuit node be zero. Further, KVL requires that the voltages around any loop of a circuit sum to zero. By linearity, the Laplace transform o f a sum is the sum o f the individual Laplace transforms. Hence, a KVL equation and a KCL equation have an j-domain formulation where elements are characterized by impedances and/or admittances. For loop analysis, one writes a KVL equation for each loop in terms o f the transformed loop cur rents and element impedances. The set o f all such equations, then, characterizes the circuit’s loop currents, which determine the ciu rents through each o f the elements. Knowledge o f the loop cur rents and the element impedances permits the computation o f any o f the element voltages. In nodal analysis, one writes a KCL equation at each node in terms o f the Laplace transform o f the node voltages with respect to a reference, the transform of the independent excitations, and the element admittances. The set o f all such equations characterizes the node voltages o f the cir cuit in the ^-domain. Solving the set o f circuit node equations yields the set o f transformed node voltages. Knowledge o f these permits the computation o f any o f the element voltages. W ith knowledge of the element admittances, one may com pute all o f the element currents. Since nodal analysis has a more extensive application than loop analysis, our focus will be on nodal analysis.

E X A M P L E 1 3 .1 5 . Figure 13.31 shows an ideal operational amplifier circuit called the

Key normalized low-pass Butterworth filter.

Sallen and

(See Chapter 19 for a full discussion o f filters.) A nor

malized low-pass filter passes frequencies below 1 rad/sec and attenuates higher frequencies. As we will see later in the text, the 1-rad/sec frequenc)' “cu toff” can be changed to any desired value by frequency-scaling the parameter values o f the circuit. (See Chapter 14 for a discussion o f frequen-


63 S

cy scaling.) The goal here is to utilize the techniques o f nodal analysis to compute the (normal ized) transfer function o f this circuit.

FIG U R E 13.31 Sallen and Key normalized Butterworth low-pass filter circuit containing an ideal operational amplifier. S

o l u t io n

T he solution proceeds in several steps that utilize nodal analysis techniques in conjunction with the properties o f an ideal op amp. Recall that for an ideal op amp, the voltage across the input ter minals is zero and the current into any o f the input terminals is also zero. Finally, note that one does not write a node equation at the output, which appears across a dependent voltage source whose value depends on other voltages in the circuit.

Step 1. Find Vy. Because the voltage across the input terminals o f an ideal operational amplifier is zero.

Step 2. 'Write a node equation at the node identified by the node voltage V^. Summing the currents leaving the node yields

(Va -Vi„) + (V a-V t) + V2.v(l/, Substituting

for

- V„,„) = 0

and grouping like terms produces

(■J2.S + 2 ) V „ - ( J 2 S + \)V,„„=V,„

(13.28)

Step 3. Write a node equation at the node identified by the node voltage Vy. By inspection, the desired node equation is

5+1

(1 3 .2 9 )

Step 4 . Write the foregoittg tivo node equations in matrix form. In matrix form, equations 1 3 .2 8 and 13.29 combine to give

- (V 2 . + I)' ■

-1

-|=5+ 1 IV 2 J

K;

■

'în 0

(1 3 .3 0 )


6.U>

Step 5. Solve equation 13.30 for

in terms ofV-^^ using Cramer's rule. From Cram ers rule,

del

'{ ■ J lS + l )

V;„

-1

0

(>/ 2 :i- +

2

)

-(V

2

.V + 1 )

del

-1

\42

The resulting transfer function is

K..

[.j2 s + 2 )lj^ s+ \ \ -[y l2 s+ \ ) IV2

s- + yf2s+\

N otice that for small values o f ^ = yto (i.e., low frequencies), the magnitude o f H{s) approximates 1, and for large values o f s = JiO (i.e., high frequencies, where |/b)| » small. Since

1), the magnitude o f H(s) is

= Myo)) ^^^(yco), such a circuit blocks high-frequency input excitations and

passes low-frequency input excitations. As mentioned at the beginning o f the example, the circuit passes low frequencies and attenuates high frequencies.

T he preceding example used matrix notation, com m on to much o f advanced circuit analysis. In one sense, matrix notation is a shorthand way of writing n simultaneous equations: the n variables are written only once. More generally, matrix notation and the associated matrix arithmetic allow engineers to handle and solve large numbers o f equations in numerically efficient ways. Further, the theory o f matrices allows one to develop insights into large circuits that would otherwise remain hidden. Hence, many o f the examples that follow will utilize the elementary properties o f matrix arithmetic. T he next example uses nodal analysis to compute the response to an initialized circuit. The exam ple combines the equivalent circuits for initialized capacitors and inductors with the technique o f nodal analysis.

E X A M P L E 1 3 .1 6 . In the circuit o f Figure 13.3 2, suppose /y^^(r) = 6(r), /^(O") = 1 A, and = 1 V. Find the voltages

V(^t)

FIG U R E

and

v^Qi~)

vît).

1 3 . 3 2

Two-node /?Z.Ccircuit for Example 13.16.

Given the indicated current direction of i[{t), what is the implied voltage polarit)' for

v^{t)}


63'

Step 1. Draw the s-domain ecjuivaletit circuit with an eye toward nodal analysis. Inserting the equiv alent current source models for the initialized capacitor and inductor in Figure 1 3 .3 0 , one obtains the 5-domain equivalent circuit shown in Figure 13 .3 3 .

V Js)

VJs)

1o

ii(O-)

,Cv,(0 )

O 1o

_

= 1

1F

1H

FIG U R E 1 3.33 5-Domain equivalent of the circuit of Figure 13.31. Step 2 . Write two node equations and put in matrix fonn. At the node labeled

V(is)

K C L implies

that (1 +

s)V(is)

+

[V(is)

-

Vj{s)] = 2

Simplifying produces the first node equation:

{s^ 2 )V ^ s)-V i{s) = l - V(is)] + {\ls)Vj{s) = - (1/i), or, equivalently,

At the node labeled

.9+1 -V c(^ ) + —

1 = —

S

5

T he matrix form o f these uvo node equations is .v + 2

,

2 1

-1

_i

.9+1

s Step 3 . Solve the matiix equation o f step 2 for the desired voltages. Using C ram ers rule, computing the inverse, or simultaneously solving the equations gives

2{s+\)-\ 2

Vcis)-

5“+ 2 . 9 + 2

[5 +1

1 1 ■2 ■

s 1

5+2

(s+Vr + \ (1 3 .3 1 )

_i

s

C v + D -3 (.V+ 1r +1

Step 4 , Take the inverse Laplace tratisfonn to obtain time domain voltages. Breaking up equation 13.31 into its components yields

Chapter 13 • L iplacc Transform Analysis II: Circuit Applications

638

2(^ + 1)________ 1

Vcis) =

(5 + 1)- + !

(5+ 1)^ + 1

in which case

V(^t)

=

e ^[2 cos(t)

- sin(/)]/^(r)

Also,

(.^ + 1)

3

(A-+1)^ + 1

(.V+1)^ + 1

V^(s) = leading to

=e Figure 13.34 presents plots o f

V(\t)

^[cos(^) - 3sin(f)]«(r)

and

H G U R E 13.34 Plots of the capacitor and inductor voltages for Example 13.16.

Dual to nodal analysis is loop analysis. In loop analysis, one defines loop currents and writes KVL equations in terms o f these loop currents. The following example illustrates the method o f loop analysis for computing the input impedance o f a bridged-T network.


639

E X A M P L E 1 3 .1 7 . Use loop analysis to com pute the input Impedance o f the bridged-T network illustrated in Figure 13 .3 5 .

/YY\ 1 H 2Q

20

20 0.25 F Z Js)

FIG U R E 1 3 .3 5 Bridged-T network for Example 13.17. S

o l u t io n

Define find

= / , U) in Figure 1 3 .3 4 . Since

I-^j{s)

Step 1.

the goal is to use loop analysis to

= /j(^) in terms o f

Smn the voltages around loop 1.

Dropping the specific j-dependence for convenience, one

obtains

2(/, - / 3 ) + - ( / , - / 2 ) = 2 —

s

Step 2.

Sum the voltages around loop 2.

- - / 2 - 2 /3 =

s

s

By inspection,

- ( / 2 - / | ) + 2( / , - / 3 ) + 2 / 2 = - - / , + 4 — .V

Step 3 .

s

“

Sum the voltages around loop 3.

Again by inspection,

2 (73- 1{) + sl^ + 2(73 - I^) = - 2/1 - 11^ + Step 4 .

/ 2 - 2 /3 = 0

s

Put the three loop equations in matrix form, and solve.

-K4)73 = 0

In matrix form, the three loop equa

tions are

,£ ± 2

.1 /l1

.1

s

-2

4—

s

-2

-2

5+ 4

h

0 0

Chapter 13 * I.aplacc Transform Analysis II: Circuit Applications

(viO

Using Cram ers rule to solve this equation for

in terms o f

= / ] W yields

.v“ + 4.V + 4

=2 Q

7

It may be a little surprising that

is independent of

s, despite

the appearance o f an inductance

and a capacitance inside the nerwork. Such a nerwork is called a

constant-resistance network.

A

problem at the end o f the chapter shows the conditions on the elements o f a bridged-T network for it to be a constant-resistance nerwork.

Exercise.

The loop equations o f the circuit in Figure 1 3 .3 6 are T + 25

0

-2 s

1

\h]

-Vs:

0

4 2.V + - + 2

_2

-1

h

0

-2 i'

_2

25 + 4

0

0

-I

1

-4

0

h V

s

0

Find the values o f C and A'in Figure 13.36. ANSWHRS: c ;= 0.2^ !• A’ -

M GURE 13.36

7. SWITCHING IN /?iLC CIRCUITS Switches control lighting systems, furnaces, car ignitions, traffic lights, and numerous other devices. Switching also takes place inside electronic circuits, as in switched power supplies and switched capacitor filters. All have a functional element called a

switch

that affects and, indeed,

shapes the behavior o f the circuit. T he switch inside an electronic circuit is a special device that we will model simply as an ideal on /off switch. This section investigates the behavior of switching


in simple

RLC circuits,

64

as preparation for the understanding o f switching in more elaborate and

sophisticated electronic circuits. O ur immediate task is to apply the Laplace transform method to com pute the responses of switched

RLC circuits.

The following example motivates a general pro

cedure.

EXA M PLE 1 3 .1 8 . In the circuit o f Figure 1 3 .3 7 , suppose

u{t),

= 0.

= 20 Q ,

R = A Q., C = 0 .2 5

F, and

The switch is initially in position A. T he switch S moves from

position A to position B at / = 1 sec and from position B to position A at ^ = 2 sec, and moves

V(^t)

back to B at / = 4 sec where it remains for all subsequent time. Find

for

t>

0.

+

R

v,(t) v jt)

FIG U R E 1 3 .3 7 Switched /eCcircuit {R^ = 20 Q, R = A Q . , C = 0.25 F) for Example 13.18 in which

V(^Q~) = 0;

the switch S moves to B at ^ = 1 sec, returns to A at ^ = 2 sec, and moves back to B at r = 4 sec, where it remains.

S

o l u t io n

Because o f the switching at r = 1 sec, the first step is to determine f < 1, This allows us, in turn, to find

V(-{t) over

the time interval 0 s

which will serve as the initial condition over the time

interval 1 s r < 2. This then produces

(2“), the initial condition for the interval

2<. t
etc.

Step 1. Compute the response for 0 ^ t < 1. Over the interval 0 :s r < 1 , the circuit o f Figure 13.37 is the simple

RC circuit

o f Figure 13.38.

20 0

+ V,(s) 0.2 5 F

FIG U R E 13.38 Equivalent circuit of Figure 13.37 over 0 ^

< 1.

For the circuit o f Figure 1 3 .3 7 , voltage division implies

4

2

s 20

20

(•v + 0.2)(.? + 0 . 1 ) " .v + 0.1 ~ .v + 0 .2

+ -

.V

Hence, for 0 ^ r < 1,

v^t)

=

20e-^-^^-20e~^-^

20


642

W e note that Step

2

.

(1") =

-

2 0

^0 . 1 ^ j

7 2 2

Coynpute the response over 1 <. t < 2.

V.

After the switch moves ft-om position A to position B,

the source is decoupled from the right half o f the circuit; the response then depends only on the initial condition at r = 1 “ , i.e., ^ f<

2

= 1 .7 2 2

, or equivalently, over the inten'al

0

V. T he goal is to compute

s r' <

1

where

t' = t - \. The

v^^t)

over the interval

1

^-domain equivalent cir

cuit that models the behavior o f the time domain circuit o f Figure 1 3 .3 7 over

\ ^ t < 2 has the Cvq (1“) =

form illustrated in Figure 13 .3 9 . We note that the value on the current source is 0 .4 3 0 5 .

FIG U R E

1 3 . 3 9

Equivalent circuit of Figure 13.37 for the time interval 1

^ t <2.

T he equivalent admittance seen by the current source is 0.25^ + 0 .2 5 . Hence by O hm s law,

Vc(.s) = For 0 s ^' <

1

4

s +1

1 72^ X 0 .4 3 0 5 3 = — ^ .V +

1

, taking the inverse transform yields =

or equivalently, for

1

^ r<

2

\.722e-^'ti{t)

,

V(^t) = \.722e-^^-^\i{t- 1) W e emphasize that this last equation is valid only for

1

:£ r <

2

. We note that

(2') = 1 .7 2 2 e ‘ ^ =

0 .6 3 3 5 . Step 3. Compute v(it) fo r 2 ^ t <4. For this interv'al, the capacitor is initialized at /' = t - 2 = 0. Using the parallel current-source equivalent circuit, we note that

(r')

0 .1584, as shown in Figure 13.40. We further note that iV„(r)= vv„(r'+2) = ? ;„ ( ;')= Thus

Vjfjis)

= 8 .l 8 7 .V - ° '''« ( r ')

in Figure 1 3 .4 0 is 8 .1 8 7 3

5 + 0.1

q

=

(2') =

Chapter 13 • Laplacc Transform Analysis II; Circuit Applications

643

FIG U RE 1 3 .4 0 Equivalent j-domain circuit valid for 2 s r < 4. Using superposition on the circuit o f Figure 13 .4 0 , 4 0 .1 5 8 4

0 . 2 x 8 .1 8 7 3

0.25.S + 0 .0 5

U + 0 .2 )(5 + 0 .1 )

____s 20

+ -

0 .6 3 3 5

+•

5 + 0 .2

.V

Hence, for 0 s t'< 2,

= l6 .3 7 4 6 ^ -^ -^ ''-1 5 .7 4 ll^ -0 " ^ ' or equivalently for 2 < f < 4,

v^t) = l6.3 7 4 6 < '-0 -i(^ -2 )-1 5 .7 4 llf> -0 -2 (^ -“) W e note that

v^4~) = \63746e-^ ~ -

1 5 . 7 4 1 =

2 .8 5 4 8 V.

Step 4 . Compute the response over 4 ^ t. After the switch moves from position A to position B, again, the source is again decoupled from the right half o f the circuit. According to Chapter 8, we can write the answer by inspection: the solution is simply v^^t') = 2.8548^ “^ n{t’). Equivalently, f o r / > 4 , y J r ) = 2 .8 5 4 8 f - ( '- ^ \ Step 5. Combine results into a single expression and plot. I'h e combined expressions for V({i) arc

Os/
\^t < 2

Chapter 13 * l^placc Transform Analysis II: Circuit Applications

()-i4

FIGURK 13.41 Plot of capacitor voltage for switched circuit of Figure 13.37.

T he extension o f this method to more than three switching times is straightfoi^vard. Although the preceding example uses an

RLC circuit,

the same strategy or algorithm is applicable to the calcu

lation o f switching transients in any linear dynamic circuit. T he following is a summary o f the general procedure. P rocedure for Applying Lapiace Transform M ethod to Switched C ircuits At / = 0, a switching may or may not occur. For / > 0, denote the switching time instants

sively as

, r-, ,

The goal is to com pute the response

Divide the time axis into intervals (0 ,rj),

successively for

fo r

succes

0 s r < sc.

^ ^ = co), and com pute the responses

each time interval.

P art I. Consider the first time interv'al, (O.fj).

Step I. C onstruct t = 0~, which are

the ^-domain equivalent circuit, making use o f the initial conditions at either given or calculated from the past history o f the circuit for ; < 0

(see note at the end o f the procedure). This equivalent circuit is valid for the time inter val 0 ^ r < / j .

Step 2. Step 3.

Find the response by the Laplace transform method for the circuit o f step 1. Evaluate the capacitor voltages and inductor currents at

t = T j.

Part II. For all remaining time intervals:

Step 4. Set the value of subscript / = 1. Step 5. (1) For t^^ t< t^^^, let t' = t - t- (i.e.,

consider the time interv'al 0 s /' < oo), and

construct the ^-domain equivalent circuit, making use o f the conditions at /'= 0 “ (i.e., at r =

t~).

calculated

initial

60

Chapter 13 * Liplace Iransform Analysis II: Circuit Applications

(2) Determine the proper form o f the input excitation(s) (if there are any), in terms of t ’. (3) I'incl the response by the Laplace transform method. Note that the time vari able is t' . Then obtain the solution in

t by

substituting { t - t'^ for t

.

(4) If / = w, stop. Otherwise, eN-aluate the capacitor voltages and inductor currents at (5) Increase the subscript value / by 1 and go to the beginning o f step 5. Note: In some situations, the first switching occurs at r = 0, but the capacitor voltages and induc tor currents at r = 0 “ are not given. Instead, the problem specifies that dc and sinusoidal sources have excited the circuit for a long time. If the network is passive— i.e., if it consists o f inductors, capacitors, or

lossy elcm ents-then

the circuit will have rcached a steady state at

t = 0 “ . The

proce

dure then is first to find the steady-state solution and then to evaluate the capacitor voltages and inductor currents at / = 0 “ . It is instructive to review the dc and sinusoidal steady-state (phasor) analysis methods studied in a first course. Recall that under certain stabilit)' conditions (to be stud ied in Chapter 15) on the network: 1.

For dc steady-state analysis, open-circuit all capacitances and short-circuit all inductors to find the steady-state voltages and currents.

2.

For sinusoidal steady-state analysis, use the phasor method to find the steady-state responses.

8. SWITCHED CAPACITOR CIRCUITS AND CONSERVATION OF CHARGE In addition to its many uses already described, the Laplace transform method is applicable to a special class o f circuits called

switched capacitor (abbreviated

SC) networks. These circuits contain

only capacitors, switches, independent voltage sources, and possibly some operational amplifiers. No resistors or inductors are present. One can dispense with resistors because it is possible to approximate the effect o f a resistor with rwo switches and a capacitor. Similarly, inductors can be approximated by circuits containing only switches, capacitors, and operational amplifiers. These facts, coupled with the easy and relatively inexpensive fabrication ol switches, capacitors, and operational amplifiers in M OS (metal-oxide semiconductor) technolog}^ have made switched capacitor filters an attractive alternative to classical filters. Given this scenario, the purpose of this section is to lay a foundation (i.e., introduce the principles) upon which switched capacitor cir cuit design builds. More advanced courses delve into the actual analysis and design o f real-world switched capacitor circuits. Besides the Laplace transform approach, an alternative method for analyzing switched capacitor networks builds on the principle o f conserv'ation o f charge.

Principle o f conservation o f charge: T he total charge transferred into a junction (or our o f a junction) o f a circuit at any time is zero.


646

This principle is a direct consequence o f KirchhofFs current law. For example, in Figure 1 3 .4 2 , KCL implies that

(13.32)

/■jU) + ijit) + i^{t) + i^{t) = 0

I,

l-IGUlUi 13.42 Node to which KCL applies. Since charge is the integral o f current over a time interval, integrating both sides o f equation 1 3 .3 2 from -00 to

t yields f —

li,(r) + i 2 ( 0 + h ( 0 + U U m = 0 X

or equivalently.

î(^) + ^2^^) + ^3(^) + ^4(r) = 0 which is just another expression o f the principle o f conservation o f charge. A simple switched capacitor circuit will now serve as a test bed for comparing the merits o f the foregoing analysis techniques.

EXA M PLE 1 3 .1 9 . Consider the circuit shown in Figure 13.43a. The switch S is closed at / = 0. Just before the closing o f S, the initial conditions are known to be C om pute the voltages

and

= 1 V and

for ^ > 0.

i(t) “•

^ -----

+

+

Vc, -

IF r

IF

-----

+ Vc, ^

C2

IF

IF ---- • -----

(a) t < 0

(b) t > 0

= 0.

Chapter 13 • Ijp la c e Transform Analysis II: Circuit Applications

64"

l(s)

V,C2

(C )

FIG U R E 13.43 Equivalent circuits for Example 13.19. S O L U T IO N M ethod 1. Using the parallel current-source model o f a capacitor, we obtain the ^-domain equiv alent circuit shown in Figure 1 3.43c. By inspection, 0 .5

s Therefore,

=

V(y{t) = 0.5u{t)

V, and

i{t)

s s+ s

s

= 0.56(^) A.

M ethod 2 . Using the series voltage-source model o f a capacitor, we have the ^-domain equivalent circuit shown in Figure 1 3.43d . Again, by inspection.

1

1

i + i ■' S

■'

i+ i

S

This is the same answer as obtained with method 1. M ethod 3 . Clearly,

Conservation-of-charge approach.

For r > 0 , the network is shown in Figure 13.43b .

After S is closed, some charge is transferred from C, to C2 . However,

according to the principle o f conservation o f charge, the

total charge

transferred out o f the junc

tion must be zero. Hence,

and

Solving these two equations for the two unknowns,

Vci(() ) - V c 2 ( 0

and

results in

) - ------------7 -------------------C] + C 2

Since there is no external input applied, the voltages remain constant once the equilibrium con dition has been reached. Therefore,

Clhapccr 13 * I.aplacc Transform Analysis il; Circuit Applications

(vi.S

for r > 0. For the specific capacitance values given in Figure 1 3.43a, we obtain

=

0 .5 V for r > 0.

Exercise. In the circuit o f Figure 13.43a, let C, = 4 F and C-, = 6 F,

) = 8 V, and v a m

-

3 V. Find, by at least two methods, the capacitor voltages after the switch is closed. (Rework the problem if the answers do not agree.) A N SW ER : S \'

Computationally, the Laplace transform method is more straightforward. On the other hand, the conservation-of-charge method is more basic and often provides better insight into what happens to the charges stored in various capacitors. It is particularly useful for the purpose o f checking answers obtained by other methods: the answers are correct when the conser\'ation-of-charge con dition is met at every node. E X A M P L E 1 3 .2 0 . The initial conditions at r = 0 “ o f an SC network are shown in Figure 13 .4 4 . Switches Sj and S-, are closed at / = 0, connecting the tw'o dc voltage sources to the nersvork. Find the node voltages for

t>Q.

lO V

FIG U R E 1 3 .4 4 Switched capacitor nervvork for Example 13.20 at r = 0 . S

o l u t io n

We first construct the.f-domain equivalent circuit using admittances. The result is shown in Figure 13.45.

fO

4

15

4s

3s 2s

12

+ A IQ

s

FIG URE 1 3.45 Equivalent circuit to Figure 13.44, accounting lor initial condiiions; the admittance ol each capacitor is indicated.

Chapter 13 * l^placc Transform Analysis II: Circuit Applications

6 i9

Applying KCL to node 3 generates the node equation

5

4s

from which wc derive verify that

= 9/s and

+ 3.V

10

= 9 V for

+ I s Vt. = 4 + 1 5 + 12

r > 0. Obviously, Kj = 5 V and K2 =10 V. To

= 9 V is indeed the correct solution, we check for conservation o f charge. From

t=

0~ to A= O'*’, tiic voltage of the 4 F capacitor jumps from 1 V to 4 V (note that 4 = 9 - 5 ) , indi cating that 4 X (4 - 1) = 12 coulombs o f charge have been transferred to this capacitor. The volt age o f the 2 F capacitor jumps from 6 V to 9 V, indicating that 2 x (9 - 6) = 6 coulombs o f charge have been transferred here. Finally, the voltage o f the 3 F capacitor changes from 5 V to - 1 V (note that - 1 = 9 - 1 0 ) , indicating that 3 x (-1 — 5) = - 1 8 coulombs of charge have been transferred to this capacitor. As a check for conservation o f charge, we have 12 + 6 + ( - 1 8 ) = 0, and the solution is assured to be correct.

Exercise.

Solve Example 1 3 .2 0 again, with all capacitors initially uncharged.

A N SW ER :

v^(t)

= 5 0 /‘J V, / > 0

T he preceding examples considered idealized circuits, i.e., no resistances were present. In zny prac

tical circuit, the connecting wires have some resistance. W hat is our interest in the analysis of an idealized circuit? T he analysis o f an idealized circuit is much more straightforward than that of a realistic circuit yet provides relatively accurate answers. As a case in point, reconsider Example 1 3

.2

0

. Suppose we insert a 0.1 Q resistance in series with ever)' capacitor. The resulting transform

analysis would produce a rational function with a cubic denominator polynomial (a third-order network) whose factorization would require the use of a root-finding program. In sharp contrast, the idealized circuit o f Example 13.20 was anaK'zed by writing a single first-order node equation, making a partial fraction expansion unnecessary. Idealizations o f circuit models sometimes lead to phenomena that defy intuitive explanations. An interesting case is given by Example 13.20. Before S is closed, the energ)' stored in the electric field is

0 .5

X 1^ + 0 .5 x 0 “ = 0.5 joule. After S is closed, the stored energ)' becomes 0 .5 x (0 .5 )“ + 0.5 x

(0 .5 )“ =

0 .2 5

joule. Apparently, 0 .2 5 joule o f energy has been lost. Since there is no resistance in the

circuit to dissipate the energ)', what accounts for the lost energ}'? Is energ)' not conser\'ed? An explanation o f this paradox is as follows. Instead o f considering a zero-resistance circuit, place a resistance R in series with all capacitances. Then analrze the circuit, and let R approach zero. The result shows that no matter what value R takes on, the total energ)' dissipated in the resistance for 0 < / < ^ exactly equals the difference o f the total stored energies before and after the closing o f the switch. This accounts for the apparent lost energ)'. In actuality, part o f the 0 .2 5 joule of cnerg)’ would be lost in the form o f radiated energ)'. However, the principles of field theor)- would be necessar)' to explain the radiation phenomenon. Note that for idealized SC circuits, as long as the independent voltage sources arc piecewise con stant, all capacitor voltages are piecewise constant and all currents in the circuit are impulses. These


650

properties remain valid for more general idealized SC circuits that allow the inclusion o f VCVSs, C C C Ss, and ideal op amps. T he reason is that the parameters characterizing these components are

instantaneous. lossy switched capacitor circuit,

dimensionless and hence do not result in a time constant. All voltage changes are On the other hand, if the circuit contains resistances, we have a

whose voltages arc no longer piecewise constant. The transient analysis o f a lossy SC circuit requires the usual Laplace transform analysis.

approximate a approximated by an

One reason for our interest in SC circuits is that a SC combination can be used to resistor. As a result, any

RC-op

amp circuit used for signal processing can be

S C -o p amp circuit. A study o f the general theor)' o f such S C -o p amp circuits is beyond the level of this book. We shall merely illustrate the approximation property with a simple integrator circuit.

E X A M P L E 13.21. Consider the RC-op amp integrator circuit shown in Figure 13.46. R = 5kQ

C,= 1 mF

FIG U R E 13.46 Op amp integrator of Example 13.21. Using the result o f Example 13.5 and the integration property o f the Laplace transform, it is straightforward to show that

If the input is a constant voltage function

vj,t)

v{ = £

= 5 V, and if

= 0, then the output waveform is a ramp

= - l ,0 0 0 ///( f ) (as long as the output has not reached the saturation level), as shown

in Figure 13.47.

t (msec) -

10-

F I G U I I E 1 3 .4 7 R am p o u tp u t o f th e in teg rato r in F igu re 1 3 .4 6 .


6SI

W h at we would like to do is construct an SC approximation to this circuit. T he idea is to replace the resistor with a switch and a capacitor as in Figure 13.48.

F-IGLIRF, 13.48 Switched capacitor equivalent of Figure 13.46. The switch S is operated in the following manner: 1.

At r = 0, S is at position a.

2.

At t = T, S is

3.

At r = 27", S is moved to position a.

4.

At / = 37", S is moved to position b, etc.

moved to position b.

T he output waveform may be determined ver)' easily by the principle o f conser\'ation o f charge as follows: for 0 :s / <

T,

y.

= £,(]= CE, qj-= 0,

and

= 0. At / = f , switch S is moved to position

b. Because the op amp is assumed to be ideal, the voltage across the input terminals is zero, and so is v^. Thus, C cannot store any charge. T he charge CE previously stored on C must be trans ferred out o f C. Since the op amp is ideal, the input impedance is infinity and the input current is zero. Therefore, none o f the charge can flow into the op amp. Instead, the charge must be trans ferred to the capacitor

This leads to

CE and Vgi'T') = - C £/C ^

At r = 27', switch S is moved back to position a, causing C to be charged to E volts again. Since the charge

“trapped” on

Cj-, the output voltage

remains unchanged until S is moved to posi

tion b again. At that time, another CE coulombs o f charge are transferred to Cp and

is incre

mented by -C EIC p Subsequent switching is similar. To make the output waveforms o f the circuits o f Figures 1 3 .4 6 and 13.48 the

average o f the

approximately the same,

charges transferred to (y m u s t be the same in both cases. For Figure 1 3 .4 6 , the

current flowing into Cîs at a constant value o f E/R. Therefore, every I T sec, the charge trans ferred to Cîs equal to 2TEIR. On the other hand, for Figure 1 3 .4 8 , the charge transferred to Cjevery I T sec is CE. Equating these rwo quantities, we have CE = 2TEIR, or RC = IT . Thus, there is no unique combination o f C and T that will produce the approximate effect o f a resistance. A smaller T (i.e., a higher operating frequency o f the switch) in Figure 13.48 produces a staircase output waveform that closely “hugs” the ramp output o f Figure 1 3 .4 7 . For the purpose o f com parison, the output waveform corresponding to 7 = 1 msec and C = 0 .4 pF is shown in Figure 1 3 .4 9 , together with the ramp output from the RC-op amp integrator. It is worthwhile to note

Chapter 13 • Laplace Transform Analysis

652

II: Circuit Applications

that each o f the circuits o f Figures 1 3 .4 6 and 1 3.48 drains the same average amount o f charge from the voltage source and puts the same average amount of charge on the capacitor difference is that in the former the process is

coutinuous,

Cj; The

only

whereas in the latter the process occurs

in quantized steps.

2 3 4 5 6 7 8 9 19 -I— I— I— I— I— I— I— I— I— It (msec) Output of switched capacitor circuit

FICURU 1 3 .4 9 Responses of circuits in Figures 13.46 and 13.48.

Exercise.

Plot the output waveform of the SC circuit o f Figure 1 3 .4 8 for 0 s

^ 2 msec if 7'= 0.1

msec and C = 0 .0 4 pF. Also, plot the ramp output on the same graph for comparison.

As recently as two decades ago, switched capacitor circuits were considered impractical. No longer! Advances in semiconductor technology have made the fabrication o f high-speed electronic switch es and op amps as cheap as resistors. Furthermore, large numbers of switches, capacitors, and op amps can be fabricated on a single chip. Consequently, switched capacitor circuits hold an impor tant place in future signal-processing applications. Although we cannot delve into the practical aspects o f the design of such circuits, we have at least outlined the basic principles needed for their approximate or exact analysis.

9. THE DESIGN OF GENERAL SUMMING INTEGRATORS In the design o f active filters and in the design of data acquisition equipment, general summing integrators play an important role. This section takes up the design o f such op amp circuits from the transfer function perspective. However, to simplify the presentation, we consider only multi input transfer functions o f the form

*oiit ( '0 -

(1 3 .3 3 )

s

s

s

s


where the a- and

are positive gain constants and

653

....

are ^-domain inputs. This

expression represents the ^-domain equivalent o f the following time domain equation assuming zero initial conditions at ; = 0 “ on the variables: 'W

( 0

=

- « 1 j j j 'V/1

+ ^

2 /,2(r)ciT /^ V

W ith a little cleverness, it is possible to design by inspection a general integrating operational amplifier circuit whose input-output characteristic is precisely equation 1 3 .3 3 . T he four-input op amp circuit o f Figure 1 3 .5 0 accomplishes this. The dashed lines are present because the admit tance may or may not be needed. Computation o f the values o f

and

are explained in

design step 2 below.

Design Choices for the General Summing Circuit o f Figure 13.50 The first two design steps constitute a preliminary or protorvpe design, meaning that the two capacitors are normalized to 1 F. After completing the prototype design, an engineer would scale the capacitances and resistances to more practical values without changing the gain characteristics. T he scaling procedure is explained in step 3. Design Step 1.

Prototype design.

Set C = 1 F and also set

S,

=

^2

^bl

= ^2 S. In the derivation described later on, we conclude that a simplified design requires that the total admittances incident on the inverting terminal equal the total admittance incident on the non inverting terminal by proper choice o f AG^ or AG^. The proper choices are given in design step

2. Design Step 2. Prototype design [continued): Computation o f

or AG^ such that the total admit tance incident at the inverting terminal o f the op amp equals the total admittance incident at the non inverting terminal. To achieve this equality, recall that in design step I > - «1 S, - ” 2 ^b\ = h, S. G ,, = S. Define a numerical quantity 6 = + a-^) - [b^ + b.^)

6S4

Chapter 13 • Laplacc Transform Analysis II; C ircuit Applications

W e distinguish two cases:

Case I: If 6 > 0 , set

= 6 and

Case 2:

= - 6 and AG^ = 0.

If 6 < 0 , set

= 0.

D esign Step 3 . Scaling to achieve practical element values. Multiply all the admittances incident at the inverting input terminal o f the op amp by a constant

K^.

Similarly, multiply all admittances

incident at the non-inverting terminal o f the op amp by /Q,. It is possible to choose

= /Q,, but

this is not necessary.

E X A M P L E 1 3 .2 2 . Design an op amp circuit having the input-output relationship

(1 3 .3 4 ) S

o l u t io n

Step 1. Prototype design. Using Figure 1 3 .5 0 , choose C = 1 F. W ith all admittances in S, set

Step 2 . Equalization o f total admittances at inverting and non-inverting terminals. Since 6 = (7 + 3) - (2 + 4) = 4 > 0, set type design.

= 4 S and AG^ = 0. T he circuit in Figure 13.5 1 a exemplifies the proto

K^ = K y- 10“^. This scaling leads R^2 = 3 3 .3 3 kQ, /?^, = 50 kQ, R^2 =

Step 3 . Scaling. To have practical element values, let us choose to a design with C = 10 j_iF and resistances

= 1 4 .2 8 kQ,

25 kQ, and A/?^ = 25 kQ.

75

6SS


1 4 .2 8 kQ lO p F a2 '

3 3 .3 3 kO

+ 5 0 kO

b2

o— 2 5 kO

I O mF

2 5 kO-

(b) FIGURE 13.51 (^) Prototype design of equation 13.34. (b) Final design after scaling with /T = AT^ = 10~5.

Exercises.

1. Show that the normalized realization o f

s

s

s

s

is given by the circuit of Figure 13.52.

2S

6S

M GURE 13.52

2 . Scale the circu it o f Figure 1 3 .5 1 so that the capacitors b ecom e 1 uF.


656

Derivation of Op Amp Input-Output Characteristic Referring to Figure 13.50, from the properties of an ideal op amp, V^= V_ and no current enters the inverting and non-inverting op amp input terminals. Derivation Step 1. Write a node equation at the non-inverting input terminal of the op amp. Summing the currents leaving the + node of the op amp yields

Solving for

yields

W b2 (13.35) Derivation Step 2. Write a node equation at the inverting input terminal of the op amp. Recall that

V_ = Vy Summing the currents leaving the inverting node yields . AG^V^ = 0

- V^2) Thus,

sC

sC

''al

G„i

+

Gno +

AG« +

(13.36)

sC

n\ Derivation Step 3 . Combine steps 1 and 2 to compute the general input-output relationship. Substituting equation 13.35 into 13.36 yields

1 /

_

^ a l T,

G a2 y

I I ^ a\ + ^ a l +

"^ 7 )

If we choose and AG^ to make AG^ + â\ + âl = + ^b\ + ^bV ‘-e- if the total admit tance incident on the inverting terminal is made equal to the total admittance incident on the non-inverting terminal, then equation 13.37 simplifies to

I sC

(13.38)

sC

If we let C = 1 F (to be scaled later to a practical value), equation 13.38 further simplifies to

ôut

^b\ +

^bl

(13.39)

Equation 13.39 shows that the circuit of Figure 13.50 is a general summing integrating circuit whose gains are proportional to the admittances G^- and/or % The sign o f each gain depends on whether the corresponding input is connected to the inverting or non-inverting terminal o f the op amp. This completes the derivation of the input-output characteristic of the op amp circuit and is the basis for the prototype design. The extension to more than four inputs is straightforward.

Chapter 13 * Laplacc Transform Analysis 11: Circuit Applications

6^

It remains to justify the scaling o f step 3 in the design procedure. In step 3 , it is stated that all admittances incident at the inverting input terminal o f the op amp can be multiplied by a (scal ing) constant

K^.

Similarly, all admittances incident at the non-inverting terminal may be multi

plied by a (scaling) constant /Q,. It is possible to choose

= /Q,, but, again, this is not necessar}^

The verification that these multiplications will not change the input-output characteristic follows directly from equation 13 .3 7 . After inserting the scale factors, one immediately sees that they can cel and have no effect on the overall gain.

11. SUMMARY This chapter has presented the basic principles and techniques o f circuit analysis in the j-domain. Impedance, admittance, Thevenin equivalents, superposition, linearity, voltage division, current division, source transformations, and nodal and loop analysis have ^-domain forms that allow the analysis o f complex circuits excited by a variety o f waveform types. Indeed, the simple starter cir cuit o f a fluorescent light points to the usefulness o f Laplace transform methods in the analysis and design o f simple, everyday electrical conveniences. Various op amp applications were also pre sented. As subsequent chapters will illustrate, complex circuits and advanced analysis methods build on these basic principles and techniques. T he chapter also introduced the notion o f switched capacitor circuits. Integrated circuit technol og)' has made such circuits easy and inexpensive to produce. Applications include speech process ing and other t\'pes o f signal processing. Although a full-scale analysis o f such circuits is beyond the scope o f this text, the basic principles o f their operation arc presented as a foundation for more advanced analysis tools. Finally the chapter introduced a general method for the design o f multi-input integrators having both positive and negative gains. Such circuits can be used for implementing active filters and for implementing controllers in practical situations.

65
Chapter 13 * Lnplacc Transform Analysis II: Circuit Applications

12. TERMS AND CONCEPTS Admittance: the ratio o f the Laplace transform o f the input current to the Laplace transform o f the input voltage with the two-terminal network initially relaxed.

Admittance o f capacitor: the quantity Cs. Admittance o f inductor: the quantity MLs. Admittance o f resistor: the quantity MR. Current division: a formula for determining how currents distribute through a set o f parallel admittances.

Impedance: the ratio o f the Laplace transform o f the input voltage to the Laplace transform o f the input current with the two-terminal network initially relaxed.

Impedance o f capacitor: the quantity MCs. Impedance o f inductor: the quantit}' Ls. Impedance o f resistor; the quantity R. Impulse response: the response o f a circuit having a single input excitation o f a unit impulse; equal to the inverse Laplace transform o f the transfer function.

Integrator circuit: usually an op amp circuit with transfer function K/s. Source transformation property: in the ^-domain, voltage sources in series with an impedance are equivalent to the same impedance in parallel with a current source whose value equals the transform voltage divided by the series impedance.

Steady-state analysis: analysis o f circuit behavior resulting after excitations have been on for a long time; often refers to finding the sinusoidal or constant parts o f the response when the circuit is excited by sinusoids or dc.

Switched capacitor circuit: a circuit containing switches, capacitors, independent voltage sources, and possibly op amps, but no resistors or inductors.

Time-invariant device: a device whose characteristics do not change with time. Transfer function; the ratio o f the Laplace transform o f the output quantity to the Laplace trans form o f the input quantity with the network initially relaxed.

Transient analysis: analysis o f circuit behavior for a period o f time immediately after independ ent sources have been turned on.

Voltage division: a formula for determining how voltages distribute around a series connection o f impedances.

6S9


Problems

R

6

IMPEDANCE, ADMITTANCE, VOLTAGE DIVISION, CURRENT DIVISION, SOURCE TRANSFORMATIONS, THEVENIN AND NORTON EQUIVALENTS

ZJs)

(a)

/Y Y V ^ L

I

1. Consider the circuits in Figure P I 3 .1 . (a)

{s) and Y-J^s) as

Find

rational func

,b)

tions for each o f the given circuits. (b)

Figure P I3.2

Find the ^-domain and time domain output responses to a step input for each circuit.

3. In the (relaxed) circuit o f Figure PI 3 .3 , if = (5(r), then 5 + 40

2.5 0

■6

0.2 F 1H

Y-^^{s) and Z-^^{s). = 10u{t) V, find for t>Q. v jt ) = 20e-^^û{t) V, find i j t ) for

(a)

C om pute

(b)

If

(c)

If

r > 0. (a)

(d) 0.25 H J T

Y

V

If

v jt ) 0.

=

20e~^K(t)

V, find

ijt)

for

ijt)

_

0.25 F Z(s)

v.(t) I

20

Figure P I3.3

(b, Figure P I3.1 C H E C K S : Figure P I 3 . la: +5 y,„(5) = 0 . 4 --------- ^ F i g u r e P I 3 . lb: ( .y + l ) -+ 2 "

4. For each o f the circuits in Figure PI 3 .4 , find expressions for (a)

Z Js) and YJs) and

(as indicated)

J" + 1 6 2. Find the input impedance and admittance in terms o f

R, L,

and C o f each o f the networks

shown in Figure P I 3 .2 . Then find /-domain expressions for the output in terms o f the input for each circuit.

Z„(s)

(a)

660


7. Repeat Problem 6 for /? = 6.25

C = 0.01

F, and v-J^t) = 32^“^^cos(4r)«(/) V. 8. For the circuit of Figure P 13.8, suppose R = 2 il, I = 4 H, and Z{s) = ^ £ ± 1 5 . Find Z -(j), j + 0.5 Voup\ and when

v-J,t) = 2Qe~^û{t) V, assuming zero

(a)

initial conditions

v-J^t) = 2Ae~^-^û{t) V, assuming zero

(b)

initial conditions

(0

r\

Figure P I3.4

r\

5. In the circuit o f Figure P13.5, suppose

) = 0. Find

and i^^t)

. Z.„W. = 5{t) and (b)

when (a) 50(1 -

o

n\

=

mA.

Z.(s)

Y Js)

o

Figure P I3.8

0.4 F

9. For the circuit of Figure P 13.9,

^m\ ~

^ml ~

= 100 12,

^

^2 “

= 1 mF. Figure P i3.5

(a)

Compute I 2 U) in terms of /,(j) and

6. In the circuit o f Figure PI 3.6a, 7? = 4 0 Q and

(b)

If v-JJ) = e^^û{t) V, compute /j(/).

(c)

If v^„{t) =

then compute Z-J^s) and Y-Jj) . C = 0.01 F. Find Z.„W, Y.J,s\ V^s), and for (a) v-J^t) = (5(r) and (b) v-J^t) as given in

V, compute /j(r).

l,(s)

Figure PI3.6b. +

R

JL

l,(s) R.

i

V^(t)

(a) Figure P i3.9

t vJOCV) 20-

C H EC K : Z;„(5) = - 1 0 0

5-10

o

5 + 10

101

2

3

4

10. (a)

(b)

Figure P I3.6

(b)

Find the input impedance of the cir cuit o f Figure PI 3 .10a. Consider Figure P I 3 .10b. Use your answer from part (a) to compute the responses i-JJ) and

to a step

n

0(11


input when /? = 3 0 H , Z,|

= Lj=

1 H,

iY Y V ,

and C = 10 mF.

0.1 H

10 pF

I (s)

vjt) Z,(s)

50 0 K,V,(s) 2,(s)

Y,„(s)o-

V,{s)

(b) Figure P I3.12

13.

Vj(s)

For the bridged-T network o f Figure

P I 3 .1 3 , show that if Z jU )Z ,(j) = i?", then ' V,(s)

^ ,„ w (b)

Z,(s)

Figure P I3.1 0

—

11. Find theThevenin equivalent impedance o f

— —

z (S) Z,(s)

each circuit in Figure P 1 3 .1 1 . Flint: Label the terminals

and an input current to the top

terminal as

determine

or

0.5 H

_ m

v _

Figure P I3 .13 14. Consider the circuit o f Figure P 1 3 .1 4 , in which

= 10

7^2 = 10 Q , Z = 0.1 H, and C

= 1 mR

0.5 Q

.(i)

I

(a) (b)

Find

and

If a voltage source is attached as indi cated and

(a)

Ifl(S) (c)

2H

V, find

capacitor are initially relaxed. (d)

(b)

Find the zero-input response if /^(O) = 0 and -o —

i,(t)

2 + 2.V + iind

v^{t) =

assuming the inductor and

Figure P I3.11

12. Find

V, find /^(r)

If a voltage source is attached as indi cated and

0.5 F

o-

ANSWF.R: (h)

=

are initially relaxed.

o- J \ / v 'n_T Y Y \ .

20

vî)

assuming the inductor and capacitor

for each o f the cir

cuits shown in Figure P i 3 .1 2 .

= 10 V.

R,

/Y Y V L

'•"’Q YJs)

Z,(s)

Figure P i 3 .1 4 Z,(s)

CH ECK:

K,„ = 1 0 — .y“ -h 200.V + ? ? ? ?

(a)

+

662

Chapter 13 * Liplace Transform Analysis II: Circuit Applications

15. Consider the circuit o f Figure P i 3 .1 5 , in which /^, = 16

y? 2 = 4 n , Z, = 1 H, and C =

0 .1 2 5 5 ^ + 4

(c)

Z /.(-v ) =

(d)

yM =-

2 mF.

0 .1 2 5 5 - + 4

Y.î,s) and

(a)

Find

(b)

Find an expression for of

0 .5 5

0.5 s

in terms

and /.„W .

(c)

505

zM

=

5 -+ 2 5 0 55

R.

(0

R,

ijs) ( t )

V.Js) (g) (h)

yM =

5“ + 2 5 0

5“ + 4 0 .2 5 5 “ + 4 Z ,v ,(j) = ----------+ 0 .5 5 is 5^ + 16 n > ,(^ )=

45

5

•+ '0 .2 5 5 ^ + 16

Figure P I3 . 15

TRANSFER FUNCTIONS „ , , 16. Find an

RC circuit

for each o f the follow

ing: (a)

Z:,As) = ’ 0 .2 5

(c)

Z ,„ ( 5 - ) = 1 0 +

+ 0 .0 2 5

(b) iv„,,(/) + 20v,,„ , ( / ) + 100vv,,„(r) = 2 0 0 v v „ (0 + 10v,„(O

Z M

=

('= )

s+ 2 4s~ +

2 0 . Find the transfer function 36.? + 6 4

RL circuit for each Z Js) = OAs^ 10

o f the following:

0 .1 5 + 1 .5

2 1 . Find the transfer function,

V H{s) = - ^

(t) + bî-'y,, (/) (5 ) -----

differential equations accounting for initial conditions? (a)

5+10

5+2

( 0 + «1 Vaut ( 0 + « 2 ^’out ( 0 = /^2în( 0 + /^|

models o f a physical process. In the case o f (b)

Z ,„ (5 ) = 0 .2 5 + 2 0 +

4

for the general

and (c), what is the Laplace transform of the 2 0 + 0 .0 4 .V

LC circuit for eac i of- the following: Zi„(s) = 0.05s + ------

^ + 25

( ;) + 1

5 v /„ (r)-1 0 jJ.

5+10 5+8

your ( 0

(b)

(c)

20 5

0

(T)dr =

Vi„{T)(h

iaut ( 0

+ 0.1

dinutit)

+ 4 2 .5

5

r,v,(5) = 0.025 + —

0

+40 J'/^,,,( t Wt = 1 0 v,„C/)

18. Find an

(b)

v ^

second-order differential equation

for each o f the following differential equation

5 + 240

V f \

2 2 5 0 V ,. , , ( 0 + l O

impulse response, and step response

70

yM =

v „ , „ ( / ) + 1 0 v „ „ ,(/ ) + 2 5 ..„ ,„ (/ ) =

2.V + 8

17. Find an

(a)

( A") -----,

models o f a physical process. (a) <’o,a(0+ P\Voui(0= Kvi„U)

(5 + 2)(.y + 4 )

(c)

V

for each o f the following differential equation

1

0 .2 5 5 + 0 .2

(a)

H{s) =

impulse response, and step response K J j) = 0 .b + 0 .1

(b)

(a)

19. Find the transfer function,

J —00

= 200V;„(/)+20i:';„(/)

{T )dT


663

IUU.V 2 6. (a)

.s- + 400 0

(b) + 40^^^^

- 0 . 1 / „ „ , ( 0 " ) + 4()

(c)

.V

2 2 , An integro-difFerential equation

Given your answers to part (a), com pute the step responses.

= 1 0 l’„

.V

Compute the transfer flinctions o f the op-amp circuit in Figures P i3 .2 6 a and b.

for an

R=

Assuming

1 kH and C = 0 .5 niF,

compute the zero-state response to the

active circuit driven by a current source with C

2u{t) - 4u (t- 2)

input

= 0 .5 F , Z = 1 H, ^ = 2 n i s

(d)

V.

For the circuit o f Figure P I3 .2 6 b , assuming

R=

1 kD and C = 0 .5 mF,

compute the output when the input is -X

as given in Figure P i3 .2 6 c .

(a)

Find the transfer function ^/(.y) =

(b)

Find the impulse response.

(c)

Find the step response.

. % He c

2R

+

V „(t)

2 3. Suppose two signals

y{t)

and

are

related by the equations (a)

^’out

--- r ; — +

dt~

) + 2y(/) = 4 — ^—

at

He-

and

2Vo„,(0 - fy(T)c/T = 2 J \’,„(T)i/T ()■

(a)

+ /

6

+

Assuming t/y^(0 ) = 0, find the transfer function

(b)

v.(t)

O'

He

N(s) =

(b)

v„(t)

Find the impulse and step responses.

2 4. A certain circuit has input

co^{t)u{t)

te~‘ii{t).

C onstruct the

and output

-► t

transfer function o f the circuit, assuming that

2

the circuit had no internal stored energ)- at r =

4

(0

0~ . Then compute the step response.

Figure P i 3.26 2 5. T he input to a relaxed (no initial condi

e~‘u{t)

tions) linear active circuit is

V.

2 7 .(a)

Find the transfer function o f the

T he response is measured in volts as

(ideal)

\0e~‘ s'm{2t)u{t).

P I3 .2 7 a in terms o f 7?,,

(a)

op

amp

circuit

.

Find the transfer function

it is desired to obtain with C = 1 IIF, then

Vl„{s)

d{s)

tored form.

Figure

and C . If

H{s) =

-2

—^ ' ^

(b)

Find the response to the input func tion in Figure P I3 .2 7 b assuming zero

Com pute the response o f the circuit to the new input y,„(f)=

ol

find /?j and /?->.

in which the denominator is in fac (b)

R-,,

2te~'u{t)

V.

for the initial capacitor voltage.


664

r\ (c)

Repeat part (b) assuming the initial capacitor voltage is = -4 V. -

29. (a)

Find the transfer fimaion of the (ideal) op amp circuit of Figure PI 3.29 in terms of /?,, /?2, Cp and Cj. Make the leading coefficient of s in the denominator 1.

(b)

If it is desired to obtain H(s) = - 4 -------

(c)

with € 2 = 100 jaF, find /?j, Rj, and Cj. j +2 Given H(s) = - 4 -------find the zero-

v,(t) +

5+ 2

s+4

^^

^^

s+4

state response to t',„(^)= sin(2r)«(f) V.

v,(t)

->t

(b)

Figure P I 3.2 7

28.(a)

(b) (c)

(d)

Find the transfer function o f the (ideal) op amp circuit o f Figure P I3.28 in terms of /?,, /?,, and C. s 8 If it is desired to obtain H (j;) = -----with C = 1 nF, find /?, and i?2- ^ ^ ^ Find the response to the input func tion in Figure P I3.27b assuming zero for the initial capacitor voltage. Repeat part (b) assuming the initial capacitor voltage is = - 4 V.

30. (a)

Find the transfer function o f the (ideal) op amp circuit o f Figure P I3.30 in terms of R^, Gj, C j, and C2 .

(b)

If it is desired to obtain //M = - 4

(c)

(s + 2)(5 + 10) with € 2 = 100 ^F, find 7?j, /?2»and C,. Given the answer to part (b), find the zero-state response to y,„(^)= -2.5e~^^ sin(2/)«(/) V.

Figure P I3.30 Figure P I3.28 Gi + G t

s + —^------ ^ CHECK:

H(s) = -------- f — s+

CHECK:

H(s) =

-1 C 2R 1

s+

s+ / ? ,C j

R 2C 2

665

Chapter 13 • Liplacc rransform Analysis II: Circuit Applications

3 1. (a)

C om pute the transfer functions o f each circuit in Figure P I 3.31 in terms of

(b)

C,

Suppose

R and K. C = 1 i.i¥, K = 3,

3 3 . C onstruct the transfer Kinction o f the d rcuit o f Figure P 1 3 .3 3 , assuming that all op amps are ideal, as follows:

and

R = R^

.

= 1 k n . Find the zero-state response to

2

.

(a)

Let G| = — , C t = — , G t, =

(b)

Com pute the ratio

R

sin(500r)//(r) for each circuit

1

R.

in Figure P 1 3 .3 1 . (c)

Repeat part (h) for

i’j„{t)=

2 sin(l 500/)//(r).

V'i(.v) (c)

Com pute the ratio

(d)

Com pute

(c)

Com pute

(0

Rack-substitute to eliminate interme

V'2(-v)‘

(a)

in terms o f

V:^{s)

and

in terms of V^|(i) and

diate variables and find

H{s)

=

Vi,M)

circuit o f Figure P I 3 .3 2 in terms o f /?j, (b)

R-y, Ry R^,

R-y, Ry H{s) = (c)

C p and

C-,. Figure P i 3.33

If Cj = C 2 = 1 t
R^ so

that

ANSW HR:

-5

-2

(.v+lOO)

(.v + 2 0 0 )

3 4

. (a)

Find the impulse and step response.

H(s) =

Ci\ s~ + (j^s +

G -1

For the circuit in Figure P i 3 .3 4 , find the transfer functions

(b)

H^{s) = - ! - ^ and //^ (.v) = Vinis) “ Vinis) If = 10 Q , Yj = 0.1 S, Z , = /, V, = s, and = 2 0 S, com pute the impulse and step responses.

(c)

Given the values o f part (b), compute the response to the input

-2u{t

2u{t)

- 4) V using linearit)^ and no

further calculations.

666


IJs)

2,(s)

+ Z.(s)

Y,(s)

VJs)

Y,(s)

❖

9 jY Figure P i3.34 35. (a)

Figure P13.36

For the circuit in Figure P13.35, find die transfer fiinctions w

i

“

Z,(s) V,(s)

< p

37 Consider the circuit o f Figure P13.37. Find die transfer fiinctions

■

1

(b)

(c)

If Z 2 = —

,Z^ = j + 3, Y^=s, Y^=

3. and r ^ = 20 Q, compute the impulse and step responses. Given the values of part (b), compute the response to the input Kû(t) -Kjuit — 7) V using linearity and no further calculations, assuming

as follows: (a) Compute the input impedance _ Vîs)

Zin(s) = (b)

T>0.

VJ„(j)

i(^)

Given the results from part (a), use voltage division to compute

Vln(s)

Figure P I3.35 36. (a)

(d)

in terms ofZy^(^), J 2W> Compute ^

(e)

Compute

Find the transfer functions « lM = ^ . « 2 W

Hîs) =

(b)

(c)

in terms of Do not substitute your answer to part (a) into the obtained expression. Compute v ,w

= ^ .a n d

for the circuit of

^

^tc.

V ,( 5 )‘

Figure P I3.36. Vinis) Vi„(s) Find the impulse and step responses and as products of your prior answers. associated with each transfer function if = 8, Kj = 2, IJs) s , _ 1 — z.(s) [-0 Z,{s) Z, = and Z 2 = j + 5

o

<$>

Yj(s) V,(s) Y,(s)

K.IJS)

—o

Figure P I3.37

r\

667


CHECK: Z ^ = Z i +

K 1^2 .K n + ^L


Vout = det(MM)/det(M) % You have now computed the symbolic % expression for Vout in terms o f lin so you % can now identify the transfer function as a % symbolic ftinction o f the variable s. % Now lets do some numerical work. C l= 1; C2 = 2; G l = 1; R2 = 1; G3 = 2; a = 0 .5 ; v„Jt) % Re-enter M above M = [Cl'*‘s+Gl 0 1; fill in the rest of your co ^ -

G is mhos; R is ohms

Figure P I3.38 (a)

(b)

Draw the equivalent frequency domain circuit assuming t'QCO") ^ 0 V and ^^2(0") ^ 0 V. Define a current Ij^{s) from left to right through Then write three (modified) nodal equations for the cir cuit o f part (a) similar to the text example, but accounting for the initial conditions. Put equations in matrix form (unknowns on the left side and knowns, input, and initial conditions on the right). You should have a 3 by 3 matrix of literals. The unknowns are two (node) voltages and one current,

cient fnatrix] % check: the roots o f the determinant should % be -1 and - 5 . dt = det(M) factor(dt) % Re-enter MM above % Compute the actual transfer function Vout = det(MM)/det(M) % Identify the transfer function (d)

Find the impulse response h(t) of the circuit as follows: % Rewrite the above expression for Vout with % out lin. syms s t H h H = -(1/2V1)/???????? h = ilaplace(H)

Determine the transfer ftinction H{s) = yôut^s)/IjJs) of the circuit in MATLAB using the following code:

(e)

Now use the MATLAB command “ilaplace” to compute the step response, again with zero initial condi tions.

syms M MM lin Vout s G l C l C 2 G3 G2 R2

Remark: Use your newfound MATLAB knowl

a t,H

edge o f symbolic computation to carry out the following tasks. Hand computation is not acceptable. (f) Find the response due only to the ini tial condition on Cj assuming " a (0 ) = 16 V. (g) Find the response due only to the ini tial condition on C2 assuming

(c)

M = [C l*s+G l 0 1; fill in the rest of your co ^ -

cient matrb^

'w ^

% Transfer ftinction computation assumes all % ICs are zero % Using Cramer s rule, define MM by % replacing the middle column o f M % by the right-hand side o f your matrix nodal % equation with all IC s set to zero. M M = [C l*s+G l lin 1; fill in the rest of your

matrii^ dt = det(M) dt = coUea(dt)

= -8 V .

(h)

Write down the complete response assuming the input is a step.

66S


3 9 . In the so-called feedback configuration o f Figure P I 3 .3 9 ,

E{s)

is the Laplace transform o f

the error between the reference signal

e{t) =

the response^(^), i.e.,

}i{s)ld{s)

and

-y{t).

G, W =

is called the plant o f tiie system and rep

resents a physical process or special circuit. (a) (b)

e{t) x^^,j{t) = Kfû{t)' conditions does e{t)

00

Under what 0 as / -♦

(d)

XrefU)

CH ECK :

Under what conditions does 0 as f -♦

(c)

H{s) =


F ig u re P I 3 . 4 0

E{s)

00

where F (s)G i(.s ) = — ------- —

djis)cl^(s)

from

E{s)

=

H{s)X^^jsy.

d{s) + n{s) = --------

d(s)

RESPONSE CALCULATION WITH INITIAL CONDITIONS 4 1 . For the circuit o f Figure P i 3 .4 1 , C = 0.1 F,

E(s)

0

1 + F(.v)Gi(.9)

when

Under what conditions and how could

4

_ d{s)Xr(.f{s)

Xrefis)

when

you determ ine

X..,(s)

E{s) =

.

-> Y(s)

G,(s)

V(is)

and

initialized

L = 0 .5

H,

i^{0~) = 2 A, and vîO~) = 2 V. Find vît) using the et]uivalent models for L and C in the j-domain. ijt)

Figure P I3.3 9 ch ec k

+ L

.

l-t-G](5)

d{s) + n{s)

4 0. In the so-called feedback configuration of Figure P i 3 .4 0 ,

E{s)

nj,s)ldjj)

x^^t) - y{t).

i.e.,

e{t) =

and =

is the plant o f the system (described

in the previous problem) and

F{s)

=

)tj(s)lci^s)

is

a feedback controller to be designed. (a)

00

when

that accounts for the initial condition, write a single node equation in

0 R1

Under what conditions does

x^^J(t) =

e{t) -*

0.4 H

A'y«(^)?

,

0-2 v , Y

2Ci

Under what conditions and how could you determ ine

e(0'^)

from

E{s) =

F igu re P i 3 .4 2 A N S \ V F :R :

(e)

and then find

50

e{t) -*

x^^Jit) =

^ 0 as r -♦ X when (d)

A. Draw an equivalent circuit for the inductor

H{s) = E{s)IX^^,J(s).

Under what conditions does as r

(c)

4 2 . For the circuit o f Figure P 1 3 .4 2 , /^(0“) = 2


(b)

Figure P I3.41

is the Laplace transform of

the error between the reference signal the response

v,(t)

Suppose

d^s)

had a pair o f poles on

the imaginary axis. How could you design an

F{s)

to cancel these poles?

V

669


43. For the circuit o f Figure P I 3.43, assume that i'q ( 0 “) = -4 0 0 mV and V(^{0~) = 0 V. Find t> 0. 2mF

CHECK: /q = 2.5 46. In the circuit o f Figure P I 3.46, 7? = 20 Q, C = 0.2 F, and the capacitor is initially charged at V(^Qr) = 10 V. (a) If v.„{t) = 20[«(f) - u{t - 20)] V, find I({s), V^s)y if^t) , and vîj) for r > 0. Plot V(jJ) and v-J^t) on the same graph for 0 < f < 40 sec using MATLAB or its equivalent. (b) If V, find i({t) and V(4J^ for ^> 0. Plot v^{t) and v-JJ) on the same graph for 0 < ? < 20 sec using MATLAB or its equivalent.

44. Consider the circuit o f Figure P I3.44, in which C = 20 mF and R = 100 O. (a) I f vÔ-) = 10 V and = 200e~^'^û{t) mA, find Vf^s) and V(^t) for f > 0. = 200 (b) I f v^0~) = 10 V and cos(2f)«(/) mA, find V(^s) and V(4J) for ^ > 0.

ic(t)

+ Vc(t)

Figure P I3.46 47. Consider the circuit of Figure P I3.47, in which R^=l k£2, R^ = S kH, and C = 50 ^F. (a) Find the transfer fiinction H{s). (b) Compute the step and impulse responses. Compute the response to v-^(t) = (c)

Figure P I3.44

(d) 45. Consider the circuit of Figure P13.45, in which R = 200 £2. (a) Compute the transfer function in terms o f L. (b) If the response to the input i-JJ) = Intuit) A is ijij) = {2.5t - 0.025 + Vw>

(c)

O.O25e"^®°0«(^) A, assuming /^(O") = 0, find the values of L and /q. Assuming ij{0~) = 1 A and i-JJ) = 2.5 sin(100f)«W A, find

(e)

(0

10r'2-5^cos(25^)«W V. Find the response to V(iO~) = 20 V. Using the principle of superposition, find the response to the input of pan (c) with the initial condition o f part (d). From the principle o f linearity, what would the response be to the input = 20£-“^^-^^cos(12.5/)«(/) V and y j^ 0-) = 1 0 V ? + v Jt)

lV (t)

Figure P I3.47

u « (t) C H EC K : Figure P I3.45

w

1

I

w

w

1

H{s) =

20 2 5 + 25


670

48. Consider the circuit o f Figure P I3.48, in which C = 2.5 mF, = 200 Q, = 50 £2, /?£ = 5 n , Z = 0 .2 H , and^„ = 0 .5 S . (a) Find the transfer functions

(c)

Find the response i({t) when v-JJ) = 0, V({Q) = 10 V, and /^(O) = 0.

//2 (5 ) = - ^ ^ , a n d

^

VcisV

Figure P I3.49

W Given N(s),

(b)

find the zero-state response; i.e., assume /^(0“) = 0 and = 0 when = 25(1 V. Find the zero-input response if /•^(0) = 6 A a n d t;c(0 ) = -1 0 V . Find the complete response (combine parts (b) and (c)) and identify the transient and steady-state parts o f it. (Note that the transient pan is the part that is not constant or not period ic. Usually the transient part converges to zero.)

(c) (d)

^

'

g.Vc(s) 1

C = -

F

9 R^ = 0.5 £2, = 5 £2, C2 = 0.05 F, and a = 4 . Compute Vciis), assuming diat v jt ) = 36u{t) V. /^(O-) = 0. v c ^ m = - 1 8 V, and v„ufiO~) = 0 V. (Hint; You must construct the equivalent circuit in the j-domain, account ing for initial conditions. Consider a source transformation on v-J^t), and then draw the equivalent circuit in the j-domain so that you can combine sources in the front half o f the cir cuit.)

1 L

Y

'1

50. In the circuit o f Figure P I 3.50, suppose L = 0.04 H. 1

r

1 Figure PI3.50

Figure P I3.48 49. Consider the circuit of Figure P 13.49. Note that the computation o f transfer functions pre sumes no initial internal stored energy. (a) Use a source transformation and the current divider formula to show that the transfer function between V-JJ) (the input) and I(\s) (the output) is 1

1

■s +

RC (b)

1 LC

If Z = 0.2 H, C = 25 mF, R = 4/3 Q, and v-^{t) = \Qe~^^û{t) V, find /^j) and i({t).

51. Consider the circuit of Figure P13.51. Suppose /?j = 2 £2, /?2 = 6 £2, i?3 = 3 £2, C = 0.125 F I = 1.6 H. Suppose v-JJ) = yj(/) + 1/2W = -1 0 « (-f) + 10 u (t) V (plot this input function so that you know what it looks like). (a) Compute V(iO~) and /^(O"). (b) Compute the “zero-input response” for ^> 0; i.e., assume you have the ini tial conditions computed in step (a) and that the voltage v-^(t) = 0 for r > 0. First draw the equivalent j-domain cir cuit accounting for the initial condi tions.


(c)

(d)

(e)

(0

Compute the transfer function H{s) of the circuit valid for ^ > 0. Compute the “zero-state response,” i.e., the response to the circuit o f the input v-JJ) = V2 {t) = 10«(f) assuming all initial conditions are zero. Compute the complete response. Identify the transient and steady-state parts of the complete response.

671

Plot the resulting time function in MATLAB for 0 < r < 1 sec. The partial fraction expansion is most easily com puted using MATLABs “residue” command. {Note: Practicing hand cal culation is important for the exams.) Now forget about the partial fraction expansion and instead use MATLAB s command “ilaplace” to compute the time function You should define H, Vc, s, t, lin, and vc as symbols using “syms.” The program should be something like the following: R = ? ; L = 1 ; C = ?;

52. Consider the circuit o f Figure P13.52. Suppose Z = 1 H, = 15 C = 0.01 F, and ^2 = 5 Q. Suppose is the desired output, (a) Find the input admittance

(b)

in terms of literds and then substitute numbers. Combine terms to form a rational function. Find the input impedance

syms Yin Zin H Vc s t Is vc lin = ? % H will be the transfer function defined in terms o f s, a symbol, and R, L, C, and K. % MATLAB will fill in the numbers. H = ???? Vc = H*Iin vc = ilaplace(Vc)

.( .) 0

99 L

's ^

(c)

w

(d) V^ (e) V^

5+

1 LC

in which case V^(j) = ????(write down the formula). Use voltage division to express K^j)in terms o f V^{s) and then in terms o f Z-JJ) and IfJ). Hence determine the transfer function H^{s) = V(^s)IV^{s). The leading coefficient o f the denom inator polynomial is to be 1. Compute the impulse and step responses. Suppose the input is = 2e~^û{i) A. Find the partial fi^ction expansion of V^s) and then compute V({t) assum ing zero initial conditions on L and C.

Figure P I3.52 53. Repeat Problem 52 using ij{i) as the desired output. 54. Consider the circuit in Figure P13.54, in which /?j = /?2 = Suppose V(\t) is the desired output. (a) Find the input impedance in the form Z /rt(j) —

in terms poles />j negative ances in

S-p\

+

S -P 2

of R, L, and C. Identify the and p 2 - They should have values. Recall that imped series add and two imped-


672

r>

(b)

ances in parallel satisfy the product over sum rule. Compute the transfer function

+ V^(t)

99 V,„(s)

(c)

id)

(e)

+

i,(t)

1

RC

LC

in terms of R, L, and C The roots of the characteristic equa tion (denominator o f H{s)) are to be at - 2 ±y4. If Z = 0.2 H, determine R and C. Then specify the transfer function with the proper numerical coeffi cients. Given the values o f part (c), (i) Compute the impulse response of the circuit. (ii) Compute the step response of the circuit. Suppose the input is vfi) = 10aT*%(/) A- Find the partial fraction expansion o f V({s) using MATLAB’s “residue” command. Now forget about the par tial fraction expansion and instead use MATLABs command “ilaplace” to compute the time function vît). Note you should define H, Vc, s, t, Vin, and vc as symbols using “syms”. For example,

R = ?; L = 0.2; C = ?; K = 1/(R"C) syms H Vc s t Vin vc Vin = 10/(s+ 10)^2 % H will be the transfer function defined in terms o f s, a symbol, and R, L, C, and K. % MATLAB will fill in the numbers. H = ???? Vc = H’^Vin vc = ilaplace(Vc)

Figure P 13.54

r\

55. This problem is to be solved using nodal analysis. Consider the circuit o f Figure P I 3.55, in which Gj = (^2 = 1 S, G3 = 4 S, Z = 1 H, C = 2 F, and g^ = 2 S . (a) Draw the equivalent frequency domain circuit assuming /^(0“) = 2 A and = 2 V.Since you are using nodal analysis, what type of equivalent circuit might be best to use to account for initial conditions? (b) Write two nodal equations for the cir cuit o f part (a) in terms o f the node voltages V.J,s) and and, of course, the input I-JJ) and initial con ditions. Simplify. Put equations in matrix form. Determine the transfer function of the (c) circuit. Use Cramers rule. (d) Find the step response of the circuit. Find the response due only to the ini (e) tial condition on the inductor. (0 Find the response due only to the ini tial condition on the capacitor. (g) Find the complete response if the input is a step.

r\

n

n o

r\ r> r>

o

r\ r\ Figure P I3.55 56.(a) (b)

For the circuit o f Figure P I3.56, cal culate the transfer function. Suppose i-Jit) = <5(/). Find and plot

r-s n\

^

673


(c)

Now suppose

i-p) = 16[u{t) - u {t- T)]

mA, where 7'=

1 0

If v jt) = l[u{t)

(d)

msec. Find and plot

0.3 H

5

- u { t- 0.5)]

V, y^(0") =

\/ and /yr(0~) = 0, find

0.1 H

ijt) 0 1 5 0 90 n

lO Q

Figure P I3 .56 Figure P I3.58 C H E C K : (a)

{s +

100)(5 + 3 0 0 )/[(j + 200)(^ +

4 0 0 )]

59. in the circuit o f Figure P 1 3 .5 9 , /?j = 2 Q, C, =1 F, /?, = 1.75

57. Consider the bridged-'F network in Figure

(a)

Vj„{t)

= 10(1 -

e--îi{t)

V({s)

and C , = 2 /7 F.

Show that the transfer function o f the _ ^C2 circuit in literal form is H (.v) =

P I 3 .5 7 . Let C = 0 .2 5 F. Assuming no initial conditions, find /j(s ), ^ (s ), and

Q,

when

V. You might use

R\C,

C ram er’s rule and the symbolic toolbox in M A FLA B to solve the problem. Now find

.v +

U 2C2 ^

"JO (b)

If

^

«|C|/?2C ,

= 15«(/) V and the capacitors

are initially relaxed, find

V(^[t)

for

t>

0. Plot the input and the response on the same graph using M ATLAB or the equivalent. (c) Z Js)

(d) Figure P I 3

= 0, ^^^(0“ ) = 0 , and i^Q(0~)

(e) which /?^ = 2 k n , C = 2 0 80 n , and

L

}d¥,

If

= 20

Q,

If

for

t>

0.

= 0 , t^<^'i(0“ ) = 0, and /^ ^ (0“ )

= 15 V, find

.5 7

V(^{t)

for r > 0.

= 15//(/) V, i^Q(0“ ) = 15 V, and

^(^(0“ ) = 15 V, find

58. Consider the circuit o f Figure P I 3 .5 8 , in

(a)

if

= 15 V, find

Vf^{t)

for r > 0.

=

=5 H.

Com pute the transfer function /V,(.v) =

(b)

v;,,(.v) Com pute the transfer function

(c)

Vf(.v) Com pute the transfer function

0 Figure P i3.59

,

A N SW ER S: (b) y.

Vi Js)

- ~e^ûu) V; (c,[^-0.25;^

V; (e) sum ol parts (b). (c). and (d)

674


60. In the circuit o f Figure PI 3 .6 0 , /?, = 0.5

Rj

=1 H, (a)

= 1.75 a

(c)

and Z , = 7 /8 H.

Show that the transfer function o f the circuit in literal form is

Now suppose A,

n {s) =

^

=

) = 3 V,

?>u{t) V,

R ecom pute

the

v,(t)

L jG’2 ^2 —+

>v+

v -> 0 .1 H

0.5 n

1

= \5u{t)

If

V and the inductors for

the same graph using M ATLAB or the equivalent. If

6 2 . Reconsider the circuit o f Figure Pi 3 .6 1 . (a)

If

(e)

If

= 0 , /^ ,( 0 -) = 0 , and

1 5 A, find =

an equivalent circuit accounting for the initial conditions

=

mesh equations w'ith

V, /^ ,( 0 -) = 15 A,

and /^-,(0“ ) = 15A, find

iu^t)

for

t>0.

/Y Y \ —►

=

0,

/■^(0“ ) = 3 A. Com pute the associated

lor / > 0.

\5u{t)

Apply a source transformation to the independent current source and draw

= 0, /^.(O -) = 0, and /> ,(0 " ) =

15 A, find /£-,(/) for r > 0. (d)

( t ) ' . ,(t)

1Q

Figure P I3.61

t^

0. Plot the input and the response on

(c)

Vjft).

^2^1/

are initially relaxed, find

equations

V .(t)

m

0.8 F

(b)

nodal

i^{Q) ) = 0 = 5u{t) A.

accounting for the initial capacitor

=

voltage. Then find

— - + ^C|L|

and

/^(s) as one o f the

mesh currents. (b)

If rîO-) = 3iiU) V, and

0,

i^(O-)

= 3 A,

=

/;.(/) = 3 « (/) A, find /^(r).

6 3 . In the circuit shown in Figure P I 3 .6 3 ,

i\„U)

= 12 V for r > 0 , i/^ ,(0 ) = 6 V, and z^^,(0) = 2 V.

'• 6

(a)

C onstruct the equivalent circuit in the Laplace transform domain, account ing for an initializ-ed capacitor.

Figure P i3 .60

(b)

Write a nodal equation for the circuit constructed in part (a).

A N SW FR S: (b) (30 (c)

A:

(c)

Find l/^ W .

+

(d)

Find

A; (d)

7e^‘\u{t) .A; 6 1 .(a)

for

t> 0 .

(e) sum o f parts(b). (c). and (d).

In the circuit o f Figure P 1 3 .6 1 , and

V(^t)

represent node voltages with

respect to the bottom/reference node. Show that with zero initial conditions the nodal equations are O . B r + 2.9 + 10 -1 0 (b)

- 1 0 ■ ■V'c'

• 2\ /,-

-h i

.v + 1 0

Find the response

with /.ero ini

tial conditions and with % ](/) = 3//(r) V and

i^2 ^t) = 'iu{t)

A.

Figure P i 3.63 CH ECK: form

should contain terms o f the and

K~)U{t).

6 4 . Consider the third-order RC circuit o f Figure P I 3 .6 4 . Suppose that the initial capaci-


tance voltages are t/c,(0) = 0, t/Q(0) = 6 V,

675

equations for the circuit of part (a) only in terms of the variables V({s)y

*;C3(0) = 2 V .

(a)

Construct the Laplace transform domain equivalent circuit. (b) Find Vf^{s) in terms o f V^J^s) and the initial conditions. (c) For the given initial conditions, if v-^{t) = \2u{t) V, find for r > 0. Hint: After obtaining Vq W, use the “residue” conmiand in MATLAB to obtain the partial flection expansion. You might also investigate the use o f the command “ilaplace.”

(c)

(d)

(e)

^2^’ Initisl conditions. Simplijy each equation. Put equations in matrix form. Assuming /jr(0“) = 0, (0~) = 5 V, W = 5<5(/) A, and (^) = \06{t) V, use Cramers rule to find the current (f) and then i^ {t). Now suppose that ii(0~) = 0, = OV. = OA. and «„(») = 10»(() V. Find 1^2 W-

1 0 I.

Figure P I3.64

Figure P I3.66

65. In the circuit of Figure P I3.65, = 0.25 S, C2 = 0.25 F, and /? = 1 Q. Write a set o f mod ified loop equations and solve for the three loop currents and V, assuming V(^0~) = 4 V and v jt ) = \2u{t) V.

67. Consider the circuit of Figure PI 3.67. This problem is to be solved using (modified) nodal analysis. (a) Draw the equivalent frequency domain circuit assuming /^(0“) = 0 and «'ci(0“) = 2 V. (b) Write three nodal equations for the circuit o f part (a) only in terms o f the voltages Ifis), and the initial conditions. Simplify your equa (c) (d)

tions. Put equations in atrix form. Using Cramers rule, find the transfer fimction _

H (s)^

(e)

'O

W

W

66. Consider the circuit of Figure PI 3.66. This problem is to be solved using (modified) nodal analysis. (a) Draw the best equivalent frequency domain circuit for nodal analysis accounting for the as-yet-unknown initial conditions ifiO') and i/q(0 ). (b) Following the procedure explained in the text, write three (modified) nodal

(0 (g)

I;in(s) o f the circuit. Find the impulse response h{t) of the circuit. Find the response of the circuit to i.J^t) = -8u(t) A, assuming the initial conditions are zero. Find the response due only to the ini tial condition on the capacitor. A sim ple observation leads to the answer direcdy.


6“6

= 190 a, L =20 H,

{()

and

100 V

=

for all

time.

L + V .(t) -

Figure 1M3.67 v jt)l

A N SW ERS: and

/-(/)

0 .2 5

H {s)=

=

4,s(.v+ 1) Figure P i3.69

1 )//(/)dcrivcd from

( 2s +\ )

0

0

4

7 0 . The switch in the circuit o f Figure P I 3 .7 0 -4

2.V

0

is in position A for a very long time. It moves

t=

to position B at time

0

I,

1 sec, back to A at r =

2 sec, and then back to B at r = 4 sec, where it remains forever. Suppose /?, = lOOQ,

/? 2

Q.,C^=2 mF, and C2 = 2 niF. Com pute when V- (t) = 20z/(^) V. Plot for 0 < r < 5 A

{t) = 2{t - \ + e CvÔ-)hit) = (e-‘ - \)u{t).

Finally

= 500 (/) sec.

B

and

V \ /V +

R,

r

SW ITCH IN G PROBLEM S 6 8 . The switch in the circuit o f Figure P i 3 .6 8

-

c,

'1

----- 0

Figure P i3.7 0

is in position A for a very long time and then moves to position B at time

t=I

sec, back to A

at f = 2 sec, and then back to B at / = 3 sec,

= 5 0 0 Q.,

where it remains forever. Suppose

= 1 k n . and C = 2 mF. Com pute when

= 10//(r) +

5u{t -

1) + 5 « ( / - - 2) -

7 1 . Repeat Problem 7 0 for

=

20u{t)

+

2 0 « ( f - 1) V. 7 2 . Repeat Problem 7 0 for

R^ =

500

Q. and

y-„(r) = 20//(r) + 2 0 « (r- l)V .

2 0//(r - 3)V. Plot for 0 < r < 5 sec. R

a

W

.

b

+ V „(t)

Figure P I3.68

7 3 . In the circuit o f Figure P i 3 .7 3 ,

Rs = Ri,=

lo g ,( 2 )

Cj = 1F, and

=

Cj = 1 F.

1 .4 4 2 7

iX

Suppose switches Sj and

S2 have been closed for a long time. At r = 0, S2 6 9 . The switch in the circuit o f Figure P i 3 .6 9

is opened. At / = 1 sec, $ 2 is closed and Sj is

has been in position A for a very long time. At

opened. Will any capacitance voltages change

/ = 0, the switch moves to position B. Compute

abruptly in this circuit?

/?, over [0, oo). Assume

energy dissipated in

(a)

Find i/,(0 ") and

R^ = 2 Q, R^ = 8 Q., R^

(b)

Find

and

for 0 < r < 1 sec.


(c) (d) (e)

677

Find v^{\~) and ^2( 1~)• Find and ^2( 1^). Find v-^{t) and ^2(^) for 1 ^ ^sec.

i'

+ v,(t)

Figure P I3.75 32 V

c,

c

Figure P I3.73 74. In die circuit of Figure P13.74, Vq = 10 V, = 4 Q, = 3i2, /?2 = 24Q, Zj = 3 H, and ^2 = 6 H. Suppose switch S j has been closed for a long time while S2 has been open. Ax. t = 0 S2 is closed, after which Sj is opened. At ^ = 1 sec, S j is closed and S2 is then opened. (a) Find /j(0") and (b) Find z‘2(/) and for 0 < f < 1 sec. Find Zj(l“) and (c) Find and /2(1'^). (d) and for 1 < /sec. (e) Find

76. In the circuit in Figure P I3.76, the switch has been in position A for a long time. The switch moves to position B at r = 0, back to A at ? = 2 sec, and finally back to B at r = 5 sec, where it remains. Suppose = 2.5 = 1.5625 H, R2 = 2.5 Q, ^2 = 1 H, C = 0.04 F, and = 10 V for all t. A

vjt)l

(a)

(c)

(d)

Figure P I3.74

75. Consider the switched RLC network of Figure P13.75, in which = 2.5 Z, = 1 H, and C = 0.04 E (a) Suppose the switch S has been closed for a long time and opens at r = 0 . If v.^{t) = -40«(-/) V, find V(^t). (b) Suppose the switch S has been closed for a long time, and it opens at r = 0 and recloses at r = 1 sec. If = -40u{t) + 40«(/) V, find

:\{t) Figure P I3.76

(b)

CHECK: (b) /jW has a pole at ^ = - 1 , and I2 is) has a pole at ^ = - 4 ; (e) !^{s) and /2W both have poles at j = - 2 and s = -5-

B

Find Vf^s) and for 0 < / < 2 sec. Draw the equivalent frequency domain circuit that is valid for / > 2 sec. Find an expression for with the switch in position B and then deter mine for 2 < f < 5 sec. Compute V(jJ) for / > 5 sec. Hint: This can be done without any further computation.

77. Consider the circuit o f Figure P I 3.77. Suppose R^ = 2Q, /?2 = 4 Q, = 4 Q, Cj = 0.3 F, C2 = 0.25 F, and v-JJ) = -30«(-/) + 30«(/) + 30u (t- 10) V. The switch has been in position A for a long time and moves to position B at ^ =

0. (a) (b) (c) (d)

Find and y ^ (0 “). Draw the ^-domain equivalent circuit valid for t >0 . Write a set o f nodal equations in terms of Vî W and Solve the set of nodal equations con structed in part (c) and determine

Chapter 13 * Uiplace Transform Analysis II: Circuit Applications

67H

for 0 < r < 10 sec. (c)

(e)

solving the equations o f part (e) for

Com pute the initial conditions at / = 10

sec,

i.e.,

find

^ ^ (1 0 ")

Set up the Cram ers rule equations for

and

y ^ ( 1 0 “). Also assume that at r = 10 sec, the switch moves back to position domain valid for r > 10 or r' > 0.

SW ITCH ED CAPACITOR NETW ORKS

Com pute

7 9 . Consider the circuit o f Figure PI 3 .7 9 .

A. Draw the equivalent circuit in the sfor r > 1 0 sec.

(a)

If

v^{t)

= 25 V and the 100 m F capac

itor is uncharged at r = 0 , find 2 sec < (b)

V(^t) for

t. vp) = 25 V and vît) for 2 sec < t.

Now suppose 10 V. Find

=

t=2 sec

Figure P I3.7 7 C H E C K : (d)

v^Jt) = [ 1,846^(-2-5')

-

'• " '6 7 8 . Consider the circuit in Figure P I 3 .7 8 . R

^

150 mF

100 mF

v,(t)

Figure P i 3.79

21,(s)

8 0 . In the switched capacitor circuits o f Figure P i 3 .8 0 , the switch closes at r = 0.

l.w 2V Js)l

:v,u)

ijs)

2Q

;

Suppose + ■V fs)

(a)

= 5 7 2 mV.

For the circuit o f Figure P I3 .8 0 a , compute

for ^ > 0.

(b) The circuit o f Figure P I3 .8 0 b differs from Figure P I3 .8 0 a in that the

Figure P i 3.78

v-^j^t)

-source is replaced by a 10 //F capaci

Part 1.

The switch is in position A at r = 0 and

tor with an initial voltage i^^/0“) = 5 7 2

t'Ô ") = 2 V, /^(0“) = 0 and for 0 < r < 10,

mV while all other capacitor voltages

= 2 e ^ ‘ u{t) A .

are zero at r = 0~. C om pute

(a)

Draw the frequenc)' domain equiva

r> 0.

lent circuit o f the capacitor and com pute

V(4,s).

(b)

C om pute

(c)

C om pute

Part 2. \i t =

Approximate t^^^lO).

V^Js)

(d)

lu { t -

v^Jt).

10 sec, the switch moves to posi

new

input is given by

three

(modified) nodal

tion B. Assume that the /.,(r) =

and

10) A.

W rite a set o f

equations in the ^-domain in terms o f file equa tions in matrix form. Your answer should contain parameter.

R as

an undetermined

(b) F igu re P I 3 .8 0

for

G79


8 1. For the circuit in Figure P I 3.81 ^^(0') = 0. At time

t=0'

both switches arc flipped to posi

itor voltages are zero at f = 0

. Com pute

and //2 (^) at / = O'", r = 2~ , and

v\(t)

t = 2* .

tions A. At r = 1 sec, both switches move back to their original positions, and then back to positions A at r = 2 sec. (a)

Determine the vakie o f yj(O^).

(b)

Determine the value of

lO F

85. In the switched capacitor circuit o f Figure

20 5(t) 5F

CH ECK :

5F

5F

20u(t)V

P I 3 .8 5 , all capacitor voltages are zero at r = 0~. Switches S arc moved to positions A at r = 0 s

Figure P i 3.81

and to positions B at /■= I sec, and then back to

=3 V

A at / = 2 sec. C om pute and plot

v^{t) and V2 {t)

for 0 > /“> 3 sec. 82. Repeat Problem 81 for the circuit o f Figure

^

+

V,

3 mF

-

A

P 1 3 .8 2 . 2m F

20 V

40 8(t)

(!)

2F

6 mF

Figure P I3.85

4F

2F

1 mF

> r i4 F

4F

2m F

10u(t)V

86. In Figure P I 3 .8 6 ,

v^{t) = - 2

V,

= 0,

Figure P I 3.82

and the switch S is in position A. At r = 0 , S is

83. For the circuit shown in Figure P I 3 .8 3 ,

back to A at / = 1 sec and then back to B at r =

moved to position B, after which it is moved = 0. The switch moves from position A

2 sec. Com pute and plot

for 0 > r > 3 sec.

to position B at r =1 sec, back to position A at

t =2

I mF 4e-

sec, and finally back to position B at ^ = 3

sec, where it remains forever. (a)

Com pute

(b)

Com pute t^^„;(3.5).

(c)

Plot

+ v.Jt)

for 0 > r > 4 sec. 2 mF

rlO F

lO F

lO F

f +

Figure P I3.8 6

20u(t)V

87. In Figure PI 3 .8 7 , suppose Figure P 13.83 A N SW E R : r„J3.5^) = 6.25

and

v^{t) = - 2

V, /’ = 1,

the capacitors are initially uncharged.

Suppose switch S is moved to position A at r = 0, 2, 4 , . . . (even integer values) msec and to position

84. In the switched capacitor network o f Figure P I3.84, the switches S are moved to positions A at r = 0 and to positions B at r = 2. Ail capac-

B at r = 1, 3, 5 ... (odd integer values) msec. (a)

Find

0 < t < 20

and plot the waveform for msec.

Chapter 13 • Laplace'Iransform Analysis II: Circuit Applications

680

(b)

Repear part (a) for ^ = 0 .5 .

(c)

Repeat parr (a) for

increasing bur less rhan 8 0 V, and is moved to position B when

k = 2.

v^^ reaches

80 V. (2) S remains at position B when

kC

is decreasing bur is greater rhan 5 V, and is moved to position A when reaches 5 V. Find

and make a

rough sketch of the waveform for one cycle o f operation. C H E C K : charging time s 4.?? msec; discharging time s 3? jisec

M ISCELLAN EO U S 8 8. Use the material on op amp integrator design to achieve the following input-output characteristics. In your final design, capacitors should be 100 nF. 0 .5

(a) V,u,b) = - — \
(b)

=

(C ) V ,

S

0 + - V ;b\

S

(•0 = - —

s

Figure P i3 .8 9 Switching circuit lor gen erating a sawtooth waveform.

S

S

I',h\ 9 0 . (Fluorescent light) Reconsider the 0 .7 5 + —

K .- —

s

s b\

S

fluorescent

1

S

light

starter

circuit

hi

described in the text (Example 13.8).

^

In this problem, suppose the ballast is realistically modeled as an ideal inductor in series with a 100 Q resistor,

as shown in Figure P i 3 .9 0 . Using SPIC E or 89. (Sawtooth waveform generation) This prob

some other circuit simulation program, com

lem uses some simple switching techniques in an

pute the starting voltage,

RC circuit

tial condition on the inductor, as depicted in

to generate an approximate sawtooth

due to the ini

waveform. Sawroorhs are common to a number

Figure P i 3 .9 0 . Estimate the starting voltage

o f devices, such as televisions and test equipment.

from a plot o f the response over one half-peri

Consider the circuit of Figure P i 3.8 9 .

od. Com pute the difference between the peak

(a)

Assume the circuit is initially at rest.

present. Is the lossless circuit o f Example 13.8 a

alternately closed to the left position. A,

good approximation o f the starter response?

for 1 msec and then to the right posi tion, B, for 50 |isec. Find z-'Q(r) for 0 s r ^ 1.05 msec (one q^cle of operation), and sketch the waveform. CH ECK : (b)

voltages with and without the 100 Q resistor

Beginning at time / = 0, the switch, S, is

19 V

T he circuit in the shaded box is a crude model o f a neon lamp (costing less than a dollar) and operates as fol lows: (1) S is at position A when

is

u


681

1000

0.001 pF

Li,(0)5(t) =0.088(t) Time domain representation of voltage due to initial inductor current when heat sensitive switch opens

Very high resistance prior to arcing

Figure P I3.90 Model of fluorescent light starter circuit that includes a ballast resistance of 100 Q. Note that in the time domain, the effect of the initial inductor current appears as an impulse in this model.

U

C

H

A

P

T

E

R

Laplace Transform Analysis IIITransfer Function Applications APPLICATION TO ELECTRIC M OTO R ANALYSIS Electric motors turn fans, run air conditioners, pull trains, rotate antennas, and help us in a wide variety o f ways by efficiently converting electrical energy into mechanical energy. These electro mechanical devices are o f two general types: ac and dc. D c motors are ordinarily used in electricpowered transit cars, often with rwo motors per axle to propel the car. A typical rating o f such a m otor is 140 hp, 3 1 0 V, 2 ,5 0 0 rpm. Another type o f dc m otor is a high-performance dc servo motor, found in computer disk drives and microprocessor-controlled machinery. These motors are very useful in applications where starts and stops must be made quickly and accurately. As an application, we will represent a dc m otor by an equivalent circuit-like model and analyze its oper ation using the transfer function method. O f the several kinds o f dc motors, the t)'pe most pertinent to the analysis techniques o f this chap ter is a permanent magnet type typically found in low-horsepower applications. They are reliable and efficient. Further, for a permanent magnet dc motor, a plot o f the torque produced on the rotating shaft o f the motor versus the input current to the motor is almost a linear curve. Hence, the m otor has a linear circuit-like model that most nearly describes its performance over a large range o f operating conditions. Since the output o f the m otor is a mechanical quantity, e.g., angu lar velocit)' o f the m otor shaft:, the transfer function is the only viable modeling tool available to us at this stage o f our development. Section 9 presents the circuit model o f the m otor and devel ops the transfer function analysis o f its operation. Several problems at the end o f the chapter extend the analysis. In more advanced courses, time domain analysis is developed and used.

CHAPTER O BJECTIVES 1. 2.

Characterize the transfer function of a circuit in terms of its poles, zeros, and gain constant. Use knowledge o f the pole locations o f a transfer ftjnction to categorize generic kinds of responses (steps, ramps, sinusoids, exponentials, etc.) that are due to different kinds o f terms in the partial fraction expansion o f the transfer function.

3.

Identify, categorize, define, and illustrate various classes o f circuit responses, including

fi.S-4

Chapter 14 • Laplacc Transform Analysis III: Transfer Function Applications

the zero-srate (zero initial conditions) response, the zero-input response, the step response, the impulse response, the transient and steady-state responses, and the natural and forced responses. 4.

Define the notion o f the frequency response o f a circuit, explore its meaning in terms of the transfer function, and introduce the concept o f a Bode plot, which is an asymptotic graph o f a circuit’s frequency response.

5.

Introduce the notions o f frequency and magnitude scaling of circuits.

6.

Illustrate the applicabilit)- o f the transfer function concept to a circuit model o f a dc motor.

SECTIO N HEADIN GS 1. 2. 3. 4.

5. 6. 7. 8. 9. 10. 11. 12.

Introduction Poles, Zeros, and the 5-Plane Classification o f Responses Computation o f the Sinusoidal Steady-State Response for Stable Networks and Systems Frequency Response Frequency Scaling and Magnitude Scaling Initial- and Final-Value Theorems Bode Plots Transfer Function Analysis o f a DC Motor Summary Terms and Concepts Problems

1. IN TRO D U CTIO N O ur experience o f using Laplace transforms to calculate responses makes clear that the pole-zero structure o f the transfer function sets up generic kinds o f circuit behaviors: constants, ramps, exponentials, sinusoidals, exponentially modulated sinusoids, etc. Such knowledge leads to a qual itative understanding o f the circuits response. For example, if a pole is in the right half o f the com plex plane, then we know that the response will grow with increasing

t.

Identifying this kind of

behavior allows us to define the notion o f stabilit)' o f a circuit or system. Generally, the pole-zero locations allow us to categorize and com pute various special t)'pes o f responses, including tran sient, steady-state, natural, forced, step, and impulse responses. Coupling the transfer function with the presence o f initial conditions in the circuit permits us to define two further types of responses fundamental to both this text and advanced courses in circuits, systems, and control: the zero-input response (due only to the initial conditions o f the circuit or system) and the zero-state response (due only to the input excitation, assuming that all initial conditions are zero). These time domain notions are balanced by the concept o f the frequenc)' response of the circuit or system. Briefly, the frequenc)' response is the evaluation o f the transfer function,

Jii).

Since

H{Jw)

H{s),

for

s=

has a magnitude and phase, the frequenc)' response breaks down into a magni-

Chapter 14 * Liplacc Transform Analysis 111: Transfer Function Applications

6S"S

rude response and a phase response. A technique for obtaining asymptotic (straight-line) approx imations (called

Bode plots)

is also outlined in this chapter.

As a final introductor)^ remark, unless stated otherwise, all circuits in this chapter are linear and have constant parameter values. Such circuits are said to be

linear

and

time invariant.

Also, for

convenience in this chapter, the symbol Z. will be used to denote either o f two things: (i) the angle o f a complex number, =

L{a + jb)

= arg(/z + jb) =

z^n~^{bla), with

due regard to quadrant, or (ii) Z.({)

The context will determine the actual usage o f the symbol.

2. POLES, ZEROS, AND TH E S-PLANE In all our circuits, impedances

Z{s),

}\s),

admittances

and transfer functions

H{s)

are rational

functions of^, i.e., they are ratios o f a numerator polynomial ;/(j), divided by a denominator poly nomial,

r/{s).

Mathematically,

(i{s) where

s = pj

is a finite pole o f

H{s), s = zj

{ s - p^ ) (s - P2 )...{S - P„)

transfer function. We assume that common factors satisfies M /’,) =

H{s), and A" is the gain constant of the o f n{s) and d{s) have been canceled. A finite pole

is a finite zero o f

which is shorthand for //(s) -♦ x as j “♦ />y, and a finite zero satisfies

the transfer function takes on the value zero at each

z-.

If pj = pj,

i

j,

= 0. i-e.,

the pole is a repeated pole. A

pole repeated twice is second order; one repeated three times is third order, etc. The terminology is the same for zeros. Also, transfer functions sometimes have infinite poles or infinite zeros. If m

-* -

00

, then

m at

H{s)

H{s) has a zero rn - n at infinit)'.

0, suggesting the term “zero at infinit}'.” In such cases,

infinit)'. If, on the other hand,

n < m, H{s)

has a pole of order

< n and s n

of order

O ut o f all this terminolog)' comes one striking fact: transfer functions, impedances, and admit tances are characterized by their finite poles, their finite zeros, and their gain constant.

E X A M P L E 1 4 .1 . A transfer function

H{i)

has poles at j = 0, - 2 , and - 4 , with zeros at j = —1 and

- 3 . A t x = l , / / ( j ) = 4 /3 . Fin d M ^ ). S

o l u t io n

From equation 14.1 and the given locations o f its poles and zeros, the transfer function must have the form (.S -- 1 K5 -C

2

)

^ (i-H l)(5 + 3)

H(s) = K ------------------------ ;--------- T— {s - /;, )(5 - P2 )(s - />3 ) Since

H{\) = 413,

.v(.v + 2 )(.v -I- 4 )

evaluating equation 14.2 at ^ = 1 yields

3

l(l + 2 ) ( l - h 4 )

15

This implies that K = 2 .5 . Equation 14.2 specifies the transfer function with K = 2 .5 .

^

686

Chapter 14 • Laplace Transform Analysis 111; Transfer Function Applications

Exercises. 0 0 , H{s) -*

1. Suppose a transfer function has a zero at j = 1 and a pole at x = - 1 and that as ^

H{s). His) = -Ms - 1 )!{s +

- 3 . Find

A N SW E R :

2. A transfer function known that

H{0)

H{s)

1)

has poles at j = - 1 and - 2 and zeros at j = - 3 and - 5 . It is further

= 15. Find //(oo).

A N SW E R : 2

Because the essential information about transfer functions resides with the poles and zeros, a plot of these locations in the 5-plane, called a pole-zero plot, proves informative.

E X A M P L E 1 4 .2 . Draw the pole-zero plot o f the transfer function o f Example 14.1. S

o l u t io n

T he transfer function given in equation 14.2 has the pole-zero plot shown in Figure 14.1, JW >k ■j

-o--------- ►'G -4

-3

-2

-1 ...J

FIG URE 14.1 Pole-zero plot of //(j) given by equation 14.2, where the poles arc flagged by

and

the zeros by “o.” Since j = a + ya), the real axis is labeled o and the imaginar)' axisyw.

As a second illustration, consider the transfer function (5 + 1)(.V4-3)

(1 4 .3 )

l(5 + l)“ -H l](5 + 2 ) which has the pole-zero plot shown in Figure 14.2.

X

jco A -j

-o---- ----- o-4

-3

-2

X FIG U R E 14.2 Pole-zero plot of

H{s) given

> a

-1 ...J

by equation 14.3. Again, the a-axis represents the real

part of the pole or zero and the /o -axis represents j times the imaginary part of the pole or zero.

Chapter 14 • Laplace Transform Analysis III: Transfer Function Applications

68’

Plots such as those in Figures 14.1 and 14.2 com m unicate much about the nature o f the imped ance, admittance, or transfer function o f a circuit. For example, an

RC input

impedance,

satisfies the following properties: (i) all o f its poles and zeros are on the non-positive C7-axis o f the complex plane; (ii) all o f its poles are simple (o f multiplicit)' 1) with real positive residues, i.e., the coefficients in a partial fraction expansion are real and positive; (iii)

does not have a pole at

j = CO; and (iv) poles and zeros alternate along the a-axis. Proofs o f these assertions can be found in texts on network synthesis.

Exercise.

C om pute the input impedance o f the circuit in Figure 14.3, and show the pole-zero plot

if / ? ,= / ? - ,= 1

Q,

Cj = 0 .2 5 F, and C2 = 0 .5 F. Are the poles on the non-positive a-axis? Do the

poles and zeros alternate? Is there a pole at

s=

oc?

J_ c.

A N SW E R : A’.C ,

More commonly, pole-zero locations provide important qualitative information about the response of the circuit. Pole locations determine the inherent, natural behavior of the circuit, and the poles are commonly called

tmttiralfrequencies.

Howe\'er, the complete set o f natural frequencies o f the circuit

may be larger than the set o f poles o f the transfer function. 'Fhis is because there might have been a pole-zero cancellation in constructing the transfer function. The canceled pole would amount to a nat ural frequency o f the circuit that is not present in the poles o f the transfer ftmction. T he terms in a partial fraction expansion o f the response establish the types o f behavior present in

Kh, pj real in each case. Figure 14.4 sketches each o f the associat ed responses. In Figure l4 .4 a , the term Kh leads to a dc response and KlJ to a polynomial response proportional to . In Figure l4 .4 b , the term K l{s- p-) leads to an exponential response that is increasing if pj > 0 and decreasing if pj < 0. Finally, in Figure l4 .4 c , if pj < 0, the response curve the response. Each term has only one o f several possible forms. Four ver)' com m on terms are

KjJ,

K/(^ - />y), and

K/{s -

with

has a hump. These qualitative behaviors suggest that one important application o f the transfer function is determining the “stability” o f the response; i.e., under what conditions will the circuit response remain finite for all time?

688


T ransform

(a)

T im e R esp onse

— — s

(b) ■ V-

Pi

(c) (S -

Pi)

FIGURE 14.4 Response t)'pcs common to partial fraction expansion terms, (a) The term response and

KJ^ to a polynomial

response proportional to

(b) The term

KJ{s —

Kh leads to a dc

leads to an expo

nential response diat is increasing i f > 0 and decreiising if pj < 0. (c) If pj < 0, the curve has a hump.

Chapter 14 • Laplacc'Iransform Analysis III: Transfer Function Applications

68^)

In addition to the preceding response t)'pes, there is the sinusoidal response associated with terms o f the form

, „ /4.V+ B -----------^------- T

f.v + a r +

Here, i f - o < 0, the response is an exponentially decaying sinusoidal, as sketched in Figure l4 .5 a , and if a = 0, the purely sinusoidal response o f Figure 1 4 .5b results. If - a > 0, the exponentially increasing sinusoidal response o f Figure l4 .5 c occurs.

FIG URE 14.5 Various sinusoidal responses, (a) Exponentially decaying, - a (b) Pure sinusoidal, 0 = 0. (c) Exponentially increasing, - a

> 0.

< 0.

Chapter 14 • l^placeTransform Analysis III; Transfer Function Applications

690

Referring again to Figure 14.5 , the real part o f a pole, i.e., - a , specifies the response. Often, the word

damping is

decay rate

o f the

used. I f -O < 0, then the response is damped and the oscil

lations die out. T he farther -O is to the left o f the imaginary axis, the greater the damping. If a =

sustained oscillation. I f - a > 0, unstable And increases without bound.

0, there is no damping and the response is a tively damped, i.e., the response is

the response is nega

O ne concludes that pole locations specify the type o f time domain behavior o f a circuit or system. A very important type o f circuit behavior characterized by the pole locations is

stability.

STABILITY AND BO U N D ED N ESS A circuit represented by a transfer function

H{s)

is called

stable if ever)' bounded input signal say,J{t), is bounded if |/(^)| < K <

yields a bounded response signal (B IB O stability). A signal, 00

for all t and some constant

K. In other words, a signal

is bounded if its magnitude has a max

imum finite height. Interpreting this definition in terms o f the poles o f the transfer function, one discovers that a circuit or system is stable if and only if all the poles o f the transfer func tion lie in the open left half o f the complex plane. This makes sense, because if any poles were in the right half o f the plane, the response would contain an exponentially increasing term; if any were on the imaginary axis with multiplicity 2 or higher, then the response would contain an unbounded term proportional to

fory > 2; and, finally, if there were an imaginary

axis pole with multiplicity 1, excitation o f the pole by an input o f the same frequency would yield a pole o f multiplicity 2. The corresponding response term would be proportional to fcos(o)r + 0 ), which grows with time— an unstable behavior. W hat this means is that, for exam ple, a unit step current source in parallel with a 1 F capacitor would produce a voltage pro portional to

tu{t). This

voltage grows without bound and would destroy the capacitor and pos

sibly the surrounding circuitry if left unchecked. Such phenomena are considered unstable.

Despite the need for stability, some circuits utilize an unstable-like response for a finite duration. Circuits that exhibit both stable and unstable-like responses are studied in electronics courses. A transfer function with first-order poles on the imaginary axis is sometimes called

metastable.

Such a classification has no practical or physically meaningful significance, since the ubiquitous presence o f noise would excite the mode and cause instabilit}' o f the circuit. Moreover, in power systems engineering, i.e., the study o f the generation and delivery o f electricit)' to homes and industry, transfer function poles that are in the left half-plane, but close to the imaginary axis, are highly undesirable. Such poles cause wide fluctuations in power levels. The situation is analogous to the way a car without shock absorbers would bounce. Much work has been done on how to move the poles that are close to the imaginary axis farther to the left. Moving these poles to the left increases the damping in the system and maintains more stable power levels. Summarizing, the requirement that the transfer function have no poles on the imaginar)’ axis is both theoreti cally and physically meaningful.


Exercises.

1. If

H{s)

=

691

find a bounded input that will make the response

unbounded. 2. If H{s) =

= l/(^^ + 1), find a bounded input that will make the response unbound

ed. Use M ATLAB or some other program to plot the response for 0 ^ r s 10 s. A N SW E R S: fa) I/- (.v) = —

io r k ^

l ;( b ) V;„(a) =

lor

k

> 1

(.V- + 1) E X A M P L E 1 4 .3 . During a laborator}' experiment, a student tried to build an inverting amplifi er, as shown in Figure l4 .6 a . T he student accidentally reversed the connection o f the two input terminals and obtained the circuit o f Figure l4 .6 b . The student was gready surprised that the cir cuit did not behave as expected. Explain this phenomenon in terms o f the stabilit)'^ theory just developed. R^ = 4 k n

= 4kO

(a)

(b)

FIGURE 14.6 (a) Correct wiring of op amp circuit, (b) Accidental, improper wiring of op amp circuit. S

o l u t io n

Assume that the op amp is modeled as a voltage-controlled voltage source with a finite gain o f 10"^ and that there is a very small stray capacitance o f 1 pF across the input terminals. Figure 14.7 illus trates the equivalent circuit model for each o f the circuits in Figure 14.6. = 4kO

R^ = 4kQ

FIG U R E 14,7 (a) Model of correctly wired op amp circuit, (b) Model of incorrectly wired op amp circuit.

6 ‘)2


P art 1:

Analysis o f the correctly wired op amp.

W riting a node equation at

to compute the trans

fer function o f Figure l4 .7 a yields

-av„

- v ;)+ ^ (- v , - v„) =0

A|

A2

After some algebra,

1

Y Substituting

1

v.,-— v,- — v = o

.V + - - - - - - - + •

CRi

’ ’ CR| ’ ’ C « , j

''

CR,

x 10“^ for Ky produces the stable transfer function

\ His) =

10 13

Vi(s)

R^C

1 l+ I O ' .y + ------ +

R^C

with approximate dc gain P art 2 :

^ + 2 .501 X 10

(1 4 .4 ) 12

RjC I

as expected.

Analysis o f the incorrectly wired op amp. Vj.

To com pute the transfer function o f the circuit o f

Figure l4 .7 b , we write a node equation at

C i V , , + ^ ( V ^ - V '. ) + -^ (V r f -V '„ ) = 0 Kj /<2 which produces the transfer function

H{s) =

1013

10"

Viis)

R^C

1 1 -u r 5 + -------- h

. y - 2 .4 9 9 x 1 0

12

(1 4 .5 )

T he transfer function o f equation 14.5 has a right half-plane pole, in contrast to that o f equation 14.4. This implies that the incorrectly wired circuit is unstable, which explains the students con cern over the surprising performance o f the op amp.

A brief interpretation o f the zeros o f a transfer function ends this section. This is best done in terms o f a simple example. Suppose

His) = Let

v- it)

=

e ‘ sin(/)/^(f)

( .v + l )- + l V;.,

(5 + l)U + 2)(^ + 3)

V, so that

1

Assuming that the system is initially “relaxed,” i.e., all initial conditions are zero, we obtain

V^,As) = H is )V :Js ) =

1 (.9 + l)(5 + 2 )C y + 3)

Chapter 14 • Laplace Transform Analysis 111: Transfer Function Applications

693

The time response is

for appropriate constants

A^-.

Observe that the response dies out very quickly and does not have

any term similar to the input signal. This follows because tiie input signal has transform poles, = -1

± J,

s

that coincide with the zeros o f the transfer function. One can think o f the pole locations

in the transform o f the input signal as identifying frequencies that are present in the input. Hence, the effect o f these input signal frequencies (poles) is canceled out by the transfer function zeros, eliminating them from the circuit response.

3. CLASSIFICATIO N OF RESPONSES In addition to the various response behaviors discussed in section 2, there are other general response classifications. Three fundamentally important general response classifications germane to all o f circuit and system theory are the zero-input response, the zero-state response, and the complete response.

Zero-input response: T he response o f a circuit/system to a set o f initial conditions with the input set to zero.

Zero-state response;The response o f a circuit to a specified input signal, given that the ini tial conditions are all set to zero. Figure 14.8 illustrates this idea.

Output Input F(s)

Relaxed Circuit

I

H(s)

>

V(s) = H(s)F(s) Zero-State Response

FIGURE 14.8 Relaxed circuit having transfer function H(s) and zero-state response. Complete response: The response o f a circuit/system to both a given set o f initial condi tions and a given input signal. For linear circuits, the complete response equals the sum o f the zero-input and zero-state responses.

Recall that a circuit is linear if, for any tv.'o inputs,yj(f)

and/jU),

whose zero-state responses are

and_y2 W> respectively, the response to the new input [AT|/j(r) +

K^ 2 ^t)]

is

+

where /T, and A'-, are arbitrary scalars. The circuits studied in this book are linear unless otherwise stated. T he decomposition o f the complete response into the sum o f the zero-input and zero-state responses is important for three reasons: 1.

It is defined for arbitrary input signals.

6 9 ‘»


£~^[H{s)F{s)],

lôr tiie arbitrary i-domain input

F{s).

2.

T he zero-state response is given by

3.

It illustrates a proper application o f the principle o f superposition for linear dynamic net works having initial conditions. T he following example illustrates point 3.

E X A M P L E 1 4 .4 . Consider two linear networks: (i) a linear resistive network, as sketched in Figure 14.9, and (ii) a linear dynamic network, as sketched in Figure 14.10.

= 6u(t)

FIG U R E 14 .9 Linear resistive network.

20 Va(t)

+ Vo(t);

= 3u(t)

2 0 v jt)

■ 1F

= 6u(t)

FIG U RE 1 4 .1 0 Linear dynamic network.

Response o f linear resistive network o f Figure 14.9. For the resistive network o f figure 14.9, vj^t) due to vj^t) with = 0 is v^J^t) = l//(f), and the contribution due to VjJ^t) with vj,t) = 0 is = lu{t). By superposition.

P art 1:

the contribution to

For this type o f circuit there is no initial condition and the complete response consists o f only the zero-state response, which decomposes into the superposition o f each source acting alone. P art 2 :

Response o f linear RC network o f Figure 14.10.

Now consider the dynamic network o f

Figure 14 .1 0 . Suppose the capacitor has an initial voltage o f 2 V at f = 0, i.e., ^^(0) = 2 V. Step 1. W ith

applied, y^(0) = 2 V, and

Vf^set

to zero, the response is

v j t ) = {0.5e-‘ ^ \.5)ii{t)W Step 2 . W ith the input

Vy{t) applied,

y^(0) = 2 V, and

= {-e-^

+

set to zero, the resulting response is

5)u{t)

V


Step 3 . An

incorrect application

695

o f superposition implies that

T he last answer is wrong because the response due to the initial condition has been added in twice. Step 4 . A correct application o f superposition would entail: (i)

computation o f the zero-state response due to

(ii)

computation o f the zero-state response due to

(iii)

vj^t), Vy{t), and

computation o f the zero-input response due to y^(0).

By superposition, the complete response is the sum o f all three. In particular, the zero-state response due to

is

v J t ) = \.%\-e-û{t)V and the zero-state response due to

is

Vobit)

= 3(1 -

e-^n{t)

V

Hence, the complete zero-state response, by superposition, is the sum o f the zero-input response is

le~Ui{i).

and

Further,

Hence, the complete response is = (4-5 - 2.5rO /K r) V

It is important to note that the transfer function is defined only for circuits whose input-output

linear input-output behav H{s), then H{s)[K^Fj^{s) + K2 Fi{s)] = K^H{s)F^{s) + Yf,s) = H{s)Ff^s) is the zero-state response o f the network

behavior is linear. In terms o f the zero-state response, if a circuit has a ior characterized by a transfer function

K2 H{s)p2 {s) = ATj KjU) + KjYîs), where to F^^s). This says that the zero-state response

to [A"j/j(r) +

is

l^^^^ce,

the transfer function model reflects the underlying linearit)^ o f the circuit. The complete response has a second structural decomposition in terms o f the transient and steadystate responses. T he notion o f a periodic signal is intrinsic to these classifications. A s ig n a l/r) is periodic if there exists a positive constant

T such

thaty(r)

=j{t + 7)

for all / > 0. (The restriction

to r > 0 exists because our Laplace transform analysis implicitly constrains our function class to those that are zero for t < 0.) If a signal is periodic, there are many positive constants for which j{t) =J{t + 7) for all t> 0. For example, i f / / ) =J{t + T) for some T and for all t> 0, then it is true for IT, 5T, etc. We define the fundam ental period, often simply called the period, o f / f ) to be the smallest positive constant T for w h ic h /f ) = J{t + T) for all t > 0. Sinusoids are periodic sig nals: sin(2Jtr) = sin(2Jt/^ + 2 k ) with fundamental period T= \. The square wave o f Figure 14.11 is periodic with fundamental period 7"= 2.

Chapter 14 • Liplace I'ransform Analysis ill: Transfer Funciion Applications

f(t)

>k. 1 •

1

2

3

4

5

■ -1 FIGUllH 14.11 A periodic square wave with fiindamental period 7'= 2. This notion of periodicity and, by default, non-periodicity allows us to define the transient and steady-state responses o f a circuit.

Steady-state response: Those terms o f the complete response that satisfy the definition o f periodicity for r > 0. This includes a constant response.

Transient response: Those terms o f the complete response that are not periodic for / > 0 , i.e., that do not satisfy' the definition o f a periodic function for / > 0. Note that a constant response satisfies the definition o f a periodic function.

A circuit response may have no transient part, as illustrated by the sustained sinusoidal oscillato ry response o f the circuit given in Figure 1 4 .1 2a. Further, the steady-state part o f the response may be zero, as in the circuit o f Figure 1 4 .12b, where

is a damped sinusoid. If the a circuit is

unstable, the transient response may blow up, overwhelming the constant or periodic part o f the complete response, as in the case o f

=

{e^‘ cos(10/)

+ 1 5)u{t) V, where the steady state is 15«(/)

V. Note that “transient” here does not mean something that diminishes in importance with time. Most circuits have both a transient and a steady-state response. When the input is constant or peri odic, the circuit response approaches the steady-state response asymptotically for large

t, i.e.,

as the

transient dies out, only if the circuit is stable. For such circuits, the steady state is crucial. Further, when the input is sinusoidal, the steady-state response is easily computed via the transfer function, //(s ), or by the phasor method. Details o f the calculation are presented in section 4.

1F

(a)

1 H

1F

1O

1 H

(b)

FlG U lll{ 14.12 (a) Unstable circuit illustrating the possibility of no transient response, (b) Stable circuit having a zero steady-state response.

Chapter 14 • Laplace Transfomi Analysis III: Transfer Function Applications

6^)'

E X A M P L E 1 4 .5 . Com puting the response o f the circuit o f Figure 1 4 .1 3 provides a simple illus tration o f the decomposition of the complete response into tiie sum of the zero-input and zerostate responses. Also, some rearrangement o f the terms identifies the transient and steady-state responses. -O

R . >

—

P —>

ZJs)

+

c

1 (a)

FIG U R E 14.13 /?Ccircuit for Example 14.5. (a) Time domain circuit. (b) Frequency domain equivalent, accounting for initial condition. S

o l u t io n

Step 1.

Computation o f the zero-state response.

The input impedance can be viewed as a special type

o f transfer function. For Figure 1 4 .13a the output

V(^s)

is the voltage appearing across the input

current source. Hence,

__ R Letting

i-^^{t) =

/q/K^) and

)

= 0, then zero-state response is

Rl,,

Rio .v-i-

Step 2 .

RC

= RI,

Ki t )

RC

Computation o f the zero-input and complete response.

Now, supposing that ^(^0“)

0 , the

zero-input response is the inverse transform o f //(i)[C y^ '(0“)] as per Figure 1 4 .13b. Hence, by superposition, the complete response is

RC

\

u{!)

zero-state response Step 3 .

}e

+

zero-input response

Decomposition into transient and steady-state responses. As

a final point, since the input is

dc, a step function, the complete response decomposes into its transient and steady-state parts as

''c ( 0 = ( v c ( 0 " ) - / ? /o ) ^

transient response

+

Rl(Mn

steady-state response

698

Chapter 14 • Laplacc Transform Analysis 111: Transfer Function Applications

Observe that from the above example the transient and zero-input responses are ordinarily different.

E X A M P L E 1 4 .6 . It is sometimes mistakenly said that the zero-input response contains only those frequencies represented by poles o f the transfer function. To see the fallac)' o f this statement con sider the y?C bridge circuit o f Figure 14 .1 4 . Com pute the zero-state and zero-input responses.

S

o l u t io n

The transfer fiinction o f this circuit is

H{s)

=

zero-state response is always zero. On the other hand, if 'o u t ( 0 = V c ( 0 where

= 1 Q. Thus,

= 0> which has no poles. Hence, the 0, then

)e x p

the zero-input response is a decaying exponential whenever

v^^Qr) ^

0 and

C > 0. Notice that the transfer function has no poles. As a side remark, in this case, the zero-input response is also the transient response, with the steady-state response being zero.

T he phenomenon illustrated by Example 14 .6 occurs because the symmetry o f the resistor values precludes excitation by the current source. Moving the current source to a different position, say, in parallel with one o f the resistors, or changing the value o f one o f the resistors to 0 .5 Q will result in a nonzero transfer function.

E X A M P L E 1 4 .7 . This example looks at a simple initialized series

RL

circuit driven by a cosine

wave, as show^n in Figure 1 4 .1 5 . Let /^(^) denote the circuit response. Suppose

= Acos{t)u{t)

V and /y(0~) = 1 A. The objective is to isolate the transient and steady-state responses from the zero-input and zero-state responses.


6 9 ‘)

USO-) =

1

> m v 1 H 1Q

v.(t)

(a)

FIG U R E 1 4 .1 5

RL circuit

. (b). for Example 14.7. (a) Time domain series

.

.

RL circuit,

(b) Frequency domain equivalent, accounting for initial condition. S

o l u t io n

Using Figure 1 4 .15b and the principle o f superposition leads to the response

4s

T he term the term response

1

+ l ) ( r + 1)] is the Laplace transform o f the zero-state response, and

Y-^^{s)Li^{Qr) = \!{s + 1) is the transform o f the zero-input response. Thus, is e~û{t). A partial fraction expansion o f the zero-state response yields 4.V

-2

(.s + l)(5 - + l)

5+1

the zero-input

2.V +2 r

+ 1

It follows that the zero-state response is -2e~^ti{t) + 2[cos(r) + sin(^)]«(/). Notice that both the zeroinput and the zero-state response contain a transient part, the part proportional to

e~^. A little

rear

ranging shows that the complete response is

i^{t)

=

-e~‘u{t)

+ 2[cos(r) + sin(/)]«{r) A

which implies that the transient response o f the circuit is

—e~û{t)

and that the steady-state

response o f the circuit is 2[cos(/) + sin(f)]/<(r) A.

Exercises.

1. An

RLC network

has transfer function

cos(?)«(t) A is applied, then for very large

t,

H{s)

=

= l/(^ + 1). If an input

approaches a cosine wave o f what form?

A N S W E R : 0 .7 0 7 cos(/ - 45'")

RLC network has transfer function e~^^û{t) A is applied, then for very large t, 2. An

A N SW E R : A constant

//(s ) =

=

\l{s +

approaches what?

1). If an input

= (1

'()()


natural response and e^‘, where k is possibly , in which case /. = j.

Many books on elementary circuits contain two other notions o f response: tlie

forced response.

To explain this we use the term “exponent” to mean

\

in

complex. For example, if sin(r) is part o f a response, then it comes from

Natural response: Fhe portion o f the complete response that has the same exponents as the zero-input response.

Forced response: 'I'he portion o f the complete response that has the same exponents as the input excitation, provided the input excitation has exponents different from those o f the zero-input response.

It would seem natural to try to decompose the complete response into the sum of the natural and forced responses. Unfortunately, such a decomposition applies only when the input excitation is (i) dc, (ii) real exponential, (iii) sinusoidal, or (iv) exponentially modulated or damped sinusoidal. Further, the exponent o f the input excitation, e.g.,

a m f{t)

=

e'^‘ti{t),

must be different from the

exponents appearing in the zero-inpur response. T he natural and forced responses are properly defined only under these conditions. T he decomposition o f a complete response into a natural response and a forced response is impor tant for two reasons. First, it agrees with the classical method o f solving ordinary differential equa tions having constant coefficients, where the natural response corresponds to the complementary function and the forced response corresponds to the particular integral. Students fresh from a course in differential equations feel quite at home with these concepts. The second reason is that the forced response is easily calculated for any o f the special inputs— dc, real exponential, sinu soidal, or damped sinusoidal. For example, if the transfer function is

H{s)

and the input is

then the forced response is simply

H{a)V^“

(1 4 .6 )

To justify equation 14.6, note that the Laplace transform o f the input is

V!{s —a).

Since the com

plete response is the sum of the zero-input and zero-state responses, we have

Complete response = [zero-inpnt response]

+iT '

The zero-input response terms all have exponents different from second term,

\H{s) Vl{s- //)],

V

Hi s ) -----s- a a,

the exponent o f the input. The

has only one term with exponent equal to

a.

fraction expansion o f this term yields

K H{ s) ------- = -------- + [terms corresponding to poles of H{s)] s - (I s - a Using the residue formula to calculate A'leads

H{s) — s-a

s- a

to K = H{a)V.

Thus,

+ [terms corresponding to poles o f H{s)]

Executing a partial


and

L'^[H{s)VI{s - a)]

has a term,

which we identify as the forced response.

By using exacdy the same arguments, it is possible to show that if the input is a complex exponential function

Ve¥, where

both Kand

are complex numbers, then the forced response is

simply

H{s^) \W A complex exponential such as

is a mathematical entity. It cannot be generated in the labora

tory. However, the real part, Re[V^T (or the associated imaginary part), is simply an exponential ly modulated sinusoidal signal, as shown in Figure 1 4 .5 , and is readily generated in a laboratory. A derivation similar to the preceding leads to the conclusion that, if the input is Re[V^^, then the forced response is R e[//W K ^ 1 This relationship o f the input to the forced response prompts some textbooks to define the trans fer function

H{s)

as the ratio o f the forced response to the input, under the condition that the

input is a complex exponential

Ve^‘. This,

however, is not natural and makes one wonder at the

applicability o f such a definition to the broad class o f inputs for which the transfer function is most naturally defined, as covered in Chapter 13 o f the text.

4. CO M PUTATION OF TH E SIN U SO ID AL STEADY-STATE RESPONSE FOR STABLE NETW ORKS AND SYSTEM S Suppose that a transfer function

H{s)

models a

stable linear

circuit containing a total o f

n capaci

tors and inductors. In addition, suppose there are no comm on factors in the numerator and denominator o f H{s) and that the

degree o f the denortiinator ofH{s) is ri. (This means that the effect H{s).) T he goal o f this section is to develop the fol

o f each capacitor and inductor is included in

lowing formula: if H{s) satisfies the aforementioned assumptions, and the input to the circuit has the form

A cos(to/ +

0 ), then the steady-state circu it o u tp u t response has the form

B cos(o)r

+ (})) (14.7a)

where the

magnitude o f the

response is

B = A\H{jco)\ (14.7b ) and the phase

shift is (j) = 0 + L H { f o )

Here we assume that to is some fixed, but arbitrarA^ value.

(l4 .7 c )

'

Chapter 14 • Laplacc Transform Analysis III: 'Iransfcr Function Applications

From an input-output viewpoint, these formulas imply that the frequency response is the steadystate response o f a circuit to sinusoids o f varying frequencies. To construct this formula, suppose

H{s).

again that a linear circuit has a stable transfer function model input is a s i n u s o i d w h o s e Liplace transform is

F{s)

Suppose also that the circuit

with zero-state response

Y\s), as

illustrated

in Figure 14.16.

F(s) ---------->1^

H(s)

^

-------- > Y(s)

Transfer Function FIGURU 1 4 .1 6 Frequency domain representation of hypothetical circuit. Since

H{s)

H{s) has H{s) will poles, p - <

is stable, all poles lie in the open left half o f the complex plane. Assume that

real, distinct poles labeled />j, ...

,

and complex poles labeled

-(tj ±

Consequently,

have a partial fraction expansion containing only two types o f terms; those having real 0, and those having complex poles with negative real parts, i.e.,

A\ A-) H{s) = ----- — ^ +— = -+ s -p \ S -P 2

...+

/4,,,

a- < 0.

Ci.v+D|

— — + --------- — 7 - 7 s - p ,„ (i- + a , r + ( P

real poles with negative real parts

Specifically, T+ -" i)"

(1 4 .8 )

complex poles with negative real parts

It is easy to account for higher-order poles. Suppose now that the circuit is excited by a sinusoidal input o f the form

/(/) =

Acoai OJ/ +

0) =

A-------------- ---------------

having Laplace transform

0.5 Aej^ 0 .5 ^(•v) = --------:— + ---------:— .V - yco

s + jo i

Then a partial fraction expansion o f the Laplace transform o f the zero-state response, F(j) =

H{s)F{s),

has the form

A\ At Y(s) = ----- ^ — +— — s - p\ s - P2

C|.v + D\ R\ R') + . . . + --------- ^ ^----- T + . . . + ----- ^ + (5 + a 1 j" + ( Pi j" - yw A--I- yco

real poles with negative real parts

complex poles ivith negative real parts

steady-state contribution =

In the steady state, i.e., for large

t,

the only residues o f interest are

and /?2 , because the part o f

the time response due to the other terms decays to zero with increasing o f complex variables, we obtain

t.

By the usual methods

Chapter 14 • Laplace Transform Analysis III: Transfer Funaion Applications

.

H(s)(s - j(o

J0.5Aej^

703

0.5Ae~j^\ jw s + jm

A

j

S = J(0

= 0.5AH( j(n)e^ = 0.5A\H(

and /?2 =

................... ( 0.5AeJ^

0.5Ae-j^\

H (s)(s-j(j})\ ------- ;— + -------- ^— I \ s - yo)

s + j(a }

S= J(0

= 0,5 AH (-j( 0 )e~-^ = 0.5A\H(-j(o)\e^^^~j'Û~j^

=

0.5A\H(-j(o)\e^^^^~j^^^^

But \Hi-p)\ = |//(;‘a))| and LH{-j(a) = -Z.H(j(a); hence, /?2 = 0.5A\H(J(a)

^)

Consequendy, die Laplace transform o f the steady-state response when all initial conditions of the circuit are zero is

s-

yo)

S+ jO)

In fact, this is the Laplace transform o f the actual steady-state response, provided that the zeroinput (nonzero initial conditions) response makes no additional contribution. The zero-input response makes no contribution to the steady-state response when one or more o f the following reasonable conditions on the circuit are met: 1 . The network has only practical passive elements, meaning that there are always stray resistances present. 2 . The circuit may have active elements in addition to passive elements, but remains stable in the sense that every capacitor voltage and every induaor current remains bounded for any bounded circuit excitation. 3. The circuit contains a total o f n capacitors and induaors, and the stable transfer func tion, H{s), has n poles. Under conditions 1 through 3,

yj,t) = A\H{jwi)\ cos(o)f + (Z.//(/o)) + 0)) = 5 cos(cor + )

704

Chapter 14 • Laplace Transform Analysis 111: rransfcr Function Applications

Hence, if the input to the circuit lias the form response has the form 0 +

B cos(ojr

A cos(o)r

+ 9 ), then the steady-state circuit output

+ (})), where the magnitude

The next question concerns the numerical calculation o f uate

B = A\H{Jlo)\ and

the phase shift ({) =

//(» .

H{s)

at

s = JiO.

B and

One method is simply to eval

W ith a calculator that easily accommodates complex numbers, this is quite

straightforward. An alternative method is to use the graphical technique o f the next section. At this point, it is instructive to illustrate equation 14 .7 and at the same time compare it with the phasor method studied in a first course on circuit theory.

E X A M P L E 1 4 .8 . In the circuit o f Figure 14.17, ji = 0.5 and

S

= cos(2r) V. Find

v^{t) for large t.

o l u t io n

P art 1.

Phasor method.

From the principles o f phasor analysis detailed in Chapter 10, the phasor

domain circuit o f Figure 1 4 .1 7 at (O = 2 rad/sec is given by the circuit of Figure 14.1 8 .

The single node equation lor

is 1_ =0 ./4 - y

T he phasor solution to equation 14.9 is V ,= Therefore, for large

1 + yO.75

= 0 .8 Z - 3 6 .9 '

t v^{t) = 0 .8

c o s ( 2 r - 3 6 . 9 “) V

(14.9)


P art 2 .

Laplace namform method.

OS

Tlie first step here is to construct the j-domain equivalent cir

cuit, which is given in Figure 14.1 9 .

The single node equation for

in the ^-domain is V i-V ;-

V i-0 .5 V | +■ =0 2.V + 2 2

Solving for the transfer function yields

v,(s)

(1 4 .1 0 ) (.v + ir

From equation 1 4 .1 0 , at j = yto = j l ,

Hijco) = H{jl) =j4l{J2

+ 1)2 = 0 .8 ^ - 3 6 . 9 °

According to equation 14.7, it follows that

v^(t) = 0 .8

c o s ( 2 /- 3 6 .9 ° ) V

In this example, the two methods give the same answers, as expected. Since complex numbers are easier to manipulate than rational functions, what is the motivation for such an analysis using the transfer function

H{s)^ W hy

not stay with the phasor method? T he next example answers these

questions.

E X A M P L E 1 4 .9 . In Example 14.8 with the circuit o f Figure 1 4 .1 7 , let the value o f p be increased to 1.5. Find y,(r) for large S

t.

o l u t io n

P art 1.

Phasor method.

Since only the response for large

t is desired,

the problem appears to be one

involving sinusoidal steady-state analysis. The phasor domain circuit o f Figure 1 4 .1 8 yields the single node equation

V, - 1

V, - 1.5V, = 0


Solving again for V . yields V, = ------ !------ = 0 .8 Z 3 6 .9 ° ' l - /).7 5 Therefore, for large

t, vAt)

= 0.8cos(2^ + 3 6 .9 “) V

A beginner who has just learned sinusoidal steady-state analysis by the phasor method might accept this answer. Unfortunately, the answer is not the voltage

v^{t) for

large r! T he reason is clear

from the Laplace transform analysis, which follows. Part 2:

Laplace transforyn method.

From the ^-domain equivalent circuit o f Figure 14.1 9, v j - y /- ^ 25+2. .9

2

Solving for the transfer function yields

His) =

%{s) Viis)

-2s (s-\ r

o f H{s) in the right half-plane, the circuit is unstable. As f becomes very large, v^{t) approaches infinit)', instead o f 0 .8 cos(2r + 3 6 .9 ") V, as calculated by the

Since there are poles the magnitude o f phasor method.

This analysis demonstrates that the unstable behavior o f a circuit cannot be determined by the phasor method. It is desirable to know when to use a particular method in order to avoid unnecessar)' complicated calculations. The following guidelines help: 1.

W hen the stability o f the circuit has been assured by some means, and co has a specific numerical value, the phasor method is the better method to use for computing the response for large

t,

which is also the steady-state response in this case. Circuits whose

stability is guaranteed include those with only passive elements, such as resistors, capaci tors, and inductors; and amplifier circuits o f well-established configurations. 2.

The circuit is known to be stable, but CD is variable. In this case, the rior to the phasor method. To say the least, we need only write

sL

H{s)

method is supe

instead ofy'coZ.. There

are other advantages to be gained from knowing the pole-zero plot o f

H{s)

that are not

possible with the phasor method. The examples o f frequency response calculations given in the next section clearly demonstrate this point. 3.

If the stability o f the circuit is not yet determined, then

H{s) should

be calculated and its pole

locations checked for stability. Then step 1 or step 2 should be referred to, as appropriate.

Exercises.

1. Suppose a second-order linear circuit having the transfer function /Y(^) =

_

V :Js)

S--0.5S + 5 5 - +0.5.V + 5. 7321

Chapter 14 • Laplace Transform Analysis III; Transfer Function Applications

is driven by a sinusoidal input

=

j2cos{2t + 45°)t^{t).

707

Show that the steady-state response

is given by

2. Consider the

L C circuit

o f Figure 1 4 .2 0 , in which

= 0,

= 0, and

= 100«(r).

Show that the largest voltage to appear across the capacitor for r > 0 is 2 0 0 V. Hint: Show that

V(\t)

= 100«(f) - 100cos(f)«(f).

L= 1 H

+

v,(t) C=1F

(a) FIG U R E 1 4 .2 0 (a) Time domain

LC circuit.

(b) Frequency domain equivalent, accounting for initial conditions.

5. FREQ U EN CY RESPONSE The frequency response o f a linear stable circuit having constant parameters characterizes the input-output behavior o f the circuit to unit magnitude sinusoids, cos(oj/), as ca varies from 0 to 00

. This extremely important concept plays a major role in the analysis and design o f circuits and

systems. In terms o f the transfer function, the frequency response o f a stable circuit is the evalua tion o f

H{$) zi s =jw.

In terms o f phasor analysis, studied in an introductory course, the frequen

cy response o f a circuit corresponds to the ratio o f the output phasor to the input phasor. From the steady-state analysis perspective o f the previous section, if an input has the form cos(ojQr), then the steady-state response (i.e., the response for large out) has the form

B

cos(ojQr + (j)), where

B

t, after

all transients have died

= |//(/(Dq)| and the phase shift ({) = Z.//(/Wq). Here

|//(/Wq)| is the magnitude o f the complex number ^(/W q), and Z.A/(/0)q) is the angle o f the com plex number //(/cOq). Thus, //(/(o ), for 0 s to < x , defines how a linear circuit adjusts the magni tude and phase o f an input sinusoid to produce a steady-state output sinusoid o f the same fre quency, but possibly with a different magnitude and phase. An example o f practical importance is the specification o f a stereo amplifier. Here one specifies the gain, gain-magnitude

\H{jLo)\,

to be more or less constant from 20 Hz to 2 0 kHz. W hy? Because

musical signals are composed o f sinusoids o f different frequencies. Accurate amplification o f the music requires that all component sinusoids be amplified with equal gain.

’ OS


Frequency response: The frequency response of a stable circuit or system represented by a transfer function

H{s)

is the complex-valued funaion //(/w ) for 0 s OJ < oo. The magnitude (frequency)

response is |//(^)| for 0

to < co, and the phase (frequency) response is

A complex-valued function

H{jw) a-^ +

number. A complex number,

is a function such that for each value of to, has a polar form,

(j), the phase angle o f the number. Moreover, if (« j + y /?i)(« 2

+ y' ^ 2

for 0 ^ co < co.

Hijoi)

is a complex

in which p , is the magnitude and another complex number, then

+Jb2^ =

and

a2 + p2

P2

In polar form, the frequency response as a function o f (O is H(Ju)) = p(to)e-^‘^^‘'^^

where p(to) = |//(yto)| denotes the m agnitude response and ({>((0 ) =

is the

angle or phase o f

/LH( jio) -

tan

-I

the frequency response. As in other books,

magnitude response means

the

magnitude o f the frequency response. Typically, frequency response computation requires a calcu lator or computer. We now illustrate the idea o f frequency response with two so-called band-pass transfer functions in which a band o f frequencies is passed with relatively little attenuation while frequencies outside the band are significantly attenuated. E X A M P L E 1 4 .1 0 . Consider the two transfer functions

Q.25s and

.v^-h 0 .2 5 5 + 1

^ _________________ 0 -0 6 2 5 . v^_________________ ^ ‘

/

+ 0.35355.s-^ + 2.0625.V - + 0.35355.V + I

Using the M ATLAB code »nl = 0.2 5 *[1 0]; »dl = [1 0 .2 5 1]; >>n2 = 0 .0 6 2 5 * [1 0 0]; >>d2 = [1 3 .5 3 5 5 e -0 1 2 .0 6 2 5 3 .5 3 5 5 e -0 1 1]; >)w =0.2:0.005:2; »hl = freqs(nl,dl,w ); »h2 = freqs(n2,d2,w); »plot(w,abs(hl),w,abs(h2)) »grid »xlabel(‘Frequenc}' r/s’)

Chapter 14 • Liplace Transform Analysis 111: Transfer Function Applications

"O ')

»yIabcI(‘Magnitude response’) »gtext(‘2nd O rder BP’) »gtexr(‘4th O rder BP’) we obtain the magnitude response plot given in Figure 14.21,

O) 1/1 c

o

CL

t/i O ) u. a; -o 13

‘c cn (TJ

Frequency r/s FIG U R E 14.21 Magnitude responses of a second-order and a fourth-order band-pass type transfer function. Tlie fourth-order response has steeper sides (sharper cutoff) and a flatter top. From the transfer function and by interpolation on the plots, one observes that at co = 0 the mag nitude is zero and at co = x the magnitude is also zero; at (O = 1 the magnitude peaks, and this frequenc)' is called the center frequency. This is characteristic o f a band-pass type o f response. T he fact that the response peaks at OJ = 1 rad/sec means that the transfer functions are “normalized.” Transfer functions o f practical band-pass circuits have much higher center frequencies. Such fre quencies can be obtained by the technique o f frequenq' scaling, taken up in the next section.

A very important system theoretic relationship is that o f the poles and zeros o f the transfer func tion to the magnitude and phase responses. In iJie above example for

H^{s),

the pole-zero plot is

given by Figure 1 4 .2 2 . One immediately notices that the poles are very close to the point yl on the imaginary axis with the magnitude response peaked. Further, the zeros at w = 0 and where the magnitude response is zero.

co = x

are

710


Pole-zero plot

1

- I --------------------- !--------------------- 1--------------------- :—

1^

1

!

0.8 :

0.6

:

0 .4

S. 0.2 :

I

0 -

-

■

0.2

-0 .4 -

-

0.6 :

0.8 -1

1

\

-1

-0 .5

f

1

1

0 .5

1

1.5

— I--------------------- --------------------- --------------------- :

-1 .5

0

Real part FIG U RE 1 4 .2 2 . Pole-zero plot of H^{s) in Example 14.10. To emphasize this qualitative discussion we consider the following example. E X A M P L E 1 4 .1 1 . Iwo circuits have transfer functions

H^{s)

and / / 2 W, with the pole-zero plots

shown in Figures 14 .2 3 a and b, respectively, and gain constants o f 1. Qualitatively speaking (with out doing any com putations), what can we deduce about the magnitude response o f each circuit? To verify our qualitative deductions we will use M ATLAB to construct the exact magnitude response. JCO A .. 12j

jO) A. X

-• 12j

-

8j

-

8j

-

4j

-

4j

0

----------© > II

a

-

-4j

■ -4j

-

-8j

- -8j

- .- 1 2 j T

- .-1 2 j >r

(a)

(b)

X

FIG U R E 1 4 .2 3 Pole-zero plots for Example 14.11. (a) H^{s) (b) Hjis).


S

11

o l u t io n

P art 1.

H^{s)

Qualitative analysis o f Figure 14.23a. This figure tells us that there are r\vo finite poles o f ±\2j, but in the left half o f the complex plane, and that there is a finite zero at the ori

near

gin. So there is also a zero at y = x . Thus

\H{jto)\ -♦ 0 as ( 0 -» 0 \H{j(o)\ -* 0 as (O -♦ oo (iii) \H{jco)\ = maximum value (i)

(ii)

as co -♦ 12

In the case o f point (iii), we can say in general that in some neighborhood o f to near the pole, the transfer function peaks in magnitude.

Qualitative atialysis o f Figure 14.23b. This figure tells us that there are again rvvo finite poles Hjis) near ±12y, but in the left half o f the complex plane, and that there arc two finite zeros at

P art 2 . of

the origin. So there is no zero at ; = oo. Hence

\H{jco)\ -♦ 0 as oj -» 0 \H{jo))\ constant as to -» (iii) \H{jo})\ ~ maximum value (i)

(ii)

oo as co

12

Again we cannot make a stronger general statement in point (iii) above. P art 3 .

Magnitude plots via MATLAB.

Suppose the poles are at -0 .1 ± y l2 and we desire the actu

al magnitude response plot for A/, (y’co), 0 < co < 3 0 rad/sec. To construct the plot shown in Figure 1 4 .2 4 , we use the following M ATLAB code: »w = 0:. 1:2 0 ; »n = [1 0]; »d = poly([-0.1 + 12*j -0 .1 -1 2 * )]); »h = freqs(n,d,w); »plot(w, abs(h)) »grid »xlabel(‘Frequency in rad/sec’) »ylabel(‘Magnitude H 1 (jw)’)

FIG URE 14.24 Magnitude frequenc)' plot, |//,(yo))| vs. to.

’ 12


Exercises.

H^{s)

1. Suppose the poles o f

are at -0 .1 ± j\ l and that we desire the actual magnitude

response plot for 0 < to < 3 0 rad/sec. Use iMATLAB to construct this plot. Verif)' the accuracy of the qualitative predictions. 2. Show that the transfer function

1

■> .V" +

H{s) =

o f an unknown circuit with

R=

1

1 s~ +

LC •.y+

RC

(1 4 .1 1 )

LC

Z. = 0.1 H , and C = 1 mF has the band-reject magnitude

response depicted in Figure 14 .2 5 . Use M ATLAB and the “freqs” command.

Frequency in r/s

FIGURH 1 4 .2 5 Plot of the magnitude response of the transfer function of equation 14.11.

The example below further illustrates this relationship with a pedagogicaily useful graphical tech nique. Mastering this technique helps concretize the meaning o f magnitude and phase and rein forces the qualitative discussion above on using pole and zero locations to compute the magnitude and phase.

E X A M P L E 1 4 .1 2 . 'lb better grasp the ideas o f the magnitude, |/y(/'to)|, and the phase, Z.//(ja>), o f a frequency response, suppose a transfer function has the form

H{s) = where Zj =

2j and

= -2j.

(s-Zi)(s-Z2) is+]){s + \ + j ) { s + \ - j )

Figure 14 .2 6 a shows the pole-zero plot o f

(1 4 .1 2 )

H{s).

Chapter 14 • Laplace Transform Analysis ill; Transfer Function Applications

~ 13

JO ) JO )

A

X

()

2j

■ ■

j

-1 X

I

-J

-2j

(a) FIG URE 14.26 (a) Pole-zero plot

of H{s) as given

by equation 14.12. (b) Measuring distances

graphically from zeros and poles to the point j\. S

o l u t io n

The plan o f this example is to compute the magnitude o f pute

\H{J\)\. This

H{jio) graphically

for oj =

1

, i.e., to com

computation entails the following steps.

Step 1. Observe that j\

-

plex plane. Similarly,yi -

defines a complex number. Think o f J\

Zj and J] - p-, where /> ] = —!

-

2

, as a vector in the com

and p 2 ,^ = - 1 ± 7 . define vectors. Each vec

tor has a length that can be determined either graphically, by physically measuring the distance with a ruler, or by the Pythagorean theorem. Figure 14 .2 6 b illustrates the idea. Step 2 . Following from step 1, the magnitude

\j\ -y'2| x N y i)| =

o f //(/I )

has the form

1/1 +y2|

1x3

I./1 + l|x 1/T -h 1 + ./|x 1/T + 1 - ./ I

= 0 .9 5

V I X V5 X 1

H{j\) graphically. - z - ory ( 0 - p- can be

Step 3. Suppose we wish to compute the phase or angle o f

In Figure l4 .2 6 b ,

observe that each complex number viewed as a vector jio

represented in the

form

x^\

where ip is the angle the vector makes with a horizontal line passing through its base.

For example,

{j\ - J2)

= -j\

=

From basic complex number theory, the angle o f the prod

uct o f two complex numbers is the sum o f the angles, and the angle o f the ratio of two complex numbers is the angle of the numerator minus the angle o f the denominator. Fience, from the angles shown in Figure 14 .2 6 b ,

LH ijl) = L { J \ - jl)

+

L{j\

+jl)

- a j\

+ 1) -

= _ 9 0 " + 9 0 ° - 4 5 ° - 0 ° - 6 3 .4 ° = - 1 0 8 .4 °

Uj\

+ 1

-

Uj\

+ 1 -j)

Chapter 14 ♦ Laplacc Transform Analysis III: Transfer Function Applications

To extend the ideas o f Example 1 4 .1 2 , let

=

(1 4 .1 3 )

{s - P\){S - P2 )---{s- P„)

Because the magnitude o f a product o f complex numbers is the product o f the magnitudes o f the numbers, and because the magnitude o f the ratio o f two complex numbers is the ratio o f the mag nitudes o f the numbers, the general form o f the magnitude response o f equation 14.13 is

Zx

-

|yto-C2 •••

- 1>\ j M - f h

(1 4 .1 4 a )

••• ,/ w - /^„

Similarly, since the angle o f the product o f complex numbers is the sum o f the angles o f the num bers, and since the angle o f the ratio o f two complex numbers is the difference in the angles o f the numbers, the general form o f the phase response o f equation 1 4 .1 3 is = [Z-(/to -

Thus, qualitatively speaking,

H{jn))

Zj) + ••• + /-(/CO M jco-pj) + ••' + /- ijco-p^)]

+ /-(yw -

[Lijco-p^)

+

+

LK\

(1 4 .1 4b)

tends to have a large magnitude for jco’s near poles and a small

magnitude for jco’s near zeros. As mentioned earlier, this can be used to advantage in estimating the magnitude response and phase response o f a transfer function.

Exercise.

Draw an estimate o f the general shape o f the magnitude and phase response o f the

Butterworth normalized low-pass transfer Kinction

M i> =

5“ +

Jls

-I- 1

Com pute the exact magnitude and phase at lo = 1. W hat happens to the magnitude and frequen cy response if H{s) is changed to

H^{s)

=

H{sl\Q)

and

H^{s)

= / / ( / / 1 00)?

A N SW ER S: in random order: - 4 5 . O.TO"^, the general shape is the same wich

H

= /7 i(/]()0 )

= 0 .7 0 7 .

6. FREQ U EN CY SCALIN G AND M A G N ITU D E SCALIN G Design of filters and amplifiers often begins with a design template in which almost all parameter values are normalized. In particular, a source or load resistance is often set to 1 t l . Also, a critical or im portant frequenq^ is set to 1 rad/sec. Such circuits are called

normalized.

W ith the comple

tion o f a normalized design, engineers can frequency-scale to obtain realistic frequency responses and magnitude-scale to obtain reasonable impedance levels as necessary to meet power and energ)' restrictions.


FR EQ U EN C Y SCA LIN G E X A M P L E 1 4 .1 3 . The circuit o f Figure 1 4 .2 7 realizes the transfer function o f equation 14.11 from the previous section for /? = 1 Q , Z, = 0.1 H , and C = 1 mF. We pose the following question.

j

Suppose

L

->

C

----- , C

Xf and

R remains

the same in the circuit; what happens to

H{s)

and the frequency response plot?

L

FIG U R E 1 4 .2 7 Band-reject

type circuit

that realizes the transfer function of equation 14.11.

S O L U T IO N Step 1.

Calculate the circuit transferfunction.

*

R

H{s) =

R +Cs Step 2 .

By voltage division,

LC RC

------

LC

Ls

Incorporate the effect o f /y.R

R R+

C

R+

1

■s +

J

s\

I

Kf

I

s V

+

1

LC

We conclude that when

^ andA T C

K with

R unchanged,

^

K

the new and old transfer functions are related in a ver>' simple way:


/

\ .y

A'

I )

Step 3. Plot the magjiitude response. Rather than go directly to M ATLAB, consider that

to Kf I

ôld

Table 14.1 depicts this relationship for specific values o f id. TABLE 14.1 Transfer Function Relationship

to (rad/s) 0

I 1

\

~ ôld J \ ^ Jf / ^ newU^f

Xf

2«f

)

)~

^ôUl ( / " ^ l = H .,u(j) \«f] \ ' 2K j ôld = « „ w (y 2 ) /

In general, the new transfer function is related to the original

s by s!K^. T he

H^[J,s)

through the replacement of

implication o f this relationship is that whatever happened at ^ =

must now happen at

s=

where

the term

before scaling

frequency scalhig.

In terms

o f the plot o f our specific transfer function, then, we obtain from Figure 14.25 the new plot o f Figure 14. 28 by inspection.

F I G U R t 1 4 .2 8 T h e new m ag n itu d e respon se after “freq u en cy sca lin g ” by

Kj-

Chapter 14 • Laplacc Iransform Analysis 111: Transfer Function Applications

For general linear circuits, vvc will only

state the

procedures for scaling and their effects, leaving a

rigorous mathematical justification to a more advanced text on circuit theory. Procedure for frequency scaling: To frequency-scale a linear network by a positive scale factor 1.

Divide all inductances and capacitances by / y

2.

Leave all resistance values and controlled source parameters unchanged.

Kj-

Effect o f frequency scaling: 1-

which means that 3.

has zeros and poles at

KjrZ- and

The magnitude and phase response cur\'es o f

for / = 1, . . . , w and

are those o f

k=

with the fre

quency' scale multiplied by / y Conversely, the magnitude and phase o f //(/to ) are the same as

Exercises.

1. Fill in the details o f the derivation o f /

2. Given point 3 above.

in step 2.

—K f Z \ ) " ‘{ s —K fZf/i) --------- ---------------------------U- K - K^p„)

\ //-m { s

= KlK r)

3. Given the circuit transfer function

H(s) = K

r

u

n

, verify the “converse statement in

(5 - /? l)( i’ -/> 2 )

find the new pole and zero locations if the circuit is frequency-scaled by

1000.

AN SW I-RS: l0{)0/>,. !()()()/>„ and HHHb,

E X A M P L E 1 4 .1 4 . T he circuit o f Figure 1 4 .29a has transfer function

OAs i*” + 0.1

1

Figure 14 .2 9 b shows the pole-zero plot and Figure 14 .2 9 c shows the magnitude response with peak value at

= 1 rad/sec.

(a) Frequency-scale the circuit by the factor /y = 10^ and compute the new transfer function. (b) C om pute the new pole-zero plot and the new magnitude response curve.

/YY\ -I-

V J

-

)

>

0.1 Q >

(a)

V

718


JCO

FIG U R E 14.29 (a) Series resonant circuit, (b) Pole-zero plot, (c) Magnitude response. S

o l u t io n

(a) Dividing the capacitance and inductance by

Kj-=

10*^ yields the circuit o f Figure 14. 30a.

V

JO )

(-0.05+ j0.999)10^

X for

Hnew(s) o—

{-0 .0 5 -j0 .9 9 9 )1 0 "

X

(b) FIG U R E 1 4 .3 0 Demonstration of the effects of frequency scaling, (a) Frequency-scaled circuit, (b) New pole-zero plot, (c) New magnitude response curve.


Using “effects o f frequenc)' scaling” property 2 , the new transfer function is

S

10^/

10^ /

/ 5 ^

( S

+ 0.1

+1

lio^/

UOî

(b) The ideas stated in part (a) are borne out by comparing the pole-zero plot and the magnitude response curve for

shown in Figures 14. 30b and c, with their counterparts for

Figure 14 .2 9 . Notice that the pole locations have been scaled by

Kj-^s

H{s)

in

per property 2. A direct

computation in M ATLAB verifies this: »pnew = r o o t s ([l/le l2 0 .1 /l e 6 1]) pnew = -5 .0 0 0 0 e + 0 4 + 9 .9 8 7 5 e + 0 5 i -5 .0 0 0 0 e + 0 4 - 9 .9 8 7 5 e + 0 5 i »pold = roots([l 0.1 1]) pold = -5 .0 0 0 0 e -0 2 + 9 .9 8 7 5 e -0 1 i -5 .0 0 0 0 e - 0 2 - 9 .9 8 7 5 e -0 1 i »% Multiply pold by Kf and check with pnew »le6*pold ans = -5 .0 0 0 0 e + 0 4 + 9 .9 8 7 5 e + 0 5 i -5 .0 0 0 0 e + 0 4 - 9 .9 8 7 5 e + 0 5 i Also, in Figure 1 4 .2 9 c for 3 0 c for

H{s),

the peak response occurs at co^/^= 1 rad/sec, whereas in Figure 14.

the peak response occurs at

= 10^ rad/sec.

At this point we round out our discussion o f frequency scaling by relating it to the time domain via the time/frequency scaling property o f the Laplace transform:

£ or, equivalently.

K fl

=H

( s \ \ j /

where H{s) is the circuit transfer function and h{t) is the circuit im pulse response.

'2 0


E X A M P L E 1 4 .1 5 . Consider the transfer function

H(s) =

30

10

10

(5 + 2 ) ( 5 + 5 )

{s + 2)

(5 + 5)

whose impulse response is

h{t)

\Oe-">‘)u{t)

=

Find the impulse response when the transfer function is frequency-scaled by S

Kj:

o l u t io n

T he scaled transfer function is 10^ ,

f-lneAs) =

(5 + 2 /r y )(5 + 5/C|)

ioa :,

(^ + 2A:/-)

(5

+

5

/:^ )

Taking the inverse Laplace transform yields /W (0 =

- \QKfe~^Âu{t)=Kf h[Kfi)

Suppose for the sake o f argument that / y = 10^. Let us plot Figures 1 4 .3 la and b.

t In mllli-sec (a)

h{t)

and

as has been done in


^21

t in milli-sec (b) FIG U R E 14.31 Illustration of the effect o f frequency scaling on impulse response. (a) Original h[t). (b) h„^J,t). Observe that the plots are structurally the same. However, the time scale in Figure 14.30b is in milliseconds and the magnitude has been scaled by

Exercises.

Kj~=

10-^.

1. Use the Laplace transform time scale propert)' to show that in general

2. In Example 14 .1 5 , suppose the poles are changed from - 2 and - 5 ro - 3 and - 6 , respectively. Suppose further that the transfer function is scaled by

10. Find the impulse response and the

new impulse response after scaling. A N SW ER : /;(/) -

- 10^--^’0//(/) and

h,

(/) = (lOOt’"-^"' - I

M A G N ITU D E SCA LIN G Frequency scaling has allowed us to obtain realistic frequency responses from normalized circuits. Another technique in achieving a realistic design is magnitude or impedance scaling. A simple example illustrates the idea.


E X A M P L E 1 4 .1 6 . Figure 14 .3 2 a shows a “normalized” voltage divider circuit having voltage ratio

ôuJîn = ^-5

= 2Q.

and consisting o f two 1 Q resistances. T he input impedance is

Suppose

both resistances are made 1 000 times larger, as shown in Figure 14 .3 2 b . W hat happens to the volt age ratio and input impedance?

lOOOQ

1O —

------0 +

—

—

—

^

10000

1O -O

(a)

(b)

F K iU R E 1 4 .3 2 (a) A normali7.ed voltage divider circuit, (b) The voltage divider of part (a) magnitude-scaled by 1000.

S olution By direct calculation, the voltage ratio remains unchanged, i.e.,

1000

out

\ V-in

I new

10004-1000

=0.5 =

out

\V ''in

old

T he new input impedance is z/;f"'(.y) = 1000 X 1 -h 1000 X 1 = which is 1000 times larger. T he elements in Figure 14 .3 2 b are said to be

magniti(de-scaled (by

1000) from those o f Figure 14 .3 2 a .

The above example motivates a more general discussion. Suppose each impedance in Figure •14 .3 3 a is scaled (multiplied) by

to yield the circuit o f Figure 14 .3 3 b . As in Example 14.16,

the voltage ratios remain the same for both circuits: 1

.

.

(«)

Further, the input impedances o f Figures 14 .3 3 a and b are related as

4 , (,»•) = K,„Z^ W

+

K,„Z2U) = K,„Z“„ (,v)

Similar to the case in Example 1 4 .1 6 , the input impedance is increased by the scale factor

The

network o f Figure 14 .3 3 b is said to be obtained from that o f Figure 14 .3 3 a by magnitude scaling with scale factor tance is =

If Zj is an inductance, then

mtiltipliedhy K^. = sl{CIK^, i.e.,

Z^{s)

=

Ls and

On the other hand, if Z , is a capacitance, then the capacitance is

dividedhy

i.e., the induc

Z^{s) =

1 /G and

723


Z,(s)

K .Z ,(s)

-o +

V r A J ---w Z

Z,(s)

-o (b)

(a)

FIG U R E 1 4 .3 3 . (a) A general voltage divider, (b) Magnitude scaling of the circuit in part (a) by T he above discussion suggests that magnitude scaling can be achieved by multiplying resistances and inductances by

and dividing conductances and capacitances by

This and several other

properties are stated next. P rocedure for m agnitude scaling: To magnitude-scale a linear network by a scale factor

K^.

1.

Multiply all resistances and inductances by

2.

Divide all capacitances and conductances by

3.

For current-controlled voltage sources (C C V S ), i.e., the

4.

For voltage-controlled current sources (V C C S), i.e., the^^^ typ^> divide the parameter

5.

by Parameters for voltage-controlled voltage sources and current-controlled current sources

type, multiply the parameter

remain unchanged. 6.

Ideal operational amplifiers remain unchanged.

The effect o f magnitude scaling on a transfer function is set forth below, with its verification left for more advanced courses on circuit theory. Effect o f m agnitude scaling on transfer functions: If magnitude scaling has no effect on

H{s).

H{s)

is a voltage ratio or current ratio,

If H{s) has units o f ohms, then the magnitude-scaled net

If H{s) has units o f siemens or mhos, then the magnitude-scaled net

work has work has

E X A M P L E 1 4 .1 7 . The series circuit o f Figure 14.34a has input impedance ,

Z{s)

I s 2 + 0 .I ,v + l = — i-5-k 0.1 =

s

C

and transfer function

0.l5 v;-,,

(a) Magnitude-scale the network by (b) Calculate

and

Z(5)

.r-hO.I.y+1

= 1000.

to verify the effects stated above.

/24

Chapter 14 • Liplace Transform Analysis III: Transfer Function Applications

_JYY\ Im F V

1000 H

V.

Znew(s)

(b) FIG U R E 1 4 .3 4 Magnitude scaling, (a) Original network, (b) Network magnitude scaled by 1000. So

lu t io n

(a) Following the procedure for scaling

R’s, L's,

and C s, we obtain the scaled network shown in

Figure 14 .3 4 b . (b) For the simple circuit o f Figure 14.34b , the new input impedance is

Z„e,,.(s) =

+ 1 0 0 0 s + 1 0 0 = 1 0 0 0 ‘^ + 0 - 1 ^ + ^ ^ l O O O Z (s ) =

K,„Z{s)

By the voltage divider formula, the new transfer function is

H

100______________ l l i

1000

.y“ + 0 .h + l

.V-+0.1.V + 1

= H(s)

Because the transfer function is a voltage ratio, there is no change. These results clearly illustrate the stated effects o f magnitude scaling.

Exercises.

1. Magnitude-scale the circuit o f Figure 1 4 .34a by

= 5.

A N SW ER : O.S 12. S H, 0 .2 12. Two parallel resistors

and

/? 2

are magnitude-scaled by

Hint. Use the formula for parallel resistance to compute that to

Verify that

old Rf,q =^mêq

and then relate

■

COMBINED MAGNITUDE AND FREQUENCY SCALING Moving from normalized circuit design to realistic circuit design ordinarily entails both magni tude and frequency scaling. This subsection provides several illustrative examples.


EXAM PLE 14.18. The amplifier circuit shown in Figure 14.35a consists of r\vo stages. The first stage is a Sallen and Key low-pass active filter. The transfer function of the Sallen and Key circuit is

T he gain o f the second stage is

H2(s ) =

V Ôllt____-'> V,

.v +

Therefore the transfer function o f the amplifier is

His) =

V Ôlll _ V * out. V ^(1^ = ........................... y. y y

-1 - " ^ 1 — ------------------------------------------------------------ {s + 1) (.v“ + ^ + 11 5 ' + 2 . r + 25 + 1

FIGURE 1 4 .3 5 A third-order Butterworth low-pass filter, (a) The original network, (b) The scaled network with

K^= 2000ti

and

= 10,000;

= 3 1.4 uS.

The overall transfer function o f the two-stage amplifier is a third-order, maximally flat (Butterworth) low-pass filter, which will be studied in Chapter 19. The magnitude response o f the transfer function has a 3 dB down frequency at OJ = 1 rad/sec, or l/(2ji) Hz = 0 .1 5 9 Hz— an extremely non-useful audio frequency. In fact, when the 3 dB down frequenc)' is at (o = 1 rad/sec, the magnitude response is said to be

normalized.

N ot only is 1 rad/sec not useful, but the element

values are unsuitable for practical applications. However, using both magnitude and frequency scaling, this “reference” or normalized amplifier circuit can be made into a very practical filter.


Suppose we wish to have the 3 dB down frequency a t^ ^ ^ = 5 0 0 Hz and the largest capacitor at 10 nF. Such a filter could be used to direct the low-frequency content o f a music signal to a woofer. O ur goal requires that we frequency-scale the circuit by

Kj-= IOOOtt and

the 2 F capacitor (the largest) becomes 10 nF; i.e.,

must satisfy

10 X K )-'' = Solving for

^

^

K jK ,„

lOOOrcK,,,

magnitude-scale such that

yields

K^, m = 6 .3 6 6 X lO'^ T he scaled circuit meeting the requirements is shown in Figure 14 .3 5 b , where

Exercise. A circuit has transfer function H{s) =

(5") cs — = -^ ------------- ,with Avj s + fl.v + h

After both magnitude and frequency scale changes o f and

= 4 and

= 3 1 .4 |iS.

a = Q.\, b = Aand c = ^ 0 .

Kjr= 2,

compute the new

a, b,

c.

A N SW E R : in random order: - 3 2 0 . 0 .2 , and 16

Examples 1 4 .1 5 and 1 4 .1 6 illustrate one o f several reasons for scaling a linear nersvork. Chapter 19, on elementary filter design, will derail further applications.

7. INITIAL- AND FINAL-VALUE THEOREMS In system theory, and especially in control theory, engineers want the output signal o f a circuit or system to track a given reference signal. T he idea behind the term

nack

is that for large t, the ref

erence signal and the circuit output are more or less indistinguishable. To accomplish this, design/control engineers generate an error signal, put and

e{i) = y(i) ~ yref^)^ where

is the circuit out

is a given reference signal. Since much o f the analysis is done in the frequency

domain, one often knows

E{s)

=

Y{s) -

without knowing the related time functions.

Ordinarily, the design engineer needs to know the initial error, if(0), and the final error, f(co). Available to engineers are the initial-value theorem and the final-value theorem, which permit the computation o f these quantities in the frequency domain.

Initial-value theorem: Let

= F{s)

be a strictly proper rational function o f j, i.e., the numer

ator and denominator o f F a r e both polynomials in

s,

with the degree o f the numerator less than

that o f the denominator. Then lim 5fCO = / ( 0 ^ )

Proof: The.

quantity x/'(y) suggests a derivative operation

sF {s)-f(Q -)= L

on j{t).

y/(0 dt

(1 4 .1 5 )

Specifically,


/ 27

Applying the definition o f the Laplace transform to the right-hand side and taking limits as

s

approaches infinity implies that l i m [ j f ( .v ) - / ( ( ) " ) ] =

lim s-*»

,V -» X

(1 4 .1 6 ) lim C h O e -^ 'd t v-»oc J O

-t- lim

S—-X

’ C h t)e-^ 'd t J{)

where the dot over the function/(r) indicates the derivative o f the fimction. Observe that

fiOe'^'dt

lim S -*

and that, because

e

=

/(/)

Wm (e~")dt = 0

30

is continuous at r = 0, lim

.V-*30

lim

j! r

s-*x

e -.90 f j . /(r)(/f = / ( 0 " ) - / ( 0 ).

(1 4 .1 7 )

Hence, the left-hand side o f equation 1 4 .1 6 equals the right-hand side o f equation 14.17. Equation 14.15 follows from equating these two sides and canceling th e /(0 “) terms in both.

E X A M P L E 1 4 .1 9 . T he Laplace transform o f a capacitor voltage is given by 2

1

.V

5i + 2

Find the initial capacitor voltage S o l u t io n

By direct application o f the initial-value theorem, Vf-fO

EXAMPLE 14.20.

)= lim

i->oo

.vV(^('.yJ=

lim

2-

= 2 - - = 1.8 V 5

Let the Laplace transform o f the velocity of a certain projectile be given by ,

500^ + 20

V{ s) = ----------------------------i-(5 5 -h 2 0 )(1 0 .v + l)

Find the initial velocity, y(0'^), and the initial acceleration, //(O"^). S o l u t io n

l b find the initial velocity, we directly apply the initial-value theorem: v(O^) =

lim sV(s) = lim S-^ o o

s —* x

5 0 0 ^ -f 2 0 (55 + 20)(10. v+ 1)

=0

Since acceleration is the derivative o f velocity, and the initial velocity is zero, from the time dif ferentiation propert}'^ o f Table 12.2, assuming the velocity is continuous at f = 0,

•2S


/\(.v) =

sV{s) -

500.V + 2 0

v(o^

(5.s + 2 0 )(1 0 .v + l)

From the inirial-value rheorem in standard units,

20 f/(0 ) =

Urn

.V-

5 () 0 i- +

20 s

=

(5.V + 20X10.S + 1)

Urn

i-ôo

5.V +

20 \

=

s!

.y /

Exercise.

Suppose

F{s)

= (8j +

2)1{Isr

10

IO.V+-

+ 85 + 3). Fin d y(0‘^).

.\N SW ER : 4

T he initial-value theorem and the examples that follow it illustrate the computation o f initial val ues. To compute final values vve use the next theorem. Final-Value T h eorem : Suppose

F{s) has

poles only in the open left half o f the complex plane, with

the possible exception o f a single-order pole at ^ = 0. Then

lim sF{s) = Urn / ( / ) . (1 4 .1 8 )

Proof:T\\Q condition

o f the theorem, i.e.. the condition that

F{s)

has poles only in the left half of

the complex plane, with the possible exception o f a first-order pole at the origin, guarantees that the limit on the right side o f equation 1 4.18 exists. This is because a partial fraction expansion of

F{s)

leads to a time fu n c tio n //) that is a sum o f exponentially decaying signals and at most one

constant signal. Since the right-hand limit is well defined, /l//l . v F ( .v ) - / ( 0 s-^0

) = /mi r " / ( f k - ^ ' c / r s-*Q JO

= { lim

\/-»X

/ ( / ) ''- / ( ( ) ■ ) . /

This implies equation 14.18.

E X A M P L E 1 4 .2 1 .

As in Example 1 4 .2 0 , suppose the velocity,

v{t),

Laplace transform V ( .) =

500^ + 20 .v(55 + 20)(10.v-h l)

with the Laplace transform o f the acceleration, a(r), given by 500.V + 20 ( 5 i + 2 0 ) ( 1 0 .v + l ) Find the final values o f iÛ) and a(r) if possible.

o f a certain projectile has

’29


S o l u t io n

K(y) and

A{s)

have poles that meet the conditions o f the final-value theorem. Hence,

lUn \'(t)= Urn ^ y ( 5 ) = lim I-*X .v-^0 ,v-*-0

5005+20

= 1

(55 + 2 0 X 1 0 5 + 1 )

and

lini a(t)= lim i-* x >

5y\(.0 =

.v-»()

lim .v-ô

5 0 0 5 “ + 20 5

= 0 ;5 5 + 2 0 )( 1 0 5 + 1)

Observe that a constant final velocity implies a zero acceleration as these expressions indicate.

E X A M P L E 1 4 .2 2 . W hat if the conditions o f the fmal-value theorem are not met? W hat would go wrong? A simple example illustrates the problem. Let

Fis) = -

5” + 1

which corresponds

toJ{t)

= sin(r)//(r). Then lim

sF{s) =

X

lim ^ 5 ----- = 0 , 1

,v-*0 5 " +

but lim

/{[) =

lim t-*x

sm(t)ii{t)

is undefined, i.e., it does not exist. The theorem, however, presupposes that both limits exist. Again, the condition of poles in the leh half complex o f the plane with at most one pole at the origin is necessary and sufficient for both limits to exist.

Exercises.

1. If

F{s)

= (6y + 10)/(2^^ + 4^), then, by the final value theorem,

value for large r? A N S W E R : 2.5 2. The Laplace transform o f a signal,

5y{t- 2)u{t-

2), is

1 5 t ^ - - - '( 5 - + 5 - 2 ) .v(5" + 5 5 + 6 )

Find the value of^(f) for very large A N SW E R : - 5

t.

f{t)

approaches what

"30


8. BODE PLOTS Section 5 described the use o f the poles and zeros o f N(s) to compute the frequency response o f a circuit. In this regard, Hendrik Bode developed a technique for computing approximate or asymp totic frequency response curves. These so-called Bode plots can be quickly drawn by hand. A description o f the technique requires the introduction o f some terms widely used in the engineer ing literature. Let //(s) be a transfer function that is a dimensionless voltage ratio or a current ratio. As explained in section 4 , for sinusoidal steady-state analysis, one replaces s by y’oj to study the circuit’s magni tude response, |//(/w)|, and phase response, Z .//(/co). For convenience, let |//(/to)| be a voltage gain, |K,/Kj|. T he ga/;/

in dB

(decibels), denoted by

is defined by the equation

= 2 0 log,„|//(/o)|

(14.19)

For convenience, whenever we write log(.v), we will mean log,Q(>;). Solving for |//(/w)| in equation 1 4 .1 9 yields the inverse relationship (1 4 .2 0 ) Table 14.1 presents some pairs o f |/y| and we may say that

is 2 0 dB

above

Thus, instead o f saying that |K,| is 10 times |Kj,

K,, or that Kj is 2 0 dB

below ¥2 - Similarly,

to say that

V-,

is 3

dB above K, means that |K,| is 1.414 times |Kj|. TABLE 14.1 Transler Function Gain in Magnitude and in Decibels

^dB

Exercise.

1

V2

2

10

100

1000

0

s3

^6

20

40

60

A certain amplifier has a dc gain o f 80 dB. W h at is the actual voltage gain?

A N S W E R : H)-'

O ne of the reasons for using the dB terminology is that it simplifies the analysis and design of multistage amplifiers. Suppose an amplifier has three stages with voltage gains equal to 2 0 , 100, and 10, respectively. The overall voltage gain is the

product o f

the gains of each individual stage,

which is 2 0 X 100 x 10 = 2 0 ,0 0 0 . Using the dB specification, the overall gain in dB is the

sum of

the dB gains o f the individual stages. This is (2 6 + 4 0 + 2 0 ) = 8 6 dB. It is easy to justify this claim. First, |H| = |//,| X |//,| X |//,


731

Taking the logarithm o f both sides and multiplying by 2 0 yields 20 log|/^ = 20 log|//J + 20 logl/Zjl + 20 log|//3| or

^dB

^2,dB + ^5,dB

This summation has pronounced advantages for repetitive calculations at many frequency points, as when plotting a magnitude response such as equation I4 .l4 a . We could, of course, convert this equation to an equation having all terms in dB. However, with an eye toward a further simplifi cation, it is desirable to first rewrite H{s) in a slighdy different, but equally general, form, namely + 1

H(s) = Ks'‘ j

'

(14.21)

-Z 2

^ - + l' V -P i

where \p^, p 2 >—} are those poles o f H{s) that are not at the origin and {zp 2^, ...} are those zeros of H{s) that are not at the origin; if a is positive (negative), then H{s) has a zeros (poles) at the origin. For example, a transfer function (^ + 5 0 ) (5 + 2 0 0 ) (5 + 8)( j + 6 0 0 ) has the equivalent form O '

H {s) = 2.5

^50

+ 1

Uoo

(f-l ^,600 Observing that

+ 1

(14.22)

+ 1

= 10 log^^|//(70))p, setting s = /a) in equation 14.21, and noting that the mag

nitude squared o f a complex number is the imaginary part squared plus the real part squared yields 0.5

HdBi(o) = \K dB

0)

(0

dB +

dB 2

,0.5

0.5

v2

\Z2/

+ 1

dB

(14.23)

-,0.5

( co\

dB Equation 14.23 suggests that we may compute the dB vs. (O curve for each term on the right-hand side and graphically sum the curves to obtain the desired H jg vs. (O curve. However, each indi vidual curve is reasonably sketched by using log(o)) instead o f 0) as the independent variable. This amounts to using semi-log paper to plot the dB vs. (O curves. Such a plot, with a linear scale for the dB values on the vertical axis and a logarithmic scale for O) on the horizontal axis, is called a Bode magnitude plot, in honor o f its inventor. Similarly, a plot of LHijiiS) vs. O), with a linear scale for Z.//(/co) and a log scale for co, is called a Bode phase plot. Note that, because of the log arithmic scale for the (o-axis, the actual distance on the paper between (O = 1 and O) = 10 is the same as that between o) = 0.1 and O) = 1. (See Figure 14.36a.) Note also that the (O = 0 point will not appear on the graph, because log(co) approaches -oo as O) approaches 0. W

"3 2

Chapter 14 • LiplaceTransform Analysis III: IVansfer Function Applications

With log(oj) chosen as the independent variable, the plot of each term in equation 14.23 either is exact ly a straight line or is a curve having two straight line asymptotes. I'his is illustrated in Figure 14.36.

FlG lIR Ii 1 4.36 Bode magnitude plots for three basic transfer functions, (a)

H{s) = K> 0.

(b)

H{s)

=

sin, a > 0.

(c)

H{s) - s!a +\, a > Q.


733

In Figures 15.36b and 15.36c, the rising asymptote has a slope of 20 dB/decade, which means that along this line, an increase in frequency by a factor of 10 causes an increase in gain of 20 dB. Here, the word decade (abbreviated dec) simply means “10 times.” Another way to express the same slope is to indicate the increase of gain in dB when the frequency is doubled, or increased by an octave, in music terminology. It is easy to see that 2 0 dB/dec is equivalent to 6 dB/octave. In gen eral, if a frequency, (Oj, is d decades above another frequency, coj, then, by definition, (cOj/cOj) = lO'^. Conversely, if (0)2/t0 j) = r, then we say that (Oj is log(r) decades above (Oj. In Figure 14.36c, the left asymptote is a horizontal line and hence has a slope of 0 dB/dec. The point where the two asymptotes intersect is called the breakpoint, and the corresponding fre quency is called the l>reakfrequency or comer frequency. The derivations o f the true curves and asymptotes in Figures 14.36a and 14.36b are very simple and are left as exercises. For figure 14.36c,

in which case +1

■

B

+ 1

and | H (» | ^ =

For 0) «

201o g

( l) '

+1

a. |H(j
indicating that \H{j(a)\^g approaches the left asymptote in the figure. For co »

|«Oo))|dB = 201og ,1 - )

V\a )

(I)'

+1 as 20 log -

a

a,

=201og(co)-201og(a)

indicating that |^(/(o)|^^ approaches the right asymptote in the figure. The two asymptotes inter sect at the point ( cd = a,

= 0). At this corner frequency, the largest error, 3 dB, occurs

between the true value of |//(/co)|^^ and the value read from the asymptotic curve. The error at twice or half the corner frequency is about 1 dB. The following variations of the three basic Bode magnitude plots o f Figure 14.36 are easily derived: 1.

If H{s) = {sld)^y the Bode magnitude is similar to that shown in Figure 14.36b, except that the slope is now 2 0 « dB/dec. The curve still passes through the point (co = a,

2.

If H{s) = {sla + 1 ) ” , the Bode magnitude is similar to that shown in figure 14.36c, except that the right asymptote has a slope o f 2 0 « dB/dec. The breakpoint is still at (o) = a, |//(/co)|^^ = 0), and the error at the comer frequency is 3 « dB. If n is negative, the right asymptote points downward.

734


Let us now consider the Bode plot for a general transfer function

H{s).

After expressing

H{s)

in

the form of equation 1 4 .2 1 , we can draw the asymptotes for each term in equation 14.23 with the aid o f Figure 1 4 .2 6 . The asymptotes for the

log(o)) curve can then be constructed very eas

ily by graphically summing the individual asymptotes. Since the asymptote for each term in equa tion 14.2 3 is a piecewise linear curvey the graphical sum o f all the asymptotes is also a piecewise lin ear curve. Accordingly, it is not necessary to calculate the sum o f dB values at a large number o f frequencies. If there are

n break frequencies,

then the summation need only be carried out at these

frequencies and for the slopes o f the leftmost and the rightmost segments o f the piecewise linear asymptote. T he following example illustrates this procedure.

E X A M P L E 1 4 .2 3 . Obtain the asymptotes for the Bode plot o f

H{s)

o f equation 1 4 .2 2 rewritten

below:

I s

— +1

H (s)= 2.5

V50

-M

/ V200

/ + 1

V8

Solution

There are four break frequencies. Rewriting cies appearing in

H{s)

/ V6 0 0 as a product o f five factors with break frequen

ascending order yields -1

H{s) =

2.5 - + 1 .8

— + 1 .5 0

.2 0 0

+1

■ .V .6 0 0

+1

i}4.24)

= / / , X / / . X //3 X //4 X Figure 14.3 7 a shows the asymptotes for the five individual terms in equation 1 4 .2 2 , and Figure 14 .3 7 b shows the asymptotes for


^^5

H

CO

(rad/sec)

(b) FIGURE 14.37 Asymptote and Bode magnitude for H{s) of equation 14.22 or, equivalendy, equadon 14.24. (a) Bode plots for / / , through

Hy

(b) Bode plot for

equal to sum of those for

through

T he calculation o f the asymptote in Figure 14 .3 7 b proceeds as follows: 1.

T he slope o f the leftmost segment, i.e., the segment to the left o f co = 8, is obviously zero,

2.

There is a breakpoint Pj at co = 8. T he only factor contributing to a dB value at this fre

3.

There is a breakpoint ? 2 at OJ = 50. Since

from Figure 14.3 7 a . quency is //j(s ) = 2 .5 for which

= 2 0 log(2.5) se 8.

H^{s)

contributes - 2 0 dB/dec to the slope for

to > 8, and since 50 rad/sec is 0 .7 9 6 decade above 8 rad/sec (from lo g (50/8) = 0 .7 9 6 ), the dB value corresponding to

Pj

is

8 - 2 0 X 0 .7 9 6 = - 7 .9 2 4.

There is a breakpoint

5.

There is a breakpoint

the slope for co > 50, slope for

0) > 6 0 0 ,

additional 2 0 dB/dec to resulting in a slope o f zero, we have - 7 . 9 dB for P^ at co = 6 0 0 . Since H^{s) contributes an additional 2 0 dB/dec to the at co = 2 0 0 . Since

contributes an

resulting in a slope o f 20 dB/dec, and since 6 0 0 rad/sec is 0 .4 7 7 decade

above 2 0 0 rad/sec (because lo g (600/200) = 0 .4 7 7 ), the dB value corresponding to

P^ is

- 7 .9 2 + 2 0 X 0 .4 7 7 = 1.63 6.

Finally, consider the slope o f the rightmost segment. Since

tional-20

contributes an

addi

dB/dec to the slope for co > 6 0 0 , the resulting slope o f the rightmost segment

is zero. T h e com p lete specification o f the piecewise linear asym ptote is shown in Figure 14 .3 7 b .

'3 6

Exercises.

Chapter 14 • Laplacc Transform Analysis 111; Transfer Function Applications

1. Given a firsi-order low-pass characteristic o f tiie form _

«(■') =

show that (i) the dc gain is

K and

(ii) for

K

out _

în

s = ja,

1 4 -£ i.e., at co =

a,

the gain is 3 dB down from the dc

H{s). function H{s)

value. I his to is called the 3 dB frequency or 3 dB bandwidth of 2. Construct the piecewise linear asymptote for the transfer

= 40j-/[(i- +

2){s +

2 0 )].

O nce the piecewise linear asymptote for the Bode plot has been constructed, the true curve can be sketched approximately by noticing that the error at each corner frequency is about 3w dB (pro vided that the two neighboring corner frequencies are more than five times larger or smaller). Figure 14 .3 7 b shows such a rough sketch. In the pre-com puter era, the ability to draw such a curve by hand— even a crude approximation— was considered valuable. Nowadays, with the wide avail ability o f personal computers and C A D sofuvare, one might just as well get the exact plot with out bothering to look at the straight-line asymptotes. From the perspective o f circuit analysis, the value of the ability to construct the asymptotes for a Bode plot is greatly diminished, but the tech nique is still important for its application in the design o f feedback control systems. Such an appli cation utilizes both the Bode magnitude plot and the Bode pha.se plot. Some background in con trol systems is required for one to understand the use o f the Bode technique in this kind o f appli cation. We will relegate the discussion of the topic to a more advanced course in feedback control. O ur objective in mentioning it in this text is twofold: (1) to introduce the definitions o f some com m oni)’ used terms, such as

decibels, decade,

and

octave,

and (2) to demonstrate a highly sys

tematic procedure for adding up several piecewise linear curves to obtain a desired curve, as described in Figure 14.37.

9. TRANSFER FUNCTION ANALYSIS OF A DC MOTOR A permanent magnet dc motor, an electromechanical device, converts direct current or voltage into mechanical eêrg^^ The shaft of the motor rotates freely on bearings. Mounted on the shaft within the housing is a wire coil called the

armature winding o f the

motor. Surrounding the coil

are permanent magnets that interact with a magnetic field produced when a current flows through the armature winding. This interaction o f magnetic fields forces the shaft to rotate in a process o f energ)' conversion. Here current flowing through the armature coils rotating through the magnetic field produced by the permanent magnets encasing the coils produces a torque on the motor shaft, which drives a load. Power is delivered from the source to the load. The electromechanical characteristics o f the permanent magnet dc m otor have a simple circuit like model amenable to Laplace transform analysis. The model is given in Figure 14.38 and con sists o f an adjustable dc voltage source in series with a resistor labeled “m otor.” Here,

an inductance

and a device

represents the resistance present in the armature winding, and

resents the equivalent inductance o f the wire coil. The device labeled “m otor” has the current

rep

i J^t)

as an input and the angular velocity co(r) o f the rotating shaft as an output. T he interaction


between the electrical part o f the model and the mechanical part o f the model occurs at the loca tion o f this symbol. T he voltage

is an induced voltage proportional to the angular velocity

oj(r). Because co(/) is not a circuit variable, classical notions o f impedance, admittance, voltage gain, etc., do not fit the problem, whereas the more general notion o f transfer function does, forc ing us to move slightly beyond the confines o f circuit theory proper to analyze the system.

The torque

J\t)

produced on the rotating shaft by the current flowing through the armature coils

is proportional to the armature current /^(/), i.e..

Tit)

=

(1 4 .2 5 )

T he mechanical rotation o f the m otor affects the electrical portion of the system by inducing a voltage

This voltage is proportional to the m otor’s rotational speed, or angular velocity, co(r),

i.e.,

=

(1 4 .2 6 )

Since the m otor converts electrical energy to mechanical energ^^ conservation of energy dictates that the constant o f proportionalit)' be equal to the same constant that relates torque and current for a lossless motor. Specifically, electrical power-in must equal mechanical power-out or, in equa tion form.

= m coit) which forces

=

k.

O ur first goal is to find the transfer function o f the motor,

H{s)

=

£[o){t)]/V^{s).

For convenience,

let Q(^) denote X [o)(r)]. As a first step, sum the voltages around the loop of elements in the cir cuit model of Figure 14.3 8 . This results in the differential equation

V (/)

=

)

at

(1 4 .2 7 )

A ssum ing zero initial con ditions, the Laplace transform of equation 1 4 .2 7 is

V^{s) = {R^ + Lj)I^{s)

f

kQis)

(1 4 .2 8 )


738

From basic mechanics, the differential equation governing the mechanical portion o f the system

r(o = y „ ^ + B < o (o

at

where

is the moment o f inertia of the combined armature, rotating shaft, and load; B is the

coefficient of friction; and 7\t) is the torque produced by the motor to turn the load. Recalling equation 14.25, T\t) =

the Laplace transform o f equation 14.29, assuming zero initial con

ditions, is (14.30) Solving equation 14.30 for IJis), substituting the result into equation 14.28, and then solving for Q(j) leads to the expression

k 2 V,{s)

Q(s) =

(14.31)

Equation 14.31 characterizes the pertinent dynamics of the permanent magnet dc motor and allows one to find the angular velocity o f the motor shaft as a function of time for given inputs. To see how the motor responds to step inputs, suppose vjit) = Ku{t). As objectives, let us find (1) the steady-state value, or final angular velocity, o f the shaft and (2) the steady-state value o f the armature current. The final speed o f the shaft is important because, for example, one needs a fixed speed for the rotation o f a compact disk or a fen. The final or steady-state current is important for determining the power needed from the source. If Vj(t) = Ku{t), then V(^s) = KIs. It follows from equation 14.31 that

k

S +

S+

RgS + k

2\

K

(14.32)

J Applying the final value theorem to equation 14.32 implies that (Ovv =

lini

k

co(r) = ------- y K

To isolate the armature current IJJ), again apply the final-value theorem to determine the steadystate value of i {t). Combine equations 14.30 and 14.31 to obtain 1

— 5+

B

4 (^ ) =

2 ’'s

LaJm

( 14.33 )


Equation 1 4 .3 3 allows us to find the armature current as a function o f time for a given input volt age. As above, if the input is a step, i.e., if

v^{t) = Ku{r),

1 — =

-

^

B s + --------^

------------ t t K.

S+ \

it follows that

m

R,,B + k~\

(1 4 .3 4 )

1

O nce again, application o f the final-value theorem to this expression leads to the value o f the steady-state armature current:

Q hi . ss ~

t —►«

^Cl

~

B+ k

^

T he preceding analysis suggests the utility o f the Laplace transform as a tool for analyzing the dynamic behavior o f electromechanical systems. In fact, system transfer functions o f the form of equations 14.31 and 14.33 are often starting points for further analysis. Extensions of the above analysis can be found in the homework exercises.

10. SUMMARY This chapter has expanded the notion o f transfer function from its definition in Cliapter 13 into a tool for modeling not only circuits, but other practical systems, such as a dc motor. The trans fer function characterizes circuit or system behavior by the location o f its poles and zeros. For example, if a transfer function has a pole on the imaginary axis or in the right half-plane, the asso ciated circuit or system is said to be

unstable,

because there is an input or, possibly, simply an ini

tial condition (as in the case o f a second-order pole on the y'oj-axis) that will excite the pole and cause the response to grow without bound. Further, the ubiquitous presence o f noise will always excite poles on the imaginary axis and cause the response to be unstable. This chapter categorized various types o f responses: zero-state and zero-input responses, natural and forced responses, transient and steady-state responses, etc. Recall that the zero-state response is the response to an input assuming zero initial conditions, which is the inverse Laplace transform o f the product o f the transfer function and the Laplace transform o f the input excitation. Recall that the zero-input response is due only to initial conditions on the capacitors and/or inductors o f the circuit. The complete response for linear circuits having constant parameter values is simply the sum o f the zero-input and zero-state responses. This decomposition generalizes to the broad class o f linear systems studied in advanced courses. Under reasonable conditions, other decom po sitions are possible, such as a decomposition into the natural and forced responses or transient and steady-state responses. Other important responses are the impulse and step responses. For stable circuits, the frequency response provides important information about the circuit. Recall that frequency response is a plot o f the magnitude and phase o f //(/w ) as to varies from 0 to cc. T he Bode plot is a plot o f gain in dB vs. frequency represented using a log scale. In this con text it is relatively straightforward to construct an asymptotic approximation using straight line segments, from which the actual plot is easily sketched by hand. T he information in such a plot tells us how a circuit behaves when excited by sinusoids o f different frequencies.

(0

Chapter 14 • Ijp la c c Transform Analysis III: Transfer Function Applications

Lastly, this chapter has introduced the initial- and final-value theorems, which provide a quick means of computing the initial and final values o f a time function from knowledge o f its Laplace transform. Such theorems have wide application in control system analysis, as evidenced in our analysis o f the dc servo motor.

TERMS AND CONCEPTS 3 d B frequency for low-pass characteristics: the frequency at which the gain is 3 dB down from the dc gain. Asym ptote: a limiting straight-line approximation to a curve. B ode m agnitude plot: a plot o f gain in dB vs. frequency represented on a log scale. B ode phase plot: a plot o f phase in degrees vs. frequency represented on a log scale. B ounded: the condition where a signal

satisfies [/(r)|

< K < for

all

f,

i.e., it has a maximum,

finite height. B reak point: the point at which two asymptotes o f a Bode plot intersect. C om p lete response: the total response o f a circuit to a given set o f initial conditions and a given input signal. C o rn er (break) frequency: frequency at which two asymptotes o f a Bode plot intersect. D ecade: a frequency band whose endpoint is a factor o f 10 larger than its beginning point. D ecibel (d B ): a log-based measure o f gain equal to 2 0 logjQ|//(yw)|. Final-value theorem : a theorem stating the following: suppose

F{s)

has poles only in the open left

half o f the complex plane, with the possible exception o f a single-order pole at ^ = 0. Then the limit

oij{t)

as /

oo equals the limit o f sF{s) as i “♦ 0.

Forced response: the portion o f the complete response that has the same exponent as the input excitation, provided the input excitation has exponents different from those o f the zeroinput response. Frequency response: measure o f circuit behavior to unit magnitude sinusoids, cos(w t), as O) varies from 0 to

00

. Equal to the evaluation o f the transfer function

H{s)

at ^

for all co.

Frequency scaling: lor a linear passive network, dividing all inductances and capacitances by a factor, Ayr while keeping all resistance values unchanged. Fundam ental period o f p e rio d icy (r): the smallest positive number

T such

that J{t)

=j{t + T).

Im pulse response: assuming all initial conditions are zero, if the circuit or system input is 6(r), then the resulting^(r) is called the

impulse response.

I he inverse transform o f the transfer

function equals the impulse response. Initial-value theorem : a theorem stating the following: let

L\J{t)] = F{s) be a strictly proper F{s) are both polynomials

rational function o f s\ i.e., the numerator and denominator o f in

s,

with the degree o f the numerator less than that o f the denominator . Then /(O’*") is

the limiting value o f sF{s) as ^

oo.

L in ear circuit: circuit such that lor any rwo in p u ts/j(r) an d/^ (f), whose zero-state responses are ^ j(/) andjsCO. respectively, the response to the new input A'2 >'2 (^)]. vvhere ATj and

Kj

[Kjf^{t)

+ A'2 /S(r)] is [A',^,(/) +

s

are arbitrar)' scalars.

M agnitude response: the magnitude o f the frequenc)' response as a function o f O). M agnitude scaling: for a linear network, multiplying all resistances and inductances by a factor, and dividing all conductances and capacitances by

for dependent sources, this

^


means all parameters having ohms as units would be multiplied by siemens as units would be divided by

and those having

dimensionless parameters are left alone.

N atural response: the portion o f the complete response that has the same exponents as the zeroinput response. Octave: a frequenq^ band whose endpoint is twice as large as its beginning point. Op amp open-loop gain: the gain of the op amp when no feedback paths to the input terminals are present. Periodicy(/); function satisfying the condition that there exists a positive constant, T, such that / f ) =fijt+ 7) for all f > 0.* Phase response: the angle o f the frequency response as a ftinction o f (O. Piecewise linear curve: an unbroken curve composed of straight-line segments. Pole (simple) o f rational function: zero of order 1 in denominator polynomial. Poles (finite) o f rational {unction: zeros o f denominator polynomial. Ramp function, K^)j integral o f unit step function having the form Ktu{i) for some constant K. Rational fim cdon: ratio o f two polynomials; a polynomial is a rational function. Stable transfer function: a transfer function for which every bounded input signal yields a bounded response signal; i.e., all poles are in open left half o f the complex plane. Steady-state response: that part of the complete response which either is constant or satisfies the definition of periodicity for t> 0 . Step response: the response of the circuit to a step function, assuming all initial conditions are zero. Transient response: those terms o f the complete response that are neither constant nor periodic for ^ > 0; i.e., the transient response does not satisfy the definition of a periodic function for f > 0. Zero-input response: the response of a circuit to a set of initial conditions with the input set to zero. Zero-state response: the response of a circuit to a specified input signal, given that the initial con ditions are all set to zero. Zeros o f rational function: values that make the numerator polynomial zero.

O '

O '

* This (nonstandard) definition has been adapted for one-sided Laplace transform problems.

O '

742

Chapter 14 * Liplacc IVansform Analysis llh lr a n s fe r Function Applications

PROBLEMS

3. Consider the pole-zero plot o f the transfer function

POLES AND ZEROS 1. For the pole-zero diagram shown in Figure P 1 4 .1 ,

C om pute the transfer function

(b)

C om p ute

the

impulse

H{s).

and

step

(b)

C om p ute

response. (c)

If the input is itive number

find the pos

a

such that the response

impulse

response.

parts o f the response. (c)

Com pute the step response. Identify the steady-state and transient parts o f the response.

CH ECK .- Your answer to (b) should be the derivative o f your answer to (c), since the delta function is the derivative o f the step function. (d)

does not have a term o f tlie form

Ke~‘’‘u{t).

the

H{s).

Identify the steady-state and transient

responses. Check that the impulse response is the derivative of the step

given in Figure P I 4 .3 .

If the dc gain is 2 , find

HiO) = 8.

(a)

H{s)

(a)

itive number

Find the response under this

\Qe~‘^‘ti{t)y find the pos a such that the response

If the input is

does not have a term o f the form

condition.

Ke~‘'û{t). jto

Find the zero-state response

under this condition.

Identify the

transient and steady-state responses.

(double pole)

jO)

-2

A

-1

)C4j

Figure P I4.1 -q>—►a

1

-2

2. Consider the pole-zero plot of a transfer function

H{s)

given in Figure P I 4 .2 .

(a)

If the dc gain i s - 1 0 , find/yW -

(b)

C om pute the impulse response.

(c)

C om pute the step response.

-A]

Figure P I4.3

C H E C K ; Your answer to (b) should be the

4. Consider the pole-zero plot o f a transfer

derivative o f your answer to (c), since the delta

function

function is the derivative o f the step km ction.

(a)

(d)

If the input is itive number

a

find the pos such that the response

H{s)

given in Figure P i 4 .4 .

If the gain

H{s)

is 4 .8 at ^ = 1, find

His). (b)

C om p ute

the

impulse

response.

does not have a term of the form

Identify^ the steady-state and transient

Ke~‘^‘u{t).

parts o f the response.

Find the zero-state response


(c)

C om pute the step response. Identify' the steady-state and transient parts of

jO)

the response. C H ECK .- Your answer to (b) should be the -e —►o -2

derivative o f your answer to (c), since the delta function is the derivative o f the step function. (d)

:-4j

Figu re P I 4 .2

If the input is 10.9

find w, such that the response does not have a purely sinusoidal term.

743


Find the zero-state response under this

(c)

If the input to the circuit is

=

condition. Identify the transient and

2tu{t), use MATLAB to compute the

steady-state responses.

partial fraction expansion and the resulting time response. Sketch the

jw

approximate response for large t.

ik

oj

(d) ------- 1—> o

-2

Repeat part (c) for v-J^t) = lp-u{t).

7. The input impedance o f each network shown in Figure P I4 .7 is 2 (5) =

s+a

O-j

with a > Q. Which network is stable and why? Figure P14.4

Compute the step response of the stable net work. Identify the transient and steady-state

5. The pole-zero plot of the transfer function

parts o f the response.

His) is given in Figure P 14.5. (a)

If the dc gain is - 1 , find H{s).

(b)

Compute the impulse response.

(c)

Compute the step response. Identify the

^

Z (s)

J

V

(a)

Compute the response to the input sin(4/)«(r).

(e)

A

Z (s)

transient and steady-state responses. (d)

\

V

J

(b)

Figure P I4.7

If the input is e~^û{t), find the posi tive number a such that the response

8 . Reconsider the linear active networks of

does not have a term of the form

Figure P I 4.7, which have input impedance

Ke~^û{t). Find the zero-state response

function Z{s) with poles at j = - 1, - 3 ± y 4 and


zeros at 5 = 2, ± p .. It is known that Z(0) = 8 .

jco >

(a)

Write Z{s) as the ratio of two polyno mials in s.

(b)

If the network of Figure P I4.7b has zero initial conditions and

— e------ -----®->»o -2 2

=

lu{t) A, write down the general form of writing down the forms, simply leave the con

C-4j

stants as literals without calculating Figure P14.5 6 . The transfer function H{s) of a particular

numbers. Does the output remain finite as co> (c)

If the network of Figure PI 4.7a has zero

active circuit has poles at - 1 ± 2j and - 2 . At j

initial conditions with input

= - 1, the transfer function gain is found to be

20u(t) V, find the general form of

8 , and //(oo) is known to be finite. When this = sin(2r)tt(/) V and v-JJ) = u{t) V, it is found

and then Again leave the con stants in these forms as literals. Does the output remain finite as f 00?

that the output is zero after a long time. (a) Find the zeros of the transfer function.

9. Repeat Problem 8 for when the zeros of Z{s)

Construct H{s) and then compute the

are j = -2, ± j2 . Also, answer the following ques

step response.

tions:

circuit is excited by the input waveforms

(b)

=

^44

Chapter 14 • Laplacc'I'ransform Analysis 111: Transfer Function Applications

(a)

W hich configuration shows a stable circuit?

(b)

If one o f the configurations is an unstable circuit, what input will cause the zero-state response to grow with out bound as /

x?

10. Reconsider Figure P l4 .7 a with impedance

Z{s) having poles at ^ = 2, 6 ± y‘8 and zeros at s = - 1 , - 2 , —4. It is known that Z (0 ) = 3 2 Q. (a)

Write Z{s) as the ratio o f two polyno mials o f s.

(b)

If the network has zero initial condi tions, and

=

û(t)

V is applied to

the two terminals, find the current /

it)

flowing into the network for

t > 0.

11. Show the pole-zero plot o f each o f the fol lowing

transfer

functions,

and

determine

whether the system (circuit) is stable. Hint: Use

Figure P I4.1 2

the M A rij\ B command “roots” to obtain the

7X . T O S

.ind poles of « k I, transfer ainction, even though there are mediods that can determine sta-

,,

Consider the circuit o f Figure P I 4 .I 3 . (a)

Show that the transfer function is

bilit)^ without calculating the exact pole locations. (a) /Y,(.v) =

(b)

Is -

s

I^

R^C2

H(s) =

s -+ ^s +\2

^ ( /?,Ci

1.5.V- -Q .5 .V -H .2 5

H2{ s ) =

1

2

(b)

r V 2 . v “ + 7.5.vH-6.5

R2C2 ^ R\C2 j

If Cj =

I

S+

= 1 mF, find

K\C\R2C2 and

Rj

so

that the poles o f the transfer function (c)

H-^{s) =

are at 5 = - 0 .2 1 9 2 2 and - 2 .2 8 0 8 . The

1.5.V- -H2.25.S + 4. 4062

problem o f finding element values to

5-V4.3I25.V + 5.3125

2s^ + 2.5s-

- 17.V + 220.5

/ - h 7 r V 5 5 . v - + 5 6 9 5 + 520

realize a given transfer function

synthesis problem. R.) = (4 k£^, 0 .5 kI2)

A N SW ER ; 2

kLl)

12. For each o f the op amp circuits o f Figure P I4 .1 2 , find the transfer functions and the poles

is

called the

He

and zeros in terms o f the indicated literals.

+ v.„(t)

Figu re P I 4 .1 3

or (1 k il,

Chapter 14 • iîplacc Transform Analysis 111: Transfer Function Applications

STABILITY PROBLEMS 14. For each of the circuits below, compute the indicated transfer function and determine the range

or

for which the circuit as m od

eled by the transfer function is stable in the BIBO sense. (a)

For the circuit of Figure P I 4 .1 4a the transfer function is

R =4a, (b)

His)

=

Vi s )

16. (a)

,

The pole-zero plot o f a transfer func tion

c = 0 . 2 5 F.

is given

in

Figure

P i 4 . 16a.

Construct an input that will cause the

For the circuit o f Figure P I 4 .1 4b the V (.v) transfer function is His) = .

response to be unbounded (unstable), (b)

Repeat part (a) for the pole-zero plot o f Figure P i 4 .1 6b.

(c)

For the circuit o f Figure P i 4 .1 4 c the transfer function is

H{s) =

Imag Axis Ojw,

finis) Real Axis

------o - >

-2

®-jco, (a) Imag Axis

(a)

Figure P I4 .1 6 17. The circuit in Figure P i 4 .1 7 is to be stabi lized in the BIBO sense. (a)

If y? = 2 range o f

Figure P i4.14

(b)

\/

15. Given that

H(s) = - ^ V *-111

complete

cuit is stable.

range of

a

in the circuit in

R for

= 0 .5

S,

which the cir

required for stability.

= 0 .5 , determine the complete

range o f

Figure P 1 4 .1 5, where C = 2 F and find the

If

determine the complete

a

R required

for stabilit}'.

746


r-s. (1) Find the zero-state response, the

in tiie circuit o f Figure P i 4 . 18; then find the range

plete response.

for whicii the circuit is sta

Now suppose tiiat

il. = -2

complete range R for

which the circuit

ble, assuming (b)

zero-input response, and the com

complete

R=

1

(2) Find the steady-state response and the transient response.

S. Find the

is stable.

(3) Find the forced response and the natural response. (c)

Repeat parr (b) for V and

v-^^{t) = 20e~^^û{t)

= 0.

Is the forced

response well defined?

H in t:

M A T L A B ’s

“ [r,p,k]

com m and

Use =

residue(n,d)” to com pute the partial fraction expansions.

R.

+

)

19. For the circuit o f Figiu-e P i 4 .1 9 find the

-C

R, <

transfer function „ v;-(.v) and the range o f a for which the transfer func tion is stable.

Figure 1P14.20 2 1 . Find the complete response for r > 0 for the circuit o f Figure PI 4 .2 0 for each o f the follow ing circuit conditions. (If you properly utilize the results o f Problem 20 and the principle o f linearity, the answers to this problem can be written down directly.) (a)

= 6 V,

(b) Figure P I4 .1 9

response


has been found to be zero-input response =

(a)

Determine the transfer function. If v-^^{t) = 10 co s(10f)«(/) and I'Ô”) is changed to 8 V, com pute the complete

8

response

i + io = 50

/?, = 2 0 0 Q ,

Suppose

v^Q-)

Vj^^{t)

= 8 V.

for /■ > 0. Identify the

transient and steady-state parts o f the

R^^^= R^

C = 2 . 5 niF. (b)

e~^û{t) V 4e~~û{t) V

(b)

response. when

V, and ^f;(0")

zero-state response = 10(1 -

2 0 . Consider the circuit in Figure P I4 .2 0 .

RC

v-^j{t) = 10«(^)

= 4 V; a decomposition o f the complete

RESPONSES AND CLASSIFICATIONS

H{s) =

5 sin(5r)w(r) V

2 2 . Consider the network o f Figure P I 4 .2 2 . Suppose C = 0.1 F,

(a)

v-^j{t) =

I 6 V > . , W = 8 f - '» M r t V

= 10 sin(5/)//{/) V, and

74

Chapter 14 • Laplacc Transform Analysis III: Transfer F u n aio n Applications

+

Pure resistive network

vJt)

sC + — + —

Figure P I4 .22 2 3 . In the circuit shown in Figure P I 4 .2 3 , = 20 a

v jt ) = 25u{t) V(^0~) = 10 V.

C = 0 .0 5 F,

v^2 i^) = 20e~^‘‘u{t)

V, and

.v-f-

/?2

5-t- 2

CR.(1/

Draw its pole-zero plot. Express the

(b )

/? 2

.V+ 1.6

C/?,

H (i) = /?1

5 a

.v +

A’C H------

v,(t)

given

in Figure

=

P24.b as a sum o f possibly shifted sim

V,

ple functions such as steps, ramps, etc. Then find the zero-state response. (c)

Draw the equivalent frequency domain circuit, and find the zero-input response when

''• ■ 6

V(^0~) = -1

V. Hint: The response

RC circuit R^^C.

of an undriven first-order where x =

6

(d) (e)

is

Com pute the complete response. If

were changed to the input shown

in Figure P l4 .2 4 c , what would be the

Figure P I4.2 3

response of the circuit? (Hint: W hat is (a) (b)

(c)

Find

for r > 0 by the Laplace

relationship

between

P I4.24b and P I4.24c? Note that differ

Write

as the simi o f two com

entiation and integration are linear oper

ponents: the zero-state response and

ations when the initial conditions are

the zero-input response.

zero and all circuit parameters are con

Separate the zero-state response into

stant. For linear constant parameter cir

one term due to

cuits, then, a linear operation on the

and another term

input induces the same linear operadon

v^.

Separate the complete response into

on the response, provided that the initial

the steady-state response and the tran

conditions are zero and the circuit

sient response.

parameter values are constant. Combine

Utilize the answers o f previous parts to

this concept with the structure o f the

write down the complete response for

decomposition to obtain the answer.)

(e)

r > 0 for the following two cases (all values in volts):

c

Case 1 Case 2

V(;iO )

v jt)

Vs2it)

20

\Ou{t)

\5e^û{t)

5

lOuit)

dOe^ûit)

R. R, < (a)

2 4 . Consider the circuit o f Figure P I4 ,2 4 a . Suppose C = 25 mF, /?, = 25 n , and

=

100

n,

A v„(t) (V)

20

Figu re P I 4 .2 4

R^^ = R^ 11 R j.

(a)

Figures

transform method.

due to (d)

the


-►t (c)

7AH

Chapter 14 • Lapbcc IVansform Analysis 111: Transfer Function Applications

25. For the circuit o f Figure P 1 4 .2 5 , Cj = 0 .2 mF and

C-,

= 0 .5 mF. The initial conditions arc

;-'q (0 “) = 8 0 mV and

v jt ) = OAu{t)

v^{0~)

i,(t)

= 0. The input is

V.

V ,.(t)

(a)


(b)

Find the zero-state response.

K.U'I (c)

Find the zero-input response.

(d)

Find the complete response.

(e)

Specify the general form of the natural

2 7 . Consider the circuit o f Figure P i 4 .2 7 . (a)


response. (0

Specify the transient and

H{s) =

is) KM

steady-state responses.

LC .9“ -I-

L

- v„ +

R2 C}

.v +

LC

R j)

16 r +

4.V + 20

when /?, = 1 Q , F, and (b)

L

=4

LX C =

0 .1 2 5

Find the zero-state response to the

=

input (c)

/? 2

= 0 .5 H. ^ IO«(r) V.

Draw an equivalent j-plane circuit that accounts for the initial condi tions. There are four possibilities, but

Figure 1M4.25

one is superior. (d)

2 6 . Consider the circuit in Figure P i 4 .2 6 . Suppose /e, = 10

C= 1 /40 F, + i>2 (() =

Q,

= 3 0 LX

= 15

il,

(e)

function so that you know its shape). C om pute Com pute

and /^(0“).

(0

response

vc{0~)

for

= 4 V.

Find the complete response for

5u{t),

=

= 2 V, and /^(O") = - 2 A.

Hint: Use linearit}'.

and /^(f) for r > 0; this ------- _ _ o

is the complete response. (c)

zero-input

Find the zero-input response for /^(0~) = 0 and

V (plot this input

(a)

the

/Y-(0“) = 0 and /^(0“) = 1 A.

and Z. = 8 H. Suppose

(b)

Find

Identify' the part o f the response due

i,(t)

only to the initial conditions at r = 0 “ that result from think

of

this

one could as

the

zero-input

Figure P I4 .2 7

response. (d)

Identify' the part o f the response due only to the part o f the input for r > 0, i.e., due to 25//(^) assuming zero initial conditions; one could think o f this as the zero-state response.

(e)

Identif)’ the transient and steady-state parts o f the complete response.

28. The purpose o f this problem is to illustrate the computation o f the zero-input response using the transfer Rmction concept. This requires proper application of the equivalent initial condition cir cuits and the use o f a transfer fiinction for each initialized dynamic element. Consider the circuit of Figure P I4.28a.

Chapter 14 • Liiplacc Iransform Analysis III: Transfer Function Applications

(a)

Find che transfer function

H\ (j)

jco

if the

input is a properly directed current source connected between nodes

1

(double pole)

and 2. T he value o f the current source, accounts

for

the

initial

-> o

-2

capacitor voltage y ^ 0 “). (b)

-2j Figure P 14.29

Find the transfer function / / 2 W if the input is a properly directed current source connected between nodes

1

and 3 . T he value o f this current

3 0 . For the pole-zero diagram o f

source,

H(s) =

s accounts for the initial inductor current. (c)

For the circuit o f Figure P I4 .2 8 b , uti lize the transfer functions obtained in

shown in Figure P I 4 .3 0

H{2)

cos{4t)u{t)

=2

V. jO)

parts (a) and (b) to find

>k. 0 4j

the zero-input response for Vq^^{)

(i)

= 10. Com pute

the sinusoidal steady-state response to

for / > 0, assuming initial condi tions

= 4 V and /^(0“) = 5 A;

■>a

(ii) the zero-state response for for ^ > 0; and 0 -4 j

(iii) the complete response for for r > 0.

Figure P I4.3 0 ANS\\T-:R: 0 /

Y

Y

V

2H 3 1

4S <

,2 S

4v.

2F

. Find the phase and magnitude in sinusoidal

steady state for each o f the given transfer func tions when excited by the indicated sinusoidal mput.

(a)

(a)

H{s) =

2s+ 6

Vi„{s)

.V--1-4.9-h i6

= 2 cos{2t

is excited by the input + 4 5 °) V. 4S <

(b)

Suppose 5

=3

output (b)

( 6’ + 4 )

has

cos(3f + 4 5 °) V.

Find the sinusoidal input

that

gives rise to this output.

Figure P i4.2 8 (c)

Let Z = 0 .2 H, C = 0.2 5 F, and /?j =

STEADY-.STATE CALCULATION

= 7? = 1 n

2 9. For the pole-zero diagram o f

P i 4 .3

1

in the circuit of Figure

. Compute the transfer function

H{s) = shown in Figure

Vi„U) P i 4 .2 9 H{0) = 4.

9

Com pute

the sinusoidal steady-state response to sin(2f)«(f) V.

=

2

1

.V“ -I-------5 -i-

RC

LC


750

Find the phase and magnitude o f the output Vfî) in sinusoidal steady state to (d)

= 20 cos(4?) V.

Repeat pan (c) for the output ijjJ). +

Figure P I4.32

V ,( t )

33. Consider the circuit o f Figure P I4.33. Given that C2 = 1 F, find Cp R^„ and /?2 so that the transfer function is H (s) = - 5 (a)

Figure P I4.31

Find the steady-state response

Kcos{2t+ 0) when 2 sin (2t)u{t).

32. Find the phase and magnitude in sinusoidal steady state for each of the circuit transfer func

(b)

^

s+4 =

=

Now find the complete response to

tions (or circuits) below when excited by the

the input

indicated sinusoidal input.

identify the transient and steady-state parts, assuming no initial conditions

16^ + 96

(a)

Vi„(s)

are present.

s^ + Ss + 64

is excited by the input

= 2 sin(2^)«(^) and

= 4 cos(4?

(c)

Verify that your answer to part (a) and your steady-state answer to part (b)

(b)

coincide.

Suppose H (s) = 4>/5 — — (5 + 2 r + 16

has output = 25 cos(3^ + 100°) V. Find the sinusoidal input =K cos(fiW + 0) that gives rise to this out put. (c)

G>nsider the circuit o f Figure P I 4 .32. Suppose Z = 1 H , F, and R^ =

= 15 n , C = 0.01

O.. Suppose

voltage + to -

the

across the current

source, is the desired output. The

Figure P I4.33

transfer function _

'in hnis)

? 1 + — 5 + -----

work transfer function H(s) =

is shown

in Figure P 14.34. If the input is a sinusoidal

L LC Find the phase and magnitude o f the

voltage o f 1 V amplitude, then, in the steady

= Kcos(cot + 6),

state, the output voltage has the greatest ampli

output cosine,

in sinusoidal steady state to (d)

34. The pole-zero plot o f a certain RLC net

99

= Z/n(^) =

=

tude at approximately what CO?

At what

200 cos(10/) mA.

approximate value o f (0 does the amplitude of

Repeat part (c) for the output

the output voltage dip to its lowest value? Give qualitative, not quantitative, justification for your answers.

n

Chapter 14 • biplacc Transform Analysis III: Transfer Function Applications

1

C om pute the phase and the magnitude o f the

jco

steady-state response for each o f the following

jlO

inputs:

= 20

(a) j3

---- ©-5

cos(4r + 4 5 ° ) V

(b)

y •„(/■) = 4 co s(4 0 ^ ) V

(c)

v,„{t) = 4cos{2t)V

Suggestion: One can use the com m and “freqs(

-2

O ••

n,d,w )” for a rational function. 3 9 . Following are the transfer functions o f

X -

some linear networks that contain controlled sourccs. The networks are initially at rest, i.e.,

Figure P i 4.3 4

with zero initial conditions. An input

cos{2r)u{t) 35. A linear circuit witii a transfer function 2v+ 4

V

H{s) = ^ = v;„

r

residue comm and in M ATLAB to generate the partial

fraction

expansion,

=4

determ ine

the

approximate form o f the output,

+ 5^ + 6

very large values o f has an input

=8

V is applied at f = 0. Using the

cos(2f + 4 5 ° ). Com pute

t.

for

Hint: For stable circuits,

what part o f the response is meaningful, and for

the magnitude and phase (in degrees) o f the

unstable circuits, what part o f the response is

output o f the circuit in the steady state.

dominant?

3 6 . A stable active circuit has the transfer func

(a)

- 3 .7 5 j + 5

//,(^) = ^ — ---.9 "-1 -4 5 -I-3

tion

H(s) =

V.

V:.

.

5“ + 4

(b )

- ^ “ ’5— — : 5-“ + 2.V + :)

2.5s- 3

H2(s) = -^

s" -t- 2 i ’ 4- 5

Com pute the phase and the magnitude o f the steady-state response for each o f the following

. .

(c)

Ht,( s )= -^ s

7 r + .v -h 4

7—

-l- 5 ~ + 9 i ’ + 9

inputs: (a)

= 4 cos(2^ ) V

(b)

U ^ ) = 4 c o s (4 0 V

(d )

H^(s) = -^

75^ + 5 + 7.75

(e)

5 ' V 3 5 “ + 1 1 . 2 5 5 + 18.5

^ --0 -5 5 + 5 5 “ -h0 .5 5 + 5 .7 3 2 1

=2^2

The input to the circuit is

cos(2r +

3 0 ° ) V. Knowing that the response has the form

A cos{2t

+ (j)), find

A and

(j).

(f)

5^ - 5 . 5 5 - +

A N SW E R : (b) I'h c result follows from the fol lowing X'lATLAB code:

3 8. A stable active circuit has the transfer func

»nin = 8*[1 0 ];

tion

»n = [2.5 -3]:

V:Js)

165^ + 4 4 5 + 1 2 8

+

85- + 365 +

80

145-12

^6(*v) = — 5-’ + 5 .5 5 “ + 1 4 5 + 12

»din = [1 0 4];

K,ur(^)

=5

. s ' + r + 7 .2 5 5 4 -1 8 .5

3 7 . A stable circuit has the transfer function

v^îs)

14,s-2-23.V + 20

»d = [l 2 51; »nn = conv(n,nin); »dd = conv(d,din);


'5 2

»[r,p,k] = residue(nn,dd)

(a)

Find the transfer function H{s) =

(b)

If the circuit is initially at rest, find an

r=

V^{s)IV^{sY

-4.0000e+ 00 - 7.0000e+00i -4.0000e+ 00 + 7.0000e+00i

expression for Vj (/) that is valid for all

4.0000e+ 00 + 4.0000e+00i

t (small or large). Use the residue com

4.0000e+ 00 - 4.0000e+00i

mand in MATLAB to do the needed partial fraction expansion.

P= -l.OOOOe+00 + 2.0000e+00i -1 .0 0 0 0 e + 0 0 -2 .0 0 0 0 e + 0 0 i

42. Consider the circuit of Figure P14.42.

-7 .7 7 l6 e -l6 + 2.0000e+00i

(a)

- 7 .7 7 l6 e -l6 - 2.0000e+00i

Compute the transfer function. (It should be first order.)

k=D

(b)

Find an expression for the natural

(c)

Suppose that for t = 0“ , each capacitor

>>K = 2*abs(r(4))

response o f the circuit.

K = 1.13l4e+01 »Phi = atan2(imag(r(4)),real(r(4)))* 180/pi

voltage is 1 V and the inductor current

Phi = -4.5000e+01

is zero.

Compute the zero-input

response. Observe that this response (Apply item 19 o f Table 12.1.)

has a steady-state part. This phenome non illustrates why equations 14.7

12140. Following are the transfer functions of

require more than just a stable transfer

some linear networks that contain controlled

fijnction as an underlying assumption.

sources. The networks are initially at rest, i.e.,

It is necessary that the transfer fiinc-

with zero initial conditions. An input

=8

tion include all the dynamics of the

sin(40f)«(r) V is applied at f = 0. Using the

circuit, which is not the case in the cir

residue command in MATLAB to generate the partial fraction expansion, determine the

cuit of Figure P I4.42.

approximate form o f the output,

for

very large values of t. Hint: For stable circuits, what part o f the response is meaningful, and for unstable circuits, what part o f the response is dominant? (a )

(b )

H^(s ) = - y H2(s) = ^

80 +4^ + 3

80 43. The circuit shown in Figure P I4.43 has zero initial conditions, yÔ") = 0, and //^(0“) = 0). 80

(a)

(c )


s + 2 j + 160l5 (b)

(d)

If

= K(j«(r), find V({s) and

5 ^ + 8 5 + 1 616

and determine the largest voltage that

5^ + 155-^ + 8 3 5 ^ + 1 9 9 5 + 170

can appear across the capacitor. At what times does this occur? This phe

H{s) = - , ------- g------- 2-----------------

y 4 l . In Figure 14.17 o f Example 14.8, let ji=

nomenon

1.4 and recall that v-{t) = cos(2r) V.

design o f insulators that support trans

has application

in the

mission lines. For example, a 100 kV

Chapter 14 • I^placc rransform Analysis III: Transfer Function Applications

voltage transmission line would need insulators that can handle at least

1) V, find

for r > 0 by the

Laplace transform method.

2 0 0

kV.

(c)

If the impulses are applied to the cir cuit ever}^ second, i.e., if

/

Y

Y

= [6(^)

V

+ 6 (r - 1) + 6 (r - 2) + - ] V, find the

+

L

output

for w < t < (« + 1) when

the integer ;/ becomes very large (i.e., after the circuit reaches a steady state). Sketch the waveform.

Figure P I4.4 3 4 4 . Suppose the voltage source in Problem 4 3 is changed to

q

9

For

t >0,

\

ti{t)

10 sin

V.

determine a tight upper bound on the

largest voltage that am appeiir across the capacitor. ANSWHR: 111.2 V

4 5

. In the circuit o f Figure P i 4 .4 5 , all stored

energ)^ is zero at r = 0. The current source is =

A. Determine

Vgn,{t). W hat

Figure P I4.4 7

is the largest

voltage that can appear across the capacitor?

FREQUENCY RESPONSE 4 8 . Draw to scale the pole-zero plot for

H{s)

=

(i + l)/[i(j^ + ^ + 10], and graphically compute the magnitude and phase o f //(/co) for to = 0 .2 , 0

Figure P i4.45 46. At r = 0, the energy stored in the

LC elements

in Figure P I4 .4 6 is nonzero. The current source is

=

cos{lt)ti{t)

A. Is the circuit stable?

Determine the behavior o f

. 5 , 1, and 10. W hat are the limiting values of

the magnitude and phase, i.e., for co = ± 0 0 ?

4 9

. Draw to scale (on graph paper) the pole-

zero plot o f the transfer function

for large t.

((.v+ l)2 + 1 6 ) ( . r - l )

His) =

( ( a--I-1)“ + 4)((.v + l)“ -1-36) and determine graphically the magnitude and phase o f

H(joj)

for co = 0, 2, 4 , 6, and oc. Do

your answers make sense physically? Explain.

Figure P i4 .4 6

. (This problem focuses on a qualitative under

4 7 . This problem illustrates the condition o f

5 0

steady state, but not sinusoidal steady state. For

standing o f poles and zeros o f a frequenc)-

the circuit shown in Figure P 1 4 .4 7 , the op amp

response.) You have determined the pole-zero plot

is assumed to be ideal.

of a band-reject filter as shown in Figure P I4 .5 0

(a) (b)

Find the transfer function If

H{s).

= 0 and y- (f) = 6(/) + 6(r -

(or so you think, according to die qualitative boardwalk suggestions o f the “professor”).

'^ 4

Chapter 14 » Laplace Transform Analysis III: Transfer Function Applications

jco

jw

J X

<>8j

'l6j

X

-> a

---- 1-------0.2

*-8 j

-2

-1

X

■ 8j II1fV ^W ■

-Bj

-16j

Figure P I4.51

Figure P I4.5 0 To make the gain o f the transfer func

5 2. Figure P 1 4 .5 2 shows the magnitude fre

tion 1 for large frequency you must

quency response o f a transfer function

choose what value o f A" in

whose numerator and denominator are both

( .9 -

fourth order. Qualitatively determine the struc ture o f

( . V - /?,)•■• •

(b)

1I X

X

(a)

1 I

H{s)

H{s)

approximately.

Using M ATLAB, compute the frequency response (magnitude and phase) over the range 0 < (o < 2 0 rad/sec. W h at range o f fre quencies are rejected, approxi mately?

51. (This problem focuses on a qualitative understanding o f the poles and zeros o f a frequency response.) You have come up with the pole-zero plot o f Figure P 14.51 for a possible band-pass filter with a sec ond-order zero at the origin. (a)

Figure P i4.5 2 1 ->

To make the gain o f the transfer func tion 1 for

CO

= 6 you must choose what

value o f A'in

ANSWI-H

(.v~ + l)~

(.v + O . l r - f l

(5 - z, )•••

5 3. Figure P i 4 .5 3 shows the magnitude fre (b)

Using M ATLAB, compute the fre quency

response

(m agnitude

quency response o f a transfer function

and

phase) over the range 0 < co < 2 0 rad/sec. (c)

Is the filter o f the band-pass type? W hat range o f frequencies are passed, approximately? W h at ranges o f fre quencies are rejected, approximately?

Frequency-rad/s

Figure P i4.5 3

H{s).

Chapter 14 • Laplacc Transform Analysis III; Transfer Function Applications

7SS

W hich o f the following is the best candidate for

equivalent, the magnitude and phase

H{s)>

responses 0 < to < IO.OOOti rad/sec for

( 1) K

(2) K (3) K (4) K (5) K

(6 )

K

s

the second-order low-pass Butterworth

(^ + 0 .5 + ;5 ) ( 5 + 0 . 5 - y 5 )

filter

l(s) = --------------- , --------^------------------------ . (5 + 0 .5 + y5)(i- + 0 . 5 - y 5 )

+ 1.414

{s + j l ) { s - j 2 ) j(5 + 0 .5 + y5)(.? + 0 . 5 -

which has a 3 dB down

j5)

____

the

(5 + 0 .3 + y4)(.v + 0 . 3 - y 4 ) ( i + 0 .8 + y6)(.v + 0 .8 - y6)

{ s + j 2 ) { s - j 2 ) ___________________

+

0

.3 + y 4 ) ( 5 + 0 . 3 - 7 ’4 )(5 + ().8 + y6)(.v + 0 . 8 -

(s+j2)(s-j2)

(7) K 5

( 5 + 0 .8 +

7 6

X5 +

0

7 6

steady-state

Butterw orth

phase

and

filter when

the

input is (i) cos(5007rr), (ii)

(5 + 0 .3 + y 4 )(5 + 0 . 3 - y 4 ) ( 5 + 0 .8 + y 6 ) ( 5 + 0 . 8 - y 6 )

5 (5

at 1000 Hz.

magnitude o f the output o f the

s { s + j 2 ) ( s - j 2 ) __________________

___________________

[X )in t

(c) Referring to part (b), find

(s+j2){s-j2) __________________

+ 1

2000:1/

\ 2 0 0 0 Ji/

cos(20007rf),

and

(iii)

cos(40007tr). ) Do your answers to part (c)

make sense? Why?

.8 -

7 6

Remark: T he

)

of A N SW E R : (5)

gain,

or the magnitude

is t}^pically given in decibels

and computed as gain (dB) = 20

54. T he circuit in Figure P I 4 .5 4 is to be fre quency and magnitude scaled so that the value o f H{s)

ats= j5

s = yiOOO.

becomes the value o f

56. (a)

at

Plot, using M ATLAB or the equiva lent,

If the final value o f the capacitor is to

the

magnitude

and

phase

responses for the second-order nor

be 1 mF, then what is the new final value o f the

malized high-pass Butterworth filter

inductor,

transfer function

(in H)?

H {S ) =

1Q

5

- + 1 .4 1 4 5 + 1

4H

over the range 0 < (O < 5 rad/sec.

1F

(b)

Repeat part (a) for the denormalized high-pass filter

Figure P I4.54

1

55.(a)


the

magnitude

and

2000;t

H(s) =

phase

+ 1.414

responses for 0 < to < 5 rad/sec for the 2

second-order

normalized

low-pass (c)

5

(b)

" + 1 .4 1 4 5 + 1

+ 1 \ 2 0 0 0 jt/

over the range 0 < (jl) < 10,000jr rad/sec.

Butterworth filter transfer function H(.5) =

OOOJ1 /

Referring to part (b), find the steadystate phase and magnitude o f the out put o f the Butterworth filter when the

Denormalization to achieve a different 3

input

dB down point is done by frequency

cos(20007tf), and (iii) cos(4000Tr/).

scaling. Plot, using MATLAB or the

is

(i)

cos(5007ir),

(ii)

~S6


57. Frequenc)' responses are typically plotred as

malized high-pass Chebyshev filter

\H{j(S))\ vs.

transfer function

logjQicol or as gain = 2 0 log,()l//(yo))l

clB vs. logjQCO. These plots are com m only called (a)

~T

0 .7 0 7 9 .S -

4 - 0 .6 4 4 9 5 +

1

Repeat parts (a) and (b) o f Problem 55, except plot only the gain in dB vs. logjolcol.

(b )

0 .5 0 1 2 .r

H{s) =

Bode plots.

over the range 0

< f<

1 Hz.

Repeat part (a) for the denormalized

(b)

Repeat parts (a) and (b) o f Problem

high-pass Chebyshev filter

5 6 , except plot only the gain in dB vs. 0 .5 0 1 2

logjolcol.

H{s) = 5 8 .(a)


the

m agnitude

responses for 0 < second-order

CO

and

0 .7 0 7 9

\200 0 7 1 /

phase

low-pass

(c)

Chebyshev filter transfer function

(b)

put o f the Chebyshev filter when the

0 .5 0 1 2

r

Referring to part (b), find the steadystate phase and magnitude o f the out

---------------

H{s) =

200071/

over the range 0 < / < 1 0 0 0 Hz.

< 5 rad/sec for the

normalized

\ 2 0 0 0 Jt/

input

+ 0.6449.s + 0 .7 0 7 9

is

(i)

cos(5007rr),

(ii)

cos(20007t^), and (iii) cos(40007i^).

Denormalization to achieve a different 3 dB down point is done by frequency

6 0 . The circuit in Figure P I 4 .6 0 , with ^ = 1 Q,

scaling. Plot, using M ATLAB or the

C = I F, and Z, = 2 H, has a so-called third-order

equivalent, the magnitude and phase

Butterworth low-pass characteristic. T he trans

responses 0 <

fer function o f the filter is

CO

< 1G.OOOtt rad/sec for

the second-order low-pass Chebyshev filter

s +2i“ +25' + l 0 .5 0 1 2

H{s) =

(a) -h 0 .6 4 4 9

+ 0 .7 0 7 9

H{s)

. Verify

that their magnitude is 1, meaning

2000 Jt

200071/

Calculate the poles o f

that they lie on a circle o f radius 1 in which has a 3 dB down point at 1000 Hz. (c)

Referring to part (b), find the steady-

Using MA'l'LAB or the equivalent,

state phase and magnitude o f the out

plot the frequency response for 0

put o f the Chebyshev filter when the

1 Hz.

input (d)

the complex plane. (b)

is

(i)

cos(5007tf),

(ii)

< f<

\NSWH

cos(20007tr), and (iii) cos(40007rr).

- [0:0, 01:

Do your answers to part (c) make

= 0 .5 ;

sense? Why? Remark; The

gain,

or the magnitude

o f //(/C O ), is t)'pically given in decibels

= freqsi n,d,\v); p lot(w /(2’ pi). abs(h))

and computed as gain (dB) = 2 0 (c)

If

is chosen so that

Kjr= (??);rso S ]59.(a) Plot, using M ATLAB or the equiva lent,

the

magnitude

and

phase

responses for the second-order nor

R=

10

find

th a tC = 1 5 .9 1 5 /iE W h at

is the new value o f Z? The filter trans fer function becomes


U.5

H (s ) =

(a)

\3

+2

+2

(ss)

^■>/

Assuming an ideal op amp, show that the transfer function is

+1

s-

Plot the new frequency response

s+

(magnitude and phase) for 0 s / s 1 0 0 0 //« .

(d)

(b)

From your plots in part (c), determine

1 RC 1 RC

Assuming that R = 20 kQ and C = 1 fiF, use MATLAB or the equivalent to

approximately the magnitude and phase

plot the magnitude and phase (in

of the steady-state output voltage when

degrees) responses for 0.1 < (o < 1000

the input is (i) 10cos(20;cf),

rad/sec.

(ii) 10cos(200;rz), (iii) 10cos(1000;ir),

response, it is apparent why this cir

(iv) 10 cos(2000;iy), (v) 10cos(5000;cf).

cuit is called an all pass network.

/

Y

Y

From

the

magnitude

V

L

R

'.w Q

Figure P I4.60 61. The circuit of Figure P I 4.61 has transfer W

s

fiinction

H{s) =

R\Ci 1 ^/?lCi

1 R i^ i

1

s+ ^ iQ /

0 6 3 . Figure P I4 .6 3 is a circuit that realizes a

^ iQ ^ 2 ^ 2 band-pass characteristic. Use SPICE to deter

With Cl = C2 = 1 mF, R^=A kH, and /?2 = 0.5 kQ, plot the magnitude o f the frequency response; i.e., plot |//(/to)| as a function of 0) over the range 0 < O) < 5 rad/sec and 0 < O) < 50 rad/sec

mine the frequency response o f the circuit and verify that it is a band-pass characteristic. Note that the computation o f the transfer function for this circuit is quite involved.

+ V .Jt)

'" “ 6

Figure P I4.61 62. The circuit of Figure P 14.62 is called an all

pass network This means that the circuit does not alter the magnitude of a sinusoidal wave form, but it does introduce a new phase shift depending on the frequency of the sinusoid.

Figure P i4.63 Si64. A certain band-pass circuit has transfer flmction

758


+ 1 3 9 .9 0 f - ^ ) VlOO^

^VlOO/

H{s) =

5 5^ + 2 .7 9 9 6 f — )

UOO/

+ 0 .9 9 8 9 f ;^ ) ^ VlOO/^

4 + 1 3 .0 5 l f — )

3 + 1 7 .2 7 lf — )

\m J

VlOO/

+ 1 3 .0 5 1

+ 2.7991 ( —

) + 1

UOO/

100^

Plot the magnitude frequency response in dB

tie attenuation. This can easily be seen, since

on a semilog scale to verify a band-pass charac

the transfer function has the form o

teristic. Find an appropriate range of (0.

1

65. The circuit o f Figure P I 4.65 is the Sallen

H{s) =

Vo(s) + —

and Key low-pass filter, which can be used to

RC

eliminate unwanted high-frequency noise. (a)

5+

In l R^C o


V. v:-in

1

___ K

Observe that if s = ycOg = //RC, then

RxR'i C\C2 \-K\ 1

//(/COq) = 0, i.e., Vo(/(Oq) = 0 for

+• /?2 Q

=

input sinusoids o f the form

s+

y4sin(cOo^ + Cp).

o

^2^2/

where K = 1+/?^^^. Hint: You will (a)

Use nodal analysis to derive the trans

the indicated nodes.

fer function.

Now let Cj = C2 = 1 F, /?j = = 1 Q, and R^ = 0.8 Q. These values

(b)

Show that the poles (zeros o f the

represent a normalized design. Use

(c)

need to write three node equations at (b)

o

denominator) are ( - 2 ± S y R C .

o

L e t / ? = 5 / J t £ 1 .5 9 1 5 £ l a n d C = 0 .0 1

MATLAB or the equivalent to plot the

F. Plot the magnitude and phase

magnitude and phase responses for 0 <

responses over the range 0 to 2 0 0 Hz.

CO< 5 rad/sec.

o

---------- r ^ ( -

K

0

"

---- 1--1 R

o

+

r\

r> 0.5R v^( 1 2C < -----L---------- n

r\

Figure P I4.66 Twin-T RC circuit.

n

67. Suppose the impulse response of the net work in Figure P I4 .6 7 is Figure P i4.65 Sallen and Key low-pass filter.

= [2e^^+ 3cos(2/) - 1.5sin(2/)]«(r) V

66, Figure P I4 .6 6 depicts the so-called twin-T

(a)

Compute the transfer function, H{s).

RC circuit. This network is often used as a

(b)

Find all poles and zeros, and draw the

band-reject filter, i.e., a circuit that stops or rejects signals at certain frequencies and allows signals at all other frequencies to pass with lit-

pole-zero plot. (c)

If there is no initial stored energy in the network, i.e., no initial conditions.

o

n

759

Chapter 14 • Laplace Transrorm Analysis 111: Transfer Function Applications

com pute the network response to

(b)

Find the response o f this same relaxed

(d)

= b(t)

circuit to the input

= sin(r)//(/) V. Does your answer (c)

make sense? W hy or why not?

V.

Find the response o f this same relaxed

= 5b(t-

cos{lt)n{t) V. For large t, show that Vg„,{t) ~ Kt cos(2t + (j)) V for appropriate K and (J).

70. For this problem you are to use the proper

Does the response remain finite as r

ties o f Laplace transforms as listed in Chapter

Now suppose

00

circuit to the input

=

? Is the network stable or unstable?

2) V.

12. Suppose the Laplace transform o f the impulse response,

h{t),

W(.v) = v„Jt)

o f a circuit is given by

s+2 (,s + l)(.v + 3)

H^{s) and Hf{s) h^{t)=h{At) (b) hît) =

Find

when

(a)

-

8)

Figure P i4 .6 7

property that its zero-state output is always

INITIAL AND FINAL VALUE PROBLEMS

equal to the derivative o f its input shifted in

7 1 . (a) The Laplace transform o f a signal y(r) is

68. A time-shift differentiator circuit has the

T > 0. Hence, its impulse response is h{t) = h'{t - 7 ), a shifted version o f the derivative o f the delta function. Suppose T

time by some

( 5 .v - l ) ( 4 .v - 5 ) ( 6 .v - 2 ) 5 (2 5 -h l)(3 .s -F 2 )(5 i + 4 )

= 1 and the input to the circuit is given byy(f), as sketched in Figure P I 4 .6 8 . Com pute the

Find the values o f /(O^) and /(co). (b)

zero-state response^Cr) o f the circuit at ^ = 2 .5 and f = 6 seconds.

Repeat part (a) when

the Laplace

transform o f a s i g n a l / r -

2)u{t-

2) is

^_ 2 , (l(Xs - l ) ( 8 . v - 5 ) ( 1 2 . v - 2 ) ^

.v(5.v+l)(3.s'-h2)(i- + 4 )

7 2 . The output,

o f a particular circuit is

engineered to track a reference signal, After the circuit overheated, it was found that 1

1

s

s

4s- - 4s s-

4

+ 2i' + 2

Figure P i4.6 8 A N SW ER ; - 2 , 1

Find the error, error(r) = large

6 9. The Laplace transform o f the response o f a relaxed circuit to a ramp

v^{t)

t.

for

W hat was the initial error, error(0"’)?

C H E C K : for large

t,

error is 3

= r ( f - 1) V is 73. Given the following fimctions

F-{s),

find

and/y(3o) for those cases where the initial(a)

Find the response o f this same relaxed circuit to the input

v-Û) = u{t)

V.

value and final-value theorems are applicable (a root-finding program is useful in this endeavor).

Chapter 14 • U place Transform Analysis 111: Transfer Function Applications

APPLICATIONS AND BODE TECHNIQUES

2 l.v + 5 O.I.v*- +4.V + 3

7 6 . Obtain the asymptotes for the Bode plot o f 2.v^ + 7.v“ + .y + 4

F,{s) =

(b)

s{s^

+ .V " + 7 . V

H(s)= (c)

F,is) =

.y(.v +

■+ 1

+ 6 )

s~ +4.9 + 3 5s^ + 5s^ + 4.V +

iO-

4)

100 .V — + V16

\ 400

1200

+ 1

Then verify your asymptotic plots by using 7 4 . The capacitor voltages,

o f various

M ATLAB to generate the true Bode plot,

networks have the Laplace transforms given below. Use only the initial- and fmal-value theorems to determine

and

7 7 . Obtain the asymptotes for the Bode plot of

for each

o f the transforms. If either theorem is not

H(s)=

applicable, explain why.

10

— + 1 V16

1200

+ 1

(a)

(i 4 .v - i r

100

s~(ls+2)

+ 1 •+ 1

V400

Then verify your asymptotic plots by using M AI LAB to generate the true Bode plot.

(b )

(c )

(d )

7 8 . Obtain the asymptotes for the Bode plot o f , ^ ^ a ) . l . v - l X 0 . 2 i + lJ zUU---------- ^----------------.y(.y“ + 144j (1 6 5 -1 )

T

(2 5 + 1 ) -

H(s)= 25-

25+1

V25 + 1

10-

+ 1

+ 1 \ 1200

+ I

4 5 -1 \ -'

+

16

100

/

5

+ 1

V 400

V 25+ I/ I'hen verify your asymptotic plots by using

7 5 . The capacitor voltages,

o f various

MA'I'LAB to generate the true Bode plot. Note

nerworks have the Laplace transforms given

that the shape o f your asymptotic Bode plot

below. Use only the initial-and fmal-value theorems to find /f;;(0‘'’) and

i(^^)

for each o f the

resembles that o f Problem 77except that the slopes are steeper,

transforms given that C = 0 .2 F. If either theo rem is not applicable, explain why.

7 9 . Obtain the asymptotes for the Bode plot of

10

(a) 5(5

+ 1

+ 2)

H(s}= (b ) 5“ (25 + 2 5 )

(c )

20

(^ ■ -2 ) 5(25 + 25) “

10

\ 20

200

2400

+ 1

+1

+1 800

Then verify your asymptotic plots by using MA^’LAB to generate the true Bode plot.


> “80. Suppose a dc motor is modeled as per

(c)

= 25 Q . / „ = 0.005 kg-m^, k = 0.0 2 N -

m/A, B = e~^ N-m-s, and

= 100 mH.

W hat effect does R^ have on the steady-state speed for a step input of a

Figure 14.38 o f the text. The motor parameters are

"^61

fixed amplitude? (d)

W hat effect does R^ have on the rate at

Calculate the steady-state angular velocity of

which the armature approaches a

the motor and the steady-state armature cur

steady state for a step input o f fixed

rent for the input 10«(r) V.

amplitude?

> “81. Consider the dc motor modeled in

> “8 2 .(a) For the dc motor of Figure 14.38 of

Figure 14.38 o f the text. (a) Using the same parameters as in

the text, suppose k = 0.0 5 N-m/A and

Problem 80 with

R^ = 50 Q. Plot the steady-state cur

= 5u{t) V, com

rent as a function of 5 as 5 ranges from 0 to c».

pute the armature current and the angular velocity as a function of time. (b)

Change

(b)

Using /?^ = 25 Q and B = e~^ N-m-s,

to 50 Q, and recompute

plot 0)^ as ^ ranges from 0 to oo. Recall

the armature current as a function of

that increasing k increases the torque

time.

per ampere o f armature current. Explain why increasing k reduces co^^.

O '

w

C H A P T E R Time Domain Circuit Responsi Computations: The Convolution Method AVERAGING BY A FINITE TIME INTEGRATOR CIRCUIT Sometimes one must compute the average o f some quantity, sucii as the average vakie o f the light intensity on a solar cell over the last fifteen minutes, or the average value o f the temperature in a room over the last hour, or the average value o f a voltage over the last 50 milliseconds. W hen such an average is updated continuously, it is termed a running average. The idea is that the readout o f the device that averages these quantities always produces an updated value valid over a specified prior time interval. If a voltage represents the value o f the quantity to be averaged, then one can build a circuit whose output voltage is the required average. This is done by observing that an aver age value o f a continuous time variable is simply the integral o f the variable over the proper time interval, divided by the length o f the time interval. A device that integrates a variable over the last, say,

T seconds

is called a finite time integrator. As an application o f the ideas o f this chapter, we will

look at a finite time integrator circuit and how it can be used to compute the average value o f a quantit)^ The convolution concept directly leads to the required transfer function o f such a circuit.


Introduce the notion o f the convolution o f two signals.

2.

Using the notion of convolution, develop a technique o f time domain circuit response computation that is the counterpart o f the transfer function approach in the frequency domain, presented in Chapter 14. In particular, we seek to show that the convolution o f an input excitation with the impulse response o f a circuit or system produces the zerostate circuit or system response.

3.

Develop objective 2 from r\vo angles: first, from a strict time domain viewpoint, and sec ond, as a formal theorem relating convolution to the transfer function approach.

4.

Develop graphical and analytical methods— in particular, an algebra— for evaluating the convolution o f two signals.

5.

Illustrate various properties o f convolution that are pertinent to block diagrams o f series, parallel, and cascade interconnections o f circuits or systems modeled by transfer func tions.

^6'i

Chapter 15 • T im e Domain Circuit Response Computations: T h e Convokition Method

SECTION HEADINGS 1.

Introduction

2. 3. 4. 5. 6. 7. 8.

Definition, Basic Properties, and Simple Examples Convolution and Laplace Transforms Graphical Convolution and Circuit Response Computation Convolution Algebra Circuit Response Computations Using Convolution Convolution Properties Revisited Time Domain Derivation o f the Convolution Integral for Linear Time-Invariant Circuits

9. Summary 10. Terms and Concepts IL

Problems

1. INTRODUCTION At the beginning o f Chapter 12, we claimed that circuit response computations could take place in either the time domain or the i-dom ain. Yet, except for the solution o f some very elementary differential equations, circuit response computations have relied almost exclusively on the Laplace transform technique. This chapter develops and explores the time domain counterpart o f the Laplace transform method by introducing the notion o f the convolution o f two signals to produce a third signal. We then show that the time domain convolution o f an input excitation with the circuits impulse response yields the zero-state circuit response. Pictorially, the idea is expressed in Figure 15.1, which is an update ot Figure 12.3.

f(t)

Circuit Impulse

-

>

Response h(t)

Input Signal

h

y(t) = h^f Output Signal

T

Transform of Circuit Transfer

Transformed-

Function H(s)

Input Signal F(s)

Output Signal Y(s) = H(s)F(s)

riCnJRF, 15.1 Diagram showing the symmetric relationship of time and hequenc)'^ domain analyses. The upper part of the diagram asserts that the convolution, denoted with the circuit impulse response

h{t)

h * f, of the

input signal/(f)

produces a third signal, ^(r). This third signal is the zero-state

response, which equals

Chapter 15 • Time Domain Circuit Response Computations: The Convolution Method

Justification of the diagram o f Figure 15.1 begins with the formal introduaion o f the notion of convolution. Mathematically, the convolution o f two fiinctions h{t) andy(f), denoted hy h * f or

h{t) * j{t)i results in a third function through the integration process, >-(0 =

j^ ^ h {t-x )f{T )d T

(1 5 .1 )

The second step in verifying the diagram is proving that the convolution of the input signal with the impulse response produces the zero-state circuit response. There are two approaches. One is to work stricdy in the time domain and construct the actual zero-state response from the impulse response and an arbitrary input excitation. The second approach is to prove that

is

the convolution of the signal y(/) with the impulse response h{t). Because o f the Laplace transform development o f the last three chapters, this direction seems the most painless and will be taken up V—^

in section 3, after an introduction to the basic ideas o f convolution in section 2. Section 4 will look at the graphical method of convolution, which helps in visually grasping the definition. Section 5 will describe a convolution algebra, which yields a harvest o f shortcuts for evaluating the convolution o f certain types o f signals. Section 6 looks at the computation o f circuit responses using the convolution method, and here the averaging circuit is developed. Section 7 overviews various properties o f convolution. The respective properties lend themselves to different structures (e.g., parallel or cascade) for designing interconnected circuits. Section 8 describes the rudiments o f the construaion o f the zero-state response o f a circuit working stricdy in the time domain toward the development of a convolution integral. Before closing this introduction, we should consider the question, why is convolution important^ One reason is that it allows engineers to directly model the input-output behavior of circuits and general systems in the time domain, just as transfer functions model circuit behavior in the sdomain. A second reason is that circuit diagrams are sometimes unavailable or even get lost. How would one generate a circuit model for analysis? One way is to display the impulse response on a CRT and approximate

O '

by some interpolation function such as the staircase approximation illus

trated in Figure 15.2. The process o f constructing an approximate impulse response o f an unknown circuit or system is called system identification and is a vibrant area o f research. By storing the measured impulse response data as a table in a computer, one can numerically compute the zero-state response of the circuit or system to an arbitrary input signal. This numerical simulation process lets an engineer investigate a circuits behavior off-line. For example, simulating a circuit destined for use in a hazardous environment provides a cost-effective means of evaluating its performance in a simulated environment. Such performance evaluations often identify needed design improvements prior to the constructing and testing of prototypes.

766

FIG UIIE

Chapter 15 * T im e Domain Circuit Response Computations: T h e Convolution Method

1 5 .2

A rectangular approximation to a hypothetical impulse response obtained from meas ured data on an unknown circuit or system.

A third important reason for studying convolution in the context o f circuits relates to a deficiency o f the one-sided Laplace transform: function segments that are nonzero for f < 0 are ignored by the one-sided Laplace transform technique. Hence, time domain convolution offers a direct means of computing the circuit response when signals are nonzero over the entire time interval, -<»

< t< cc.

2. DEFINITION, BASIC PROPERTIES, AND SIMPLE EXAMPLES As mentioned, convolution is an integral operation between two functions to produce a third; i.e., the convolution o f two functions

h(t) and J{t)

produces a third function j = h* f. One might expect

that the convolution h * /e q u a ls the co n v o lu tio n /* h, i.e., the operation o f convolution is com mutative. In fact, this is the case. To emphasize this property, we restate equation 15.1 in its more general form:

CONVOLUTION T he convolution o f two sig n a ls/r) and

h{t)

produces a third signal, ^(/^), defined according to

the formula

v (/ ) = J

J A t - r)f{T)dT =

-

T)dx

(1 5 .2 )

which is well defined, provided that the integral exists. This formula emphasizes the proper ty that convolution is a commutative operation, i.e.,

h * f = f * h.

T he equivalence expressed in equation 15.2 comes about in a straightfonvard manner by the change o f variable t ' = r - t . In addition to being commutative, convolution is associative,

h * { r g ) = { h- f i * g

(1 5 .3 )

and distributive, (1 5 .4 )

Chapter 15 • T im e Domain Circuit Response Com putations: T h e Convolution Method

Exercises.

1. Verify equation 15.2 using tiie change o f variable T

'=t

767

- t .

2. Illustrate the associative property o f equation 15.3 with a specific example. 3. Verify the distributive property o f equation 15.4. Hint: Integration is distributive. 4. Verify that

a{h *f) = {af) *h =f*{ah).

Hint: Use equation 15.2.

Another very useful property o f convolution is the so-called tim e shift property: if

h{t) * j[t)

=

then

h{t-T^)^j{t-T^)=g^t-T,-T,)

(1 5 .5 )

Some simple examples serve to demonstrate the actual calculation process. These examples will utilize the sifting p rop erty o f the delta function: if h{t) is continuous at / =

//(7') = J*^_ h(r)d{T - r ) d T =

h{T

T,

then

-T)d{T)dT

E X A M P L E 1 5 .1 . C om pute the convolution o f an arbitrary continuous function y{r) with So

6{t) .

lu t io n

By the definition o f equation 15.2 and the sifting property o f the delta function,

y{t) = f * 5 =

-T)d(r)dT

= fit

= f it )

(1 5 .6 )

Example 15.1 makes the point that 6 {t) acts like an identity' for convolution, i.e., it always returns the function with which it is convolved. T he next example indicates that the convolution o f an arbitrar)' continuous function y{/) with

ad(t - T)

produces a scaled and shifted version

as

given below in equation 15.7.

E X A M P L E 1 5 .2 . Compute the convolution o f an arbitrar)' continuous fTinctiony(/) with So

a d {t-

7).

lu t io n

By the definition o f equation 15.2 and the sifting property o f the delta function.

y(t) = f it ) * [adit - T)] = C lJ i t - T)diT - T)cIt = a f i t -

t)^ ^ =

Exercise. Check the associative property o f convolution by ii)fxit) = d i t - 1) ,^ ( r ) = u{t) - ui t-2 ) , a n d ^ (f) = dit -

co m p u tin g /j

(/;■)/; (r) =

+ T^)

d it-

7 , ) , / , ( / ) = u i t ) - u { t - 2 ) , a n d / 3(r) = d i t

3)

*fi*

afit - T)

(1 5 .7 )

:"6 8

Chapter 15 • 'Fimc Domain Circuit Response Com putations; T h e Convohition Method

Ler us a|3ply the results o f the above two examples to a simple staircase function.

EXA M PLE

1 5

. 3 . Com pute the convolution o f the functiony(/) shown in Figure 15.3a with h { t )

^ ld {t) -2 d {t -

1).

f(t) iL 2 --

1-

H -------

1

2

(a) FIG URE 15.3 (a) Functiony(f) for Example 15.3. (h ) 7 = / ‘ h. So

lu t io n

By the definition of equation 15.1 and the sifting propert)' of the delta function,

v(/) = / * / / = 2 j =

2

- T) d( T - l)r/r

- t ) 6 ( t ) cI t - 2 j

( 15. 8)

/ < r - T - 2 / ( / - T)]^,| = 2 / ( / ) + ( - 2 / ( / - 1))

Graphically combining the result given in equation 15.8 yields the relation in Figure 15.3b.

Exercise. C om pute 25{t-2).

the convolution o f the functiony(r) shown in Figure 15.3a with

h{t)

=

26{t) -

AN SW 'FR: l f { t ) - l / { r - 2) =

EXA M PLE

1 5

. 4 . Find the convolution o f u { t ) with itself, i.e., n { t ) * u { ( ) .

S o lu tio n

By the definition o f equation 15.2,

ii(i)*u {t)= C

J -o c

u(t-T )u(T)dT =

f

Jo

n (t-T )d r

(1 5 .9 )

Chapter 15 • T im e Domain Circuit Response Com putations: T h e Convolution Methotl

"6 9

since TÔ |0

T< 0

We now break equation 1 5.9 into two regions: (i) t > r, in which case tion 15.9 zero; and (ii) 0 s t s /, in which case

u{t- t)

= 1. W hen 0

u{t- t) = 0 , making equa t, equation 15.9 reduces

to

u ( / ) * i ( ( f ) = f'i/(r - t ) c/ t = r' (It = tu{t) = r{t) %/()

In other words,

u{t)

*

u{t) = r(t),

(1 5 .1 0 )

•/()

the ramp function.

W e now apply the time shift propert}' o f convolution to the result o f Example 15.4.

E X A M P L E 1 5 .5 . Find

y = h ‘ /w h e n h{t) = ii{t)

a n d /r) =

ii{t + 1)

u{t-

-

1), as shown in Figures

15.4a and b.

u(t+ D - u(t-l) >k

>^ u(t)

1

1

W

--- 1 -1

1 1

1 2

> t

^ 1 1

(b)

(a) y(t)

FIG U R E 15.4 (a)

h{t)

= h(/), the step function. ( b ) / r ) =

u{t +

(c) Resulting y(r) for the convolution /;*/.'

1) - «(r - 1).

Chapter 15 * T im e Dom ain Circuit Response Com putations: T h e Convolution M ethod

So

lu t io n

Using the distributive law o f convolution,

y{t) = u{tY[u{t^

1) -

ii{t-

1)] =

u{tYu{t^

1) -

u{tYu{t-

1)

(1 5 .1 1 )

Now using the time shift property o f convolution and equation 1 5 .1 0 , we conclude that

y{t) = u{t) * «(r

+ 1) -

1) = K / + 1) -

r(t-

1)

which is plotted in Figure 15.4c. We note that this result was achieved without any direct inte gration.

E X A M P L E 1 5 .6 . Com pute the convolution^ =

h */ôf the

two waveforms

/j(t) and/(t)

in Figure

15.5.

h(t)

(a)

(b) y(t)

> X

FIG U R E 15.5 (a), (b) Waveforms for Example 15.6.

{c)y = h * f S o Lu n o N

We first note t h a t / r ) = tive law o f convolution.

ii{t)

+ «(r - 1 ) -

lu{t -

2) and

h{t)

=

u{t) - ii{t -

1). Using the distribu

Chapter 15 ‘ T im e Domain Circuit Response Com putations: T h e Convolution Method

y{t)

=

[u{t)

= [«(^) + « ( / - 1) -

+

u{t-

1) -

l i i i t - 2)Yti{t)

2u{t-2)V[u{t) - u{t- [//(r) +

ti{t-

1) -

1)]

2)]*//(/■- I)

lu {t-

Now using the time shift property o f convolution and equation 1 5 .1 0 , with a further application o f the distributive law, wc conclude that

y{t) = [r{t) + r { t - \ ) - 2r{t- 2)] - [r(t- \) + r{t- 2) -

2 r ( r - 3)] =

r{t) - M t - 2 )

+

2 r it-

3)

which is plotted in Figure 15.5c. We again note that this result was achieved without any direct integration.

E X A M P L E 1 5 .7 . C om pute the convolution

—u{t-

h * /o f

the signals

h{t)

=

a n d /(r ) =

e ^[u{t)

7 )], as given in Figure 15.6.

h(t) >k

1

w

0 (a)

FIG U R E 15 .6 (a), (b) Fu n ction s/r) and

h{t)

for the convolution of Example 15.7.

(c) The resulting convolution. So

lu t io n

Step 1.

Apply definition and adjust limits o f integration.

Applying the definition o f equation 15.2

yields

\{t) = h * f =

-T)f{T)dT =

i/(t

-T)]dT

Chapter 15 • T im e Domain Circuit Response Com putations; T h e Convolution Method

Since ;/(x) - z/(x - 7 ) is nonzero only for 0 ^ x :s T, the lower and upper limits o f integration

T,

bccome 0 and

respectively:

y{t) = h * f = ^ ^ Step 2.

e~û{T-t)clT

(1 5 .1 2 )

Detennirie the regions o f t over luhich the integral is to be evaluated. From t), there are three regions o f interest: (i) r < 0, (ii) 0 s ^ s

gration and the w(x -

^

the limits o f intex <

T, and

(iii)

T

^ t. Step 3.

Evaluate the convolution integral, equation 15.12, over the given regions.

Case I: t < 0. Z and

Here, r < 0 implies that x - f > 0 over 0 s x s 7". Hence, u(t - /) = 1 over 0 s x ^

j. y( 0 =

Case 2: 0 ^ t < T.

For this c;ise, u { t

- t)

e~'" c/t = 1- e~^

in equation 1 5 .1 2 is nonzero only when

i >t.

Hence, in

the region 0 ^ f < 7 , it must also be true that 0 < r s x < 7 for the integral o f equation 1 5 .1 2 to be nonzero. Thus, the lower limit o f integration with respect to the variable x becomes v (/) = J ^

Case 3: t>T. A simple Step 4 .

Plot y{t).

J

.

e 'û{T-t)dT =

calculation shows that

r

_^ n 7 -

=e

- e

f.

j

= 0 in this region.

Combining the results o f step 3 with 7 = 1 implies that y(r) has the graph

sketched in Figure 15.6c.

Exercise.

Find the convolution, say, y(/), o f the signal J{t) =

A N SW E R :

Ka

'//(- /) f

e ^hi{t)

with

h{t)

=

Ku{-t).

Ka

3. CONVOLUTION AND LAPLACE TRANSFORMS

^

As claimed in the introduction, circuit analysis in the time domain by convolution and circuit analysis in the frequenc)' domain by Laplace transformation are equivalent in terms o f zero-state response calculations. T he purpose o f this section is to rigorize the equivalence between the time and frequenc)' domain analysis methods by formally show'ing that presents a time domain rendition.

L[h ^j] = H{s)F{s).

Section 8


CONVOLUTION THEOREM h{t)

Supposey(/) = 0 and

= 0 for / < 0. Then (1 5 .1 3a )

i.e., convolurion in the rime domain is equivalent to multiplication o f transforms in the fre quency domain; or equivalently,

hit) ^j{t)

=

£r^[H{s)F{s)]

(1 5 .1 3 b )

T he justification o f this theorem proceeds as follows: Step 1. Given equation 15.13a, the first step is to write down the transform o f

h[t) * J{t).

Specifically,

£ f /;(/)*/(Ol =

-T)//(/ -T)/(T)^/rje ‘" J/

(1 5 .1 4 )

A couple of points are in order: (i) the inner integral, surrounded by parentheses, represents the convolution

h{t) * J{t);

emphasize the fact that

and (ii) the presence o f the step function

h{t) = 0

u{t - r)

is added as an aid to

for r < 0.

Step 2 . The goal at this point is to manipulate the integral o f equation 1 5.14 into a form that can be identified as the product ol the Laplace transforms o f two functions, i.e., as

H{s)F{s). The

only

possible strateg)' is to interchange the order o f integration and group appropriate terms. Note that the Re[^] must be chosen sufficiently large to ensure the existence o f the Laplace transforms of both

h{r)

Under certain conditions that are t\'pically met, it is possible to interchange the

order o f integration within a com m on domain of convergence of

H{s)

and

F{s).

Interchanging the

order and regrouping the f-dependent terms inside a single parenthetical expression produces

£ ( / ? { / ) * / ( / ) ] = J ^ _ / ( t ) ^J^_/;(/ - T) / / ( /

(15. 15)

Step 3 . Observe that the interior integral, surrounded by parentheses, in equation 15.15 is sim ply the Laplace transform o f a time-shifted

e~^H{s)

i.e.,

£[M{t -

r)] =

Substituting

for the interior integral leads to the desired equivalence: £ |/ i« ) « / ( O I = j ; ” f ( T ) H U ) e - ” 'dr =

= H ( . v ) f ( ,v )

This theorem asserts the equivalence o f convolution o f one-sided signals with multiplication of their transforms in the /-dom ain. For our purposes,

h{t)

assumes the role o f the impulse response

o f our circuit andy(r) the role of the input excitation. Accordingly, the convolution o f the impulse response o f a circuit or system with an input signal, a time domain com putation, equals the inverse transform o f the product o f the respective Laplace transforms. In other words, the diagram o f Figure 15.1 is correct, as claimed under the conditions o f the theorem.

Chapter 15 * T im e Domain Circuit Response Com putations: T h e Convolution Method

Exercises.

1. An unknown relaxed linear system has impulse response

the response to the input signaly(f) = l u { t

- 1)

b{t)

=

n{t) - u{t -

1). Find

using convolution and check your answer using

the Laplace transform method. A N SW E R :

h it -

1) - 2 r ( / - 2)

2. An unknown relaxed linear system has impulse response sig n a l/r) =

e~^‘u{t).

Find the response ^(f) =

h{t) = (a - b)e~‘^‘u{t),aj? > 0

and input

h{t) * fit).

A N SW ER :

T he conditions o f the theorem are somewhat restrictive in terms o f computing circuit responses strictly in the time domain. Specifically, it is the one-sided Laplace transform that does not rec ognize function segments over the negative real axis— hence the condition on the input excitation t h a t / r ) = 0 for /■< 0. This restriction does not lend itself to the computation o f initial conditions and circuit responses due to input signals extending back in time to / =

In general, the con

volution o f an input excitation with a circuits impulse response presupposes no such restriction. However, justification o f the computation o f zero-state responses due to input excitations extend ing back in time to / = -oo cannot be based on the convolution theorem o f the one-sided Laplace transform. A justification o f the time domain convolution approach to computing circuit response is reserved for the last section o f this chapter, due to its complexit)^ Nevertheless we will use the result as necessary, such as in the next exercise.

Exercise, T he

transfer function o f a particular system is

2

H{s)^

.9+ 0.2

C om pute the convolution o f the impulse response with the input A N SW ER : lOlI -

^

+ 1) - 1()[1 -

')],,(/

v{t),

shown in Figure 15.7.

d

v(t) J < ,1

-1

FIG U RE 15.7

4. GRAPHICAL CONVOLUTION AND CIRCUIT RESPONSE COMPUTATION T he convolution integration formula, although explicit, has many layers of interpretation. G raphical convolution is a pen-and-pencil technique for determining the convolution integral o f


simple, squarish waveforms. T he technique often leads to a more penetrating insight into the con volution integral. There are four key ideas in the graphical procedure: flip, shift, multiply, area. The following is a detailed description o f the procedure. To compute

t = T,

given by the convolution integral o f equation 15.1 or 15.2, for a specific value o f

perform the following steps:

1.

Plot

h{x)

vs. T andy(T) vs. T curves.

2.

Flip t h e / r ) curve about the vertical axis (t = 0) to obtain t h e / - r ) vs. T curve.

3.

T to obtain t h e y ( r Multiply. Plot the product h{z)j{T - r) vs. x curve. Area. Calculate the area beneath the hir)j{T - r) vs. x curve for —x < t Shift th e y (- t) curve to the right by the am ount

4. 5.

t)

t curve.

v s.

< oo. The result

is7(T ). We make the following remarks with regard to the above steps: (1)

The roles o f /;( t ) andy(T) may be interchanged because o f the commutative propert)' of convolution, equation 15.2. In other words, we may flip and shift /?(t) instead oF /( t). Usually we flip and shift the simpler waveform.

(2)

In step 3 , if T is negative, the shift is actually to the left.

(3)

As r i s varied from -oo to +oo, a complete plot of^(^) is obtained. In shifting and find ing the area, we often have to divide T in to separate inter\'als, becausc each interval may require a different formula to com pute the area beneath the

t, h{t-T)

vs. x curve.

t moves

from

to cc,

x) moves along the x-axis from x = -oo to x = oo; a simple illustration is h(t - x) =

u{t -

x), w'hich

With regard to step 2, for each

h{t -

h{T)j{T- r)

is a shifted horizontal flip o f/;(x): as

is sketched in Figure 15.8. Part (a) o f the figure shows «(x), part (b) depicts

u{t-T),

which slides to the right along the x-axis as

ftuKtion

h{t-x)

=

u{t-

t increases.

u{-x),

and part (c) plots

For comparison, we can consider the

1 - t), whose three forms are given in Figures I5.8d through f

u(-l) ><

>k u(x)

u(t-x) >k increasing t —

----w -

w

---------------- ^ ^ (b)

(a)

(C)

increasing t

u(-T-l) A

A u (x -l)

1

t- 1

-1 (d)

(e)

h{t-x) h{t-x) = u{t-x - 1),

(f)

FIG URE 15.8 (a)-(c) For the function

= « ( r - x ) , plots of (a) «(x), (b) « ( - t ) , and (c) « ( r - x ) .

(d)-(O For the function

plots of (d)

tt{x -

1), (e) //(-x - 1), and ( 0 «(^-x - 1).

Chapter 15 • T im e Domain Circuit Response Com putations: 'I'he Convolution Method

E X A M P L E 1 5 .8 . Graphically compute the convolution y = / / / o f the two waveforms

t)

and JU) =

2u{-t)

h{t) = u{-

shown in Figure I 5.9.

f(t)

"

k.

f(T)

I h(t-T) = u (t-1 ) h(t)

1

increasing t —

— y (a)

(b )

FIGURE 15.9 Waveforms for Example 15.8. So

lu t io n

T he graphical solution to this convolution depends on a partitioning o f the time line into special segments over which the graphical convolution is easily done. There are two regions to consider: (1)

-0 0

< / :s 0 and (2)

t > 0.

Step 1. Consider the region r > 0. Figure 15.9b shows that h{t - t ) J { t ) = 0 for all r > 0 and all T . Thus the convolution integral is zero and

= 0, r > 0.

Step 2. Consider the region -oo < ^ ^ 0. From Figure 15.9b , the product h{t

=1 x 2 = 2 for

r ^ T s 0 and zero elsewhere. Consequently, the area under the nonzero portion o f the product functions

h{t -

is y (0 = £

hit - T )f{ T ) d r =

dr = -It

Step 3. Combine the foregoing calculations into a plotted waveform. Figure 15.10 shows the func tion j/(r) resulting from the convolution.

y(t)

F IG U R E 1 5 .1 0 P lo t o f th e resu ltin g fu n c tio n , ^(/).

Chapter 15 ’ T im e Domain Circuit Response Computations: The Convolution Method

Exercise.

In Example 15.8, supposey(/) is changed

ioj{t)

=

2u{t) - 2ii{t —1).

M ndj'(r) at r = 0 .5 ,

1.5, and 2 .5 sec by the graphical convolution method. A N SW ER : 1 , 1 , 0

Another, more complex example will end our illustration o f the graphical convolution technique.

E X A M P L E 15.9. Com pute the convolution

j{t)

o f the triangular pulse

h{t)

with the square pulse

as sketched in Figure 15 .1 1 .

f(t)

T (b)

/ \ / \ \ /I \ / / /

/

-1

0

1

2

3

V

4

.

5

6

7

Time (seconds) (c)

FIG U R E 15.11 Convolution of triangular signal (a) with square pulse (b) to produce the signal (c) for T'^ 2 for Example 15.9. T he goal, o f course, is to graphically evaluate the convolution integral o f equation 15.2 using the following steps:

Chapter 15 • Tim e Domain Circuit Response Com putations: T h e Convolution Method

h{t-T)

1.

Since x is the variable o f inregrarion, draw

2.

Evaluate the product

3.

Com pute the area under the product curve for each region determined in step 2.

Step 1.

Drauf h{t -

h{t -

a n d / x ) on the x-axis.

x)J{t ) for various regions o f

t.

x) audJ{T) on the T-axis for f < 0 and compute the area o f their product. Figure

15.12 shows h{t - x) andy(x) on the X-axis. From the figure, it is clear that h{t 0; hence,

x )fj)

= 0 for ? <

= 0 in the first region.

FIG U R E 1 5 .1 2 Graph of h{t-x) and/(x) on x-axis for r < 0. Step 2.

Consider the region 0

^ r <

as illustrated in Figure 15.13. h(t-x)f(T) A

-- 2T Area 2T-t

t-2T

t

T

FIG U R E 1 5.13 Graph of h{t-x)fix) on the x-axis for 0 s / <

T.

T he shaded area o f the figure is the difference between the area o f the large triangle, defined as Area A = 0 .5 (2 7 )^ and the area o f the smaller triangle to the left o f the vertical axis, defined as Area B = 0 . 5 ( 2 7 -

Chapter 15 •Tim e Domain Circuit Response Com putations: T h e Convolution Method

Hence, j/(r) = Area = Area A - Area B = 2 Tr - 0 . 5 r for 0 :s r <

T.

Alternatively, one may use the formula for the area o f a trapezoid, i.e., the average

height times the base, in which case one immediately obtains 0 . 5 ( 2 7 +

I T - t)t

= Area =

ITt-

0.5/^. Step 3 .

Now consider the region

t <2T, as depicted in Figure 15.14. h(t-T)f(T) A -- 2T

Area

T

t-2T

t

■> I

FIG U R E 1 5.14 Graph of h{t-x)j{T) on the x-axis for

T^ t <1T.

In this figure, the shaded area, which determines j'(f), is again the difference o f two triangular areas, specifically.

y{t) = 0 . 5 [ r - { t - I T ) ] for

- 0 . 5 [ 2 r - r]2 = 2 .5 7 ^ -

Tt

T^t<2T.

Step 4 . Figure 1 5 .1 5 shows the next region,

2T ^ t < 3T.

Another straightforward calculation

yields

y{t) = 0 .5 ( 3 r - /)2 = 4.5T^ - 5 T t ^ 0.5t^

for 2T^ t < 3T. h(t-T)f(x) A

-■ 2T

Area

A

> T

t-2T F IG U R E 1 5 . 1 5 Graph o f

h{t-T)j{T)

on the x-axis

ioxlT^ t < 5 T

780

Step 5.


Consider the region 5T^ t.

Here, the product

h{t-T )/(t)

= 0 , in which

caseyit)

= 0 for r

>37: In sum,

0,

v(0 =

/< 0

2Tt-0.5r,

0^t
2.5T^-Tt,

T^[<2T.

4.5T- -3T f + 0.5r, 2 T ^ t < 3 T 0. A plot of^(r) appears in Figure 15.1 Ic For

T=

2.

Exercises. 1. Repeat the calculations o f the preceding example, except flip and shift J{t) instead of h{t). Here, one looks zx. j{t - r) sliding through h{x). The calculations should be easier and the result the same. 2, Find the output o f a linear and relaxed circuit with in p u t //) and impulse response

h{t)

(shown

in Figures 15 .1 6 a and b) at time r = 2 .5 seconds. Hint: Use graphical convolution. A N SW E R : !

MGURF. 15.16 3. T he impulse response o f a particular circuit is approximately measured on a scope as illustrat ed in Figure 1 5 .1 7 . If the input to the circuit \s J{t) = u{t - 1 ) -u{t response, y{t), at / = 3 sec and r = 3 sec. /VN’SW ER S: 2, 1

-

2), find the value o f the

’ 81

Chapter 15 * Tim e Domain Circuit Response Com putations: T h e Convolution Method

4. For

h{t)

andy{r) as sketched in Figure 1 5 .1 8 , fin d ^ (l) for the convolution ^(r) =

h{t) j{t).

A N SW ER : 4

h(t)

^

t

-1 FIG U RE 1 5 .1 8

Sometimes the foregoing graphical techniques prove difficult to execute. Nevertheless, an under standing o f them offers fundamental insight into the meaning o f the convolution integral. A use ful set o f techniques for quickly evaluating convolution integrals arises from the properties o f a convolution algebra, discussed in the next section.

5. CONVOLUTION ALGEBRA A set o f functions, together with operations called addition and multiplication, is called an

bra,

alge

assuming certain conditions are satisfied. The set o f all functions that can be convolved with

each other also constitutes an algebra with respect to the operations o f addition (+), and convolu tion (*). This set, together with the two operations, is called a convolution algebra. In this con text, operations such as differentiation and integration are inverses o f each other. For example, integrating a function and then differentiating the result returns the original function. W ithin the convolution algebra, the co n v o lu tio n /* the derivative o f

g is equivalent

to the convolution o f the integral o f/w ith

The advantage here is that, by successive integration and differentiation, it is

often possible to reduce an apparently difficult convolution to a simpler one. For a set o f functions to be an algebra with respect to + and *, several arithmetic operations must hold. In particular, + and • must be both commutative and associative. The commutativit}^ and


associativity o f + is clear:/ + g = ^ + / a n d / + ciativity o f * is equally clear: f *

g =g * / a n d

functions that are mutually convolvable must /*

h.

(g + h) = ( f ^ g) + h. T he commutativity and asso {g* h) = {/* g) * h. To be an algebra, the set o f all also satisfy the distributive la w ,/* {g+ h) = f * g +

In addition to obeying these laws, algebras o f functions must contain identity elements. For

+, the zero function ser\'es as the identity. For convolution, the delta function plays this role. The delta function is an identity element because o f the sifting property as set forth in Example 15.2 and equation 15.7. For our purposes, the interesting aspects o f a convolution algebra o f functions rests with the inter relationship o f convolution, differentiation, and integration. To map out this kinship, we use the following notation /

/(t)A

(1 5 .1 6 )

and /,< !)(,) = ^

(1 5 .1 7 )

dt It can be shown that / * / 7 = / ( • ) * // - I ) if /(-o o ) = 0 and

t=

(1 5 .1 8 )

exists. T he co n stra in t/(-o o ) = 0 implies that the derivative o f / / ) is zero at

-oo and the constraint that

exists means that the integral o f

semi-infinite interval (-oo, /] for all finite

t.

h{t)

has finite area over the

Similarly,

/* /,= /(- • )

(1 5 .1 9 )

if /;(-oo) = 0 a n d /^ “ '^ exists.

u{t)

+

u{t

T'he goal is to use equation 1 5 .1 8 to evaluate the convolution, i .e .,/ *

h

=/^'^

E X A M P L E 1 5 .1 0 . Find the convolution

y =

h

f o r //) =

- 1 ) and

h{t) = uU) -

//(/ - 1 ) , as sketched in Figures 15.5a and b. So

lu t io n

= 5(r) + 6 ( / - I)

-2 6 { t -

2) and

=

r{t) -

, where/^*^(/)

K ^-1), as presented in Figure 15.19.

Chapter 15 * T u n c Domain Circuit Response Com putations: T h e Convolution Method

/H3

t

(b) FIG U R E 1 5 .1 9 (a) The derivative ofy(r) = (b) The integral of h{t) = Since f^\t) =

+ 6{t - \) -26

u{t) - u {t -

n{t)

+ « (r-l) -

2u{t -

2),

1), as given in Example 15.10.

(/ - 2 ), a sum o f impulse functions, the sifting propert)-’ o f the

impulse function implies

1) -

yU) =f^ W ith the picture o f g i v e n

1)

(1 5 .2 0 )

in Figure 15.19b , the right-hand side o f equation 1 5 .2 0 can be

interpreted as a graphical sum o f (shifted) versions o f //~ '^ (/), as illustrated in Figure 15.20.

FIG URE 1 5 .2 0 y(r) equal to superposition of shifted replicas of

Equations 1 5 .1 8 and 1 5 .1 9 , as illustrated in the preceding example, are special cases o f more gen eral formulas. Specifically, let

y =f * h- Then yij-^ k) -J-iJ) - fj{k)

(15.21)

Chapter 15 • Tim e Domain Circuit Response Com putations: riie Convolution Method

where j and

k are

integers and the n o t a t i o n m e a n s theyth integral o f/o v e r [ - 0 0 ,

the y’th derivative if j where j =

-k

EXA M PLE

> 0.

t] if j < 0,

and

O f c o u r s e , = /. An application o f this formula to the special case

with y = 2 is given in the following example.

1 5

. 1 1. Find the convolution

^t)

= :r

cos{m)u{t)

withy(f) =

r{t) - r{t-

2 ), as sketched

in Figure 15 .2 1 .

S o lution Some preliminary thought suggests that evaluation o f the convolution integral might proceed more smoothly via equation 1 5 .2 1 ; i.e.,

where =

TC~

J0

cos(jTq)dq = JTsmiJTt)u(t)

and

f _n\n(jtq)dq = {\ - coîJtt)]u{[) J0 Differentiatingy(r) twice leads to

Hence,

f*g =f^-'> *

But 1 1 - cos(M ^~ 2))]u (r -

2)

= [1 - cos(jrr)]K(r) - [1 - co s(7 r(r-

2))\ii{t -

2)

is just a right shift by one period o f [1- cos(;rr)]«(f). Therefore, = [1 - c o s (;rr)]« (/)//(2 -

T h is function is plotted in Figure 1 5 . 22 .

t)

Chapter 15 • T im e Domain Circuic Response Com putations: T h e Convolution Method

Tim e (seconds) FIG URE 15.22 Graph of the convolution

for Example 15.11.

Convolution algebra and graphical convolution lend themselves to a second application o f the convolution technique: the computation o f circuit responses from a staircase approximation to a circuit impulse response. If a circuit schematic is lost, such an approximation could result from a C R T readout o f the circuit impulse response measured in a laboratory. The following example illustrates this application.

EXA M PLE 15.12. Suppose the schematic diagram o f a very old linear circuit is lost. However, the circuit impulse response is measured in the laboratory and approximated by the staircase wave form o f Figure 15.23. If the input to the circuit is at r = 0 and at r = 0.5 sec.

= 100«(^), compute the output voltage,

h t)

J

1.0

0 .8

--

0 .6

-

0.4 -

0.2

-

0.2

0.4

0.6

■f—I— h H — I— h 0.8 1.0 1.2

H— I— I— I— I— h -l1.4

1.6

1.8

2.0

FIGURE 15.2.3 Staircase approximation to circuit impulse response.

'


S olution The objective is to convolve the input with the impulse response. The technique o f convolution algebra where one differentiates the impulse response to obtain a sum o f shifted impulse functions and integrates the input to obtain a ramp function seems to be the most straightforward approach for this calculation. First, observe that

h{t) = 0.1

- 0.1) +

- 0.2) + 0.1u{t - 0.3) + 0.2«(r - 0.4)

(15.22)

+ 0 .2 « U - 0.5) + 0 .1 « (r -0 .6 ) - 0 .1 « ( r - 1) - Q>.2u{t- 1.3) - 0.2//(r - 1.5) - 0.2«(/ - 1.7) - 0.2«(^ - 2) - 0.1 u{t - 2.2) Hence,

h'{t) = 0 . l 6 ( r - 0.1) + 0 .2 6 ( r - 0.2) + 0 .2 6 ( r - 0.3) + 0 .2 6 ( f - 0.4) + 0 .2 6 U - 0 .5 ) + 0 . l 8 ( r - 0 . 6 ) - 0 . l 6 ( r - l ) - 0 . 2 6 ( r - 1.3) - 0 .2 6 ( t - 1 .5 ) - 0 .2 5 ( f - 1 .7 ) - 0 .2 6 ( r - 2 ) - 0 .l 6 ( f - 2 .2 ) Now, since the integral o f the input is 100r(^), we can compute the output voltage as = 1Or(? - 0.1) + 20r(r - 0.2) + 20r(r - 0.3) + 20r(r - 0.4) + 20K/^- 0.5) + 10r(^ - 0.6) - 1 0 r ( r - 1) - 2 0 ;-(r- 1.3) - 20Kr - 1.5) - 20K r - 1.7) - 20r(^ - 2) - 1OKr - 2.2) At f = 0,

= 0, and at / = 0.5, ôiJO.5) = 4 + 6 + 4 + 2 = 16.

O f course, it is possible to obtain the solution just as easily in this case using the graphical method. Simply flip the

curve and slide it through the /?(f) curve. The area under the product curve

is simply the sum o f the rectangular areas, which are easy to compute.

Exercises. 1. Compute

for Example 15.12 at

1 and

1.5 sec.

A N SW ERS: 65, 106 2. As an alternative to the convolution algebra, find the convolution

j{t) = u{t) with h{t) given

in equation 15.22. Hint: Use the result o f Example 15.4 or equation 15.10, and the time shift the orem o f convolution, equation 15.5. ANSWF.R: Replace the step functions in equation 15.22 with ramp functions.

6. CIRCUIT RESPONSE COMPUTATIONS USING CONVOLUTION This section contains a series o f examples that illustrate the convolution approach to computing zero-state circuit responses.

Chapter 15 • T im e Dom ain Circuit Response Com putations: T h e Convolution Method

/87

EXA M PLE 15.13. Consider the RC circuit o f Figure 15.24, whose impulse response is h{t)

e~û{t). If the input is

find

when 7" = 0 and T > 0 for <7 = 0 and <7 > 0.

10 IF

FIG URE 15.24 /?C circuit for Example 15.13. So

lu t io n

Case 1: a = 0. Here,

= u{-t) V. Since the capacitor voltage,

is the convolution o f the

input with the impulse response,

^'oia (7’) = r

J —00

= f

J-00

h{T -

t)v ,„

{T)dT = f

J —oc

/Kt)v,„(T - T)dr

e~îf(T)i/(T - T )dr = f e~û{T-T)dT

Jo

Consider the case o f 7'= 0. Then ^ W (0 )=

e~'î((T)dT = 1

This makes sense because with a = 0, = u{-t) V is a 1 V constant, and the capacitor looks like an open circuit at r = 0. On the other hand if r > 0,

- T)dx = e~^

^ W (^ ) = Jq

This also makes sense physically because at 7’ = 0, the initial capacitor voltage is 1 V and the capac itor discharges with a time constant o f 1 sec since there is no further nonzero input.

Case 2: a > ^. Similarly to case 1, 'W ( ^ ) =

- r)dT =

- T)dx

Consider the case o f T = 0. Then .-0^^)rdT = 1+ a This initial voltage depends on the history o f the excitation and can be computed only by convo lution because the circuit is not in steady state at T = 0.


For 7’ > 0, then, since the input is zero, we know that

e-'^

•r

Vout^T) = Vc(0)e-û{T) = -------u{T) \+ a

Exercises. 1. For case 1 in Example 15.13, find

for 7 > 0.

AN SW l-R: I 2. For case 2 in Example 15.13, find

for T > 0.

AN SWF. R: ^ 1+i/ 3. In Figure 15.24, suppose the resistor has value /? > 0 and the capacitor C > 0. Show that the impulse response o f the circuit is

^

h{t) = ^ e aC

4. For the general impulse response computed in Exercise 3, suppose

= e^‘'u{-t) V. Compute

1C

1

RC

ANSWF.RS; 1, 1

----- + a

RC

E X A M PLE 15 .1 4 . The goal o f this example is to design a circuit that computes the running aver age o f a voltage over the interval [/^- T, t] given a specification o f the necessary impulse response, h{t). S o l u t io n

A circuit that computes a running average must have the input-output relation

From our development o f convolution, such a circuit must have an impulse response h{t) satisfy ing the relationship

Vnu,iO = Now, h{t - t) must be a window function that captures the portion o f Ts

T

^ r. Figure 15.25 depicts the proper forms o f

-

t)

and /;(?).

over the interval t -

Chapter 15 * T im e Domain Circuit Response Compurations: T h e Convolution Method

789

>^h(t)

h(t-i) T

1 T -

-

--t-T FIG U RE 15.25 (a) The window function h{t - r). (b) The impulse response h{t). The circuit design problem reduces to developing a circuit that integrates the function segment

V- {t) over t - T <, T

To achieve this integration, note that v,„ { T ) d r =

J

V,-,, { T ) d T +

V,.„ { T ) d T

This relationship leads to (15.23) For the second integral on the right-hand side o f equation 15.23, let /I = r + 7 ’. Then T

=dr,

= A + r , and (15.24)

Since both X and x are dummy variables o f integration, we may replace A by ra n d rewrite equa tion 15.23 as (15.25) where Vj^^{T - 7) is simply a delayed replica o f t^/„(x) and where, for practical reasons, we can replace the lower limit of-oo by /q, the time the actual circuit turns on. As a convenience, we will define a device called an ideal delay o f T seconds, whose input is yy,/^) and whose output is

T). Figure 15.26 shows the ideal delay as a device having infinite input

impedance combined with a dependent voltage source whose output is a delayed version o f the input. Such a device can be achieved by storing the values of

in a digital computer or, for

small T, by the use o f an analog delay line.

F lG U llE 15.26 Depiction of a device called an ideal delay of T seconds whose output is a delayed replica of the input.

790

Chapcer 15 • T im e Domain Circuit Response Computations: T h e Convolution Method

All the pieces are now available; it is merely a matter o f putting them together. Integration can occur using an inverting ideal op amp circuit having a capacitive feedback and resistive input. The input to this ideal op amp integrator can then scale and sum the voltages produce the desired running average,

and

T) to

by setting RC = T. This will guarantee the correct

scaling to achieve the desired average, since the transfer function o f the integrator will be M{RCs). Figure 15.27 shows a circuit that will realize the desired running average.

FIG U RE 15.27 Op amp circuit that produces a running average of the input waveform, provided that RC = T. For the circuit o f Figure 15.27, the input drives the first inverting op amp circuit and also feeds the ideal delay. T he second op amp circuit is an ideal inverting integrator whose inputs are and

T). With RC = T the output is the required running average.

7. CONVOLUTION PROPERTIES REVISITED From the perspective o f the impulse response theorem, the convolution properties o f commutativir)^ associativity, and distributivity have important implications in terms o f circuit and system configurations. For example, if /?, {t) and hjit) are the impulse responses o f two systems, then com mutativity says that

means that the order o f a cascade o f circuits or systems is

mathematically irrelevant, provided that there is no loading between the circuits. The idea is illus trated theoretically in Figure 15.28.

Chapter 15 * T im c Domain Circuit Response Com putations; T h e Convolution Method

h,(t)

}

h,(t)

'9 1

}

(a)

h,(t)

}

h,(t)

}

(b)

FIGURE 15.28 Interchanging the order of impulse responses, in which the equivalence of parts (a) and (b) follows from the commutativity of convolution. Although mathematically parts (a) and (b) o f Figure 15.28 produce the same result, practical con siderations often dictate a more careful realization process. For example, one circuit may have a high input impedance w'hile the other has both a low input impedance and a low output imped ance. In this case we would put the first high-input-impcdance circuit at the front end and the other circuit at the output end. A cascade op amp realization o f a transfer function,

H {s) =

100 (15.26)

(.y-Kl)(.9-H20)

illustrates commutativity nicely. A designer may use either Figure 15.28a or Figure 15.28b to real ize H{s). Magnitude scaling, say by

= 10“^, will yield more realistic resistor and capacitor val

ues. Recall that in magnitude scaling,

and

20

0.05 0

(a)

•92

Chapter 15 • I1 m c Domain Circuit Response Com putations: T h e Convolution Method

20

0.05 Q

(b) FIG U RE 15.29 Realization of interchanging the order oFcascaded circuits.

Exercise. Verify that each o f the circuits o f Figure 15.29 realizes the transfer function o f equation 15.26. Magnitude-scale by = lO'’ to obtain more realistic parameter values. If only 1 [.iF capac itors were available, you would need two scale factors—

for the first op amp stage and

for the second op amp stage. What are the rwo scale factors? AN SW ERS: Multiply each resistor by I O'’ and di\'ide each capacitor by 10'' to obtain the new val ues. II only 1 uF capacitors are available, then for Figure 15.29b, For Figure 15.29a, the two scale factors are interchanged.

Theoretically speaking, the associative propert)', /;, * (A, *

= 10*^’ and

= 0.5 x 10^’.

= {h^ * h-y) * hy means that multi

ple cascades o f circuits or systems can be combined or realized in whatever order the designer chooses. This assumes that there is no loading between the circuits or systems. Op amp circuits called buffers or voltage followers having gains o f 1 can be used to isolate stages. On the other hand, practical constraints may impose a condition on the realization o f a circuit for which the mathematics o f the associative property does not account. Finally, we consider the distributive property o f convolution: /;, * {hj +

= {h^

(/?j * h^. One interpretation o f this property is that the superposition o f the input signals

h-^) + and

is valid. Flowever, Figure 15.30 presents rwo block diagrams with different interpretations. Here one sees rwo possible topologies for realizing a system.


~93

y(t)

FIGURE 15.30 Two possible block diagram interpretations of the distributive law For convolution.

Exercise. In terms o f reliability, i.e., possibly continued partial operation in the face o f a circuit failure, suggest reasons why one realization in Figure 15.30 would be superior to the other. In terms o f minimum number o f components, suggest reasons why one realization would be better than the other.

8. TIM E DOMAIN DERIVATION OF THE CONVOLUTION INTEGRAL FOR LINEAR TIME-INVARIANT CIRCUITS As mentioned, a deficiency in the one-sided Laplace transform technique is its inability to deal with signals whose time dependence may extend back to -<». This section develops the zero-state system response as the convolution o f a not necessarily one-sided input excitation with the impulse response o f the circuit or system. Throughout the development, we will assume that the circuit or system under consideration is linear, i.e., composed of linear circuit elements. This implies that the zero-state response o f the circuit satisfies the conditions o f linearity; i.e., if the zero-state response to the cxcitation fj{t) \s yj{t) for / = 1 ,2 , then the zero-state response to the input excitation

addition, we assume that the circuit or sys

tem is time invariant, i.e., each circuit element is characterized by constant parameter values. Mathematically, this means that

j{t) is the input to a circuit element and y(/) the zero-state 1 0 j{t - T) for all T and all possible

response o f the circuit element, then y{t - 7) is the response

794


input signals y(^). intuitively speaking, time invariance means that if we shift the input in time, then the associated zero-state response is shifted in time by a like amount. These properties under lie the development that follows. Rectangular Approximations to Signals Let us define a pulse o f width A and height 1/A as 6^(r). In particular. 4A

0 < /< A (15.27)

othenvise

0 Mgure 15.31 sketches

for several A values.

u(t) - u(t-A) , , . 6A(t) = ---- 7---- for several As

1 A.

■> t 1

2

3

FIG URE 15..31 6^(/) for several As. Shifting the pulse o f equation 15.27 yields

1

as represented in Figure 15.32.

5^(t-kA)

A

kA

FIGURE 15.32

- M ) = b(^{t -

(k-Hl)A

is a shifted version of the pulse 6a(^)> where

= M.

For convenience, let = M so that 6^ (r - M ) = 6/^(/‘ - tj^. Figure 15.33 shows a rectangular approximation, v{t), to a continuous waveform v{t).


79 S

^ v (t)

t

FIGURE 15.33 Rectangular approximation, v{t), to a continuous signal v(t). Expressing the rectangular approximation indicated in the figure analytically, using shifted ver sions o f the pulse functions defined in equation 15.27, leads to the infinite summation

HD= 2 v(A-A)d^(r-A'A)A= ^ v{ti. )d^(t - tf^)A k=-oa k=-x

(15.28)

Hence, for sufficiently small A, it follows from equation 15.28 that X

v{t) «

V(/) =

v{tf, )6^ (/

2

-

tf, )A

(15.29)

k=-oo One concludes that if v{t) is continuous, then X

y A-*{) A'=-00 ^

- ti.)A= C

v'(/)= lim

J-oo

\iT)d{l - t ) ( I t

(15.30)

where we have interpreted the delta function as a limit o f short-duration pulses whose height is inverse to the width so that the area is constant: d(t)=

lim d ^ ( { ) AÔ

Observe that the right-hand integral o f equation 15.30 is precisely the convolution v{t) * S(r) C om putation o f Response for Linear Tim e-Invariant Systems Suppose

is the zero-state response o f a linear time-invariant circuit to the pulse

If the

circuits impulse response satisfies smoothness conditions, i.e., if it has sufilcient continuous deriv atives, then the circuits impulse response is the limit o f the /ja U) values as A goes to zero. In par ticular, //(/)= lim//^(/)= lim f A - * ()

/?(f -

T )d^ (T )c/r

= f

A - * ()JO

where the fact that h i^ {t)= ^ ^ _ h {t -T )d f ^ {T )d T

h{t -T)d{T)dT,

(15.31)


follows from the convolution theorem. Equation 15.31 restates the law that the zero-state response o f an input to a linear time-invariant circuit is the convolution o f the input with the impulse response. Now, by the assumption o f time invariance, linear time-invariant circuit to

kA) is the zero-state response o f a well-behaved

- M ). Suppose further thatj'(f) is the zero-state response of

the same circuit to v{t). (See Figure 15.34.)

LinearTlme 8,(t)

Invariant Circuit

LinearTime

8.(t-kA)

I

!< = -CC

Invariant Circuit

(v(gA)8,(t-tJ

LinearTime k = -cc

Invariant Circuit

h

LinearTime

v(t)

y(t)- / v(t) h(t-x) di

Invariant Circuit

/-oc

h FlCirilF. 15.34 Zero-state responses to a particular linear time-invariant s)^stem, showing the framework of the derivation. Note that the bottom condition follows because as A

0, A

di,

x, and H

j.

It is now possible to use the approximation for v{t) given in equation 15.31 to generate an approx imation to^(^) in terms o f a summation o f terms o f the form v{kA)hi^{t - kA)A. Taking the limit as A

0 will yield

as the convolution o f v{t) and h{t).

To derive this, note tliat for each k\ v{kA) = v{ti^ is a scalar. Hence, the zero-state response to

tfÂ is

tf^)A. By the linearity assumption, which implies superposition, the zero-state circuit response to v{t), equation 15.29, is (15.32) k--x

In the limit, as A approaches zero,

approaches a continuous variable t and A -♦ ch. Hence, if

the impulse response is sufficiently smooth— i.e., if it has sufficient continuous derivatives— then


v ( / ) = lim

y

v(A-A)/2 a ( / - A'a )A = r

v{T)h{t-T)ch

Thus, we conclude that for a linear time-invariant circuit, the zero-state response

(1 5.33 )

to an input

excitation v{t) is the convolution o f the input v{t) with the impulse response h{t). We will refer to equation 15.33 as the impulse response theorem,* which says that the zero-state response o f a lin

ear time-'mvariant circuit or system to a (possibly tivo-sided) input signal is the convolution o f the input with the impube response o f the circuit.

9 . SUMMARY This chapter has introduced the concept o f the convolution o f two signals. The convolution can be evaluated analytically (by direct computation o f the convolution integral) or graphically. Often, by applying the properties o f convolution algebra, it is possible to implement shortcuts for calcu lating the convolution o f two signals, resulting in the simplification o f the analytical calculation or o f the graphical calculation. Using the notion o f convolution, the chapter developed a technique o f computing circuit respons es in the time domain. This technique is the direct counterpart o f computing the transfer func tion in the frequency domain, the approach presented in Chapter 14. Using the convolution approach, one can compute the zero-state response o f a circuit excited by signals that extend back in time to -oo, something not directly possible with the one-sided Laplace transform. However, for one-sided signals— which constitute the great majority o f signals that are relevant to circuit analysis— the convolution and Laplace transform approaches are completely equivalent, as demonstrated by the convolution theorem. The chapter gave an example o f designing a circuit to compute a running average. In addition, it presented an application of the convolution technique to the computation o f circuit responses for a circuit whose impulse response is approximated on a CRT. Future courses will expand the seeds planted in this chapter. For example, convolution is pertinent to an understanding o f radar techniques, commonly used to identify speeding motorists.

10. TERMS AND CONCEPTS Algebra: a set o f functions with respect to two operations, + and *, satisfying the commutative, associative, and distributive laws. In addition, there must be an identit)' with respect to each operation. For addition, the zero function serves as the identity. For convolution, the delta function plays this role. The delta function is an identit)' element because o f its sifting property. Associativity: for convolution, the property h * (/* ^ = {h * f } * g. Com m utativity: for convolution, the property h ' f = f * h. Convolution: an integration process between two functions to produce a third, new function in accordance with equation 15.1 or 15.2. Convolution algebra: the algebra o f functions with respect to the operations o f addition (+) and convolution (*).


798

Convoladon theorem: for.one-sided waveforms

ancIy(^), L{h{j^ * J{t) ] = H{s)F{s).

Distributivity: for convolution, the property h * {/■¥ ^ = h* f + h * g. Graphical convolution (fiip-and-shift method): a pen-and-pencil technique for determining the convolution integral o f simple, squarish waveforms.

^

Ideal delay o f T seconds: waveform with inputy(/) and o u t p u t 7), a delayed replica ofy(f). Impulse response theorem: theorem stating that the zero-state response o f a linear time-invariant circuit or system to a (possibly two-sided) input signal is the convolution of the input with the impulse response o f the circuit. Linearity: property whereby, if the zero-state response to the excitation f^t) is then the zero-state response to the input excitation

for / = 1 ,2 ,

‘s

Running average: average that is updated continuously. Sifidng property: the property o f a delta function for simplifying an integral, i.e., 4-00 f{T )= - T )d t , provided t h a t / 7) is continuous at T. * Time invariance: property such that, i f / / ) is the input to a circuit element and^(^) is the zero-

f

state response to the circuit element, th e n ^ (/- 7) is the response

^__

7) for all T and

all possible input signals//). Zero-state response: response o f a circuit or system to an input when all initial stored energy is zero, i.e., all initial conditions are zero.

o

o

n

* T h e derivation o f this result is, o f course, not rigorous. A rigorous justification is given as theorem 4 o f Sandbergs “Linear Maps and Impulse Responses,” pp. 2 0 1 -2 0 6 .

IEEE Transactions, on Circuits and Systems, vol. 3 5 , no. 2, February 1988,


Problems

799

A

A

CONVOLUTION BY INTEGRAL 1. Leijit) = K^5{t - r , ) , plot in terms o f A', and

> 0. Compute and

-► t

the results o f the fol Figure P i5.5

lowing convolutions: (a) (b)

P )'u {t^ 2 T ^ ) J{t)'r{t^ T ^ )

6. L et/ f) = Ke~‘“u{t)y ^ > 0. Use the convolu

(c)

r ,)- « (r -4 r i)]

tion integral to compute the convolution in

(d)

/ f ) M K r + 2 r , ) - r ( ^ - 2 7 ', ) ]

part (a). Then use the results for (a) along with

(e)

7 ' , ) - ; < r - 2 7 ’,)]

the various convolution properties, especially the time shift property, to compute the remain

2. Let^(f) = 25(/ + 2) - 2 5 { f - 2). Compute and plot the following convolutions:

J{t)'u {t) (b) j{t)-r {t) (c) J { t ) - [ u { t ) - u { t - A ) ] id) J{t)-[r{t)-r{t-2)] (e) J { t ) - [ r { t ) - r { t - 2 ) - u { t - 4 ) ] (a)

ing convolutions. (a )

t({r) * f i t )

(b)

u { t ^ T) * J{ t) [u{t^ T ) - u { t - T ) ] ' P ) [uir^T^)-u{r-T,)]*J{t-T^),T^ > 0, Tj > 0, >0 uir^ l)*[J{t)-e-^'% -T)]

(c) (d) (e)

plot the following convolutions:

=Kê-^‘ii{t), a > Q, and f^{t) -K-,e~Ûiii), b > 0. Use the Laplace transform

(a)

J{t) - cos{m)u{t)

method to compute the following convolutions:

(b)

J{t)

3. Let//) = 25{t + 2) - 2 6 {t- 2). Compute and

- sin(;rr)«(r)

7. Let f i t )

(a )

fir)* fir)

4. Compute and plot the results o f each o f the

(b) T iW / z W Now use the results for parts (a) and (b) and the

following convolutions:

various convolution properties to compute the

(d)

u{t) * u(t - 2) n { t - 2) u(^)*[u(t)-;^(t-4)] u(t^2)*lu(r)-u(r-2)]

(e)

[ « ( f + 2 ) - ;v U ) ] * [ / / W - « ( r - 2 ) ]

(a) (b)

(c)

following convolutions: (c) f i r ^ T ) * f i t - T ) , T > 0 (d) (e)

fit^ T )*f^ [t-T ) fit+T^)-f^{r-T,),T^,T^>Q

5. Use the convolution integral to compute the

8. Compute and draw the convolution y{t) = f t ) * v{t) o f Figure Pi 5.8 with v{t) = 5{t + 4) -

convolutions in parts (a) and (b). Then use the

5 (f+ 2 ).

results for (a) and (b) to compute the remain

f(t)

ing convolutions. (Parts (a) and (b) can be used to solve many o f the subsequent problems also.) (a) (b) (c) (d) (e)

«(/) * Hr) Hr) • r(r) u(r)*[r(r)-r(t-4)] r (r )* [ r ( r ) - K r - 4 ) ] M r + 2 )-/ / W ] * l u ( r ) - u ( r - 2 ) ]

(0

[K ^ + 2 )-K r)]*[K r)-K r-2 )]

(g)

/^(r) * fjir) for the functions in Figure P15.5

>k 3 2

-

Figu re P i 5 .8

Chapter 1 5 * Time Domain Circuit Response Computations: The Convolution Method

800

9. (a)

Compute the convolution y{t) = f j ) *

(b)

Repeat the calculation in part (a)

v{t) for the functions in Figure PI 5.9 using the fact that r{i) = «(/) * u{t) and

using graphical convolution for K = 2 .

the properties o f convolution. (b)

Repeat the calculation in part (a) using graphical convolution for 7’ = 1 and/r= 1.

. v(t) 2K K (a)

H------h 1

2

3 (b)

2K

Figure P15.11 2T

12. (a) Compute the convolution j>(/) = j{t) *

■>t

-T

v{t) o f the functions in Figure PI 5.12.

--K

r\

Hint: Use the results o f Problem 5 for

(b)

u{t) * r{t). Figure P I5.9

(b)

Repeat the calculation in part (a) using graphical convolution for

lO.(a)

Compute the convolution oiy{t) = fj)

*

v{t) in Figure PI 5 . 1 0 using the fact ,f(t)

that lit) = «(/) * u{t) and the proper Repeat the calculation in part (a) • 1*1 1 ■ /* usmg graphical convolution ror 7 = 1.

i. v(t)

A

ties o f convolution. (b )

=2

and 7’ = 1.

2A

r

T

.f(t)

/N 2T

(a)

4T

(b)

2+

Figure P i 5.12

1

13. ■> t

■> t -T

T

2T

2T

PI 5 .1 3 . Hint: Use the results o f

Figure P I 5.10

Problem 5. (b)

1 l.(a)

Compute the convolution o^y{t) = fjj * v{t) in Figure P I 5 . 11 using the fact that r(/) = u{t) * «(/) and the properties o f convolution.

(a) Compute the convolution

J{t) * v{t) o f the functions in Figure

Repeat the calculation in part (a) using graphical convolution for A = 2 and T= I.

=


17. I f / r ) = Kû{-t), and h{t) = 0, a > 0, compute j'(^) = h{t) * j{t).

f (t )

a >

t 18. Compute the impulse response,

■>t

-T

801

2T

h{t),

of the

circuit of Figure PI 5.18. Express h{t) in terms of

b = l/(/?Q . Supposeyj(/) = K^^û{—t)y a > 0,

(a)

andy^(^) = K 2 e~^*u{t), a > 0 and a

Figure P I5.13

b. Compute

each o f the indicated convolutions using equa 14. Find the convolution of j{t) in Figure P I 5.14 with

v{t) = d{t) + 5(f + 2) (Hint: Use graph

(a)

ical convolution.)

h{t) =

(b)

tion 15.2. (a)

^j(r) = h(t) * u{-t), and sketch for < r < 00.

(b)

yjit) = h{t) * ^<

+1) + 5(? - 1) (Hint: Use (c)

graphical convolution.)

and sketch for - o o <

00.

y^{t) =

h{t)

*^ (/ ).

and sketch for - o o

<

r < 00.

/(^) = u{t + 3) - «(? + 1)

(c)

-o o

(d) (e)

y^{t) = h{t) *[/i(r) ^ and sketch for -CO < t < 0 0 . yÛ) = h{t) *[ u{-t) +^(^)], and sketch for -oo < / < 00.

'" 6 Figure P I5.18 19. A particular active circuit has the transfer function Let h{t) = 4«(/). Find^(/) = h{t) * fît)

15.(a)

H (s) = — s+a

w ith ^ (/) as shown in Figure PI 5 .15a.

w

(b)

Sketch y{t). Verify your answer using

where a> 0. Suppose the input to the circuit is

the Laplace transform method. Repeat part (a) for /j ( /) in Figure

v{t), shown in Figure P I 5.1 9 with 7’ > 0. Using

PI 5 .15b. A

convolution techniques, compute the response

yit)-

f ,( t ) V (t)

>

t

>«t

-► t

-T

Figure PI 5.19 Figure P I 5.15 16. I f / / ) = Kêûi-t), ^ > 0, and h{i) = K2 u(t), compute ^(/) = h(t)

20. Consider the RLC circuit in Figure PI 5.20. (a) Compute the transfer fiinaion //(j), and express H {s) = — —

s+ a

—— s+ b

802


the circuit is

Then compute the impulse response

response ^(r) using MATLAB as follows: tstep = 1;

hit). (b)

Compute y{t) = h{t) *

where

= v{t) is given in Figure P I 5.19. (c)

If the input voltage is

V,

com pute y{t) = h{t) *

(d)

= ti{t) -u {t - 1) V, plot the

using the indicated element values.

vin = [1]; h = [0. 2, 3, 1, 1];

y = tstep*conv(vin, h) y = [0 y 0];

Repeat (c) if the input voltage is

t = 0:tstep:tstep’ (length(vin)+length(h));

= 5e^ûH) V.

plot(t,y) grid /

Y

Y

—o +

V

2H

5Q

0.5F

h(t)

Vou,(t)

3 --

—0

2-

Figure P I5.20 21. The transfer function o f a particular time-

■> t

invariant linear circuit is

H{s) = (a) (b)

+ .? + 1

i- + 2

Figure P I5.22

A•+ 4

Find the impulse response o f the cir

23. Repeat Problem 22, for the input

=

cuit.

u{t) + u { t - 2) - l u { t - A ) V. Hint: You need to

Find the transfer function o f the cir

change only the specification o f vin in the

cuit as a ratio o f polynomials by using

MATLAB code o f Problem 22.

the residue command o f MATLAB as follows:

24. This problem formalizes the use o f the

(i)

MATLAB code above to compute the convolu

define the array p = [-1 , - 2 ,- 4 ] ;

(ii) define the array r = [2 ,-2 ,4 ];

tion o f two piecewise constant time functions.

(iii) define k = 0;

The steps below develop a formula for the con =

volution o f two staircase time functions v{t)

residue(r,p,k)” to obtain the coef

and h{t) (shown in Figures P I5 .2 4 a and b)

(iv) use

(c)

the

command

“[n,d]

ficients o f the numerator and

using polynomial multiplication. In general, we

denominator polynomials.

assume h{t) and v{t) are zero for negative t and

Find the step response o f the circuit by the convolution method.

(d)

Compute the zero-state response o f

have a finite nonzero duration. (a)

Let y{t) = h[t) * v{t) and let the time step T = \. Using Example 15.4 and

the circuit to the inputy(r) = 8 « (-f) +

the convolution properties, argue that

?,u{t-T)y T > 0.

y{t) has a piecewise linear structure as shown in Figure P i5 .2 4 c. Find the breakpoint coordinates y i, y2 y3 and

GRAPHICAL CONVOLUTION 22. The impulse response o f a particular circuit

y4, first in terms o f the literal levels vO, v l, ho, hi, and h2, and then in terms

is measured approximately on an oscilloscope,

o f their actual numerical values.

as illustrated in Figure P I 5.22. If the input to

(b) Define three polynomials in

a-

803


using the coefficients from Figures

h(t)

*<

P I5.24a, b, and c as follows:

5 -hO

p(x) = vO X + vl q(x) = ho x“ + h 1 X + h2

h2

r(x) = yl x^ + y2 x^ + y3 x +y4 Show that rix) = p{x)(]ix) with yl through y4

0

assigned the values computed in part (a).

2

3

- 2 -1 -

Remark: The result o f step (b) indicates that the

(b)

coefficients [yl, y2, y3, y4] in K-v) or the break

y(t)

point values in y{t) o f Figure P I5 .2 4 c can be obtained from the coefficients [vO vl] and [hO hi

h2] by polynomial multiplication. The

M ATLAB command “conv” performs the desired polynomial multiplication. In particu lar, the code for obtaining the plot o f Figure

> t

P I5 .2 4 c is V = [ 2 4]; h= [ 3 - 2 1]; T = 1; tstep = T;

Figure P I5.24

y = [0 conv(v,h)*tstep 0]; % The additional beginning and ending zeros are added to indicate % that the initial and final values o f the convolution are zero, due to the % finite duration assumption, t = 0: tstep : tstep* (length(v) + length(h));

A N SW ERS: (a) yl = vO x hO = 6, y2 = vO x h 1 + vl X ho = 8 , v3 = vO X h2 + vl x h 1 = - 6 . and v4 = vl x h2 = 4 25. Repeat Problem 24 for the waveforms in Figure PI 5.25. When using the MATLAB code, it is necessary to account for/j(/) being

plot(t, y)

nonzero for negative t. This is easily accom

grid

plished simply by shifting/J(r) to the right by one unit and then doing the indicated convo lution; the proper result is obtained by a left

v(t)

A

shift o f one unit.

5 -vl

442

f,(t)

vO

1

2

(a)

-4..

(b)

(a) Figure P i 5 .2 5

.so^


26. The schematic diagram o f a very old linear circuit is lost. The impulse response, h{t), is measured in the laboratory and approximated by the staircase waveform o f Figure P I 5.26. Based on the available information, solve cach o f the following problems. (a)

If the input is output

= 100«(r), find the (b)

at r = 0, 0.5, 1, and 1.5

sec by the convolution method. (b)

Figure P I5.27

If the input is 100f«(/), find the out put at r = 1 by the method o f your

(c)

choice.

CONVOLUTION ALGEBRA

Use the MATLAB code o f Problem 24

28. (a) Using the techniques o f convolution

to solve part (a). Hint: Although the

algebra, find j^(r) = /j(r) *

step function is not o f finite duration,

plot^ (r) vs. t, where/j(r) and^(^) are

after 2.2 seconds the output,^(r) = h{t)

as shown in Figures P i5 .2 8 a and b.

* v-^j{t)y does not change; also, the time step 7'= 0.1 sec.

and

Hint: Graphically integratey^(r). (b)

Repeat part (a) with/j(/) changed to the waveform o f Figure Pi 5.28c for T = 2 and /I = 2. A

Figure P i5.26 27. A crude approximation to the impulse response o f an active RLC circuit is given by h{t)y as sketched in Figure Pi 5.27a. II the input signal is v{t), as given in Figure P i5 .2 7 b , find the response y{t) using MATLAB and the

■> t

method developed in Problem 24. Plot the response for 0 < ? < 6 sec. Verify using the tech

-4 . .

nique o f Example 15.4.

(b)

A

-T

2T (c)

Figu re P i 5 .2 8

Chapter 15 * Time Domain Circuit Response Computations: The Convolution Method

29. Repeat Problem 28(a) w ith^(t) as given in

O '

Figure P I 5.29. Hint: Graphically int^rate^(/).

32. Consider Figure P I 5.32. Find the transfer function H{s) and

(a)

the impulse response h(t) of the circuit

fj(t)

in Figure PI 5.32a.

w

(b)

Find the step response using convolu tion algebra methods.

(c)

Find

due to the rectangular

pulse input in Figure P I5.32b . Hint:

■> t

Again, use convolution algebra meth ods, and sketch the output waveform

(or T = 2n^l(LC) .■

-4 --

iYYV

Figure P i 5.29

L

30. Suppose the impulse response of a circuit is given by pulse

ôuM^~

= (P'e~^û{i) and the input is a

T

= u{t+ T) - u {t- T), T > 0 V. Find

(a)

using convolution algebra

(b)

Figure P I5.32

techniques. 31. Use convolution algebra techniques to determine y / ) = h{t) * j{t) for each of the func tio n s //) given in Figure P I 5.31 where h{i) =

7^ cos{m)u{t). Plot the resulting j^(/) using MATLAB or its equivalent.

33. For the circuit o f Figure PI 5.32a, find the convolution^/) = h{t) * J{t) for eachJ{t) given in Figures 15.31b and c. 34. Consider the circuit presented in Figure PI 5.34a.

A«t)

w

■>t

(a)

(a)

Find the transfer function H{s).

(b)

Compute the impulse response h{t).

(c)

Use only convolution algebra tech

(d)

for the input in Figure P I5.34b. Plot your answer.

niques to compute the response

w w

(a)

A(.)A w Vw^ Vw> 'O

■> t

2 1 > t

/

-► tse c 2n

4n

6n

(b)

Figure P I 5 .3 4

806


35. Compute the convolution of^(/) = h{t) *

37. This problem repeats the trick o f Problem

for each j { t ) below, given the h { t) o f Figure

36 of using the Laplace transform method to

j{t)

P 15.35. (a) j{t) = o r cos{ojt)u{t) (b) (c)

compute responses when the input is nonzero

for - T < t < 0 when T > 0. Suppose a circuit

J{t) = (ur s\n{(Ot)u{t) J{t) = aê-‘^'u{i)

has the transfer function

H(s)=

V

-

Vin

A s +a

Suppose the input to the circuit is

shown

in Figure P I 5.37. (a) Define (/) = Vj„(t - T )and compute

(b)

Compute

due to the input

(c)

To compute

V/„(.y) and then

CONVOLUTION BY INVERSE LAPLACE TRANSFORM

). due to

use

time invariance, i.e.,

36. This problem illustrates a trick for using the Laplace transform

responses when the input is nonzero for - T / < 0 when

A

method for computing q

<

> 0. Suppose a circuit has the

transfer function

■>t

-T

V Figure P I5.37

s+2

Vin

Suppose the input to the circuit is in Figure P I 5.36. (a) Define

shown

38. The ideas presented in Problems 36 and 37 can be generalized as developed in this prob

(t) =

V/„( t

- T )

and compute

lem. Starting with the definition of convolu tion integral, equation 15.2, prove the time

(b)

Compute

) due to the input

(a) f i t ) *

Vjnis) and then (c)

To compute

shift properties for convolution as follows:

due to

use (b)

time invariance, i.e.,

Vout(^)=^’out(l + T')-

=

- T^) * }>(t)

/«)«^,'(0 = [ / «

f=t+T,

- r , ) * g(r - r ,

39. Two active circuits with impulse responses

h^{t) = 2e~^‘u{t) and h-,{t) =

v.(t)

are cas

caded as shown in Figure P I 5.39. No loading occurs between stages. Compute the zero-state response for

■>t

-T

here?

Figure P I5.36 ANSWHK: (c)

= ti{t + 2) V. Would it be

advantageous to use the convolution method

= (I-

+T ) \’

^


v.Jt)

SO

h,(t)

h,(t)

'« 6

Figure P I 5.39 40. This problem shows the advantage o f the

Figure P I5.41

Laplace transform method over the convolu tion method in the context o f a very common

42. Consider the circuit o f Figure P I 5.42,

situation. After solving this problem, try evalu

which has the transfer function

ating the convolution integrals, but don’t spend

s R,C

much time on this. Consider Figure P i 5.40, which shows the cascade o f three circuits with indicated impulse responses h^{t) =

2u{t), , h2 {t) = 2Qe~'^û{t), and (a)

= 2tu{t).

S~ +

R\C\R2C2

Find the impulse response o f the cascade, i.e., h{t) = h^{t) * h-y{t) *

(b)

H(s) =

Let C, = 0.5 mF, Cj = 1 niF, /?, = 2

(a)

Find the step response o f the cascade.

kQ, and /?2 = 1 kH. Find the poles and

Hint: You might want to use the residue com

zeros of H(s), a partial fraction expan

mand in MATLAB to compute the partial frac

sion of H{s), and the resulting impulse

tion expansions.

response, h(t). (b)

v.(t)

h,(t)

h,(t)

j—

h3(t)

Find

= 54tr’û{-t)

when

V. Then find the (zero-input) response

j— ►

due to VoutÔ~). (c)

Find the zero-state response when = 54te~^'u{t) V. Would the time

Figure P I5.40

domain convolution method be desir able for this calculation?

41. Consider the circuit o f Figure P 15.41. (a)

Determine the value o f

from

(d)

Find the complete response, given vour answers to parts (b) and (c).

= \Qe^^‘u{-t) V. Can

the input

were doubled and

this be done using Liplace transform

72te~^‘u{t) V, find

techniques? It so, how? (b)

the

=

complete

response without doing any further

Find the transfer fijnction

calculations. C,

H(s) =

ê and the impulse response h{t) in terms v„(t)

o f R and C. (c)

Suppose R = 10 n and C = 10 mF.

6

Given

(0") computed in part (a), find the response due to the input

Fig u re P i 5 . 4 2

te~‘u{t) using the Laplace transform

method.

Evaluate

the

advantages o f this method over the

MISCELLANEOUS

time domain convolution method. 43. Figure PI 5.43 shows a configuration for an interconnection o f active circuits. The function


inside each box is the impulse response o f the sub-circuit or subsystem. Suppose A,(r) = Au{t),

h-î) = 4 « ( f - 2 ), and (a)

(b)

47. Repeat Problem 4 5 for h^(t) = 4«(/), ^2(^) = 56(^), hît) = 5 6 { t - 2 ) , and h^{t) =

= k cos{,m)u{t).

Using any of the convolution tech

4 8 . Figure P I 5.48 shows a configuration for an

niques you have learned, compute and

interconnection o f active circuits. The function

sketch the overall impulse response.

inside each box is the impulse response of the

Using any of the convolution tech

circuit. Suppose h^(t) = 46{t), hjit) = 4 8 {t - 1),

niques you have learned, find y{t) for

hît) = 4d{t - 2 ). h^(t) = 45{t - 4), and h^{t) =

j{t) = 6 «(r).

2 cos(;rf)tt(r).

h,(t)

(a)

h

Using any of the convolution tech niques you have learned, compute and sketch the overall impulse response of each configuration.

f(t)-

(b)

Using any of the convolution tech niques you have learned, find y{t) for

h,(t)

M = 6u{t). Figure P I5.43

44. Repeat Problem 43 for ^2(^) = 4 6 {t-2 ) and with the other impulse responses remaining the same. 45. Figure P I 5.45 shows a configuration for an interconnection of active circuits. The function inside each box is the impulse response of the sub-circuit or subsystem. Suppose h^{t) = 4«(/),

hjit) =

h^{t) =

and //4(f) =

6 (r). Using any o f the convolution techniques you have learned, compute and plot (using MATLAB) the composite impulse response of the configuration.

4 9 . Lety(f) be an arbitrary signal having a welldefined Laplace transform. Let 00

K t)= y > y(t)

f(t)

6{t-kT)

itti be a so-called impulse train. Find L Note that if

z, a complex variable, then

your answer should be of the form Figure P I5.45 46. Repeat Problem 45 with h^{t) changed to h^{t) = 4[u{t) - u (t- 4)]. Having done Problem 4 5 , can you do this problem without any fur ther calculations?

00

2

f(k T ) z -'‘

it=0

which is the so called Z-transform of the sequence \J{kT) |^ = 0 ,1 ,2 ,...}.

809

C!liapccr 15 * Tim e Domain Circuit Response Com putations: T h e Convolution Method

50. The circuit o f Figure Pi 5.50 is initially at

51. Consider the circuit o f Figure P I 5.50

rest. The input

again. Let the input

v ,„ (0 = ^ 6 { t + kT)

A=() is a periodic impulse train with T = 1 sec. (a)

be the impulse train

shown in Figure PI 5.51, i.e., analytically

cc '■ ,„«)= ' ^ d U - k T )

k=0

Show that the impulse response is h{t)

with r = 1. Find

for r > 0 and plot the

resulting waveform. (b)

Find the exact solution o f

for 0

Find the exact solution o f < /< 2

for 1

(d)

Repeat part (c) for the interval 4 < r < 5. You should wind up with the expression

VoJt) = 0.5^-0-5('-'^)(1 + ^-0-5 + ( ,.- 0 .5 ) 2 ^ (^ -0 .5 )3 ^ (^ -0 .5 )4 )

(e)

Simplif}' the expression computed in part (d) by making use o f the sum for mula for a geometric series: //-I =

1 -A ”

A -O

\.

provided (f)

1 -A

Find the solution o f

for « < / <

(;; + 1). K4ake use o f the techniques set forth in parts (d) and (e) to simplify your expression. (g) Sketch the waveform for large >i. This is the so-called steady-state solu tion.

'"6

20

IF

Figure P i 5 .5 0

(1) -4

(1) -3

(1) -2

(1) -1

(1)

-> t

0

Figure P I5.51 AN SW ER: i>Jr) = 1.27t'“*'-^^ for r > 0

C

H

A

P

T

E

R

Band-Pass Circuits and Resonance HOW A TOUCH-TONE PHONE SIGNALS THE NUMBERS DIALED Calling friends and others by phone occurs daily. When we dial a number, the information is sent to the central office by one o f two methods: fast tone dialing or the much slower pulse dialing. For example, electronic processing o f the pulse-dialed long-distance number 555-555-5555 requires about 11 seconds, while electronic processing o f the tone-dialed takes only about 1 second. The keypad o f an ordinary touch-tone phone has 12 buttons arranged in four rows and three columns, as shown in the following diagram:

row 1

697 Hz

row 2

770 Hz

row 3

852 Hz

row 4 941 Hz

colum n 1

colum n 2

colum n 3

1209 Hz

1336 Hz

1477 Hz

keypad

“

Chapter 16 • Band-Pass Circuits and Resonance

Pressing any one button generates two tones, with the frequencies selected by an electronic circuit inside the telephone set. For example, pressing the number 5 generates tones at 770 and 1336 Hz. The row and column arrangements and the dual-tone method permit the representation o f 10 digits (0 ,... , 9) and two symbols (*, #) using only seven tones. These seven tones are divided into two groups: the low-frequency group, from 6 9 7 to 941 Hz, and the high-frequenc)' group, from 1209 to 1477 Hz. " Such tones are easily produced by an LC resonant circuit. The four tones in the low-frequency group are produced by connecting a capacitor to four different taps o f a single coil (inductor). A similar connection generates the three tones in the high-frequenq' group. When a button is pressed to the halRvay point, a dc current from the central office is sent through the coil in the tank circuit. When the button is fully pressed, the dc current is interrupted. This action initiates sinusoidal oscillations in the LC resonant tank circuit at a frequency inversely proportional to V L C . The presence o f small resistances causes the oscillations o f the tank circuit to die out. However, pressing the button fully also connects the tank circuit to a transistor circuit that replenishes the lost energy and sustains the oscillations. At the central office, the equipment used to detect the presence ol the tones and to identify their frequencies is much more sophisticated. Two filters are required, one for each ol the frequency

^

groups. Each filter must pass the frequencies within ±2% o f their nominal values (697 to 941 Hz for one filter and 1209 to 1477 Hz for the other) and reject the signal if the frequencies are out-

^

side o f ±3% limits. The output tone from each filter is then proce.ssed digitally to determine its frequency. The concepts and methods developed in this chapter will allow us to understand the properties o f resonant circuits and the design o f various basic t}^pes o f band-pass circuits, as used in the touchtone telephone system.

CHAPTER OBJECTIVES 1. 2.

Describe and characterize the ideal band-pass filter. Understand band-pass circuits from the viewpoint o f transfer functions and pole-zero

3.

plots. Investigate the basic band-pass transfer function and its realization as a parallel or series

^

^

RLC circuit or op amp circuit. 4.

Investigate band-pass circuits having practical capacitors and inductors in contrast to

5.

ideal capacitors and inductors Describe the phenomenon o f resonance and investigate the properties and applications

6.

o f resonant circuits. Investigate general second-order transfer functions having a band-pass type o f frequency response.

^


SECTION HEADINGS 1.

Introduction

2. 3.

The Ideal Band-Pass Filter The Basic Band-Pass Transfer Function and Its Circuit Realizations

4. 5. 6.

Band-Pass Circuits with Practical Components The Resonance Phenomenon and Resonant Circuits General Structure of the Band-Pass Transfer Function with One Pair o f Complex

7. 8.

Poles Summary Terms and Concepts

9.

Problems

1. INTRODUCTION How is it possible to listen to a favorite radio station by merely pushing a button or two or sim ply turning a dial? Why do some very expensive receivers have very clear reception, while with some cheaper models other stations chatter in the background? What circuitry inside the radio makes this difference? The ability to clearly select a particular broadcast station depends on the design o f an internal band-pass circuit. Such a circuit will pass signals within a narrow band o f fre quencies while rejecting or significantly attenuating signals outside o f that band. To understand why this is important, note that audio signals have significant frequency components up to about 3 kHz for voice and up to about 15 kHz for high-fidelity music. These frequencies are far too low for wireless transmission. In (wireless) AM radio transmission, the audio signal modulates the amplitude o f a carrier signal that is suitable for wireless transmission. The carrier signal is a highfrequency sinusoidal waveform bet\veen 500 kHz and 1650 kHz. The modulated waveform con tains many frequency components centered about the carrier signal frequency, but extending over a range o f frequencies equal to twice the highest audio frequency. For example, the radio station WBAA, at Purdue University, has a carrier frequency o f 920 kHz and occupies a band from approximately 915 to 925 kHz. To select WBAA from all the carrier signals received by a radio requires a good band-pass filter to pass the frequency band o f 915 to 925 kHz while rejecting sig nals outside this band. This chapter introduces the idea and properties o f a band-pass filter. In its simplest form, a band-pass circuit consists o f only one capacitor, one inductor, and one resis tor, connected either in series or in parallel. In the first half o f the text, we analyze simple RLC cir cuits where we emphasize (1) transient behavior under dc excitation and (2) sinusoidal steady-state behavior at a single frequency. This chapter investigates the behavior o f circuits over bands o f sinu soidal frequencies. iMany useful results may be obtained with the phasor and impedance concepts studied earlier. However, rapid advances in technology have made it possible to have a band-pass circuit without any o f the usual RLC circuit components. Therefore, a study o f the band-pass propert)' o f a transfer function H{s) dominates the material o f this chapter. The resulting analysis is readily applicable to general linear systems, whether they be electrical, mechanical, or otherwise. Although transfer functions underlie our approach, circuit realizations with ideal and practical components illustrate all the basic concepts and properties.

S14


2. THE IDEAL BAND-PASS FILTER This section sets forth the ideal band-pass characteristic and shows how a basic second-order trans fer function can be used to approximate the ideal. Specifically, let H{s) be a voltage gain or some other type o f network function. As we know, the curves for \H{j(a)\ vs. (O and /.//(/□) vs. OJ are called the magnitude (frequency) response and phase response, respectively. Figure 16.1a shows an

ideal band-pass (rectangular) magnitude response curve. Here, “ideal” means that all frequency components o f the input signal within the range 0)j < to < (o^ are amplified with equal gain (in magnitude), and all frequency components outside o f the range are totally eliminated from the output. Actually, for a band-pass circuit to pass a signal with frequency components in the range

0)j < CO < 0J 2 without distortion, there is also a requirement on the phase response that is ordi narily studied in a course on signal analysis.

(b )

S IS


(C)

FIG URE 16.1 Definitions o f peak frequency, co^, and bandwidth,

(a) Ideal band-pass character

istic. (b) Approximate band-pass characteristic of simple RLC circuit, (c) Band-pass characteristic of a more complex circuit. Unfortunately, an ideal rectangular band-pass characteristic is not realizable by a rational transfer function or any circuit. One can, however, approximate the ideal characteristic with a simple tuned circuit whose transfer function produces a bell-shaped magnitude response curve as illus trated in Figure 1 6 .lb. With more complex circuits and more complex transfer functions, one can improve the approximation as shown in Figure I6 .1 c. (How to do so is a topic to be studied in advanced courses.) The bell-shaped curve o f Figure 16.1b has several important features. The fre quency at which |//(/w)| reaches its maximum value, denoted by

is called the peak frequency and is

The two side frequencies at which |//(/w)| is 1/V 2 o f its maximum value are

called the half-power frequencies, denoted tO| and 0^2 - The term “half-power” comes from the fact that if the output is a voltage across a fixed resistance, then a drop in voltage by the factor

1/-v/2 means a drop in power equal to ( l/ yfl] = 0 .5 . Half-power frequencies are also called 3 dB (down) frequencies. Recall from Chapter 14 that dB gain is defined as dB {gain) = 201ogjQ |Myw)| If a gain o f is reduced by the factor \/j2 , then the dB reduction in gain is given by - 3 d B = 2 0 log 10

H en ce the terminolog}^ “3 dB d o w n .”

71

Chapter 16 • Rand-Pass Circuits and Resonance

For obvious reasons, 0)| is called the lower half-power frequency and frequency, and their difference

the upper half-power

= CO2 - co, is the half-power bandwidth (or simply bandwidth)

o f the band-pass circuit/transfer function. Band-pass circuits are designed so that all frequencies o f interest fall within the pass band [(Op co^]. One way o f categorizing and comparing different band-pass circuits/transfer functions is by their selectivity, i.e., a circuits relative capability to discriminate between frequencies inside the pass band and signals outside the pass band. The selectivit)' is measured by the quality factor, Q, o f a band-pass circuit/transfer function. The quality factor, Q, is the ratio o f the (geometric) center fre quency {yj(JJiC0 2 ) to the bandwidth,

. For second-order circuits and transfer functions with a

bell-shaped magnitude response (Figure 16.1 b), the center frequency and peak frequency cide, as we will show. In this case Q =

coin

=

where Q denotes the pole Q, to be defined in equation 16.1. For the circuit realization o f the transfer function, Q is sometimes denoted by

or

narrow band o f frequencies relative to yJcOiOh or

A high-Q circuit passes only a very whereas a low-Q circuit has a broad (pass)

band and a less selective characteristic. It is important to note that the concepts o f

B^^, and Q o f a circuit are all based on the magnitude

function |//(/o)| and, therefore, on how the transfer function H{s) is defined. Even for the same circuit, these values are different when the output is associated with different branches or when the input is changed from a voltage source to a currcnt source. Further, for the investigation o f the frequency-selec tive characteristic of the circuit, the foregoing definition o f Q is most appropriate because it directly assesses the sharpness o f the magnitude response cur\'e. As such the definition allows Q to be experi mentally determined. In certain other applications, where only one fixed frequency is o f interest, there is another definition o f a circuits Q based on an energ)' relationship that is inadequate for general band pass circuit design.

3 . THE BASIC BAND-PASS TRANSFER FUNCTION AND ITS CIRCUIT REALIZATIONS The most basic second-order band-pass transfer function having the bell-shaped curve o f Figure 16. lb has a pair o f complex poles and a single zero at the origin, i.e.,

(.V - p\){s- P2)

(s + G p - j w j ){s + Op + jco^j) (16.1)

+ 2apS + a l + m ; ,

r + 2ct,,.v + (o^

+

+

Qp'

"

Figure 16.2 illustrates the pole-zero plot o f the transfer function o f equation 16.1 as well as the relationships among the various parameters. Note that there is a finite zero at the origin. Further, because the numerator has degree strictly less than the denominator, there is also a zero at ^ = co. In both cases the magnitude o f the transfer fijnction is zero.

SI

Chapter 16 • Band-Pass Circuits and Rcsonancc

JO)

FIGIJRK 16.2 Pole-zero plot of a band-pass H{s) with a single zero at origin as per equation 16.1. Our next goal is to derive the peak frequency the half-power frequencies tOj and t02,

the maximum gain

=|

the circuit Q, and the frequency at which the angle of

/y(/to) is zero. The first step in this derivation is to write down

K (16.2a) 2 o » + 7 ----------- (JO

whose peak magnitude response is

m ax

(I)

1^1

H(Jco)

/

_

2

I CO., - 0 )

l\~\ I 4a“+

mm

0)

(0

/

■) 9 \(OZ, - U>~ \

\K 2a,

(16.2b)

CO

Note that since the numerator is a constant, the maximum occurs when the denominator is a min imum, i.e., when co = (O^. Therefore we conclude that

m

p

(16.3)

COI,

Noting that Qp ~

, we further conclude

H(jco)

(16.4)

To find the half-power frequencies lOj and co-,, the maximum gain must be reduced by the factor l/V 2 . Considering equation l6 .2 b , this occurs at those co’s for which

818


(col-co-) CO

(16.5)

= 4a“

This fourth-degree equation in to reduces to two quadratic equations defined by cOp - c o “ = ±2opO) => co^ + 2o^,(o -co^, = 0

(16.6)

We denote the two positive solutions by cOj and a>2. with CO2 > o jj. Solving equation 16.6 using the quadratic formula, we have

yJcTp+OJ-p = CO.

( (16.7)

To compute the bandwidth, we have CO,

B,^=co2-C0^=20p=-^.

Exercise. Show that for high-Q^^ transfer functions/circuits (say

(16.8)

> 8), (16.9)

We can deduce one more important property from the above equations. When equation 16.5 holds, the real and imaginary parts o f the denominator in equation 16.2a are equal. Hence the denominator has an angle o f ±45^^. Thus the angles o f

^

H(jcOi) = ----2CT. + 7

(cor-col\ CO:

(/ = 1, 2) are ±45^" for K> 0, and ± 4 5 ° - 180° (or K < 0.

Exercises. 1. Given the transfer function H{s) = —

--------- , find 5 ^ + 2 ^ -h 2 5 6

A N SW I'RS: 2. Suppose that

= 16 r.id/sec, //,,, = 0.5,

, (0,^ ,

, and a>i 2 .

= 2, o), = 15.031. and (o^ = 17.031

= 8 and 10^ = 1000 rad/sec. Find the exact to^ ^ from equation 16.7, the

from equation 16.8, and the approximate Wj 2 from equation 16.9. Now compute the magnitude o f the percent errors between the approximate and exact values o f tOj 2A N SW ERS: in random order; 125. 939.^^. 1064.5, 1062.5. 937.5. 6 .1833, 0 .2077 3. Given equation 16.9, show that for a high-(^ transfer function/circuit (say

to^ s 0.5(cOi + W2)

> 8),

819


i.e., for the high-Q case the geometric and arithmetic means of 0)j and 0J 2 are approximately equal.

The next two questions are: (i) what is the geometric center frequency o f the magnitude response and (ii) is it equal to to ? Here, using the square o f the geometric center frequency,

CO1 CO2 = -O p + J a j - ¥ c o l

Op +

p+<4

2

2

^

= - Gp + Op + ( O p = ( O p

we conclude that for the above transfer function and associated circuit realizations, the geometric center frequency, the pole frequency, and the peak frequency coincide, i.e..

Îoj\co2 = COp = CO. As mentioned earlier, the circuit/transfer function Q is the ratio o f the center frequency to the bandwidth. Given equations 16.8 and 16.10, we have

Q = Q ..- Q ,, = ^

(16.11)

Finally, the nonzero frequency for zero phase shift occurs when the transfer hmction is purely real, i.e, when the imaginary part is zero. This occurs at CO = CO^; hence (16.12)

zero phase-shift frequency = co

E XA M PLE 16.1. Compute

|

= |Myto^^^)|, Q,

O),, (1)2 (exact and approximate

values) for the simple parallel RLC circuit shown in Figure 16.3a where the current source is the input and the voltage across the input nodes is the output. Then compute and label the magni tude response and verify that it has the bell shape o f Figure 1 6 .lb.

S20


.(b ) FICiURE 16.3 (a) A parallel /?/,C circuit, (b) Magnitude response curve. So

lu t io n

The key to the solution o f the example is the computation o f the transfer function as follows: 1

1

Y(s) ^

—s C

1

^

1

1

C5 + — + —

R

^

Li

1

1

.V + - — .y-l- —

RC

LC

Then, according to equation 16.1,

—s

H{s) = K —

C

=K

we conclude that Op = ^

, cOp = j-— , and 2 R C " " '’ - 1 W iOn RC Q = Q, = ^ = 4^ = R

c L

(16.13a)

Thus from equations 16.3 and 16.4, (16.13b) and

2a^

( 1 6 .1 3 c )

Chapter 16 • Bund-Pass Circuirs and Resonance

i '- 1

= R.

Note that this impHes that To compute the bandwidth

equation 16.8 implies that

li,„ = 2 0 p = - ^

(16.14)

Similarly from equations 16.7 and 16.9, the exact half-power frequencies are

tO\ —H---------- 1- 4 IR C V \1RC}

+

1

LC

(16.15)

while the approximate half-power frequencies for the high-(^ case are f/Ji s —p = ----------, VZC IRC

W o s —j^=H --------- .

~

ylLC

(16.16)

IR C

Finally, to obtain the frequency response we note the values o f the above computations, and the fact that at

CO

= 0 and

co

= oo the magnitude o f the frequency response is zero, and the bell-shaped

curve o f Figure 16.3b results.

Exercises. 1. Compute the exact values o f

tOp O),,

and Q for the circuit o f Example 16.1

when R - 2.5 Z, = 0.1 H, and C = 0.1 uF. AN SW ERS: (0,,, = 10"^ rad/sec, o , = 12.198 x lO 'l to. = 8.198 x 10\

= 4 x \o\ and Q = 2.S

The preceding example and exercise demonstrate that for high-Q circuits (Q > 8), there is really no need to use the exact equation 16.15 to compute cOj and co-,, as the much simpler estimates given by equation 16.16 are sufficiently close to the true answers. Indeed, sometimes it is con venient to use the approximate formulas when Q > 6. In many practical circuits, the independent source could be a voltage source in series with a resis tor. Before applying any o f the foregoing formulas, it is necessary to transform the circuit into the form o f Figure l6.3a by the use o f the Norton equivalent circuit studied in a first course on cir cuits. The resistance R in Figure l6.3a then is not a physical resistor, but rather, the equivalent resistance of several resistances in parallel. The following example illustrates this reformulation.

EXA M PLE 16.2. In the circuit o f Figure 16.4a, an independent voltage source

in series with

an internal resistance R^ = 40 kQ models a real-world sinusoidal excitation. Suppose L = 20 mH, C = 0.05 l i F, and Rj = \0 kQ. (a) Find the exact values o f co^^^, B^^, and Q. (b)

Estimate the values o f OOj and co-,.

(c)

Find I

S22


+ V

(a)

FIGURE 16.4 (a) Tuned circuit driven by a practical voltage source. (b) Equivalent circuit showing Norton equivalent for the source. So

lu t io n

The solution proceeds by using the formulas developed in Example 16.1 after replacing the prac source by its Norton equivalent and identifying R in Example 16.1 as the parallel com

tical bination o f

and /?^. By the usual formula for two parallel resistors,

R,R, 4 0 .0 0 0 x 1 0 ,0 0 0 _ R= — =— :------ = 8 ,0 0 0 Q Rs + R l 4 0 ,0 0 0 + 10,000 This results in the transfer function

—s C 'm

\

(a)

0

R,C \

1

RC

LC

s " + -------s ' - + -------------------------------------- S + —

RC Notice the new value: K =

R

LC

R,C

From equations 16.13b, 16.14, and 1 6 .13c,

1

W,„=(Op = ^

= 3 1 ,6 2 2 .7 7 rad/s

V 0 .0 2 x 0 .0 5 x 1 0 " ^ 1

Bco =

1

RC

1

8,000x0.05x10

-6

= 2,500 rad/s


823

and 6

. ^

.

3 1 .6 2 2 .7 7 ^ , , ^ Z50 0

(b)

Using the approximate equations for high-Q circuits, we obtain the half-power frequen cies as ^

= 31,622.77 - 1,250 = 30,372.77 rad/s

+^

= 31,622.77 + 1,250 = 32,872.77 rad/s

and "2

(c)

^

Now we use equation 16.4 to obtain a . 2cr^

A

.

R,CB^

,

, /?,

,

Notice that we did not use equation 1 6 .13c because the input in Example 16.1 is a current source and not the voltage source o f Figure l6.4a. Hence the maximum value is not equal to R.

This example demonstrates that putting an external resistance in parallel with the LC tank circuit reduces the value o f R, which in turn causes a larger bandwidth and thus a lower circuit Q while keeping the peak frequency

unaffected.

Exercise. Repeat Example 16.2 with the element values changed to R^ = 36 kQ, L = 40 mH, C = 0.25 pF, and R^ = 4 kH. A N SW ERS: 10,000 rad/sec, 1111.11 rad/sec, 9, 9 4 44.44 rad/sec, 10,555.55 rad/sec.

The examples so far have illustrated only the analysis o f parallel IU.C circuits. In the design of a par allel-tuned circuit, we must also pay attention to other factors, such as available component sizes, desired voltage gain, and cost. In practice, design specifications ordinarily impose a small number of constraints relative to the number o f circuit parameters to be determined. Consequendy, realistic design problems usually do not have a unique answer, as illustrated in the next example.

E X A M PLE 16.3. Design a parallel RLC circuit, as shown in Figure l6 .4 a , to have a magnitude response with^^^ = 200 kHz and a bandwidth o f 20 kHz. Only inductors in the range 1 to 5 mH are available. The source has a resistance R^ = 50 kH. So

lu t io n

For the circuit o f Figure l6 .4 a there is a restriction on the available inductors. Hence, we keep L as a variable, subject to the condition that 0.001 < L < 0.005 H. Using the specified peak fre-


.S2-»

quency and rhc fact that

2

=

2

^

LC = 6 .3 3 x 1 0 "'^ L

C = - r r = - -r From the specified bandwidth and equation 16.14,

= ±L^2li^ liC ^ =l.2 5 7

« = ^

x l0 U

2 ;r x 2 x lO '^ As explained in Example 16.2, R is the parallel combination ol

or

1 -_L

_L

R ~ Ri_

R, '

1

I

I

R[

R

R,

and R^, i.e.,

Once a specific value o f L is chosen, we can calculate successively the values of C, R, and Rj. Since R, which is the parallel combination ot R^ and R^, must be no greater than = 5 x lO'^ f2, the upper limit for L is 5 0 .0 0 0

R.

^ma\ ~

1 .2 5 7 x 1 0 '

= 0 .0 0 3 9 8 H .

1.257 x 1 0 '

Numerical values corresponding to the extreme values ot L are given in Table 16.1. TABLE 16.1

L (mH)

C (p F )

R(VQ)

Rl i ^ )

1

633

12.57

16.77

3.98

159

50

infinite

Table 16.1 clearly shows that there is no unique answer to the design problem. 'Fhe freedom in choosing a value for L in the range 1 to 3.98 mH can be utilized to accommodate another design specification, such as a value for

Exercises. 1. In Example 16.3, find the maximum and minimum values o f |//(/to^^^)|. AN SW ER: 0.251, 1.0 2. In Example 16.3, if the bandwidth requirement is changed to 10 kHz, determine the minimum and maximum possible values o f L. AN SW ER: 1.0 mH and 1.99 mH

H2S


Dual to the parallel /?/.Cof Figure 16.3a is the series y?Z.Cof Figure 16.5, which has a voltage input and a current output. Although we can use clualit)', to infer frequency response behavior, we pre fer the direct transfer function approach.

E XA M PLE 16.4. For the series RLC circuit o f Figure 16.5, let 1^ be the desired output. Find (a)

The transfer function H (>y) =

(b)

The peak frequency

(c)

The bandwidth,

,

(d)

The circuit Q

(e)

The half-power frequencies, to, and (o-, (exact and approximate values)

(0 (g)

W„, = |M;co)|„,„,= |//(/0| The frequency at which the angle o f H{jio) is zero

L

R V

FIGURK 16.5 Series RLC circuit having a band-pass transfer function. SO L U T IO N (a) Find the t r a n s f e r By inspection,

z ,„ ( .)

(16.17)

-)

1

.y“ + — .v-l- -

C ,.'

= K S~ + l O n S + Clj:

=K 5“ + —^S + W~ Q „

Thus, A-= ML,. 2a^ = «//.,, co^ =

and

'

= «/£,

(b )-(f) Inspecting the transfer function o f equation 16.17, we immediately have Peak frequency: (o„, = Wp =

C ircu it Q:

_ „

_

=Qp =

1

J

lc

(16.18)

,

^p^S

_

1

( 1 6 .1 9 )

Cs

826


Half-power bandwidth:

(16.20)

Half-power frequencies (see equation 16.7):

(0 ,2 = 0 3 ^ , iV

4Q -

2Qpj

^

2L,

where the approximation for the high-Q^ case is

-B co

C0\ ') •=' (0,„ H--------— • ,

1

i -------

^ l^s

2L,

2

/, /'

0 6 -2 2 )

From equation 16.4, _ |/f| _ \K\Q,, _ 2
(o,„

1 (16.23)

R,

Finally, from equation 16.12, zero phase-shift frequency = iOp

Exercise. For Example 16.4, suppose nitude. Find o)^^, AN SW ER:

= 5 O.,

= 1 mH,

= 0.1 |.iF, and

has a fixed mag

and the approximate half-power frequencies.

= 10’’ rad/scc.

= SOOO rad/sec, Q.-^ = 20. 97,500 rad/sec, and 102,500 rad/sec.

The above derivations and calculations yield formulas and numbers. To add some meaning to the concept o f circuit Q and its relationship to bandwidth, we provide a plot o f the normalized |//(yco)| vs. (0 curve for different values o f

in Figure 16.6. The ordinate is the ratio

|//(yco)|/|//(yo))|^^^^^, while the abscissa shows the ratio to / to T h e se curves are called universal res onance curves because they are applicable to parallel RLC circuits, to series RLC circuits, or to any system having a transfer function o f the form o f equation 16.1. Observe that as Q increases, the bandwidth decreases, indicating a better selectivit)^


82'

X

s

(O/Wm FIGURE 16.6 Normalized magnitude response for equation 16.1. To conclude this section, we present an example o f an active band-pass circuit that avoids the use o f inductors. The band-pass circuit illustrated here is only one o f more than a dozen configura tions in use. This example illustrates the possibility o f eliminating inductances while producing the same kind o f frequency response as the parallel or series RLC circuit. You can learn a lot more about these active filters in a more advanced course.

E X A M P L E 16.5. The operational amplifier in Figure 16.7 is assumed to be ideal. 1. 2.

Find the transfer function A band-pass circuit is to have peak frequency

= 1000 rad/sec and a bandwidth

=

100 rad/sec. Find Q for the desired transfer function. Then realize (find values for /?p Rj, Cp and C-y) for the normalized transfer function

H.

Ks

,{s) = S

2

+

*

—5 +

, 1

Q

3.

under the condition that Cj = C2. Magnitude- and frequency-scale the circuit to achieve the correct center frequency with

4.

q = C2 = 1 1-iF Verify the frequency response with a SPIC E simulation o f the real circuit.


82.S

MGURE 16. S o lu t io n 1.

An active band-pass circuit without inductance.

Find the circuit transfer function

=

From the assumption that the operational amplifier is ideal, we have K_ = 0 (virtual ground) and /_ = 0 (infinite input impedance). Applying KCL to the inverting input node V_, we obtain ■ 'C , K , + - ^ = o , A-7

which yields

v;. = -

Next, we apply KCL to node

to obtain

V ^'ci - V * -in /e,

+ iC | V „ + iC 2 (K „ -V „ „ ,)= ()

Substituting the previous expression for

ôut

ou t

into this equation, regrouping terms, and solving for

in the desired transfer function:

-s ~RxC^ +

(16.24)

1 ,/?2C,

R iC iV

I^\Ri C\C2

This is precisely the form o f equation 16.1.

l.Computation o f normalized transfer fimction to be realized. The desired transfer function is Ks

H{s) = V'/„(^V)

+ Ba) S + 1 0 ; „

Ks 2 i-" +

Ks S + iO

Q

}

2

.v^ + 100 .v + (1000)“


We now frequency-scale s -*

= 1000 rad/sec. Thus uhe normalized transfer function

with

is

_ = —

s

K

X ------- r r ^ --------= TTrrr x lei

?>.Computation o f element valuesfor normalized transferfiinction. Equating the circuit transfer func tion with the normalized transfer function, we have --------------- = 1 rad /sec

(16.25)

R\

and H---------- --- 0.1 rad /s e c .

(16.26)

Under the condition that C, = C, = 1 F we have, from equation 16.26,

From equation 16.25 we obtain

Ri R2C\C2

R[20

Note also that

= I H U co,„ ) I =

I

I

= ^

=^

= 200

RjC^ ^ RoSi 4. Frequency and magnitude scaling. As per the problem statement, we desire Hence we frequency-scale with Kj- = 1000. It is further required that

= 1000 rad/sec. = 1 liF. Hence,

we solve the following for K^^-. C„„„. = f f -

KfjjKj

-

K„, =

10

X 10

= 1000

It follows that c ,„ „ „ = 1

50n,

iOOOR2M- 2 0 n

The ratio o f /?, to R^ is very large and unrealistic. It turns out that this circuit is best suited to lowQ transfer functions. 5. Verif' frequency response. To verify the frequency response for a realistic implementation o f the above circuit, we consider a SPIC E simulation using the standard 740 operational amplifier as shown in Figure 16.8. Observe that the maximum value o f the magnitude response is 200, as expected, and that the bandwidth is about 16 Hz, which translates to about 100 rad/sec.


830

Exl 6.5-Small Signal AC-3 +50.000

+100.000

+150.000

+200.000

Frequency (Hz) +250.000

+300.000

FIG URE 16.8 Magnitude response plot resulting from a SPICE simulation o f the 740 operational amplifier.

Exercises. 1. In Example 16.5, find the circuit Q and the approximate half-power frequencies CO

1,2AN SW ER. 10,950 rad/sec and 1050 rad/sec 2. In Example 16.5, if Cj = 0.5 uF and C j = 1 f.iF, find the values o f /?, and Rj required to meet the specifications on AN SW ER: A’, = 66.67

and

R,= 30 kU

Throughout this section we have assumed the use o f ideal inductors and capacitors. Practical inductors have complex circuit models to account for real-world behavior. The next section takes up an approximate analysis o f circuits containing simple models o f practical inductors and capac itors.

4. BAND-PASS CIRCUITS WITH PRACTICAL COMPONENTS Quality Factor o fL and C Components How can we analyze band-pass circuits in the presence o f practical (non-ideal) inductors and capacitors? Practical inductors and capacitors have models consisting o f their ideal cousins and other ideal (“parasitic”) elements to account tor losses and coupling effects. Figure 16. 9a illustrates

8 31


a simple model o f a practical inductor for low to medium frequencies while Figure l6 .9 b is a rea sonable model for high frequencies.

FIGURE 16.9 Two models of an inductor, (a) For low to medium frequencies, (b) For high frequencies. The primary parameter is, o f course, the inductance L. The remaining elements, R^,

and C^,

account for undesirable yet unavoidable practical effects and are called parasitic. Since an inductor usually consists o f a coil o f wire, R^ represents the wires resistance. Also, a capacitance is present between adjacent turns o f wire. Hence models this parasitic capacitance. The resistance R accounts for the energy loss in the magnetic core material (if present) inside the coil. Complex mod els such as Figure 16.9b, although important, if used for every inductor would unduly complicate the analysis o f a band-pass circuit. Fortunately, for low to medium frequencies (up to a few mega hertz), the simpler model o f Figure l6.9a suffices and hence underlies the material that follows. Figure 16.10 shows two models o f a practical capacitor. Again, the primary parameter here is the capac itance C; R , Rp and

are “parasitics.” The leakage resistance, R^, accounts for the energy loss in the

dielectric; tne inductance At frequencies above \j

and resistance R^ are due mainly to the connecting wires o f the capacitor. , the capacitor actually behaves as an inductor! For frequencies of up to

a few megahertz, the simpler model o f Figure 16.10a suffices, and it is used for the analyses o f this text.

(a) FIGURE 16.10 Two models of a practical capacitor, (a) For low to medium frequencies. (b) For high frequencies.

832


How close is a pracrical inductor (Figure l6.9a) or a practical capacitor (Figure 1 6 .10a) to the ideal? The so-called element quality factor provides a quantitative measure. To develop the qualit)' factor, consider that each practical inductor or capacitor has an impedance Z(;(o) = RejZ(>))} + j Im {Z(>))} = /^(oj) + >V{(o) For both the practical inductor and capacitor, the frequenc)-dependent reactance X is the primary parameter of concern, whereas R represents the unavoidable “parasitic effect.” In the ideal case, R is not present. Hence, the ratio provides a measure of how close the network model is to an ideal inductor or capacitor. The larg er the ratio, the better the element behaves as an ideal inductor or capacitor, (ideally, R = 0 and the ratio is infinite.) This suggests defining the quality factor associated with a practical inductor or capacitor having impedance Z(yco) as ,

^

UCw)!

The inclusion o f co in equation 16.27 is to emphasize the fact that Qy depends on the frequency' o f operation. The subscript Z indicates a generic impedance and may be replaced by more specif ic descriptors such as “coil” or “capacitor.” Unlike the Q o f a circuit, which depends on the values of the elements of the circuit and on the circuits configuration, the quality factor Qj/oi a practi cal inductor or capacitor varies with the operating frequenc)' OJ and remains unchanged irrespec tive o f its connection in the circuit. Any element with a finite ratio given in equation 16.27 is termed a lossy component, as are all real-world components. For the practical inductor model o f Figure l6.9a, equation 16.27 reduces to

R^

^ where we may synonymously denote

as

(16.28)

As mentioned, higher

implies a better-qual-

it)' coil in the sense that the energ)' loss in the component is smaller. Infinite inductance L, which is lossless. In the audio frequency^ range,

represents a pure

may vary from 5 to 20, where

as in the radio frequency range, it may exceed 100 in practical applications. Although R^ here varies with frequency (due to the skin effect), at the level o f this text, we treat R^ as a constant inde pendent o f (0 . Similarly, the capacitor model o f Figure 1 6 .10a has impedance I

Z ( » = - j -----------=

Hence, the factor Q^ or

R

/

=

,

R

"

' , (l - JioK c ) » m,o)+JX(cu)

calculated in accordance with equation 16.27 is

=S =

(’6.29)

S33


A higher

implies a better-qualiry capacitor, in the sense that the energ)' loss in the device is

smaller and closer to ideal. Infinite Q^-represents an idealized, lossless capacitor modeled by a pure capacitance. In practice, Q^;;is usually much greater than Q^, meaning that

is more critical for

circuit performance, i.e., Q(;;is often assumed to be infinite. The reciprocal o f Q^^ is called the dis

sipation factor o f the capacitor and is denoted by cIq A lower dissipation factor means a betterquality capacitor. The determination o f

and

requires specification o f the operating frequency co. If the value

o f 0) is unspecified, the analysis o f a band-pass (or tuned) circuit proceeds under the assumption that

= Q^(cOq) and Q ^= QôOq), where C0q = M \[l C . We will discuss the meaning o f oJq in

section 5, on resonance.

Reductioit o f Band-Pass Circuits to Approximate Series or Parallel RLC Circuits When one uses the practical inductor and capacitor models o f Figures 16.9a and 16.10a, the series

RLC and parallel /?ZC band-pass circuits with practical sources have the more complex configura tions o f Figures 16.1 la and b, respectively. Since they are no longer series or parallel RLC, their transfer function is not the ideal band-pass type of equation 16.1. Hence the associated formulas for peak frequency, bandwidth, etc. are not directly applicable.

FIG URE 16.11 (a) Model of a parallel tuned circuit using practical inductor and capacitor models, (b) Model of a series tuned circuit using practical inductor and capacitor models.


834

Topologically speaking, the circuits o f Figure 16.11 are series-parallel because the input imped ance “seen” by the source consists o f a sequence o f series connections and parallel connections o f simple networks. Exact analysis o f such series-parallel band-pass circuits to obtain

and

is

cumbersome, especially in light o f a simpler, more efficient method widely used by engineers to compute approximate solutions. The approximate analysis relies on the conversion between a series circuit and an equivalent parallel circuit at a particular frequency. This conversion process depends on the component quality factors developed in the previous subsection. See Problem 85 for the development of this equivalence. The next example illustrates the conversion process for an inductor.

EXA M PLE 16.6. When the ideal components o f a parallel RLC circuit are modeled with a prac tical inductor and a practical capacitor, the circuit is no longer parallel. However, by converting the series inductor model to an “equivalent” parallel model, we can proceed w'ith our standard analysis. To illustrate this conversion, consider the practical inductor model o f Figure 16.12a and the “equivalent” parallel configuration in Figure 1 6 .12b. The goal o f this example is to find

L in terms o f

and

and

at a particular O).

(a)

(b)

FIG URE 16.12 Conversion o f an inductor model from (a) a series connection to (b) a parallel connection at a fixed frequency. So

lu t io n

For Figure 16.12a,

Y
yrU (o) = — - j Rp Equating the real and imaginary parts o f the above t\vo admittances, we obtain

Rp = and

ly

= Rs['‘ +Q h
(16.30a)


^

83^

co"-L,

\

q I uo)]

where

(oL

At any particular frequency (O, if Qy (to) is sufficiently large (say Qj (o)) > 8), then equations 16.30a and b suggest that

R., a

{(o)Rs = Q i( co ) x( oL

(16.31a)

^p-^s

(16.31b)

and

Conclusion: In a "paraller RLC with a practical inductor, we can replace the practical inductor by its parallel counterpart valid in a neighborhood o f a single frequency and analyze the circuit to obtain approximate values o f peak frequency, bandiuidth, etc.

Exercise. A 2 mH coil purchased from an electronic parts store has a

50 at 100 kHz. Find

the element values in the series representation and the parallel representation at 100 kHz as shown in Figure 16,35. A N SW ERS: l.^ = 2 mH, R^ = 25.13 LX

s 2 mH, R^ = 62.83 \dl

A similar derivation can be done for the practical capacitor model o f Figure 16.13a. The details of the derivation are left as a homework problem. The exact conversion equations for a specific to are

(,

\

\

and

(

1

\ (16.32b)

As before, for high

(say >8), these equations reduce to the simpler forms o f (16.33a)

and 1 Q c (c o )

^

(1 6 .3 3 b )


(a) (b) F-KiriU-. 16.13 Conversion of a capacitor model from (a) a parallel connection to (b) a series connection exact at a fixed fi-equency. Table 16.2 summarizes the various conversion formulas for both parallel to series and series to par allel inductor and capacitor models. I'AMI.F 16.2 Conversion of Models for Practical Inductors and Capacitors Approximate Original Circuit

Exact Equivalent Circuit at (d,,

Equivalent Circuit, for High

{Q, > 8

Q

> 8)

and

and u) within (1 ± 0 .0 5 ) oJq

Q^R. =Q,.o)oL

R (i+ Q l)

Hf-

Q,((i)^ = o)„R,C

Q;

CL

1 + 1/Q' 1

Qc(®o) = (o„R,C

Q,

S '- ‘


The next example illustrates the application o f the conversion formulas to a “practical” circuit to obtain approximate band-pass characteristics.

EXA M PLE 16.7. The circuit o f Figure 16.14a contains practical components that make the cir cuit not amenable to the formulas for ideal series and parallel RLCs developed earlier. For exam ple, the sinusoidal source is represented by an independent voltage source internal resistance tion factor o f

= 40

in series with an

The practical capacitor has a value o f C = 0.05 |.iF and a dissipa

0.01 at cOq. Here cOq represents the undamped (no resistance) natural frequen

cy o f the circuit, i.e.,

^ “'» = 7 Z c -

The practical coil has an inductance o f 20 mH and Q/(cOq) = 40. The external load resistance is /?^ = 10 k n . Our goal is to find approximate values o f

cOj, CO2, and

V

(a)

leakage

signal source (b)

FIG URE 16.14 Approximate analysis of a high-Q circuit, (a) Original circuit, (b) Approximate parallel /?Z,C circuit. Solution Step 1. Compute cOq.

= 31,623 rad /s

COq =

V 0 .0 2 x 0 .0 5 x 10"^

83R


Step 2. Find the parallel equivalent circuit values for the practical inductor. Since Q^(cOq) = 40, from equations 16.31 and column 3 o f Table 16.1,

Rp = Q/(Wo)WoZ. = 40 X 3 1 ,623 x 0.02 = 25,298 Q. and

L r-L Step 3. Represent the capacitor model by a parallel RC. First, the dissipation fiictor tells us that

Qc ^^q) - ~T~ - 100 dc From row 3 and column 4 of Table 16.1, we note that the parallel resistance associated with the capacitor is the so-called leakage resistance

,

^ ^ ‘

^^

clcCo^C

^

^ ------------ ^ 0 ---------- ^ ^

o jj:

^

3 1 6 2 3 x 0 .0 5 x 1 0 " ^ ’

Step 4. Replace the practical source with its Norton eqidvalent and compute the eqtdvalent parallel

resistance, denoted as R. Replacing the practical source with its Norton equivalent and incorporat ing the results o f steps 2 and 3 produces the network o f Figure 1 6 .14b. The parallel combination o f R^, R^, ^leakage* and R^ is (using the notation “//” to indicate the parallel combination)

R = 40,000//25,298//63,247//10,000 = 5545 n Step 5. Approximate analysis o f the circuit o f Figure 16.14a using Figure 16.14b. Our formulas from the ideal parallel RLC case now apply, but the results are, of course, approximate for the circuit o f Figure 16.14a:

0)^^^ = 1

= —

RC

co^ = 3 1 ,6 2 3 rad/sec

=— — — ------- = 3 ,6 0 7 rad /s , 5 , 5 4 5 x 0 . 0 5 x 1 0 “^

3,607 and since Q^.-^ is sufficiently large. - - ^ = 3 1 ,6 2 3 - 1 ,8 0 3 .5 = 2 9 ,8 1 9 rad/s,

0 )2

s

+ - ^ = 3 1 ,6 2 3 + 1,803.5 = 3 3 ,4 2 6 rad / s

Finally,

^max =

---- = ------------ 5------------------- Z----------- = 0 .1 3 8 6 2 4 0 X 10^ X 0.05 x 10"^ x 3607

R,CB,,


839

5. THE RESONANCE PHENOMENON AND RESONANT CIRCUITS The term “resonance” has different meanings in different disciplines. From Webster's Collegiate

Dictionary, in the field o f engineering, resonance refers to the phenomenon o f “a vibration o f large amplitude in a mechanical or electrical system caused by a relatively small periodic stimulus o f the same or nearly the same period as the natural vibration period o f the system.” In this section, we shall investigate this notion o f resonance and its manifestation in RLC circuits, in both cases uti lizing the theories studied in previous chapters. The main applications o f resonant circuits are for the filtering and tuning purposes. Additionally, resonant circuits can be used to transform a resist ance from one value to another value (at a single frequency) to achieve maximum power transfer. This “matching” application is discussed in the last subsection.

The Resonance Phenomenon A child on a swing knows by instinct when to flex his or her knees in synchrony with the swings pendulum motion to make the swing go higher and higher. This activity illustrates the phenom enon o f resonance, in which small and quick leg movements at Just the right moment produce the large pendulum-like motion o f the swing. If the child stops his or her leg movements, the swing ing motion gradually dies down. Let us approximate the pendulum motion o f the swing using a linear system with transfer function +

(16.34a)

h it)= K e~ ''’'' s\nico,it)ii{i)

(16.34b)

and having the associated impulse response

Mathematically, the periodic leg movements are modeled by the periodic impulse function x(r) = /l [ 6 ( t )

+d{t-T)-^6{t-n)^

...]

(1 6 .35 )

where T = 2 7 t/ o )T h is choice coincides with the period o f the natural response o f the system transfer function. The build-up in magnitude can be seen quickly through superposition. The contribution to the output, sayj/(r), due to the first Impulse at / = 0 in equation 16.35 Is

\\){t)=ÂKe

sln(f/j^//)//(/)

(16 .3 6)

Figure 1 6 .15a shows the form o f this equation. Note that successive positive peaks {Kq, aV^, a^V^, a^Vf^, ...} decrease geometrically by a = e~^P^. In fact, the waveform o f each period replicates the waveform o f the prior period scaled by the factor a. From time invariance, the contribution to the output ^(/) due to the second impulse at r = 7'in equation 16.35 is simply J'q(/‘ - 7), as shown in Figure 16.1 5b. Similarly, for the third input impulse zi t =1T, the response 1s7q(^- 2T), as shown In Figure 1 6 .15c.

8-10


From superposition, the response over [0, 3T) \s simply the sum

i.e., for 0 < r < 37'. This sum is illustrated in Figure I6.15cl.

a^)

F IG U R E 1 6 .1 5

Chapter 16 • Rand-Pass Circuits and Resonance

In Figure l6.15cl \vc observe thar successive positive peaks arc given by

(1 +

(1 + a +

In general one can show that the «th positive peak is given by

n

= 1

rt-1 11 + ^/+ t/“ H------ ~

^

(16.37)

“ l-« "

k=0

1

For our purposes n =

< 1 since

-

a

> 0. Hence as n gets large,

\-a

1- «

This means that the waveform reaches a steady-state periodic response in which the positive peaks have value

v \-a

Exercises. 1. Consider the slightly damped second-order system with transfer function

H(s) = ------ ----(s + 0.1)" + 4 having a periodic input

x{t) = 2[6(f) + 6 ( r - . t ) + 6 { r - 2;r) + — ] (a) Find the first five positive peaks. (b) Find the positive peak in steady state. ANSWHRS; (a) 3.698, 7.116, 10.28, 13.2, 15.9: (b) 48 .9 6 2. Suppose in the previous exercise

is changed from 0.1 to zero, i.e.,

5“ +

4

(a) Find the first five peaks for the input o f the previous exercise. (b) Is the system stable? Why or why not? ANSWHRS; (a) 4, 8, 12, 16, 20. (b) The system is unstable bccausc there is a pole on the imagi nary axis of the complex plane.

Another interesting application o f resonance is in product security in stores. The securit)' tag is an

RLC circuit with very small R. The circuit is excited by a “radio” wave at its resonant frequenc}' by the security panels in front o f the exit doors. If the circuit has not been destroyed at the checkout counter, it begins to resonate as one approaches the exit and transmits a signal back to a detecting device at an amplitude much higher than the original transmitted signal. Fhis sets ofi an alarm. The easiest way to understand this phenomenon is by way o f frequency response. Let us consider the second-order transfer function

^

H(s) = --------- ---------(5 -h 0 .1 )-+ 4

84 2


The magnitude frequency response is given in Figure 16.16. Notice the sharp peak at about 2 rad/sec, which is 10 times the dc gain o f about 1. Thus sinusoidal inputs at frequencies close to 2 rad/sec produce a steady-state response with magnitude almost 10 times larger. This is precisely the type o f resonance phenomenon that occurs with the security tags at stores.

Frequency in rad/s F IG U R E 16.16

Series and Parallel Resojiant Circuits The simplest stable linear circuits capable o f producing the resonance phenomenon are the series and parallel resonant circuits shown in Figures 16.17a and b, respectively.

/Y Y ^ L

R V

(a)

:D

L >R ^

c

(b) F I G U R E 1 6 .1 7 Series (a) and parallel (b) reson an t circu its.


843

Using the frequency response approach o f Figure 16.16, we see that small inputs in these circuits can produce large outputs when the conditions are right: (i) a high-Q circuit and (ii) input signal fre quency and the peak frequency coincide. To develop a circuit-theoretic perspective on resonance we take a more basic approach using the sinusoidal steady-state analysis ideas o f Chapter 10. For any circuit containing one inductance L and one capacitance C, denote by (Oq the frequency at which the two reactances

= -1/oJqC and

= (OqZ have equal magnitudes, i.e.,

(16.38) in which case 0)^ =

J

(16.39)

lc

In Chapter 9, on second-order RLC circuits, cOq = I I JT C is called the undamped natural fre quency. The name stems from the fact that if all resistive elements are absent (i.e., the circuit is undamped), then a parallel or series connection o f L and C produces a natural response o f the form K cos((Oq^ + 0). In the jargon, the parallel LC circuit o f Figure 1 6 .17b is called a tank cir

cuit, and 0)q is called the tank frequency. Since

and jXj^ =

have opposite signs and equal magnitudes, if the elements L

and C are connected in series, then at (Oq the resulting impedance,

=

0 and hence is equivalent to a short circuit. Similarly, if the elements L and C are connected in parallel, then at (Oq the resulting admittance is zero and hence equivalent to an open circuit. These properties are illustrated in Figure 16.18.

OZ(jO)o)

Z(jCOo)

a

O(a)

o-

O-

Y(jcOo)

Y(jcOo)

o-

O(b)

FIG URE 16.18 (a) Series- and (b) parallel-connected ZCelements with coq =

>/Zc ■

With the short circuit and open circuit equivalents sho\vn in Figure 16.18, we can easily deduce die fol lowing properties for the series and parallel resonant circuits of Figure 16.17 operating a t : coq =

■.

8 ''il

Chapter 16 * Band-Pass Circuits and Resonance*

(1)

The magnitude o f the impedance seen by the voltage source in the series RLC o f Figure 16.17a is minimum, and the impedance is a pure resistance equal to R.

(2)

The magnitude o f the impedance seen by the current source in the parallel RLC o f Figure

(3)

The magnitude o f the voltage across C o r I in the series /?ZCof Figure 16.17a is Q^./;.(Wq)

1 6 .17b is maximum, and the impedance is a pure resistance equal to R. = oJqZ.//? times the magnitude o f the source voltage. (4)

The magnitude o f the current through Z, or C in the parallel RLC o f Figure 1 6 .17b is O^.y^(ojQ) = cOq RC times the magnitude o f the source current.

To establish property' (3), we use voltage division and propert)’ (1):

> ()C

M )L

COqL

R

R

R

^ rO o )

A similar derivation yields property' (4).

Exercise. Derive property (4) using current division and property (2).

Properties (1) and (2) lead to a general definition o f resonant frequency, denoted to^.. Specifically is the frequenq' at which the source sees an impedance or an admittance that is purely real, i.e., pure ly resistive, despite the presence o f capacitors and inductors. For the series and parallel RLC circuits • For more general circuits,

oJ q.

Circuits operating at ojâre said to be at resonance.

Exercise. A sinusoidal voltage source at 1 MHz is applied to a series RLC circuit. If Z. = 300 uH, /? = 5 n , and C is adjustable, what value o f C produces resonance? A N SW ER: 84.4

p 1-

Using property (2) above, the calculation o f

in many o f the earlier examples can be done with

very little effort. For example, in Example 16.2, at resonance.

RL

10'

R, + Ri^

4 x 1 0 -^ + 10-^

Exercise. Use propert}' (2) to calculate A N S W l - K : 1//^

=

0.2

in Example 16.4 for Figure 16.5.

*

The resonance condition co^ = (O^ = 1/V LC can be achieved by vary'ing one o f the three parameters oJq, C, or L. When C or Z, or both arc adjusted to achieve resonance, the circuit is often called a tuned circuit. The next example illustrates how this tuning can be used in a practical application.

84^


EXA M PLE 16.8. Figure 16.19 displays an amplifier model containing a V CC S with g^ = 2 mS (milli-siemens) and

= 20 k fi. The applied sinusoidal voltage,

V at 10 MHz. The load is modeled by the parallel combination o f

has a magnitude o f 0.1 and the 40 pF capacitor;

the capacitance accounts for such real-world phenomena as wiring capacitance, the device input capacitance, and other embedded capacitances. This capacitance cannot be removed from the cir cuit and often has deleterious effects on the amplifier performance.

FIGURE 16.19 Amplifier circuit illustrating the application of the tuned circuit conccpt to elimi nate undesirable capacitive effccts. The example objectives are (a)

With the load connected directly as shown (without I ) , find the magnitude o f the out put voltage.

(b)

If an inductance L is connected across the load to tune out the effect o f the capacitance, find the value o f L and the resulting

that will show that the amplifier gain at 10

M Hz is greatly increased. So

lu t io n

(a)

At 10 MHz, the load impedance is

1

3 p r = 3 9 7 . 8 Z - 8 8 .9 ° Q .

0 .00005 + y 2 jrx 10 X 4 0 x 1 0 Therefore, since jV'il =

|= 0.1 V, the magnitude o f the output voltage is I

^ 0-002 X 3 9 7 .8 = 0.0 7 9 6

Here the voltage gain is 0.0796/0.1 = 0.796 due to the low impedance o f C at the high operating frequency.

(b)

By tuning out the effect o f the capacitance, this poor gain response can be eliminated. The inductance needed to tune out the capacitance is calculatcd from equation 16.39:

L = A - = = — 7------- n-^------------ T T T = 6.33 X 10"^’ H ojqC

4 ji”

x

10

X 4 0 x 10

W ith a 6.33 uH inductor connected across the load, the parallel Z.C behaves like an open circuit at 10 MHz and the load looks like a pure resistance o f 20 kQ to the amplifier. The new output voltage magnitude is

846


Iôut

1^ 0-002 X 2 0 ,0 0 0 = 4 V

wich a resultant voltage gain o f 4/0.1 = 40.

Exercises. I. Suppose the frequency o f the input is changed from 10 M Hz to 5 M Hz in Example 16.8. Find (a) the gain without the inductor connected, (b) the value o f the inductor to tune out the capacitance, and (c) the resulting gain with the capacitance tuned out. ANSWHRS: (a) 3.18, (h) 2S.33 ul 1. (c) -40 2. In Example 16.8, if the embedded capacitance is 63.3 pF instead o f 40 pF, what inductance is needed to tune out the capacitive effects? AN SW ER: 4 IIH

Series-ParalelResonantCircuits For parallel and series RLC circuits, co^ = cOq. For series-parallel circuits containing only one L and one C when it exists to^ ^ (jd^j in general, but it may not exist at all. The next example illustrates the point.

E X A M P L E 16.9. Find the resonant frequency,

and the input impedance at co^for the circuit

shown in Figure 16.20.

/YYY L

Z(jo))

R

FIG URE 16.20 A variation of a parallel resonant circuit.

SO L U T IO N Step 1. Calculate the admittance "looking into ” the input node pair o f the circuit o f Figure 16.20. By the usual techniques. (16.40)

K(/oj^) = ----- !----- = M C + ------- !-------= Z O ,) R + ji^ ,L

y ? ^ + (a ),L r

Step 2. Set the imaginary part to zero. Resonance occurs when Kis real, i.e., when

\m{Y}= o ) ,C ----- ^ = 0 . R~ -t- (oifL) Solving for

and then expressing it as a function o f Wq = \JyjLC yields

0).

=

1

R-

1-

CR-

(16.41)

84‘


The rightmost term shows how the resonant frequenc)'’ is scaled away from the parallel or series ideal cases where to^ = cOq. Step 3. To obtain the values o f the admittance and impedance at resonance, substitute this value o f (Of into equation 16.40 to obtain

Yij0^r) =

RC RC

(16.42)

Three conclusions can be drawn from equations 16.40 through 16.42: 1.

If {CR^)IL > 1, then there is no real solution for (0^ this means that the source voltage and source current cannot be in phase at any frequency.

2.

If {CR~)IL < 1, then there is a unique nonzero resonant frequency O)^ that is strictly small er than Wq.

3.

When O)^ exists, i.e., there is a real solution to equation 16.41, then at (O = CO^the source “sees” a pure resistance, the value o f which equals {LIRQ and is greater than R.

Exercises. 1. Fill in the details o f step 3 above. 2. In Figure 16.20, let Z. = 1 H and C = 1 F. Find the resonant frequency co,. for (a) R = 0.8 Q, and (b) 7? = 2

Compute the resistance seen by the “source” in each case.

AN SW ERS: (a) 0.6 rad/scc, 1.23 H; (b) (O^. docs not exist.

Results similar to equations 16.40 through 16.42 can be derived for the circuit o f Figure 16.21.

/YY\ Z(jco)

L R

FIGURE 16.21 A variation o f a series resonant circuit.

Exercises. 1. For the circuit o f Figure 16.21, let G = MR. Show that the input impedance is ./ G " + ( ojC )

coC

coZ-- —^------7 G~ + (coC)“

2. Again for Figure 16.21, show that the resonant frequency is 0),. = to,

1 - ^

(16.43)

CR'

3. Finally, show that the input impedance at the resonant frequency is

Z(jo),) =

L RC

( 1 6 .4 4 )

848

Chapter 16 • Band-I’ass Circuits and Rcsonancc

In the above example and exercise, vve concluded that When co-exists, i.e., there is a real solution to equation 16.41, then at to =co^ the source “sees” a pure resistance, the value o f which equals {URQ. This property finds application in maximum power transfer from source to load in fixed-frequency situations, as illustrated by the next example.

E XA M PLE 16.10. The output stage o f a certain radio transmitter is represented by a 1 MHz sinu soidal voltage source having a fixed magnitude o f 50

and an internal resistance o f 100 Q as

shown in Figure 16.22a. A load resistance Rj models an antenna connected to the transmitter also shown in the figure. The purpose of this example is to shov/ how a m atching network based on the principle o f resonance can be designed to maximize the power delivered to the antenna. (a)

If Rj is adjustable, find the value o f

(b)

by the load. What is the value o f (/*jr)niax^ If Rj = 2Q Q. in Figure 16.22a, find the value o f P^.

(c)

Suppose that Rj^ is fixed at 20 O., but a coupling net%vork consisting o f LC elements is

yielding the maximum average power

absorbed

inserted between the source and load to increase the power P^ as shown in Figure 16.22b. Choose values for L and C in the circuit so that {Pi),j,^j^ o f part (a) is again obtained.

1000

lOOO

-

0

K

/YYV L

50 V

20 0 1MHz

IM Hz

(b)

(a)

FIG URE 16.22 Matching load to source using a resonant circuit, (a) Load connected direcdy to source, (b) A coupling network designed to maximize the load power. So

lu t io n

(a)

From the maximum power transfer theorem,

Ri = R^ = 100 n and 2,500

AR, (b)

400

= 6.25 W.

With R^ = 20 Q., by voltage division

20 in which case

100 + 20

50 = 8.33 V,

3 .4 7 2 w . 20


(c)

S h9

If we can make the impedance at the input terminals o f the LC coupling network equal to (100 + yO) LI, then maximum power will be drawn from the source. Since Z.Celements consume zero average power, the same maximum power will be delivered to the load resistance. The resonant circuit shown in Figure 16.22b provides a possible design. The

“LCR" load circuit is the one analyzed in Example 16.9. Hence, to calculate the element values, we use equations 16.41 and 16.42 as follows: 7 6\“ 1 20“ (oT = 2 ji X 10 j -----------(from equation 16.41, squared)

LC

and

Z{jiD^)= 100 =

(from equation 16.42) 20 C

Solving these equations simultaneously results in Z = 6.37 uH and C = 3.18 nF. Some remarks about this design are in order: 1. fg= \I{2ii J T c ) = 1.12 MHz, and fj. = 1 M Hz i f j . The 20 D. resistance is transformed into a 100 Q resistance at/ = fj. (not at/ = J^). 2.

Should the source resistance be smaller than the fixed load resistance, C is moved to be in parallel with

In that case, equations 16.43 and 16.44 are used in place o f equations

16.41 and 16.42.

Exercise. Redesign the coupling network in Example 16.10 if the resistors are

= 300 Q and Rj

= 50 n . AN SW ERS: 17.79 ul 1 .nui 1.18 nl-

A variation on the computation o f the resonant frequency is the design o f a circuit to achieve a desired resonant frequency using a variable capacitor; this is the design that underlies the tuning o f many AM radios.

E X A M PLE 16.11. Consider the series RLC o f Figure 16.23. Here, the voltage source has a fixed magnitude |V^| and a fixedfrequency O). W ith R and L fixed, we seek the value o f the variable capac itance C that maximizes the magnitude o f the voltage across the capacitor.

F I G U R E 1 6 .2 3 A d ju stin g

C fo r

m a x im u m o u tp u t voltage.


850

The first step is to compute the magnitude o f the voltage across the capacitor using vohage divi sion: I, , I

^

Vc o

V

/

R -+ U)L-

1 coC/

\/ Maximizing |K j is equivalent to maximizing \V(^ or minimizing------- . To obtain this last expres sion, we square the above expression for \V^ and rewrite it as

/ (C ) = ------ ^ = (coCRr + (CO^LC - 1)l^ c l" To m in im izey(0, s e t / '( Q to zero, i.e., / '( O

=

+ 2 ( lo ^ L C - \ W L

= 0

Solving for C produces

C=

1

R~ +(coL)

Zxo‘

+1

Although this value o f C produces a maximum capacitor voltage, the circuit is not hi resonance., as the value o f 1/VLC is not equal to the signal source frequency o). However, given this equation for C, for a high-Q circuit (ojZ//? > 8), the condition is practically the same as l/VTC = to.

E X A M PLE 16.12, In Figure 16.23, let Find I

= 1 V, (O = 10^ rad/sec, ^ = 5 H. Let C b e variable.

and the corresponding value o f C for each o f the following cases: (a) Z, = 1 mH; (b)

L = 100 uH. SO L U T IO N (a)

For L = 1 mH,

= toZ,//? = 10^ x 10"^/5 = 20. This is a high-Q circuit. Therefore, C is

given approximately by

C = - \ - = — ^ ------ r = ( ) . l x l O ”^’ F , to -L

and I Vy

lO '^ x lO " '

is given approxmiately by l^ c U « = Q Jn i = 2ov

Exact solutions follow from equations 16.48 and 16.47:

0 .00 !

an d

x-6 . ^ = 0 .0 9 9 7 5 x 1 0 ”" F , 5^ +(10^ X 0.001 r


n J5 “ +

7 T 1 0 ^ x 0 .0 0 1 -

8S

^

^

10-'’ X 0 .0 9 9 7 5 X 10

10^ X 0 .0 9 9 7 5 X 10"^ J Plainly, the approximate solutions are very close to the exact solutions. (b)

For £ = 100 uH,

= coZ//? = 10^ x 10“^/5 = 2. This is a low-Q circuit, requiring the

use o f equations 16.48 and 16.47. Here, 0-0001 C=— 5 5 2 + ( 1 0 ^ x 0 .0 0 0 ir

,^ _ 6 ^ 0 .8 x 1 0 F,

and = 2 .2 3 6 V . 2 5^+

lO-'^ X 0 . 8 x 1 0 “^

iO-'^x 0 .0 0 0 1 1 0 ^ x 0 .8 x 1 0 " ^ /

Exercise. For part (b) o f Example 16.12, compute cOq = 1/ V lC . Is this value equal to the signal frequency? Why or why not? AN SW I:R. {.)(, = 1 .1 1 8 x 1 0 ^ rad/scc

6. GENERAL STRUCTURE OF THE BAND-PASS TRANSFER FUNCTION WITH ONE PAIR OF COMPLEX POLES W ith the experience gained from the analysis o f the first five sections, we now present a general transfer function approach to the analysis o f band-pass type circuits. The circuits considered in this section will contain only one inductance and one capacitance. Hence any associated network function will have at most a second-degree polynomial in s as the denominator and numerator. Thus, the general transfer fiinction H{s) (which includes the impedance function Z{s) and the admittance fiinction F(j) as special cases) has the biquadratic form

u( A -

+

d(s)

s~ + 2 o pS + io~p

(16.45)

A reasonably sharp band-pass characteristic requires that H{s) have complex poles, i.e.,

<

O)^.

The finite zeros o f H{s), which are roots o f n{s) = 0, may be real or complex. The case o f complex zeros, corresponding to more advanced filter characteristics such as the inverse Chebyshev or ellip tic types, is beyond the scope o f this text. For practical reasons we focus on the case where H{s) has one real zero or no zero. When H{s) has one real zero and complex poles, then equation 16.45 reduces to

+ _ s- + 2 o ^ s + (o l

a\S + ao +

e„

"

Chapter 16 • Band-Pass Circuics and Resonance

SS2

with the pole-zero plot shown in Figure 16.24. jco

A p, = -^ p + i« d \“p s-plane

(0p2 = ap2 + C02

-a

(0

-e -

-a

Q =

P2 = -^ p -j« d

-

FIG UIIE 16.24 Pole-zero plot of the transfer function H{s) of equation 16.46. Equation 16.46 contains the usual qualit)' factor, Q , mathematically called the pole

defined as

CO,

(16.47a)

= arcsin

(16.47b)

\^P/ where the angle 0 is as shown in Figure 16.24. Relative to the pole-zero plot,

measures how

close the pole is to the/o-axis: a higher Q means a smaller 0, implying a pole closer to they'coaxis. As we already know, is related to the circuit Q, i.e., the sharpness o f the response curve. For some special cases, In Figure 16.24, another new quantity,

and serves as a quick estimate o f

appears. To ascertain the meaning o f to^y, recall from

Chapter 15 that the impulse response o f a system characterized by equation 16.46 has the form hit) =

r^

{/ / (.V )} =

Ke'""'’" c o s ( w , y

+

d)

The waveform h{t) is a damped sinusoid, and the quantit)^ (not co^ specifies the frequency o f oscillation. For this reason, co^ is referred to as the damped oscillation frequency. For a transfer function o f the form o f equation 16.46, our goal is to determine several key quan tities that are indicative o f the circuits behavior: frequencies.

the bandwidth

and the half-power


Case 1. No finite zeros. When the transfer function o f equation 16.46 has no finite zero, then it reduces to

K

H{s) =

K (16.48)

.V" +

a

The pole-zero plot o f this H(s) differs from that o f Figure 16.24 only in that the single zero is now absent. The series RLC circuit o f Figure 16.25a with the capacitor voltage as the output and the par allel RLC o f Figure 16.25b with the inductor current as the output both fall into this category.

fYYVV.

(b)

(a)

FIGURE 16.25 (a) Scries RLC with capacitor voltage as output. (b) Parallel RLC with inductor currcnt as output. The transfer function o f equation 16.48 can represent either a low-pass or a band-pass character istic depending on the value o f

and the value o f H{0). In fact, the transfer function can display

both characteristics as well as intermediate behaviors. For example, if o ,,,

_________OOOO)-

1

■V“ +

1000

VI

then the maximally flat low-pass characteristic o f Figure 16.26a results. On the other hand, if (1000)^

H{5) =

.v“ +

1.31

5 + (1 0 0 0 )-

then something between a low-pass and a band-pass characteristic results, as shown in Figure 16.26b. Here low frequencies are still passed, yet the characteristic has a selectivit)' propert}^ result ing from the pole

o f 1.31. The ratio o f the maximum gain to dc gain is yfl . This means that

the peak is 3 dB above the dc gain. Now, with increasing pole Q the selectivity goes up, as does the maximum gain, and the characteristic looks more and more like a pure band-pass. Finally, if H{s) =

.

(1000)*-

1000

.v“ + -------- s

10

(1000)*

then the approximate band-pass characteristic o f Figure 16.26c results. Here the pole Q is 10 and the ratio o f the maximum gain to dc gain is also 10. Although low frequencies are not attenuat-

854


ed, the characteristic is highly selective and frequencies near co^ are highly amplified, so that for all practical purposes the characteristic is identified as band-pass.

Frequency in rad/sec (a)

(b)

Frequency in rad/sec (c) FIGURK 16.26 (a) Low-pass characteristic, (b) Moderate Q^, resulting in characteristic exhibiting both low-pass and band-pass behavior, (c) High-Q^ case, showing a marked band-pass characteristic.


8^5

The important aspects of the curves in Figure 16.26 are co^, and then

and

We first compute O)^

At the peak value, the derivative o f |//(/o))| is zero. Setting the derivative of |//(/to)|

to zero and solving for O) produces the exact formula ^

< B „= 0)p W

l - ^

=

‘^ P ^ co s(2 0 )

1

To compute

consider

'm

2

2

^(Op - ct)^ co s(2 0 )j + (U p 4sin ^ (0)co s(20) (o „ = o )p y Jco s(2 d )

^

ct)p4sin (0 )c o s (0 )

(Op sin (26)

Hence

K ____ ^/»=

2 .

in(2d)

H (0) sin(20

0 6 .5 0 )

Similar derivations with more complex algebra yield the half-power frequencies, <^1,2 = (Opyl cos(2d) T sin(20)

(16.51)

^(o = « p ^ 2 (c o s ( 2 0 ) - 7 c o s ( 4 0 ) )

(16.52)

and the bandwidth,

provided there exist real solutions. 's-/ Referring back to equation 16.49, if Q® < l/V 2 = 0 .7 0 7 or, equivalently, if 0 > 4 5 °, then there is no real solution for (o^. In this case, almough the poles are complex, the magnitude response dis plays no peak at any firequency. Rather,

= |//(/))| and |//(/o))| decreases monotonically to zero

as (0 increases, as demonstrated in Figure 16.26a. When W

is greater than l/V 2 or, equivalendy, if 0 < 4 5 °, the magnitude response of equation

16.48 starts fi’om a nonzero value at co = 0, rises to the peak value at co = O)^, and finally decreas es to zero as CO -► 00. This behavior is illustrated in Figures 16.26b and c. If

is only slighdy

greater than 0.707, then the magnitude response is essentially that of the low-pass type, with a small hump in the pass band. For the high-(^ case ( ( ^ > 8), the magnitude response near co^ approximates that of a pure band pass circuit. The preceding exact expression for (o^ reduces to O)^ a o)^ for the high-(^ case.


856

Similarly, for high Q^,

t^\Qp

s 2a^,

. and ojj 2 -

I'"*

error o f these

approximations is less than 0 .5 % .

Exercise. For

= 8, (o^ = 1000 rad/sec, and A^= in equation 16.48, computc the exact and approx

imate values o f

and

Then compute the percentage error in the approximation:

\exact - approximate

X 100

\exacf\ AN SW liRS: Exact values arc Approximate values are

= 99 6 .0 9 rad/sec, B^^^ - 125.5 rad/sec, and

1000 rad/sec, B^^^ = 125 rad/sec. and

- 8.015^. = 8. Percentage errors are,

respectively, 0 .3 9 ‘^(), O-.V)'*/!), and 0.196% .

EXA M PLE 16.13. Consider the Sallen and K eyzcw e network o f Figure 16.27, which can be used to realize the transfer function o f equation 16.48. As per a homework exercise in Chapter 14, the transfer function is

1 H(,v) =

Ri R2C[C2

' nut

1

U ,c ,

R2 C J

R\R2<îC2

Suppose Cj = 2 uF, C , = 5 nF, /?j = /?2 = 10 kQ. and K o f equation 16.48.

(a)

Compute co^,

(b)

Compute exact values o f

(c)

Compute approximate values o f

(d) (e)

Plot the magnitude response curve o f the transfer function. Simulate the circuit using SPICE.

and 0)j, and co,.

FIGURE 16.27 Sallen and Key active net\vork for realizing the transfer function o f equation 16.48.


So

8^'

lu t io n

For the given parameter values, _ V„„, _

(1 0 0 0 )2

'm

+

(a)

By inspection (o^ = 1000 rad/sec,

(b).

From equation 16.48,

I --------V =

(JO,,, = o)p

11 Numerically evaluating N(s) at s =

5 + (1 0 0 0 )“ 10 = 10, and K = = 10^’.

9 9 7 .5 rad/.sec.

~^P

we obtain

= 10.013 via the following MATLAB code:

»n = wp^2; »d = [1 w p / Q p w p ^ 2 ]; »Hm = abs(pol\âl(n,j*wm))/abs(pol)âl(d,j'wm)) Hm = 1.0013e+01 (c)

The approximate values are given by s

2 a ^

=

1 0 0 , 0 ),

a

co ^ -

= 9 0 0 , CO

2^

=

1 1 0 0 ,

all in rad/sec. (d)

'lb obtain the magnitude response for the given transfer function, as shown in Figure 16.28, we use the following MATLAB code:

»n = wp^2; »d = [1 w p / Q p w p '^ 2]; »f = logspace(l ,3,600); »w = 2*pi*f; »h = freqs(n,d,w); »semilogx(f,abs(h)) »grid »xlabel(‘Frequency in Hz’) »ylabel(‘Magnitude H(jw)’)


858

Frequency in Hz FIG URE 16.28 Magnitude response o f high-Q^ active circuit of Figure 16.27. (e) A SPIC E simulation yields the corresponding plot in Figure 16.29.

Sallen and Key-Small Signal AC-2 (V)

+10.000 -H 5.849

Frequency(Hz)

+25A 19+39.811 + 6 3 .0 9 6 + 10 0.00 0+ 1 58 .48 9+ 2 51 .1 89 + 3 98 .107+ 630.957

F I G U R E 1 6 .2 9 SPIC E sim u la tio n o f activ e circu it o f F igu re 1 6 .2 7 .

8S9


Exercise. Suppose Cj = 0.2 j.iF, Cj = 0.5 nF, and /?j = ^2 ~ tion 16.48, along with exact values o f A N SW ERS:

A"of equa Find approximate values o f B^, Wp and Wj.

and

= 10, K = I0 ‘\

= 10^ rad/sec,

= 9975 rad/sec. /-/,,, = 10.013, /?,„ =

10-^

rad/sec, 10, 2 9000 rad/scc, to, = 1 1,000 rad/scc

Case 2. A single zero o jf the origin. In equation 16.46 with a^^

0,

i=- 0, and

and <7, o f the

same sign, a zero is present in the left half-plane. A second form o f the band-pass transfer func tion o f equation 16.16 is

s +a

H{s) = K

=K

CO,

s +• Qn

s+ a s~ + 2 0 p.v + oj“

(16.53)

Figure 16.32 sketches the pole-zero plot for this transfer function. Intuitively speaking, the closer the zero is to the origin, the more the magnitude response resembles the response o f the case with a single zero at the origin. For the transfer function o f equation 16.53, derivations o f of

values

and the zero-phase-shift frequency are possible. (See the homework problems.) The results

are + J(o )l + a -y -(2 o^ ay ,

to.., =

CO (for zero phase shift) =

(16.54)

- 2o^,a

(16.55)

No exact expressions are available for the half-power frequencies and the bandwidth case o f high

and a «

For the

approximate answers are 1^1 ,a n d f i ^ ^ 2 a ^ , 2o,

for

>8

(16.56)

To see why, observe that jc o +

=

A.

a

9 (

»

“

+j2oxj^^

+

jo3 + a

K IS.

-)

9

(yco)“ + j 2 o i( j j , + co“

co“

K

X

(16.57a)

>

1+

/ c o - - c o “ '\

(16.57b)

2 o „ + ./ 03

The second factor is approximately 1 for frequencies near co^ by virtue o f our assumption a « OJ^. Hence the properties o f the magnitude response reduce approximately to those o f the first factor in equation 16.57b. These properties are those o f a single zero at the origin. Hence for high-Q^^ circuits, the relations o f equation 16.56 approximate the single zero at the origin case.

(S60


1^1

Exercise. Use equation 16.57b to show that when

> 8 and a « to , H

■P

P

~

?Q

■

When exact values are desired, one must resort to SPIC E or MATLAB to obtain the frequency response from which values can be graphically determined. SPIC E has the advantage o f not hav ing to compute the transfer function o f the circuit; MATLAB requires this computation.

7. SUMMARY This chapter began with a study o f simple series and parallel

band-pass circuits. Because o f

its generality, we set forth a transfer function approach to the analysis and design o f these band pass circuits. Specifically, we first investigated a second-order transfer function with a single zero at the origin for which

His) = K ---------------------= K - ------------------- ^ ,v’- + 2 o „ .v + co^,

Qp ' For this transfer function we derived various formulas for determining band-pass parameters, such as the peak frequency,

the peak value,

IG p

and the bandwidth,

to„,

a~p These formulas have a special form for the parallel RLO. to,,, = 0)^ =

4 lc

which is also the resonant frequency of the associated impedance; the bandwidth is found to be ^ (0 = t0 2 -0 J, = - ^ while the circuit Q is 0 = “

= u .„ R C = « ^

The resonance phenomenon of a second-order RLC circuit was then set forth from a frequency domain perspective. One application was the design o f a matching network that produces maxi mum power transfer, at a single frequency, from a source with fixed internal resistance to a fixed resistance load.


iSOl

After this we cook up the consideration o f more general second-order band-pass transfer func tions— for example, those with a single zero o ff the origin,

H(s) = K ------ ------------ = K — ^ , 2 5 “ + 2 a„.9 + Q)“ .y + —‘- s + oi~ P P a

or with no finite zero:

^

Qp

"

Other cases are left to more advanced courses on filter design.

8. TERMS AND CONCEPTS Active bajid-pass circuit: a circuit containing operational amplifiers and no inductors that achieves a band-pass characteristic. Band-pass circuit: a circuit that passes signals within a band o f frequencies while rejecting other frequency components outside o f the band. Bandwidth (3 dB bandwidth),

CO2 - cOp the difference between the two half-power fre

quencies. Dam ped oscillation frequency, (0^ frequency given by the condition that if the transfer func tion o f a second-order linear circuit has complex poles at j = -a ^ ± jio^, then the impulse response has the form K cos(co^f + 0). The constant (jo^ is called the dayyiped namralfre

quency. Half-power frequencies; see lower and upper half-power frequencies.

L C resonance frequency: frequenc)' at which the reactances o f L and Chave the same magnitude; equals M-JLC rad/sec. (Sec also tank frequency.) Lower half-power frequency, COj: the radian frequency below the center frequency at which the magnitude response is 0.707 times the maximum value. M atching network: an LC net\vork that transforms a resistance

into a resistance o f a differ

ent, specified value at one frequenc)- or a band o f frequencies. Peak frequency (center frequency),

the radian frequency at which the magnitude response

curve reaches its peak.

Qp (pole Q): for a pair o f complex poles j = Qp = 1/(2 sin 0). Quality factor Q (Qr/>) o f a band-pass circuit: the ratio o f the center frequency to the band width, i.e., Q = Quality factor Q£ {Qcoit) o f a coil: for a coil modeled by an inductance L in series with a resist ance Rs y Qi = and is frequenc}^ dependent. Q uality factor Q c iQcap) ® capacitor: for a capacitor modeled by a capacitance C in parallel with a resistance R^^, Q c~ frequency dependent. Q uality factor o f a reactive com ponent: for an impedance expressed zs Z = R + jX, Q^ = Reactance: in sinusoidal steady-state analysis, the imaginary part o f an impcdance. For Z,, the reac tance is

for C the reactance is -l/(coC).

862

Chapter 16

• Band-Pass Circuits and Resonance

-----------------------------------------------------------------------------------------------------------------------

^

Resonance frequency, 00^ the unique radian frequenq^ at which the input impedance of a twoterminal linear circuit becomes purely resistive. Selectivity of a band-pass circuit: The circuit Q, defined as the ratio of the center frequency to the bandwidth. A higher Q corresponds to better selectivity. Susceptance: in sinusoidal steady-state analysis, the imaginary part of an admittance. For C, the susceptance is coC For Z, the susceptance is -l/(coZ). Tank circuit: the parallel connection of an inductor and a capacitor. In the idealized case (no resistance), the total enei^ stored in a tank circuit remains constant, although there is a continuous interchange of the energy stored in the various components. Tank frequency, (Dq: defined as H j W in this text, regardless of the connection of the single L and single C with other components in the circuit. Tuned circuit: a second-order circuit containing one inductance and one capacitance, at least one of which is adjustable to reach a condition of near resonance. Undamped natural frequency: the natural fi-equency of a circuit consisting of lossless inductors and capacitors. For the case of one inductor and one capacitor connected together, this frequency is the same as the LC resonance frequency or the tank firequency and is equal to m J l c . Universal resonance curve: a normalized magnitude response curve of a band-pass transfer func tion having one pair of complex poles and a single zero at the origin. The magnitude is normalized with respect to the maximum gain, and the frequency is normalized with respect to the center frequency. Upper half-power frequency, (Oj: the radian frequency above the center frequency at which the magnitude response is 0.707 times the maximum value.

r> .

n o

o

* W e would like to thank Les Axelrod o f the Illinois Institute o f Technology for providing derivations that led to this formula and those o f the bandwidth and half-power frequencies.

863


C H EC K ; poles are p, 2 =

'Problems

5 . Suppose a basic band-pass transfer function

BASIC BAND-PASS TRANSFER FUNCTION PROBLEMS

has

= V s 200 rad/sec, (a) (b)

p

= ^ p

i+ -

1

4G,-;

= 10, and

=

8 rad/sec.

1. Fill in the details o f the derivation of

0) 1.2 = ±a^; +

±

Find H{s), its poles, and (X. Find approximate values For cOj and (O2. Using MATLAB, plot the magni tude response and verify your calcula

22,

tions. from

, find a parallel

N(Jco) = K — ^ T i j o i ) + j l w o p+CO p

K =

2 o „ -h y

/ 2 _ 2\ circuit realization o f H{s). I UJ uJ p \ _____ _ C H EC K : Q^= 11.32

transfer function o f equation 16.1 has poles at

6 . Suppose a series RLC circuit has = 80 rad/sec, = 8 , C = 1/80 F, and output l^{s)

-1 ± j5 and the gain at

with voltage source input

2. Suppose the basic second-order band-pass is 2 . Find K, H{s),

^1’ ^2quency response using MATLAB and verify

(a)

Write down the transfer fijnction for the

your answers.

(b)

Find the values o f R and L.

C H EC K : cOj = 2.831 rad/sec

(c)

Find

3. Suppose the basic second-order band-pass

(d)

series RLC circuit in terms o f R, L, and C.

m

and approximate values for

OJj and 0)^.

answers to part (c).

- 8 0 ± y l5 9 9 and the gain at co^ is 6.25. Find

K, H{s)y

Q,

exact and approximate

Use SPIC E to obtain the magnitude (frequency) response and verify your

transfer function o f equation 16.1 has poles at

C H EC K : /?= 0.125 n

values for tUj and Wj, and the relative percent error in the approximate computation o f CO2.

7. Suppose a series RLC circuit has

= 100

Using MATLAB, plot the magnitude response

rad/sec,

and verify your calculations.

with voltage source input (a) Write down the transfer function for

C H EC K : t02 = 1681 rad/sec while the approx

= 4, C = 0.01 F, and output

the series RLC circuit in terms o f R, L,

imate value o f CO2 is 1679 rad/sec

and C Find the values o f R and L.

4. Suppose a basic band-pass transfer function has

(b)

= 10, and B = 2 rad/sec.

(c)

Find

(d)

cOj and CO2. Use SPIC E to obtain the magnitude

to

= V S 2 rad/sec, (a) (b)

Find H{s) and its poles. Find tOj and CO2. Using MATLAB, plot the magnitude response and veri fy' your calculations.

Vguri-^) , find a parallel

(d)

■m(^) circuit realization o f M{s). Use SPIC E to obtain the magnitude frequency response and compare it with your answer in part (b).

and approximate values for

(frequenqO response and verif}' your answers to part (c). C H EC K : I = 0.01 H and =4 S

8 . Consider the circuit o f Figure P I 6.8 for which R^ = 40 k n , R, = 10 and C = 1 liF.

/• = 10 mH,


S64

(a)

Find the exact value o f the maximum

P I 6.10, w h e r e = 2 mS,

voltage gain and the corresponding

= 10 \dl.

frequenc)' (in Hz). (b)

Find the exact 3 dB bandwidth

(in

100 Q, and

(a)

Find H{s) in literal form.

(b)

If A = 100 u H ,^ = 1040 kHz, and Bjr = 10 kHz, find C, R, and

Hz). (c)

(c)

Find the circuit Q.

(d)

Find approximate values o f the upper

If^^^ = 920 kHz, Bj= 10 kHz, and C = 250 pF, find A, /c, and

(cOo) and lower ((o,) half-power fre quencies (in Hz). (e)

Use iVIATLAB or its equivalent to v plot

(0

|//(_/0))|

vs.

CO.

What is the new bandwidth if the input is changed from

to an inde

pendent current source (g)

If the circuit is frequency scaled by = 10, find the new values of

to-,,

and (Oj.

AN SW ERS; ui) 234.2 pi-, 6~.96 k ii; (b) 119 7 pH, 6 3 .66 k n 11. For this problem consider the circuit o f

R

Figure P 1 6 .1 1, in which A = 300 uH. The fol lowing additional ideal components are avail able;

Variable capacitors (in pF); 2 0 -2 0 0 , 3 0 -2 0 0 ,

Figure P I6.8

3 0 -3 0 0 , 4 0 -4 0 0 ANSWFR^.: (a) 0.20 at 1591 Hz; (b) 19.‘) Hz; (c) 1601 M Hz. I 5 8 1.6 11/.; (d) 80; (0 1S.‘) 1 Hz

Resistors; all values (a)

Using these components, design a par allel resonant circuit such that; (i)

9. I'he parallel /ÂC circuit o f Figure 1M6.9 has to^^^ = 1 Mrad/sec,

The circuit can be tuned from 550 to 1650 kHz (standard AivI

= 20 kH, and

broadcast band).

|Z(0.9(o^^^)| = 10 k n . Find the transfer func tion, M{s) = Z{s), R, A, C, B^^, Q, and approxi

(ii) When the circuit is tuned to 920 kHz (WBAA, at Purdue), the

mate values for to, and (1)2.

bandwidth is 20 kHz. (b)

With the components as selected in part (a), find the bandwidth when the

ZIjco)

■> R

circuit is tuned to the low end and then the high end o f the AM broad cast band.

Figure P I6.9 AN SW ERS; 20 kLl. 2.438 mH, ^10.2 pF. 121.9 krad/sec. 8.2 10. 'Fhe equivalent circuit o f a radio frequenq-

Figure P 16.11 AN SW ERS; 3 0 -3 0 0 pF. “ 9 .'’8 kLl, 7.1 6 kHz.

amplifier in an AM receiver is shown in Figure

64.2 kHz

865


12. For this problem again consider the circuit o f Figure P 16.11. The following ideal compo

Z(s)

nents are available:

Variable capacitor (in pF): 3 0 -3 0 0 Figure P i6.13

Resistors: all values (a)

Using these components, find the

A N SW ER: (a) 2.S mH .uui 530 12

range o f allowable inductance so that the circuit can be tuned from 550 to

14.

1650 kHz (standard AM broadcast

P 16.14, H{s) = Z{s), the peak f r e q u e n c y = 10

band) as follows:

kFIz, and the bandwidth B^ = 3 kHz. If C = 0.1

(i)

For the two-terminal circuit in Figure

Find the largest value o f L

l-iF, find the corresponding values o f L, R, and

allowable. Hint: For the smallest

the circuit Q.

value of C, the inductor must be

•—

chosen so that the circuit can be

—TYYY R

L

tuned to 1650 kHz. Z(s)

(ii) Find the smallest value o f Z. allowable. Hint: For the largest value o f C, the inductor must be

Figure P i6.14

chosen so that the circuit can be

C H EC K : Q = 3.333

tuned to 550 kHz. (b) (c)

For

find the range o f capaci

tance utilized in the tuning.

15. Consider the circuit o f Figure P i 6.15, for

For

which R^ = 40 kQ, Rj = \0 kH, /. = 10 mH,

find the range o f capaci

and C = 1 uF.

tance utilized in the tuning.

(a)


kHz

(b)

Find (o,„,

(WBAA, Purdue), the bandwidth is

(c)

C H E C K : 30 pF s C s 270 pF (d)

For L = 295 pH, find R so that when the circuit is tuned to 920

With the components as selected as in

(d)

(the corresponding

rad/sec and Hz.

the circuit is tuned to the low end and (e)

Use MATLAB or the equivalent to plot |//(yto)| vs. (I).

cast band?

What is the new bandwidth if the

(f)

input is changed from to an inde pendent current source / ?

13. For the rwo-terminal parallel /?Z,Ccircuit in Figure P i 6.13, H{s) = Z{s), the peak frequency 10 kHz, and the bandwidth Z^y= 3 kHz. If

R

C = 0.1 uF, find the corresponding values o f L,

R, and the circuit Q.

and

Find approximate values o f the upper and lower half-power frequencies in

part (d), what is the bandwidth when then to the high end o f the AM broad

and

quantities, in Hz, to

20 kHz. (e)

Find

Q, and

6 Figure P i 6 .1 5

866


peak frequency (in Hz).

AN SW ERS: O),,, = 10^' and Q = 80 (c) 16. Again consider the circuit o f Figure P 16.15. Now suppose

= 100

Q = 10, and^^ = 10

kHz.

Find the exact 3 dB bandwidth (in Hz).

(d)

Find Q o f the circuit.

(e)

Find approximate values o f the upper and lower half-power frequencies.

(a)

If /?^ = 00, find C and L.

(b)

Now suppose Rj^ = 1000 Q. Again

(f)

Plot the magnitude response using MATLAB or the equivalent.

find C and L. C H EC K : (a) C = 1.59 uF

C H EC K : (e) 9.9 and

1 0 .1

krad/sec.

17. For the series resonant circuit shown in

20.

For the circuit in Figure P I 6 . 2 0

Figure P16.17, /?, = 40 ^ Z- = 0.8 H, C= 1.25

Z. =

10

uF, and R j= 160 Q.

op amp is ideal. (a)

niH, C =

0 .1

mF,

=

100

1

kQ,

and the

Construct the transfer function H{s) =

(a)


(b)

Find the exact value o f the maxirnum

= -Zj(s)IZ,,{s) in terms o f the circuit elements R-^^, Rp L, and

voltage gain and the corresponding

C, and put it in the general form

peak frequency (in Hz).; (c)

Find the exact 3 dB bandwidth (in

s ~ ■ \-la p S + {H~p

Hz). (b)

Find the values o f K, 0)_, the circuit

(c)

Q, and = |M;co,„)|. Compute the value o f the half-power

(d)

Find the exact values o f the upper and

(e)

Find Q o f the circuit.

(0

Plot the magnitude response using

bandwidth

MATLAB or the equivalent.

quencies

lower half-power frequencies.

m R.

CO,

and the half-powcr fre and oyj. (Approximate

values are acceptable.)

v

(d)

L

Sketch the pole-zero diagram that rep resents the circuit, and note the exact locations o f all the poles and zeros.

(e)

the magnitude o f Vg^^f{t) in steady

Figure P i6.17

state.

A N SW ERS: (b) 0.8 at 159.15 Hz: (c) 39.79 Hz; (d) 180.29, 140.51 Hz; (c) 4 18. Repeat Problem 17 for /?, = 80

Z = 0.25

= 3200 rad/sec.

19. Reconsider the series circuit shown in Figure P16.17. Let /^, ^ A iX L = 0.1 H, C = 0.1 liF, and /?2 = 16 Q. (a) Find the transfer function H{s) = (b)

(0

Use SPIC E or the equivalent to gener ate the magnitude response plot for 0 < / < 500 Hz.

H , C = 0 . 5 u F , and/^2 = -'^20

C H EC K :

If v-J^t) = lOOsin(lO^f) mV, determine

Find the exact value o f the maximum voltage gain and the corresponding

R,


867

C H EC K : Q = 100

circuit should be 0.1 uF. Hint: Follow the pro cedure described in Example 16.5. Use SPIC E

21. For the circuit shown in Figure P i 6.21: (a)

or its equivalent to generate the magnitude

I (^) response plot for 900 < / < 1200 Hz.

Find the transfer flinCTion H {s) = --------

(b) (c)

Now supposey^^ = 200 Hz and Br= 20 Hz are desired. One has available a 1 H inductor, a 10 jiF capacitor, and arbitrar)’’ resistors. (i)

Determine the necessary value o f (3.

(ii) Determine the value o f R .

ANSWER: (b) w, -

f£ ± r V LC ' Figure P I6.23 24. For the circuit transfer function o f Problem 23,

''out +

Figure P I6.21

1

s +^2^2/

22. For the circuit shown in Figure P I6.22: (a)


(b)

Find

(c)

Now supposey^^^ = 200 Hz and Bj-=20

and Q.

^ 1^ 2Q ^ 2

show' that (i) H.„ =

C,

Co

-hi

and (ii)

Q- ..c, R' ^2

Hi

C,

Hz are desired. One has available a 1

Hence, for high-Q transfer functions, this cir

H inductor, a 10 jxF capacitor, and

cuit is undesirable.

arbitrary resistors. (i)

Determine the necessary value o f j.i.

(ii) Determine the value o f R.

25. Reconsider the active band-pass circuit o f Figure P16.23. The filter is to pick out the midrange o f a typical audio speaker, in which case/; = 500 Hz ,/2 = 3200 Hz, a n d ^ = 1265 Hz. The gain

is to be 10. Find Bj'd.nd the

circuit parameters. Hint: Follow the procedure described in Example 16.5. Use SPIC E or its equivalent to generate the magnitude response plot for 1 < / < 5000 Hz. Figure P I6.22 C H E C K : R = 795.77

C H EC K : Q < 1 and R^ > 2, slightly 26. Consider the band-pass circuit o f Figure P I 6.26, which contains two op amps rather

23. Design the active band-pass circuit in Figure P I 6.23 to have^^^^ = 1000 Hz and Bj- = 12.5 Hz. The final value o f Cj for the actual

than the one in Problem 25(a) Show that the transfer function o f the circuit is


86S

28. Consider the circuit in Figure P i 6.28.

2G: H{s) =

c

Model of Coil

Hints; (i) Use the properties o f ideal op amps, ' •

(ii) Use voltage division across Gy (iii) Write

0

two nodal equations and solve with unknowns K_...and K,. out (b)

(c)

Figure P i6.28

Design an active band-pass circuit to have^^^ = 1000 Hz and B^= 12.5 Hz.

(a)

Find Hîs) = 7^/1/and H^{s) =

The final value o f C should be 0.1 jdF.

(b)

For //j W, find the exact values of

(c)

For

Compare the resulting resistor values with those computed in Problem 25.

■^(0' Q-consider the case where the

ac voltage source has fixed

and

CO,

but the capacitance C is adjustable. Find the exact value o f C (in terms o f

R, L, and co) such that I i s

maxi

mized. Show that if the coil has a high Q, then

This result provides a practical way o f measuring

It also provides a practical means for

generating a very high short-duration voltage from a relatively small voltage; this has applica tion in ignition circuits.

27. Consider again the band-pass circuit of Figure P I6.26. In Problem 16.26 it was shown that 2^3 „

H{s) =

C J9 i\

\c)

(a)

Design an active band-pass circuit to have peak frequency and bandwidth

= 1000 rad/sec

= 100 rad/sec. The

final value of C should be 1 Ii/". (b)

Compare the resulting resistor values with those computed in Example 16.5.

BAND-PASS CIRCUITS WITH PRACTICAL COMPONENTS 29. Consider the /?/.Ccircuit in Figure P16.29, in which 7?^ = 50 Z, = 1/9 H, = 0.08 Q, F and C = 1/9 F. 1

(a)

Find the coil Q at Wg =

(b)

Find an approximate parallel represen tation o f the coil near (Oq .

(c)

Find the circuit Q.

(d)

If the circuit is high Q, compute approx

(e)

Suppose = 2cos(0),^/) A. Find, approximately, (in volts) in steady

imate values for

state.

y fiC '

CO,, and CO2.

869


=

A N SW ER S: (a) 12.S; (c) 10; (cl)

20

(a) (b)

Find the capacitor Q at Wq = j-----. Find an approximate series representation o f the capacitor near ( jO q .

(c) (d)

Find the approximate value o f Q. If the circuit is high Q, compute approximate values for

0)j, and CO2.

,

CYY\ L

Coil

Figure P I6.29

= 3750 Q., L = =2

30. Repeat Problem 29 for 0.1 mH, 7?^ = 4 a

T

Capacitor

C = 10 nF, and

cos(oj^^^r) mA.

Figure P I 6.33

C H EC K ; Q = 15 and

= 3 cos(co^^^^) V

31. Consider the RLC circuit in Figure P 16.31, in which

L = 0.5 mH,

= 20

Q., and C = 0.5 nF.

= 20 ^

C H EC K ; CO, = 9975 rad/sec

34 . Consider the circuit o f Figure P I 6.34. Suppose R^ = 0.6 O., L = \ mH, R^- = 2.5 kT2, and C = 0.1

------V LC

(a)

Find the coil Q at wq =

(b)

Find an approximate parallel represen tation o f the coil near cOq. Find the approximate circuit Q.

(d)

If the circuit is high Q, compute

(b)

Find an approximate series

-------

representation o f the capacitor near (d)

o jq

.

Find the approximate circuit Q. If the circuit is high Q, compute approximate values for

approximate values for

(Oj, and to^.

o jj, and t02AN SW ERS; (a) 12.5; (c) 10; (d)

j

Find the capacitor Q at (Oq -

(c)

(c)

l i F.

(a)

= 20

c o s ((t)„ / ) \ '

R.

'» 6

Figure P I6.34 C H EC K ;

100

Figure P I6.31

3 5 . Repeat Problem 34 for R^ = 0.32 D., L = \6 32. Repeat Problem 31 for R^ = 40 k ii, L = 4 mH,

= 100 a , and C = 0.25 nE

C H EC K ; 1 >

l i H , /?c= 800 a , and C = 0.25 uE C H EC K ; = 20

> 0.5 36. Consider the circuit of Figure P i 6.36.

33. Consider the circuit o f Figure P i 6.33.

Suppose L = 0.5 mH, C = 1.25 uF> Q_i^ = 20,

Suppose R^=2 k n , L = 1 mH, and C = 10 tiF.

and

= 80.

870

Chapter 16 <>Band-Pass Circuits and Resonance

r>

(a)

Find

(b)

Convert the capacitor to a series

and

model at

model at (Oq. Denote the resulting

oO q.

Denote the resulting

(c)

series resistance by 1 series Find the approximate circuit Q.

(d)

If the circuit is high Q, compute approximate values for

/\ (e)

Lossy Inductor

If the circuit is high Q, compute approximate values for

co^^,

cOpandcDj. n ,0icap -r I n ^cotl^c Verify that Qdr ® Q coil

(c) (d)

series resistance by RpamFind the approximate circuit Q.

Qcap

0)p and (02-

q

(e) Verify that C H EC K : = 20

, O)^, q

+ Qcap

40. (Design) Consider once more the circuit of

Lossy Capacitor

Figure P I 6.38. Recall that for high-Qcompo-

n n ^coil^cap ~n -I- n WjcoiY ^cap ;c

Suppose you have been asked by your supervi sor to use this circuit in a band-pass design with

f „ = ^ M Hz and a bandwidth o f 20 kHz and a

Figure P I6.36

lossy capacitor with C = 2 0 0 pF and having

C H EC K : Q „ ,= 16 37. Again consider the circuit o f Figure PI 6 .36.

(a)

Determine the circuit Q, i.e.,

Verify that for high

(b)

Find the necessary

and high

at cOq,

to achieve the

desired Q coilQ cap

^

(c) F

P16.38.

38. Now consider the circuit o f Figure P I 6.38. Verify that for high

and high

ind the inductance L o f the lossy coil and then find R^ as shown in Figure

Qcoil + Qcap

at (Oq,

*(d)

Suppose we now desire to double the bandwidth by adding a resistor, R^g^„^

QcoilQcap

a Q coil

Qcap

in parallel with the lossy coil (i.e., in parallel with the current source). W hat is the proper value of R^source' ^ ^

Lossy Inductor

C H EC K : (c) R^
= 200.

42. (Design) Consider the circuit o f Figure Figure P I6.38

P I 6.42. Recall that for high-Q components, ^

3 9 . Consider again the circuit o f Figure P I 6.38. Suppose L = 0.5 m H , C = 1.25 jiF, = 30, and = 60 at cDq. (a) Find R^ and R^. (b)

Conven the inductor to a parallel

QcoilQcap

Qdr = „ .f / ^coil *•*^cap Suppose you have been asked by your supervi sor to use this circuit in a band-pass design with = 1 MHz, a bandwidth o f 2 0 kHz, and a

8 71


lossy inductor I = 100 uH having Q^gj(= 100. Determine the circuit Q, i.e., Q^j^. (a) Find the necessary to achieve the (b) and the corresponding frequency (in Hz),

desired (c)

Find the capacitance C o f the lossy capacitor and then find

(d)

Check your results by doing a SPICE (or equivalent) simulation of your circuit.

as given

in Figure P I 6.42. (d)

Suppose we now desire to double the bandwidth by adding a resistor,

^source’

series with the lossy coil (i.e., in series with the voltage source).

What is the proper value of Lossy Inductor

amplifier

Lossy Capacitor

coil

capacitor

Figure P I6.44 A N SW ERS: (a) 50; (b) 1 mH, 50 kQ; (c) 6 3 6 .6 Hz. 10, 15.91 kHz

:C

45. Repeat Problem 44 for the circuit shown in Figure P I 6.45, where /?, = 500 Q., = 2 kQ, R^ = 5 k n , R^ = 2Q., L = \ mH, C = 1 uH, and a = 25.

Figure P I6.42 C H E C K : (c) 130 k n > /?^ > 115 kQ 43. (a)

Repeat Problem 42 (except for part (d)) under the conditions

= 1.6

MHz, a bandwidth o f 25 kHz, and a lossy inductor Z, = 100 j.iH hav (b)

ing ^ oii= ^00. Now suppose you were only able to purchase a capacitor having Figure P I6.45

= 200; you must achieve the required bandwidth by adding a resis

C H E C K : 650 Hz > Bj- > 630 Hz

tor in parallel with the lossy capacitor. What is the value o f this new parallel resistance? C H EC K :

46. Consider the circuit in Figure P i 6.46. (a)

Find the input admittance o f the cir cuit, ^y„(^), and compute its poles and

= 64

zeros. 44 Consider the amplifier circuit shown in Figure P I 6.44. Find Q o f the coil at to = HP rad/sec. (a) Represent the coil by a parallel RL cir (b) cuit that is valid for frequencies near 10^ rad/sec using column 3 o f Table (c)

(b)

Determine

and the approximate 3

dB bandwidth and half-power fre (c)

quencies o f the circuit. Determine and at cOq. Then use column 3 o f Table 16.2 to find the approximate bandwidth and half-

16.2.

power frequencies. Compare the result

Find approximate values of the 3 dB

with that obtained in part (b).

bandwidth.


8 "’ 2

(cl)

Cheek your results by doing a SPICE (or I. r~

______

10.000 rad/scc.

=£ 10^’ rad/sec. (O, a

H). ‘)S0.0()0

rad/sec. (u-, s 1,050,000 rad/sec; (c)

coii 0.01 H

A N SW ERS: (b) (i»^ s

equiv;ilent) simulation of your circuit.

=

0.1 sin(oj,-r) V 750

10’“F

3000

RESONANT CIRCUITS WITH APPLICATIONS

SOOkO

48. capacitor

For each two-terminal circuit in Figure

P I6.48, the resonant frequency/^ = 10 kHz. If C = 0.1 iiF and R = 1 kH, find the corre

Figure P I6.46

sponding values o f L. For each circuit, deter

47. Consider the circuit in Figure P i6.47, which

mine the input impedance at resonance.

contains a non-ideal capacitor, a non-ideal induc tor, and a meter to measure the current response,

Z(j(o)

io„fU)- The 1 Q resistor representing the meter is a precision resistor. The voltage across the resistor, equals the current through the resistor.

(a)

'Fhus, a practical way o f measuring current is by

•—

measuring the voltage across a small resistance in

—TY"YV R

the circuit. If the resistance o f the meter is suffi

L

Z(jto)

ciently small, it should have little effect on the behavior o f the circuit. Nevertheless, in analyzing the circuit, account must be taken o f the resist

(b)

ance o f the meter. (a)

Use the approximation techniques o f column 3 ofTiible 16.2 to develop an

Figure P i6.48 AN SW ERS: (a) 2.S niH; (b) 2.5 mi l

approximating series RLC circuit for the given circuit.

49. For each circuit in Figure P i 6.49,

(b)

Compute approximate values for

(c)

Q, to,, and co,. At resonance, fmd the approximate

(a)

L, and C. Verif}' that 10^ = iOq. (This always follows when L and C are either in parallel or in

steady-state current response, when

the

input

voltage

is

10

sin(to^r)«(r). (d)

series.) (b)

Check your results by doing a SPICE (or

that the input impedance is independ ent of L and C at co^.. (Again, this

inductor ^

Now find the input impedance at to^ in terms o f /?,, R-,, L, and C. Verif)'

equivalent) simulation of your circuit. m eter

Find the resonant frequency', co^, in terms o f /?j,

^

capacitor

always follow's when L and C are either in parallel or in series.) (c)

In the case o f Figure P I6.49a, find

Vo,„{^) in steady state in terms o f /^,, R^, and if v-^^{t) = sin(w/). In the case o f Figure Pi 6.49b, find /„,^,(/) in steady state if Figure P i 6 .4 7

sin(co^).


shown, find the value o f L to maxi mize the voltage gain / Vy^^(/to)| at/= 10 MHz. Verify that the

+ Z(jw)

amplifier gain at 10 M Hz is greatly increased. (c)

(a)

—

fyy\L

(!)

Plot the frequency responses o f the cir cuit over 1 M H z < / < 10 M Hz with L and without L.

i „(t)

Z(j(o)

(b)

Figure P I6.49 amplifier

50. Figure P I 6.5 0 displays an amplifier model containingo a V CC S with ofn ?•,„ = 4 mS (milli-siemens) and R/ = 40 kf2. Suppose the applied sinusoidal voltage,

has a mag

nitude o f 0.2 V at 20 MHz. (a)

W ith the load connected directly as

load

Figure P i6.51 52. Find the resonant frequency (O^., in rad/sec, and Z{jio) o f the circuit in Figure P i6.52 for R

= 2.8

C= 0.2 mF, and L = 20 mH. Verify that

shown (without L), find the magni

■ lie '

tude o f the output voltage. (b)

If an inductance L is connected across

Now compute

the load to tune out the effect o f the

and phase responses from 0.75co^ to

capacitance, find the value o f L and

Verify that at co^, the phase angle o f

the resulting

zero.

that will show that

the amplifier gain at 10 MHz is great

and plot the magnitude is R

(t)

ly Increased.

1 . 2 5 c d ^.

Figure P I6.52 AN SW ERS: 480 rad/sec, 35.71 LI Figure P I6.50

53. Consider the circuit o f Figure PI 6.53. The

51. Figure P I 6 .5 1 displays an amplifier model containing a V C C S with = 4 mS (milli-

matching network when the values are proper

CL part o f the circuit can be thought o f as a

siemens) and Rj^ = 20 k^2. Observe that the

ly chosen. In case 1 we will see that maximum power transfer is achieved, wherea.s in case 2 it

amplifier has an input capacitance of 0.2 nF.

is not.

The inductor L is used to tune out this effect. (a) (b)

(a)

Find the resonant frequenc\\

Without L find the magnitude o f the

for the cases

voltage gain at 10 MHz.

(i)

If an inductance L is connected as

(ii) /^^‘ = 6 0 £ 2

Rj =

80

LI

and


874

(b)

Now suppose

= 250

with co =

L

0)^.. Then compute the average power

/Y Y V

delivered to Ri for each o f the cases in Z(j(o)

part (a). 0.1 H

125Q

/Y Y V Figure P i6.55 V cos(o)t) «)

0

Z(jw)

—►

IO mF

56. Find the resonant frequenc)'

in rad/sec,

and Z(Jo}j-) o f the circuit in Figure Pi 6.56 for R = 12.5 Q, C= 0.2 mF, and L = 20 mH. Verify that

Figure P I6.53 54. For the circuit o f Figure P I 6.54, R = 800

Q, L = 0.2 H, and C = 0.25 liF. For the two cases (i) Rj = 1 k fl and (ii) = 800 find co^ and at resonance, find

the voltage across

y jlc ' Now compute Z-^^{s) and plot the magnitude and phase responses from 0.5w^ to 2to^. Verify that at (0^, the phase angle o f

the capacitor in the steady state, due to the input

/Y Y V

= 10 cos(o);-f) niA. Finally, find the

average power delivered to R^ in each case.

is zero.

L

VJj(o)

o

Z(j(o)

—►

Figure P i6.56 57. Consider the circuit o f Figure Pi 6.57 for R^ = 80 a , R^ = 125 n , C = 2 uF, and Z. = 20 mH. The LC part o f the circuit can be thought o f as a matching network when the values are prop erly chosen. (a)

Figure P I6.54 ANSWHR: Case I: 201)0 rad/sec, ^ cos(2000/) V, 12.5 m\V

Find

the

resonant

frequency

CD^,

Z,(;cop, and Zijia). (b)

Now suppose

= 250 V^^ with ti) =

Compute the average power deliv 55. This problem develops formulas similar to

ered to

equations 16.41 and 16.42 for the

more power to

circuit

by varying the val

ues o f L and C?

o f Figure P i 6.55. (a)

Is it possible to deliver

Prove that, for the circuit shown in Figure P I 6.55,

:..................... -•— ^ - J

LC

I ’ '------L

Z(jco) c

<

CR-

and

Z,(jw)

Z{j(0,.) =

Figure P i6.57

CR 58. A two-terminal network has input imped ance


61. This problem uses an LC coupling network

s~ + 4 s+ S = 1, find co^ and

to maximize power to a load. The LC coupling network always has a series connection for L\ the “parallel” capacitor is always closest to the

(a)

If

(b)

Find the range o f /7 > 0 for which the

larger resistance— in this case the load resist

impedance has a real resonant fre-

ance. Problem 62 will consider the general case.

quenc)'.

Suppose the voltage soQrce in Figure P I6.61 has value

= 1 0 0 V 2 cos(27i x lO^r) V.

ANSWHRS: (a) 2 rad/scc and 0.25 i h (b) 0 <

Compute the values o f L and C such that the

a <1

average power delivered to the load resistance

RtL is maximized. What is Pttuix 59. For the circuit shown in Figure P I 6.59, where . 205 y i s ) = —--------- , 5 - + 100 find the value o f C that makes the circuit reso nant at

CO

/Y Y \ son v jt )

300 0

6

= 30 rad/sec and then find the value

ofr.,(/-30).

Figure P i6.61

OY„(s)

AN SW FRS: P .9 uH. 1,186 pF 50 W Y(s)

62. Equations 16.41 through 16.44, or those derived in Problem 55, can be combined to produce a set o f design formulas for a lossless

Figure P i6.59

network that matches two unequal resistances at a single frequenc)^ The matching network

60. This problem uses an LC coupling network to maximize power to a load. The LC coupling

consists o f only one capacitance C and one inductance L, as shown in Figure P I6.62. At a

network always has a series connection for L\

specified frequency OJ, it is desired to have

the “parallel” capacitor is always closest to the

matching at both ends, i.e.,

larger resistance— in this case the source resist ance. Problem 62 will consider the general case.

Z , (yo)) = /?, + /O and Zjijiyit) = R2 + P

Suppose the voltage source in Figure P I 6.60 has value

= 100V 2cos(27t x 10^/) V.

Compute the values o f L and C such that the average power delivered to the load resistance

Let R^„jîi denote the smaller o f (/?p /?-,) and the larger. Prove that (a)

is maximized. What is

Cshould be connected in parallel with

^large'

^ should be connected between the rsvo top terminals and

(b)

thus in series with R^„j^nThe element values are given by

large

—Rsmall

and

Figure P I6.60 A N SW l-RS: 17.79 uH, 1.186 pF, 8.3.^33 W

C=

(oRla/ge

Rsmall

8:^6


AN SW ERS: 12.5 V I /S = 21.65 rad/s cfc 22.2 Q. 5\/21 = 22.91 rad/s. 4.44 Q.

BAND-PASS TRANSFER FUNCTIONS WITH NO ZEROS OR A SINGLE ZERO OFF THE ORIGIN

Figure P i6.62 63. The purpose o f this problem is to show that the resonant frequenc)' (0^ depends on the

65. Consider the circuit o f Figure P i 6.65. (a)

choice o f the input terminals and that (0^^ (Oq = 0.25

His). (b)

I f / ? = 5 a , Z = 0.1 H ,an d C = 10|iF,

(c)

Compute

rad/sec. Find the resonant frequency (0^ if the input is connected across (a)

the

output, compute the transfer function

in general. Consider the circuit o f Figure P 1 6 .6 3 , which has Wq = l/ V LC

With V^{s) as the input and

verify that the circuit is high Q^. exactly and verify that

A and B

0)^^ = CO^. Then compute the maxi

(b)

B and C

mum gain.

(c)

A and C

(d)

40 n

With the values given in part (b), compute approximate values for

10Q

CO,, and CO2. '0 .8 H 2m F

Figure P I6.63 ANSWURS: 2S, I2.5\/3 = 2 1 . 2 8 . 8 6 . all in rad/sfc 64.

66. Consider the circuit o f Figure P i 6.66.

The

circuit

= l/ 'Tl C

of

Figure

Pi 6.64

has

(a)

depends on the choice ot the input terminals and 03^ in general. Find the resonant frc-

quenc)' 03^ and the equi\'alcnt impedance seen at (a)

E and F D and E

as the input and I^{s) the

His). (b)

I f / ? = 4 k a Z . = 0.1 H ,an d C = 10).lF, verify that the circuit is high

(c)

Compute (0^^^ exactly and verify that = CO^. Then compute the maxi

the terminals if the source is connected across (b)

With

output, compute the transfer function

= 25 rad/sec. The purpose of this

problem is to show that the resonant frequency (0,. that (0^

Figure P I6.65

mum gain, H^^j. (d)

With the values given in part (b), compute approximate values for

100

CO,,

and 0) 2-

. 0.8 H 2m F

50 0 /

F

Figure PI 6.64

F igu re P I 6 .6 6


67. Consider the Sallen and Key active network of Figure P I6.67, which realizes the transfer function

H(s)= f— UiC

____1

H{s) = —

LC ]

l

_

1

1

+■ /?2Q/

s+ =

1

Recall that cOq = lly J h C . By the use of

(b)

equation 16.49, show that

= 50

kfl.

1 --

Compute (O^ and H^.

(c)

Find approximate values of

1-

0)j,

4lc\

and (Oj.

R^C^+s \

1

2LC

S

4

c2 +

\

}

^

4 .

Plot the magnitude response curve of the transfer fimction.

(e)

1

2LC

Compute O) , Qp, and K of equation 16.48.

(b)

(d)

LC

/?l/?2C|C2

Suppose Cj = 0 .2 |jiF, C2 = 0.5 nP, (a)

S +■

/YY\

Simulate the circuit using SPICE.

L

o— + Figure P I6.69 70. Consider a transfer function with a single zero off the origin, i.e., Figure P I6.67

s+2OpS+(0p

H{s) = K - 2

------j -

68. Consider the Sallen and Key circuit of Problem 67. Suppose Cj = 50 nF, Cj = 3.125 nF, = y?2 = 8 1 ^ . (a)

Compute O) , Qp, and K o f equation

(a)

Derive equation 16.54, i.e.,

o )„ =

16.48. Verify that the circuit is not high Q. (b)

Compute the exact values o f (O^ and

Hint: Instead o f maximizing |//(/(o)|, try to maximize l//(/to)p, and consider co^ as the

(c)

m Find the exact values o f 0)o.

(d)

Plot the magnitude response curve of

o)j, and

the transfer fimction. 69. For the circuit shown in Figure P I 6.69: (a) Show that the transfer function is

independent variable. (b)

Derive equation 16.55, i.e.,

0) (for zero phase shift) = ^(Op - 2opO

Hint: Write \/H(s) as [{l/K + Im{/^(/0))} = 0, and solve for co.

set

878



73. The y?/.C circuit o f Figure PI 6.71 has trans-

P 16.71.

fer function

(a)

\/


H {s) = Zi„{s) = ------- ^ '

s+

H {s) = Z;As) =

+

L}

^

[X pC

+ ‘''S s + 11+ L \

LC

D \ 1

Rf,C (b)

R.

s+— L)

Let /?. = 100 k^2, L = 0.225 H, R, = 1000 Q, s + l+^.s_ RPI LC and = 0.5 [.iF.

L

(a)

By the use o f equation 16.54 or part

Compute to and verify that the cir cuit is low Q^.

(a) o f Problem 65, show that

(b)

Compute

exactly and verify that

is quite different from 1

0).,, =

(c)

1+-

Plot the frequency response for 0 < to < 40 0 0 rad/sec. Compute the maxi

R r

mum gain and verify the result from the plot. Also compute the zero o f the transfer function as -Zy (d)

Define the coefficients o f the numera tor and denominator o f your transfer function in MATLAB as n = [1 zl]/C and d = [1 ? ?]. Now use the com mands below to compute the impulse

Coil

and step responses o f your transfer Figure P I6.71

function.

72. The RLC circuit o f Figure P 16.71 has trans

sys=TF(n,d)

fer function

impulsc(sys) pause

. s '-

(a)

step(sys)

L}

H {s) = Z U s) = (

-I-

1_

AN 'SW FRS: wp = 2 .9 9 6 3 e+ 0 3 ,

s+

K„jLC

If /?^ = 50 a , Z- = 119 H, R^ = 0.08 £2, and C = 1/9 F, verify that the circuit is high

(b)

and that a «

Compute

to^.

exactly and verify that

to^^ a (i)^. Then, compute the maxi mum gam. (c)

Compute approximate values for

(d)

tOj, and C02Suppose diat at to^, |/,„(/‘toJ | = 1 A. Determine, approximately, the value

ANSW'HRS: (b) to,,^ = 9.0071 rad / sec; (d) 10 \'

Q

=

6.7115C -01, 7.1 = 4 .4 4 4 4 c+03, wm = 1.0l28e+ 03

74. Again consider the circuit o f Figure P 16.71, in which Z, = 1 mH. This inductor has

=

40 at 100 kHz. The magnitude response is to have a peak at

s 100 kHz.

(a)

Specify the value o f the capacitor C.

(b)

Specify the value o f Rp so that the bandwidth is approximately 10 kHz.

(c)

What is

A N SW ERS: 2533 pF, 83‘'8 Q, 10

8~9


element values and repeat parts (c)

75. Consider the circuit in Figure P i 6.75. Let = 0.08 Q, R c = 0.02 a

Z. = 1 H, C = 1 F,

and (d) for the scaled circuit.

and Rs = 40 ^2. (a) Find the transfer function H{s) =

R, = 0.05 n

o u r ''" ' nt'

(b) (c)

Find

and approximate answers for

O btain

/Y Y V

" P “ ‘1 “ 2magnitude frequency

a

,0

response plot to graphically verify

R, = 50 O

L=1 H C =1 F R, = 0.05 n

your answers. Figure P I6.76 R

MISCELLANEOUS 77. Consider the idealized (tank) circuit o f

'■ 6

Figure P I 6,77. The moment the inductor cur rent passes through zero with positive slope is taken as the reference point, r = 0. At this time instant the capacitance voltage is E volts. Find V(\t) and i^{t) for / > 0 by the

Figure P i6.75

(a)

76. The analysis o f the non-series-paraliel cir

(b)

Find the energy stored in C as a func tion o f t.

(c )

Find the energy stored in L as a func tion o f t.

(d)

Show that the total energy stored in

Laplace transform method. cuit shown in Figure P i 6.76 requires writing node or loop equations. Because there are no series-parallel connections, one cannot apply the equivalents o f Table 16.2. (a)

Let r , = 1//?,, ^2 = i//?2,

= \/L^s,

the LC tank is constant and is equal to 0.5C £2.

= i//?3,

= C 5J. Using nodal

analysis, show that the transfer func tion is

His) =

V

-

hn

>2 (>i + y4){y3+>5 ) + y?>Y4{y\+>5 ) + y ^ y s + 1^4)

Figure P I6.77

.y-hO.05 0 .1 2 0 3 8 5 + 1.009 (b)

Obtain the pole-zero plot o f H{s).

(c)

Find

(d)

bigb Q^Find approximate values o f (O^,

(e)

Check your results by doing a SPIC E

and verify that the circuit is

78. Consider the circuit in Figure P I6.78, which contains a non-ideal inductor and a vari able capacitor. Suppose Z = 100 iiH and 7?^ = 5.4 Q.

o jj, and oj-,. (or equivalent) simulation o f your cir cuit. (0

Frequency-scale the circuit by ^ = 1000 and = 100. Show the new

Figu re P I 6 .7 8

880


(a)

Determine the range of the capaci

80. Consider the circuit o f Figure P I6 .80, in

tance C, i.e., Cq < C < C j, such that

which R^= \ 0 n , L= 1 H, /e/= 0.8 n , and C

the circuit can be tuned to resonance

= 1 F.

over the AM radio band (from 550

(a)

Show that the transfer function H{s) is

kHz to 1650 kHz). (b)

that given in Problem 71.

When the circuit is tuned to 550 kHz

Find the exact values o f to^, coy,,

(b)

and 1650 kHz, determine the circuit

CO,, and co^^^. Since this is a low-Q cir

Q's, the two bandwidths, and the

cuit, do not use the high-Q approxi

lower and upper halF-power frequen

mations.

cies for each bandwidth. 7 9 . 1'his problem illustrates the conceptual dif ferences

among

the

various

frequencies

encountered in this chapter. For practical high er circuits, the numerical values of these fre quencies are all very close. To see the differ ences, we choose a low-Q circuit. Consider the

Figure P i6.80

circuit shown in Figure P I 6.79. (a)

Show that

1

R ,C [

R\

AN SW FKS; (b) = 1, co^ = 1.0392. = 0 .9367, (-),. = 0.6, and = 0 .9602, all in

L)

rad/sec

.V H------

.V-l- 1+ (b)

R .) LC

81. Suppose the current source in the circuit o f Problem 80 is /^^(/) = 2 cos{t) A and that C is

I f = 5/3 £ 2 ,1 = 1 H, R = 0.8 Q., and

variable. Find the value o f C such that |

C = 1 F, show that:

maximum. Verif}' that Cj3q is not equal to the

The LC tank frequency

= 1 rad/sec.

is

source frequency in this case.

The resonant frequency co^ = 0.6 rad/sec.

82.

The peak frequency

Consider the circuit in Figure P I 6.82, in which

= yjl9 / 25 = 1.077 rad/sec.

(Experimental

C = 0.1 uF, L = 1 niH, and R^ =

of

Q)

Q.

The pole frequency

(a)


^p ~ = 1.2165 rad/sec. The natural damped frequency co^y =

(b)

Compute the poles and zeros o f H{s). From these, approximately compute

0.995 rad/sec.

, and

co^^^,

Hints:

1. To find 0

measurement

use equation 16.41

To find OJ , use equation 16.54.

(c)

Is this a high-Q cir

cuit? rhe impulse response for the circuit is o f the form

h{t) = Ae~‘" cos(cor) +

•6

Show that \A\ »

|i9| for the high-Q case and,

therefore, h{t) s (d) Figu re P I 6 .7 9

Plot

sin(co/).

cos((i)r).

h{t)

=

Ae~^‘cos{iOt)

using

MATLAB or its equivalent. Show that

ss;

Chapter 16 • Band'Pass Circuits and Resonance

the peak will decrease to 1/^ o f the first peak in approximately QIti cycles of oscillations. (This is how one experi mentally determines Q.) (e)

Find approximate values of O)^, and Q by changing the series L-R Figure P I6.83

to an approximate parallel L-R con nection. Do the results agree with those obtained in part (a) by the trans fer function approach?

84. The opening seaion of this chapter dis cussed the generation of dial tones by resonant circuits. The circuit for generating the three

O '

tones in the high-frequency group is shown in Figure P 16.84. Using the results of Problem 71, explain how the tone generation circuit works. Your explanation will include an oscillation of an undriven LC circuit that is coupled to the transistor circuit inside the box by transformers w

Coil

studied in the next chapter.

Figure P i 6.82

frequency determining resonant circuit

83. The switch S in Figure P I6.83 has been closed for a long time and is opened at ^ = 0. (a)

Use the Laplace transform method of Chapter 14 to show that if Q = 0 ) ^ C > 0.5, then for / > 0, the capacitance voltage is sin (o )/ + 0)

where a = and

to.

2(2 1-

1 4(2^

(b)

Show that the peak amplitude o f the damped sinusoidal waveform of part

•all switches open when no button is pressed •pressing a button in column k closes the switch SCK •when any button is pressed halfway, S is closed to contact H; when the button Is fully pressed, S moves away from H and makes contact with F

(a) decreases to Me = 0.368 of the

Figure P i6.84

highest peak approximately after Q/ji cycles if Q is large (and, hence, 'w '

co^).

(O^ es

85. For the circuit o f Figure P I6.85a, we have the impedance

W

and for the circuit of Figure 16.85b, we have the admittance W

882


r\ YijiO) = Find expressions for the conductance the susceptance

in terms o f

and

and

r\

so that

the two circuits have the same impedance at a

r\

single frequency O). Then find expressions for and

in terms of G^ and

under the same

condition.

■>G.

iBp

JX.

r>

•-------- J o

(b)

(a)

Figure P i6.85 Conversion of (a) a series RfX^ combination to (b) a parallel G^-B^ combina tion. No specific functional form is imposed upon X^ or B^ here. 86. Reconsider Problem 85 for the case where

jCsO) andy5^ = jC^fa. Find formulas for

and R^ in

terms of C and 7?^ and thus specialize the for mulas o f Problem 85 to the case of a trans forming a parallel RC to a series RC at a single frequency (o. Now do the converse.

C H A P T E R Magnetically Coupled Circuits and Transformers WHAT IS INSIDE THE AC ADAPTOR? Most electronic equipment operates with dc power sources. For portable equipment, such as a cordless phone and a cordless electric drill, batteries supply the dc power. Using non-rechargeable batteries becomes expensive. Furthermore, replacing batteries in special equipment is a task not easily handled by ordinary consumers. These two factors have prompted manufacturers to install rechargeable lithium-ion batteries in portable equipment. By connecting several batteries in series, the available dc voltage may range from 1.5 to 12 V. Whenever the battery runs low, it must be recharged. Recharging a battery requires a low dc voltage source (1.5—12V). An adaptor houses a device called a transformer that changes the 110 V ac voltage at the household outlet to a much lower ac volt age. The lower ac voltage is then rectified to become a dc voltage that charges the battery. Some adaptors contain the transformer only, while others may also contain the rectifier circuit. Typical specifications appearing on the casing o f an adaptor may be as follows:

model: AC9131

model: KX-AIO

input: ac 1 2 0 V ,6 0 H z,6 W

input:

output:

output:

ac 3.3 V. 500 mA

ac 1 2 0 V ,6 0 H z,5 W dc 12V, 100m A

The concepts and methods developed in this chapter will allow us to understand how a trans former works to change the ac voltage level and also to perform some other important functions in electronic equipment.

8iS4

Chapter 17 • Magnetically Coupled Circuits and Transformers


Understand how the mutual Inductance M, between two inductances

and Z ,,

accounts for an induced voltage in each inductor due to the change o f current in the other inductor. 2.

Develop a systematic method for writing time domain and frequency domain equations for circuits containing mutual inductances.

3.

Understand why the mutual inductance is less than or equal to the geometric mean of the individual self-inductances using an energy perspective.

4.

Expand the repertoire o f basic circuit elements to include ideal transformers, and learn how to analyze circuits containing ideal transformers.

5.

Learn how to model a pair o f coupled inductors by an ideal transformer and at most two self-inductances.

6.

Learn how a practical transformer can be modeled by an ideal transformer and some additional RL elements.

7.

Investigate some important applications o f transformers and coupled inductors in power engineering and communication engineering.

SECTION HEADINGS 1.

Introduction

2. 3. 4. 5.

Mutual Inductance and the D ot Convention Differential Equation, Laplace Transform, and Phasor Models o f Coupled Inductors Analysis o f Coupled Circuits with Open-Circuited Secondary Analysis o f Coupled Circuits with Terminated Secondary

6.

Coefficient o f Coupling and Energy Calculations

7. 8. 9. 10.

Ideal Transformers Models for Practical Transformers Coupled Inductors Modeled with an Ideal Transformer Summary

11. 12.


1. INTRODUCTION You may recall from your high school or grade school science class that if iron filings are sprinkled on a piece o|- paper and a magnet is moved around beneath the paper, the iron filings move in con cert with the magnet because the magnetic field induces a force on the iron filings. Similar to the magnet and the iron filings, a changing current in one coil that is ver)' close to another coil induces a voltage across the terminals o f the other coil. Figure 17.1a shows rwo unconnected coils o f wire in close proximity. Figure 17.1b shows two unconnected wire coils wound around a single ferromagnetic core, hi both cases, a voltage source excites coil 1 while coil 2 is left open-circuited. Experimental evidence shows that a change in the current z, generates a voltage Vj, called the induced voltage, across the open circuit; the induced


voltage is proportional to the rate o f change of/ j. Each pair o f coils in Figures 17.1a and 17.1b is said to be magnetically coupled. Coll 1

-•

+ V ,( t )

Coil 2 (a)

(b )

FIGURE 17.1 Induced voltage in coupled coils, (a) Two coils in close proximity, (b) Two coils wound on the same ferromagnetic core. How does one quantitatively account for magnetic coupling? The strategv' is to introduce a new circuit quantity called mutual inductance for coupled coils; specifically, similar to the v-i rela tionship o f a single coil, the induced voltage satisfies the equation V-7 = ± M 21 where

cli^

dt

(17.1)

> 0 is the proportionality constant callcd the mutual inductance from coil 1 to coil

2, and the sign, here ±, depends on the relative winding directions o f the coils. Dot markings indi cate the relative winding directions. With reference to Figure 17.1, a dot is placed on coil 1 for reference; if the dot on coil 2 is in position A, the sign on equation 17.1 is +, and if the dot is in position B, the sign is - . A description of the general dot convention is presented in the next sec tion. The situation illustrated in Figure 17.1b is motivated by an extremely important magnetically coupled device called a transform er, which is used to transform voltages and currents from one level to another. In electric power systems, transformers are used to step up ac voltages from 10 kV at a generating station to over 240 kV for the purpose o f transmitting electric power efficiently over long distances. At a customer’s site, such as a home, transformers step these high voltage lev els down to 220 V or 110 V for safe, everyday uses. In addition, transformers have numerous uses in electronic systems, including (1) stepping ac voltages up or down, (2) isolating parts o f a cir cuit from dc voltages, and (3) providing impedance level changes to achieve maximum power transfer berween devices, and tuning circuits to achieve a resonant behavior at a particular frequenc)'. After the basic analysis methods are set forth, some examples will illustrate these uses.

886


2. MUTUAL INDUCTANCE AND THE DOT CONVENTION Experimental evidence demonstrates that if the two coils in Figure 17.1 are stationary, the induced voltage,

is proportional to the rate o f change o f i . e . , the induced voltage

\>2{t) = ±M2\

du{t) dt

as set forth in equation 17.1. Note, however, that coil 1 with inductance Z,, continues to act as an inductor for which v,(/) = Li

di\{t) dt

There are two effects present in the circuits o f Figure 17.1: an induced effect and the usual v-i relationship o f an inductor. Similarly, Figure 17.2 shows the reverse coupling to that o f Figure 17.1. For the circuit o f Figure 17.2, with the reference dot placed at the top o f coil 2, v, ( 0

= ± M ,2

d ilit)

(17.2)

dt

where + would be used if the dot on coil 1 were in position A and - if in position B. As in figure 17.1, the dots indicate the relative directions o f the windings o f the two coils. Also as before, the second coil continues to act as an inductor, for which

viit) —Lo

d ilit) dt

Secondary Coil 2

v,(t)

(l )

FIGURE 17.2. Coupling from coil 2 to coil 1 (winding directions not explicidy shown). As will be verified in a later section,

=

^ \ 2

>

Hence we designate the positive constant (17.3)

as the mutual inductance (in henries) o f the coupled inductors. Figure 17.3 shows a composite o f Figures 17.1a and 17.2 where currents are present in both coil 1 and coil 2. Four effects are now present in the circuit: two induced effects and the tw'O usual self inductance effects. Linearly superimposing the effects (superposition), we obtain the equations o f the mutually coupled inductors:


(It dt

887

(17.4a)

dt

(17.4b)

~ dt

Two questions remain: (i) When is the sign positive and when is the sign negative? (ii) How is the value o f M determined (experimentally)?

Primary + i,(t) 0

Secondary

A v,(t)

v,(t) B •

FIGURE 17.3. Coupled coils with current excitations present on primar}' (coil 1) and secondar}' (coil 2). The following rule, identified with equation numbers, governs the choice o f sign lor the induced voltage.

RULE FOR THE INDUCED VOLTAGE DROP DUE TO MUTUAL INDUCTANCE The voltage drop across one coil, from the dotted terminal to the undotted terminal, equals M times the derivative o f the current through

(17.5a)

the other coil, from the dotted terminal to the undotted terminal. Or, equivalently, The voltage drop across one coil, from the undotted terminal to the dotted terminal, equals M times the derivative o f the current

(17.5b)

through the other coil, from the undotted terminal to the dotted terminal.

With reference to Figure 17.3, if the dot is in position A, all signs are positive, whereas if the dot is in position B, the sign on M is negative. This rule gives the voltage drop due to the mutual inductance. To obtain the total voltage drop across an inductor that is coupled to another, one must add in the voltage drop induced by the self-inductance o f the individual coil, which depends on whether the labeling is consistent with the passive sign convention.


8 S8

EXA M PLE 17.1. For the configurations o f Figure 17.4, determine the pair of equations that spec ify the relationship betAveen the voltages and currents.

M

M

+ 1 +

2

+

•-

FIGURE 17.4. Two scenarios for setting up coil equations. S o l u t io n

First we consider Figure 17.4a. The voltage z^,(r) and the current /, (r) as well as the voltage V2 U) and the current ijit) satisfy the passive convention. For each coil acting alone,

cji^ (It for k = 1 ,2 . However, the voltage induced in coil 1 by the current in coil 2 is negative relative to the indicated polarity on v^{t) as per rule 17.5b, i.e., /-,(r) enters the dotted terminal so that-/'2W can be viewed as entering the undotted terminal. Fience Vt(/)= L |—- + M ------ ^ = L i — ‘ (If (it ' (li

(It

Using rule 17.5a, the same arguments apply to coil 2, in which case

U )= M -------+ L-) — = -M — + L-y ^ (It

~ (It

dt

~ (It

Now we consider Figure 17.4b. Here observe that neither pair (i^j, /,) nor {vj, h) satisfy the pas sive sign convention; hence

Vk(t) = -L^

(iik dt

for ^ = 1 ,2 . However, the voltage induced in coil 1 by the current /-,(/) satisfies rule 17.5a. Hence

‘

‘ dt

dt

On the other hand, the voltage induced in coil 2 by iAt) satisfies rule 17.5b. Hence

8S9


Exercise. For the configurations o f Figure 17.5, determine the pair o f equations that specify the relationship between the voltages and currents. M

M

+ V,

(a) F IG U R E 17.5. Two more scenarios for setting up coil equations. A N SW E R S: (a)

di] vAr) = - L —- + di

—

— and di

(//

di-, di

= -/V/— +

(If

^ ( b )

dt

, , , di] , di-> and v^(t) = - M —- + 1.-,^^ "

dt

~ dt

EX A M P L E 17.2. This example presents the procedures for marking the dots on an unmarked pair o f coupled inductors and for determining the value o f M. Consider the configuration o f Figure 17.6, in which a current source is exciting terminal A ot coil 1 o f the coupled inductors with unknown M. DVM

Also marked com, neg, low increasing current

F IG U R E 1 7 .6 . D iag ram fo r d e te rm in in g d o t p lacem en t and m u tu al in d u cta n ce.

890


Part I: Dot Placement Step 1. Place a dot on the terminal at which /j (r) enters. Observe that we have inserted a dot at terminal A.

Step 2. Apply an increasing current /j(r), i.e., a current for which

> q fo r all t. clt

For example, one could set /,(/) = 10/«(/) mA for 10 seconds. We know that /,(/) induces a volt age at the terminals o f coil 2 according to

(It

where M > 0. If we put the leads o f a voltmeter across the terminals C -D as suggested in Figure 17.6, the reading will either be Vjit) > 0 or v^{t) < 0. Suppose the reading is V2 {t) < 0. If we reverse the leads o f the voltmeter by putting them across the terminals D -C , the reading will have the opposite sign, i.e., v-,{t) > 0.

Step 3. Reconnect the voltmeter leads until V2 {t) > 0, i.e., the reading is positive. Place a dot on the terminal o f coil 2 for which the voltmeter lead is marked + {or "plus" or “high” or “pos'), i.e., at the terminal o f higher potential. For the situation described above, the dot would be placed at terminal D. However, the dots on coils 1 and 2 could be simultaneously moved to the opposite terminals o f each coil without chang ing the relative information they convey. (Problem 1 confirms this statement.)

Part 2: Determining M Again, \'2(0 v-){t) = ±M implies that

(If

where M > 0. Taking absolute values o f both sides o f this equation

M=

^2(0

clidt) dt

Exercise. In figure 17.6, /j(/) = 2tu{t) A and a voltage the placement o f the dots and the value o f M. AN SW ERS: Dots are at terminals A and

= \Ou{t) mV is measured. Determine j

or at B and D: M = ------ = 0.005 H.

Compared with the ramp o f Example 17.2, a more practical input signal and measurement scheme uses a triangular waveform, as displayed in Figure 17.7.

891

Chapter 17 * Magnetically Coupled Circuits and Transformers

E X A M PLE 17.3. Figure 17.7 shows a circuit and two waveforms, /,(r) and V2 {t), as might be dis played on an oscilloscope. Determine the placement o f a dot either at position C or at position D, and the value o f the mutual inductance. i,(A)

FIG URE 17.7 Circuit and waveforms for Example 17.3. So

lu t io n

For the time interval 0 < r < 0.5 msec, /,(/) is a ramp function and Vjit) is constant. The infor mation is similar to that given in Example 17.2 and the solution method is the same. First we place the dot at terminal A as the current enters A. The current /, is increasing, and ^2 is positive. We must now determine if the dot goes at terminal C or D. Since the current i^{t) enters the dotted terminal and is increasing over 0 < r < 0.5 msec, its derivative is positive over 0 < r < 0.5 msec. Also, the voltage iQ^t) is positive for 0 < r < 0.5 msec with the indicated polarities. Hence accord ing to rule 17.5a, the dot goes at terminal C and

di^ dt To determine M, consider that the measured values during 0 < r < 0.5 msec give Vj =

and

di^ldt= 1/0.0005 = 2000 A/sec. Thus, M = 2/2000 = 0.001 H.

EXA M PLE 17.4. In the circuit o f Figure 17.7, suppose the dot positions are at A and C. If /jW A, find Vjit).

= 2(1 So

lu t io n

In this case, — = 200t^"“^"//(/;. From Example 17.3, A/= 0.001 H. Hence

dt

v’9(/) = M — = 0 .2 e “ ’^'//(/) V

dt


Exercise, In the circuit o f Figure 17.7 with the dots in positions A and C, if /j(t) = 0.01 sin(1000?)«(^) A, find v~,{i) tor r > 0. ANSW ER: = 0.0 Tcosi 1000.^)/,u) A

The preceding treatment o f the mutual inductance has not referred to the physical construction o f the coils, although we have described procedures for measuring M when the coupled inductors are assumed to be enclosed in a sealed box. However, for designing a pair ot coupled inductors, or for a better understanding o f mutual inductance, one must relate the coil construction to the val ues o f Z.J, Ljy and M. A rigorous study o f this problem requires a background in field theory and magnetic circuits, which are covered in advanced texts or physics courses. Nevertheless, we set forth here a few basic properties with reference to Figure 17.8. M Coil 1

M

Coil 2 (b) FIGURE 17.8. (a) Coupled coils in close proximity, (b) Coils couplcd through magnetic core. 1.

In Figures 17.8a and b, the number o f turns for each o f coils 1 and 2, respectively, is TVj and Nj. Then the self- and mutual inductances have approximately the ratio

L, :L 2 :M = n } \NI\N^N2 2.

If two coils/inductors are placed in a nonmagnetic medium (e.g., air), bringing the inductors closer together increases the value o f M.

3.

If one inductor o f a pair is rotated, then a larger value o f M results when the axes o f the inductors are parallel to each other. The smallest value o f M occurs when the axes are per pendicular to each other.

4.

Changing the core on which the two inductors are wound from a nonmagnetic material (e.g., air, plastic) to a ferromagnetic material may increase the values o f Z,j, L-,, and A/by a factor o f several thousand.


As a final note, the development above presupposes linearity. If the rvvo inductors are placed in a nonmagnetic medium, this holds true. If the inductors are coupled through a ferromagnetic medi um (e.g., an iron core), then the linear relationships o f equation 17.4 hold only if both currents are sufficiently small that the magnetic medium avoids saturation, a phenomenon discussed in other courses or more advanced texts. Our investigations consider only the linear case.

3. DIFFERENTIAL EQUATION, LAPLACE TRANSFORM, AND PHASOR MODELS OF COUPLED INDUCTORS Figure 17.9 shows a pair o f coupled inductors. As developed in section 2, since each coil contin ues to act as an inductor, the terminal voltage depends on the derivative o f the current through the coil plus an additional induced voltage due to the changing current in the other coil. This led to the set o f differential equations 17.4, repeated below as 17.6, where the plus sign is used when the secondary dot is in position A and the minus sign is used when the secondary dot is in posi tion B.

dt

dt

dh dh V2(t) = ± M —^ + L 2 ^ dt dt

(17.6a)

(17.6b)

M

FIGURE 17.9 A pair of couplcd inductors. Assuming that there is no internal stored energ}^ in the coupled inductors at time r = 0, then the Laplace transform o f equations 17.6 yields

Vîs) =

± Msl. is)

Vîs) = ± Msl^ is) + L^sUs)

(17.7a) (17.7b)

Equations 17.7 represent the x-domain model o f the coupled inductors. Further, if one is con cerned w'ith the sinusoidal steady state, then replacing s by jui yields the following equations for analysis using the phasor method:

894


(17.8a)

V j = yojz.ji, ±yojyv/i2

(17.8b)

V 2 = ± y to M j +

For equations 17.7 and 17.8, the plus sign is used when the secondar)' dot is in position A and the negative sign otherwise. These three sets o f equations constitute the core o f the examples and analyses presented in the remainder o f the chapter. Throughout, the j-domain method is the pre ferred method, but each pair o f equations has its specific uses.

EXA M PLE 17 . 5 . A pair o f coupled inductors are connected in two different ways, as shown in Figure 17.10. Find the input impedances, inductances,

and -^,„2(^)> and the corresponding equivalent

and L^^2 '

(b) FIGURK 17.10 Impcdanccs and equivalent inductances of two series-connected inductors. (a)

+ Z2 + 2 ^ - (b)

+ >^2 - 2 ^ -

S o lution

For Figure 1 7 .10a, label the inductor voltages o f each inductor. Observe that

j and

with positive reference on the left side

enters the dotted terminals o f both coils. Hence

= I-jis). Direcdy applying equations 17.7, we obtain

în (-^) =

(-^) +

^L2 (^) =

(^ 1 +

= {L ^ + L 2 + 2 M )sIi„ {s) It follows that

im plying that

Ms) Ii,j (^) + {Ms +

) I i n (^)

= I^{s)


êq\ -

89S

+ ^2 = I^{s)

Similarly, for Figure 17.10b, with the same voltage definitions, we observe that again

= Ijis). Also note that /2W enters the undotted terminal. Applying equations 17.7, we obtain y-m is) = ^ L \(-^) + ^ L 2 (-^ ) = (^1^- - M s ) I i „ { s ) + ( - M s + U s ) l i „ { s )

= (L , + L o -2M ).9/ ,.„(5)

It follows that

Z,»2(^>=

ImU)

= [L\ + h - 2 « ) s =

implying that

Exercise. Suppose the circuits o f Figures 17.10a and b are connected in parallel. If Z,, = 40 mH, Z,2 = 60 mH, and A/ = 25 mH, find the equivalent inductance o f the parallel connection. AN SW ER: . r . 5 ml I

The preceding example implies that =

0 7 -9 )

This relationship suggests another way o f determining M and dot markings from measurements. If an instrument for measuring self-inductance is available, we can use the instrument to measure Z.^^j and L^^2} from equation 17.9, the difference is AM.

4. ANALYSIS OF COUPLED CIRCUITS WITH OPEN-CIRCUITED SECONDARY In the next example we compute the voltage across the secondary o f a coupled inductive circuit using a high-resistance voltmeter.

EXA M PLE 17.6. In the circuit o f Figure 17.11, assume that the meter resistance

is ver)' large

and looks like an open circuit to the secondary. The switch S is closed at r = 0. Find ij{t) and Vy{t).


8%

M

FIGURU 1 7 .11 A coLiplcd induccor circuit that might be lotind in a lab to determine dot position. So

lu t io n

Since the secondary looks like an open circuit, /-> = 0 and K, = Msly Hence if we find /j we can easily find K,. Applying KYL ro the primary loop, using equation 17.7a and the fact that /, = 0 produces

sms+R)

.V

s

+

Therefore

R /l(0 =

R

\ -e

Since V/^ = yW.s7, = Ms

£q

_

MZiq

_ iW/To/Li

s { L^s +R) ~ m s + R ) ~

It follows that

=

H(/)

h Note that u-y(t) > 0 by virtue o f the way the meter leads were connected to the secondary. Hence we conclude that the dot is indeed on the upper terminal of the secondary.

89'

Chapter 17 • Magnetically Coupled Circuits and IVansformers

Exeixise. Suppose in the circuit o f Example 17.6 is 1 msec, and the voltage

at

= 10 V, /^= 2 Q, the time constant o f the circuit

= 1 msec is 3679 V. Find L^, M , and the ratio M : Lîo see the

factor by which the small, safe voltage

is amplified when the switch is closed. The point here is that

when switching with inductors occurs, a small voltage ma)' produce a dangerously iiigh voltage. AN SW ERS: /., = 2 mi l. M = 2 H, and M : A, = 1000

An interesting application to older car ignition systems also uses an open-circuited secondary to produce a very high voltage from a small one to fire a spark plug.

EXA M PLE 17.7. Figure 17.12a shows an automobile ignition system found on older cars while figure 17.12b shows a simplified equivalent circuit model. Todays ignition systems use electronic switching. Specifically, the block with the condensor (capacitor) and ignition point is replaced by something referred to as an “ignition module.” The module contains a power transistor circuit to perform the switching action electronically, with the trigger timing typically actuated by a sensor that measures the position of the cam shaft cither optically or magnetically.

CONDENSOR SPARK PLUG

fL IGNITION f SWITCH

AL

i IGNITION POINT

+ ---- n

IGNIT10N(f COIL

o)

12V

DISTRIBUTOR

n---H I

t J

BATTERY (a) point

— +

I mF

primary

to spark plugs through a distributor

secondary

(b) F I G U R E 1 7 .1 2 (a) A n a u to m o b ile ig n itio n system , (b) S im p lified eq u ivalen t c ircu it m od el.

898


In Figure 17.12a, the ignition coil is a pair o f inductors wound on the same iron core, which cre ates a strong coupling between the coils. The prim ary coil is connected to the battery, while the secondary coil is connected to the spark plugs or load. The primary has a few hundred turns o f heavy wire, the secondary about 20,000 turns o f ver)' fine wire. When the ignition point (or con tact) opens by cam action, a voltage exceeding 20 ,0 0 0 V is induced across the secondary, causing the spark plug to fire. The generation o f a high voltage to cause the spark plug to fire is accom plished by a basic RLC circuit containing a switch that represents the point o f the ignition system. Since the secondary is open-circuited, it has no effect on the solution for the primary current. Let us do the analysis using the equivalent circuit model o f Figure 17.12b. Suppose that the switch has been closed for a long time. Accordingly, at r = 0~, we have /j = ElR= 12 A. Using the model for an initialized inductor given in Figure 14.18 results in the ^-domain equivalent circuit o f fig ure 17.13.

FIGURE 17.13 ^-Domain circuit for Example 17.7. The primary current is simply the net driving voltage divided by the total impedance in the series circuit o f Figure 17.13, i.e..

R

------

— t- L i —

R+

sC

+ sL

(17.10) .V“ + .9

‘

R Li

+L ,C

Suppose now that E = 12 V, /? = 2 Q, Z.j = 2 mH, M = 0.5 H,

L~,

= 130 H, and C = 5 lîF.

Substituting the given component values into equation 17.10 yields 6 (5 + 1 0 0 0 )

.v“ + 10()0i + 10^

6

(.V + 5 0 0 ) + 0 . 0 0 5 X 9 , 9 8 7 . 5

(.v + 5 0 0 )- + ( 9 ,9 8 7 .5 ^

(17.11)

Taking the inverse Laplace transform o f I^{s) yields /j(r) = 6^-500^ c o s (9 9 8 7 .5 f ) + 0.03^ ’-5 « « M n (9 9 9 8 7 .5 r ) s

cos(lO.OOOr)

(17.12a) (1 7 .1 2 b )

899


Having obtained

we calculate V2 {t) from the basic relationship o f equation 17.6, using the

plus sign for the dot positions: V2(/) = M

di^ (It

= 0.5 X 6

cos(10,000/) - 1

0

^

si n(10, 000/)

s - 3 0 ,0 0 0 e " ^ ^ ' sin(l 0,000/ ) V, for t > 0.

From this expression, the voltage

reaches a magnitude o f 3 0 ,0 0 0 V in about 157 psec (one-

fourth o f a cycle o f the oscillations). This voltage is high enough to cause the spark plug to fire. After the spark plug fires, the secondary is no longer an open circuit; the above methodology ceas es to hold during the firing o f the spark plug.

Exercise. Compute the energy stored in

just prior to the switch across the capacitor opening.

AN SW ER. 0.036 1

In going from equation 17.12a to 17.12b, we neglected the second term in 17.12a, retaining only the first. W ith the practical component values used in ignition circuits, this approximation is usu ally valid. In terms o f equation 17.11, the approximation is as follows:

R \

I

S H------- I

R

H----------------

2 lJ

2L , “

R

2 R S + ----- 5

1

+ ---------

L]

R

5 +

E R

—

X

L,C

(17.13)

1 R 1 s~ + — s + -----L, L]C

Applying the inverse Laplace transform to equation 17.13 yields

/ j(/ ) ai

—e R

cos[o)jt)u (t)

where

R

and

^d =

\ z .,c

I

r

V

U tJ

I ” ^/z;c

For the first few cycles o f oscillations, the value o f is nearly zero and the value o f the expo nential factor is nearly 1. Therefore, /j(/) may be further simplified to

')()()


£

/|(/) s — COS

(

r \

R

for small t > 0

Finally, we compute V2 {t) from equation 17.6: -1 . i \'o{t) = M — = M — ^sm dt R ■ Ju c

t

\

Hit)

for small r > 0. Thus, V-7(/) -

E 1 a A/----- f =

M

= — QE Z.|

(17.14)

approximates the maximum value o f t'2 >where

Q=

R

is the quality factor o f the series RL\ C circuit. (See Chapter 16.) Later we will show that M!L\ is approximately equal to the ratio o f the number o f turns o f the secondary to that o f the primary. Equation 17.14 is a simple formula for estimating the maximum voltage that occurs at the sec ondary. It shows that although the battery voltage is only 12 V, what appears at the secondary for a brief moment is very much higher due to the switching action. The voltage is stepped up due to two factors: (1) the Q o f the series RL^C circuit and (2) the turns ratio o f the ignition coil. From equation 17.14, a smaller R produces a higher voltage across the secondar)\ But a small R causes a larger current to flow in the primary circuit and therefore shortens the life o f the breaker point. In practice, when the engine is running, a resistance wire or an actual resistor is placed in the pri mary circuit to limit the amount o f current flow through the breaker point. The capacitor (con denser) serves a similar purpose— that o f protecting the breaker point by suppressing the arc that results when the point opens.

Exercise. The ignition circuit o f Figure 17.12b has £ = 12 V, Z,j = 1 mH, and C = 0.01 [J.F. If the total resistance in the primary circuit is /? = 8 Q and M/Zj = 50, estimate the maximum voltage appearing at the open-circuited secondary when the s\vitch opens. AN SW ER; Alniui l .s r k\'

')()!

Chapter 17 • Magnetically Coupled Circuits and Iransform ers

5. ANALYSIS OF COUPLED CIRCUITS WITH TERMINATED SECONDARY To begin this section, we would like to develop some general formulas that make the analysis of doubly terminated coupled inductors straightforward. By “doubly terminated” wc mean the con figuration o f Figure 17.14 where there is a source and source impedance connected to a load impcdance through a pair o f coupled inductors. EX A M PLE 17.8. Consider the circuit o f Figure 17.14. Show that (i) Z i„ {s)= L ^ s(ii)

M~s~ L'fS + Z-)

is) = Z, (.V) +

Z i„

(s) = Z, (.9) + L^s Z ,M

(iii) G ,2 = ^ =

L-)S + Zt

—M~ s + L.\Z~)

VI

' V\ (iv) the voltage gain Gj,| = — =

Z;„

(v) the overall voltage gain Gy = — =

Z.v

ZoM

An + 2 ]

(^1^2 - M -).v + L,Z2

M

FICU RE 17.14. Coupled inductor circuit. So

lu t io n

'1 V / Vi(^)

Part 1: Find Z,„(5) = -------

by two loop equations in the y-domain, converting to matrix form,

and solving. Step 1. For the primary loop, K, =

+ Msl^

Fo r the secondary loop

0 = ^2 + Z ,/ , = Msl^ + L^sl, + Z’. A =MsI^ + m j + Z p /2

902


In matrix notation, L ji’

Ms-

■ ■ /r

My

L^s + Z 2

■V,0

h

Step 2. Solve for /j and /j. Using the well-known formula for the inverse o f a 2 * 2 matrix, we have ■ /r

L,.v

/2

A/s

-1 p

Mv + Z-)

L|.v(L2.v + Z 2 ) - M~s~

L-) s + Z-)

—Ms

rv,

-M s

L^s

0

Hence

L 2 S + Z2 ^1 ------------------------------- T T M

LiS{L2S + Z 2 )-M \ s-

which implies that V'l

,

W^.v-

L 2 S + Z2

(17.15a)

(a) = Z, (.v) + Z,„(.v) = Z ,( 5 ) + L,5 -

(17.15b)

/|

Part 2: Obviously, then,

Also, we note that

-M L\(L2S

(17.16)

Z 2 ) - M~s

Part 3: Find C ^.2 = — •Using expression 17.16 and the fact that Vi = - •^2-^2’ “

Vi

^

V2 _ -Z y l.

^v2 -

7

7

Ki

-

V,

conclude that

Z-yM

Z^M

m L2S + Z 2 ) - M - s

( l , L 2 - M “ ).v + L,Z2

(17.17)

V'l

Part 4: Find Gj.i = — .B y voltage division Z;.,

Z,„ z,-„ + z ,

Part 5: Find

(17.18)

- Gv^Gv^7. G y = G ^ lG y 2 =

ZoM (1 7 .1 9 )

903


Exercise. In Example 17.8 find the currcnt gain ratio,

.

-M v

Note that the answer can be obtained directly be writing a loop equation for the secondar\'.

The next example applies the foregoing development to a specific case and at the same time expands the developed formulas to new cases.

EXA M PLE 17.9 Consider the circuit o f Figure 17.15, in which \’,„(/) = 60\/2 cos(10/) V , /?y = 10 Q

, = 5 Q,

C = 0.01 F, L, = 2 H, I , = 0.5 H, and M = 0.5 H. Find (i) Z.„(/10), (ii) V,(/IO), (iii) V.CylO), (iv) Ij(/10), (v) l 2(/10), and (vi) the average power Pi{j\0) delivered to the C-R^ load. M

—

L

V,

R,

©

-------------------•----------+ • 1

------------- •----------• +

FIGURE 17.15 Coupled inductor circuit. So

lu t io n

First we note that co = 10 rad/sec. AJso, using the effective value for the phasor voltage o f the source yields W = 60 Z. 0 " V. Thus all calculated voltage and current phasors are effective values. Step 1. F

i n

d

Observe that

Z2(yiO ) = — CylO +

10

}_ R,

O.ly + 0.1

7 = 5 - 5 ./ 1+ 7

From equation 17.15 and the fact that M = 0.5 H, we have

Z„,(ylO) = L|ylO- — - ^ . . = Uj\{) + 2 0 M - = 5 + y2() 0.5 X ylO + 5 - ^ 7

904

Chapter 17 • Magnetically Coupled Circuits and'Fratisformcrs

Step 2 . /vW Vj(yiO ). Using voltage division,

= i ,S l o T 4 .

= 5 4 ^ jl2 ^ 2 1 .8 7 Z 1 2 .5 .r

Step 3. /vW V 2(/'10). From equation 17.17, V 2 = G .oV , =

ZyM

V, =

- M ‘ )./10+L|Z2

0 .5 ( 5 - 5 ./ )

- ( 5 4 + j\ 2)= 1 8 - 6 7 = 1 8 .9 7 / 1 -1 8 .4 4 '' 7.5y + 10 - IO7

Step 4. /7W lT(ylO ). From equation 17.16,

u =

-M

-0.5 V, = ------- ^----- (54 + y l2 ) = - 2 . 4 - 1 . 2 7 = 2 .6 8 Z .- 1 5 3 .4 " 10 S ;■ (L ,L 2 - M - ) 7 lO + L ,Z 2 1 0 -— 92.57

Step 5. F/W Ij(yiO ). From Ohm’s law,

I. =

^1

-"54+12./

Z,„(ylO )

5 + y20

= 1 . 2 - 2.4./ = 2.683Z. - 63.44'^

Step 6 . Fitid the average power delivered to the load. The complex power delivered to the load is S(;10) = V.OlO) X [-I^l/lO)]* = (18 - 6;) X (2.4 - 1.2;) = 36 - ; 3 6 'Faking the real part yields an average power o f /’/(ylO) = 36 watts.

Sometimes the application o f the formulas developed in the previous rsvo examples, although straightforward, is not the simplest route to the answer. The following example is a case in point.

EXA M PLE 17.10. Find the steady-state components o f z/j(r) and V2 {t) at the frequency 1 rad/sec for the circuit o f Figure 17.16, in which ^^^^(t) = cos{t)u{t) V. Note that because a resistance is pres ent, the circuit responses will contain both a transient and a steady-state component. 1Q

M = 2H ------ O +

V

2F

V, IH

5H -

-o F I G U R E 1 7 .1 6 C irc u it fo r loo p analysis o f co u p led in d u ctor.

90S

Chapter 17 * Magnetically ('ouplcd Circuits aiulTransform ers

So

lu t io n

Since only the sceady-stare responses are required, we use rlie pliasor method and write two loop equations given that co =1 rad/sec and

= 1 Z. 0^^ V.

The two loop equations are

and 0 =—

12 + ./5I 2 + y2I,

= y2I, + y4.5l2

Putting this in matrix form, we have T

■|+y

0

jl

j l - ■*ll y-4.5

I2

II

I

■74.5

h

- 0 .5 + y'4.5

-j-

-P -

T

1

;4 .5 -

0

- 0 .5 + y4.5

.-j-.

It follows that

v,=v,,-i, =i-

J4.5

0. 5

- 0 .5 + ./4.5

0.5 - ./4.5

= 0 .1 1 0 4 3 Z .8 3 .6 6 "

-./•9

1

_9

- 0 .5 + J4.5

1 - ./9

-j2

v, = - - L i , = _ _ x jl “ j2 - 0 .5 + 74.5

1

= 0.22086Z . - 96.34"'

Observe that V , = -2 V , in which case = 0 .2 2 0 8 6 c o s (r - 9 6 .3 4 °) V and

= 0 .1 1 0 4 3 cos(; + 8 3 .6 6 °) V

Exercise. In Example 17.7, write the simultaneous equations in matrix form and then solve by Cramer’s rule. Which method is easier?

Example 17.10 is a drastic case contrived to bring up an underlying propert)': the dot markings for coupled inductors (/.,, A,, M) determine the ± sign in the equation relating v-^ to eli^ldt and the equation relating

to di-)ldt. No intrinsic relations ber^veen the polarities o f

and v-, are

conveyed by the dot convention. For most practical circuits, hoAvever, it is true that the voltage drops o f coupled inductors from the dotted terminal to the undotted terminal are in phase or nearly in phase.

906


As a practical example ro end this section, we analyze the circuitry typical o f the front end o f an AM radio receiver.

EXA M PLE 17 .1 1 . Figure 17.17a shows circuitry typical o f the front end o f an AM radio. Figure 17.17b shows a simplified RLCM model o f this circuitr}'. Here,

and the resistance

= 300 ^2

together represent the antenna. Typical parameter values might be /?2 = Rj„ = 14,7 k fl, Z,j = 50 nH, L-, = 2450 nH, M = 350 nH, and C = 104.5 pF. Find the transfer function

H {s) = rhen using the methods developed in Chapter 16, compute

|, Q,

coj, (JL)-,, and the maximum value and the bandwidth o f the magnitude response (i.e., the curve o f vs. to). antenna

RF Amplifier M •

coupling coils

■ tuning 1capacitor

SR X

in

'■ 6

C

(b)

(a)

R = 300 Q

'■ 6

L

c

/

R = 14.7 kO 2

= 98nH = 2612 pF (c)

FIG U RE 17.17 Circuits for Example 17.11. (a) The original circuit, (b) A simplified circuit model in which L^L-y = M~. (c) A design without couplcd inductors. So

lu t io n

The first step in the solution is to observe that L^L-, = Afi. Then from equation 17.19, the trans fer function is

M

Z-yM V ,{s)

where Z , = /?, and

Z ,„ + Z ,

(z,,L 2 - M “ ) j + L , Z 2

^m + ^l

Z^y = R-y I !C = ----- — . Also, from equation 17.15, s+ R-yC

^1

(1 7 .2 0 )


90

2 ,2 V

L2 S + Z 2

Loi + Z j

Substituting the expression for Z-^^{s) in equation 17.21 into equation 17.20, then dividing through by Z j and simplifying, we obtain

M

CR^Lo

H (s) =

s

_

_y2 ^ / _ _ _ _ h ^ _ L _ \ ^ ^ _ _ L

\CRiL2

R2 C )

L2 C

4 .5568x10^ 9

5^ + 1 .3 0 2 0 X 10^^ + f 6 .2497 x 1 0 ^ ( 1 7 . 2 2 ) '

Ks S

+ B^^.S + fO,„

Ks ^2 ^

^^ ^ 2

Q By inspection, then, co„, = 6.2497 X 10"^ rad/s, TU

W

= 13.02 x 10^ rad/sec,

=— P

and

= 3.5 ’, 0 = 4 8 ,' XI.

w (0, 2 =co,„ + 0.55(^ = 6 1 .8 5 x 1 0 ^ , 6 3 .1 5 x 1 0 ^ rad/s.

Instead o f the coupled inductors o f Figures 17.17a and b, the same

and

can be obtained

with the parallel resonant circuit o f Figure 17.17c. Following the design method described in Example 16.3, we find the required element values to be Z, = 98 nH and C = 2612 pF. However, the maximum voltage gain would have a much lower value, l^(;w,„)| = 14,700/(14,700 + 300) = 0.98, compared with 3.5 for the coupled circuit o f Figure 17.17a. The higher voltage gain achieved in Figure 17.17b can be explained by the concept o f nuiximum power transfer. A routine analysis would reveal that at to = to^^, the input impedance “seen” by the source is a pure resistance equal to the source resistance R^, i.e.. Z ,v ,0 ,„ ) =

Hence the sources impedance o f /?j sees a load

= 300 Q

equal to itself. Thus, maximum power has been

extracted from the source. Since the inductors and the capacitor together form a lossless coupling net\vork (see the homework problems), the same maximum power is transferred to the load resis tor Rj. In Figure 17.17c, the reflected load impedance at/o^^^ is not 300

and hence maximum

power is not transferred. Figure 17.17b contains two inductors and one capacitor. Accordingly, one would normally con sider it a third-order circuit. Yet the denominator o f equation 17.22 is only a second-degree poly nomial in s. This simplification is due to the condition M~ =

The significance o f this con

dition will be discussed in the next section. The general analysis given above is quite useful for complex circuits. However, for many applica tions the rwo coils have a common terminal, as shown in Figure 17.18a. Figures 17.18b and c show

908


rhe T- and Tt-equivalent circuits char often allow the application o f series-parallel techniques to simplify the analysis. With the common terminal the coupled inductors have only three accessible terminals instead o f four. Such an arrangement is called a three-terminal device. The reader might observe that if A/> Z, or M > L~,y one o f the three inductances in the T- or Tt-equivalent circuit may have a negative value. This negative inductance appears in a mathematical model and is not the inductance o f a physical component. Also, the equivalent circuits shown in Figures 17.18b and c are for the specific dot locations indicated in Figure 1 7 .18a. A change o f one dot location in Figure 17.18a will result in a change in the sign in front of M in Figures 17.18b and c.

2 L ,- M

L ,- M

L^-M

A = L,L^-M2 (3) (b) (c) FIGURE 17.18 (a) Coupled coils with a common terminal, (b) T-cquivalent circuit. (c) 7t-equivalent circuit. Verification o f the parameter values in the equivalent circuits in terms o l l j , Z.,, and M is left to the homework problems. We now illustrate their use with a simple equivalent inductance example.

EXA M PLE 17.12. In the circuit o f Figure 17.19, all initial conditions are zero. If i-^(^t) = u { t - 1) A, find the response, 0.5H

2H

10

IH 2.5H

3H

FIG URE 17.19 RLM circuit to be simplified using the T-cquivalent. SO L U T IO N The key is to find the input impedance seen by the current source. Apply the T-equivalent circuit to the coupled inductors o f Figure 17.18b. Using series-parallel techniques, the equivalent induc tance in parallel with the 1 H resistance is

25+ 1

T h e re fo re

v„,ît) =

~^h({t -\)

V.

= 2 H. T'hus

2.V -Hi

5 + 0.5

909

Chapter 17 * Magnetically Coupled Circuits and 'Iransformers

Exercises. 1. Repeat Example 17.12 with the dot on the 3 H inductor moved to the bottom. AN SW ER: Same as in example. 2. Find the input impedance

for the circuit o f Figure 17.20. In the circuit, k =

is called the coupling coefficient, to be studied in the next section. ANSW ER:

.

M

v ^ l^ 2

2.V+ 0.5

3H

3H

0.50

o— oFIGURE 17.20

6. COEFFICIENT OF COUPLING AND ENERGY CALCULATIONS The first part o f this section justifies our assumption that

One can justify that

A/j2 = A/t] = M by the principles o f magnetic circuits, but this is beyond the scope o f this text. Our justification stems from the physical property that a pair o f stationary coupled coils cannot generate average power. We will also show that the mutual inductance M has upper bound

’■> i.e., the mutual inductance can never exceed the geometric mean o f the self-inductances. Justification o f Myj = ^ 2 1 ~ ^ This property is a consequence o f the principles o f electromagnetic field theory, which are beyond the scope o f this text. To make our approach accessible to the beginning student, we build our jus tification on the passivit)' principle for inductors.

THE PASSIVITY PRINCIPLE FOR INDUCTORS A pair o f stationary coupled inductors is a passive system; i.e., they cannot generate energy and, hence, cannot deliver average power to any external network.

Our technique is a so-called proof by contradiction; i.e., to show that = ^ i\ = ‘s equiva lent to showing that if A/, 2 then the passivit}' principle is violated. As a first step, suppose that yV/p M-yy Then, instead o f having equations 17.6, the differential equations for the cou pled inductors with the standard dot locations must take the form

10


, du

d ii (17.23a)

„ =

+

~ dt

(17.23b)

~ dt

Let us apply /j = sin(/) and /, = cos(r) to the inductors. From equations 17.23, the terminal volt ages are

d

d

dt

" dt

Vj = Li — s in (/ )+ M p — c o s ( / ) = Li c o s ( / ) - M p s in (0

and

Vo = yUî — s i n ( 0 + c o s ( 0

~

~ dt

~ dt

= M ^ i c o s i t ) - Losin(t)

The total instantaneous power delivered to the coupled inductors is the sum o f the powers deliv ered to the inputs, i.e.,/>(r) = v^{t)i^{t) + V2 {t)i2 it)- Therefore

p{t) = yj/j + Vjio = To calculate

cos(f)sin(t) -

sin^(r) + M 21 cos^(r) - L-, sin(r)cos(t)

(17.24)

the average power delivered to the coupled inductors, we use the identities

sin(r) cos(f) = 0.5 sin(2f), sin^(r) = 0.5[1 - cos(2r)], and cos^(r) = 0.5[1 + cos(2/)]. It follows imme diately that the first and the last terms in equation 17.24 make no contribution to the terms involving

whereas

ô =

(17.25)

This result shows very clearly that if M j 2 > •^21’ excitations /, = sin(f) and ij = cos(r) will lead to a negative violating the passivity principle. Similarly, if M p < then the new (transposed) excitations Zj = cos(f) and

12

= sin(r) will again lead to a negative

violating the

passivity principle. Therefore, we conclude that A/p = J^ 2 r

W ith

= A/-7] = M, the average power, P^^^, is always zero for arbitrary sinusoidal excitations.

CALCULATION OF STORED ENERGY Having proved that A/j2 =

= M, we shall now show that there is a limit to the value o f M

that is attainable once I j and L 2 are specified. Again we use the passivity principle and the fact that stored energy is the integral o f the instantaneous power.


11

Consider the coupled inductors shown in Figure 17.9. The voltage-current relationships at the terminals are given by ^

v.(r) = L , 5 ± M &

(17.26a)

V2(/) = ± M ^ + t 2 ^

(17.26b)

and ^

with the upper sign (+) for the dot in position A and the lower sign ( -) for the dot in position B. O '

Let us assume that the inductor currents are initially zero (at r = 0). In this state, there is no ener gy stored in the system.

o Recall that the energy stored by any device over a time period [0, 7] is the integral o f the instan taneous power over the interval. For a pair o f coupled inductors with no initial stored energy, we

o

have

o

dt

o = i L „ f (r )+ i z „ | (r ) ± j [ M (/,f + ,-,5 ), o ^

^

4 L ,f ( r ) 4 i ,/ | ( r ) ± M j ;'3 '^ ( ,V ,) 2 “ 2 J/,(0)/2(0)

= U ,i f ( T ) + ^ L ^ il ( T ) ± M i t { T M T )

^

O '

(17.27)

where ^(/|/2) is the total derivative o f the product /j /2 and is equal to /jc/f^ + For the specific scenario above, we apply driving sources to the inductors to bring the currents up

VO

to /|(7) = /j and i2{1) = /j . Then, the energy delivered to the inductors during the time interval (0, 7) is, by equation 17.27,

W (J) = ^ W \

± ^ hh

(17.28a)

A couple of things about this result are worth noting: 1.

The final integral in equation 17.27 depends only on the final values, which in the case of equation 17.28a are 7j and /j . The exact waveforms o f /j(f) and t2 it) during

0

< t< T

are immaterial. 2.

The energy W{ T) delivered by the sources during 0 < f < T is not lost, but merely stored in the system.

To grasp property 2, we may adjust the sources so that the currents are brought back from 7j and / j at / = r to zero at some t = T > T. Then the energy delivered by the sources to the inductors during T < t < T may be calculated in a similar manner, to obtain

o

'

912

Chapter 17 * Magnetically Coupled Circuits and Iransformers

W iT') - \\\T) = - - L d } 7

9

“ “

‘ ~

(17.28b)

with the upper (-) sign for the dot in position A and the lower (+) sign For the dot in position B in Figure 17.9. Equation 17.28b is precisely the negative ol equation 17.28a. Thus, all o f the ener gy delivered by the sources during 0 < t < 7’ has been returned to the sources during T ^ t ^ T\ For this reason, the energy given by equation 17.27 is called the stored energy. Another way of recovering the stored energ)' is described in Problem 48. The physics o f the situation shows that the energ)- is stored in the magnetic field produced by the currents in the inductors. Upper Bound for M and the Coefficient o f Coupling The energ)' W[T) must be nonnegative for arbitrar)’ values o f /, and /-,. Otherwise, the inductors will ht g e n e r a t i n g during the time interval 0 < t< T , which would violate the passivity prin ciple. To ensure a nonnegative W{T) for all /] and I 2 , the values o f Zp L j, and M must satisfy the inequalit)' (17.29a) or

M ^ 7 ^ ,L 2

(17.29b)

To show this, we rewrite equation 17.27 in the form 2

W(T) = - l where x = ^\l^2

2

U2J

[h )

-

2 “

L|.v“ i 1 ± 2M.X + L ^

(17.30)

^ current ratio. Equation 17.30 shows that Wi'f) is negative whenever^(a;) is

negative. Now,y(,v) is a second-degree polynomial in a: with a positive coefficient for the

term.

Consequendy, the curv'e ofy(.v) vs. a: will be a parabola opening upward. From analytic geometr)', depending on the sign o f the discriminant D = {M~ - L^L-,), the curve may or may not intersect they{.v) = 0 axis, as illustrated in Figure 17.21. From the figure, it is obvious that if D > 0, there will be some current ratio that yields a negative^(a:) and hence a negative W{T), again violating the passivity principle. Therefore, D = {M~ - L^L^) < 0, which yields equation 17.29.

f(x)


13

The degree co which M approaches its upper bound yJL\l^ is expressed by a positive number called the coefficient o f coupling, defined as

M

k ^ - j= .

(17.31)

yjL\L2

From equations 17.30 and 17.31, (17.32) When /^ = 0, Af is also zero, and the inductors are uncoupled. When ^ = 1, the inductors have

uuity coupling, an idealized situation impossible to realize in practice.

E XA M PLE 17.13. For both situations o f Figure 17.9, suppose that H, /,(7) = /, = 2 A. and

= 20 H, A/ =

7) = I^= 4 A. Find:

1.

The coupling coefficient k.

2.

The stored energy- at r = T.

S o lu tio n

1.

= 5 H,

^

For both dot positions, k

^

= - 7- -------- =

,

=

0.8

.

V 5x20 2.

For the dot in position A in Figure 17.9,

Stored energy = = 0.5 X 5 X 2^ + 0.5 X 20 X 4^ + 8 X 2 X 4 = 234 joules For the dot in position B, Stored energ)' = 0.5/.j/|“ +

- A/Zj/j

= 0.5 X 5 X 2^ + 0.5 X 20 X 4^ - 8 X 2 X 4 = 106 joules

Exercise. Repeat Example 17.13 with Z,j = 4 H, AN SW ERS: in random order: 112, 0.~5. 16

= 16 H, A/ = 6 Fi, 7j = 4 A, and 7, = - 2 A.

914


EX A M PLE 17.1 4 . In the circuit o f Figure 17.22, /j(7) = /j= 6 A. Find the minimum value o f the stored energy and the corresponding value o f /2(7) = A. M = 3H 9H 4H

FIGURE 17.22 Coupled inductors for calculating the stored energy in Example 17.14. S o lu tio n

From equation 17.27, W = 0 .5 L ,/ f+ 0 .5 L 2 / | -M / ,/ 2 = 0.5 X 4 X 36 + 0.5 X 9 X /2^ - 3 X 6 X /2 = 4 .5 / 2“ - 18/2 + 7 2 = /(/2 ) Following the standard method in calculus for finding the maximum and minimum, we set

dW Idlj to zero and solve for Ij-

d l2

clh

This yields 12 = 2 A, and the corresponding minimum stored energy

= 18 - 36 + 72 = 54 J.

Exercise. Repeat Example 17.14 for /j = 12 A. AN SW FRS: 4 A. 216 1

7. IDEAL TRANSFORMERS Developing the Equations for an Ideal Transformer Two coupled dijfereutial equations containing three parameters Z.j, Z<2, and M characterize the cou pled coils of Figure 17.9. By imposing two idealized conditions on these parameters, the pair o f differential equations can be approximated by a pair o f algebraic equations. Idealization 1: The coupled inductors have unity coupling; i.e., M~ =

and the cou

pling coefficient ^ = 1.

Effect o f Idealization 1: With ^ = 1, the coupled coils have the voltage transform ation property

chapter 17 * Magnetically Coupled Circuits and Transfonners

915

where a\ s 2i constant and both v-^ and V2 are the voltage dropsfrom dotted {position A) to

undotted terminals o f the coils, as per figure 17.9. To derive the condition of equation 17.33, note that the constraint

= ZjZ2 implies that Lj/M

= M /Zj. Let the ratio M Then Zj = oM and M = aLj. Substitute these relationships into equations 17.7. This leads to the following sequence of equalities:

vi(t) = Li — +M —^ = aM—- + aL2—^ = a M—- + L2—- =av2(t) dt

dt

dt

dt

dt

d t}

Although the idealization o f a unity coupling coefficient is not achievable in practice, coupling coefficients near unity are achievable by winding the turns o f two inductors very closely together so that n^ l y all the flux that Unks one coil also links the other coil. The constant a = L^/M = M/L^ = V V Z ^ in equation 17.33 reduces to the ratio of the numbers of turns, denoted by N-^ and N 2 , of the two coils, i.e., a = //j/A^2* Thus a is called the turns ratio. When ^ = 1, it is possi ble to show that

W ith unity coupled coils, equation 17.33 indicates that the voltages v^{t) and V2 {t), both from dot

ted {position A o f Figure 17.9) to undotted terminals, always have the same polarity. W ith coupling less than unity, it is possible for v-^ and V2 to have opposite polarities at some time instants, as shown in Example 17-10. Idealization 2: In addition to unity coupling, the coupled coils have infinite mutual and self-inductances.

O

Effect o f Idealizations 1 and 2: W ith k= \ and idealization 2, the pair o f coils has the cur rent transformation property

i2 {t) = -ai^{t) where both / j and

(17.34)

are the currents entering the dotted terminals of the coils, as per Figure

17.9, dot in position A. To derive equation 17.34, we again make use of the unity coupling condition L^/M = M/L 2 = a^ and rewrite equations 17.7 as

d(Ly.

.

dt\M ^

^

-------- ^/l + /9

(17.35a)

916


Reasonably assuming that the voltage vît) is bounded, i.e., there is some finite constant that |i'|(f)|s for all t, then

lim

M-*oo

such

\M

A similar relationship holds for (17.35b)

V2

where if |y2(/)|^

for all t. lim

L^-*o In other words, as Zj “ ►<», Z2

<», and Af -► <», equations 17.35a and 17.35b each lead to

^ (a ii + / 2 ) - 0

(17.36a)

ai^{t) + i2 {t) = C

(17.36b)

whose solution is

for some constant C. Now assume that the coils are unenergized prior to the application o f exci tations; i.e., at some time f = Zq in the past, /j(Zq) =

which must be true for any real cir

cuit. It follows that the constant C in equation 17.36b is zero, and consequently /2(^) = -^^|(^)We remark that this derivation is valid for non-dc voltages and currents. The negative sign in equa tion 17.33 implies that at any time, if a current enters one coil at the dotted terminal, then the current in the other coil must leave the dotted terminal. The condition o f infinite inductances (Z,p Lj, M) is another idealization that is not realizable in practice. However, this condition can be approximatedhy using a magnetic material with very high permeability as the common core for the two coils.

r\

917

Chapter 17 * Magnetically Coupled Circuits anti Transformers

THE IDEAL TRANSFORMER Two coupled coils satisfying the relationships

V’2 (0

N2

/tCO

Ni

a

are said to be an ideal transformer, as shown in Figure 17.23a.

FIG U RE 17.23 Symbol and defining equations for an ideal transformer, (a) Both currents enter the dotted terminals, (b)

enters the dotted terminal and

leaves the dotted terminal.

In Figure 17.23 two vertical bars serve as a reminder o f the presence o f a ferromagnetic core in the physical device. The word “ideal” may or may not appear in the schematic diagram. Again, the mathematical model o f an ideal transformer depends only on the turns ratio a : 1 and the relative dot positions. To avoid the negative sign in the current relationship, an alternative labeling o f volt ages and currents as shown in Figure 17.23b may be used. The subscript p stands for the primary coil, which is connected to a power source, and s for the secondary coil, which is connected to a load. Note that

is entering at the dotted terminal and

is leaving the dotted terminal. The nota

tion o f Figure 17.23b is more commonly used in the study o f electric power flow. One important simplification resulting from the idealizations is that an ideal transformer is char acterized by two algebraic cc[\.\7iX.\ons in terms o f its terminal voltages and currents through a sin gle parameter a, the turns ratio. This is to be contrasted with a pair o f coupled coils, characterized by two differential equations containing three parameters, Z,p L-,, and M.

018

Chapter 17 • Magnetically Coupled Circuits and Iransformers

EXA M PLE 17 .1 5 . For the circuit o f Figure 17.24, suppose /?^ = 10 0.,a = 0.1,/ ?^ = 1000 Q, and

v^{t) = 20 cos(300;rr) V. Find /?,,

/j(r),

R

and /-,(/).

i,(t)

ijlt)

a: 1

+

FIG URE 17.24, Doubly terminated circuit for Example 17.15. SO LU TIO N From the definition o f resistance, „

V ,(0

a V 2 (/ )

= —— = —

'l(')

2 V’2 ( 0

2n

n n .

in n n

,n

= a — ----- = a Ri = 0.01 x 1000 = 10 Q

-'2 (')

Using voltage division,

R

V](/) = ------ -— /?_, + /?, •

= — X 20cos(300> t/) = 1 0 co s(3 0 0 ;r/) V 2

It follows that

vAr)

10cos(300;rr)

/,(/) = -^— ----------------------- = cos(300;r/) A

10

Using the ideal transformer equations,

viU) =

a

= 100co s(3 0 0 ;r/) V and /,(/) =

Exercises. 1. If in Example 17.15

= -0 .1 cos(300;rr) A

= 0, find v^{t), /j(f), V2 {t), , and

AN SW ERS. 10 i l , 20 cos(300.t/) V, 2 cos(300.Tr) A, 200 ôs{Mnnt) V, - 0 .2 cos(300.t/) A 2. If in Example \7.\5 R^= 10 ^2 and = 400 0 find /?,, /j(r), V2 {t), and ijit). A N SW ERS. 40 £2. 16 cos(300.t/) V, O.-i cos(300.Tr) A. U)0 a)s(3()0;r/) V,-0.04cos(300.t/) A 3. In Example 17.15, find the average power absorbed by R^. ANSW ER. 5 watts

919


The next example generalizes Example 17.15. E X A M P L E 17.1 6 . For the circuit o f Figure 17.25, find input impedance Z-JS) and then the transfer functions

V,(s)

V,(s)

ViCs)

Vj(.s)

a:1

— <— • + Z,(s)

: ideal

FIG URE 17.25 Doubly terminated circuit for Example 17.16. SO L U T IO N Using the definition o f an impedance and the properties o f the ideal transformer.

^ ^ ^ V^(s) aVjis) 2 . Z:,,{s) = —— = — = a — ---= a ZAs) /|Cv) J i i s ) -hAs) ^

Therefore, by voltage division,

c rz ^ is)

Vi„(s)

Z,(>v) + Zi„{s)

z,{s) + a -Z ^s )

Further, by the properties o f the ideal transformer,

V|(5)

a

in which case

Vsis)

Z ,(s) + crZ L (s)

Finally,

, , , ( , ) . Z2<£) =

= ----------=1^^--------Z i,(.?)V j(s)

Zi_U)

-

Z,(.s) + a-Z^(s)

920


Exercises. 1. Repeat the derivation

Z-^p) in Example 17.16 For the case where the dot on the

secondary is at the bottom. ANSW ER: Z J s ) = a"Z,[s) and Z^{s) = R + L^. Find the frequency oJq at

2. In Example 17.16, suppose Z/ (.v) =

which the primary is resonant, i.e., the source sees a pure resistance. ANSW ER.

a J lc

3. In Example 17.16, suppose a = 2, Zîs)

Zp) = 10 12, and ^

Find/jW and/j(r).

=— • ^

ANSW ER: — ' s + 0.4

The calculation o f

in Example 17.16 is so important (especially in maximum power trans

fer calculations) that we call it the im pedance transform ation property o f an ideal transformer.

IMPEDANCE TRANSFORMATION PROPERTY If the secondary o f an ideal transformer is terminated in a load impedance Zj{s), then the impedance looking into the primary is

Z Js ) = a^~Z^{s)

(17.37)

where a is the turns ratio taken in the direction from source to load. (See Figure 17.26; the dots are not marked on the figure because their positions are immaterial for this application.)

Chapter 17 • Magnetically Coupled Circuits aiul Transformers

921

10.V

Exercise. In Figure 17.26, suppose n = 0.1 and

Find

ANSW ER. .s- + 10

E X A M PLE 17.17. For the circuit o f Figure 17.25, find the instantaneous power delivered by the ideal transformer and the energ)' stored by the ideal transformer. SO L U T IO N With conversion o f the voltages and currents to the time domain, the power delivered to ideal transformer is by definition

pit) = V j/|

i't

+ v’o/^ = (c /i'-)) — =- +

~ \

= 0

a}

Hence, considered as a single unit, an ideal transformer neither generates nor consumes instanta neous power; whatever instantaneous power is received at one side must transfer to the other side. Furthermore, since p{t) is identically zero, its integral with respect to t is also zero. Thus, an ideal transformer cannot store any energy. In summary, the instantaneous power dehvered to one side o f an ideal transformer is transferred to whatever loads the other side, and as a result no energy is stored in the ideal transformer.

Exercises. 1. In Figure 17.26 suppose the primar)' o f the transformer has <7 = 0.1 and is connect ed to a source having = 10 Q. Find Zj{s) for maximum power transfer to the load. ANSWHR. ZÛ) = 1 kLl 2. In Figure 17.26 suppose the primary o f the transformer again has a = 0.1 and is connected to a source having Z^^(/100) = 10 + J5 0 Cl. Find ZîJlOO) for maximum power transfer to the load. AN SW ER. Z^(/100) = 1000 -y '5000 LI

Limitations o f the Idealization Although mathematically direct, the derivation o f equation 17.34 does not provide insight into the limits o f practical transformer design and operation. Let us reconsider the second idealization using the 5-domain coupled transformer equations, Vîs) = L^slîs) + Mslîs)


922

Divide rhe first equation by Ms and the second equation by L 2 S. To observe the frequenc)'-dependent behavior, set s =j(x) to obtain

Mjw

L 2 J (0

= ^ I \ i j ( o ) + I- , ( j co ) = a I i { j c o ) + 12 (j co )

M

= - ^ /1 (yw ) + / 2 (7 w ) = ^//1 (yw ) + / 2 ( )

Lj

Practical transformers have rsvo customary properties: (i) the frequency content o f the voltages and currents are band limited, meaning that the transformer is guaranteed to operate only over a restricted frequency range, 0 <

^ O) :S

ages are specified, i.e., |Vj(/to)| :s

that docs not include dc; and (ii) maximum volt

and |K,(/a))| rs

Under these practical conditions we

see that it is necessar)' to make M and L-> (and since a = frequency range

^ co :s

I M,

also) sufficiendy large over the

so that

Viij(o) Mco

0 and

V2 O )

Li(o

sO

Thus we will achieve the current transformation property /|{/(jd) =

over the frequency

range o f interest provided the voltage magnitudes are less than Som e Practical Applications o f Ideal Transformers We now indicate some practical applications o f ideal transformer properties.

E X A M PLE 17 .1 8 . In Figure 17.27 an ideal transformer steps down the voltage o f a 2400 source to pow'er 10 incandescent lamps in parallel, each drawing 0.5 A. (a)

Find the voltage (magnitude) across each lamp, and

(b)

find the current (magnitude) delivered by the source.

transformer source 2400 V

rms

0

load

10

1760

turns

88 turns

10 incandescent lamps represented by resistors

FIGURE 17.27 A transformer is used to step down a source voltage to meet the lamp specs. S olution Since only magnitudes are involved in this problem, dot positions on the transformer and refer ence directions for voltages and currents arc immaterial. The turns ratio is a =1760/88, Since 2400 =

the voltage across each lamp is

923


88

^’load

1,760

2 .4 0 0 = 120 V,

Since 10 lamps use 0.5 A each, the total current used is 5 A. Using the current transformation property o f the ideal transformer, /.V

1,760

X 50 = 0.25 A

Exercise. In the circuit o f Figure 17.26, represent each lamp by a 200 H resistance. Suppose the source is limited to delivering a maximum o f 3 kVA. How many lamps in parallel can be con nected to the secondary o f the transformer? ANSWER: 41

E X A M PLE 17.19. Figure 17.28 shows a simplified model o f an audio amplifier containing an ideal transformer. The input voltage is 1

at 2 kHz. T he load is a loudspeaker, represented by

z 4 Q resistance. (a)

Find the average power delivered to the 4

load if it is connected directly to the ampli

fier (i.e., with the transformer removed and the resistor connected across A-B). (b)

With the transformer and load connected as per Figure 17.28, with turns ratio a= 5, find the average power delivered to the load.

(c)

If the turns ratio a is adjustable, what value allows maximum power transfer to the load? W hat is the value o f the maximum power?

a:1

speaker FIGURE 17.28 An ideal transformer used for maximum power transfer. So

lu t io n

(a) No transformer and resistor connected across A-B. First observe that the magnitude o f the current through the 4 Q resistor is

200 4Q

9 0 0 -^ 4

= 0 .2 2 1 2 A

T h e re fo re , the average pow er absorbed is

/^4fi = (0.2212)‘ x4 = 0.1958W


‘) 2 i

(b) Transformer reconnected, a - b . K\ the primary o f the transformer, Z^„. = //2 x 4 = 100 From the voltage divider formula, the voltage across the primary is

100 pri

(200|V ,„|)=20 V

900+100

T he voltage across the secondary is = 351=4v 5 Therefore, 4“ — = 4 W

(c) For the maximum power transfer, the turns ratio a should match the secondary impedance to that o f the primary; i.e., the 4

resistance reflected back to the primary should be 900

to match

the internal resistance o f the amplifier. Hence 900 = W ith

X

4

= 15, the reflected impedance is 900 Q. and 100 15

^ = 15 |= 100 V, meaning that

f A ^ = 6.667 V. Thus

--------= 1 1 .1 1 W.

Exercise. Repeat all parts o f Example 17.19 for a loudspeaker with a resistance o f 16 ohms. ANSW ERS: i).-(,28 W.

W, 11.11 W

8. MODELS FOR PRACTICAL TRANSFORMERS The ideal dc voltage source was one o f the first basic circuit elements studied in this text. “Ideal” means that it maintains a constant voltage for arbitrary loads connected across its terminals. Such a voltage source does not exist in the real world. However, because an ideal dc source in series with a resistance will approximately represent practical sources such as a battery, its use is very impor tant. Similarly, although an ideal transformer does not exist in the real world, it can approximate ly represent a practical transformer with the addition o f some other ideal circuit elements.

92 S


Because o f the ferromagnetic core used in its construction, a practical transformer is inherently a nonlinear device. Nevertheless, a first level o f approximation represents a practical transformer, by the circuit o f either Figure 17.29b or Figure 17.29c. Figure 17.29a is ideal, whereas Figure 17.29b contains extra inductances to account for leakage fluxes and other magnetic phenomena. These inductances give rise to a finite usable bandwidth for a practical transformer. Figure 17.29c shows additional resistances to account for internal power losses. Thus the ideal transform in conjunc tion with other circuit elements can be used to approximate a real transformer. Specifically, trans formers do not work for dc or for very high-frequency signals. The circuit model o f Figure 17.29b makes this behavior clear: the inductance and

causes a short circuit at dc, and the inductances

produce open circuits at high frequencies. A second level o f approximation would begin to

account for nonlinearities, a topic left for more advanced courses.

N ,:N , +

(a)

-I-

N.

V.

N,

N.

ideal N ,:N ,

L» /Y Y \ -h

(b)

+

V,

ideal

• L.

R

+ V,

Ideal F IG IJR F 17.29 Linear transformer models, (a) Ideal case, (b) Extra inductances to account for leak age fluxes and other magnetic phenomena, (c) Extra resistances to account for internal power losses.

In Figures 17.29b and c, the parameters

and

are called primary and secondary’ winding leak

age inductances, respectively. They are due to that part o f the magnetic flux that links w'ith one winding but not the other.

is called the magnetizing inductance and is due to that part o f the

magnetic flux that links both windings. Practically speaking,

and

926


The model o f Figure 17.29c incorporates the resistances

and

which are the resistances o f the

primary and secondary windings, respectively. R^^ accounts for the power loss in the iron core due

L^, and

to hysteresis and eddy currents. These resistances as well as the inductances

can be

calculated from knowledge o f the physical layout and materials properties used in the construc tion o f the transformer. This task usually requires a large set o f design formulas and (empirical) charts. After a transformer is constructed, these parameters can be experimentally determined. Books on ac machinery set forth such experimental techniques. From a circuits perspective, once the model parameters in Figure 17.30 are known, the steadystate analysis o f a circuit containing practical transformers requires no more than the phasor method or the Laplace transform method for its analysis. l\vo examples will now be given. Example 17.20 shows the effect these parameters have on frequency response.

E XA M PLE 17.2 0 . Example 17.19 considered an audio amplifier with an ideal coupling trans former. Figure 17.30 shows the same amplifier with a practical transformer in place o f the ideal one. O ur objective is to illustrate the effect a practical transformer has on the power delivered to the 4 Q. speaker as a function o f frequency. We assume our turns ratio ^ = 15, which produced maximum power transfer in Example 17.19. We use the model o f Figure 17.29c with (equivalent winding resistance), ^ ^ = 8 mH (effective leakage inductance), and

= 40 Q

1 M ii (core loss),

= 1 H (magnetizing inductance). For simplicit)^ we set Z,^= /?^= 0 in the model o f Figure

17.29c.

Lp= 8mH 1 5 :1

40

FIGURE 17.30 Audio amplifier with practical transformer. S o l u t io n

Our analysis begins with the writing o f two loop equations in terms o f /,(;) and l 2 {s):

-20()K „ = [ « , + /? . + L,,^ + " M

‘

N ote that since R^^^= 1 M^^ and

"

] /, - - V V L / , R ,„ ^ L ,„ s } R,„ + L,„s -

= 1 H, j

1 -I-

y

R,„

92-7


for s =yto and w in the audio frequency range 0 to 2 ;r x 20 krad/sec. Hence

- 2 0 W ,„ .( R , . For the second loop, again neglecting

we have

In matrix form - 2 0 0 V/,,-

(/?! + Rp + {Lp +

0

L„jS

-L,„s

/l1

{R^q + L,„s)

From Cramers rule - 2 0 0 L„, 5

h =

LpL,„S^ +[{R\+ Rp)L,„ + Req{i-p +

■Vin

+ (Ri + f
Our purpose is to show how average power to the load varies as a function o f frequency in a prac tical transformer. To this end, for v,„(/) = yj2cos(cot) V, the power delivered to the load is the power delivered to the primary o f the ideal transformer, which is /> ,> ) = «,,|/2(/“ )P Recall that the effective value o f the input voltage is 1 V. To obtain the desired plot o f P iif) (dis played in Figure 17.31), consider the following MATLAB code. »R1 = 900; Rp = 40; Lm = 1; Lp = 0.016; »Req = 900; »num = [-200 0]; »den = [Lp*Lm ((Rl+Rp)*Lm +Req*(Lp+Lm )) (Rl+Rp)*R eq]; »w = linspace(0,2*pi*le4,500); »I2 = freqs(num,den,w); »I2mag = abs(I2); »PL = Req*I2mag.^2; »plot(w/(2*pi),PL) »grid »xlabel(‘Frequency in Hz’) »ylabel(‘Average Power to Load’)

928

Chapter 17 • Magnetically Coupled Circuits and Iransformcrs

TD

TO O

—I

o

4-'

$ 0 Q. q;

cn

(Z I—

01 > <

Frequency in Hz (a)

"O fD

o

OJ 5

o

Cl.

0)

cn

2 oj <

Frequency in Hz (b) FIG URE 17.31 Plot of the power delivered to the speaker load as a function of frequency in Hz. (a) Overall response, showing sharp decrease with increasing frequency, (b) Low-frequency response, showing significant reduction in power transfer. The plot shows that the power is down from the peak by about 0.5 W at 4 5 0 0 Hz, and, at 10 kHz, the power delivered to the load is down about 2.5 W. Also note that a practical transformer does not operate at dc and at very low frequencies, as demonstrated in Figure 17.31b.

In the next example we consider voltage drops in commercial power lines using practical trans formers.

‘)2 9


EX A M PLE 17.21. Fhe pniccical transformer in Figure 17.32a is designed for operation at 60 Hz and 1100/220 V, i.e., ideally a 5:1 step-down transformer. Using the model o f Figure 17.29c, we represent the circuit in Figure 17.32b with the following parameters: winding resistance), age reactance),

= 0.002 1^2 (secondary winding resistance),

= 0 .050 H (the primar)^

= coZ,^ = 0.4 Q (primary leak

= (oZ,^ = 0 .016 ^2 (secondary leakage reactance), and

= 250 Q (mag

netization reactance). R^^îs very large and its effect is neglected. If the load draws 100 A at a power factor o f 0.6 lagging, i.e., I 2 = 100 Z. -53.13^^ A, and the load voltage is V , = 220 Z. 0^’ V, what is the magnitude o f the needed source voltage to achieve the desired load voltage-current values in the presence o f non-idealities o f the transformer?

ac High Voltage Source

Low Voltage Load

(a)

Load

5:1 ac High Voltage Source

©

2

ideal (b) FIGURE 17.32 Analysis of a circuit containing a practical transformer, (a) A practical transformer for stepping down ac voltage, (b) A circuit model utilizing an ideal transformer. So

lu t io n

Using phasors for all voltages and currents, we have Load voltage = V 2 = 220 Z.0° V Load current = I 9 = 100 Z .-53.13^ = 60 - ySO A (from cos(53.13°) = 0 .6 , the given power factor). Voltage drop across the impedance

R^ + jX^ = 0 .0 0 2

+ y'0.016 Q

= (60 - y 8 0 )(0 .0 0 2 +y0 .0 1 6 ) = 1.4 + >0.8 = 1 .6 1 2 Z 2 9 .7 ° V

930


Voltage across the secondary o f the ideal transformer = (1.4 + J0 .8 ) + 220 = 221.4 + yO.8 = 221.4^/0.207° V c^1 J’ lOO .. . Since — !- = --------= 5 (eiven turns ratio), we have yV2 220 Voltage across the primary o f the ideal transformer = 5 X 2 2 1 .4 ^ 0 .2 0 7 ° = 1 1 0 7 ^ 0 .2 0 7 ° V Current through the primary o f the ideal transformer = 0 .2 x ( 6 0 - ; 8 0 ) = 1 2 - y l 6 A Current through the magnetizing inductance:

=

1 107Z 0

^------- = 4 .4 2 8 Z. - 8 9.8° = 0 .015 - y 4.428 A J2 5 0

Current through the primary winding: = (12 - J 1 6 ) + (0.015 - J 4 .4 2 8 ) = 2 3 .7 ii- 5 9 .5 4 0 A Voltage drop across the impedance

+ jX^ = 0.05 + yO.4 Q.

= 2 3 .7 ^ -5 9 .5 4 ° x (0.05 + ;0 .4 ) = 9 .5 5 ^ 2 3 .3 4 ° V Therefore, the source voltage is 9.55Z -23.34° + 1 107Z .0.207° = 1 1 15 .8 z l0 .4 ° V Thus, the magnitude o f the source voltage is 1115.8 V, which is 15.8 V higher (to overcome the non-ideal effects) than what would be needed for an ideal transformer.

9. COUPLED INDUCTORS MODELED WITH AN IDEAL TRANSFORMER When a circuit contains coupled inductors, loop analysis is a natural way to analyze the circuit. One then solves the resulting simultaneous equations by any o f the techniques studied earlier. Although very general and systematic, such methods have extensive mathematical operations that obscure the essential physical properties o f the circuit. In this section, we shall present some mod els for a pair o f coupled inductors that utilize an ideal transformer. Since the three basic prop erties (voltage, current, and impedance transformations) o f an ideal transformer are relatively easy to comprehend, using such models in place o f a pair o f coupled inductors helps us to more easily understand the circuit behavior, without complicated mathematics.


931

As a first case, consider a pair o f inductors with unity coupling (i.e., k = 1, or

as

shown in Figure 17.33a. Figures 17.33b and 17.33c show two equivalent circuits, each consisting o f one inductor and one ideal transformer whose turns ratio depends on the values o f Z,j and Z,,.

k=1

1

1:

o+

+ V,

ideal (b)

FIGURE 17.33 Ideal transformer models for unity coupled inductors. (a) A pair of unity coupled inductors, (b) A model consisting of one inductor and one ideal trans former. (c) An alternative model. T he proof o f the equivalence is easy. The v-i relationships for the circuit o f Figure 17.33a are given by differential equations 17.6. We need only show that the circuits o f Figures 17.33b and 17.33c have the same v-i relationships. Consider Figure 17.33b first. In this case, A/= JL]L-> . Using the current transformation property o f the ideal transformer, we obtain

d Zo . ■ Vj = L — i\ + , — h —L\------ JL\L-t — = L i -------- h M — ' ' dt cit ^ dt ‘ di dt V Next, using the voltage transformation propert)',

^+U — = M — + U “ dt

dt

-

dt

These two equations are exacdy the same as equations 17.6. Hence, the circuits o f Figure 17.33a and Figure 17.33b are equivalent. A similar derivation proves the equivalence o f the circuits o f Figures 17.33a and 17.33c.


Exercises. 1. Verify the equivalence o f the circuits o f Figures 17.33a and 17.33c. 2. In Figures 17.34a and b, let Z, = 2.4 H, Z-2 = 6 mH, L j = 3 mH, and

= 0.2 H. Find

for each circuit using the equivalences in Figure 17.33. ANS WK KS . O.Ss, 0 . 1 6s.

k=1

k=1

z in

^ --

— > L

,

C

L

,

L

,

C

L

.

(a)

(a) FIGURE 17.34

E X A M PLE 17.2 2 . The circuit o f Figure 17.35a has a unity coupling coefficient. Find the band width, the center frequency, and the maximum voltage gain.

300 Q

« .©

50 nH

M = 350nH

2450 nH 104.5 pF (a)

soon

1 :7

14.7 kn


300 0 5 1 2 0 .5 pF + V

©

5 0 nH

V.

300 0

(0 FIGURE 17.35 Analysis of a unit)' coupled circuit, (a) A circuit containing unit)' coupled inductors, (b) An equivalent circuit utilizing ideal transformers, (c) Equivalent circuit after reflecting load impedances to primary. So

lu t io n

The coefFicient o f coupling is = M l -y/^1^2 = 3501 yjSO x 2 ,4 5 0 = 1. Replacing the coupled coils with the equivalent circuit o f Figure 17.33b yields the circuit o f Figure 17.35b. From the imped ance transformation property o f an ideal transformer, looking into the primary we see an imped ance 1/49 times the load impedance. Therefore, looking into the primary, we see a resistance o f 14,700/49 = 300 Q. in parallel with a capacitance o f 49 x 104.5 pF = 5120.5 pF. Figure 17.35c captures the new equivalent circuit. Th e band-pass characteristics o f H {s) =

Vl(s) Vsis)

follow the analysis done in Example 16.2. Consistent with the notation o f Example 16.2, we have /?= 300//300 = 150 0 , 1 = 50 nH, C = 5,120.5 pF

CO„,=Wp =

^

^

* = 62.5 X 10^ rad/sec V 5 0 x 10"^ X 5 ,1 2 0 .5 X 1 0 "*“

Q _ ___ _ ___________ RC

150 X 5 ,1 2 0 .5 x l 0 ” ' “

\H(/ I 'io)\ 'huav = — ID = 0 .5 From the voltage transformation propert)' o f an ideal transformer, only quantities affected for the transfer function

are the maximum values:

is simply 7 times Vy So the

‘)34


The above example illustrates how a simple substitution o f an equivalent model for coupled inductors can reduce a circuit to a simpler form for which the analysis is straightforward.

Exercise. Solve Example 17.22 again using the equivalent circuit o f Figure 17.33c. ANSW'HRS: I he same as given in Example 17.22, of course.

As mentioned earlier, a unity coupling coefficient is an ideal condition impossible to achieve in practice. It is desirable, therefore, to modify the models o f Figure 17.33 to account for a coupling coefficient k < \. The resulting equivalent circuits are shown in Figure 17.36.

M :L

M :L

■<— o +

I

k'L. ideal

ideal

(b)

(c)

o— + i.

O— +

L.:M

/YY\

■<— o +

V, k'L ideal

ideal

(d)

(e)

FIG U RF 17.36 Four different models for coupled coils with /'< 1, cach consisting of one ideal transformer and rwo self-inductances, (a) Coupled inductor with k < 1. (b) One model using an ideal transformer and two inductances, (c), (d), (e) Three alternative models.

Exercise. In Figures 17.37a and b, Zj = 2.4 H, Find Z -is) for the circuits. A N ^ W 'E R S . 1.6>-. O .S j

= 6 mH, L^ = 5 mFI,

= 1.6 H, and k =

—j = .

93

Chapter 17 • iMagnctically Coupled Circuits and Transformers

(a)

(b) FIGURE 17.37

The equivalent circuits of Figure 17.36 are derived as follows. Since

it is possible to subtract

a small inductance from Z,j such that the remaining inductance, Z,'|, satisfies the condition L ’

= M".

In other words, the new inductance, L' ^ together with L-, and M, forms a unity coupling system. Since Z,'j = MÎL) = k~L^, the induaance to be subtracted is equal to Zj - Z .'j = (1 -

This inductance

must be added back in series with Z,', to obtain a model for the original coupled inductors. The mod els o f Figures 17.36b and c result. Repe-ating the process on

yields the models o f Figures 17.36d and

e. A total o f four equivalent circuits is possible. Each consists o f two uncoupled inductors and one ideal transformer. Clearly, when k= 1, the models in Figure 17.36 reduce to those in Figure 17.33. As an application o f the models o f Figure 17.36, let us reconsider Example 17.10. and Vjit) at the frequency 1 rad/sec

E XA M PLE 17.23. Find the steady-state components o f

= cos{t)u{t) V. Note that because a resistance is pres

for the circuit o f Figure 17.38, in which

ent, the circuit responses will contain both a transient and a steady-state component. 1Q ----------O ----------+

M = 2H

•

(a) i

2 :5 + V,

2F

(b) F I G U R E 1 7 .3 8 Stead y -state analysis o f co u p led in d u cto rs u sing th e ideal tra n sfo rm er m od el.


S olu tion

1

,.2

M

For the circuit o f Figure 17.38a, k" = ------- = 0 .8 . Substituting the model o f Figure 17.36b L| Li for the coupled inductors in Figure 17.38a, we obtain the equivalent circuit o f Figure 17.38b. From the impedance transformation property o f the ideal transformer,

It follows that z . , , a ) = 0 .8 y , i z , ,, ,( ;) = 5

g

; j ^

= - 0 .0 8 8 9 ,Q

Using the voltage divider propert)', V,, = ---------yOX)889-----^ I + 0.2./ - yO.0889

^ 0 .0 8 8 3 4 5 Z - 96.34'^ V

Further, ^ ‘

0.2y 7 0 .0 8 8 9 1 + 0.2./ - 7X).0889

^ ^ ^ I0 4 2 Z 8 3 .6 6 ^ V

Using the ideal transformer properc)', V , = 2.5V^, = 0 .2 2 0 8 6 ^ - 96.34^’ V I'his result agrees with the solution given in Example 17.10.

SUMMARY This chapter has examined the phenomenon o f induced voltage in one inductor caused by a change o f current in another inductor. A new parameter called the tnutttal inductance (A/) between the coils was introduced. M was defined as a constant and is present in the coupled differential and 5-domain equation models o f the coupled inductors. As illustrated in the chapter, M can be determined experimentally. This treatment has avoided digressing into the study o f magnetic cir cuits, which in fact underlies a rigorous development o f the concepts in this chapter. From the cir cuit analysis perspective, this mathematical treatment is adequate. However, for a deeper under standing o f the physical phenomena, one must study the principles o f magnetic circuits. O f foremost importance in analyzing a circuit containing mutual inductance is the formulation o f correct time domain or frequency domain equations— in particular, the correct signs for the mutual terms. For this reason, considerable time was spent on the dot convention. Once the equa tions are correctly written, we may use any o f the techniques (loop or node analysis) studied ear lier to anal)'ze the circuit.


A proof o f M p =

= M was given that made use o f the passivit}' principle and some trigono

metric identities. With the establishment o f this equalit)', computing the energ)' stored in the cou pled inductors follows from simple integration. The expression for the stored energ)' and the passiyitv principle then led to an upper bound for the value ot the mutual inductance, namely M <

yj^ ^2 • 1 he coefficient of coupling was then defined as k = M

l .

An ideal transformer was defined as a device satisfying both the voltage transformation and cur rent transformation properties. For practical transformers, these two properties hold only approx imately— for example, over specified voltage and current ranges as well as over specified frequen cy ranges. Transformers have broad applicability, h'or example, in power engineering, transformers are used to step up or step down the voltages. In communication engineering, transformers are used to change a load impedance for the purpose o f maximum power transfer. Although ideal transformers can only be approximated in the real world, they nevertheless remain an important basic circuit element because a practical transformer or a pair o f coupled inductors can be modeled by an ideal transformer and some additional R, L, and/or C elements. The use o f such models simplifies many analysis problems and gives physical insight into the operation o f a circuit.

TERMS AND CONCEPTS CoefFicient o f coupling: usually denoted by /r, equal to M /yj ^1^2 • Coupled inductors (coils): rwo inductors having a mutual inductance M ^ 0. Current transformation property: for unity coupled inductors having infinite inductances, the ratio |/|(/‘) : /2(r)l is a constant equal to the turns ratio D ot convention: a commonly used marking scheme for determining the polarit)' o f induced volt ages. Energy stored in a pair o f coupled inductors: 0.5/.]/|“ +

± A//,/-, joules, with the +

sign for the case where both currents enter (or leave) dotted terminals and the —sign for the case where one current enters a dotted terminal and the other leaves the dotted ter minal. Ideal transformer: rwo network branches satisfying both the voltage transformation and current transformation properties exactly. Impedance transformation property: when the secondary o f an ideal transformer is terminated in an impedance Z(;), the input impedance across the primary is equal to {N^lN-,)^Z{s) = a^Z{s). Models for a pair of coupled inductors: a pair o f coupled inductors can be represented by one ideal transformer and one inductance for the case ^ = 1. For coupling coefficients k less than 1, the representation requires one ideal transformer and two self-inductances. Mutual inductance: a real number, usually denoted by M, that determines the induced voltage in one coil due to the change o f current in another coil.

r> Chapter 17 • Magnetically Coupled Circiiits and Transformers

938

71-equivalent for coupled inductors: if two inductors have one terminal in common, then the three-terminal coupled inductors are equivalent to three uncoupled induaors (one of which may have a negative inductance) connected in the 7t-form. Primary: The winding (coil) in a transformer that is connected to a power source. Secondary: The winding (coil) in a transformer that is connected to a load. T-equivalent for coupled inductors: if two inductors have one terminal in common, then the three-terminal coupled inductors are equivalent to three uncoupled inductors (one of which may have a negative inductance) connected in the T-form. Transformer: a practical device that satisfies approximately the voltage transformation and cur rent transformation properties. Unity coupling: coefficient of coupling k= I. Voltage transformation property: for unity coupled inductors, the ratio |i/j(/) : V2 {t)\ is a con stant equal to the turns ratio NÎNj =

r>

o

n o r>

939


3. For each circuit shown in Figure P i 7.3:

Problems

(a)

Express vj^t) and

in terms o f /^(r)

and i^{t) for the dot in position C.

DOT PLACEMENT, M, AND BASIC EQUATIONS

(b)

Obtain the ^-domain equations that contain the initial currents /'^(0“)and

1. For each o f the circuits in Figure P i 7.1,

/'^(0“) by taking the Laplace transform

determine the proper signs on the equations for coupled inductors:

o f the equations o f part (a). (c)

Repeat parts (a) and (b) with the dot moved to position D.

(It

V9 (r) = ± M

^

M = 2H

20

(It

± L9 ^

dt

~ dt

4H

3H

Observe that some o f the labelings do not con form to the passive sign convention. M

Figure P I7.3 M 4. For the circuit shown in Figure P17.4; (a) Express vj^t) and v^{t) in terms o f /^(^) and /^(r) for the dot in position C. (b) Obtain the ^-domain equations containing the initial /^(0~) and by taking the Laplace trans

M

form o f the equations o f part (a). (c) Repeat parts (a) and (b) with the dot moved to position D.

Figure P i7.1 2. For the circuit o f Figure P I7.2, A/ = 2 H; compute v^{t) when

= I t cos(20/)//(f) mA

for the dot in position A and then in position B. M

Figure P i7.4 5. In the circuit o f Figure PI 7.5, if/j(^) = Atu{t) A, a voltage = 40//(r) mV is observ'ed. Determine the placement o f the dots and the value o f M. Repeat if observed.

Figure P I 7.2

= -80u{t) mV is

940


(b)

Compute V2 {t) for /* > 0. M

i,(t)

■>

Figure P i7.5 6. Consider the laboratory setup o f Figure PI 7.6a. (a)

(b)

Figure P i7.7 ANSW1-:RS; (a) 4 H: (b)

V

The two waveforms, /j(r) and v-^{t), in Figures P I7.6b and c are shown as dis

8. Write three mesh equations in the time

played on an oscilloscope attached to

domain for the circuit shown in Figure P i 7.8.

the circuit o f Figure PI 7.6a. Determine

Be particularly careful about the signs o f

the placement o f the dots and the value

induced voltages due to the mutual inductance.

o f the mutual inductance.

Apply the rule given by equation 17.6 if you

Now suppose the signal source pro duces /jCr) = 2(1 -

have any doubt about the signs.

A. Find

D (a) i,(A)

C H E C K : Coefficients for derivative terms in

1> t (msec)

1\ / ^ 2

\

^

Kc

loop a, ( I , , - Zp -

loop b,

(-Z p - Z j, - A/), loop c,

-1 (b) v,(V) t (msec)

---- ► -2 (c)

COMPUTING EQUIVALENT INDUCTANCES, Z,.^(s), AND RESPONSES FOR SIMPLE CIRCUITS

Figure P I7.6 9. (a) 7. Consider the circuit shown in Figure P i 7.7.

to find the impulse and step respons

I'h e 2 H inductor is short-circuited. The cur

es.

rents in the coupled inductors for t> 0 are /,(r) = 2e~‘ A and i^{t) = e~^ A. (a)

Find the mutual inductance M (in henries).

Find the equivalent impedance o f the circuit o f Figure PI 7.9a. Use the result

(b)

Repeat part (a) for the circuit o f Figure P I 7.9b .

‘M l


M = 2H

j

T Y V -^ Y Y V SW

3H ^

v.(t)

6

M = 2.5H

30 n

2H

r m

IF

''.w ©

.

Figure P I7.11 (a)

AN SW ERS: (a)

.(t) /Y Y V — TYYV 3H ^

(D

^

= 1 H, (c) i^ J i) = U’-'uU)

A, (d) 1 rad/sec, (e) B

= 2 rad/sec, Q = 0.5

5H

M = 2.5 H 30 0

12. Consider the circuit shown in Figure 2H

/Y Y V

P17.12. (a)

Determine the input impedance and input admittance.

(b)


C H E C K : admittance poles at - 0 .5 and - 3 (b)

Determine the transfer function

Find rhe equivalent impedance o f the

H{s) =

circuit o f Figure PI 7.10 when (i)

the dot is in position A

(ii) the dot is in position B (b)

(c)

= 30(1 -

If

V, com

Use the result o f part (a) to find the

pute

using the “ilaplace” com

transfer function with the dot in posi

mand in MATLAB.

tion A. (c)

Vin(s)

0.5H

Use the result o f part (b) to find the impulse and step responses with the dot in position A.

(d)

If v^t) = 2 cos(2f) V, find the steady-

v ,Jt )

(n

0.1 H

state value o f /Y Y \

2H

Figure P I7.12

'y

.w o

IY Y \

13. Figure P I 7.13 shows the two ways to con 2H

Figure P i7.10

nect a pair o f coupled inductors in parallel. For each case, find 0---

M

o—

M

^•

►•

11. For the circuit shown in Figure P 17.11, find (a) the equivalent inductance, Z,^^, o f the coupled coils; (b) the transfer Rmction o f the circuit; (c) the step response; (d) the resonant frequenq' o f the circuit; and (e) the bandwidth and Q o f the circuit.

(a)

(b)

Figure P I 7.13

942


A N SW ERS; (a) (b) A/-)/(A, + / . . + 2/\/)

- 2A/);

dotted terminals connected together. If it is known that the primary has the larger induc tance, find Z,j, Ljy and M.

14. Find the input impedance o f each circuit in

A N SW ERS: 2.6 H. 0.5 H. 0.3 H

Figure P I 7.14. Hint: Make use o f each 17. Find

found in Problem 13.

for each circuit o f Figure P i 7.17.

Does the answer depend on the positions o f M

the dots? Z„(s)

(b)

(a)

Figure P I7.14 AN SW ERS: (a) Z,„(.v) =

RL.„s -cir

(a)

(b) Same as (a) with cilHcrcnt 15. For each o f the circuits in Figure P 17.15, R = 1 n , I j = 2 H, I , = 10 H, M = 2 H. For fig ure P I 7 . 15a, C = 0.8 mF; for figure P i 7 . 15b, C = 2 F. Find the values o f O) at which reso

(b)

nance occurs. For each case find

Figure P I7.17 i f

ANSW1{RS: (a) L\ -

ZJs)

Z Js )

*

M-

independent oi dor positions: L . + 7-3 independent ol dot positions

(a)

Figure P I7.15 18. Find the input admittance o f each o f the AN SW ERS: (a) 25 rad/sec, 0 12; (b) I rad/sec, Z j j i u ) ^ 0.5 Q

coupled inductor circuits in Figure P I7.18. M = 4H

16. A handy dandy henry counter is used to measure the inductance o f a pair o f coupled

OZ,„(s)

inductors in various configurations. After three experiments the results are as follows: equiva lent inductances are (i) 3.7 H after series-aiding connection; (ii) 2.5 H after series-opposing connection (dotted to dotted connection); and (iii) 121/250 H after parallel connection with

6H

8H (a)

943


A N SW ER: C.v-(. v -!-4) + .V+2 C = 1 / 6 F ; Z;„(.v) = Cv(.v + 2)

M = 2H

OZ,„(s)

22. For the circuit o f Figure PI 7.8, find 3H

4H

Zin(s) =

(b) when

Figure P I7.18

= /?^ =

= 1 n , Z,j = Z,-, = 2 H, and

A/ = 1 H. Hint: After writing loop equations with numbers, consider using the symbolic

AN SW ERS: (;,) — , (b) — 4.S i.v

toolbox in MATLAB (or the equivalent) to

19. Find the input impedance

o f the cir

evaluate determinants in a Cramers rule solu-

cuit shown in Figure P I 7.19. Does the answer

non.

depend on the positions o f the dots?

AN SW ER: Z,„(.v) = - ^

9.V- +6s+\

3.S- + 10.V + 3

GENERAL ANALYSIS OF CIRCUITS WITH COUPLED INDUCTORS 23. For the circuit shown in Figure P i 7.23,

= 1 AN SW ER;

(a)

M-

.v

V. If a dot is placed at position A, com

pute the zero-state response.

Li + Ljj (b)

independent o f the dot positions

If a dot is placed at position B, com pute the zero-state response. M= 1

20. Find the zero-state response input

to the

V for the circuit o f Figure

P17.20.

2H

2H

24. The switch S in the circuit o f Figure PI 7.24 is closed at f = 0. If

= 10 V, find

Figure P i 7.20 21. For the circuit shown in Figure P I 7.21, find

(j) and the value o f C (in F) that caus

es resonance to occur at

oj

= 2 rad/sec.

v.(t)

Figure P i7.24 25. If a 900 Q resistor is connected across the F igu re P i 7 .2 1

secondary o f the circuit o f Figure P I 7.24 and


28. For the circuit shown in Figure P i 7 .28,

= 10 V, find the voltage gain

compute the zero-state response for

V2(.v)

G ,{s) =

and then compute Vjit), t > 0. Use the formu

C H EC K :

(a) and (b).

mesh equations and solve by

Cramers rule.

(d)

9000.V

C^.=

= 30e-^',Ki) mA /^^(/) = 60(1 - e~^û{t) niA. Hint; Apply linearit)' to the answers to parts

las developed in Example 17.8 or, alternatively, write two

= 30u{t) mA

(a) (b) (c)

Suppose Vf^Q~) = 30 V and all other initial conditions are zero. Find the

( a + 500)(5 + 9000)

zero-input response.

26. Consider the circuit shown in Figure PI 7.26. (a)

Suppose the circuit is relaxed at r = 0~. If = \2n{t) V, find the time con stant o f the circuit and

(b)

Suppose

(c)

Suppose

0.625 H

-L

for / > 0.

= - \lu{-t) V; find v^{t)

<^10

0.1 F

0

for r > 0.

= - 6 u{-t) + Guit) V;

find V2 U) for /^> 0. Hint: Use lineari ty.

Figure P I7.28 AN SW ER rd): v^ Jt) = 1.6<---'//(r) - GAe-^'u{t)

\ 200 0

M = 0.5 H

—o +

vJt)

6

0.2 H'

v,(t)

29 . In the circuit shown in Figure P I7.29, all initial conditions are zero at r = 0~. (a)

5H —o

If

= 2u{t - 1) A, compute the

response,

Hint: Use the result

o f Problem 19.

Figure P i7.26 27. For the circuit shown in Figure P i 7.27,

(b)

Find /,(r).

(c)

Find /,(r)

i j t ) = 100«(r) niA. (a)

If a dot is placed at position A, com pute the zero-state response.

(b)

If a dot is placed at position B, com IH

pute the zero-state response. M = 0 .i H

o+ v Jt)

ANSWl-R: (b) /,(/) = 2(1 -

A

0.1 H B

1•

50 mF

Figure P i7.27 A N S W E R (b ):

'»)Ma-1 ) A

\0.2 H ''

= 0 .1

30 , Consider the circuit shown in Figure P I 7 .30 , having zero initial stored energ)'. Let /?! = 9 a

sin ( l()/)//(/) \’

(a) (b)

Find F in d Z J ^ ) = /? + Z .,W .

(c)

Find

945

Chapter 17 •Magnetically Coupled Circuits and Transformers

(d)

Y = .......... H = det(Y)/det(A)

F in d G „ 2 = ^

simple(H) (e)

Find Gy =

(f)

Find the response,

to

input

v j t ) = \6 .8 le^^û{t) V.

(b)

(^ + 1X^ + 3)

Find the impulse response, ^(f), again using the symbolic toolbox in MAT LAB:

R.

+ V,(t)

'.w Q

-3 s

C H EC K : H (s) =

syms t h 200

h = ilaplace(H)

0.2 H

(c)

ZJs)

Using

the

residue

command

or

“ilaplace”, show that the zero-state response, V2 (f), is

Figure P17.30 31. Suppose that

= 3yf20cos{t)u(t) A

V2 (t) = 18A12e~^û(t) - 10.06e~û(t) + 9 co s(r-1 5 3 .4 °)« W

and the initial stored energy is zero in the cir cuit shown in Figure P17.31. (a)

(d)

Compute the transfer function,

Identify the steady-state and transient responses.

using the symbolic toolbox in MAT-

1.5 0

LAB as follows: (i)

I

1H

Write down the mutual induc tance equations in s, a set o f two equations in

Kj,

Kj, / p

l2-

Figure P I7.31

Rewrite each equation in the formO = ? X Kj + ? X /j + ? X K2 +

32. Consider the circuit o f Figure P I 7 .32.

• ^ ^2(ii) Use KCL and Ohms law to write

output volt^ e is zero. The input voltage is

the relationship in the form ? x = ? x Vi + F x / j . (iii) Similarly construct the terminal

Before the closing of the switch at r = 0, the v ^ (0 = 1 2 V 2 V . (a) Write two loop equations and find the matrix Z(s) such that

constraint equation of the form 0 = ? X V^2 + ? x / 2.

Z(s)

(iv) Write the preceding four equa

h (s y

■? ?■ h is y

h (s )

? ?

h is )

Vinisy 0

tions in matrix form, Ax= b. (v) Use MATLAB as follows:

(b)

Using Cramers rule find / 2 (s). Then determine K2W.

syms s lin A b Y H A = “Coefficient Matrix in part 4 ” % Y is the coefficient matrix A with the column corresponding to V2 replaced by the % vector b with /y^ (s) set to 1.

(c) Find V2 (f). C H EC K : V2 (f) =


946

+ V,(t)

33. Consider the RLC circuit shown in Figure

3 5 . Consider the circuit o f Figure P 17.35, in

P 17.33. (a) Find

which Z, = 0.9 H , Z2 = 0 .4 H , Af = 0.2 H , and

(b)

andZ.„W .

Find the poles and zeros of Z-JJ). Find the impulse response.

(c)

T?2 = 80 Q. Suppose V. (a)

Compute the currents in the two

(b)

Write two differential equations for

Find the zero-input response if

inductors at ^ = 0“ .

= 10 V with all other initial conditions zero. (d)

the coupled circuits in terms o f /j(r)

=

Find the zero-state response if cos(100/)«(/) A.

(e)

= 120«(-r) - I 2 0 u{t)

and (c)

Take the Laplace transform of the

Find the complete response of the cir

equations computed

cuit.

accounting for the initial condition. (d)

Find V2(t) for t > 0 .

(e)

Now suppose V, Find

• j /2 5 0

in part (b),

= \2u{-t) for / > 0.

-----

L (t)( S

0 .2 m F

0.2 H ' L

z

v jt )i

Figure P I7.33 34. Consider the circuit o f Figure P I 7 .3 4 , in which = 0.9 H , £2 = 0-6 H, 0.2 H , and /?2 = 100 Q. Let = 120 V. The switch S has been closed for a long time (i.e, the circuit has reached equilibrium) and is then opened at f= 0. (a)

After the switch is opened, write two differential equations for the coupled circuits in terms of /j(t) and ijit).

(c)

Take the Laplace transform o f the equations

(d)

36. The two circuits shown in Figure P I7 .3 6 have the same transfer function

Compute the currents in the two inductors at ^ = 0“ .

(b)

Figure P17.35

computed

in

part

(b)

accounting for the initial condition. Find V2 {t) for f > 0.

Find the values o f the three uncoupled induc tances assuming A / = 2 H. Hint: Use the Tequivalent circuit of Figure 17.18 with the indicated labeling in Figure P I7.36a; node 3 is the common terminal of the coupled inductors.

94'


Figure P I7.38 .ANSWI-R: k = 0.2S 30

fY Y Y jy Y V 39. Consider the circuit o f Figure PI 7.39. Find

Q

the value o f Csuch that the voltage gain is zero at

20

CO

= 3 3 3.33 rad/sec. Use the 7i-equivalent

circuit o f Figure 17.18c. no coupling betw een inductances

c

(b)

Hh k=0.5

Figure P i7.36 20 0

AN SW ER; L^.= 2 H. A., = 3 H, and

= 1H

COUPLING COEFFICIENT PROBLEMS AND ENERGY CALCULATIONS

”0

40 0 0.27 H

0 .1 2 H .

200 0

,

Figure P I7.39 A N SW l'R: 3 3 .33 ul-

37. Consider the circuit in Figure P i 7.37. Find

40.

the value o f Csuch that the voltage gain is zero

find the coupling coefficient k such that at co =

at

CO

= 333.33 rad/sec. Hint: Use theT-equiva-

lent circuit.

For the circuit shown in Figure P I 7.40,

10"^ rad/sec, the voltage gain

is zero.

Use the Ti-equivalent circuit o f Figure 17.18c. 0.5 pF

k=0.5

20 0

40 0 0.27 H

0.12

0.6 mF

T l Figure P I7.37 AN SW ER: 0.1 ml-

Figure P I7.40 AN SW ER: k = 0.946

38. For the circuit shown in Figure P i 7.38, determine the coupling coefficient k such that

41. Consider the circuit o f Figure P 17.41.

at

CO

=10"^ rad/sec, the voltage gain

(a)

Find the mutual inductance M so that the coupling coefficient k = \.

(b)

Obtain the transfer function using the

is

zero. Hint; Use theT-equivalent circuit for cou pled coils and recall that a series LC behaves as

formulas o f Example 17.8. Make use

a short circuit at w = \NLC.

o f your answer to part (a).

9i.S


(c)

Q, and H

Obtain values for (u ,

40 n >V

(s)

Figure P i7.42 43. Let /(r) = 2 cos(10r + 3 5 °) A in the circuit

Figure P i7.41

in Figure P i 7.43. Find the maximum instanta 0) .

ANSW l-K: (c) - 2.5 rad/scc, (I) = 10 rad/sec./-/= 17.H«8

= —

= -^ neous stored energy.

42. In die circuit shown in Figure P 17.42, the coupling coefficient k has been adjusted for the circuit to be resonant at OJ = 10 rad/sec. /?^ = 0.1

Q., L^=2 H, and L-, = \ H while C = 20 niF. (a)

Find k.

(b)

Construct the transfer function H{s) =

Figure P i7.43

K ..,(5 )

44. In the circuit o f Figure P 17.44, Z.j = 0.8 H, and then determine its poles and

(c)

zeros. Hint: Use the T-equivalent for

Z.2 = 0.45 H ,M = 0.175 H ,/?= 12 ^ Ify/r) = 30cos( 1Or) V, find the maximum instantaneous

coupled coils to simplify calculations.

steady-state power delivered to R when

For a sense o f the circuit behavior,

(a)

A dot is in position A.

obtain

(b)

A dot is in position B.

a

magnitude

frequency

response (using MATLAB) o f the cir

fv c t

cuit for 1 rad/sec < (U < 20 rad/sec. Observe the band-pass characteristic behavior. (d)

To obtain approximate values o f 2 ’ ^^^ctor H{s) =

Q> where

H^{s)= *

Figure P i7.44 A+ 0.05

Note that |//, (/10)| 2 2 ”^ order

A N SW l'R: (a) 48 w atts, (b) 27 w atts

.

45 . Consider the coupled inductors shown in 1 so that the

characteristic

o f //2W

Figure P I7.45. (a) If /, = 2 A and

approximately characterizes the band pass behavior.

= - 3 A, find the

stored energy. (b)

If /j = 2 A, find the value o f I 2 that will minimize the stored

A N SW ERS: (a) ^ 2 . , (d) Q a 200. = 100

= (^-05 rad/scc, (c)

energ)' W. What is the value o f Plot W as a function o f ^ for - 3 < ^ < 3 A using MATLAB or the equiva lent.

W


(d)

Determine the coefficient o f coupling k.

currents are /j(0) = 7| and /-,(0) = Ij- Also sup pose /j(7) = 0 =

Between 0 and 7', /,(a)

and i-y(t) have arbitrary waveforms. Show that the energy delivered by the inductors to the current sources during the interval 0 < t < T 4H

equals ( 0 .5 7 ,7j2

9H

0 .5 ^ 2 ^ +

dots are now moved to A-C, show that the final result is (0.57.j7,2 + Q.5L,l2^ Figure P i7.45 48. In contrast to Problem 47, where the cou pled inductors delivered power to the sources,

ANSWHRS; (a) 6 6 .S j . (h) 2/3 A and 6 J, (d) U.S

we can connect two resistors 7?, and Rj to the

46.

inductors and let the current decrease exponen tially to zero. In particular, for the circuit o f

For the coupled inductors o f Figure = B

Figure P I7.48, it is possible to show that the

cos(o)/ + q)). What is the period T in terms o f

energ)' delivered by the inductors to the resis

P I7.46, /,(r) = A cos(o)/ +6) and CO?

Regardless o f whether the dots are at A-B or

tors during 0 < ^ < x

the coupled inductors is zero; i.e., show that

equals (0.57,,7,^ +

± A/7,7-,) depending on the position

A-C, sliovv that the average power delivered to

o f the dots. Demonstrate the validity o f this assertion for the specific parameter values and

-Lii\ ( 0 +

initial currents 7.j = 10 H, 7-, = 2 H, A/ = 3 H, /?, = /e, = 1 Q., /,(0) = 1 A,“uid /2(0) = - 3 A, with dots in position A-B, as follows:

=

0

(a)

Calculate the stored energ)^ at r = 0.

(b)

Calculate /, (r) and

(c)

Evaluate the integral

for / > 0.

Hints: 1.

+V'2/2)^//

Write down the coupled inductor equations in the time domain with ±

and comparc with the result o f part (a).

for dots in B and C. 2.

Recall that

1 J = — f (vi/, + v^h)dt\ “

CHECK.- (a) 5 J, (b) /,(r) = e Ui{t) A and i^{t) = -?>e-‘u{t) A

substitute and simplify.

JL ,{t)

L,

L,

R, >

v,(t)

L,

v,(,) 0 1 , ( 0 Figure P I7.48

Figure P I7.46

49. Mr. A claimed he had construaed a pair o f

4 7 . Reconsider the circuit o f Figure P i 7 .4 6

A7= 9 H. Rebut his claim by showing a specific set

with dots at A -B. At r = 0, suppose the initial

o f (7,, I-,) for which the stored energ\^ is negative.

coupled inductors with Z, = 10 H, L, = 8 H, and

950


ANALYSIS OF CIRCUITS CONTAINING IDEAL TRANSFORMERS

vJt)

50. In the circuit o f Figure P i 7.50,

R^=\50 Q, /?, = 600 D .,R^=\2 kQ, and v,„(/) = 5 yj2 c 0 s{(0 t) V,where O) = 200071 rad/sec. (a)

mum power transfer to Rj^. (b)

Figure P I7.52

Find the turns ratio, Wj : ft-, for maxi Given your answer to part (a), find

ANSW l-R; (a) Z , = 10 k il, Z, = 500 k il: (h)

vît) and v^ Jt).

0 .625. 0.1, 0.1, 0.0 0 6 2 5 ; (c) 5/1, -50/1, (d)

(c)

Given your answer to part (a), find

0.5

and /2(r). Find the power delivered to the load.

53. Repeat Problem 52 for ?n = 20 and n = 5.

(d)

CH EC K :

0.2

54. Repeat Problem 52 for

= 300 kH, R =

10 k n , R^ = 100 Q, ti = 0.2, and m = 0.1. C H E C k 'z , = 80 n ,

= -0 .0 1 3 3 3 3

55 In the audio amplifier circuit shown in Figure P i 7.55, both transformers are ideal, R^ = 50 Q, and v^.(0 = O.1V 2 cos(w/) V. The 4 Cl resistor represents a loudspeaker load. Figure P I7.50

(a)

Find the turns ratios a and b such that the average power delivered to the 4 Q

51. Repeat Problem 50 for R^ = 400 Q, R^ = 1200 D.,R^ = 5 iX and (/) = 10>/2 cos(cot ) V, where oj = 20007T rad/sec.

loudspeaker is as large as possible. What is the maximum power? (b)

The turns ratios are those determined in part (a) for maximum power trans

52. In the circuit o f Figure P I 7.52, Rj = 100

fer. Now suppose that a loudspeaker

n , R = \ 0 kH; R^ = 300 kI2, and w = n = 10. Compute Z j and Z ,. (a)

with 16

(b)

Compute the voltage gains

resistance is used in place o f

the 4 Q speaker. What is the power consumed by the 16

speaker?

Vjlvxy v-ijv2 , and (c)

Com pute

the current gains

(d)

Compute the power gain.

_ V3/3 ’po w e r Vi/ l'//i phono

input

cartridge

transformer

Amplifier

output

loud

transformer

speaker

Figure P I7.55 A N S W E R : (a) 2 0 , 5, 100 W ; (b) 6 4 W

O '


95 1

o 56. The circuit shown in Figure P I 7.5 6 crude

such, it can be represented approximately by an

ly models an audio amplifier circuit. Set

ideal transformer in conjunaion with some

20

inductances and resistances. This problem illus

3

trates the use of such a model for analysis pur poses.

K =— . Sw/

Each 8 Cl resistor models a tweeter, and each 16 n resistor models a woofer. Suppose the left

A certain calculator that normally uses four 1.5

and right speakers each consume 80 watts of

V batteries comes with an adapter whose

average power. Determine:

approximate

(a) w

P I7.57a, where

average power delivered to the speak

and R = 200 Q.

ers is as large as possible. (b)

w

(c) (d) (e) (f)

The

voltage

across

and

(a)

is shown

= 0.9 H,

in

Figure

= 9.6 H, « = 20,

If the calculator is not connected to

current

the adapter output, but the adapter is

through (rms values) each woofer and

plugged into a household ac outiet of

tweeter.

110 V ^ , what average power is con

i and then (rms values). The power delivered by the dependent

sumed by the adapter? (b)

If the output o f the adapter is acciden

source.

tally short-circuited, what are the

The power delivered by the independ

magnitudes (rms) o f the ac current

ent source.

drawn from the wall outlet and

The power gain o f the circuit, i.e., the ratio of the power delivered to the

O '

model

The turns ratios a and b such that the

through the short circuit? (c)

If a typical load, represented by a 50 Q

speaker load to the power delivered by

resistor, is conneaed to the output of

the independent source.

the adapter, what are the approximate Right Speaker

magnitudes (rms values) of the voltage across and the current through the load? (d)

Source

If the

adapter

is

'O

treated as a pair of coupled

O '

inductors as shown in Figure P I 7 .57b, deter

O

L2> and M.

mine the parameters Zj, Figure P I7.56 O

C H EC K : i = 4 . . ^ ^ = 8 V ^ ,

Pl.m = 27.778 57. Today almost every electronic O

gadget comes with its own ac adapter. An ac adapters main func

O

tion is to step down the household

O

ac voltage o f 110 to, say, 3 rms or 6 V__ rmsbefore conversion to a low dc voltage by the use of a rectifier. An adapter is basically a (practical) transformer. As

O O

o

AC adapter as a pair of coupled inductors (b)

AC Adapter as a practical transformer (a)

Figure P i7.57

952


AN SW ER: (a) 0.154 warts; (b) 0.279 A, 5.59 A, 15.6 watts; (c) 4.98 V, 0.0 9 9 6 A; (d) 10.5 H,

(c)

24 mi l, 0.48 H

=10/^^) V. Find the steady-state response to = 20 cos(20r) V using the phasor method.


(d)

P I 7.58 with dots in positions A-B. ( b)

Compute the impulse response. Compute the step response.

(c)

If the capacitor is initialized at

(d)

Compute the zero-state response to

(a)

Repeat steps (a) through (c) with the dots in positions A-C. lO m F

1 :10

lo n

= 16 V, find the zero-input response. 0.1 F

ZA^(t) = 20 cos(2t) V. Identify the tran sient and steady-state responses.

Figure P i7.60

Repeat steps (a)-(d) with the dots in

(e)

position A-C.

61. Figure PI 7.61 contains a linear circuit with a pair o f coupled inductors. Suppose k = 1 (unity coupling) and

i l = ^ = « = O.I M

ideal

L2

The circuit is in the sinusoidal steady state and

Figure P I7.58

has known phasor values V| = 1 V and I| = —j A.


(a)

former) and 0) = 10 rad/sec, find the

P 17.59. (a)

If A'f is infinite (i.e., an ideal trans

Compute the impulse response.

magnitude o f the current ratio

(b)

Compute the step response.

Compare your answer with the value

(c)

Compute the zero-state response to

= 0.1.

= 26 cos(2t) V. Identify the tran

ANSW ER: 0.1

sient and steady-state responses. (b)

1:10

10 rad/sec. Hint: Use the phasor equa

20 v„(s)

Repeat part (a) for M = 10 H and to = tions in section 7 to calculate the cur

©

1F

rent ratio I^/Ip AN SW ER: 0.09

0.01 F

Figure P I7.59 (c) 60. Consider the circuit o f Figure P I 7.60. Suppose the transformer dots are in positions A-B. (a)


H {s) = (b)

C o m p u te

the

V,„U) zero-state

response,

Repeat part (a) for A/ = 10 H and (O = 1207T rad/sec.

A N SW liR: 0.0 9 9 7 3 5 . praciicall\' the same as for ^/ = 0 . 1

953


Consider the circuit shown in Figure

64.

Rest of Circuit

PI 7 . 6 4 .

(a)

Find the output impedance, as a function o f the turns ratio a and the resistance R.

(b)

Find the Thevenin equivalent circuit in terms o f a, R, and

(c) Figure P I7.61

and ^ = 10.

Compute the transfer function H{s) =

Compute the transfer function. Note

(d)

locations of the transfer function. From the pole locations determine the

the

impulse

and

step

responses. (c)

the transient and steady-state responses. C H EC K : W ( 0 = ( - 3 . 2 2 9 8 f ^ “ ^^^^' + 5 . 1 1 8 1 f - - - " " ' -

and Q^. Is

Find

Given the parameter values of part (c) with = 3^/io sin(VIO 0 V, find the zero-state response, Identify

cally damped, overdamped, or under Com pute

for v-Jt) = 15u(t).

toolbox o f MATLAB. (e)

t)'pe (impulse/step) o f response (criti damped).

Compute

Hint: Use “ilaplace” in the symbolic

there are no finite zeros. Find the pole

(b)

and a = 2. Suppose the

with Z = 0.1 H and C = 0.1 mF.

62. In the circuit o f Figure Pi 7.62, /?. = 0.2 Q (a)

Let /? = 13

output is terminated in a series LC

high? If

highQ^,, find approximate values for (0^^, and the half-power frequen

1. 8 8 8 3 c o s ( 1 0 V i 0 r ) -h 9 . 0 9 4 8 sin(10>/i0r)]//(O

cies OJ, 2- Recall equations 16.47,

Q = Plot the magnitude response o f the transfer

function

to

verif}'

your

answers.

65. For the circuit shown in Figure P i 7.65, find circuit. 63. Repeat Problem 62 for R^= 2 Cl. This will be a low-Q^ case. Recall equation 16.47: e

.

lo ,

'

2sin(^^)

A 6a = arcsm ■ and

to PJ

and for exact answers to part (c) recall equa tions 16.49 through 16.52.

and the Thevenin equivalent

9S4


B

D

Figure P I7.67 Figure 1’ 17.65 AN SW ER: Vv(.) (H =

^th (

it

~

68. hi the circuit o f Figure P I7.68, and

U

144 /?^ = ^(/)= 10>/2cos{l0-‘^/) V. (a)

73.5

Suppose

= 25 H, and

= 1 (equivalent to a

straight-through connection with dc

^ "I-------

isolation).

Find

66. In the circuit shown in Figure P I7.66,

^2, find

ijjit), and /^(r).

(b)

maximum

^ 2

power transfer to

= 5 cos(2/) V. Find

256

For this value o f

and |V^ |.

Suppose = 1 (equivalent to a straight-through connection with dc isolation).

Find

for maximum

power transfer to /?y. For this value o f (c)

Suppose

and

^^re both adjustable.

Find a set o f values o f

Pi

and a-j so that

is greater than the values com

puted in parts (a) and (b). Find this (d)

B

ideal transformer

value o f PL.ave' What is the maximum value o f P Lave if//, and a-^ are both adjustable?

d

Figure P I7.66 67. In the circuit o f Figure P 17.67, suppose = 10 n , (a)

(b)

= 40 U, and R^ = 2 0..

Figure P I7.68 ANSWL'.R: (a) 0.1479 watt, (b) 0.64 watt, (c)

- 2,

for

= 0.1463. 0.87671 watt, (d) 1 watt

If the load is identified as the termi nals A-B, find a for maximum power

69. Consider the network configuration in

transfer.

Figure PI 7 .6 9 . The source voltage v^(t) = y / l C 0 S ((0 / ) V, where to = 10^ rad/sec.

If the load is identified as terminals CD (i.e., is the load), find a for max imum power transfer.

(a)

Design a lossless network N such that maximum average power is transferred to the 10

load at to = 10^ rad/sec for

Chapter 17 • Magnetically G>upled Circuits and Transformers

955

O ' each o f the cases cited below.

Case 1 : N consists of one inductor L and one ideal transformer. You will need to determine the configuration of the inductor and ideal transformer as well as the values o f L and the turns ratio a.

Figure P I7.70

Case 2 : N consists of a pair of coupled inductors, in which case you must determine the values o f

L^, and M.

Case 3: N is simply an LC network. (b)

For each case compute the transfer

71. The front end o f a radio receiver uses a band-pass circuit as shown in Figure P I 7 .7 la consisting of a capacitor and coupled coils whose mutual inductance is determined by the

fimction

tap position. For analysis purposes, this circuit is modeled using the equivalent circuit of Figure P I 7 .7 lb. Since the circuit is connected

Vsis)

o

Plot the magnitude response for 10^ s (0 £ 10^ rad/sec using a semilog scale.

to an RF amplifier, we will assume that the amplifier input impedance is infinite to avoid

Which circuit has the better perform

loading. Our goal is to obtain the frequency

ance over the range o f frequencies and

response o f this circuit and verify that it is a

why?

band-pass circuit. Assume C = 100 pF and

=

500 a 1 kO ^

O '

''.W

Ak. Ak.

(p

a

Lossless N

10 nF

^

(a)

Compute the coupling coefficient k.

(b)

Find the transfer fimction H{s) = Suggestion: Use one o f

100

the models of Figure 17.33 for the coupled coils. (c)

Figure P I7.69 70. Consider the resonant circuit shown in Figure P I7 .7 0 containing a pair of coupled inductors. Suppose O

= 20

Zj = 1.5 mH, M

= 3 m H, Z.2 = 6 m H , and R i = 320 Q. Find the coupling coefficient k. (a) (b)

O

Model the coupled inductors using one o f the equivalent circuits in Figure 17.33. Find the impedance yj(f) and

O

then Kj(/o)) as a function of C (c)

O o o

< J

Determine the value o f C that leads to a resonant frequency of O)^ = 10

O (d)

krad/sec. Find the voltage gain at resonance; i.e., find

J| ,

and Q of

the circuit.

O ' W

Compute (O^,

Remark: We could have designed a parallel res onant circuit to achieve the frequency-selective property of the circuit in Figure P I 7 .7 la. However, because of the mutual inductance, we can achieve a much larger voltage gain than that achievable by a simple parallel RLC. This becomes clear when one uses a model for the coupled inductors o f Figure P I 7 .7 lb; this model contains an ideal (step-up) transformer having a turns ratio of 1:10.


(C H EC K : 600 nH) in the model con structed in part (b) can be neglected. Find

approximate

|//(/(0^^^)|,

values

of

and Q for the voltage

gain H{s) =

Plot the magni

tude frequency response for 2 M Hz < / < 16 MHz. ’ Hint: Reflect the voltage source and 1.2 M= IOmH

resistance to the right side o f the ideal trans former in the equivalent circuit model o f part (b). That is to say, replace the circuit to the left o f the ideal transformer by its Thevenin equiv alent.

(b)

(d)

Figure P I7.71

Including the effects o f

and the

6 00 nH shunt inductance, use a cir A N SW ERS: (a) 1: (c)

1

X I t P nitl/scc,

Q

cuit simulation program to find more

= lo " rad/scc. li.. =

= SO, ; i i k1

accurate values for

= 10

and Q o f part (c) and compare.

APPLICATIONS 72. Recently, coupled coils have been manufac tured on printed circuit boards by printing two spiral coils on opposite sides o f a board. Such coils are called coreless PCB transformers and can be used to effect electrical isolation as well as energ}' and signal transfer. This problem illus trates the use o f the models developed in this chapter for an approximate analysis o f the fre quency response o f coreless PCB transformers.

AN SW ERS; k - 0.5. (b) l,slmni= 600 n il.

Figure P i 7.72 shows an equivalent circuit o f a

6 .0 8 6 Mrad/:,CC. 7^,., = 3.333 .Vlrad/j,ec. Q =

PCB transformer for which Z., = /-2 “ = Rj = 1.2 which rep

18.26.

l.sfrics= 4^0 nH, turns ratio a = 2, (c)

=

= 9.13

M = 300 nH, and

resent the winding resistances. Suppose tiie sec

73. Figure P I 7.73 shows two series RLC cir

ond coil is terminated in a capacitance o f C =

cuits physically isolated but magnetically cou

600 pF in parallel with a resistance of /?y =

pled. The input is a sinusoidal voltage sotirce cos(cor). We will investigate some

2000 a .

steady-state voltage gains. This investigation (a)

Find the coupling coefficient k.

(b)

Construct a model for the coupled

the coupled inductors represent a magnetic

inductors using Figure 17.35 of the

coupling between an implant and an external

text. For a first analysis, assume that the

circuit.

(c)

effect o f Rj^ and the shunt inductance

has apphcation to biomedical implants, where


M

V;„

(c) Suppose L = 1 mH, R = 5 O., C = 0 . l/.< F, 0) = 10 0 krad/sec, and

= 0 .1

V. Find the value o f the mutual induc tance M needed to maximize

Figure P I7.73 If the coupling coefficient ^ = 0 (or Af = 0), then the steady-state output magnitude is = 0. Further, as shown in Chapter 16, [V qI = Ql

V^. As k increases from zero to its maximum

value o f 1, |Vq| decreases monotonically, whereas lV^,J increases to a maximum value and then decreases. Exact formulas for several critical values will now be derived. (a)

|VJ.

What is the maximum value o f |Vout\' (c)

Q = 20. M

= 0 . 05

mi l .

r nttijnux = 1

74 . Example 17.7 provided an approximate analysis o f a car ignition system. This problem asks for a SPIC E simulation o f the example. Reconsider the circuit o f Figure 17.12b with out the switch. Show that the initial current through the primary o f the coupled inductors is

Show that if the rsvo series RLC cir cuits in the primary and secondary are identical, i.e., Z,j = Z2 = C, 2nd

ANSWTRS:

Cj = C2 =

/■^l(0~) = 12 A. Since the secondary is open-circuited, there is no initial current there. Generate a SPIC E simulation for 0 < f < 1 msec. Verify that |^'2(^)|,„^v - 36 x 10^ V.

7 5 . In the car ignition circuit o f Figure 17.12b, suppose the resistance and capacitance values are changed to 2 Q and 0.25 uF. Find approxi

then

mately the maximum voltage appearing cross

kQ^

'out

1+ (A Q )and

76. Again consider the car ignition circuit o f

Q

*c\

i+{kQf (b)

the spark plug. A N S W E R ; 3 6 . 0 0 0 \'

Show that if k is the only adjustable parameter and Q > 1, then

max

'out

k

when

k=~,

Q

and at this degree o f coupling

Figure 1 7 .12b. Suppose the resistance is 2 = 1 i-tF, and

C

= 10 mH; find the approximate

value o f the mutual inductance M to achieve = 38

X

10^ V. W hat is the approxi

mate minimum corresponding value o f L j, the inductance o f the secondary o f the coil? Verify your circuit operation using SPICE.

C H A P T E R Two-Ports

THE AMPLIFIER: A PRACTICAL TWO-PORT An actress speaks into a microphone. Speakers instantly replicate her voice, which resounds throughout the auditorium. W hat happens between the speaking and the hearing? A microphone produces a voltage signal that changes in proportion to the tenor and loudness o f the voice o f the actress. Amplifiers magnify this changing voltage signal perhaps a hundred or a thousand times to drive speakers whose cones reverberate in proportion to the changing voltage signal. The cones then cause the air to vibrate intensely, also in proportion to the tenor and loudness o f the actress’s voice. Her words become heard by thousands because o f the amplifier. Amplifier circuits are found in instrumentation and a huge number o f appliances. In radios, radio frequency amplifiers first magnify signals from an antenna. Special circuitry then extracts the audio portion from these antenna signals. Other circuitry amplifies the audio to drive speakers. Video signals from a video cassette recorder are amplified by special circuits for connection to a monitor. Amplifier circuitry enhances signals coming from sensors in various manufacturing processes. Repeater circuits, among other things, amplify phone signals whose magnitudes have attenuated during microwave transmission. There are a large number o f other applications o f amplifiers. From the preceding discussion, one can surmise that an amplifier circuit has an input signal and an output signal. This configuration is represented by a device called a two-port that has an input port for the input signal and an output port for the output signal. The following figure represents the idea (one o f the homework problems asks for an analysis o f this amplifier):

960

Chapter 18 • Two-Ports

TRANSISTOR AMPLIFIER CIRCUIT FOR MICROPHONE

Input Signal

Output Signal

Often the circuit berween the ports is highly complex. This chapter looks at shorthand methods for analyzing the input-output properties o f two-ports without having to deal directly with a pos sibly highly complex circuit internal to the two-port. The chapter will provide a variet)- o f meth ods for analyzing amplifier circuits.


Provide a general setting for one-port analysis.

2.

Extend one-port analysis to the analysis o f two-ports.

3.

Describe the input-output properties ot two-ports in terms of four sets of characteristic parameters: impedance, or z-parameters; admittance, or ^'-parameters; hybrid, or hparameters; and transmission, or r-parameters.

4.

Develop specific formulas for the input impedance, output impedance, and voltage gain o f two-ports driven by a practical source voltage and terminated by a load impedance.

5.

Introduce and interpret the notion of reciprocity in terms o f the various two-port parameters.

SECTION HEADINGS 1.

Introduction

2.

O ne-Port Networks

3. 4. 5.

Tw o-Port Admittance Parameters Parameter Analysis o f Terminated Two-Ports Two-Port Impedancc Parameters

6.

Impedance and Gain Calculations o f Terminated Two-Ports Modeled by zParameters

%1

Chapter 18 • 'Iwo-Ports

7.

Hybrid Parameters

8.

Transmission Parameters

9.

General Relations am ong Two-Port Parameters

10.

Reciprocity

11.

Parallel, Series, and Cascaded Connections o f Two-Ports

12.

Summary

13.

Terms and Concepts

14.

Problems

1. INTRODUCTION Figure 18.1 shows a general one-port, whose two terminals satisfy the propert)' that for any volt age Kj across them, the current entering one terminal, say, /,, equals the current leaving the sec ond terminal. A resistor is a one-port: the currcnt entering one terminal equals the current leav ing the other terminal. A capacitor and an inductor are also one-ports. K general one-port contains any number o f interconnected resistors, capacitors, inductors, and other devices. In a one-port, only the relationship between the port voltage and current is o f interest. For example, the port

Cdv^Jdt

voltage and current in a resistor, capacitor, and inductor satisfy the relationships

- Iq and Ldi^ldt=

respectively. Practically speaking, one-ports are macroscopic device models

emphasizing input-output properties rather than detailed internal models. Thevenin and Norton equivalent circuits determine one-port models when only a pair o f terminals o f a network is o f interest.

O+

V.

One Port

O+

V,

Two Port

-O +

V,

-o (a)

(b)

FIGURE 18.1 (a) General one-port, (b) General two-port. A two-port is a linear network having two pairs o f terminals, as illustrated in Figure 18.1b. Each terminal pair behaves as a port; i.e., the current entering one terminal o f a port equals the current leaving the second terminal o f the same port for all voltages across the port. Coupling networks such as transformers have two pairs o f terminals, each o f which behaves as a one-port. Hence, transformers are two-port devices. In modeling a rsvo-port, one must define a relationship among four variables. Different groupings o f current and voltage variables lead to different t}^pes o f char acterizing parameters. For example, adm ittance parameters (termed ^-parameters) relate the two (input) voltages,

and V-,, to the (output) port currents, /j and 7^. Impedance parameters

(termed 2-parameters) relate the two-port (input) currents to the two-port (output) voltages. Other t}’pes o f parameters investigated in this chapter are hybrid or A-parametcrs, and trans mission or /-parameters.

962

Chapter 18 • Two-Pons

Source One-Port

FIG U RE 18.2 A general rwo-port connccced to a source one-port (represented by a Thevenin equiv alent) and terminated in a load one-port (represented by a Thevenin equivalent impedance). In practice, two-ports often represent coupling devices in which a source delivers energy to a load through the two-port network as suggested in Figure 18.2. For example, stereo amplifiers take a small low-powcr audio signal and increase its power so that it will drive a speaker system. Determining and knowing ratios such as the voltage gain, current gain, and power gain o f a twoport is ver\' important in dealing with a source that delivers power through a t\vo-port to a load. This chapter develops various formulas for computing these gains for each type o f two-port parameter.

2. ONE-PORT NETWORKS Basic Im pedance Calculations As mentioned in the introduction, the current entering one terminal o f a one-port, illustrated in Figure 18.1, must equal the current leaving the second terminal for any voltage

across the ter

minals. We begin study o f one-ports by exploring two impedance calculations o f a transistor ampli fier circuit that are pertinent to ba.sic electronic analysis. The impedance or admittance seen at a port is fundamental to the behavior o f a network to which a one-port or two-port is connected. EXA M PLE 18.1. The circuit o f Figure 18.3 is pertinent to a simplified model o f a common-collector stage o f a transistor amplifier circuit. Specifically, the one-port o f Figure 18.3 models the input impedance. In a common-collector amplifier stage, the impedance Z j is very large and is often neg lected, i.e., we assume |Z,| = =c over the useful bandwidth o f the circuit. The goal o f this example is to compute Z

and interpret Zy^^ in terms o f a transistor current gain parameter denoted by (3.

FIG U llE 18.3 One-port represennng the input characteristic of a common-collector transistor amplifier circuit.

963


So

lu t io n

We attach a hypothetical vohage source, say,

across the port terminals, top to bottom being

plus to minus, to induce a hypothetical current ly Neglecting Z ,, from KCL,

Since the input impedance is the ratio o f the input voltage to the input current,

Zi„=-^=(P+l)Zi, Thus, for a large P, say, 150, the input impedance can reach very high levels for reasonably sized impedances Z^. When amplifying voltage signals, we desire to have the ratio o f the internal source impedance to the amplifier input impedance be small. Conversely, the amplifier input impedancc should be large relative to the internal source impedance.

Exercises. 1. In the circuit o f Figure 18.3, suppose P = 100, Z j = oo, and the load resistance Z^ = 16 Q. If a sinusoidal voltage the average powers

and

= 15^ 2 sin(5000r) V is connected to the input terminals, find and the power gain.

AN SW ER: 139 mW. 14.06 W, 101 2. Consider the circuit o f Figure 18.4. T he one-port shown here models the output impedance of a common-collector stage o f a transistor amplifier circuit. Show that 7

_

'o i l ! _

Io„,

(P + l ) '

FIGURE 18.4 One-port representing the output characteristic o f a common-collector transistor amplifier circuit. 3. In the circuit o f Figure 18.4, suppose P = 99 and Z j = 1 kH. If a current source is connected to the one-port terminals, find ANSW ER: 2 V. 2 mA

and |/j|.

= 200 niA


A second example that is common to basic electronic analysis depicts a circuit used in the analy sis o f field-cfFect transistors (FETs). E X A M P L E 18.2. Figure 18.5a represents a simplified model o f the input impedance o f a fieldeffect transistor circuit. As in Example 18.1, we neglect the large impedance in parallel with^^^K,. Compute the input impedance

in terms o f Z j, Z 2, Z 3, and

FIC'il’RE 18.5. Simplified input impedance model of a field-effect transistor (FET) circuit. S o lu tio n

As illustrated in Figure 18.5, assume a hypothetical voltage source

(plus to minus is right to

left) has been attached across the port terminals. At the bottom node, /, + g„,V-[ + ^ = 0. Since K, = Z,/, and /2 =

Now, since Z,„ =

+ K,)/Z2, it follows that

VAvi

Z ;„ = Z , + (1 + g „ Z | ) Z2 = Z j + (1 +

Exercise. In the circuit o f Figure 18.5, suppose

Z j) z ,.

= 0 .002 S, Z-,= 500 Q, and Z j is the imped

ance o f a 0.1 l-iF capacitor. Find the frequency (in Hz) at which the magnitude ofZy^^ is 707 Q.. ANSW i-R; 1(I03.J 11/

Thevenin and N orton Equivalent Circuits Recall from Chapters 6 and 14 that the Thevenin equivalent o f a network, as seen from a pair o f terminals (i.e., from a one-port), is a voltage source in series with the Thevenin equivalent imped ance Z,!,is simply the equivalent impedance o f the one-port when all internal independent sources are set equal to zero. The value o f the voltage source, equals that voltage appearing at the open-circuited port terminals. A source transformation on the Thevenin equivalent produces the N orton equivalent, which is simply a current source in parallel with the Thevenin impedance. T he value o f the current source is 7^^ = VgJZ^f^. This is the current that would pass through a short circuit on the port terminals.

96 S


EXAMPLE 18.3. Consider the circuit ofFfigiire 18.6. Find the Tlievenin and Norton equivalent circuits. I. I.

1 :a

----->--------+ • 9

>

J

+ C

A +

■

FIGURK 18.6 Transformer circuit for Example 18.3. S o l u t io n

The voltage across A-B satisfies

Kw = - ^ I a + V 2 = ^ I a * « V i Cs

Cs

To determine the Thevenin equivalent, we express Kj in terms o f the current relationship /j = -nl^,

and /^. From Ohm’s law and

Therefore,

yAB =

+ R cr

Cs

^) Thus Z,|^{s) = — + R cr, V^J,s) = aRIi^j, and /^(jr) = V ( -------------------— , as set frorthL •m cFigures

18.7a and b.

Cs

Cs 1 + Ra^

1

h

aRCs RCa^s + 1

(a)

Cs

+ Ra2

J

(b)

FIGURE 18.7 (a) Thevenin and (b) Norton equivalent circuits of Figure 18.6.

Exercise. Suppose R = 100 and zeros o fZ ,,,

= 5, C = 10 mF, and

AN SW hR: I 'j. = lO/.v; pole at x = 0 ,ukI zero ai

= U.04.

= Au{t) mA. Find

and the poles

966


A technique known as matrix partitioning often simplifies the calculation o f Thevenin and Norton equivalent circuits. Example 18.4 illustrates the technique.

EXA M PLE 18.4. Find the Thevenin equivalent circuit for the network o f Figure 18.8.

So

lu t io n

Step 1. Construct the loop equations. By inspection, the loop equations for the circuit satisfy

• v, ■

— + 18 2.V

-8

-10

h

-8

18

-4

h

-10

-4

16

/3 .

13 6

s 0

The goal is to find

(18.1)

in terms o f /p the matrix equation 18.1 is partitioned accordingly. To avoid

dealing with numbers, let us rewrite this partitioned matrix equation as

r\v,i

\V,2 vvv>

where

= 136/j,

(18.2)

= 0, and the ^Vjjs represent their analogous (partitioned) counterparts in

equation 18.1. In general, Vj is the net voltage in loop 2 and

is the net voltage in loop 3.

Step 2. Solve the partitioned matrix equations for K, in terms o f /j arid the vector [ K,,

The

matrix equation 18.2 may be rewritten as a set o f two equations, namely V|

and

+'^12

h h

(1 8 .3 )

967


V2 I

= W21/1 + VV22

(18.4)

^3

Since

is invertible (for all passive networks), solving equation 18.4 for the vector [/j

yields V2 I (18.5) Substituting equation 18.5 into equation 18.3 produces [V2

U/,, - W^2W22^V2^ ] /i + U'12^^22'

(18.6)

^3

MATLAB or its equivalent, or possibly a symbolic manipulation software package, allows the matrices in this equation to be conveniently computed. Comparing equation 18.6 with the struc ture o f a Thevenin equivalent circuit,

allows us to conclude that the Thevenin impedance is iV,, -VV,2VV27VV2|

(18.7)

and the open-circuit voltage is V,

Step 3. Compute Z I and

-

Z ,„ ( .) =

(18.8)

From equation 18.7,

Is

-1 -8

-1 0

■18

-4 ‘

-4

18

18

-4 ‘

-4

16

-1

•- 8 -1 0

. _ .V+ 0 .0 9 5 = 5 .2 6 5 --------------

s

and from equation 18.8, —

- 10 ]

-1 •1361

s 0

Figure 18.9 shows the resulting Thevenin equivalent circuit.

84

s

^ Q

968

Chapter 18 •Two-Ports

5.3

s + 0.095

h

+ \- M

s

FIGURE 18.9Tlicvcnin equivalent circuit of Figure 18.8.

Exercise. Suppose the capacitor in Example 18.4 is changed to a 2 H resistor. Compute the new Z,,[s). AN SW l-ll: Z^,,{s) = 7.2647 i l via the MA'I'LAB code AVI 1 = 20; W 12 = [-8 -10]: »\V21 = [-8: -101: \V22 = [18 -4:-4 16]: vZth = \V11 - \Vl2'inv(\X/22)*\X'21

3. TWO-PORT ADMITTANCE PARAMETERS B asic D efinitions a n d Exam ples As mentioned earlier, rwo-port representations allow us to deal only with the terminal voltages and currents o f the two-port, depicted in Figure 18.1b and repeated in Figure 18.10. 1. O+

V. o

L

Two Port

-O +

V, -o

FIG URE 18.10 Standard two-port configuration having four external variables: /j, I^, Kj, and V-,.

969

Chapter 18 ‘ Two-Ports

ADMITTANCE PARAMETERS 'rhroughout this and later discussions, assume that the two-port ot Figure 18.10 has no inter nal independent sources and that all dynamic elements are initially relaxed, i.e., have no ini tial conditions. Under these assumptions, the admittance parameters or ^-parameters o f a two-port are expressions for the terminal currents, /j and I 2 , in terms o f the port voltages, and ^2’

/r

■-Vli

V|2' -\/,-

^2

.^21

>’22

(18.9a)

^2

or, in scalar notation, ^1 “ A’l 1^1 "*■.'’12^2

(I8 .9 h )

I 2 = V21V 1 -I- >22^2 Using either o f these sets o f equations, one can define each admittance parameter, lows:

-II

V,

as fol

( 18. 10) vs=o

Vl=0

V2 I = V, =0

Since each admittance is defined with regard to a shorted terminal voltage

= 0, the^^y are

often called short-circuit adm ittance parameters. I'he unit o f an admittance parameter is S.

Some examples will illustrate convenient methods for computing the ^-parameters.

EXA M PLE 18.5. Compute the short-circuit admittance parameters o f the circuit in Figure 18.'

FIGURE 18.11 Circuit for Example 18.5. S o l u t io n

The overall strateg}' is to write equations for /, and /, in terms o f Accordingly, by Inspection,

and

/, = K, I/, + ^3 (Kj - V,) = (K, + Y,) K, - K, 1/ and

using nodal analysis.

970


* y 2 V ._ * K, ( K , - K,) = ( g ^ - r ,) . (K j * K,) In matrix form, these equations are h

■K, + K3

-

I2

A ' -^3

Yl

•V/,-

A ■-Vii V’21 .^2.

V,2‘

V/,-

V’22

vs.

in which case

y\ \ - ^1 + ^3>^12 -

^V>'21 -

~ ^3’ ‘'*^‘^>'22 " ^2

^3'

Exercise. Suppose the controlled current source in Figure 18.11 is reconnected across

with the

arrow pointing to the left. Find the new ^-parameters, A N S W E R : n , = K, +

= g „ ,~

and y ,, = )\ +

In the next example, we combine the method o f matrix partitioning with the use o f nodal equa tions to compute the y-parameters.

EXA M PLE 18.6. Compute the ^-parameters for the circuit o f Figure 18.12.

^TT

FIGURE 18.12 Three-node circuit for Example 18.6. S o l u t io n

Computation o f the j'-parameters will again proceed by the method o f nodal analysis. Consider, for example, node 1, in which the current /, must be determined, /j has the form

where each variable is understood to be a function o f s. The coefficient K j, is simply the sum o f the admittances incident at node 1. The coefficient admittances between nodes 1 and 2, and nodes 1 and 3. Similarly, I 2 = V-y]

is simply the negative o f the sum o f the

^ is the negative o f the sum o f the admittances between + Y-^^Vy where Yjj is the sum o f the admittances

incident at node 2, etc. Hence, in matrix form, the node equations o f the circuit o f Figure 18.12 are

97


.v+ 3

w =

h

-s

: -2

- .V

a

5 + 21 -1

0’

r\y,i iM 11 : M/pi VV2V V w ji

V2

This nodal equation matrix is symmetric, because the /^LC network contains no dependent sources. Using the method o f matrix partitioning introduced by equation 18.6 yields ■/r /2.

-\/,-

(

■ .^ 2 .

[

-li -

■5 + 3 - 5

i - + 2.2

-(5 + 0 .4 )

-(5 + 0 .4 )

5 + 1.8

-

5

s

'

1 T

+ 2 ” 5

Exercise. The circuit o f Figure 18.12 is modified by adding a 2

1

[2

1]]

•V,-

resistor in parallel with the

capacitor. Find the new ^/-parameters. r .ANSWFR:

5

+ 2 .7

- ( 5

- ( .s + 0 .9 )

+ 0 .9 )!

.s + 2.3

A last example couples a transformer with a resistive n-net\vork. EX A M P L E 18.7. Compute the ^-parameters o f the circuit in Figure 18.13.

FIGURE 18.13 A resistive :r-network coupled to an ideal transformer circuit for Example 18.7. S o l u t io n

Find the y-parameters by using nodal analysis in conjunction with the ideal transformer equations. First, at port 1, /, = K, + (V , - V,) = 2Vi -

= 2V , - - ^ 2 (1 8 .1 2 )

Now considering that al 2 = -/, a node equation at the primary o f the transformer yields


(V i-V i)

h =— h = -a a

= - - V ,+ 4 - V - ,.

a

a~

(18.13)

where the last equality uses the relationship aV^ = V2. Putting equations 18.12 and 18.13 in matrix form yields the ^'-parameter relationship

r\/,i a

a

Exercise. In the circuit o f Figure 18.13, the top resistor is changed from 1 Q. to 0.25 turns ratio a =2 , find the new ^/-parameters. A N SW I- R ;

-2

1.25

With the

.S.

Tw o-D ependent Source E quivalen t Circuit The key to engineering analysis rests with the interpretation o f appropriate mathematical equa tions. The key to two-port analysis is the interpretation o f the two-port equations. Take, for exam ple, the first admittance equation, 1 =.^11^1 +>'12^2 One circuit-theoretic interpretation of this equation has the port current /j equal to the port volt age Kj times an a d m i t t a n c e ] in parallel with a voltage-controlled current source^, 2 ^^2- ^ ilar interpretation o f the equation A =^21 yields an admittance branch

+J'22 ^2

in parallel with a voltage-controlled current source^ 9, K ,.

These interpretations lead to the tw o-dependent source equivalent circuit o f a rwo-port repre sented by the short-circuit admittance parameters. (See Figure 18.14b.) This equivalent circuit aids the computation o f input and output impedances and voltage gain formulas. Note: in this chapter and later chapters the standard resistance symbol often is used to designate a general impedance or admittance.


-o

O+ V.

o-

+ V,

y„ y.2 L

V2,

V22 J

(a) FIGURE 18.14 (a) Short-circuit admittance parameters, (b) Their two-dependent source equivalent circuit. The dotted line at the bottom of (b) indicates that the two halves are not necessarily connected.

4. Y-PARAMETER ANALYSIS OF TERMINATED TWO-PORTS This section takes up the task o f analyzing term inated tw o-ports modeled by ^-parameters. A rvvo-port is terminated when it has source/load admittances. Any circuit or system in which a source provides an excitation signal to an interconnection network that modifies the signal and drives a load impedance can be represented by a terminated rwo-port. Such a scenario is common to numerous real-world systems. For example, the utility industry delivers power to a home from a generating facility through a transmission network, and a telephone system delivers voice infor mation from the phone through a transmission network to a receiver.

Input and Output Admittance Calculations The input and output admittances o f a terminated two-port are important for determining power transfer and for various gain computations. In what follows, we will show two different methods for computing the input admittance o f the terminated two-port illustrated in Figure 18.15. Computation o f the output admittance is left as an exercise.

O +

-o + V,

’’ Vn y ,2 L V21 V22 J

O

■O

FIG U RF 18.15 Two-port terminated by a load admittance The first method for computing

is a matrix method. Recall equation 18.9a, ■/l'

Vll

>'12

■\V

h

.'21

V22

V2

Using the terminal conditions imposed by the load

in Figure 1 8 .1 5 , we obtain


97-1

Incorporating this terminal condition into the two-port ^-parameter equation yields ■

■

■>’11

>’12 '

V/,-

■/l'

>’11

Vl2

>’21

>’22

.^^2.

0

>’21

y ii +

Using Cramers rule to solve for

det V. =

Because

^2

in terms o f /j results in

■/l 0

>’12

>’22 + y'L >’12 det ■>’11 >’21 >’22 +

is the ratio o f /, to

rv,

(>’22 + ^l ) ^1 >’l 1(>’22 + " >’12>’21

the input adm ittance o f the two-port o f Figure 18.15 is Vl2>'21

(18.14)

>’22 + E X A M P L E 18.8. Derive the input admittance o f equation 18.14 using the two-dependent source equivalent o f Figure 18.16. Here we avoid the solution o f simultaneous equations while increasing insight into the operation o f the two-port.

I.

L

FIG URll 18.16 Input admittance calculation using two—dependent source equivalent circuit. S o l u t io n

W ith regard to the right side o f Figure 18.16, the current ^^21

iTiust equal the current through

the parallel admittances j /22 ->'21^. = - ( 7 2 2 - ^ / .) ^2 Hence, v-, = -

.V2 I (18.15)

Equation 18.15 says that the voltage V2 equals the current, -^21 Now consider the left side o f Figure 18.16. Here >’l I -

impedance,

>'12>'2I >’22 +

Again, y,>, = —

leads to the same formula as equation 18.14.

Vi /

1 + K^)

9“ 5


Exercise. Let

denote a source admittance. Show that the output admittance o f Figure 18.17 is J 12J 2 I

(18.16)

V-

-o

-O+

+

Vu Vl2

V.

V,

V2, V22

-o 2

FIGURK 18.17 Input-terminated two-port for output admittance calculation.

G ain C alculations Our objective now is to derive a formula for the voltage gain of a doubly terminated r\vo-port, as illustrated in Figure 18.18. Again, t h e s y m b o l s denote general admittances rather than the tra ditional conductances.

+

+

yi2^2

V.

V, y„

'22

V2/,

FIG URE 18.18 Doubly terminated rwo-port driven by voltage source,

symbols denote general

admittances. The specific goal is to derive the voltage gain formula

{

Ys

\

->’21

(18.17)

The overall gain calculation breaks down into two cascaded gain calculations as follows: ^

V,

V/.

V/,

Gi/ = — = — X — v; V, V/,

(18.18)

Computation o f the gain, ( 7^^ = l l ^ follows direcdy from voltage division: (1 8 .1 9 a )

Chapter 18 • Two-Forts

VS To compute the gain G ^.2 = — from directly from equation 18.15, V /.=------>’22 + Yl

which implies

^ V21 ^ v2= — = M >22 +

(I8 .1 9 b )

Equation 18.17 follovv.s by substituting equations 18.19a and 18.19b into equation 18.18.

Exercises. 1. In the circuit o f Figure 18 . 18, 7 ,, = 5 ,^ p = -0 .2 , y 21= 50, j/2-) = 1 (all *n mS), 1 k n and /?^ = 2 kQ. Find Z-„ and G y = VJV^. A N SW ER:

= 8 5 .' U. 6\ -

2. Compute the current gain,

=

2.(>32 o f the circuit o f Figure 18.18.

ANSW^HR;

5. TWO-PORT IMPEDANCE PARAMETERS Definition and Examples The 2-parameters, or impedance parameters, relate ^-domain currents to ^-domain voltages, as one would expect. The z-parameters are the inverse o f the ^'-parameters in most cases.

IMPEDANCE PARAMETERS For the nuo-port o f Figure 18.10, the z-parameters zij relate the port currents to the port voltages according to the mati ix equation ■Vj-

’^11

-12'

w

V2

.^21

^22_

h

(18.20)

under the assumption o f zero initial conditions and no internal independent sources. Therefore, from equation 18.20, each individual z-parameter is defined according to the formulas

h h=0

-1 2

-

/,=0 (18.21)

Z2\ =

^22

h=0

/,=0

Since each zij is defined with one o f the ports open-circuited, i.e., /j = 0 or /-, = 0, the z ip are called open-circiiit im pedance param eters. Their unit is the ohm.


EXA M PLE 18.9 Compute the 2-paranieters for the circuit o f Figure 18.19.

FIGURE 18.19 A simple 7'-circuit for computation o f z-parameters in Example 18.9. S o l u t io n

Rather than apply the z-parameter definitions o f equation 18.21, we will use mesh analysis to obtain equation 18.20 directly. Step 1. A loop equation at port 1 yields

.y

“

.V

\

s/

(18.22)

Step 2. Similarly, a loop equation at port 2 produces V9 =

~

“

+ — (/] + h ) = — /i + f 3 + — ] h

s

~

s

\

s /

(18.23)

Step 3 . By inspection of the right-hand sides o f equations 18.22 and 18.23, 10 Z=

’^11

- 12'

^21

~22

s ]0

.y ay + 10

s

s

In the next example, we utilize the technique o f matrix partitioning to compute the z-parameters o f a n:-network. E X A M PLE 18.10. Compute the z-parameters o f the Jt-network o f Figure 18.20.

F I G U R E 18.20. Ji-network for Exam ple 18.10.

9~S


S o l u t io n

It is straightforward to write the following three loop equations: l/ ,= Z ,/ ,- Z ,/ 3 = Z 2/2 + Z 2/3 and 0 = - Z / i + Z 2/2 + (Z, + Z 2 + Z^)I^ Putting these equations in matrix form and partitioning the matrix appropriately yields

^2 0

‘ z,

0 1

-z ,

h

0

Z, j

Z2

^2

=

—Z|

Z2 j Z| + Z2 + Z3

h

Hence, using the matrix partitioning formula derived in Example 18.4, we obtain ■^11

^12 '

=

^22

1 0

■z,1

0-

1

0

^ 2.

Zi + Z 2 + Z 3

0 ■

1

- Z iZ i'

■

Z] + Z 2 + Z 3

^2

^2

-Z ,Z 2

^2

Z | (Z 2 + Z 3 )

^ 1^2

Z|Z2

Z 2 (Z ]+ Z 3 )

z, + z , + z .

Exercises. 1. In Example 18.10, suppose Z j = Z-, = 1 H and Z 3 is a 1 H inductor. Find the zparameters. AN SW ER:

-11

" i:‘

-21

-22

1

■.V+ 1

.v+2

1

1 ■ .v + 1

2. Now suppose that Z , and Z 2 are changed to 1 F capacitors and Z 3 is a 1 H inductor. Find the new 2-parameters. A.N'SWER:

’-II

- 12'

1

"21

-22

.v(.v-+2]

’ .V -

+1 1

I .V -

■

+ 1

979


Relationship to j-Paxam eters Since the z-parameters relate port currents to port voltages and the j/-parameters relate port volt ages to port currents, one might expect that the ^-parameter matrix and the ^-parameter matrix are inverses o f each other. Specifically, if •V,V2_

Z21

^12'

w

^22

h

then perhaps it follows that

h'

’-11

^12'

h

>21

Z22

^2

~

This relationship is valid provided Zj ’^11 ■^21

and

-1 ■Vi'

^12'

'y\\

3’12‘ -Vi-

.>’21

.''22 .^2.

^ ^ and^j -1

^22_

V|1

>’12'

>’21

>’22

■^11

^12'

■>’11

>’12'

>21

-22.

>’21

>’22.

Exercise. A certain two-port has z-parameters ’^11 ^21

^

which case

(18.24a)

-1

(18.24b)

212'

1

■5 + 1

^22

.9-^2

1

1 ■ 5+ 1

Find the ^-parameters. Can you construct a three-element passive circuit that has these ^/-parameters? ANSWER: ; H inductor.

>11

>12‘ _ 1 '.v + l

>'21

>22

.V -1

-1 • .v + l

: sec 1-igure 18.20 with Z, = Z t = 1 £2 and Z^ changcd to a 1

Despite this inverse relationship, some circuits have z-parameters but not ^-parameters, and vice versa, as illustrated by the following example.

EXAM PLE 18.11. Compute the z-parameters o f the circuit of Figure 18.21. Do the/-parameters exist?

O+

-O

V.

V,

o-

-o

+

FIG U R E 18.21 Resistive two-port having z-parameters but not ^parameters.

980


S o l u t io n

By inspection, uhe 2-parameters o f the circuit o f Figure 18.21 are V,-

R

R-

vs.

R

R

A ^2

The 2-parameter matrix, [z-^, is singular, since d e t[2 j =

I2 = 0. Because the [z-^ matrix does

not have an inverse, the circuit has no y-parameters. One can check j/, j directly to verify this claim. Consider Figure 18.22. Because Vj = 0, there is also a short circuit across Kj, making the ratio

vi 1 =

Vi

^-,=0

undefined.

O-

V,

R

v ,= o

OFIGURE 18.22. Equivalent circuit for c o m p u t i n g i n which port 2 is shorted, so that

= 0.

The Two-Dependent Source Equivalent Circuit As with thej)/-parameters, the z-parameters have a two-dependent source equivalent circuit inter pretation, illustrated in Figure 18.23b. Consider first the equation

Here,

equals the sum o f two voltages: z^ j/j plus the voltage due to a current-controlled volt

age source given by z^-Îj- This is precisely the left-hand portion o f Figure 18.23b. A similar inter pretation follows for Vj = -221A

O + V,

^2, ^22

(a)

Mr-

0

-0

+

+

^12 -

’ yielding the right-hand side o f Figure 18.23b.

V,2 .

-0

— o ^22

V,

+ V,

a

"211

(b)

F IG U R E 1 8 .2 3 (a) z-paramcters for modeled nervvork. (b) Tw o-dependent source equivalent circuit.

98


This equivalent circuit proves useful for computing voltage gains and input and output imped ances o f terminated two-ports. It should also be noted that there are other, equivalent circuits that interpret the 2-parameters. A similar remark can be made for j/-parameters.

6. IMPEDANCE AND GAIN CALCULATIONS OF TERMINATED TWO-PORTS MODELED BY Z-PARAMETERS Input and O utput Impedance Com putations Earlier we derived the formula for the input admittance o f a terminated two-port in terms o f the V-parameters, leaving the output admittance calculation as an exercise. Here, using ^-parameters, we derive the output impedance of the terminated two-port o f Figure 18.24 and leave the input impedance calculation as an exercise. Specifically, our first step is to derive the output impedance formula

ôiit ~ ^22

-12-21 ! 1 + •^v

(18.25)

as a function o f the net^vork z-parameters and the source impedance.

Beginning on the right-hand side o f Figure 18.24,

Vj = Z22^2 In calculating source,

^21^1

(18.26)

which is the Thevenin impedance seen by the load, the independent voltage

is set to zero. Hence, -(Z^ + Z j,) /j = 2,2 h

Solving for /, and substituting into equation 18.26 yields the output impedance formula o f equa tion 18.25.


Exercises. 1. Repeat the preceding derivation using a matrix method and Cramer’s rule. 2. Derive the following formula for the input impedance o f a terminated rwo-port:

V

^12^21

A n = ^ \ ---------(18.27)

<•22 +

Gain Calculations The next phase o f our rwo-port analysis is to repeat the ^'-parameter derivation o f the voltage gain o f the rwo-port in the context o f 2-parameters. Specifically, our aim is to compute

r

The ratio

^2

^2

^

^

C 8-28)

follows from voltage division: V^l _ Gvi = ^ =

To compute the gain Q ^ = Yl. V,

Z/,

(18.29)

first apply voltage division to obtain

-----^21^1 + ^22

^2

From the definition o f input impedance, /j =

n

Hence,

+ <-22 în

Substituting equations 18.29 and 18.30 into equation 18.28 yields the voltage gain:

-21

Gy = Gy2Gy\ =

\Z22 + ^ L / \ în +

(18.31) /

An application o f this formula and its derivation to a cascaded network of two-ports (two transis tor amplifier circuits) appears in the next example.

E X A M PLE 18.12. Consider the nerwork o f Figure 18.25, which represents a two-stage (transis tor) amplifier configuration. Each stage utilizes the same transistor in a difiêrent circuit configu ration. The first stage is an amplification stage that will amplify a small source voltage to a much larger one. The second stage is an impedance-matching stage used to match the load to the out put impedance o f the amplifier circuitry to achieve maximum power transfer, at least approxi mately. The 2-parameters, in ohms, for each stage in Figure 18.25 are given by

983


•350

2.667-

-1 0 ^

6667

21 =

; Z2 =

’ 1.0262 X 10^’ 1.0258

x

K/’

6 ,7 9 1 '

(18.32)

6 ,7 9 4

(a)

Compute the input impedances,

and

(b)

Compute the voltage gain,

(c)

Check the matching o f the load and output impedance o f the amplifier circuit.

160

V

FIG URE 1 8.25 Two-stage amplifier network. S o l u t io n

Step 1. Determine ^ straightforward application o f equation 18.27 using the z-parameters o f stage 2 produces the impedance

An! = ^11----- = 1.0262 X 10^ -

6

6 7 9 1 x 1 .0 2 5 8 x10^

+ Z,

Step 2. Compute the voltage gain, ticular, K..„ >^2

~ ^0 1 1 / ^ 2 ’ Z,

z,|

Z,,_ + Z22 Z/,.2

(18.33)

6794+16

18.30 applies here. In par I 6 ( i .0 2 5 8 x I0*>) (16 + 6 ,7 9 4 )3 ,2 6 2

(18.34)

The gain here is small, but remember that this stage’s real purpose is impedance matching, not amplification. By proper choice o f the z-parameters, the output impedance will approximately match that o f the load. This allows us to dispense with an expensive and bulky impedance-match ing transformer. Step 3. Compute

fo r stage 1. Here, observe that

(i-’quation 18.33) in parallel with the 2

kD resistor between the two stages acts as a load to stage 1. Z-^^j the load to stage 1, denoted Zj^ = 1239.8 H.

parallel with 2 kQ becomes

The input impedance seen at the front end o f stage 1 follows from equation 18.27:

4 , =

- ^ 1 2 ^ 1 - = 350 + Z22+^

li

^ 6 . 6 6 7 + 1 ,2 3 9 .8

= 6 8 7.3 n (1 8 .3 5 )

Chapter 18 • Two-I’orts

Step 4. Compute the voltage gain, G,,, applying equation 18.29 yields

^vl - T 7 -

=

V ,/VsJ o r stage 1. Using the result o f equation 18.35 and

j" ^21 ^ \ Z^. + Zj,j) \^/J + ^22/

i

-10^’

^(

1,239.8

[ 7 5 + 687 .3 ] 1 1 ,2 3 9 .8 + 6.6 6 7 ;

(18.36)

= -2 0 5 .7 .

Here the large gain of stage 1 leads to significant amplification o f the input signal. For example, a —40 mV sine wave would be amplified to a little over 8 V, which can drive a small speaker. Step 5 . Compute the overall voltage gain, Gy=

The desired gain is simply the product o f

equations 18.34 and 18.36, i.e., G’^/= x = “ 2 0 5.7 x 0.7388 = - 1 5 2 , which remains fair ly large. Indeed, a —40 mV sine wave would be amplified to approximately 6 V. Other amplifica tion stages could be added to further increase the overall gain. Step 6. Verify that the load matches the amplifier circuitjy to a reasonable degree. In this task, one first computes the output impedance, nation of

o f stage 1 using equation 18.25. The parallel combi

with 2 k n , denoted Z p = 1.732 kH, becomes the source impedance to stage 2. It

is then easy to compute the output impedance o f stage 2, again using equation 18.25. The answer is

= 17 Q. The details are left as an exercise.

Exercises. 1. For the circuit o f Figure 18.25, verify that the output impedance equals 17 Q. 2. In Figure 18.25, call the 2 kQ resistor R-,. Find a new value of Rj so that exactly equals 16 Q. ANSW ER: 1.783 k il 3 . Suppose the 75 O. source resistance in Example 18.12 is changed to 300

which would represent a

twin line connection bet\veen an ideal voltage source and the first amplifier stage. Redo the example. ANSW ER: Numerical values are obtained using the h)llowing M AFI.AB code; 1^ = 300; z z ll = 1.0262e6: zzl2 = 6791; 7.7.21 = 1.0258e6; /.z22 = 6794; x.ll = 350; vM = 2 .6 6 "; z21 = -le 6 ; z22 = 6667; R2 = 2e3; ZL = 16; Zin2 = zzl 1 - z/.I 2'/.z21/(/z22 + ZI.) (;2 = (zz21/Zin2)-ZL/(Zl.+zz22) ZLl = R 2 ’ Zin2/(R2+Zin2) Zin = z ll - zl2'z21/(z22 + Z L l ) G1 = (z21/(Rs+Zin))-ZLl/(ZI.Uz22) Gv = G 1 *G2 Zoutl = z22-zl2*z21/ (7ll+ R s) Zs2 = R 2*Zoutl/(R 2+Zouil) Zout = zz22-zz 12' zzl I /(zz 1 1+Zs2)

985


7. HYBRID PARAMETERS Basic D efinitions and Equivalences As we have seen, some circuits have ^-parameters but not z-parameters, and vice versa. A circuit element that has neither is the ideal transformer.

EXA M PLE 18.13. This example shows that the ideal transformer o f Figure 18.26 has neither znor/-parameters. From the definition o f an ideal transformer (Chapter 17), V2 = a V and /j =

-al-). Clearly, K, and

cannot be expressed as functions o f /j and I 2 , nor can /, and

be

expressed as functions o f K, and Vj- Hence, an ideal transformer has neither j/- nor z-parameters.

1 :a o+

+

V,

FIGURE 18.26 Ideal transformer, having neither j'- nor z-parameters.

Two-port circuits having neither z- nor/- parameters require an alternative modeling technique. The hybrid parameters offer one o f several alternatives.

HYBRID PARAMETERS Hybrid parameters, h-jy are a cross between y- and 2-parameters: a voltage

and a current

are outputs, with /, and Vj as inputs. Specifically, if the rwo-porr o f Figure 18.10 contains no internal independent sources and has no initial stored energy, then the hybrid parameters are defined by the matrix equation Vi-

•/'II

im -

h

/'21

hi

V2

(18.37)

This mixture o f variables actually arises as a simplification o f a model o f a common emitter configuration o f a bipolar transistor. From the context o f equation 18.37, />], has units o f ohms,

and ^21 are dimension less, and h i has units o f S.

As with both/- and 2-parameters, we interpret equation 18.37 as a two-dependent source equiv alent circuit, as illustrated in Figure 18.27.

‘)86

Chapter 18 'Two-Ports

'2 + V.

a

+

-o +

h,2~ _ h21 h22 _ ■

V. o-

-o

(a) FIGURE 18.27 (a) Hybrid parameters, (b) Two-dependent source equivalent circuit.

Exercise. Justify the two-dependent source equivalent circuit interpretation o f Figure 18.27; i.e., apply KVL and KCL to the circuit o f Figure 18.27b to derive equation 18.37.

The definition o f each /^-parameter follows directly from either the preceding equivalent circuit or from equation 18.37. For example,

(18.38a)

V'-,=0

Because Aj| is the ratio o f an input voltage to an input current, it is an input impedance. Since

^2 = 0, /?], is termed the short-circuit input impedance. Notice, however, that A, j is simply related to both the and the z-parameters as follows: I

/ ,„ = ^

'I I

-12^21 (18.38b)

The second //-parameter, />2 p is called the short-circuit forward current gain, since it is the ratio o f I j to /] under the condition K, = 0, i.e.,

(18.39a)

V^=0 From the 2-parameter equation Vj = ^21A pretation.

^'>2-^2’

^2 “

*2 , has a simple z-parameter inter

(18.39b) ^-,=0

^22

The third //-parameter is

/,=0

(18.40a)

Since it is the ratio o f Kj to Vj under the condition that port 1 is open-circuited, i.e., /j = 0, it is called the reverse open-circuit voltage gain. Interpreting //p in terms o f the y-parameters, we obtain

987


>'12

I,=0

(18.40b)

'"'I

Finally, we note that the open-circuit output adm ittance is

h l' n = -

(18.41a) /,=0

and has units o f S. The word “open-circuit” suggests a z-parameter interpretation. Considering that V-y = 221-^1

h ~ 1 _ V12J 21 — - >'22-----------:22 3’11

/,=0

(18.41b)

This relationship is similar (notice the subscripts) to equation 18.38b, which determines the shortcircuit input impedance. We will return to these equivalences later, after we gain some computational experience.

Com putation o f /^-Parameters The first example o f this subsection demonstrates a circuit that has neither z- nor ^/-parameters.

E XA M PLE 18.14. Consider the two-port o f Figure 18.28, whose front end is a short circuit and whose secondary is an open circuit. Thus it has neither z- nor ^-parameters. Our objective is to compute the /^-parameters. I

L -O +

O+

V,

-o FIG URE 18.28 Simple two-port with /^-parameters but neither z- nor ^-parameters. SO L U T IO N . By inspection, the A-parameters are V,-

0

O' •/r

/2.

0

0

V2

T he second example illustrates the computation o f /^-parameters for an ideal transformer circuit.

Chapter 18 • Two-Porcs

EXA M PLE 18.15. Find the /^-parameters o f the c%vo-port in Figure 18.29.

1 :a

O--------+ •

------o *+ A

V.

+

R

V,

V,

FIGURE 18.29 ideal transformer circuit for Example 18.15. S o l u t io n

Step 1. Construct an equation fo r K,. From the primar}' and secondary voltage relationship o f an ideal transformer,

V, = - v - , By KVL at port 2, =

—RI->, in which case n V, + - / o

1

a ~

Step 2. Construct an equation fo r

a ~

(18.42)

in terms o f the other variables. From the primary and second

ary current relationship o f an ideal transformer,

/ 2 = -/ , Cl

(18.43)

Step 3. Write equations 18.42 an d 18.43 in matrix form, and solve fo r Kj and I-, in terms ofl^ and

Vj. In matrix form, equations 18.42 and 18.43 are

Solving for the vector [ Kj

rv/,1

0

i

/o

- a

0

a

/ll

(18.44)

produces the /^-parameter equation -1

1 -V,-

a

h 0

1

r 0n - -

a

-M

a

0

h' VS

\R —r a~

\' — Cl

-1

0

a

^2


Exercise. In the circuit o f Figure 18.29, suppose the resistor R is connected in parallel (instead o f in series) with the secondar)' winding o f the ideal trans former. Find the new />parameters. ANSWMR: /»,, = 0. /»,, = Mu. A,, = -\hi,

= MR

EX A M PLE 18.16. Find the /;-paramcters o f the circuit o f Figure 18.30. L

R. O----+ V,

'2

o-

V,

-o

FIGURE 18.30 Simple circuit for illustrating A-parameter definitions. S o l u t io n

Step 1. Find A, j using equation 18.38a. With V2 = 0, i.e., port 2 shorted,

= R^ v-,=o Step 2. Find

using equation 18.39a. Again with

= 0, i.e., port 2 shorted, the current I 2 =

-/ j since all current flows through the short circuit. Hence

V-,=0 Step 3. Find A,2 using equation 18.40a. W ith /j = 0, i.e., port 1 open-circuited, K, = V2 since there is no current through Ry Hence,

/,=0 Step 4. Find A22 using equation 18.4 la. Again with /j = 0 , /2 = (C^ +

= Cs + Gt /,=0 In summar\-,

\Rx

1

-1

Cs + G^

in which case


990

Exercises. 1. If all parameters in Example 18.16 are 1 with proper units, find the /;-parametcrs. ANSW ER: 2. If the capacitor in Example 18.16 becomes an Z. H inductor, find the new A-parameters. ANSW ER:

-1

— + G-. Lv

Impedance and Gain C alculations o f Terminated Two-Ports This subsection analyzes the impedance and gain properties o f a terminated two-port character ized by /^-parameters as in Figure 18.31, similar to the analysis done with both the z- and )'-parameters.

V

FIG URE 18.31 Doubly terminated hybrid equivalent circuit of a two-port. Vi Recall that Zj„{s) = — . From the right half o f Figure 18.31,

~

~

h It follows immediately that /hi

From the left-hand side o f Figure 18.31,

(18.45)

= A|j/| +

Substituting for Vj using equation

18.45 implies that the input impedance is

1,22+ Yl

/,

(18.46)

Exercise. Show that the output admittance is given by I2 K)ut ~

~ ^*22 “

//|o/h 1

h\ 1 +

(18.47)


991

Knowledge o f the input and output impedances/admittances permits us to derive various gain for mulas in terms o f the //-parameters. For example, consider again the left half o f Figure 18.31. Since the input impedance is known from equation 18.46,

h=^V \

(18.48a)

Thus, from equations 18.45 and 18.48a,

^2 = ------ ^ / | = ------ V,

1122+Yl

în

(18.48b)

implying the voltage gain formula, V2

^

^v2 = 7 7 =

+ W

^1

(18.49)

Voltage division at the front end o f Figure 18.31 yields the other voltage gain formula, G, _

Zi„ + Z^

(18.50)

The overall voltage gain is the product o f equations 18.49 and 18.50, i.e., ^ ^ 7 7 = C j.|Gi,2 = -

Exercise. Compute the current gain A N SW ER:

1

/?21

•

^2 ^ ^lM\ /1 h->'^ + )' /

8. TRANSMISSION PARAMETERS Transmission or f-parameters were first used by power system engineers for transmission line analysis and are still so used today. They are sometimes called ABCD parameters.

V92

Chapter 18 • Two-I’orts

TRANSMISSION PARAMETERS T he /-parameter representation has the matrix relationship

with the matrix, T =

Vj-

■^11

hi

■V2 ■

/ l.

fl\

t il

-I2

( 18.52)

called the /-parameter matrix. As with th e ^ , z-, and /^-parameters,

the entries t-j are defined as follows:

^ " ■ ^ 2 /, = 0

^-,=0 (18.53)

v^=o

/,=0

The matrix equation 18.52 leads directly to the relationships o f equations 18.53 by setting the appropriate quantity, I 2 or Vj, to zero.

In computing a single t-j vs'ith equations 18.53, some care must be exercised in exciting the circuit. By definition,

h= ()

The ordinary interpretation o f this equation is: apply an input V-, and find an output the condition that

= 0. 'I'hen

j is the ratio o f the Laplace transform o f the response

under to that

o f the input K,, i.e., a reverse voltage gain when port 2 is open-circuited. This situation causes a predicament: an independent voltage source for Vj causes a current

to flow. To circumvent this

predicament, we use the slightly modified formula 1

^11 /7 = 0

The quantit)'

^

forward voltage gain when port 2 is open-circuited) suggests that

we excite port 1 by Kj with port 2 open-circuited, which forces /-, = 0. It is then straightforward to calculate /j j as the inverse o f

Similar interpretations must be made with regard to

the other defining formulas in 18.53. A simple example illustrates r-parameter calculations.

EXA M PLE 18 .1 7 . Consider again an ideal transformer circuit, shown in Figure 18.32. Here V-,

=

and /j =

This leads to the /-parameter matrix

0 \ V2 a 0

a

-/o


1 :a

O +

-o +

V.

FICiURt 18.32 Simple cransformer circuit for Example 18.17.

Input and output impcdance calculations for terminated two-ports modeled by f-parameters do not follow the usual pattern. Nevertheless, the t\s'o-port of Figure 18.33 has input impedance 7

_ h\'^L +^12 (18.54)

and output impedajice 7

ôut

= ^22^v *^^12 ^21^5 +^11

(18.55)

The derivation o f these results is left as a homework problem.

+ V,

FIGURE 18.33 Terminated r\vo-port modeled by /-parameters.

Exercise. The rwo-port o f Figure 18.33 has r-parameters r, j = 0 .0025, ^p= 500 ^ ,^ 21= 3 .1 2 5 x 1 0 -8 S, and = 0.00625. IfZ^ = 200 kQ. and = 20 kH, find AN SW ER: 80 kL X 200 k L l

One o f the most important characteristics o f /-parameters is the ease with which one can use them to determine the overall /-parameters o f cascaded rwo-ports, as illustrated in the next example.

994


EXA M PLE 18 .1 8 . Compute the f-parameters o f the cascaded two-port o f Figure 18.34 in terms of

and Tj, the t^vo-port parameter matrices o f the first and second sections, respectively.

FIGURE 18.34 Cascade of two two-ports modeled by f-parameters. S o l u t io n

From the definition o f the f-parameters for each two-port, 1

1

/ l. But since V2 =

r ^21

o

and

= 7’-,

[ K;*T 1

h

.-^ 2 .

.-^ 4 .

and -1-, - ly it follows that

1

•

r^ ii =

■V3-

■

= T\1

7 ’i1

-h

■V

= T\Tl 1 L h

4

■

.-^ 4 .

implying that the f-parameter matrix o f the cascade o f Figure 18.35 is simply

Tj.

Exercise. The f-parameters o f the two-ports in Figure 18.34 are (in standard units) 0.1

T\1 =

0

0 • 10

and Ti =

•2 0.1

10' 1

Find the f-parameters o f the cascade (T j comes first, followed on the right by Tj )• Then find Vj when a voltage source o f 4 V is applied to port 1 o f the cascade and port 2 o f the cascade is opencircuited. A N S W I- R : T =

0.2 10

9. GENERAL RELATIONS AMONG TWO-PORT PARAMETERS The h-.y-, t-, and z-parameters are interrelated. This subsection derives several relationships, with others left to the homework problems. The complete set o f results is summarized in Table 18.1 for reference purposes. To express the /7-parameters in terms of the z-parameters, first note that

995


Vl-

■^11 hi2

h'

.^2.

.^1

.'^2.

and

■Vl'

Zii

Z\2

.'^2.

Z21

^22. h .

The trick is to rewrite the jzf-parameter equations so that Vj and I2 are on the left with /j and Vj on the right: ~ ^ \ lh - ^\\h Z2 2 I 2 - ^ \ h ~ ^2

Writing these two equations in matrix form yields O' •/r

1

-Zi2

y ,-

Zii

0

Z22

h.

-Z 21

(18.56)

1

Solving equation 18.56 for the vector [V^ /j]ûnder the proviso that ^^2 T ,-

h

■1 -Zl2 0 Z22

-1

■Zii

O'

-Z2\

1

ZIIZ22 “ 212^21

Z22

0 yields

h' V2

(18.57)

Z12 r/i

-Z 2 1

1

V2

Thus, we have used matrix methods to direcdy compute the A-parameters in terms of the 2^-parameters under the condition that 222 ^ hu

h i'

1

’211^22 “ Zl2^21

Z\2

h\

^22,

Z22

-Z 2I

1

(18.58)

All other relationships are derived in a similar manner. For example, to express j^-parameters in terms of /^-parameters, one must rewrite the A-parameter equations so that and /j appear on the left-hand side with and on the right. Then using matrix form and inverting the appropri ate matrix under the condition of a nonzero determinant produces the desired result. Vw>

w

Table 18.1 specifies the interrelationships among all the parameters studied thus far.

Chapter 18 * Two-Ports

FABLE 18.1. Interrelations among Two-Port Parameters, Where y Can Stand for z,y, /;, or t in A/ = y ,, ^22 " /12 >'21

2-Paranietcrs

z-Parameters

y-Parameter

^11

^12

Z21

Z2 2

^22 A-■^21

/;-Parameter

/-Parameter

j'-Paramcters

>'22 Aj -> ’21 Av

~>'12 Ay >’11 Av

A/;

>’11 >’21

/!21

hn.

/222

>’12 >22

!h i I1 2 2

1

J_

■12 Ac

/-Parameters

//2i

_- h 12

^21

^21

1

ti2

/21

h\

[^22

-A t ■

^12

^12

^

-1

^11

/12

^12

^12

A/ ■

^11 Az

■ A'

^12

^22

Z2 2

>’I1

>’11

^21

1

>’21

A>-

Z22

Z2 2

>’l 1

>^l 1

•^11

A zl

-> ’22

-1

-A/2

- h 11

^21

^21

>’21

V21

/?2i

1

^22

-^ > ’

-> ’11

-/?22

_^21

^21

>’21

J 21

/2.21

/Ml

Ar ■

/Ml

A::

fh 1

[^11

1

/M2

^22

/22

-1

^21

^22

^22

Il2\

^11

^12

-1

^21

^22

h')')

Chapter 18 * Two-Ports

Exercises. 1. Use the code below to create an m-file in MATLAB for conversion o f 2-parameters to /^-parameters. Verify that Z = [1 2;3 4] produces H = [-0 .5 0 .5 ;-0 .7 5 0.25]. % convert z parameters to h parameters function [h, hi I,h l2 ,h 2 1 ,h 2 2 ] = ztoh(z) zl 1 = z ( l,l) ; zl2= z(l,2);z21= z(2,l);z22= z(2,2); deltaz = z ir z 2 2 -z l2 " z 2 1 ; h 11 = deltaz/z22; hi 2 =zl2/z22; h21 = -z21/z22; h22 = l/z22; h = [ h l l h l2 ; h21 h22]; 2. Use the code below to create an m-file in MATLAB for conversion o f z-parameters to ^-parameters. Verify that Z = [1 2;3 4] produces T = [1/3 -2/ 3;l/ 3 4/3]. %converting z to t paramters function [t,tl I,tl2 ,t2 1 ,t2 2 ] = ztot(z) zl 1 = z (l,l); z l2 = z (l,2 ); z21=z(2,l); z22=z(2,2); deltaz = z ll * z 2 2 - z l2 * z 2 1 : tl 1 = zll/z21; t l2 = deltaz/z21; t21 = l/z21; t22 = z22/z21; t = [ t i l tl2 ; t21 t22]; 3. Write m-files for the remaining items in the conversion table for your own future use.

10. RECIPROCITY Writing node equations for an ordinar)' linear circuit leads to a matrix equation having the form A'

>11

•••

y\n

Vi'

■vî =Y

y'n1 ■■■ y'nn Often the node admittance matrix Y = [y^^] is symmetric, i-e-.7,y = yjj for ; ^ J. Such networks are termed “reciprocal.”

998

Chapter 18 •Two-l’orts

RECIPROCAL NETWORKS Any circuit that has a symmetric coefficient matrix either in a nodal equation or loop equa tion representation is said to be reciprocal. Further, a two-port represented by either ^-param eters or ^-parameters is said to be reciprocal if Z j 2 = Z21

~^2r

From Chapter 3 we know that circuits without dependent sources have symmetric coefficient matrices in both the nodal and loop equation representations. On the other hand, the symmetr)' o f the z-parameters and j'-parameters is typically lost when dependent sources are present. In gen eral, we can prove that a circuit containing R ’s, L's, C ’s, and transformers, but no dependent sources or

independent sources, is a reciprocal network. From the definition o f a reciprocal two-port, we can conclude further that if the hybrid and/or f-parameters exist, then from Table 18.1 = “ ^21 and/or

Conversely, any two-port that has parameters satisfying these conditions is said to be reciprocal.

Exercise. Recall that the relationship between the ^-parameters and z-parameters is

-11

-12

'II

A/

^21

^21

1

ho

^21

^21

^21

Show that if 2 j 2 = ^21’ if A/ = 1, then Zp = z^j.

Ar = 1, where A denotes “determinant.” Then show the converse, i.e.

Proving that any two-port created from a reciprocal network has symmetric z-parameters is straightforward. We write the loop equations in matrix form with

/p

and /, being the volt

ages and currents o f ports 1 and 2. Since the underlying netw'ork is reciprocal by assumption, its loop equation has a symmetric coefficient matrix partitioned as shown:

■Vi-

’ ^11

-12

-1 3

vs

-12

-22

^23

-In

^13

^ 23

^33

-3/j

'•1/j

'-2/1

-3/?

VV

=

999


Solving for [Kj ^-5]^ by the method o f matrix partitioning yields ■V,-

^^2

r

-li \V,| - VV,2VV2VVV2i ■

h

where the W-j matrices are defined in the obvious way.

^11

^12

/l1

C21

Z2 2

h

j is symmetric. Since the inverse o f the

symmetric matrix Wj-y is symmetric, and since the sum o f two symmetric matrices is symmetric, the resulting z-parameters are also symmetric. The symmetry o f the ^-parameters follows by the symmetry o f the inverse o f a symmetric matrix. We now set forth physical interpretations o f reciprocit)^

Reciprocity Interpretation 1: Consider a reciprocal two-port, N. If the voltage inputs,

are

the same as in Figures 18.35a and b, then by reciprocit)' the zero-state short-circuit responses, I 2 a and l-^y, coincide.

FIG URE 18.35 Equivalence o f short-circuit zero-state responses induced by voltage sources for reciprocal netw'orks.

What reciprocity interpretation 1 says is that if one applies a voltage at port 1 and measures the short-circuit current at port 2 with an ideal ammeter (zero meter resistance), then applying the same voltage at port 2 would result in measurement o f the same short-circuit current at port 1. Conversely, we can show that if reciprocit}' interpretation 1 is true, then the j'-parameters are sym metric. To see this, observe that the configuration o f Figure 18.35a implies

1()()()


_ ^<>“1 Vo I —----„

ha

and the configuration o f Figure 18.35b implies

_ Ôllt » 2 - —

în r,„=o

Thus one must havej/p = 72 i>

^2h

symmetric j^-parameters.

Reciprocity Interpretation 2: Consider a reciprocal rvs'o-port, N. As illustrated in Figure 18.36, is the same in Figures 18.36a and 18.36b, then the open-circuit zero-state responses V-^ and

coincide.

FIGURE 18.36 Equivalence oF open-circuit zero-state responses for a reciprocal network.

Reciprocit)' interpretation 2 says that if one injects a current at port 1 and measures the voltage at port 2 with an ideal voltmeter (infinite input resistance), injecting the same current at port 2 would result in measurement o f the same voltage at port 1.

Exercise. Assuming reciprocit)' interpretation 2 is true, show that it follows that

2-parameters are symmetric.

^2V

1001


Reciprocity Interpretation 3 : Consider a reciprocal two-port, N. As illustrated in Figure 18.37, leads to a zero-state response denoted ~ hii‘ as shown in Figure 18.37a, and leads to a zero-state response denoted

as in Figure 18.37b. For a reciprocal two-port, the short-cir

cuit current ratio in Figure 18.37a is equal to the open-circuit voltage ratio in Figure 18.37b.

Exercise. Show that if reciprocity interpretation 3 is assumed true, it follows that A|, = the forw ard short-circuit current gain,

în

, equals the reverse open-circuit voltage gain, —

\i.e., .

V;«

The three reciprocity interpretations have rigorous proofs. They however, are beyond the scope ol the text. Finally, a two-port that is reciprocal has an equivalent circuit representation with no dependent sources. For example, suppose a reciprocal two-port has the z-parameter representation V\ - 2|I /j + 2|2 A and

~22

^2

1002


in which case (18.59a) and (18.59b) liquations 18.59 have the so-called T-equivalent circuit interpretation given by Figure 18.38. I. ^22’ ^12

-o +

+

V.

V,

oFIGURE 18.38 2-parametcr T-equivalent circuit of a reciprocal 2-port. The resistors represent general impedances.

Exercise. Suppose the r\vo-port o f Figure 18.39 is reciprocal and modeled by ^param eter equa tions. Compute K, ^ Yj, and in terms o f the ^/-parameters, |, j/p, AN SW ER:

1 .J'l2’ -^22

O-

-'12

-o

-I-

-I-

V.

V,

oFIGURE 18.39 ^parameter jt-equivalent circuit of a reciprocal rwo-port. The resistors represent general impedances.

11. PARALLEL, SERIES, AND CASCADED CONNECTIONS OF TWO-PORTS A general linear two-port has four external terminals for connection to other networks as illus trated in Figure 18.40a. When the input and output ports have a common terminal, only three external terminals are available, as indicated in Figure 18.40b. Such a rwo-port is typically called a common-ground tivo-port, although the common terminal is not necessarily grounded in the sense o f being connected to earth.

1003


+

V.

-•

•-

General two-port

A

+

+

V,

V.

'^Common'^ ground two-port

A

+

V,

W 1.

L (b)

(a)

FIG URE 18.40 (a) A general rwo-port. (b) A cornmon-ground r\vo-port. Interconnecting two common-ground rvvo-ports

and

forms a new two-port N. Figure

18.41 shows three typical interconnection structures: parallel, series, and cascade. To avoid over crowding o f symbols, Figure 18.41 omits all voltage and current reference labels. These labels are understood to conform to those in Figure 18.40a. In particular, note that at each port the current entering one terminal must equal the current leaving the other terminal for the two-port param eters to be valid or meaningful. An interconnection o f rwo-ports has a new set o f z-, y-, /;-, or ^-parameters obtained very simply from the individual t^vo-port parameters. The interconnected rwo-ports o f Figure 18.41 have new parameters computed from those o f 1.

and

as follows:

For the parallel connection o f Figure 18.41 a. (18.60)

2.

For the series connection o f Figure 1 8 .4 lb , (18.61)

3.

For the cascade connection o f Figure 1 8 .4 lc, (18.62)

(a)

(b)

1004

Chapter 18 • Two-Porrs

•-

-• Na -#

#■ N

(c) FIGURE 18.41 Three typical interconnections of rvvo-ports. (a) Parallel connection, (b) Series connection, (c) Cascade connection. where Y, Z, and T denote the admittance, impedance, and transmission parameter matrices, respectively, and the subscripts a and b refer to the networks

and N^, respectively.

A derivation o f these formulas is straightforward. For example, from the definitions o f

and N^,

where Vi.

r/i.

ha and similarly for the voltage and current vectors o f N^. From Figure 1 8 .4 la,

^“

K ^ h - ^ substitution yields I - (K^+ Y^Vy w'hich verifies equation 18.60. To verify equation 18.61, consider Figure 1 8 .4 lb. Here = /. By direct substitution V = (Z^+

and

and

= Z^/^. But V=

verifying equation 18.61.

Equation 18.62 w'as derived earlier in this chapter. The derivation o f equations 18.60 and 18.62 is easily extended to the case o f more than two rwoports; 1.

If two or more common-ground t\vo-ports are connected in parallel, then K=

2.

k. h-

-

(18.63)

If two or more two-ports, common-ground or not, are connected in cascade, then

r = r ,r ,r ,...

(18.64)

Equations 18.62 and 18.64 for the cascade connection hold whether or not the component rwoports are o f the common-ground t)'pe. However, equations 18.60 and 18.63 for parallel connec tions in general \\o\(\ only for common-ground two-port connections, as shown in Figure 18.4 la. Similarly, the series connection equation 18.61 holds only for the case illustrated in Figure 1 8 .4 lb. Series connection o f two general rwo-ports (Figure 18.40a) or series connection o f more than t\vo common-ground two-ports (Figure 18.40b) requires an ideal transformer for coupling, as demon strated in the homework problems. Examples 18.19 and 18.20 explain why equations 18.60 and 18.61 fail when two non-common-ground two-ports are connected together.


100=)

E X A M PLE 18 .1 9 . This example illustrates the difficult)^ with a non-common-ground series con nection. Consider the two-port shown in Figure 18.42, which is a series connection o f two com ponent two-ports. The z-parameters o f the individual two-ports are given by ■2

r

1

and

2

Zf,u -

Show that Z 7^:

+ Zy when

(b)

Show that Z =

+ Zy if

(c)

Justify the statements o f parts (a) and (b).

(a)

=6

■Rx+2

1

1

/?3 + 2

■

and R-^ = 3 Q..

= Rj = 0.

FIGURE 18.42 A series connection that causes difficulty. S o l u t io n

(a) Observe that the parallel connection o f the 6 H and 3 ^ resistors is 2 Q. Thus, by direct cal culation, the z-parameter matrix o f the interconnected two-port is Z=

■6 4

4' 6

•2 ^ Zn £i + Z;, I) =

1

r 2

-1-

■8

r

10

2-

2

7

—

1

5

(b) With R ,= R . = 0,

Z,, =

On the other hand, by direct calculation, the z-parameter matrix o f the interconnected two-port is •4 Z=

2

24

=

+ Ay =

•4

1

2

4

(c) In part (a), Z ^ Z^ + Z^ because, after the interconnection, neither

nor

acts as a two-

port, as defined in Figure 18.40. This can be understood by inspecting Figure 18.43, with the indicated the loop currents.


1006

------- O-

0

-oV

V,

-------o-

o ---------^ s / s A FIGURE 18.43 A nonzero W ith y?j and Rj nonzero, the mesh current /|. Hence, for the left terminal pair o f

leads to /j ^ /*.

is in general nonzero. Observe that 1*^ = 1^ the current entering the top terminal does not equal

the current leaving the bottom terminal. With unequal terminal currents,

no longer has a z-

parameter characterization, because the 2-parameter definition requires equal currents entering and leaving the terminal pair, as per Figure 18.40. On the other hand, if

= /?, = 0 in Figure

18.43, then the third mesh equation is satisfied for arbitrary values o f the mesh currents /j, Ij, and 7^, as there is no resistance at all in the third mesh. In particular, let and the 2-parameter characterization o f

= 0, in which case /|j = 7j,

is valid. Similar arguments hold for N^, If

= R^ =

0, equation 18.61 holds because Figure 18.43 now has the same interconnection as depicted in Figure 18.4 lb.

E X A M PLE 18.20. This example illustrates the problem o f a non-common-ground parallel con nection. Figure 18.44 shows rwo two-ports connected in parallel. Before the connection, each two-port has ^-parameters 0.7

- 0 .2

- 0 .2

0.7

S.

After the connection, by direct calculation, the new two-port has ^-parameter matrix

Y=

Clearly,

Y^+ Yy = 2Y^\n this case.

r 1.625 -0 .6 2 5

-0 .6 2 5 1.625

S.


lOO'

FIG URE 18.44 New rvvo-port N subject to invalid application of equation 18.60 for non-commonground rwo-ports in parallel. The reason for the failure o f equation 18.60 under these circumstances is the same as for the cir cuit o f Example 18.19. If a voltage source is applied to port 1 o f N, we find that the currents and /j'^^ are not equal. Thus, eters in forming N.

cannot continue to be characterized by a set o f rwo-port^-param

There are, however, some special cases o f non-common-ground two-ports for which equation 18.60 holds for a parallel connection. The following is one example.

E X A M PLE 18.21. fhis example illustrates how to achieve a parallel interconnection o f two gen eral rwo-ports o f Figure 18.40a so that equation 18.63 remains valid. Reconsider the tw'o-ports and

o f Example 18.20, which have y-parameter matrices

a 1:1 ideal transformer is placed at the front end o f

and Y^, respectively. Suppose

in Figure 18.45; call this new two-port

(a)

Show that the ^-parameter matrix o f N^* is Y^.

(b)

Show that the j-param eter matrix o f the interconnection o f even though N^* and

are not common-ground two-ports.

and

is Y = Y^ + Y^,


1 008

FIG URE 18.45 Equation 18.63 holds for these non-common-ground rwo-ports. (a) Parallel connec tion of rwo general rvvo-ports. (b) Justification of equal currents at rw'o terminals of each port. S o l u t io n

(a) The )'-paramcrers o f N j arc rhe same as for ed dor positions forces N * and if N * and

because rhe ideal 1:1 transformer with indicat

to have the same port currents and voltages at port 1. In fact,

were enclosed in a box with only the leads observable, the two-port properties would

be identical. (b) To show that the ^/-parameters of N (the interconnection o f N * and must first show that

(ii)

and 12 1 , - / ,j

are Y =

+ Y^, we


l()()‘)

To confirm these equalities, consider Figure 18.45b, which represents Figure 18.45a with all the nonessential components removed to avoid overcrowding figure. Four Gaussian surfaces are drawn,

5,^,

and 5,^^. Note that all Gaussian surfaces go through the core o f the trans

former. Recall that KCL holds for a Gaussian surface: the algebraic sum o f the currents entering (leaving) the surface is zero. This immediately asserts the validit)' o f statements (i) and (ii) above. Hence, within the interconnection, the two-ports N * and continue to act as individual twoports. Therefore, equation 18.60 remains valid, i.e., n .4 Y = Y a -fY b =

- 0 .4

- 0 .4 1.4

This example extends directly to multiple parallel interconnections. Although the precise condi tions for the applicability o f equations 18.60 and 18.61 to the parallel and series connections o f non-common-ground two-ports are known, they are not practical enough to be included here. Our emphasis is on interconnections o f common-ground two-ports, which occur most often in practice.

12. SUMMARY This chapter presented a unified setting for one-port analysis while providing a comprehensive extension to two-ports. Two-ports are common to numerous real-world systems such as the utili ty power grid that delivers power to a home from a generating facility through a transmission net work. Another representative two-port is a telephone system that delivers a speaker’s voice to a lis tener by sending a converted electrical signal through a transmission network. The characteriza tion o f a two-port for such systems is done through their input-output properties. Four sets o f characterizing parameters were developed: impedance or z-parameters, adm ittance or ^-parame ters, hybrid or //-parameters, and transmission or f-parameters. In order to analyze various aspects o f a system characterized by a two-port, formulas for computing the input impedance/admittance, the output impedance/admittance, the voltage gain, etc. were derived. Quantities such as voltage and power gain are very important aspects o f amplifier analysis and design, as illustrated in Example 18.12, which depicts a two-stage transistor amplifier configura tion. Although Example 18.12 utilized the medium o f z-parameters, the more customar)' medi um for transistor amplifier design is //-parameters. Conditions and formulas for parallel connection o f two-ports were presented in terms of^-parameters while series connections were studied using 2-parameters. Formulas for determining the transmission parameters o f cascades o f rwo-ports were also developed. In addition, the chapter introduced and interpreted the notion o f reciprocit)' in terms o f the different two-port parame ters. Reciprocal circuits generally contain only R's, Z.s, Cs and transformers. Under certain restrict ed conditions a reciprocal network may contain a dependent source, as the homework problems will investigate.

101 U


13. TERMS AND CONCEPTS Admittance or j-parameters: descriptive two-port parameters in which the port currents are functions o f the port voltages.

Hybrid parameters or /^-parameters: descriptive two-port parameters in which Kj and

are

expressed as functions o f /j and K,.

Impedance or z-parameters: descriptive two-port parameters in which the port voltages are func tions o f the port currents.

Input admittance; the admittance seen at port 1 o f a possibly terminated two-port. Input impedance; the impedance seen at port 1 o f a possibly terminated two-port. Matrix partitioning: the partitioning o f a matrix set o f equations into groups to obtain a simpli fied solution in terms o f the partitioned submatrices.

Norton equivalent of one-port: a current source in parallel with the Thevenin impedance. Open-circuit impedance parameters; the impedance or z-parameters. Open-circuit output admittance: the hybrid parameter hj-,Output admittance: the admittance seen at port 2 o f a rvvo-port possibly terminated by a source impedancc.

Output impedance: the impedance seen at port 2 o f a two-port possibly terminated by a source impedance.

Partitioned matrix; a matrix that is partitioned into submatrices for easier solution o f sets o f equations.

TU-equivalent circuit: equivalent circuit o f a reciprocal two-port containing three general imped ances in the form o f k, as in Figure 18.39.

Reciprocal network: a network whose node equations or loop equations have a symmetric coef ficient matrix.

Reciprocal two-port; Z]2 = ^21 ~J2V Reverse open-circuit voltage gain; the hybrid parameter A jj. Short-circuit admittance parameters; the admittance or ^'-parameters. Short-circuit forward current gain; the hybrid parameter Short-circuit input impedance; the hybrid parameter T-equivalent circuit; equivalent circuit o f a reciprocal two-port having three general impedances in a T shape as in Figure 18.38.

Terminated two-port: a two-port attached to a load impedance and a source with, in general, a nonzero impedance.

Thevenin equivalent of a one-port: Voltage source in series with the Thevenin impedance. Transmission or ^-parameters: parameters where and are expressed as functions o f Vj and -h -

Two-dependent source equivalent circuit: equivalent circuit tor a two-port containing imped ances/admittances and two dependent sources.


P r o b le m s

+ V -

ONE-PORTS 1. In Figure PI 8.1, suppose n , /?^ = 1 k n , and (3 = 49. Find port. Then find

100 V,Z^ = 20 o f rhe one-

and the power to the one-

port. Does Z j have any effect on the answers?

Figure P I8.3 C H EC K : (a) 1 1 k n 4. Consider the circuit o f Figure P I 8.4.

= \ 0 j2 cos(lOOOf) A, a = \0, R

Suppose

= i o n , and C = 1 îF. Figure P I8.1

(a)

Find the Thevenin equivalent seen by

(b)

Find the load impedance for maxi mum average power transfer and com

the load.

C H E C K : 50 V 2. In Figure P I 8.2, suppose 2500

= 100 V, Z j =

pute the resulting average power.

= 50 Q, and (3 = 49. Find Z-^ o f the

one-port. Then find

and the power deliv

(c)

Find a series RL load that achieves maximum power transfer.

ered to the one-port. Does Z^ have any effect on rhe answers?

1 :a

'»(t) Figure P i8.4 C H EC K ; (c) 1 kQ resistor in series with 1 H inductor Figure P I8.2 C H EC K : 50 W

5. T h e loop equations in standard units describing the one-port o f Figure P I 8.5 are

3. Consider the circuit o f Figure P I 8 .3 .

given by

Suppose C = 0.1 mF, y(;;(0“) = 10 V, Z j = 1 kH, Z2 = 2

Z 3 = 1 kl^, and

vr

= 4 mS,

(a)

Find the input impedance Zy^^.

0

(b)

Find v^r) for t > 0 .

0 (a)

=

■10«

1

-a

h

a

0.5

0

h

-1

0

0.5

h

Find the input impedance o f the oneport as a function o f a. Hint: Consider

(b)

Cramers rule. If = 2 and a voltage source is con nected so that V|(/)= 12V 2 cos( 2 /) V, find the average power consumed


1012

8. Consider the circuit o f Figure Pi 8.8. In solv

by the one-port. (c)

Repeat part (b) for a = (1+ ;4/3) at

CO

ing this problem, fully utilize the properties o f an ideal transformer.

=2 rad/sec.

(a)

Find the Thevenin equivalent circuit as a function o f the turns ratio, b, and the resistance, R.

i-Port

(b)

If

= cos(10r + 45°) A, find the

Norton equivalent circuit with the Norton source represented as a phasor.

Figure P I8.5 AN SW ER: (a) Ga U, (b) 12 W. (c) 8.64 \V

(c)

If /? = 25

^ = 2, and the output is

terminated in a parallel LC, compute

L and C so that the bandwidth is 10

6. Consider the circuit in Figure P I8.6. (a) (b)

Find the Thevenin equivalent circuit

and the resonant frequency CO^ = 5

using the method o f matrix partitioning.

rad/sec. (d)

If the circuit is terminated in a 1.2 Q resistor, find

Compute the zero-state response with the above input and element values.

for r > 0 assuming

Identify the steady-state part. I f the

the initial inductor current is zero.

input cosine frequency were changed to 100 rad/sec, what would happen to the steady-state magnitude? R

-------syx/'-----b :1

Figure P i8.6 ANSWKR; (a) Z,;, = .s- + 0.8. I

Figure P I8.8

= -2 /

hR 2h~ - 2 h + \

7. Consider the one-port circuit o f Figure P I8.7. (a) Find the Norton equivalent using the methods of matrbc partitioning and node analysis. (b)

Making use of the information obtained in part (a), compute the impulse and step responses o f the circuit.

40

^tUII

2h~ - 2 h + \

C= 0.02 F and /. = 2 H

/-P A R A M ET ER S 9. Find the v-parameters o f the two-port in Figure P i 8.9.

40

in

R ^lll

— I— "\/V^ I 20 '2 0

J_ ' 16

Figure P I 8.7 CH ECK:

= 0.5

Figure P i 8.9


10.

1013

Consider the circuit o f Figure P i 8.10, and G 2 denote conductances.

where (a)

Compute the short-circuit admittance parameters.

(b)

Suppose port 1 is short-circuited and a voltage Vjit) is applied to port 2. Find

Figure P I8.12.

I^{s) and I~,{s), in terms o f the literals and

AN.SWl-’RS: in random order: ()',| - Vp). (V| | 0'22 ■^.>■■12^

+ 1: a V.

13 . Consider the circuit o f Figure PI 8.13. (a)

G

Compute the short-circuit admittance parameters.

O------------(b)

Figure P I8.10

>’11

>’12

>'21

>’22

Reverse the process and compute K,,

Yj, Yy and g^^^ in terms o f the yparameters.

11. Consider the circuit o f Figure P18.11, in which and (a)

= 2, /?, = 2 Q,

= 16 Q,

= 320 Q,

(c)

Suppose K| = K2 = K3 = (^ + 1),

= 80 Q. Compute the short-circuit admittance

is+\r

parameters. (b)

V/, is) = (c)

and port 2 is short-circuited. Find

If port 2 is short-circuited and

/,W, AW , /jCr), and /^C/).

64 ------- , find /, is) and I,{s). 5 “ + 16

If port 2 is terminated in a 240 H resistor, /?j = 6 Q, R2 ~ 64 V] ( 5 ) = —------- , find /, W and .v“ + 16

Figure P I 8.13 14. In the circuit o f Figure P 18.14, /?, = 2 0 ., = 2 Q, /?3 = 2 Q, C, = 0.5 F, Cj = 0.25 F, and C3 = 0.5 F. (a)

Compute the v-parameters using the method o f matrix partitioning

Figure P I8.11

(b)

If port 2 is short-circuited, find /,(j), /2W, i]U), and ;-,(/) assuming v^{t) =

12. The two-port o f Figure P i 8.12 has yparameter matrix

yu^=

".V li

.Vi 2

>21

}'22

Compute Kj, V-,, Yy and^^^^ in terms o f th e jparameters.

\Ou{t)V.

1014


17. Reconsider the circuit o f Figure Pi 8.15, in

1n

which He

He

2F

1F

10

2F

in

.'^’11

.’^’12'

•4

>’21

.'’22.

60

= 1 k n , and

- o .r 1

m S,

= 2 kn .

(a)

Compute K,,,. Z^. Y^,, and Z„,„.

(b)

Find the voltage gain

Figure P I8.14

G,. = ^ . K,

15. The terminated two-port configuration in (c)

Figure PI 8.15 is characterized by ^-parameters.

If v^{t) = 9u{t) V, find the power absorbed by R^.

Suppose

V/ -

+

ANSW FRSM a) );„ = 8 m S ,Z ,,= 1 2 5 a = 2.2 mS; (b) - 4 .4 4 ; (c) 0.8 wart

(a)

Find^, j i f = 10 ^^,721 = - 1 S .^,2

(b)

= 0.03 S ,y 2 2 = 0.2 S, and R ^ = \ O n . If v^{t) = 3 0 i4 {t) V, find the power,

pÛ), absorbed by the 10 Q load at port 2.

18. Reconsider the two-port o f Figure P I 8.15, in which R^= 10 H ,^ 2i = 2 S ,^ p = 0.02 S ,^22 = 0.2 S, R^ = i o n , and the voltage gain is

^

= 0 .6 . (a)

Findj/j,.

(b)

If v^{t) = 10«(r) V, find the power absorbed by Rj^.

C H EC K : (b) 160 watts Figure P i8.15

19. Consider the circuit o f Figure P i 8.19, in which R^ = 2 Q , Rj = 8 Q, R^ = 52 Ol, P = 3,

16. Consider the circuit o f Figure PI 8.16.

and /U = 4.

(a)

Com pute the ^-parameters o f the

(a)

Com pute the >'-parameters o f the

(b)

boxed rwo-port. If port 2 is terminated in a 2 S resistor, compute and the voltage gain

(b)

Compute the input admittance,

(c)

Compute

(d)

If i^{t) = 5u{t) A, compute y,(r).

_ V jis)

G .,=

Vds)

boxed two-port. o f the complete circuit.

.(s)

Figure P i 8.16

20. In a laboratory, you are asked to determine the admittance parameters o f a circuit. You


1015

decide to shon-circuit port 2, place a unit step current source at port 1, and measure the port voltage, t/j = (1 V, and the pon 2 current, /2 (^) = A. Knowing that this is sufficient to determine at most two of the parameters, you then break the short circuit and terminate port 2 with a 1 resistor and measure the new step responses as = (1 + V and ij{t) = A. (a) Compute the/-parameters of the twoport. (b) If port 2 is terminated in a 1 Q resis tor, find the input impedance seen at port 1. (c) If port 2 is terminated in a 1 resis tor and driven at port 1 by a current source i^{t) = cos{t)u{t) A, compute the steady-state magnitude of the gain,

network assuming y2 (/) is the output. If vj^t) =10 s'm{2t)u{t) V, compute the steady-state and transient responses assuming V2(t) is the output. Use the residue command in MATLAB to compute the partial fi^ction expansion.

(d)

Figure PI8.22 CHECK: (b) 0.25 / (s + 1), (d) transient response e~û{t) V, steady-state response (-cos(2f) + 0.5 sin(2/))«(f) V

Yz 23. Figure P I8.23 represents a two-stage ampli fier. Suppose

h

21. This problem shows that one can simulate an inductor using an active two-port terminat ed by a capacitor. For the circuit of Figure P18.21, Cj = 125 juF, C2 = 0.8 F,/u =>^22 = 0’ -JVi2 = 4 S. (a) Compute the input impedance Z-JJ). What is the equivalent L seen at port 1? (b) Determine the resonant frequency CO^ (c) If /? = 2 5 is placed in parallel with Cj, find the new resonant fi^juency 0)^ o-n eY=

CHECK: (a) 240 rad/sec

y.. y.j

Figure PI 8.21 = 50 mH, (b) 400 rad/sec, (c)

22. For the circuit of Figure PI 8.22, R^= I Q, =1 Q, and q = 1 E (a) Find the ^parameters. ^2 (5 ) (b) Find the voltage gain, Gy2 = y • (c) Compute the impulse ^ and step responses of the terminated

-0.64'

•2

r, =

25

1

mS, Y2 =

0.4

- 0 .0 0 r

7.5

0.025

= 150 Q, = 2 kn, and /?! = 2 kQ. Find the voltê gain z-' _ VOM/

Although the solution may be obtained by solv ing a set of six simultaneous equations, a much better method that also gives more insight into the performance of the amplifier, and works for any number of stages, is to proceed as follows: (a) Find the input admittances y.nl and i^»i-

(b)

Find the voltage gains of the stages successively, starting from the source end. Use this information to find the overall voltage gain, G, = ^

în

.


101(>

z-PARAMETERS Stage 1

a

25. (a)

Y.

Compute the z-parameters o f the mutually coupled inductors in Figure P I 8.25 when (i) the dots are in posi tions A-B and (ii) the dots are in posi tions A-C,

Figure P I8.23 (b) AN SW ERS: (a) (b )6 V = S 0 0

port by inverting the z-parameters

= 500 LlS:

= 10 inS,

Compute the y-parameters o f the twomatrix. Do the y-parameters exist if the coupling coefficient ^ = 1?

24. Consider the switched circuit o f Figure P i 8.24, in which C = 0.25 F and R^= \ Q.. For

t < 0.25 sec the switch is in position A, and for t>

0.25 sec the switch is in position B. Suppose

the y-parameter matrix is S.

K(.v) = (a)

Figure P I8.25

With the switch in position A, find

26. Consider the circuit o f Figure P i 8.26. (a)

V2(-v) V'|(5)

Compute the open-circuit impedance parameters.

(b)

(b)

If

(c)

0.25 sec. Is rhe circuit stable? Explain your rea

= -4u{t) V, find

0 < r<

If port 2 is terminated in Z-,(s), find the input impedance,

(c)

If port 1 is open-circuited, Zj(s) is a 1 H inductor, Z-,(s) is a parallel combi

soning.

nation o f a I Q resistor and a 0.5 F

id)

Find i/-)(0.25~).

capacitor, and

(e)

For t > 0.25, after the switch has moved to position B, draw and label

2K

the frequency domain equivalent cir

S-+4

cuit that will allow one to compute

(0

y^(s). Compute Zj(s) if ^ = 2.

(g)

Compute I, A

-^ s/ V -

for /> 0.25. B

•-

find f,(r) and v>y(r) in steady state. 1 :a

z.

b;1

Y=

Q v ,

Vm y,2

Figure P I8.26 C H EC K : (c) i^2 sss(^) = 0.707A>/lsin(2/-45°) V

F'igure P i8.24 C H EC K : (d) i^M 25) = 3.463 V, (g) 3.436 f-('-'^ -5 ),‘; ( / - 0 . 2 5 ) V

=

27. Consider the circuit o f Figure P I 8.27, in which /?, = 35 Q, /?7 = 50 Q, 7?^ = 800 Q, = 100 Q, and a = 4.


1017

W

(a) (b)

Compute the z-parameters. Find the ^parameters by matrix inver sion using, for example, MATLAB, If port 1 is open-circuited and

(c)

------ ^ r

h

1£1

It 0.5 F

r

1

-------

\f 1 < 1 _ 0.25 F > *

r\

+

0.5 F

8

----- 0

D-----

Figure P i8.29

1:4

-> / W -

—o

R.

K

60

??1

-100

??

10

+ Zîs) = — , Z^ = 10 Q, and V, tS^

(a) (b) (c)

Figure PI8.27 CHECK: (a) Z =

30. Consider the circuit of Figure PI8.30, in which Zj(s) = 5 0 , Z2(s) = 10 n ,

Compute the z-parameters. Compute Z-J,s). If v-^{t) = 10tt(/) V, find /j(/) and /Û).

28. Consider the two-port of Figure PI8.28. (a) Compute the open-circuit impedance parameters in terms of the Z- and r^. (b) If the two-part has z-parameter matrix '^11

m '

^21

^22

find Z j, Z2 , Z3 , and the ^-parameters. (c)

3 1 . The

in terms of

port 2 is short-circuited, and v^{t) = 10«W V, find /j(?) and z,

circuit of Figure PI 8.31 hasZ^= 10 Q, and 2 -parameter matrix Uij] =

V,

V.

-o Figure PI8.28 CHECK:

(a)

(b)

+

(c)

n

1

5^ + 100 1

1

205

79

Q

Use the result of Problem 28 part (b) to draw an equivalent circuit having these z-parameters. The circuit is resonant at what value of Compute the Q of the circuit. Plot the magnitude and phase response of Zy^(/co), using a program such as MATLAB.

Z] H- Z3 Z3 99

29. Use the method of matrix partitioning to find the z-parameter matrix of the circuit in Figure P I8.29.

O '

Figure PI8.30

If Z iW = Z 2 W = ^,Z3W = l . r ^ = - 5 ,

-------

= 10 .

Figure P I8.31


1018

output o f the two-port to the load.

32. Consider rhe circuit o f Figure P i 8.32. Suppose Vy(0 = 20 V 2 co.s(w/);/(0 V, = 1 Q,

(b)

= 1 Q, and the z-parameter matrix is

s+

Z=

(a) (b)

0.5

0.5

s

s

0.5

,v +

What is the maximum average power absorbed by = 2 when /^(/) = 2>/2 cos(2/) A, a = 2 , and/?Q = 50

0.5

Find Z-„W and Find Vjis) and the average power

b :l

o

aR„ 2-Port

absorbed by Z^. For CO = 0 and cu = 1 rad/sec, find the

(c)

Figure P I8.33

average powers absorbed by Z^. (d)

Using the equivalent circuit o f Figure

CHECK: (a) (??),

‘79

18.38, construct a circuit having the given 2-parameters.

5a

'V R,

; (b) 2 watts

34 . T he two-port shown in Figure P i 8.34 is called a “g)Tator.” A g)'rator terminated in a capacitor beiiaves like an inductor at its input terminals. This is one o f the many ways to sim

A

''■O

’

ulate inductors in the design o f an “active fil

2-Port

ter.” Suppose the active two-port in Figure P I 8 .34 , built with resistors and ideal opera tional amplifiers, has z-parameters Figure P I8.32

ANSW ERS: 2.v' + 2.V- + 2.V + I

0

1000

-1000

0

(a)

0.25 1 + uy ■

a surprise to you? (b)

•V + 2.V" + 2.V +

and

: (c) 100 w and .50W

If C-, = 10 nF, find Z (j), the imped ance looking into port 1. Is the result

(?).v- + (?).v-(-(•?) 0.5

Q and C, = 100 pF.

If

= R j= 100 kQ and rhe output is

Kj(/), find OJ,;; and the bandwidth o f the circuit. (c)

If

=/?-,= 10 kQ and the output is find the impulse response, h{t),

o f the circuit. (c) 100 W an d SOW

(d)

Find thej'-parameters o f the two-port, using the property that theZ-matrix is

3 3 . Consider the two-port configuration of

the inverse o f the K-matrix (and vice

Figure P18.33 having z-parameter matrix

Z=

aRn

a~Rv

2a/?o

2 6 a “/e„

versa). (e)

If /^] = /?2 = 1^^ kD, and the output is

I'lU), what is the impulse response? Hint: Make use o f the results o f parts (c) and (d), and the two-dependent

(a)

Determine the value o f b that yields maximum power transfer from the

source equivalent circuit in terms o f the v-parameters.


1019

The two-port hasj^parameters and the twoport Ny has ^-parameters as given below (with standard port labeling and units):

2-Port

'“O

■0.1 2 (s) Figure PI8.34

- 0.2

35. The stages in the circuit of Figure P I8.35 have i^parameter matrices Z, =

2 -1 0 ^

01 20

62.582

1.20751

63.75

1.25

respectively. (a) Compute the input impedances, Z -^2 andZ;„,. (b) Compute the voltage gain, VgJVj(c) Compute the power gain, p

_

^ o u t^ o u t

(a) (b) (c) kQ

(d)

0 . 1'

‘36

0.1

40 4

2

FindZ^, FindZ,.„. Find the gains V, G .= ^ ,G 2 = ^ ,G 3 = ^ ,a n d Ha Hfe

Find the power gain of the circuit, i.e., the ratio p _ S o in -

^ o u t^ o u t ..

vi/,

37. ^ equivalent circuits for a pair of coupled put impedance of the amplifier circuit. inductors using z-parameters. Each equivalent circuit consists of one ideal transformer and I. I, L two inductances. The analysis of a coupled circuit with the use of such equivalent circuits 'len is very often more illuminating than writing and solving simultaneous equations. (a) Find the z-parameters of the two-port Nj of Figure PI 8,37a. (b) Show that two-port N j of Figure Figure P18.35 P I 8 .3 7 b has the same z-parameters as N j. CHECK: m = 2 kii and ^y = -9.8147 (c) Show that two-port N3 of Figure ''s P18.37C has the same z-parameters as 36. Consider the cascaded two-port in Figure N,. P I8.36, in which = 10 Q, /?| = 20 Q, and Rj^ (d) Use the equivalent circuit N2 of pan (b) =4Q . and the properties of an ideal trans former to find 0)^, the bandwidth, and ' ^

Vi„ max

approximate values for the half-power frequencies for the coupled tuned cir cuit of Figure P18.37d. Figure P I8.36

This problem shows


10 2 0

1 :n

(d) (a)Two-port N,

Figure P I8.38

m v_o

39. Find the /;-parameters for each two-port

(1-k^) L,

shown in Figure P I 8.39, assuming standard port voltage and port current labeling, with R =

2 kf2, C = 0.1 mF, and « = 0.1. ideal transformer

1 :n

(b) Two-port Nj k y ~ L ,: >/l .

o -fY Y \ (1-k^) L, NT

N2

Ideal Transformer

Figure P I8.39

ideal transformer (c) Two-port Nj

40. Find the /^-parameters for each two-port

k= 1

shown in Figure P i 8.40, assuming standard port voltage and port current labeling, with R = 2 k n , C = 0.1 mF, and w = 10.

(d) Coupled tuned circuit

Figure P I8.37 (a)

/i-PARAMETERS 38. Find the /^-parameters for each two-port shown in Figure P I 8.38 assuming the standard labeling and units.

----- o

o----- r^ s/ V '------ o


-o

o-

(b)

(a)

1: n

Find the /^-parameters for the two-port of Figure PI 8 .4 la assuming that the hparameters o f the two-port N, are

Z,

A 111I (0

A


W

(b)

1021

Hint: What does the transformer do to the output variables of the boxed circuit? Break the problem up into two separate parts. Apply the result of part (a) to the cir cuit of Figure Pl8.4lb.

rv / W -o z

+ 2-Port

Figure PI8.42 43. Consider the amplifier network of Figure P I8.43. Stage 1 is a common-emitter stage that drives stage 2 , the common-collector stage. Such an amplifier combination might be used to drive a low-impedance load. Suppose Z^ = 2 Idi, Z^ = 3 kn, Z^ = 64 £2, and the A-parameters in standard units of the two stages are

O '

(a)

H ,=

Ho =

Figure P18.41 42. The /f-parameters of the two-port of Figure P18.32 are = 250 SI, h -^2 - 2-5 x 10“^, />2 i = 125 , h22 = 2.25 mS, -2^ = 1 kH, and = 500 a (a) FindZ.„ andZ^^,.

W

Find the gain

(c)

Find the power gain of the circuit, i.e., the ratio

=

■

Suppose a capacitance of 5 |AF is con nected across Dort 2, Suppose V j(/ ) = 10V2cos(400f) V. FindKjW in steady state and the average power absorbed by Z^ = 500 Q. Hint: Obtain the Thevenin equivalent circuit seen by the Z^ and C combination.

0

50

0.05x10-3

nooo

0.966

-51

0.8x10-3

■

Find the input impedances of stages successively, starting from load end. Find the output impedances of stages successively, staning firom source end. Find the overall voltage gain, ^

_

''out

C .- —

(d)

V,/,

(d)

(b)

(c)

(b)

p

(a)

2000

the the the the

.

A 1 pF capacitor is inserted in series with Z^ to prevent dc voltage in the power supply (not shown in the dia gram) from entering the signal source. Because of this capacitance, low-fre quency signals will be amplified less. Determine the frequency (in Hz) at which the magnitude ôut - 0.707 X Max Value

Hint: Analyze the simple circuit con sisting of Vp Zp C, and Z- .

O

O


1022

'• 6

Common

Common

emitter

collector

stage-1

stage-2

Figure PI8.43 44. Reconsider Problem 43. If-2^ is adjustable, find the value of so that = Z^ = 64 Q. Then find Z^-„ and G. = ^

.

(a) Compute ^22 i*' terms of and pos sibly other /^-parameters. Explain your reasoning. (b) Derive a formula for the current gain, Gp'm terms of the A-parameters and Y^. (c) Find />2 i(d) Suppose now that the source, - Z^, at the front end of the two-port is briefly disconnected and a voltage, V2{t) = 10«(^) V is applied to port 2. The measurement v^(t) = -5u{t) V is made. Compute h^2' (e) Suppose the source is reconnected to the two-port. If the current gain

45. For Figure P I8.45, the and ^-parameter matrices (in standard units) of two-ports Nj and N j, respectively, are 0.02

0.001

2.5

0.2

-

H;V2 =

10

0.051

-125

0.5

■ ^ = 0.8

h

and if YXt#= 0.25 S, determine Z-in and A „.

If Zjr = 8 Q and Z^ = 25 12, find the input impedance and the voltage gains,

for the cascaded two-pon. p y s/ V -o -

2-port N1

CHECK:

02-port N2

Figure PI8.45 = 0.8 and G^2 ~ ~ 1 0 0

46. Consider the terminated two-port configu ration in Figure P I8.46, with h-- indicating the two-port /^-parameters. Suppose that (i) the current through the admittance, Yj^ equals the current through ^22 ? the current gain

Figure PI8.46 CHECK: />2i = 200,

= 1800

•L Q ,Z }„ = 2 k Q 4 7 . In this problem you are to design an ampli fier circuit represented by the doubly terminated equivalent circuit shown in Figure PI8.47. This means you will be given certain amplifier speci fications that will allow you to determine the parameters of the amplifier circuit.

Amplifier specifications: R^ = 2 Q and

(i)

= 40 Q.

When /j is zero, the ratio

G, = -^ = 100; and (iii) the source resistance is Z^ = 8 Idi. (ii)

when a source is applied to port 2. There must be maximum power trans fer from the amplifier output to the


1023

load under the condition that

-

(b)

800 Q.. ..... (ill) (iv)

Verify that the /-parameters o f the cir cuit o f Figure PI 8.49b arc

V.

25

V,

26

_L = —

1

O'

>2

1

T =

The voltage gain

Vo — = -1 0 0 .

(c)

Compute the /-parameters o f Figures P I8 .4 9 c and d.

Given these specifications; (a) (b)

Compute Z/j 2Compute

-o

o-

-O

O-

^îd the turns ratio

a. (c)

Compute//jj. Compute the input impedance

(d) (e)

(a)

Compue *2 , Compute the ratio o f the power deliv

(b)

-o

o-

-o

o-

ered to the load to the power delivered to Rjj (d)

(C )

Figure P i8.49 ANSWHRS: ■1

2i

: (d) 1

>2

5 0 .(a)

Figure P I8.47

Z.

Show that the /-parameters o f the ideal transformer o f Figure P I8 .5 0 a are

48. Repeat Problem 47 with the amplifier spec T =

ification

^

(b)

= 0.01

■//

0

0

\ h i

Given the answer to part (a), compute the /-parameters o f the circuits o f Figure P I8 .5 0 b and c.

when /j = 0. 2 xlO -’

0.01

n: 1

AN SW ER: l/?,/l = 4W .3S

3.6538 x lO " '

(in standard units) (a)

n: 1

t-PARAMETERS 4 9 .(a)

Verify that the /-parameters o f the cir cuit o f Figure P i 8.49a are T =

(b)

0

1


1024

n: 1 c^

JY Y Y -6 4H

(0 Figure P i8.50 Figure P I8.53 51. Find the /-parameters o f the ner\vork in Figure P I 8.51.

54. The rwo-port o f Figure PI 8.54 is described by r-parameters.

4; 1

(a)

2H

Derive the input impedance relationship ^ll^L +^12

ti\^L Figure P I8.51 -4 ANSWlUl:

0

0 ■ ’1 1 0 — 4_

(b)

2.v^

-4

1

0

Derive the output impedance relationship

-S.v ■

ôut ~

t~>2^s

network in Figure P I 8.52. If a 14 volt source is applied at port 1,

(c)

V, Derive the voltage gain Gj.] = — . Kv

(d)

Derive the voltage gain

Figure P I8.52

Figure P I8.54 7

12

4

7

53. Use the results o f previous problems to obtain the r-parameters o f two-ports A, B, C, and D in Figure P I 8.53. Then use matrix mul

55. Suppose a t\vo-port has both 2-parameters and /“-parameters. Compute the r-parameters in terms of the z-parameters. Hint: Rewrite the zparameter equations in the form

Ml

= Mo

tiplication to obtain the r-parameters o f the overall two-port.

= Ir. ^1

compute Vjit)-

ANSW l-R: (a)

hi

-0 .2 5

5 2 .(a) Find the r-parameters o f tiie cascaded (b)

^22

-h Vi

and invert the appropriate matrix to obtain

<•21

-21

1

o.

'•21

-21

102S


57.

- ZpZ-,j.

where Az =

A certain high-voltage transmission line

operating at 60 Hz is represented by a two-port 56. For the circuits of Figure P i8.56, suppose the r-

with the following r-parameters:

parameters in standard units o f the two-port N j are

y0.022, r ,2 = 40

■

-

-

0.1

-

-0 .0 0 1 5

-0 .0 0 5

current -I-, = 3 6 1 Z 0 A at V-, = 1 1 5 ,2 0 0 Z 0 V, find the sending end

Find the /-parameters o f the cascaded

(port 1) voltage Vj^ the current Ip the

two-port for each configuration.

power delivered by the source, and the

Find the steady-state /j(/) under the

(b)

j = 0.895 +

= - 2 .6 1 7 x 10“5

+ y i.l0 2 X 10-3 =“o.895 + y0.022. (a) If the receiving end (port 2) draws a

0.1

R = 200 Q.,L = - 0 .5 H, and ;/ = 5. (a)

y l8 0 iX

conditions that

= 10 cos(lOOr) V

power loss in the transmission line. (b)

II V j = 1 3 4 .0 0 0 Z 0 V, and a resistive

and each rwo-port is terminated in a

load represented by a 500 Q resistor is

parallel RC circuit with ^? = 10 Q and

connected to the receiving end (port

C = 1 mF.

2), find the power delivered by the source and the magnitude o f the load voltage.

1 :n

—

0

+

2-port

V.I

NT —

ANSW l-K; (a) V, = 135,550Z29..S7<' V, I, = 3 4 7 .3 6 Z 2 2 .8 6 ° A. 4 6 .733 megawatts. 5.146 megawatts; (b) 35.331 megawatts, 12"’.96 k\’.

0

(a)

PARAMETER CONVERSION AND INTERCONNECTION OF TWO-PORTS 2-port

58. A two-port N has known 2-, y-, and h-

N1

parameters. A new two-port

is formed by

adding one single impedance Z or admittance

Y to N in various ways. Prove the following

(b) I.

relationships bervveen the old and new -rwo-

'

w

port parameters. +

L

(a)

2-port

V,

-

If Z is connected in series with port 1 o f N, then Zn

N1

^^\\,new =

= 2Ml ,, + Z and

+ Z. Other z- and h-

parameters remain the same. (0

(b)

Figure P I8.56

- 0 . 0 0 l.v-0 .1

\i„\ =

- 0 .0 0 2

If Z is connected in series with port 2 o f N, then 2^2 n rtv = -2=22 + parameters remain the same.

■ 0.02

0.02 ■

0.0075

0.025

■ -0.1

-0.1 ■

-0 .0 0 2 -0 .0 0 5 5

0.(K )275.s -0. 1 -0 .( )() 5 5

; (b )

= 2 . 3 0 5 c o s ( l 0 0 ^ - 1 2 . 5 3 " ) A.

1026

(c)

(d)


If K is connected in parallel with port 2 of N, then =^22+^ ^21,new = ^22 + ^ ■ ^ther y- and hparameters remain the same. If Y is connected in parallel with port 1 of N, then;»„ + K Other ^parameters remain the same.

59. A common-ground two-port N has known is formed by adding one single impedance Z or admittance Kto N in two different ways. Prove the following relationships between the old and new -two-port parameters. (a) If Kis conneaed across the top termi nals of N, then = jVn + Y,

61. Compute the^-parameters of the two-port of Figure Pi 8.61.

z- and ^parameters. A new two-port

Figure PI8.61 62. Compute the j/-parameters of the two-port of Figure PI 8.62.

yil,new " yil'*’ ^ ’ y\2 ,new = >^12 “ ^ ^ y 2 h n c w = y 2 l-^ '

(b)

If Z is connected in series with the common terminal of N, then =

2 ] 1 + Z, Z22,new ~ ^ 2 '^ ^ ’ ^12,t

12

r )

60. Assume standard labeling in Figure P I8.60, in which the inner two-port labeled N has «parameter matrix

(a)

(b)

63. Assume standard port labeling for the cir cuits in Figure P I8.63. Find the z-parameters of each of these circuits. For the interconnec tions shown in Figures PI8.63c and d, when do T4 5' Q. the overall ^-parameters equal the sum of the 3 4 individual two-port parameters? When this is Find the /-parameters of the overall not the case, explain why not. two-port N* and then find the z- o-^ys/ V —o o------------ 1------------o 1o 1n parameter matrix. Now suppose the ports are connected 1o to current sources, in which case /j(/) = 12(f) = I5u(t) A. Assuming zero ini (a) (b) tial conditions, find and —o o— —I—s/X/v—o What is the response if ij(t) = = o —\/\y\/—I— 10 ) 10 5u(t) A? Hint: Use linearity. 2F

10

port 1 port 2

1O

------- •-------

10

N‘

O

O—n / \ / V

(0 Figure P I8.60

< ^

10

O

(d)

Figure PI 8.63

r ^


102'

64. Compute any one set o f the z-, y-, and iparameters for each of the circuits o f Figure P I 8.64. Then obtain the remaining two sets by the use of the conversion table (Table 18.1).

yYY\ 160

1 H Ij 0.5 F

A

'

67. Find the z-parameters of the interconnect ed circuit shown in Figure P i 8.67, in which L = 1 H , assuming standard units for the yparameters.

1#

m \

Figure P I 8.64

Y=

"4

-1 “

.-2

1

_

65. For the circuit o f Figure P 18.65, find the zparameters o f each rvvo-port,

and N^, and

2-parameters

of the intercon

then the overall

Figure P I 8.67

nected two-port. AN SW ER:

\-ij

■v-l-0.5

5 4-0.5

.v+l.O

.v-h2.()

Q.

68. Repeat Problem 67 with the inductor changed to a 0.5 F capacitor. 69. Find the z- and ^'-parameters of the inter connected network shown in Figure P I 8.69, in which ;?, = 2 n ,

Z = Figure P I 8.65 66. For the circuit o f Figure P I 8.66, find the^parameters o f each two-port,

and N^, and

2-parameters

o f the intercon

then the overall nected two-port.

= 2 iX

2

1

0

-2

= 1 iX and

a.

102.S


Figure P I 8.69 A N SW ER:

[3.2

!' //■Im'ii- ~ (^)

l.S

Q.

- 0 .8 72. Two rwo-ports

70. The rwo-port

is connected to a 1:1 ideal

transformer to form N a *. Then N^* a and N 0/ are connected in series to form a new rwo-port N

and N^âre connected in

series to form a new two-port N , as shown in Figure P I 8.72. Find the new z-parameters.

as shown in Figure P I 8.70, in which /?j = 2

/?2 = 2 Q , /?3 = 3

and R^ = 1Q .. Find the z-

parameters of N .

Figure P I 8.72 73. Consider the two-port in Figure P i 8.73, which depicts a transistor amplifier stage for a microphone. Suppose that the transistor has h71. Two two-ports

and

are connected in

series to form a new two-port N , as shown in Figure P I 8.71, in which « = 2, /?j = 4 Q, R j = 4 i i , /?3 = 2 n ,

^7 = 1 Q, and

= 8 n . /?5 = 2 n ,

parameters given as Aj j = 4.2

~

^h\

= 150, and h ^2 = 0-1 mS. (a)

Assume the input and output capaci

=2h.

tors are short circuits at the frequency

= 4 Q, Find the new z-param-

of interest and that the capacitor

eters.

across the 470 Q resistor is also a short circuit. Determine the /;-parametcrs of the overall two-port. (b)

Given your answer to part (a), deter mine the value of n for maximum power transfer.

(c)

Given your answer to part (a), deter-

1029


(cl)

mine rhe voltage and power gain of

A N SW ER:

rlie overall c\vo-port.

30.15“) V

(h)

/'| ^^{[) = 2.505 a)s(5()0r -

Plot the frequency response o f the

^

ampliher as a function ot / = 271(0 for ^ ^ , ,, . ^ . 0 < /< 20,000 kHz using Spice. ° ^

which u = 0.2, /?, = 1 Q, and R . = 4 Q. /\ c l i c • l (a) For what value or is the rwo-port

r

■

r r

^

r

.. , . • r tn,o-rc • Consider the circuit or Figure P I S . 75, m

jn

reciprocal? (b)

For the value of

found in part (a),

compute tlie circuits z- and />parameters.

Figure P i 8.73

(c)

If port 2 is terminated in a 2 Q-1 mF parallel

RC

combination,

and

RECIPROCITY

\’l(/) = 20>/2 cos(20000 V ,

74. The two-port in Figure P I 8.74 has port 1

v-,{t) in steady state. Use the value of

attached to an ideal current source. It is known

r

find

found in part (b).

that the two-port consists only o f R's, L's, Cs, and transformers. If /|(r) = 5(^), your oscillo scope shows a waveform having the analytic expression V2U) =

cos{t - 30°) V for

f > 0. Figure PI8.74b shows the same t\vo-port with different terminations. (a)

If

= u{t) A, find the zero-state

response, (b)

If the input is changed so that ;-,(/) = 10 cos(500/)//(f) A, find the steadystate response, i^j(f).

Figure P I 8.75 A N S W ’FR; (a) r

i,(t) vjt)

—• (a)

v,W

7.5

5.0 ■

5.0

4.0

■ 2-Port

v/t)

2-Port

(b)

Figure P i 8.74

(b)

= 1 Lh ■ 1.25

;c' r,: r) = 10 co.s(200(); - n “) \’.

-1 .2 5

1.25 0.25

1030


76. Consider the circuit of Fieure PI 8.76, in which R^ = I Q, L - y/l H and C = ^^2 F.

o

The transfer function

H (s) =

n

njs) ^ ____ d(s)

+ ypis + l

Use all the theorems you have learned in this course (including reciprocity) to construa an alternative circuit that has the same d{s). r>

yyy\

R.

L

Figure P I8.76 77. Again consider the circuit o f Figure PI 8.76, in which 7?^ = 1 Q,

= 0.5 £2, Z = 1.673 H

and C = 0 .8 9 6 6 F, or Z = 0.4 4 8 3 H and C = 3 .3 4 6 6 F. The transfer function is

în (^)

5^ + >/2 J + 1

Use the theorems you learned in this course (source transformations, reciprocity, scaling) to construct two new circuits in which

= 1 Q,

= 2 Q, and the new transfer function is HngJ^^ = KHiJ) for an appropriate K. W hat is the value o f K that is obtained? The result of

r\

this problem shows how one circuit with a

r s

resistance ratio

R

r\

R. can be derived from another with

Rc

a n

C H EC K : L = 3 .3 4 6 6 H or I = 0.8 9 6 6 H in series R^ with AT= 2

n

C

H

A

P

T

E

R

Principles of Basic Filtering

LOUDSPEAKERS AND CROSSOVER NETWORK In a stereo system, a power amplifier is connected to a pair of speakers. The most common type of speak er system consists of one or more drivers enclosed in a wooden box. The amplifier feeds a signal to each drivers voice coil. The voice coil sits within a field produced by a permanent magnet and is attached to a heavy paper or plastic cone. When excited by a current from the power amplifier, the coil interacts with the magnetic field of the permanent magnet and vibrates. The coil pushes and pulls the speakers cone, which, in turn, proportionately moves the air, making sound waves. The high and low frequencies that make up music place opposite requirements on a loudspeaker. A good low-frequency speaker should be large, in order to push a lot of air. A good high-frequency speaker should be light, in order to move back and forth rapidly. A rwo-way speaker system consists of a small, light tweeter to handle the treble signals and a large woofer to handle the bass. A better system is a three-way system, with a third, midrange speaker to handle the f-requencies in the middle. The magnitude response o f a typical rwo-way system is shown in the following figure.

lÔ’ ltoa; U

II

OI CT fO 'O

:=

e E k. O C

C

OJ

1032

Chapter 19 • Principles of Basic Filtering

A crossover netiuork (low-pass or high-pass filter) separates the frequencies so that the woofer receives the low-frequency content of the music and the tweeter the higher-frequency content. In the magnitude response plot illustrated in the figure, both curves have the same 3 dB frequency at 2000 Hz. This frequenq’’ is called the crossover frequency. This chapter explores the basic design principles and realizations o f low-pass and high-pass filters. Some band-pass filtering is also dis cussed. Some simple crossover circuits are considered in the problems section.


Introduce the meaning and (brickwall) specification of low-pass filters in terms of dB loss.

2.

Set forth the maximally flat Butterworth magnitude response and associated Butterworth transfer function.

3.

Present a step-by-step design algorithm for finding the filter order and associated

4.

Present basic passive and active circuits that realize a Butterworth transfer function.

Butterworth transfer function. 5.

Set forth the properties of the Butterworth transfer function.

6.

hitroduce high-pass filter design through frequency transformation.

7.

Present basic passive and active circuits that realize a Butterworth high-pass filter.

8.

Introduce an algorithm for the design of a band-pass filter using frequency transformations.

SECTION HEADINGS 1.

Introduction and Basic Terminology

2.

Low-Pass Filter Basics

3. 4. 5. 6. 7. 8.

Butterworth Solution to the Approximation Problem Butterworth Passive Realization Active Realization o f Low-Pass Butterworth Filters Input Attenuation for Active Circuit Design Properties of the Butterworth Loss Function Basic High-Pass Filter Design with Passive Realization

9. 10. 11.

Active Realization o f High-Pass Filters Band-Pass Filter Design An Algorithm for Singly Terminated Butterworth Low-Pass Networks

12.

Summary

13. 14.


1. INTRODUCTION AND BASIC TERMINOLOGY Types o f Filtering Filtering plays an important role in circuit theory, as well as in communication theory, image pro cessing, and control. There are three fundamental types of filters: analog, digital, and switched capacitor. Digital filter analysis and design requires knowledge of the Nyquist sampling theorem.

Chapter 19 * Principles o f Basic Filtering

1033

Switched capacitor networks/filters, an idea introduced in Chapter 13, are something of a hybrid between analog and digital filters. Both are beyond the scope o f this chapter, which takes a cir cuits viewpoint on some basic analog filtering concepts and techniques. Analog filters process the actual input waveform with circuits composed of discrete components such as resistors, capacitors, inductors, and op amps. Analog filters are of two types, passive and active. Passive analog filters are composed only of resistors, capacitors, and inductors. Active ana log filters consist o f resistors, capacitors, and op amps or other types of active elements.

B a s ic T erm in olog y A filter is a device (often an electrical circuit) that shapes or modifies the fi^quency content (spearum) of a signal or waveform. We represent a filter by a transfer ftmction H{s) whose frequency response is The gain, gain magnitude, or frequency response magnitude is |//(^)|. If

is normal

ized so that its maximum gain is 1, then the gain in dB (decibels) is Gjg{vi) = 20 log]Q|//(/ci))|. In this chapter we will assume that the maximum value of |//(/o))| has been normalized to 1. An important frequency called the cutofF frequency (or frequencies), also called the half-power point (or points), is that frequency, denoted O)^ for which 2

H (j(Oc)

1

1

'

= 2 ’

Since power is proportional to voltage squared or current squared (gain squared) for a fixed load resistance, we arrive at the terminology of half-power. This is also called the 3d B down point O '

because

= 10 logiol//(/copP = 10 logjo(0.5) = - 3 dB.

Often, design specifications for a filter are expressed in terms of attenuation or loss rather than gain. This results in several definitions that are dual to the gain-related definitions above. The loss fiinction, denoted H (/w), is the reciprocal o f the transfer ftmction. Thus, the attenuation, or loss magnitude, is

H (jw )

'

HiJco)

It follows that the filter loss or attenuation in dB is

=

/o55 = -201ogio |//(yw)| = 201ogio

(19.1)

With these definitions established, the chapter will examine low-pass (LP), high-pass (HP), andband pass (BP) filters, leaving the study o f band-reject filters to higher-level texts. A low-pass fil ter is a device (typically a circuit) offering very little attenuation to the low-frequency content (low-frequency spectrum) of signals while significandy attenuating (blocking) the high-frequency content o f those signals. High-pass filters do the opposite: they block the low-frequency content and allow the high-frequency content of a signal to pass through. Finally, band-pass filters, as described in Chapter 16, allow a band o f frequencies to pass while significantly attenuating those outside the band. Interestingly, the general practical design of all such filters, especially for the high-order case, is generally based on a low-pass prototype design; the low-pass prototype is trans formed into a high-pass or band-pass type using a frequency transformation.

H )34

Chapter 19 • Principles o f Basic Filtering

As a firsr step in exploring the design o f such filters, vve describe Butterworth (low-pass) transfer functions. Butterworth was the name o f the English engineer who first developed this special class

^

o f transfer functions in his paper “ O n the Theory of Filter Amplifiers.” ' The next step is presen tation of an algorithm that adapts these basic Butterworth transfer functions into ones that meet a given set of L P filter design specifications. Once the transfer function is known, an engineer must implement this transfer function as a passive or active circuit. This chapter will also outline

/—

some techniques for generating passive and active reaHzations.

2. LOW-PASS FILTER BASICS As mentioned, a low-pass filter allows the low-frequency content o f a signal to pass with little attenuation while significantly blocking the high-frequency content. The immediate question is, why use a low-pass filter? One possible answer is that the noise in a noisy signal often has most of its energy in the high-frequency range. For example, so-called white noise has a constant frequenq- spectrum. Hence, low-pass filtering a sinusoidal signal corrupted by white noise will generally result in a “cleaned-up” information signal, as illustrated in Figure 19.1. In Figure 19.1, the thickness o f the curves represents the infiltration of noise. By using the low-pass filter, the curve is “sharpened” by reducing the high-frequency noise content. The suspension system o f a car is a low-pass filter: slow, rhythmic (low-frequency) road variations are permitted, while the effects o f chuckholes and bumps (high frequencies) are “ filtered out.”

Lowpass Filter

F IG U R E 19.1 The effect of low-pass filtering on a noisy signal. Whenever LP filters are needed, an engineer must provide design specifications. Historically this was done using a low-pass filter (brickwaJl) specification, as illustrated in Figure 19.2. Two pairs of numbers (d)^,, 1. 2.

and (to^,

characterize this brickwall specification, where

equals the pass-band edge frequency, 0 < 0) <

3.

is called the pass-band,

niaximum dB attenuation permitted in the pass-band,

4.

to^ is the stop band edge frequency.

5.

0)^

<

6. 7.

CD d e f i n e s t h e s t o p b a n d ,

is the minimum allowable dB attenuation in the stop band, and co^ <

(0 <

defines the transition band.

^

103s


A{(d) = Loss (db) A

F IG U R E

19.2 Brickwall specification of low-pass filter.

The shaded region in Figure 19.2 represents a brickwall. The attenuation of the filter in dB, i.e., /l(co) =-20 log,Q|//(/w)| = 20 log,o|/^(/w)|, must reside outside the shaded region. Finding a (nor malized) filter transfer function that meets the brickwall specs is callcd the approximation prob lem. Once the approximation problem is solved by identification o f the proper normalized trans fer function, the next step is to construct a circuit realization of the normalized transfer function and then to frequency-scale to obtain the proper pass-band edge frequency, and finally to magnitude-scale to obtain the proper impedance levels. The simplest technique for solving the approximation problem is with 3dB

normalized

Butterworth transfer functions whose squared magnitude responses arc given by

H ( jc o )

( (ti )

(19.2)

1+ U c/

where

(tip

is the 3 dB down point, or cutoff frequency, o f the filter, and e is a to-be

specified constant. As mentioned earlier, the term “3 dB down point” arises here because, for all w,

10

lO g lQ

1+

( coY

= 10 1og,ol2j = 3 dB

i.e., there is 3 dB of loss at (O = co^ W hen e = 1 in equation 19.2, the pass-band edge fre quency

and cutoff frequency

coincide. If we now define the normalizedfrequency as Q =

CO,.

the magnitude response of equation 19.2 becomes

^3dBNLP ( J ^ ) ~ =

(0

1 + Q^

( 1 9 .3 )

1036


Equation 19.3 denotes the wtli-order 3 dB normalized Butterworth magnitude response. The words “ 3 dB normalized” refer to the hict that at Q = 1, the loss is 3 dB (the gain is -3 dB). Remember that the actual filter transfer function depends on a proper choice of

or f , which we will clar

ify shortly. Before proceeding further, we ask a critical question: Does this kind o f representation make sensed To answer this question, note that the dB loss of equation 19.3 is

A{Q) = lossidB) =

2 0

logiQ H^dBNLpU^) =

1 0

logSlO 1 + Q^

( 1 9 .4 )

Plotting this function for various ;/s as a function of normali/xd frequency, Q, for a normalized stop band edge frequency,

= 3.5,

= 3 dB, and

= 20 dB yields the polynomial curves

in Figure 19.3. Clearly, for h i g h e r w e can have higher values o f

These cur\'es can be made

to lie outside the brickwalls o f Figure 19.2 and hence can validly be used to meet a L P filter spec ification.

-XD

c

Normalized Frequency, Q FIGURE 19.3 Plot of the normalized Butterworth magnitude responses for n = 2, 3,


1()3‘

3. BUTTERWORTH SOLUTION TO THE APPROXIMATION PROBLEM From Figure 19.3, vve see that as n increases, the pass-band magnitude response becomes flatter and the transition to stop band becomes steeper. This property suggests that there is an appropri ate value of w to meet a given set of brick\vall specifications (ci3^,

and (o)^,

The idea

is to use equation 19.2 and these specs to determine a proper value for the filter order «, a prop er cutoff frequency (o^., and the normalized transfer function.

Computation o f Filter Order n an d CO^ To compute the filter order, observe that n must satisfy the pass-band and stop band edge fre quency constraints. From equations 19.1 and 19.4, at the stop band edge frequency,

1 -I-

(^ s\

After some algebra, we obtain (19.5a)

2: 1 0 ^ - 1

Similarly, at the pass-band edge frequency.

Again, after some algebra we obtain

/'" p V

_

1

(19.5b)

Dividing the left and right sides of equation 19.5a by the left and right sides o f equation 19.5b (this maintains the inequalit)' o f equation 19.5a) yields

(w X

(19.6)

Solving for n produces the order form u la

•«gio (19.7)

na ( COs^ lo g lO

(Or, J

\ ^ P

Thus n can be any integer satisfying inequalit}' 19.7. Usually one takes the smallest such n.

103S


Once n is picked, there is a permissible range of co^ given by

(M,

LOcjnm

CO.

2 ,^ A A min

= COc,max

(19.8)

__ j

To derive the range given in equation 19.8, we reconsider equations 19.5. Taking the 2«th root of both sides of equation 19.5a and solving for

yields the right side of equation 19.8. On the other

hand, taking the 2«th root of equation 19.5b and solving for to^ yields the left side of equation 19.8.

E xercise. Verify the mathematical details in the above paragraph on the derivation of equation 19.8.

EXA M PLE 19.1. Suppose we are given the brickwall specs

= (500 Hz,2 d B ) and

- (2000 Hz,3 dB). Find (a) the minimum filter order «, (b)

and co^

(c) and

the normalized L P squared magnitude function, and then (d) plot the magnitude responses for the cases where O)^ =

and a)^ =

over the frequency range 0 :s

I n x 2500.

Step 1. Find the minimum filter order n from equation 19.7. Here /

1 0 ^ ^ - _1 '

iogio

_ I

( log 10

l O - '- l ^ 10° “ - 1 - = 2.68

•ogio(4)

1 •ogio \

P/

implying that the minimum filter order is 3. This can also be accomplished in M A T L A B as follows: »n=buttord(wp,ws,Amax,Amin,V) n =3

Step 2. Using equation 19.8 with w = 3, I n X 500

CD, COc \ m

m

= 2;r X 547 = 3435 rad/s

2 ^ 1 A "max _ I

and

2jt X 2000

CO, COc , m

ax

min _

J

= 2;r X 633 = 3975 rad/s

^ 1 0 ^ -1

Step 3. Since the order is 3, we have from equation 19.2 ^ (> )

2

1 1+

(O co^

1039


Step 4. Plotting this function over the frequenc)' range 0 ^ 2jvf^ 2 ttx 2500 can be achieved with the following M A T L A B code. The resulting plot is displayed in Figure 19.4. »f= 0:4:2500; »hl = sqrt(l. ./(I + (2*pi*f ./wcmin).^6)); »h2 = sqrt(l. ./(I + (2*pi*f ./wcmiix),^6)); »plot(f,-20*logl 0(h 1),f,-20*logl 0(h2)) »grid

Frequency in Hz F IG U R E 19.4. Plot of third-order LP filter response using Observe that with the choice o f

= (jo^

the magnitude response curve passes through the

pass-band edge frequency with loss exactly equal to O n the other hand, with the choice of co^ = co^

leaving more than adequate loss at (O^. the magnitude response curve passes through

the stop band edge frequency with loss exactly equal to loss at

and

with less than the maximal allowable

. In practice, to allow for element tolerances, one would choose O)^ somewhere in

between.

Exercises. 1. Verify that if co^ = 2. Verify that if co^ =

the loss

and ^(co^) >

and A { lo^) s

1 04


-7.071 le-Ol + 7.071 le-Oli -7.071 le-01 -7.071 le-Oli 7.071 le-Ol + 7.071 le-Oli 7.071 le-Ol -7.071 le-Oli Note that rvvo zeros are in the right lialf complex plane and two in the left. The two in the left half plane determine the Butterworth loss function. For illustration we provide the zero plot of Figure 19.5. Again, note the symmetry o f the zeros with respect to the imaginary axis.

1 .......... i.............. i ..............

............•........... ;

0.8

X

:

i

I /'

:

i

i/

!

i

0.6 0.4

0.2

*

J

0

I**• •

-0.2 -0.4

-0.6 -0.8

i \

i

: i

!

/ X

!

-1

-1

•

-0.5

1 0.5

0

1 1

Real part F IG U R E 19.5 Plot of the zeros of H (s) H {-s) = 1 + As per our earlier statement, from stability considerations we pick the left half plane zeros to form

i-e., »Hhat = poly(zeros(l :2)) Hhat = l.OOOOe+00 1.4l42e+00 l.OOOOe+00 W e conclude that

r + y[2s+\ For higher-order cases, the procedure of Example 9.3 can again be used. The results are given in Table 19.1, which presents the 3 dB normalized Buttenvorth loss functions. In practice, one never computes the actual transfer function o f a filter that meets a set of non-normalized specs. Rather, one realizes the normalized loss or transfer function and then magnitude- and frequency-scales to obtain the proper circuit. This process is illustrated in the next section.

1042

C h ap te r 19 • Principles o f Basic Filtering

T A B L E 1 9 .1 N orm alized Bu tterw o rth Lo s s F u n c t i o n s , « = 1,

n

5.

^3 d B N L p (^)

1

S+ 1

2

S~ + y / l s + 1

3

+ 2 r + 2s + 1

{s + l ) ( j ^ + s + \ ) =

{ r + 0 . 7 6 5 3 7 s + 1)(^2 + 1 . 8 4 7 7 6 i + 1 )

4

= / + 2 .6 1 3 1 P + 3 .4 1 4 2 j2 ^ 2 . 6 1 3 b + 1

{s +

5 = P

l)(j^

+ 0 . 6 1 8 0 3 ^ + l ) ( j ^ + 1 . 6 1 8 0 3 j + I)

+ 3 .2 3 6 1 /

+ 5 .2 3 6 1 P

+ 5.2 3 6 1 ^ 2 + 3 . 2 3 6 b

+ 1

4. BUTTERWORTH PASSIVE REALIZATION W e begin this section with an example of a third-order passive realization. E X A M P L E 19.4 The circuit of Figure 19.6 must realize a Butter\vorth filter meeting the low-pass brickwall speci fication of Figure 19.7. In the final design we require R^= 100 Q.

R.

LC out

Network

-o

F IG U R K 19.6 Structure of a filter driven by a practical source.

40 db

2db >• f(Hz) 100 Hz

1000 Hz

F I G U R E 19.7 Low-pass brickw all specification o f Exam ple 19.4.


So

1043

lu t io n

From the brickwall specification of Figure 19. 7 , we have (oj^ = 2 0 0 k , 20007T,

= 2 d B) and (o)^ =

= 40 dB).

Step 1. Compute the filter order. Here _ 1

loglO

1 0 ® - _ 1 2 0 0 0 .T

•o g lO

- =

2.1164

200 jt

This implies that the minimum filter order is n = 3. Step 2. Compute

Recall equation 19.8;

COn

CO.

By historical convention, we will use

0j^=(0 c.min

in which case

(O,

2 0 0 JZ

= 687 = 21 SJjT rad/s ec

Step 3. Looking up the third-order normalized Butterworth loss fiinction and inverting to obtain the transfer function yields

H MBNLP

(19.12)

Step 4. Choose a candidate passive circuit and obtain its transfer function, W e wind up with the circuit o f Figure 19.8. Note that we will adjust the

is) . Vouti^) Vin(s)

source resistance later with magnitude scaling.

/VY\ in

c

L

V.

V

F IG U R E 19.8 Third-order filter circuit for Example 19.4. W e obtain the transfer function using nodal analysis. At the top node o f C j,

Vc - - ! - K u, = Ls A t the o u tp u t node.

v,„

(19.13a)

1044


--V /C + Lv

C-iS H-----

(1 9 .1 3 b )

Ls

Solving equation 19.13b for Kq and eliminating K q from equation 19.13a yields

" - ( ■ 'J - i r M ‘'//il-V

3

+-

I

0

C|

Step 4. Equate the coefficieuts o f

C 1 + C 2------------ i ~ ------LC jC t LC^C2

s ~ + -^ -------^ s +

Equating the

an d

coefficients in equations 19.14 and 19.12 yields

• . 1 , . Cl + . — = 2 , ------ = 1, and — ---- = 2 C| LC^C') LC\C~) 2 Clearly, Cj = 0.5 F and

LCi

= 1 implies LC^ = 2. Then

C 1+ C 2

0 .5 + C 9

LC ,C 2

1

1c

1r

r-

— ---- = 2 => C ? = 1.5 F and L = — H

In summary, C, = 0.5 F,

^

■^11 3

= 1.5 F, and L = 4/3 FI.

Step 5. Frequency-scale the circuit to obtain the desired, a n d magnitude-scale to obtain the specified

source resistance o/lOO Q. W e frequency-scale by Kj-= co^ = scale by

= 100. Thus, Q

= 687 = 218.7TT and magnitude-

= 100 /?^ = 100 Q

.......... ^ ^ = 7.2773 HF.

C.

______ ^ . K ,„L = 2 1.832 ^F, and L ,,,, = ^ = 0.19406 H.

This completes the design.

E xercise. Reconsider Example 19.4. I f

is to be 50 Q and O)^ =

recompute the final ele

ment values. .\ N S W F R S : 7.38~2 iiK 22.162 uF. and 0.0492 H

In the above example, there is no load attached to the filter. A situation in which there is no source resistance but a load resistance is given in the problems at the end of the chapter. In addition, there is a problem containing both a source and a load resistance. Here we have used a coefficient matching technique that is manageable for orders 1, 2, and 3. For higher orders the method is unwieldy and we must use the methods o f nerwork synthesis studied in other courses or resort to normalized filter tabulations given in filter handbooks.

1045


5. ACTIVE REALIZATION OF LOW-PASS BUTTERWORTH FILTERS For active realization of Buttervvorth filters we use the factored form of the transfer functions given in Table 19.1. For example, we would represent the third-order normalized Butterworth transfer function as the product

K.

H?iclBNLp{^) = —-- — S~ + .V + 1

where

= 1. The gains

K

= H„{s)Hi,{s)

(19.15)

i’ + 1

and /iQ, are present in equation 19.15 to allow the individual stages

to have dc gains different from 1, as is sometimes necessary. The overall transfer function is real ized as a cascade of

^

^

= — — -------- a n d H ij { s ) =

s~ + .s -I" 1 with the constraint

.v+1

= 1. All transfer functions beyond first order require one or more sec-

ond-order stages in cascade for their realization. The use of a basic second-order active circuit is key to the realization of any of these second-order sections.

The Sallen and Key Second-Order Section One of the common single-op amp second-order circuits is the Sallen and Key (S & K ) configura tion illustrated in Figure 19.9. As part of the realization process we must first compute its secondorder transfer function. Denote the circuit transfer function by

F IG U llE 19.9 Sallen and Key low-pass second-order section. To derive

note that the properties of an ideal op amp force the voltage across 7?^ to be

Using voltage division,

■Vo, u t o r equivalently,

1046


This observation permits us to simplify the diagram of Figure 19.9 to the equivalent circuit of Figure 19.10.

F IG U R E 19.10 Equivalent circuit of the Sallen and Key low-pass filter of Figure 19.9;

identifies node b and

identifies node a.

Construction o f H { s ) now proceeds by nodal analysis. At node a— the node between C , and

Ri, 1

( 1

(19.16)

Similarly at node b, (19.17)

Eliminating

and solving for the ratio

V

( 5) ^

yields the transfer function of the Sallen and

Key low-pass filter:

K R\R2C\C2

K ,A ^ )

Vi„U)

2

M

S + 1 ------------ h —

1

]-K \

H-------

-I

R 2 C 2 )

U ^ lC ,

1

(19.18)

H-----------------------R\Ri C iC 2

w h e r e / r = l / , , / l / = 1 + /? ^ //? ^

Butterworth LP Design Algorithm Based on Sallefi and Key Circuit For practical reasons, design engineers have developed S & K circuit design templates based only on the circuit Q and dc gain K for a transfer function of the form

H [s) =

K s^ + — s+ \

(19.19)

Q

The design template derives from equating the denominator coefficients of equations 19.18 and

19. 19. The dc gain can be adjusted using other methods such as input attenuation (to be described later). In particular, equating the denominator coefficients o f equations 19.18 and 19.19 requires that


104'

Q

1

1

1- K

/?,C,

/?2Ci

R 2C 2

(19.20a)

and (19.20b)

R\R2C\C2

The solutions to equations 19.20a and 19.20b are not unique because there are t%vo equations in five unknowns. This means that we can impose up to three additional constraints to produce dif ferent solutions. Different solutions produce the proper filtering action, but have different behav iors in terms o f the sensitivit)' of the frequenq' response to variations nominal resistance and capacitor values. Also, different designs have different ratios of clement values that may have prac tical significance. One popular and robust design is the Saraga design. For the (normali/ed) Saraga design, the three addi-

R.

Q

R]

v3

tional constraints imposed on the solution of equations 19.20 are C, = 1, Cj = >/3Q, and — =

,

which were chosen to minimize certain sensitivities in the circuit performance. Solving equations 19.20a

and

19.20b

for

R ,,

C 2 = IC \ = S q . R] = Q~\ R2 =

/?2,

and

3

I/V ,

K

using

these

additional

K = 4 / 3 , Rs = R^/3.

constraints

yields

The final circuit realization of

equation 1 9 . 1 9 is given in Figure 1 9 . 1 1 . Again we note that the dc gain K = 413 will need to be adjusted by other means, to be described. Frequency and magnitude scaling are necessary to achieve proper cutoff frequencies and impedance levels.

F IG U R F. 19.11 Normalized Saraga design of Sallen and Key circuit for realizing the transfer function of equation 19.19. W ith this design in hand, we now redo Example 19.4 using an active realization.

1048


E X A M P L E 19.5. This example presents an active realization of a third-order Butterworth filter with 0)^ = 687 = 218.7;r rad/s ec. Recall from equation 19.15 that we factor the third-order Butterworth transfer function as a product:

K

K

^7,clBNLp{^) = -------------- ^ ------- - H^,{s )H ij{ s ) S~ + .V -I- I

where

5 + 1

= 1. Later, we will frequency- and magnitude-scale to obtain proper cutoffs and

impedance levels. For

S~ + .S• + I

S~ +

.V +

Q

we have Q = 1. The Saraga design element values are C-, = 1 F, Cj = >/3 F, /?, = I Q, 7?2 “

Q., K = K j = 4/3, and

^

These element values realize the transfer function

.V“ + .V+ 1 To obtain the correct

realize the transfer function

K,, _ 3/4 s -t-1

(19.21)

.V+

Observe that the choice o f /Q, makes A'/Q, = U the correct overall dc gain. The transfer function of equation 19.21 is realized with the leaky integrator circuit o f Figure 19.12, whose transfer func tion

set equal to H(^{s) is

1

Hl,âr(s) =

3/4

R ,C .y-l-

1

5+1

R'yC

To complete the normalized design, choose C = \ V, Rj - \

and /^| = 4/3 Q.

R,

F I G U R E 19.12. Leaky integrator circuit.

1049


Combining the second-order and first-order sections o f the normalized design produces the cir cuit of Figure 19.13.

F IG U R E 19.13. Realization of third-order 3 dB N L P Butterworth transfer function. In the last step o f our design we frequenc}'-scale by /y = 218.7;r. If the smallest capacitor is to be 100 nF, then

K,n =

c o ld

_

K jC ,,,,,

= 14.555 x 1 0 ^ 2 I 8 . 7 ; r l O “ ^

Hence the final parameter values for the second-order section are

Q = 100 tiF, q = 173.2 uF,

= 14.56 kQ, Rj = 8.4 kQ, R^ = 30 kQ, ami Rf^ = 10 K2

Similarly, for the first-order section C = 100 nF,

/ ?2

= 14.55

and R^ = 19.4 kQ. This leads to

the final circuit design given in Figure 19.14. 173.2nF

14 .5 5 kO


1050

This design assumes the fiirer is driven by a volrage source with a very small resistance; otherwise the source impedance must be considered part o f the front end series resistance. Also, we could have magnitude-scaled each section separately (but chose not to).

Other Sallen and Key Designs in addition to the Saraga design, there are two other S & K normalized designs that we will set forth here. The first is termed Design A: /?^ = oo, To obtain design A we normalize K =

1 and

= 0, /T = 1, /^, = /?2 = 1 = /?2 = 1

j

Cj = 2Q, and C j = ---

Using these three constraints

^

and equating the coefficients o f equations 19.19 and 19.18 leads to

K s~ +

— .V

C\C2 +1

+ --

C,

Q

.V +

(19.22)

--------

C ,C 2

Matching the denominator coefficients produces Cj = 2Q and C j = 1/2Q. Since the dc gain with

K = 1 is 1, there is no need to modify the overall circuit gain. Figure 19.15 illustrates design A. C, = 2Q

F IG U R E 19.15 Design A for realizing the transfer function of equation 19.19. Notice, however, that for Design A the ratio C^/C, = 4Q^ is 400 for a circuit with Q = 10. Such a large variation may be undesirable in a practical circuit. 2, C, = 1 F, C 2 = 1/Q F,

A third design we term Design B: /?^ = /?^ = 1 Q, = Q L2. This design is computed using the constraints and equal time constant, and /^2 -

Q- Thus,

/?jCj =

= ^2^2'

= I Q (yielding /f= 2), C, =

these choices,--------- =

1 = RjC2 implies that C ,

= 1 Q, and

1implies that /?|

=

1 F, 1

= — . Here the maximum parameter ratio is

= C j/ G ) = Q, w h ic h is better than the 4Q^ ratio of D esign A . Figure 19.16 shows the result in g circu it.

lO r


C =1

F IG U R E 19.16 Design B for realizing the normalized transfer function of equation 19.19, assuming standard units.

6. INPUT ATTENUATION FOR ACTIVE CIRCUIT DESIGN In design A of the previous section, the gain o f the filter is A"= 1. In design B, the filter gain is K = 2. In Example 19.5, the gain of the Saraga design {K = 4/3) was corrected by the subsequent first-order section. I f the filter were second order with a dc gain of 1, then the circuit gain would need to be modified for both Design B and the Saraga design. In order to reduce a high gain, maintain the filtering properties, and keep as many parameter values as possible at their original design values, we use a technique known as input attenuation. In this technique, the front-end resistor is replaced with a voltage divider circuit, as illustrated in Figure 19.17.

(b) F IG U R E 19.17 Illustration of input attenuation concept, (a) Original circuit, (b) Circuit with input attenuation.

10 ^ 2


Figure 19.17a represents the original active network, while Figure 19.17b represents the modified network. For the filtering characteristics to remain invariant, the impedance looking into the dashed boxes must remain at /?, for both circuits. Thus, the parallel combination of

and

must equal /?j, i.e., '

+

(19.23)

In addition, if the new gain is to be a K , a < 1, instead o f K, then

a =

^4

/?3 + /?4

(19.24)

Solving equations 19.23 and 19.24 for R^ and R^ yields

■

=^ a

and

^

(19.25)

\ -a

Thus, one can reduce the gain o f the Sallen and Key low-pass circuit via the simple technique of input attenuation.

7. PROPERTIES OF THE BUTTERWORTH LOSS FUNCTION As mentioned earlier, the Butterworth loss function has a maximally flat pass-band response in the sense that as many derivatives as possible o f the loss magnitude response are zero at Q = 0. This is often termed m axim ally flat. In the case o f the ;zth-order Butterworth loss function, it is possible to show that

H ( j Q ) = 0,

for A: = 1,2,...,

cIQ'^

2/2-1

(19.26)

at Q = 0 and

cf dQ^’’

H (jQ )\ = 0 .5 { 2 n ) \ * 0

( , 9.27)

at Q = 0. This is consistent with the notion of being maximally flat. To verify equations 19.26 and 19.27, observe that if

«

1, then

Equations 19.26 and 19.27 follow after differentiation of equation 19.28 2u times and evaluation of each derivative at Q = 0. Several closing remarks are now in order. First, the cutoff frequency is the half-power point, or the 3 dB down point. The terminology follows because |MyO)| = 1 (the gain at dc is unity) and


10^ 3

H (j\ ) = - ^ = 0.707, V2 yielding half power for normalized Butterworth gain functions H{s). Second, for large Q,

Hence as Q becomes large, the gain G^g{Q) decreases according to

-n

[20 logjQ(Q)]

indicating that the gain rolls off with a slope proportional to the number of poles, n, o f the gain function. Specifically, the slope equals -20« dB per decade. These statements are also valid if to replaces Q.

8. BASIC HIGH-PASS FILTER DESIGN WITH PASSIVE REALIZATION High-pass filters invert the frequency characteristics of low-pass filters. A high-pass filter is a device— usually a circuit— that significantly attenuates the low-frequency content o f a signal while passing the high-frequency content with minimal attenuation. Figure 19.18 illustrates typical brickwall specifications for a high-pass filter. Here frequencies above the pass-band edge frequen-

cy co^ have little attenuation, while frequencies below

are significantly attenuated— precisely

the inverse function o f a low-pass filter. In fact, low-pass and high-pass specifications are related by a simple inversion of O). In particular, we define a H P (high-pass) to L P (low-pass) frequen cy transformation as

^ = —

(19.29)

This frequency transformation applied to the briclc\vall specifications o f Figure 19.18 yields the set of normalized low-pass specifications given by Figure 19.19.

10 V i


A(co)

min

3dB A_..> (0 0

CO.

%

F IG U R E 19.18 Typical brickwall characteristic of a high-pass filter.

A(co)

F IG U R E 19.19 Normalized low-pass equivalent specifications derived from high-pass specifications. Given the normalized L P (N L P ) equivalent specs, one finds the Buttenvorth filter order accord ing to the usual formula, equation 19. 7, reproduced as equation 19.30:

]

(

1''VniH _ I

log clO

10

•o g lO _

lo2 10

1

(19.30)

( Q ,\

By convention we take the filter order to be the smallest integer n satisfying equation 19.30. For a passive realization, once n is known, one would realize the resulting N L P Butterworth transfer

I05S


function. For example, suppose « = 2; the second-order 3 dB N L P Buttervvorth transfer function could be realized by (among others) the circuit of Figure 19.20, whose transfer function is J_

J_

Cs Ls + /?<• +

_ 1

Cs

LC S

1

+ — S+ -

LC

V

F IG U R F 19.20 Circuit to realize second-order N L P Butterworth filter. As with the development o f equation 19.8, once n is picked, there is a permissible range of nor malized

given by

Q.

Qc ju i n

where

(19.31)

Once

is chosen, the Butterworth transfer function that meets the N L P

equivalent specs is found as

^NLP i^) = ^3dBNLP | ^ j This is, of course, equivalent to frequency scaling a realization of the Butterworth filter by Ay = Q^, as illustrated in Figure 19.2L

/YY\ u o .C

V,

C/Q,

V

F IG U R E 19.21 Circuit that realizes The actual H P transfer function is related to the other “ normalized” transfer functions as follows:

10„ H H p{s) = H N Lpi— \ = H This is equivalent to doing a special frequenq' transformation on the circuit elements o f Figure 19.21. Specifically

1056


(O. (19.32) is called the L P to H P frequency transformation and has a natural interpretation in terms of the inductors and capacitors of a passive network. In particular, a capacitor C with impedance

Cs becomes an inductor o f value

, I.e.,

C(0:

Cs

C

(0),

C oj,

A similar substitution shows that an Inductor L with impedance Ls becomes a capacitor of value Figure 19.22 illustrates the transformation.

^

/YY\

-O

■O

1 L= Cco

o- fY Y \

-o

-o 1

c = Leo. F IG U R E 19.22 Circuit element change under the LP to H P transformation. Hence the L P circuit of Figure 19.21 becomes the H P circuit o f Figure 19.23.

Cnew= — L CO

F IG U R H 19.23. H P circuit derived from the N L P reali/aiion of L'igure 19.21. Observe that

to.

It remains only to magnitude-scale this circuit to obtain an acceptable passive H P filter. W e illus trate the above ideas in the following example.

lOS"


E X A M P L E 19.6. Design a minimum-order Buttenvorth filter meeting the high-pass brickwall specifications of Figure 19.24, in which the loss magnitude curve is to pass through the pass-band corner.

>0) 2000n

200n

F IG U R E 19.24 High-pass filter specifications tor Example 19.6.

Solution

9000 t

Step 1. Compute the equivalent N LP specificatiom . Using

Q =

the H P to N L P trans

formation, the N L P specifications are easily computed as

(H/. ' ' ■

= 2 dB) and (Q^ = 10, A„,.„ =40 dB)

Step 2. Detennine the Butterworth transfer function. As these specifications correspond to those of Example 19.3, one computes the filter order n = 3. From Tiible 19.1, the third-order Butterworth transfer function is

s~ -I- .V -I- 1 ](.v + 1)

•v-' + 2 .v “ + 2.V + 1

Step 3. Realize the third-order 3 dB N LP transfer fin ctio n . According to Example 19.4, the 3 dB N L P Butterworth transfer function can be reali7.ed by the passive RLC circuit o f Figure 19.8, with values assigned as in Figure 19.25.

/YY\ 1n V,

C^=1.5F

-O + V

-y C =0.5 F

-o

---- c

F I G U R E 19.25 Third-order 3 d B N L P filter circuit o f Example 19.4 having 3 d B down point at 1 rad/sec.

1058

Chapter 19 * Principles of Basic Filtering

Step 4. Compute the norm alized 3 dB down frequency. The

3 dB

frequency

of the actual N L P

equivalent filter is calculated from equation 19.31, using the left inequality:

1/6

Step 5. Frequency-scale by

= 1.0935

to obtain the N L P circuit. The N L P circuit o f Figure 19.25 is fre-

quency-scaled by the factor Kj-= 1.0935 to produce the network o f Figure 19.26.

/YY\ V.

-O +

1.219H

1o

- J

'' ■N

0.4573 F

r

V

1.3717 F t

F I G U R t 19.26 Normalized low-pass circuit of Figure 19.25 frequency-scaled by /y-= L0935. Step

6. Apply the LP to H P circuit element transformation detailed in Figure 19.22. Converting the

circuit of Figure 19.26 to the required H P circuit yields the network of Figure 19.27.

0.1305 mF -O +

1Q V.

V 0.3481 mH

0.116mH

F IG U R E 19.27 High-pass circuit meeting the specifications of Figure 19.24. Step 7. Impedance-scale. To obtain a 10

source resistance, we impedance-scale by the factor

= 10 to obtain the final realization shown in Figure 19.28.

13.1 mF

-O +

lOO

V.

1.16 mH

V

3.481 mH -o F IG U R I: 19.28 Passive circuit realizing the high-pass specifications of Figure 19.24 with 10 source impedance.

E x e r c is e . Use SPIC E to confirm the high-pass characteristic o f the circuit o f Figure 19.28.

10^‘)


The preceding example completes our basic discussion o f passive high-pass filtering.

9. ACTIVE REALIZATION OF HIGH-PASS FILTERS Active realization o f high-pass filters ordinarily proceeds with the cascading o f first- and secondorder sections, as done earlier. There is a Sallen and Key second-order high-pass section obtained with an interchange of resistors and capacitors in the S & K low-pass configuration. As with all active circuits, inductors are not used. W e illustrate this interchange of resistors and capacitors as an approach to high-pass design with the following example. E X A M P L E 19.7. Let us reconsider Example 19.6, in which

= (2000;r, 2 dB) and

= (200;r, 40 dB). The filter design proceeded by converting to equivalent N L P specs, which led to a 3 dB N L P third-order transfer function,

H 3dBNLP {s) =

I

( i - +l ) |. v“

ij

+ 2 a’“ + 2.S + 1

One approach to H P design first realizes this 3 dB N L P transfer function as a cascade o f a secondorder active circuit with a first-order circuitin Example 19.5, this resulted in the circuit o f Figure 19.13, redrawn as Figure 19.29.

1O

F IG U R E 19.29. Realization of third-order 3 dB N L P Butterworth transfer function. The second step in the completion o f a H P design is to replace each resistor with a capacitor whose value is the reciprocal of the resistance; likewise, each capacitor is replaced by a resistor whose value is the reciprocal o f the resistance value:

R (19.33) C

1060

Chapter 19 • Principles of Basic I*iltering

Applying the transformation o f equation 19.33 to the circuit of Figure 19.29 produces the circuit o f Figure 19.30, with one exception to the rule: the feedback resistors on the second-order section that connect to the inverting input node are not changed to capacitors; these resistors only set up a voltage division, so it is unnecessar)' to replace them.

F IG U R E 19.30. 3 dB normalized third-order high-pass filter in which

1.

In terms of transfer functions, the operation o f equation 19.33 changes s in the 3 dB N L P trans fer function to \/s. Specifically,

H 7.CIBNHP

HM BNLP W =

s M BNLP

X

(^ + 0

s

2

^“ -f-5-f-l

Notice that each section of the transfer Rinction has a (normalized) H P characteristic. For example,

5

l i m ------ = 1 and lim 5 —*o c

s+ 1

5

------- = I whereas li m ---- = 0 and lim

The final design is achieved by frequency-scaling by Kj-= tor K

= 0.

and impedance-scaling by a fac

. The counterpart of equation 19.8 for the H P case is given by

In

-1 5

- 1

(19.34)

(The derivation o f equation 19.34 is assigned as a homework problem.) By convention, frequen cy scaling places the magnitude response through the pass-band edge, in which case

(o, Hp = w "^ 1 0 ° '^ " - - 1 = 5 7 45.9 rad/sec and K j- = 5745.9 .

1061


Let us suppose that the smallest capacitor is to be 10 nF. Then

^

= 13.05

t
13.3 nF 7.54 kO

M G L IR E 19.31. Final design of third-order high-pass filter. The circuit o f Figure 19.31 presumes that an ideal voltage source is input to the filter. I f the source has an internal impedance, this impedance will affect the pole locations and the overall perform ance o f the high-pass circuit. To circumvent this problem, one may insert a voltage follower cir cuit between the practical source and the input to the filtering circuit o f Figure 19.31, as illus trated in Figure 19.32.

Practical

Voltage Follower Circuit

F I G U R E 19.32. Practical source in series w ith voltage follower circuit.

1062


It is possible to justify the RC to CR transformation through the use o f well-known networks theorems. However, such justification, which is found in more advanced texts on filter design, is beyond the scope o f our current endeavors. There are other H P design algorithms that do not uti lize the transformation o f 19.33, but these are left to texts on filter design.

10. BAND-PASS FILTER DESIGN Band-pass circuits were first considered in Chapter 16, mostly from an analysis perspective. In this section we will briefly oudine the design o f a B P filter to meet a given set o f brickwall specifica tions as shown in Figure 19.33.

A Loss(dB)

>► CO

0)

0).

0)

P2

FIGURK 19.33. Typical brickwall B P filter specifications. In Figure 19.33 O)^, and

10^2

pass-band edge frequencies while

and 0)^2

band edge frequencies. The center of the B P filter is defined as the geometric mean of pass-band edge frequencies, i.e.,

= yjojp\(0p2 •

Given the B P brickwall specs, the design procedure begins with a frequency transformation to convert these specs to equivalent N L P specs, i.e., (Q^ = 1,

and (Q^.,

To illustrate how

to generate these N L P specs we use a frequency transformation as follow's: define a bandwidth w ith respect to

as Z? = 0)^2 “

define two potential N L P stop band edge frequencies

as

^.v/ =

Iko,i

(19.35)

for / = 1,2. Given these two numbers, the N L P equivalent specs are

= '•

and (£2, = m in[Q ,,, Q,^], A „ J

(19.36)

To illustrate the need for the magnitude sign in equation 19.35 and the need for the minimum function in equation 19.36, let us do a simple example by breaking apart the formula o f equation

19.35 and

putting it back together. Suppose

= 2 rad/sec and 0)^2 = ^ rad/sec with to^j = 1

Chapter 19 * Principles of Basic Filrcring

rad/sec and

1063

= ^0 rad/sec. Then

rad/sec. Let us evaluate a part

- yj(0p^(0^y2 = ^

of equation 19.35 in which B is not present and the absolute value signs are gone. Thus

_ [-6

i=\

I 6

i=l

a ; ,,

[19.2

i= 2

One can view this partial transformation as the generation of two distinct L P filters: one defined on negative frequency (Q^

6,

= - 6,

=-15) and one defined on positive frequency

=

= 19.2). O f course, we cannot really have two distinct L P filters, one for negative fre

quency and another for positive frequency. O u r mathematics requires that the magnitude respons es of each filter be symmetric because magnitude is an even function of (o. Thus the only way to properly interpret the above transformation is that it implicitly generates two distinct low-pass fil ters, one of which is more stringent than the other. Specifically, the “ negative” filter is really a LP filter with edge frequencies (Q^, = 6, Q^, = 15), and the positive filter is really a L P filter with edge frequencies

~

“ 19.2). Observe that the ratio

^,v/ Q^,

|3.2

/= 2

differs for the rwo filters. The first filter is more stringent than the second, prompting the need for a minimum fiinction in equation 19.36. Since our edge frequencies are always specified on the positive axis and since the magnitude response is symmetric, w'e insert the absolute value signs in equation 19.35. Finally, in order to have Q^- =

1 we divide by

- to^j so that the edge fre

quencies correspond to a N L P filter.

EXA M PLE 19.8. A B P filter has the specs krad/sec, 0)^2 = 32.6 krad/sec,

= 26 krad/sec,

= 3 dB, and

= ^7 krad/sec,

= 30.6

= 14 d k Design a minimum-order

Butterworth passive B P filter meeting these specs, assuming the filter is driven by a voltage source in series with a

10 kQ

source resistance.

S o l u t io n

Step 1. F in d the equivalent N L P specs fo r this filter. Using M A T L A B , we have »wsl = 26e3; ws2 = 37e3; w'pl = 30.6e3; »wp2 = 32.6e3; Amax = 3; Amin = 14; »K = w pl*w p2

K= 997560000 »B = w'p2 - w pl B =

2000 »ws = [wsl ws2]; »Wsi = abs((%vs M - K) ./(B*ws))

1064


W si = 6.1838e+00 5.0195e+00 »W s = m in(W si) W s =

5.0195e+00 Step 2. Find the filter order. »n = buttord(l,W s,Am ax,Am in,V

n= 1

Step 3. Find a realiztition o f the NLP filter. Since

= 3 dB, the first-order Butterworth N L P

has transfer fûnction Hf^j^p{s) = ----,whose passive circuit realization is given in Figure 19.34.

.V+ 1 -O +

1Q 1F

F IG U R E 19.34. Rcalizjuion of the N L P transfer function. Step 4. Realize the given BP filter. The algorithm to generate the B P filter from the N L P filter is based on the frequency transformation

.v“ + (19.37)

Bs

This frequency transformation causes a change in the capacitive and inductive circuit elements o f the N L P circuit. From equation 19.37, the impedance o f an inductor changes as follows:

Ls

( s - + K\ Bs

L = —s + B

1 (19.38)

B

\LK I rhe expression on the right of equation 19.38 is a sum of impedances. This sum, then, is a cir cuit composed o f an inductor in series with a capacitor as shown in Figure 19.35.

/YY\ L

L B

3.

LK

F IG U R E 19.35. LP to BP transformation on an inductor. Similarly, from equation 19.37, the admittance of a capacitor changes as follows:

106S

Chapccr 19 • Principles o f Basic Filtering

= — .V+

Bs

B

(19.39)

f

XK)

The expression on rhe right is a sum of admittances. Hence under the frequency transformation, a capacitor becomes a parallel L C combination as illustrated in Figure 19.36.

C CK

F IG U R E 19.36. LP to B P transformation on a capacitor. Applying the transformation of Figure 19.36 to the N L P circuit of Figure 19.34 yields the circuit o f Figure 19.37, where - = lO""^ and — = 2.005 x 10'^\

B

K ------- O +

1O 1

B

B

K

'ôut

------- O F IG U R E 19.37. BP circuit without magnitude scaling. Step 5. Miignitudc-scale to obtain the correct source resistance. W ith = 50 nF and

=

10"^, we conclude that

= 20 m H . The final circuit realization is illustrated in Figure 19.38. Km

-o I

_ BK

bp

“

+

k"

-o F IG U R E 19.38. Final BP design. The final transfer function o f the circuit is

/Y gp (.v)-

Bs

This transfer function and circuit show a clear similarity to the material developed in Chapter 16. In contrast to the above procedure, the development o f Chapter 16 works only for w =

1 in

the

equivalent N L P circuit, which results in a B P circuit o f order 2. I f the order o f the N L P equiva-

'0 6 6


lent were 2 or higher, rhe B P order would be at least 4 and the background from Chapter 16 would prove inadequate. The next example illustrates how design is done for higher-order circuits.

E X A M P L E 19.9. A B P filter has the specs to^j = 1500 rad/sec, rad/sec, to^^ = 4000 rad/sec,

= 2 dB, and

= ^^00 rad/sec,

these specs that is driven by a voltage source having an internal resistance of So

= 3000

= 20 dB. Design a passive B P filter meeting

100 Q.

lu t io n

Part

1. F in d equivaletn N L P specs.

Define K =

12 x 10^ as the square o f the center fre respect to as 5 = co^2 ~ ~ 1000 rad/sec. =

quency of the filter. Define the bandwidth with

W e use the following M A T L A B code to obtain the necessary numbers. wsl = 1500; ws2 = 6000;wpl = 3000; wp 2 = 4000; B = wp2 - wpl K = w p l*w p 2 ws = [wsl ws2]; W si = abs((ws .^2 - K ) ./(B*ws)) W s = m in(W si) The M A T L A B output is B = 1000

K = 12000000 W si = 6.5000e+00 4.0000e+00 Ws =4 Therefore, the equivalent N L P specs arc = 2) and (Q^ = 4,

= 20)

These N L P specs are then used to design a N L P transfer function. Part 2. F ind an equivalent N L P Butterioortb transfer fin c tio n an d its realization. In M A T L A B , we have »W p = 1; W s = 4; Amax = 2; Amin = 20; »n = buttord(W p,W s,Am ax,Am in,V)

n=2 From Table 19.1, the second-order Butterworth filter is given by

" T T T ir r T

(19.40)

1067

Chapter 19 • Principles of Ba.sic Filtering

at Q = 1. W e will adjust this by frequency-scaling the circuit

This docs not have the correct that realizes

^ circuit that realizes the second-order Butterworth transfer function is

given in Figure 19.39.

yvYV

-o

L

+

1n

'- 0 F IG U R E 19.39. Second-order circuit. The circuit o f Figure 19.39 has transfer function

1

S

2

LC 1

(19.41)

1

H----- A + ------

L

LC

By equating the denominators of equations 19.40 and 19.41, we obtain L = ~

H and C = V2 F.

V2 spec at Q = 1. For this we use

Next we frequency-scale so that we obtain the correct

1 c,w/«

in which case

L

=—

- W - \A^mat _ I

■= 1.1435 rad/sec

C

= 0.61838 H and

=—

= 1.2368 F.

Thus the circuit that realizes the N L P characteristic with a loss of

at Q =

1 or, equivaiendy,

3 dB of loss at Q^= 1.1435 is given in Figure 19.40.

IY Y \ IQ

-O

0 .6 1 8 3 H

+ V out

1 .2 3 6 8 F

F IG U R E 19.40. Second-order circuit. Note that the circuit of Figure 19.40 has transfer function

o , - H},dBNLP u n^LPkS)

1.3076

I

■ U .1 4 3 5 J

-1-

(19.42)

\ \ A 435}

+1

+ 1.617 Ly-f-1.3076

Part 3. Realize the passive band-pass filter. To compute the desired B P transfer function, one would replace s in

as follows:

1068


+Wp\COp2 '

^

Bs

One could then attempt to realize the B P transfer function directly. This is numerically unwise and, fortunately, unnecessary. A simple, numerically sound procedure is to replace each inductor and capacitor in the N L P realization by an equivalent circuit representing the above frequency transformation. These equivalent circuits were developed in equations 19.38 and 19.39, and illus trated in Figures 19.35 and 19.36, respectively. The substitution scheme is repeated in Figure 19.41.

/YTV L

L B

_B LK

(a)

(b) F IG U R E 19.41. Illustration of the LP to B P transformation in terms of inductors and capacitors. Given the transformations of Figure 19.41, the following M A F L A B code produces the desired B P circuit parameter values, assuming the final source resistance is » % Insert L, C, B, K, and Km »L = 0.61838; C = 1.2368; »B = 1000; K = 12000000; Km = 100; » % Compute K1 » K f = sqrt(K) K f= 3.464 le+03 » % Compute B P circuit parameters associated with L » Llb p = Km *L/B L I bp = 6.1838e-02 » C lb p = B/(L*K )/Km C lb p = 1.3476e-06 » % Compute B P circuit parameters associated with C >)C2bp = C /(B*K m ) C2bp= 1.2368e-05 »L2bp = K m *B /(C ’ K) L2bp = 6.7378e-03

100 O..

1069


The final circuit is given in Figure 19.42, where R^= 100 ^2,

~

= 61.8 m H ,

- 1.35 |.iF,

Îbp ~ LI bp

Cl bp

R.

-O +

C2bp

V.

ôut

L2bp ^

F IG U R K 19.42. Final BP filter realization. This concludes our illustration of the passive B P filter design procedure.

Active B P design is beyond the scope of this text, but coverage can be found in more advanced texts.

11. AN ALGORITHM FOR SINGLY TERMINATED BUTTERWORTH LOW-PASS NETWORKS In 1937 E. L. Norton published a paper in the Bell Systems Technical Jou rn a l? The paper contains explicit formulas for the L C element values in a singly terminated 3 dB normalized Butterworth L P filter where the load or source resistance is 1 ohm. O u r goal in this section is to present these formulas without derivation. Similar formulas have been developed by other researchers in the context of network synthesis, a subject for which many texts are available for further reference. For an «th order 3 dB normalized Butterworth filter with a single

1H

termination, the sequen

tial formulas for the L C elements are ' 7T \ = s in

(19.43a)

\I n ) /3 > t ^ s in

S in

a-) =

\ 2nj

I In ]

K

co s*

sm

=

(19.43b)

sm

2/1

( 1 9 .4 3 c )

In

1070


a„ = ;jsin

JT (19.43d)

\In }

It turns out that equation 19.43d is a special case o f 19.43c with m = n. These formulas can be easily programmed into a M A T L A B m-file as follows: function buttLC (n) nn = 2*n; a = zeros(l,n); a(l)=sin(pi/nn); for m = 2:l:n ; X = a(m -l)’'(cos((m -l)’'pi/nn))^ 2; a(m) = sin(( 2*m-l)*pi/nn)*sin((2*m-3)*pi/nn)/x; end Elvalues = a’ To illustrate the use, suppose the Butterworth filter order is 4. Then »buttLC(4) Elvalues = 3.8268e-01 1.0824e+00 1.5772e+00 1.5307e+00 Given these values, it remains to interpret them as inductances or capacitances. Two rules govern the realization: (i) (ii)

There is never a shunt element in parallel with a voltage source. There is never a dangling element at the load end.

Hence for the fourth-order filter above there are two possible circuit realizations, given in Figure 19.43. 0.3827 H

1.577 H

/YYY.

/YY\

1Q V.

1.5307 F

1.0824 F

(a)

-O +

1071


0.3827 H

1.577 H

/YYY

/YY\ 1Q

1.0824 F

-O +

1.5307 F

-O (b) F IG U R E 19.43. (a) R ^ = \ Q . (b) 7?/ = 1. All that remains in the design procedure is frequency and magnitude scaHng, which are covered in earher sections o f the chapter.

12. SUMMARY This chapter has covered the basics of Butterworth LP, HP, and B P fiher design. Such design tech niques build on a set of filter specifications requiring that the desired filter magnitude response lie outside certain briclcvvall regions. Finding transfer functions that meet a set o f briclcwall con straints is called the approxim ation problem . This chapter developed algorithms for finding Butterworth transfer functions for the LP, HP, and B P cases. In addition, basic passive realizations were presented as well as active circuit realizations, except in the B P case. In the active case, the focus was the Sallen and Key low-pass and high-pass topologies. Both the passive and active t)^pes of realization are built around the coefficient-matching technique, which associates the coeffi cients in the circuit transfer function with the coefficients of the desired transfer function; one then solves the resulting equations for appropriate circuit parameter values. There are, of course, many other types of filter transfer functions; Chebyshev, inverse Chebyshev, and elliptic are other well-known t)'^pes. Also, in addition to low- and high-pass filters, there are band-pass, band-reject, and magnitude and phase equalizers. To add to the richness of the area of filtering, there are analog passive, analog active, recursive digital, non-recursive digital, and switched capacitor implementations o f all o f these filter types. The preceding exposition is mere ly a drop in a very large and fascinating bucket of filter design challenges.

13. TERMS AND CONCEPTS 3 d B bandwidth o f B P filter: the difference between the two frequencies of a B P filter at which the gain is 3 dB down from the maximum gain; if the maximum gain is 1, these fre quencies correspond to a gain of M -Jl this is the most common reference for the mean ing o f bandwidth. 3 d B bandwidth o f L P filter: the 3 dB down frequency, to Active realization: a realization consisting of op amps, /?’s, and C s. Approxim ation problem: the problem of finding a transfer function having a magnitude response that meets a given set o f brickwall filter specifications.

1072


Attenuation (dB); the loss magnitude expressed in dB, i.e., y4((o) = - 2 0 logjQ|//(/(D)|. BP to LP frequency transfonnation: the generation of equivalent normalized LP specs from a given set o f BP specs.

Bandwidth with respect to

the difference between the two frequencies o f a BP filter at

which the gain is down by a value of

from the maximum gain.

Band-pass filter: a filter that passes any frequency within the band 0)^j s O) s

while signifi-

cantiy attenuating frequency content outside this band.

Coefficient-matching technique: a method o f determining circuit parameter values by matching the coefficients of the transfer funaion of the circuit to those of the desired transfer func tion and solving the resulting equations for the circuit parameters.

Cutoff frequency: the frequency at which the magnitude response o f the filter is 3 dB down from

'

its maximum value.

Filten a circuit or device that significandy attenuates the frequency content o f signals in certain frequency bands and passes the frequency content within certain other, user-specified fre quency bands in the sinusoidal steady state.

Frequency response magnitude: magnitude of the transfer function as a function of j(a, i.e., ^

Gain in dB: 2 0 logjo|//(/w)| or 10 log,o|//(/co)p. Gain magnitude: frequency response magnitude. Half-power point: the point at which the magnitude response curve is 3 dB down. High-pass fUten a filter that significantly attenuates the low-frequency content of signals and passes the high-frequency content.

^

HP to LP frequency transformation: a transformation that converts a given set o f high-pass

■

'

brickwall specifications to an equivalent set o f low-pass specifications. 1

Loss magnitude: the inverse of the gain magnitude, i.e., ^ ^ ^ .^ y • Low-pass (brickwall) filter specification: a filter specification requiring that the desired filter magnitude response lie outside certain “brickwall” regions.

Low-pass filter: a filter that passes the low-fi-equency content o f signals and significantly attenu ates the high-frequency content.

LP to BP fi%quency transformation: a technique for converting a passive LP filter to a BP filter by converting Vs to series LC circuits and Cs to parallel LC circuits. LP to HP frequency transformation: a technique for converting a passive LP filter to a HP fil ter by converting Vs to Cs and Cs to Vs. Maximally flat: property o f a filter at a point (0 wherein the magnitude response has a maximum number o f zero derivatives.

Normalized Butterworth loss functions: a set o f Butterworth transfer functions, ordered by degree, having 3 dB loss at the normalized frequency s= jQ = j\ . Passive analog filter: a filter composed only of resistors, capacitors, inductors, and transformers. R C to CR transformation: a technique for translating a low-pass active filter to a high-pass active

'

filter in which resistors become capacitors and capacitors become resistors.

* Wireless En^neer, vol. 7, 1930, pp. 536-541. ^ E. L. Nonon, “Constant Resistance Networks with Applications to Filter Groups,” Bell Systems TechnicalJournal, vol. 16, 1937, pp. 178-196.

O

'

K r3


'Problems

2. Figures P I 9.2 (a) and (c) show an approxi

LOW-PASS BASICS 1. Filters can have magnitude

different transfer functions H{s). Figures P i 9.2

mate gain magnitude response associated with

responses quite

different from the maximally flat Butterworth magnitude response. For each of the plots in Figure P i 9.1, identif)'^ whether it is HP, LP, or

(b) and (d) show new gain magnitude respons es whose transfer functions are related to the

H{s) o f (a) and (c) by scaling. Properly label the horizontal axis of each graph in (b) and (d).

BP

Radian Frequency (a)

Radian Frequency

(b)

Figure P i 9.2 3. A N L P filter transfer function

(0

Figure P I 9.1

0.65378 . r -h 0.803816435 + 0.82306043

10 7 4


has a pass-band edge frequency at

= 2 dB)

1 rad/sec. (a)

Use M A T L A B to plot the magnitude response o f the transfer function and verify that the magnitude response is 2 dB down from its maximum of

1.

= > / i # lA.

= e.

-1

(c o A ‘ (OPJ 7. Suppose //jW and third-order N L P

are second- and

Butterworth transfer func

(b)

Use the “ roots” command in M A T L A B to compute the poles of

tions, respectively, with co. = 03^ = 1 rad/sec and

(c)

The transfer function is to be scaled so

//(0) =

1.

that the new pass-band edge frequen-

(a)

Findand//.W .

cy is 750 Hz. Compute the new trans

(b)

Using M A T L A B and the “ freqs” com mand, plot the magnitude frequency

fer function. W hat are the poles of this new transfer function? Where are the

response of H^{s) and //2W for

zeros of this transfer function?

< 5 rad/sec on

0 < O)

the same graph.

Properly label your plots. 4. Suppose H{s) is the transfer function of a sec-

(c)

Find the step responses for both sys tems using the “residue” command in

ond-order Butterworth low-pass filter, H{Qi) =

M A T L A B to compute the partial frac

1, with 3 dB frequency at 500 Hz. (a)

Determine H{s).

tion expansion, and then use M A T

(b)

Using M A T L A B , plot the magnitude

L A B to sketch the response curves.

and phase responses. (c)

Determine the magnitude of the trans fer function at 250 Hz and

(d)

1 kHz.

LOW-PASS APPROXIMATION 8. Find a minimum-order Butterworth transfer

Realize the filter using the circuit of

function meeting the brickwall specs in Figure

Problem 14 as follows:

P i 9.8 as follows:

(i)

First realize

(ii) Frequency-scale the circuit

to

(a)

Find the filter order.

(b)

Find the range of allowable

(c)

Find the poles, zeros, and gain con

range o f allowable^-

obtain the proper cutoff frequen cy for the filter.

and the

stant of the 3 dB N L P transfer func tion and verify that they lie on the

5. Fill in the details to the derivation of equa

unit circle.

tion 19.8, i.e., of the relationship (d)

to,

0), AAn

sections. (e)

6. Recall equation

Find the 3 dB N L P transfer function as a product o f second- and first-order Using co^

find the zeros, poles, and

gain constant o f the non-normalized

19.2, i.e.,

transfer function. Do the poles lie on

H iJ c o )

2

1

2n + £'

(

a circle about the origin?

.2n

1+

(0

' CO '

Accurately plot (using M A T L A B ) the magnitude response of your filter in terms of gain and gain in dB for 0

This equation specifies a relationship between

< 1.2^. Does your filter meets the

the coefficient e , (O^, and to^.. Show that for any coeff

given brickwall specs?

L P brickwall specification the range o f e brick given by

is

1075


and

Loss (db)

= (45 dB, 450 Hz) as detailed

below. Hint: M odify the M A T L A B code of Problem

(a) (b)

8.

Find the filter order. Find the range o f allowable co^ and the range of allowable/^

(c)

Find the poles, zeros, and gain con stant of the N L P transfer function and

0.3 dB

verify that they lie on the unit circle. > f ( H z ) (d )

Find the N L P transfer function as a product o f the second- and first-order sections.

Figure P I 9.8

(e)

Using (JL)^


gain constant of the non-normalized A N S W E R : Enter the following M A T L A B code

transfer function. Do they lie on the

to obtain the solution.

unit circle;

fp = 100; Fs = 1200;Amax = 0.3; Amin = .35:

(0

n = buttord(lp,fs,Ama.\,Amin. V)

Accurately plot (using M A T L A B ) the magnitude response o f your filter in

fcmin = fp/((10A(0.1*Amax)-l)^(l/(2*n)))

terms of gain and gain in d B for 0 < f

Fcmax = fs/((1 0^ (0.]*A m in)-l)''(l/(2 ‘ n)))

< 1.2^. Does your filter meets the

wcmin = 2’ pi’ Fcmin

given brickwall specs?

wcmax = 2*pi’ fcmax wc = wcmin:

11. Repeat parts (e) and (0 of Problem 10

Ic = fcmin;

using

[z,p,k] = buttap(n)

M A T L A B code of Problem 10 as necessary.

instead o f

M odify the

zplane(p) grid

12. The specs

pause

znew = z'w c pnew = p‘ \vc knew = k’ wc'^n f=

f^) = (2 dB, 100 Hz),

= (20 dB, 500 H i) , and = (40 dB, 1000 Hz) are associated with the brick%vall specifications o f Figure P i 9.12.

0:fc/50:l. 2 * f s ;

Loss (db)

h= freqs(knew* poiy(znew),poIy(pne\v),2' p i' t ): plot(f,abs(h)) grid pause plot{l',20‘ lo g l 0(abs(h))) srid 2db

9. Repeat parts (e) and (0 of Problem

^cnuix instead o f L A B code of Problem

8 using

2000n

M odify the M A T

8 as necessary.

10. Find a minimum-order Butterworth filter meeting the specs

> 0) lOOOn

= (1 dB, 75 Hz)

Figure P I 9.12 (a)

Find the minimum filter order meet ing these specs.

1076

(b)


Find the range of allowable co^ and the

L

range o f allowable (c)

/

Y

\

Find the poles, zeros, and gain con stant of the N L P transfer function.

(d)

Y

'■ " O

Find the N L P transfer function as a product of second- and first-order sec

Figure P I 9.13

tions. (e)

Using


gain constant of the non-normalized transfer fijnction. (0

14. The circuit of Figure P I 9.14 is to realize a maximally flat second-order low-pass filter with 3 dB frequency

Now find the transfer fijnction as a prod

(a)

= 500 Hz.


uct of first- and second-order sections. (g)

1

Accurately plot (using M A T L A B ) the magnitude response of your filter in

(.v) =

LC

terms o f gain and gain in dB for 0 < f

<

1.2f^. Does your filter meet the given

briclcwall specs?

LC (b)

PASSIVE LOW-PASS REALIZATION

W ith /?^ = 1 Q, compute I in H and C in F to realize the normalized secondorder Butterworth transfer function. is to be i o n in your final design,

(c)

13. The circuit o f Figure P i 9.13 is to realize a

choose a magnitude scale factor,

maximally flat second-order low-pass filter with

and the proper frequency scale factor,

3 dB frequency

Kf, to obtain the given filter specifica

(a)

= 1000 Hz.

Show that the circuit transfer function is

tions. Then compute the new values of

L and C. = s~ + (b)

LC , , -s+R^C LC

(d)

As an alternative design, suppose C is to be

1 |.iF. Find

the new values of

and Kj- to achieve the original filter specifications. W h at are the new val

W ith Rj = \ Q., compute L in H and

ues of L and R jj

C in F to realize the normalized sec ond-order Butterworth transfer func tion. (c)

(D

I f Ri is to be 1 kQ in your final design, choose a magnitude scale factor, and the proper frequency scale factor,

Figure P i 9.14

K p to obtain the given filter specifica tions. Then compute the new values of

15. Consider the circuit in Figure P19.15.

L and C (d)

As an alternative design, suppose C is to be

1 uF.

Find the new values of

(a)

Show that the transfer function o f the circuit is

and Kj- to achieve the original filter specifications. W h a t are the new val

V

ues o f L and /?^?

în

LC 1 R, 1 1 + ^sl^L s~ -f- ---- - H----- 5 + CR, L LC

1077


If R = 2 Q and Rj = 8 Q, find two solu

(b)

tions {L, Q so that H{s) has a second-

17. Consider the circuit of Figure P I 9.17. (a)

Show tiiat the transfer function is

order Butterworth (maximally flat)

I

response with 3 dB down (from its maximum gain) frequency equal to

1

R ,L C

H{s) =

T .V“ +

rad/sec.

s +■ + ! / « ,

LC

Magnitude- and frequency-scale both

(c)

net\vorks resulting from part (b) so

(b)

If Rj= ^

find two sets of parameter

that in the new networks the smaller

values that realize the second-order

resistance is 2 kQ, and the 3 dB fre

normalized Butterworth gain func

quency is 5 kHz.

tion. (c)

Magnitude- and frequency-scale both networks resulting from part (b) so that in the new nervvorks the smaller resistance is 2 k il, and the 3 dB fre quency is 5 kHz.

R,> 1 Figure P I 9.15

/

Y

Y

\

L

16. W ith

1Q

R^= Rj = \ Q .m the circuit o f Figure

P19.15, the transfer function becomes Figure P I 9.17

LC

H .A s) = s~ +

2 ---- f- -- y -I-----\C U LC

.A N S W E R S : (b) (C, L) = (0.4483, 3.3460) or (1.6730, 0.8966) in F and H; (c) ( C L) = (26.6

which can be used to realize the normalized sec

nF, 57.1 m H ) or (7.1 nF, 0.213 H ).

ond-order Butterworth characteristic

4 [d l

H i s ) = —T.V

K -I- J 2

:s

18. Consider the second-order low-pass trans

+

fer function

for K = 0.5. Suppose such a filter is to have cut

H(s) =

off f r e q u e n c y = 3500 Hz. (In this case, the fil

V'^p

I

Find values of L and C to realize the where (0^^= 10^ rad/sec.

normalized transfer function. I f the final value o f the resistors is to be

(a)

1 kt^, compute the new values o f L

Alternatively, suppose the final value o f the capacitor is to be

this case?

that realize the desired filter.

(b)

Find the impulse response and step

(c)

Using the RLC topolog)' shown in

response.

10 nF.

Compute the new values of L and C

Find |Myw)| and co^ (the 3 dB down frequency). Is it true that co^ = O)^ in

and C that realize the desired filter. (c)

( s

I—

maximum gain at f^.)

(b)

K

^ în

ter is to have magnitude 3 dB down from its (a)

becomes

Figure P i 9 .18a, realize H{s) given that /?£ = 10 k n in your final design. W hat is the value of K in vour solution?

1078


Hint: First set O)^ = quency(d)

and

1 and

then fre

magnitude-scale

gain is down 3 dB from its maximum

to

value), find the new element values.

obtain the final answer.

A N S W E K . C, „ „ „ = C , „ „ „ = 0 .106 u F and

Using the R L C topology shown in

= 0.2122 H.

Figure P i 9.18b, realize H{s) given that

R^= 100

in your final design. W hat

20. Consider Figure P i 9.20.

is the value o f A" in your solution?

(a)


1

lY T v

V

L

L^LjC p .

I \

n ----- 1— ^ ^1 ^2/

2

( L\ + Lo + C\ 2 + —------ =------ j + ---------L 1L2C } L^LjC

(a)

yvYY

(b)

L

Determine values so that the circuit realizes a third-order Butterworth gain

6

characteristic. (c) (b)

Find parameter values o f a third-order low-pass Butterworth

filter having

cutoff 20 kHz and resistor values o f Figure P I 9.18

1

kn. (d)

19. Consider the doubly terminated filter cir

Use S P IC E to verify the frequency response.

cuit o f Figure P I 9.19

10

/Y Y \

+ Vo„,

/Y Y \

L,

K p

—0

r

10

c

I > >

+

Figure P i 9.20 Figure P I 9.19 A N S W F R S : (b) A, = L , = ] H, C = 2 1-: (c) (a)

Show that the transfer function of the

15.9 nF and 15.9 niH.

circuit is

L C .C .

« c > ( .0 = -

^

=

c ,l c'-2 (b)

L C .C t

5 4-

By equating the coefficients o f the

2I.

denominator o f with the denominator o f the third-order 3 dB

shown in Fig ure P I 9.21.

Consider the two-port configurations

(a)

N L P Butterworth loss function, show = C2 =

1F

Prove that for Figure P I 9.2la

ôiit _

that the only possible realization is (c)

L C .C l '-2

^21

and Z. = 2 H.

If the source and load resistance are to be 1 k n an d^ = 1500 Hz (meaning the

(b)

Prove that for Figure P I 9 .2 lb

1079


N LP

-V21 V22 +

Vin (c)

transfer function obtained in

part (a).

Making use of the results o f part (a),

(c)

Magnitude- and frequency-scale the

design a second-order Butterworth

circuit so that the gain is 3 dB down at

low-pass filter having to^ = co^ = 1

^'Kntin

rad/sec,

q

=1

and Rj = so. Hint:

Divide the numerator and denomina

(d)

source resistance is 100

'

Plot the magnitude response in terms of dB gain vs./ (in Hz).

tor o f His) by s, and then equate the result with the gain expression to obtain two z-parameters. Next, review ing of one series element and one

ACTIVE LOW-PASS DESIGN 23. Consider the circuit of Figure P i 9.23,

shunt element (a special case o f a T- or

which has a practical source driving a leaky

7i-network).

integrator circuit.

the z-parameters of a two-port consist

Finally, put all results

together to design the two-port. (d)

(a)

u s i n g R^^^ = R^ + R ^.

Making use o f the results o f part (b), associated with Figure 19.21b, design

Compute the circuit transfer function

(b)

Realize

the

first-order

normalized

Butterworth transfer function

a second-order Butterworth low-pass filter having co^= 1 rad/sec, R = 0 Q

^ 3dBNLP^^'^) - ~rr~r7 - — r

and R^= \ Q. (e)

V in is )

Frequency- and magnitude-scale the

i.e., compute values for R , R-,, and C.

circuits o f parts (c) and (d) so that the

Hint: Let C= I E

new filters have to^ = 5000 rad/sec, and the single resistance in the circuit

(c)

I f the 3 dB down frequency is to b e ^ = 3500 Hz, find

is 1 k n .

S+\

so that the capac

itor is 1 nF in your design. Compute the new resistor values and then deter mine /?j if the source resistance is R^ = 500 Q. C H E C K : /?, = 45 kQ.

22. (a)

Find the 3 dB

N LP

Butterworth

transfer function meeting the specs = (2;o00, 2 dB) and (o)^,

Figure P i 9.23

= (2.T2000, 20 dB). Then find

24. Consider the circuit o f Figure P i 9.24 as a (b)

Using one o f the circuits with a non zero source resistance from the earlier homework problems, realize the 3 dB

candidate for realizing the second-order lowpass transfer function

1080

Chapter 19 • Principles of Basic Filtering o

m s ) =_

K

*out _

( s 0).

(a)

(b)

(c)

For a normalized design, O)^ = 1 rad/sec and A" = 10, choose element values to realize the given transfer function; i.e., find all circuit parameter values.

+1

(c)

Compute the transfer function of the circuit of Figure P I9.24.

(b)

For a normahzed design, co

and

compute the final circuit parameter

rad/sec and K = - 1 0 , choose element

values. Simulate using SPICE to verify

values to realize the given transfer

your design. (The solution is not

fijnction; i.e., find all circuit parame

unique, vy^hich is often the case for

ter values.

practical design problems.)

If (0^= 10^ rad/sec and the capacitors are to be 1 nF,’ choose AV-and K m and j compute the final circuit parameter values. Simulate using SPICE to ver

(d)

terms of (O^. C H EC K : 0)^ = 0.5098co^ r>

R.

R.

= 1 kn

the

down

o

Find the 3 dB down frequency, o)^ in

ify your design. Find

o

If 0)^= 10^ rad/sec and the capacitors are to be 1 nF, choose iSând

= 1

C H EC K :

3

n

n

dB

frequency, —0 +

0)^ in terms of (O . C H EC K : (0^ = 0 .6 4 3 6 co^

n

■ r^ n Figure P I9.25 26. A second-order normalized Butterworth fil ter can be realized by the Sallen and Key circuit described in section 5. In the final design, the filter is to have a dc gain o f 1, a 3 dB down fre quency

= 1000 Hz, and largest capacitance

of 10 nF. (a)

Figure P I9.24

Using design A and input attenuation

n n

if necessary, determine the appropriate 25. Consider the circuit of Figure P 19.25 as a candidate for realizing the third-order low-pass

(b)

transfer function

V

Sallen and Key circuit. Using design B and input attenuation

n

if necessary, determine the appropriate

n

Sallen and Key circuit.

K (c)

Using the Saraga design and input

n

attenuation if necessary, determine the appropriate Sallen and Key circuit. Hint: It is often useful to generate an Excel (a)

spreadsheet to do the relevant calculations,

Compute the transfer function = circuit of Figure P I 9.25.

of

especially for multiple designs such as these;

n o

this allows one to set forth the normalized and final designs along with

and n

1081

Chapter 19 • Principles oi Basic Filtering

27. Realize rhe filter o f Problem

8 as a cascade

(b)

M odify the circuit to achieve the

o f a Sallen and Key circuit (Saraga design) and

desired gain of the transfer function.

a first-order active circuit such as the leaky inte grator. The 3 dB down point should be at

(c)

determine the new clement values.

(H int: It is often useful to generate an Excel spreadsheet to do the relevant calculations,

If the actual pass-band edge frequency is to be 7 kHz and C j is to be 0.05 uF,

largest capacitor should be 50 nF.

^'^cmhr

W lia t is the dc gain, K, o f the circuit?

(d)

Plot rhe loss magnitude response of

/V(s) in part (a) in dB. Determine

especially for multiple designs.) Verify the fre quency response o f your design using S P IC E .

C H E C K :^ „„„,= 2 d B 28. Repeat Problem 27, except use design A instead ol rhe Saraga design for the Sallen and

32. Repeat Problem 31 using design A of the

Key portion.

Sallen and Key circuit.

29. Realize the filter o f Problem 10 as a cascade

33. Repeat Problem 31 using design B of the

of two Sallen and Key circuits. The 3 dB down

Sallen and Kev circuit.

point should be at

The largest capacitor

should be 0.1 uF.

HIGH-PASS PASSIVE DESIGN 30. Repeat Problem 29, except use design B

34. A second-order Buttenvorth H P filter has 3

instead of the Saraga design for the Sallen and

dB down point at^. = 2000 Hz. 'Fhe second-

Key portion.

order Butterworth N L P prototype is given in Figure P I 9.34 and has transfer function

31. In addition to Butterworth filter transfer functions, there are Chebyshev filters, which

LC 1 1 s~ -I- -------.V-f" R jC LC

have a faster transition from pass-band to stop

~>

band. This problem investigates the implemen tation o f a Chebyshev LP filter transfer function. Recall that a second-order transfer function

(a)

W ith R^= ] Q., compute L in H and C in F to realize the normalized second-order prototype. C H ECK:

(b)

C = l/ V 2

F

Using the results o f part (a), construct

can be realized by the Saraga design o f a Sallen

a N H P circuit with

and Key circuit.

^^cHP~ ^ rad/sec. This is the so-called

(a)

Determine the new values o f

/?,,

3 dB

a

normalized

second-order

(c) (d)

------------

0.803825+ 0.82306

Hint: Frequenc)--scale, s

(DqJ where

oj“q= 0.82306. W h at is the resulting transfer function?

Now construct the final H P design with R j =

Chebyshev L P transfer fiinction

^

filter (3 dB

N H P ).

Cj, and C j that will realize the poles of

normalized H P

\ Q. and

100

Do a S P IC E simulation to verif}- your design.

1082

Chapter 19 •Principles of Basic Filtering

(c)

Using the results of part (b), construa a N H P circuit with the source and load resistances equal to 1 Q and

^c,HP ~ ^ •■^
Figure P I9.34 (d)

35. A third-order Butterworth HP filter has 3 dB down point at

Given your answer to pan (c), con struct an appropriate high-pass filter

= 5000 Hz. The third-

circuit whose resistor values are 1 ki2.

order Butterworth NLP prototype is given in Figure P I 9.35 and has transfer function

r -W --/Y Y \ 10 •-

1

Hr

3 S

\ 2

Cj

+ —

S

LC jC j + Co

+ —

------ —

-o +

1o

vJt)( 1

S

LC\C2

+ ----------

LC\C2

Figure P I9.36 Compute Z in H and Cj and C2 in F

(a)

to realize the 3 dB normalized third-

37. The circuit of Figure P I 9 .3 7 realizes a

a CH ECK : q = 1.5 F and L = - H

a third-order high-pass filter circuit, with

Using the results of part

= 2 dB az L = 5 kHz so that the largest capaci

third-order NLP Butterworth filter. Determine

order prototype. (b)

3

tor equals 100 nF, as follows:

(a), construCT a NHP circuit with = 1 rad/sec. This is the so-called 3 dB

(a)

normalized HP filter (3 dB NHP).

(d)

max

and

for the

(b)

Compute O)^ for the desired HP filter.

the source resistance is to be 100 Q.

Does your answer make sense? Think

Do a SPICE simulation to verify your (c)

about this carefully. Realize the 3 dB NLP transfer func

Now construct the final HP design if

(c)

Determine NLP filter.

design.

tion

using the circuit of Figure

P 19.37. Set I = 2 H and make Cj = C2. Recall from Problem 19 that the transfer function of the circuit is 1

VquA^) Vin(s) 36.

A

second-order

high-pass

LC1C2

c, +c c,c 1^2

LC 1C 2

LC 1C2

Butterworth loss function has stop band edge fre quency fs= i kHz, pass-band edge frequency

=

7 kHz, and cutofi^fijequencyf(. = 5500 Hz. (a)

Determine the attenuation in dB 2x fp

(b)

and^. Realize the second-order 3 dB NLP Butterworth filter using the circuit of Figure P I9.36; i.e., compute values for L and C.

(d)

Construct the appropriate high-pass circuit with the largest capacitor equal to 100 nF.

(e)

Verify the frequency response o f your design using SPICE.

1083


41. Reuse the specs of Problem 40, except that there is to be negligible source resistance and a X

IQ

L

c.

load resistance of 75 (a)

C

Compute the transfer fianction, say o f the circuit in Figure P i 9.41 in terms o f Z j, L 2, and C

Figure P I 9.37

Determine values so that the circuit

(b)

realizes a third-order Butterworth 3

38. The brickwall specs of a H P filter are shown in Figure P i 9.38, in which 2500 Hz,

= 2 dB, and

dB N L P Butterworth gain function.

= 500 Hz, ^ =

Determine the H P filter.

(c)

= 25 dB. The

filter is to be driven by a voltage source having a source resistance o f 100

jy y

There is no load

6\

resistance. At the pass-band edge frequency the attenuation should be exactly 2 dB. Develop a passive H P filter circuit to solve this problem. A A(w)

\

L,

1o

Figure P I 9.41

HIGH-PASS ACTIVE REALIZATION 42. Consider the set of H P specs for which

J

- (2400 Hz, 3dB) and

=

(600 Hz, 30dB). Following the procedure o f

-► 0)

Example 19.7, design an active H P filter meet ing these specs so that the smallest capacitor in the final design is 10 nF.

Figure P I 9.38

43.

Consider the set of H P specs for which

39. Repeat Problem 38 assuming that the filter

= (2400 Hz. 2dB) and = Hz, 30dB). Following the procedure of

is driven by a voltage source with negligible

(600

source resistance and that there is a load resist

Example 19.7, design an active H P filter meet

ance that in the final design is to be 75 Q.

ing these specs so that the smallest capacitor in

Consider using the N L P prototype o f Problem

the final design is 10 nF. The magnitude

34.

response should go through the pass-band edge. Hint: See equation 19.34.

40. The brickwall specs o f the H P filter in Figure P i 9.38 now apply to/^' = 500 H z ,/ , =

44. The Sallen and Key Saraga design H P filter

2000 Hz,

o f Figure P i 9.44 realizes the normalized H P

= 2 dB, and

= 30 dB. The

filter is to be driven by a voltage source having

filter transfer function

a source resistance o f 250 Q. There is no load resistance. At the pass-band edge frequency the

H {s) =

K s-

attenuation should be exactly 2 dB. Develop a passive H P filter circuit to meet these specifica tions. Hint: Use the circuit o f Figure 19.8, hav

where A" = 1 +

ing transfer function of equation 19.14.

Butterworth filter is to have f ^ = 3

A second-order H P kHz.

1o s^


Compute the parameter values o f the H P filter of Figure P I 9.44. The smallest resistor should

/Y Y l

be 10 kD. Verify the frequenc)^ response o f your

L 8Q woofer

design using S P IC E .

80 tweeter

Figure P i 9.47 AN.SWl-.R: L = 63(> u H , C = 9.95 uF' 48. The crossover nersvork of the previous problem provides first-order Butterworth low-

Figure P 19.44

pass and high-pass response curves. Better45. Again consider the circuit o f Figure P i 9.44. (a)

Cascading two sections together, real ize a fourth-order normalized H P

upgrading the responses to the second-order Butterworth t}'pe. Design such a crossover net work having the same crossover frequency and

Butterworth transfer function. (b)

quality sound reproduction is achieved by

If the smallest resistor is to be 10 kH

loads as in Problem 47.

a n d ^ = 3 kHz, find the new parame A X S W l'R : For the woofer. L = 0.90032 mH,

ter values for your design.

C = "’.033*' uF. For the tweeter, /. = 0.‘)0032 46. You have a set of rweeters that feature great sound reproduction for frequencies above

5

kHz. As a lark, you decide to build an active high-pass Butterworth filter that will isolate highs from a particular audio signal. The specs you decide on are (^ = 5 kHz, (^ = 1.5 kHz,

= 3 d B) and

= 40 dB). Determine the fil

ter as a product of second-order active Sallen and Key circuits. M inim um

resistor values

should be 20 k H in a Saraga design.

MISCELLANEOUS 47. A certain audio amplifier has a very low internal resistance. It is therefore approximate ly represented by an ideal voltage sourcc. The

8

woofer and t\veeter each may be approxi

mately represented by a resistance of

8 ohms.

Design the simple crossover network shown in Figure P i 9.47 such that the crossover frequen cy is 2000 Hz.

m H. (■ = ".0337 itl-.

C

H

A

P

T

E

R

Brief Introduction to Fourier Series CHAPTER OBJECTIVES 1.

Introduce the concept o f and calculation procedure for the Fourier series of a periodic signal.

2.

Describe the relationship between the complex and the real Fourier series representations.

3.

Set forth and discuss basic properties of the Fourier series.

4.

Show how the basic Fourier series properties can be used to compute the Fourier series of a wide range of signals from a few basic ones.

SECTION HEADINGS 1. 2.

Introduction The Fourier Series: Trigonometric and Exponential Forms

3. 4.

Additional Properties and Com putational Shortcuts for the Fourier Series Representation Summary

5.

Terms and Concepts

6.

Problems

1. INTRODUCTION Non-sinusoidal periodic waveforms are an important class of signals in electrical systems. Some prominent examples are the square waveform used to clock a digital computer and the sawtooth waveform used to control the horizontal motion of the electron beam of a cathode ray T V pic ture tube. Non-sinusoidal periodic functions also have importance for non-electrical systems. In fact, the study of heat flow in a metal rod led the French mathematician J. B. J. Fourier to invent the trigonometric series representation o f a periodic function. Today the series bears his name. The Fourier series o f a periodic signal exciting a linear circuit or system leads to a simplified understanding o f the effect o f the system on the original periodic signal. This idea is briefly explained in the next few paragraphs and illustrated in Example 20.1.

1086

Chapter 20 • B rief Introduction to Fourier Scries

W hen a periodic input excites a linear circuit, there are many ways to determine the steady-state output. Using a Fourier series method of analysis, the input is first resolved into the sum o f a dc component and infinitely many ac components at harmonically related frequencies. For example, a

1 kHz

square wave voltage with zero mean and a 0.5n V peak-to-peak value, w ithy{f) = {u{t)

-2u{t -0 .57) + u {t- 7)), T = 1 msec, Vy(/) = 0 .2 5 jt ^

/(/ - n T )

( 20. 1a)

H=0 has the Fourier series representation

2/2 + 1

/i=0

( 20. 1b)

s in (w „r) - ^sin(3cu^/) + ^ sin(5w ^/) - ... u ( t )

w h e re ^Q =

2jtx

1000 rad/sec. In the Fourier series representation (which we will later develop), we

observe that the signal has a zero average dc value and harmonically related frequency components

(-l)"sin ((2 « + 1)0;^/) 2/ ; +l By linearity and superposition, the effect of the linear system on v^{t) can be determined by sum ming the time domain effects o f the linear system on each component in the Fourier series. For steady-state calculations, let Fourier series representation of is given by

be a phasor representing the (2;/ + l)tli harmonic o f the Then, in steady state, the effect o f the system on this term

= H {j{2n + 1)Wq)V^ , assuming that the transfer function H{s) is stable. Again by

linearity, the actual time domain output is then computed for each

The resulting time func

tions are then summed to obtain the steady-state part o f the complete response. Example 20.1 illustrates the particulars.

E X A M P L E 20.1. Figure 20.1b shows a simple R C circuit (/? =

1

C = 1 F) and a square wave

input voltage (Figure 20.1a) with E = 3 0 jt V and T = 4 sec. Find the first four components o f the output voltage v^{t) in the steady state. > r b ■

R • • •

(a)

v Jt)

+

v jt)

(b)

F IG U R E 20.1 Series RC circuit excited by square wave, used to demonstrate calculation of the steady-state response.

1087

Chapter 20 • B rief Introduction to Fourier Scries

S o l u t io n

Step 1. Determine the Fourier series representation o f v-^ît). The fundamental frequency o f v-^^{t) is Jq = 1/7’ = 0.25 Hz or C0q = I jifo = 0.5jt rad/sec. As shown in die next section, u-^^{t) has the infinite (Fourier) series representation

= 30;r ^ [u{t - n T ) - n{t - {/? + 0 .5 )7 ) /)=() 15;r + 6 0sin(a;o/) + 20sin(3a)oO + 1~ sin(5wQ/) +... u { t )

” s in ((2 « + ^ «=0n For this signal the average dc value is

15 ti. The

nil)

2// + 1

remaining terms in the Fourier series are harmon

ically related sinusoids of decreasing magnitude. Step

2. F in d the circuit transfer fim ction.

The stable transfer function o f the circuit is

^ s) =

Vi,

. v+1

Step 3. D eterm ine the magnitude an d phase o f H{s) at s = j( 0 . For sinusoidal steady-state analysis, set s =jo ) to obtain

Jd

H(jco) = -

JOJ + 1

=H„/

in which case

, 6 = - tan“ ^(w ) yjco^ + 1 Step 4. Find steady-state responses to a ll components o f Vj^^{t). Table

20.1

lists the steady-state

responses to several components of Vj^Jit). TA BLE

20. 1. Several Fourier Series Components of Vj„{.t) and the

Corresponding Magnitude and Phase of the Response

0)

0

Wo

Input magnitude

157:

60

5«o

20

1

12

0

0

0

1

0.5370

0.2075

0.1263

0 (degrees)

0

-57.52

-78.02

-82.74

Output magnitude

1571

32.22

4.150

1.516

Output angle

0

-57.52

-78.02

-82.74

Input angle’'

'Angles are in reference to sine functions in this example.

0

I

C'hapicr 20 • Brief Introduction to I\>iiricr Series

Step 5. Apply superposition to obtain the steady-state portion o f the complete response. Neglecting har monics o f scventii order and higher, the approximate steady-state solution is

Vg{t) = 15j*r + 3 2 .22sin((/jyr- 57.52") + 4.15 sin(3w Q /-7 8 .0 2 ‘’) + 1.516 sin(5wQ - 8 2 ,7 4 ”) + ... This response shows the effect of the system on each component of the input signal and how it in turn affects the overall steady-state output response. Since the time constant o f the circuit is 1 sec, this steady-state response more or less constitutes the actual response for r > 5 sec.

Exercise. Use MATLAB to plot the approximate waveform for vj^t) for 0 < r < 8 sec, based on Tiible 20.1. From the plot identif)'

Alternatively, you can use the max

and min commands in MATLAB. AN SW l-R: 82.013 V and 12.235 \'

In Example 20.1, the first step o f the solution was to represent a periodic waveform as a sum o f sinusoidal components, called the Fourier series. Section 2 covers the definition and basic proper ties o f Fourier series. Section 3 describes several shortcuts for computing the Fourier coefficients and identifies other important properties. Since, in practice, only a finite number o f terms can be considered, the Fourier series method yields only an approximate solution. Because many mathematical and engineering handbooks have extensive tables o f Fourier series o f different waveforms, it is convenient to use these tables in much the same way as one uses a table o f integrals or a table o f Liplace transforms. The Fourier series o f some basic signals are provid ed later, in Table 20.3. The use o f this table, together with some properties and shortcuts dis cussed in section 3, make the study o f the Fourier series method much more palatable to begin ning students o f circuit analysis.

2. THE FOURIER SERIES: TRIGONOM ETRIC AND EXPONENTIAL FORMS B a s ic s

A signaly(f) is periodic if, for some T> 0 and all t, / r + 7) = P )

(20.2)

r i s the period o f the signal. The fundamental period is the smallest positive real number Tq for which equation 20.2 holds; ^ = l/7j, is called the fundamental frequency (in hertz) o f the sig nal; cOq = I k/ q = 271/7*0 is the fundamental angular frequenc)^ (in rad/sec). T he sinusoidal wave form o f an ac power source and the square wave form used to clock a digital computer are com mon periodic signals. Figure 20.2 shows a portion o f a hypothetical periodic signal.

1089

Chapter 20 • Brief Introduction to Fourier Series

THE REAL TRIGONOM ETRIC FOURIER SERIES Under conditions tiiat are ordinarily satisfied by signals encountered in engineering practice, a periodic function y(/^) has a decomposition as a sum o f sinusoidal functions 00 / (0 = Y +

2 ! (« «co s(m v )f/ 7„sin (/ ;fo „/ )) //= 1

(20.3)

Equivalently, / (O =

coa{mo„r + d„)

(20.4)

where (20.5a) and (20.5b) Both infinite series, equations 20.3 and 20.4, are called the trigonometric Fourier series repre sentations

We note that equations 20.5a and 20.5b follow fi-om the trigonometric identit)'

A cos(.v) + fisin(.v) =

\

with due regard to quadrant. + B~ cos .Y + tan ^ ----I Aj f \

From equation 20.4, observe that is the average value ofy(r). In electric circuit analysis, refers to the dc component ofJ{r) . The first term under the summation sign, <^jCos(tOQt + ^^,), is called the fundamentaJ com ponent (or first harmonic) ofy(r), with amplitude

and phase angle

0 j. The term d-, cos(2o;(jt + 6-,) is called the second harmonic o^J{t), with amplitude d-y and phase angle

and similarly for the other terms d^^cos(;/WQt + 0^^), which are the ;/th harmonics as indi

cated by the term nô^y

Chapter 20 • Brief Introduction to Fourier Scries

As illustrated in Example 20.1, given any periodic function/r), it is important to determine the coefficients in equation 20.3 or equation 20.4. For the purpose o f easier calculation, it is advan tageous to introduce the equivalent complex Fourier series representation o f the periodic func tion y(f).

COM PLEX EXPONENTIAL FOURIER SERIES For a periodic s i g n a l t h e so-called complex exponential Fourier series is

/ (0 =

i

(20.6a)

n = -ao where it can be shown that

In + T,

c„ =

(20.6b)

Since we have two (allegedly) equivalent forms, let us now develop the relationship between the real and complex forms o f the Fourier series. Recall the Euler identities: cos( a-)

= --------------- and sm(A-) = --------------2 2j

Then equating the real and complex forms, we have

/ (')=

2

E

‘V

/I = - oc

“

/I = I

= -y + ^ n = \\ = ^

+ I

^----------- + h

(0.5(.;„ - jh„

2j

+ 0.5(«„ + jb„

]

/! = 1 Equating coefficients yields =

(20.7a)

and

c_„ = 0.5

+ jh „ ) = c*

(20.7b)

where c*, is the complex conjugate o f f,,. Equivalently, = 2 R e (0 b„ = -21m(^-„)

(20.8a)

(20.8b)

1091


(20.8c)

^0 = ^0 = ^

d„ = 2\c„ln= 1,2, 3,...

(20.8d)

= ^ c „ ,n = 1, 2, 3 ,...

(20.8e)

In the real trigonometric Fourier series equations 20.3 and 20.4, the summation is over positive intêr values o f«, whereas in the complex exponential Fourier series the summation extends over intêrs n such that -oo < « < oo. While each term in equation 20.3 or equation 20.4 has a wave form displayable on an oscilloscope, each individual term in the complex exponential Fourier series lacks such a clear physical picture. However, two conjugate terms in the complex exponen tial Fourier series always combine to yield a real-time signal d^cos{no)Qt + 6^. To develop equation 20.6b for the coefficient r > " 0 'to obtain

we multiply both sides of equation 20.6a by

* = -00

Integrating over one entire period, [/q,/q + T^y produces

Jto

k = -oo

(20.9) * = -00

Because (OQrQ = 2ti and

it = -00

= 1, the integral in equation 20.9 becomes cê

'v ^

'

'

»■ 0 c„Tq

for k ^ n

for k = n

(20.10)

Substituting equation 20.10 into equation 20.9 and dividing by Tq yields c„ =

tn + T„ Tn

(20.11)

The lower limit of integration, ?q, can be any real number, but is usually chosen to be 0 or - T qH, whichever is more convenient. In addition, Tq will sometimes be written simply as T. A hand computation of coefficients would proceed by first computing for « = 0, 1,..., using equa tion 20.11. One would then obtain and using equations 20.8a and b, and and 0^ by equa tions 20.8c, d, and e. Other formulas are available for obtaining the Fourier coefficients by integrals involving sine and cosuie functions. However, equation 20.11 is preferred because an integration involving exponential fimctions is often simpler than an int^ration involving sinusoidal flmctions.

y w

w


EXA M PLE 2 0 .2 . Find the trigonometric Fourier series for the square wave signal o f Figure 20.3.

S olution The fundamental period o f/ f) is Tq = T. By inspection, the average (dc) value

is

To calculate other Fourier coefficients, choose Tq = -7 7 2 . Equation 20.11 yields, for n ^ 0,

L

L

L

(20.12)

A .(

JT^

— sin n — , / ; = l , 2 , . . .

2}

Jtn

Thus our signal has the complex Fourier series of equation 20.6 with

given by equation 20.12 and

To obtain the real Fourier series, from equation 20.8, a „ = 2 R e ( c , J = — sm n — ,h , = 0

jTn

Substituting these coefficients into equation 20.3 yields the following trigonometric Fourier series for the square wave o f Figure 20.3:

A 2A / I I / ( 0 = T ‘*'— c o s ( o ; o / ) - - c o s ( 3 a ;o O + 7 C o s ( 5 a ;o / ) - .. . 2 7T \ 3 5

(20.13)

T W O PRO PERTIES OF THE FO URIER SERIES After computing the Fourier series of a periodic signal y(^), it is straightforward to find the Fourier series o f a related periodic signal g(t) whose plot is a translation o f the plot of/f). A translation o f a plot is


1093

a horizontal and/or vertical movement o f the plot without any rotation. A translation o f the waveform in the vertical direction causes a change in the dc level and affects only the coefficients d^y and Cq. A translation o f the waveform in the horizontal direction causes a time shift that changes only the angles 0,^ and has no effect on the amplitude

We state the relanonship formally as follows.

Translation property o f Fourier series: Let

be the coefficients o f the exponential

Fourier series o f a periodic functiony(f), and c„ be those for another periodic function g{t). If^ f) is a translation o(J{t) consisting o f a dc-level increase right by

and a time shift (delay) to the

then

g it)= p -r J)^ K

(20.14a)

^ 0= do = CQ + K = dQ + K = — + K /; = ± 1 , ± 2 , . . .

=1 = Z.C,,

, /? = ± l ± 2 , . . .

=[d„ -ncjQt^), n = 1 ,2 ,3 ,...

(2 0 .l4 b ) (20.14c) (2 0 .l4 d ) (2 0 .l4 e )

The proof o f this property is straightforward and is left as an exercise. Note that equation 2 0 .1 4c indicates that a time shift o f the signal affects only the phase angles o f the harmonics; the ampli tudes o f all harmonics remain unchanged.

Exercise. Supposey(f) in Example 20.2 has A = 30 and suppose ^ f) =J{t) - 10 . Find the coeffi c i e n t s a n d o f the Fourier series o f ^ r). AN SW I'R: = 0.5^/,, = 3; all other coefficients arc unLh.mgcd, i.e.,

= — sin n —

because h.. = 0.

We will now use the translation property to obtain the Fourier series o f a square wave that is a translation o f Figure 20.3.

EXA M PLE 2 0 .3 The square wave g{t) shown in Figure 20.4 is anti-symmetrical with respect to the origin. Find the complex and real Fourier series for^^) given the Fourier series (equation 20.12) ofJ{t) depict ed in Figure 20.3.

u rn


g(t) >

'

A/2 w

-0.5T

-I

0.5T

0

T

-A/2 FIG URE 20.4 A square wave anti-symmetrical with respcct to the origin. S olution The curve

is a translation o f the waveform y(f) o f Figure 20.3. Specifically,

S ( 0 = / ((-/ ,/ ) + a: = / By equations 20.14 and 20.13, COS ncoQt

\

4

( = cos ncoQf - n — \ 2}

resulting in the desired Fourier series,

jr 1 2A 7Z

-----CO.S 3W(,/ 3 2}

?>7Z

+ - COS 5

5 jt

(20.15)

sin(wQ/) + —sin(3wQ/)-i- jsln (5 w ()/ ) + ...

For the case o f a square wave signal, by choosing the time origin and dc level properly, the result ant plot displays a special kind of symmetry that results in the disappearance o f all sine terms or all cosine terms. The square waves o f Figures 20.3 and 20.4 are special cases o f the periodic func tions amenable to such simplifications. The general case is given by the following statement. Symmetry properties o f the Fourier series (1) If a periodic functiony(/) is an even function, i.e., has only cosine terms and possibly a constant term.

then its Fourier series

(2) If a periodic functiony{r) is an odd function,

then its Fourier series

has only sine terms. The plot o f an even function is symmetrical about the vertical axis. Examples o f even functions include

cos(cl)/)

and the square wave o f Figure 20.3. The plot o f an odd function is anti-sym-

metrical about the vertical axis. Examples o f odd functions are sin(o)/) and the square wave o f Figure 20.4. The proofs o f the symmetry properties are left as homework problems.

1095

Chapter 20 * Brief Introduction to Fourier Scries

Exercise. Suppose

in Figure 20.4 From Example 20.3 is shifted into q{t) = g { t - 0 .2 5 7 ). Find

the coefficients

and

AN SW ER:

o f the Fourier series o f q{t).

lA (h) = (i{) = 0. and hir n ^ 0. d„ = a„ = -------sin .111

-T\

n — .h„ = 0 . ~)

To simplify the calculation o f the Fourier coefficients, we should attempt to relocate the time ori gin or change the dc level so that the new function ^t) displays even or odd function symmetry. This may not be possible for an arbitrary periodic signal. When it is possible, we will calculate the Fourier coefficients o f the new function ^t), which has only cosine terms or sine terms, and then use the translation propert)' to obtain the Fourier coefficients for the original function//) . A waveform o f particular importance in signal analysis is the periodic rectangular signal shown in Figure 20.5. The fundamental period is T, and the pulse width is fiT. The constant (5 is called the duty cycle, usually expressed as a percentage o f T. The square wave o f Figure 20.3 is a spe cial ca.se o f the rectangular wave o f Figure 20.5 with a 50% duty cycle.

EXA M PLE 2 0 .4 Find the trigonometric Fourier series for the rectangular waveform o f Figure 20.5.

S olution T he procedure is almost identical to that used in Example 20,2 for a square wave. By inspection, the average value

is (20.16a)

To calculate the other Fourier coefficients, choose fg = -7 7 2 . Equation 2 0 .10 then yields

pT

fW

~E. = — sin(/7/3.T), /? = 1 ,2 ,...

Tin

( 2 0 . 16b)

10 9 6


From equation 20.7, rhe coefficients o f tiie trigonometric series are ^ /\ = 2 R e (c -„ ) = — sin(/?^.T)

(20.16c)

jzn

b.. = 0

(2 0 .l6 d )

Specifying the coefficients completes tine determination ot the Fourier series, i.e.,

2A ^ (sin(nl^jr)\

) = p/\ 1------- y

(1,1

n= \

COS(nw„t)

/I = l^

A very important conclusion about the rectangular wave can be drawn by examining equations 20.16: as the ratio o f the pulse width to the period becomes very small, the magnitudes o f the fundamental and all harmonic components converge to twice the average (dc) values. To see this, recall that sin(.v)/x approaches 1 as .v approaches 0. From equation 20.16c, we may rewrite

a„ 'n = — sin(///i.;r)= 2ftA jTn It follows that

“♦ 2fM as ft

as

nftji

0.

To give some concrete feel to this property, lable 20.2 gives the ratios o f

^'(ly

= —— and — âv

for w = 1, ..., 9 for the case o f ft = 0 .0 1 . Answers are rounded o ff to three digits after the decimal

=

point. Note that when the periodic rectangular signal is shifted vertically, the ratio is affected, but the ratio —

remains unchanged. TABLE 20.2

A m p lit u d e s o f t h e F i r s t N in e H a r m o n i c s f o r t h e C a s e P = 0 .0 1

n

F ‘ av

F ‘ av

The constant

2

3

4

5

6

7

8

9

2.000

1.999

1.998

1.995

1.992

1.989

1.984

1.979

1.974

1.000

1.000

0.999

0.998

0 .996

0.994

0.992

0 .990

0 .987

property holds approximately true when the pulse width is a very small fraction

o f the period 7q. Even i f a waveform is not rectangular, if the pulse width is very small compared

109:

Chapter 20 * Brief Introduction to Fourier Series

to its period, then the nearly constant propcrrv' o f —~ iind ^ continues to hold, as long as the «I pulse is o f a single polarit)'. For example, consider the periodic short pulse shown in Figure In calculating the Fourier coefficients gral, originally (/q,

+ 7 ),

f;;

o f this pulse using equation

are changed to

(-(^ 7 7 2 ,

^ 7 7 2 ).

As

2 0 .1 1 ,

2 0 .6 .

the limits o f the inte

approaches zero, the factor

in the integrand has a value very close to 1 in the new time interval, as long as n, the harmonic order being considered, is not very high. Therefore, for pulses o f narrow width we have (xT

aT

(20.17a) rhus in terms of the

coefficients, from equation 20.8, again for pulses o f narrow width we

have (20.17b) This result is pertinent to the approximate analysis o f a rectifier circuit covered in other texts or in the second edition o f this text.

To this point, we have calculated the Fourier coefficients only for some very simple periodic sig nals. The evaluation o f the integral in equation 20.11 becomes much more involved when the signal j{t) is not rectangular. Fortunately, many mathematical and engineering handbooks now include comprehensive tables o f Fourier scries. From a utility point o f view, one may use these tables much the same as one uses a table o f integrals or a table o f Laplace transforms. In effect, the need to carrj' out the integration in equation 20.11 is not compelling in practice. In many applications, it is important to know the average power o f a (periodic) signal and the magnitude o f its various harmonics. From equation 20.5, the effective value o f the dc compo nent is |<^q|, and those for the fundamental and various harmonics are d^j! yfl , /; = 1 , 2 , . . . . It is easy to show that the effective value, or the rms value, o f/ r) is

(20.18a) where the d- coefficients are from the Fourier series o f equation 20.4. If fit) (current or voltage) drives a 1 ohm resistor, then the average power absorbed by the resistor is

~

êff • Hence we

say that the average power o f a periodic signal y(/) represented by a Fourier series is given by

1098


r~\ 1

1

P av eJ{t)= F }ff= dl+ -di + - d l - ^ ..= 2

*

2

k„|

(20.18b)

O

n=-oo

The relationships indicated in equations 20.18 are often termed Parsevals theorem. The information on the phase angle is important when one wishes to construct the time domain response in steady state. For the time domain problem, the Fourier series method yields only an approximate solution, because one can only sum a finite number of terms in the series.

C on v erg en ce o f th e F o u r ie r S eries

Convergence of the Fourier series is an intricate mathematical problem, the details of which are beyond the scope of this text. On the other hand, it is important to be aware of the ways in which the Fourier series may or may not converge to a giveny(/). Our discussion is not comprehensive, but is adequate for our present purposes. To begin, we define a partial sum of terms of the complex Fourier series of a function ^(/) as sn « )=

2

k= -N

From our experience thus far, 5yy(f) must in some way approximatey(r). The difference between j{t) and its approximation, 5yy(r), is defined as the error ât( 0 = / ( 0 - 5 ^ ( 0

One way to get a handle on how well Sj^t) approximates is to use the so-called integral squared-error magnitude over one period, [/gi T\, defined as

r\

e ^ (l) dt

This is often called the energy in the error signal, as energy is proportional to the integral of the squared magnitude of a function. It turns out that for ftmaions having a Fourier series, the choice of the Fourier coefficients minimizes E[^ for each N. Further, for such fiinctions, 0 as 00, i.e., the energy in the error goes to zero as JV becomes large. This does not mean that at each r, fij) and its Fourier series are equal; it merely means that the energy in the error goes to zero. Continuous and piecewise continuous periodic functions have Fourier series representations. A piecewise continuous fiinction, such as a square wave, is a function that (1) has a finite number of discontinuities over each period but is otherwise continuous, and (2) has well-defined rightand left-hand limits as the function approaches a point of discontinuity. For piecewise continu ous functions, it turns out that the Fourier series converges to a value halfway between the values of the left- and right-hand limits of the function around the point of discontinuity. Even so, -* 0 as TV-» 00 for piecewise continuous functions. There are many functions that are not piecewise continuous and yet have a Fourier series. A set of conditions that is sufficient, but not necessary, for a function to have a Fourier series repre sentation is called the Dirichlet conditions.

o

r\


1099

D irichlet conditions Condition 1. Over any period, [fg, tQ+T\,J{t) must have the property that

In the language o f matliematics, this means that J{t) is absolutely integrable. The conse quence o f this property is that each o f the Fourier coefficients

is finite, i.e., the

exist.

Condition 2. Over any period o f the signal, there must be only a finite number o f min ima and maxima. In other vvôrds, functions like sin(l/^) are excluded. In the language o f mathematics, a function that has only a finite number o f maxima and minima over any finite interval is said to be o f hounded variation.

Condition 3. Over any period, y(r) can have only a finite number o f discontinuities. As mentioned, at points o f discontinuity, the Fourier series will converge to a value midway between the left- and right-hand values o f the function next to the discontinuity. There may be other differ ences as well. Despite these differences, the energy between the function representation is zero; i.e.,

and its Fourier series

with N approaching oo, goes to zero. Thus, for all practical purpos

es, the functions are identical. This practical equivalence allows us to analyze how a circuit responds to a s i g n a l b y analyzing how the circuit responds to each o f its Fourier series components.

3. ADDITIONAL PROPERTIES AND COMPUTATIONAL SHORTCUTS FOR THE FOURIER SERIES REPRESENTATION If a periodic function y(f) is known only at some sampled points, e.g., by measurements, then its Fourier coefficients must be calculated by the use o f numerical methods. On the other hand, if

j{t) has an analytic expression, then its Fourier coefficients can often be calculated from equation 20.11, The properties discussed below are o f great value in simplifying the calculation o f Fourier coefficients. Their proofs are fairly straightforward and arc left as homework problems. The linearity property: Letyj(/) and f^{t) be periodic w'ith fundamental period T. \^j{t) = KJ'^{t) + K^fyU), then the Fourier coefficients o^j{t) may be expressed in terms o f those of/j(f) and ^ (r) according to the following formulas: (20.19a) (20.19b) and (20.19c) In general,

unless the angle 0^ is the same for all n.

1 100


n s

This property allows us to easily obtain the Fourier series of a sum of periodic signals when the Fourier series of the individual signals is already known. DEFINITION A periodic function y(/) is said to be half-wave symmetric if -0.57) = -j{t) for all t

In words, y(r) is half-wave symmetric if a half-period shift of the plot combined with a flip about the horizontal axis results in the identical function y(/). Some simple examples of half-wave sym metric functions are sin(ft)r), cos{(Ot), and the square wave of Figure 20.4, The half-wave symmetry property: A half-wave symmetric periodic function f j ) con tains only odd harmonics. The waveform of Figure 20.4 illustrates this property. The dertvadve/integration proper^.

denote the ^ Ik)

functiony(f). Then the Fourier coefficients, Q s o f

ci*> = (jmoo f c „ for all n

c„ =

derivative of a periodic

(k) ’ (t) of satisfy

20a)

c<*)

Except for the constant term, all terms derive from those ofy(r) by differentiating k times; conversely, all terms oij{t) derive from those of/^^(/) by ^-fold (indefinite) integration. The exclu sion of the constant term in the relationships poses no difficulty, since the constant is simply the average value of the periodic function. Again, these properties help to simplify the calculation of Fourier coefficients. In fact, these prop erties make it possible to compute the Fourier series for the waveforms given later in Table 20.3 without carrying out the integration of equation 20.11. Achieving this, however, depends on first finding the Fourier series of a waveform for which computing by equation 20.11 is extremely easy. We will illustrate several such calculations in Examples 20.5 through 20.8. During this devel opment, the following trigonometric identities will prove useful; / ^ y\

/ _ y'

cos(jc)-cos(y) = - 2 s i n ( ^ ^ - j sin (^ -Y -j

(20.21a)

cos(x)cos()') = 0.5 [ cos(x + j ) + co s(;f-j)]

(20.21b)

n


!01

E X A M PLE 2 0 . 5 . Find rhe Fourier series for rhe periodic impulse trainy§(^) shown in Figure 20.7.

jk

A (A)

-21

A (A)

k. (A)

A (A)

(A)

-T

2T

FIG U RE 20.7 A periodic impulse train. S olution Using the sifting property' o f an impulse function together with equation 20 .1 0 yields

T

A

(20.22a)

T for all n. Hence

Equation 20.22b states that, for a periodic impulse train, all harmonic components have magni tude equal to twice the average value. This is the limiting case o f the short pulse propert)- stated in section 2. The next example uses the derivative and integral properties o f the Fourier series to develop an alternative derivation o f the Fourier series for a periodic rectangular pulse train, derived earlier by the use o f equation 20.11.

E X A M PLE 2 0 .6 . Find the Fourier series for the periodic rectangular pulses^(f) shown in Figure 20.8a. (This corresponds to item 2 in Table 20.3.) S olution Figures 20.8a and b show ^(r) and its derivative,^ (t). The latter may be written as the sum o f two shifted impulse trains:

{ f n( t ) = f6 t + — \

^ r

/

-fa

t-

/(5T «T’ \\ \ ~ r;

(2 0 .2 3 )

102


A

(A)

(A)

-T

T 2

i(-A )

(A) -f —

2

^ 2

T ,,(-A )

2

>

t

(-A)

(b) FIG U RE 20.8 (a) Periodic rectangular pulse crain and (b) its derivative. The parenthetical values,

{A) and {-A), next to each impulse in part (b) denote the weight of that impulse, i.e., its area. Using the time shift property (translation in the horizontal direction), together with equations 20.22b and 20.21a,

f'p ( 0 = Y,

X

"

E

« = I

2A

- cos{na)Qt - n^jv)) (20.24)

4/\ (sin(/7o;oOsin(«/3;r)) ^

Applying the derivative/integration property to equation 20.24 yields 00

(2 0 .2 5 )

T h e result agrees, o f course, with equation 2 0 .1 6 .


1103

E XA M PLE 2 0 .7 . Find the Fourier series for the half-wave rectified sine wave fhsi^) shown in Figure 20.9a (item 10 in Table 20.3).

A f.(t)

(0 FIGURE 20.9 (a) Half-wave rectified sine wave as the product of two functions: (b) a cosine wave and (c) a square wave. S olution The periodic fiinction j{t) o f Figure 20.9a may be viewed as the product o f the sinusoidal wave y4cos((0Qr) and the square wave^(r), shown in Figures 20.9b and c, respectively. Using the Fourier series for^(^) given by equation 20.13, we have //i.v(0= [ ^ c o s ( w o O lA ( 0 M = y4cos(o;o0

2

9 / 1 1 \1 ~ cos(ojo^) — cos(3w()/) + —cos(5o;o/) - ... ;r ^ 3 5

(20.26)

104


Applying equation 20.21b to each product in equation 20.26 yields fh s

( 0 = — cos( wq/) + — |cos(2o;()r) + cos(Oo;o/) - -cos(4w()/') + cos(2woO “ —

7z I

2

3

A lA = - + - cos(w()/) + —

IT

z

- c o s ( 2 ft ;,)0 - — cos(4coqi) Id

JT \ J

+ — cos(6w()/) + ... + -— ^------ cos(2/?W()/) + ... 35 4/;"- _ 1 A = - + - co.s((OoO + — 2 ^ 2 JT Note that the fundamental component is present in

T T — -co.s( 2 mu„/ 4 ,r _ 1

(20.27)

and the remaining terms are all even

harmonics.

E X A M PLE 2 0 ,8 . Find the Fourier series for the full-wave rectified sine wavej^(r) shown in Figure 20.10a (item 9 in 'I'able 20.3).

lOS

Chapter 20 • Ekicf Introduction to Fourier Series

■> t

FIGURK 20.10 (a) Full-wavc rcctificd sine wave as the sum of the two signals in (b) and (c). S olution One approach is to apply the same technique as in Example 20.7. Specifically,

f^ {t)= A cos{io^ t)[lfi,t)-\ ] where f^{t) is the square wave o f Figure 20.9c. Alternatively, to avoid repeating all the arithmetic, note that

is the sum o f the two waveforms shown in Figures 20.10b and c; i.e., (20.28)

Substituting equation 20.27, the Fourier series Fourier series:

lA

into equation 20.28 yields the desired

4/4

(20.29)

4 ;r -I The derivative propert}' is particularly useful for tackling periodic piecewise linear waveforms. Piecewise linear waveforms consist o f straight-line segments. From differentiating once, or at most twice, impulses appear. The integration given by equation 20.11 is trivial if the integrand contains a shifted impulse function. This fact, together with the derivative and integral properties, reduce the task o f calculating the Fourier coefficients for piecewise linear waveforms to some complex number arithmetic. These examples and several other commonly encountered periodic waveforms have Fourier series as given in Table 20.3. Engineering and mathematical handbooks contain much more comprehensive tables. O f course, when a waveform does not appear in a table, the Fourier coefficients must be computed manually or numerically.

106


TABLE 20.3

107


6. Clipped sawtooth wave ^0 =

= 0-5a/\

d n = — 2~^ sin ^ (nojr) + naji[ncm cm n~

sin(2//o;r))

0.5

: too complicated for inclusion

7. Asymmetrical triangular wave ^/o = F „ ,= 0 .5 A

d,i = ---------------^ -:rsin (« a jr) a {\ -a )T r n ~

: too complicated for inclusion

8. Symmetrical trapezoidal wave 0.5ao = Fay =^{a + P)A

n (a + P)ji

b„= 0

9. Full-wave rectified sine wave

Tq = 0.5T,

f(t) =

(Oq =

2(0

^ ^ cos(2(o?) 3

4A K

cos(4w0 ^ 15

cos(6(o/) 35 •••

(-l)"^'cos(2«wO 7

4 fr-i

"*“•••

10. Half-wave rectified sine wave

A

A

Jl

L

/ (/ ) = - + - c o s (w o / ) + n+1

M

v H L _

^ k

ave=A/n

rms=A/2

4n^ - 1

COS (inojQt)

108


4. SUMMARY Given that many mathematical and engineering handbooks have extensive lists o f the Fourier series o f common signals, this chapter has taken a practical approach to the calculation o f the Fourier series and its application to circuit analysis. The idea is to use tables such as Table 20.3 in the same way engineers have come to use integral tables. The keys to using such tables for the com putation o f the Fourier series o f a waveform are the various properties that allow one to convert a known series into one that fits a new waveform. The idea here is to express the new waveform as a translation of, a linear combination of, a ^-fold derivative of, or a ^-fold integral o f signals with known Fourier series as in Table 20.3, or any mixture o f these operations. The Fourier coefficients o f the new signal can then be expressed in terms o f the Fourier coefficients o f signals with known Fourier series. Knowledge o f the Fourier coefficients o f a signal such as the output o f an audio amplifier allows one to investigate phenomena including the distortion introduced by the amplifier. In the case o f a dc power supply, such knowledge allows us to characterize the degree o f unwanted ripple in the output o f a rectifier circuit. In addition, the application o f Fourier series plays an important role in the computation o f steady-state circuit responses to periodic input signals.

5. TERMS AND CONCEPTS Average value o f periodic/(^): the term

in equation 20.4; also referred to as the dc compo

nent ofy(r). Derivative/integration property: Let/^^H^) denote the ^ h derivative o f a periodic function y(^). Then the Fourier coefficients o f the ^ h derivative, and conversely,

satisf}'

= (/>zoJq)^c^^ for all

J{t) is the /’th integral of/^^^(^), then

for all n except

0. D uty cycle: for a rectangular signal having fundamental period T as illustrated in Figure 20.5, the duty cycle is the constant (3 that determines the pulse width ftT. Effective value (rms value); |/l=-cc coefficients are from the Fourier series o f equation 20.4 and the

are from equation

20.6b. Even function: y(/) =

The plot o f an even function is symmetrical about the vertical axis.

Exponential (complex) Fourier series; decomposition as given in equations 20.6.

into a sum of complex exponentials

Fundamental com ponent (first harm onic): the first term under the summation sign in equa tion 20.4,

cos(o)q^ + 0 j), having amplitude d^ and phase angle 0 j.

Fundamental frequency (in hertz): /o = — . where T^^ is the fundamental period. Note: _ 2:7r . ^0 Wq = 2;r/() = — is the fundamental angular frequency in rad/sec. Fundamental period: the smallest positive real number

for w h i c h + Tq) =J{t).


1109

Half-wave symmetria refers to a periodic function/(/) that satisfies 0.57) = -fJ)Half-wave symmetry property: A half-wave symmetric periodic function//) contains only odd

harmonics. Lineari^ property: let^(z) and^(/) be periodic with fundamental period T. = ATj^(/) + A!^(r), then the Fourier coefficients o^j{t) may be expressed in terms of those and fj{t) according to the formulas c^ = + K2C2„, = ^\^\n In general, the angle 0^ is the same for all n. Odd function:/r) = Periodic signal, y(f): A signal whose waveform repeats every T seconds. Mathematically, for some

7’> 0, and all

+ 7)

where Tis the period of the signal.

Symmetry properties of Fourier series: (1) If a periodic function ^(/) is an even function, then

its Fourier series has only cosine terms and possibly a constant term. (2) If a periodic function y(/) is an odd function, then its Fourier series has only sine terms. Translation of a plot: horizontal and/or vertical movement of the plot of a function without any rotation. Translation proper^ of Fourier series: if fit) is a translation of J{t) consisting of a dc level increase K and a time shift (delay) to the right by then ^/) = j { t +K. See equa tion 20.14. Trigonometric Fourier series: representation of a periodic signal f j ) in terms of sines and cosines, as given in equations 20.3 or 20.4.

10


ANSWF^R: (a)

^ ^ r o b le m s

= - 0 .7 2 1 3 0.0795

FOURIER SERIES COEFFICIENTS BY INTEGRATION

cos(4.t /- 8 6 .8 4 '’)

and write the Fourier series in the form o f equations 20.3.

■ ■

N

-0.5 T

0

f(t)

0.5 T

(1)

4. Using trig identities, compute the Fourier series coefficients

-2

(b)

j{t) = cos(4r) sin(2f) j{t) = sin^(4/^) cos^(8/)

(c)

/ f) = [2 + 1 .5 sin (5 0 0 r)

(a)

A^ f(t)

(-1) >r

and

for the func

tions

(a)

(-1) >r

N

■r

Figure P20.3

(1)

0

+ ...

>^ f(t)

out the integration given by equation 20.10

-2 -1 1

+ / 2 ; / . t ) ; (b) fit)

P20.3.

impulse trains shown in Figure P20.1. Carry

-3

2

3. Repeat Problem 2 for the signal o f Figure

1. Find the Fourier series for the periodic

(1)

= 0 .5 /(Io g ^ .

+ 0 . 1 5 8 c o s ( 2 . t / - 8 3 . 7 ' ’) +

1 2 (-1) >r

i

3

w

■ >

(b) Figure P20.1

- 2 cos(2000r)]cos(10^V)

FOURIER SERIES USING PROPERTIES W ITHOUT INTEGRATION

AN SW ER: (a) 0.5 + cos(.7r) + cos(2rr/) +

5. Construct the Fourier series o f item 5 in

cos(3-t/‘) + cos (4.t;) + ...: (b) - 0 .5 + cos(.Tr) -

co^{2m) + coa{5m) - cos (4m) + cos(5-t/) ...

Table 20.3;, i.e., find the Fourier series o f the sawtooth waveform in Figure P20.5. Flint:

2. For the periodic function J{t) shown in

20.22b for/^(r).

Note that f{ t ) = A -

Use equation

Figure P20.2, T = 1 second and a - log^ = 0.693. (a)

Find the coefficient

by carrying out

the integration given by equation 20.6b. (b)

Write the first three terms o f the Fourier scries (see equation 20.4). f(t) e** 6. Find the Fourier coefficients

and 6^ o f the

periodic function shown in Figure P20.6. Flint: Make use o f the result o f Problem 5.

Figure P20.2

1111

Chapter 20 • Brief Introduction to F-'ouricr Series

form o f item 2 with proper height and (2) the waveform o f item 5 shifted down by a suitable amount.

-1

0

1

2

3

Figure P20.6 -►t

ANSWF.R: /(/) = 0.75 + 0.159 sin(2Tut) 0.0795sin(47rt) + Figure P20.9

7. Construct item 3 of Table 20.3; i.e., com pute the Fourier scries for the triangular wave shown in Figure P20.7. Hint: f'{t) is a square

10.

w'ave whose Fourier series was calculated

shown in Figure P 20.10. Hint: Make use o f the

Find the Fourier series for the waveform

before.

result o f Problem 9.

>

Figure P20.10 8. Consider the isosceles triangular wave shown in Figure P20.8, which is item 4 ofTable 20.3.

11. Consider the asymmetrical triangular wave

Compute the Fourier series o f this waveform

shown in Figure P 2 0 .l l, which is item 7 o f

utilizing the results for items 2 and 3 ofTable

Fable 20.3. Let 7 = 1 and u = 0.25. Find the

20.3. Hint:y(^) is the product o f (1) the wave

Fourier coefficients

form o f item 2 with proper height and (2) the

Differentiating J{t) twice results in periodic

waveform o f item 3 shifted down by a suitable

impulse trains.

amount.

and d-^. Hint:

f(t)

> t

Figure P20.8 12. Find the Fourier series coefficients Tq, c’p 9. Consider the clipped sawtooth wave shown in Figure P20.9, which is item 6 ofTable 20.3.

and C-, for the periodic waveform shown in Figure P20.12. Hint: The period T = 4, and

Compute the Fourier scries o f this waveform

you should express J{t) as the sum o f several

utilizing the results o f items 2 and 5 ofT able

shifted square waves.

20.3. Hint:y(r) is the product o f (1) the wave

1 112


oj= 10^ rad/secand 10-^ rad/scc. For the frequency range990 x 10-^ < w< 1010

A

X 10-^ rad/sec, assume that the band-p;iss amplifier has the following ideal magni tude and phase characteristics;

\H{joj)\ = 10 and

-2 Figure P 20.12

/LH ( jco) = - ——

5000

Express

(a )

SINUSOIDAL STEADY-STATE ANALYSIS WITH MULITFREQUENCY INPUTS

X 45^'

as the sum o f all o f its

frequency components.

13. The input to the low-pass filter circuit

(b)

Express

(c )

frequency components. If is expressed in the form

shown in Figure P 20.13 is

as the sum o f all o f its

^ cosCwq/) V, show that

g{t) = \Oj{t - t j , where v,„(/) = 200 -h 200n/^cos(377/) + 6 0 V 2 cos(3 x 377/ + 30'^ ;

sec. The interpretation o f this result is that the envelope o f

+ 8 0 V 2 co s(5 x 377/ + 5 0 ^ )V

= 78.54 a is an ampli

fied and delayed version ol the enve (a)

Find

(b)

Find the rms value o f

at steady state.

lope o f and the

the amplification is 10

and time delay is 78 .5 4 [.isec.

average power absorbed by the 10 kQ resistor.

jy r \

v„(t)

10H

6

4 pF Fig P20.14 Figure P20.13 ANS\X'ER; (a)

ANSW ER; (a)

0.1 cos((
+ r.... (M = 200 + >/2

42.6cos((o/ - 1 7 5 .4 ") +

+ 0.1 cos(((/^ +

2 cos((oj. -

1.192cos(3n;/ - 1 4 S .7 " )+ 0 .5 6 COM5o>/ - 129.2^)

= 0.2 cos(((/;. -

10

Jt) +

+ 2 cos(a;^./) + 0 .2 c o s ((o ;.

V; (b)

+ 4.5°) + cos((«;. -

y 9°) + 20 cos(oj^j') + 2 cos((o>. +

= + -

4.5") + cos((o;,+ 2 r o J / - 9 " ) V

with u) - 377 rad/sec; lb) 204.5 V. 4.18 W

15 . Consider the circuit o f Figure P20.15, in 14. Consider the circuit o f Figure P20.14. The

which C = I F (initially relaxed), R = 1.443 =

input to this ideal band-pass amplifier is an

\Un{l) n , and

amplitude-modulated waveform given by

currents Q b[t—nT), T - 1 sec, Q = 1 coulomb,

=J{t)x cos(cu^r) V, where

and ;; = 0 , 1 , 2 , — (a) Show that the response due to the first

J{t)

=

2[1 - 0.2 cos(w,/) + 0.1 cos(2(t;,/)]

is a sequence o f impulse

hnpitlse current alone is

=

13


^-0.693r ^ (0.5)^ V for 0 < ^ < 00; v ^ Jt)

(b)

'w '

(c) (d)

(e)

starts with 1 V at ^ = 0 and then is halved for every second of elapsed time. Making use of linearity and the result of part (a), show that for 1 < r < 2, v^Jt) = 1.5(0.5)<'- '> V. Note that only the first and second impulses have an effect on this time interval. Show that for « < /< « + 1, = [2 -(0.5 )"](0 .5 )('-”^V. For very large « < r < « + 1, show that v^Jt) = 2(0.5)(' V and that the waveforms for the time intervals («, n + 1) and (« + 1, « + 2) are identical except for a time delay of n. This means has reached the steady state. Show that for arbitrary values of /?, C, Q, and T the output in steady state is 'ôut ( 0 =

G

1

Ôllt ( 0 = (b)

(c)

----- u

(a)

,I

ijt )

k

*

(Q)

(Q) -T

1r

•••

C = 1 F, Q = 1 coulomb, and T = 1 sec. Show that the steady-state response for the first cycle is = 2(0.5)^ V, for 0 < r < 1. Sketch the waveform for the first cycle. Using the transfer function 1

hnis)

2T

Figure P20.15 Consider again the circuit of Figure

P20.15. If the impulse current train has been

R

and the Fourier series for equa tion 20.22, find the Fourier series for Write the answer to include up to the third harmonic. Find the average power (approximate value) dissipated the resistor. Since only a finite number of the terms in the infinite series are included in the calculation, the answer is only approx imate. What is the error when only harmonics up to the third are consid ered? To determine this error, compute the exact average power using the result of part (b).

CHECK: (c) v^Jt) = 1.4427 + 0.3164 cos(2;rf - 87.7047°) + 0.1589 cos(4;rf - 86.8428“) + 0.1060 cos(6;w - 87.88“) V; (d) 2.109 watts, exact answer is 2.164 watts, error = 2.53% 17. The LC resonant circuit in Figure P20.17a is initially relaxed and the impulse train

applied for a long time (theoretically since t = - oo), that is, « = - 00,..., -2, -1,0, 1,2, ...

(a)

C5 H----

(Q)

(b)

16.

For this and the remaining parts, use the special set of values

H {s) =

(d)

V„Jt)

RC

_Z_

« = — !— = I.4 4 2 Q M 2)

\ - e RC

—O +

C \ -e

RC

for n T
1

Show that for the time interval (0, T) /i=0

14


shown in Figure P20.17b is applied. The natu ral frequency o f the tank circuit is f.. = ------\-----

2 h /lC

I

Tq - — = 2vtVZc .

and the period is (a)

shown in P 2 0 .I8 b is applied. This problem (and circuit) differs from the previous one only in the presence o f the resistance R. When the value o f R is large enough, the impulse response

Jo

is a damped sinusoid. The purpose o f this prob

Show that if the frequency o f the

lem is to show that if the damping effect o f i s

impulse train, / = — , is the same as 7' , the resonant frequency / q = then

slight and the input impulse train “synchro nizes” with the damped sinusoids o f the impulse response, then the output voltage mag

In y fL C ' nitude, although finite, can reach a ver\' large

/y(r) grows without bound.

value.

Specifically, show that the inductor

(a)

current in the time interval {n - \ ) T <

o f poles at

t < 7i T \s given by i^{t) = n x Show that iff=fQlm for any integer m,

CJand

the inductor current again grows unbounded. (c)

±

interval o f length

Show that in general if T = (2w +

Wq =

2ji Tj = ---- ) differs (X)j

from its adjacent cycle by only a mul tiplicative constant a =

2m + 1 then /^(/) is periodic. Sketch t\vo

then the

(ii) the waveform o f every cycle (time

two cycles o f /^(r). 1)(0.57'q) or, equivalently, 2/o

CJ

in terms o f R, L, and C.

Note that (i)w j

Show that if/ = 2/q, then i^{t) is peri odic and remains bounded. Sketch

(d)

^= -

impulse response has the form h{t) = (,-ot cos{ioj) + 5sin(w^)]//(f). Find

sin(w^r). (b)

¥\n6H{s) = - ^ ---- .If//(^) has a pair

(b)

.

Show that if the input impulse train syn chronizes with the damped oscillations, meaning that T = T j = 2nUo^ then

cycles of

the output due to the impulse train o f Figure P 20.18b is

for (;/ - 1) (c)

(a)

if the impulse current has been applied for a long time (theoretically from t =

>I >

W

(Q)

(Q) -T

< r < ;/

Deduce from the result o f part (b) that

T

-oc), then the steady-state response

•

during the first cycle is given by

(Q) 2T

= 7 ^ / K O for 0 < r < 7;y 1- a which shows that as a comes very

(b)

close to 1 (meaning very little damp

Figure P 20.17 18. The LC resonant circuit in Figure P 2 0 .l8 a is initially relaxed and the impulse train

iin(0 = Q ^ H r - n T )

/i=(i

ing), the magnitude o f the output can reach a very large value. (d)

If ^ = 0 .7213 a

C = 1 F, and L =

0.025 H, in which case a = 0.5, find

H{s) and the impulse response h{t).

1115


Further, if Q = 1 coulomb and T = 1 sec, show that v^JJ) in steady state is given by (consider item 18 of Table

(a)

12.1) ■'.»,W = 2<'-«®3I<[cos(2 w ) -

• — 1/ I V ''max - V*mm ôul,mm ~ *min o .5 7

0.1103 sin(2;»)] V for 0 < r < 1

L,

\+ e

Plot this waveform using MATLAB or its equivalent. Justify that this is indeed the steady-state value by showing v<.„,(0-^)-v„„,(D = | = l V. (e)

w

I f /? = 9.747 a , C= 1 F, I = 0.02533 H, Q = 1 coulomb and T = \ sec, in which case a = 0.95, find the steadystate response for 0 < ^< 1 sec.

If the input has been applied for a long time, then the circuit reaches steady state and as shown in Figure P20.19c. Using this fact and equation 8.19a, show that in steady state.

V . = V^max ôut,max

(b)

(c)

u » (t)

V ■f ' nt/TT - V' »tff» 0.5T \+ g RC

In Example 20.1, find the exact values of '“ '"g results of part (a). Then plot for one complete cycle using MATLAB or its equivalent. In the exercise following Example 20.1, four frequency components of the Fourier series are used to approxi mate the input, and it is found that ôut.min ~ 12-235 V and ôuumax ~ 82.013 V. What are the percentage errors of these approximate answers?

(a) U t)

ik (C8

(Q)

-T

(Q) 2T

(a)

(b)

.v jt) V

Figure P20.18 CHECK: (d) //(5) = 's-^

(e)/ / (5 ) = _ ----------

fflin

r + 1.3863^ + 39.96 s

r +0.102586j + 34.481049 and = 20tf“°*°5’^'[cos(2;«) 0.0082 sin(2;r^)] V for 0 < ^< 1 19. Consider the first-order RC circuit of Figure P20.19a, where the input v-JJ) is a square wave with period T, as shown in Figure P20.19b. From the analysis methods discussed in Chapter 8, the output consists of exponential rises and decays, as illustrated in Figure P20.19c.

w

w

■

...

-0.5T

0.5T

T

0.5T

T

(b)

(c) Figure P20.19

1.5T

16


ANS\X^:RS:(h) 1 1.235V. 83.013 V, (c) H.9 % . -1 .2 %

22. Consider the circuit o f Figure P20.22, in which C = I F, 7? = \Un{2) = 1.443 T=A sec, I/, = 15 V.

= 0.25, v^Q-) =

20. The method o f Problem 19 can be extend

V, and

ed to computation o f the steady-state response

as shown in Figure P20.22b.

o f any first-order linear network to a square

(a)

= 1

is a sequence o f rectangular pulses Show that for the first cycle, 0 < r < 4

wave input. Let the transfer function o f the

sec, the complete response is V(^t) =

first-order linear network be

[15 - 1 5 (0 .5 )1 + (0 .5 )r = 15 14(0.5)^ V for 0 < / < 1 sec and V(^t) =

H(s) =

/ ’[Output(/)]

H{0)

/ ’[InputC/)]

1+T5

8(0.5)^ V, for 1 < t < A sec. Sketch the

v^t) waveform for the first c)'cle 0 < r < 4 sec.

and the input be the square wave of Figure P 20.19b. Show that in steady state,

(b)

Show that v^JA') = = 1 V.

Output(/),„,„ = H{0)

w

,

V * max - V^min

(c)

cycle, n T < t < {n + U T is identical to

l+ e ^

that o f the first cycle except for a time

and 0

Use the result of part (b) to verify that the V(\t) waveform for any subsequent

0.57

shift, i.e., the first-cycle waveform

V ■ ''max - V* mm

utput(0 ,„av =

found above is the steady-state wave form (d)

Using the transfer function

21. Consider the leaky integrator circuit o f Figure P20.21. The input

is a 1 kHz

H{s) =

square wave with zero mean and 2 V peak to

RC s+

peak. The circuit has reached steady state.

RC

(a)

Find the transfer function H(s).

and the Fourier series for

(b)

Find the maximum and minimum

o f la b le 20.3 , find the magnitudes o f the harmonics o f the Fourier series for

values o f

using the results o f

Vo,^,{t), up to the third harmonic.

Problem 20. (c)

item 2

Use MATLAB to plot the waveform o f for one complete cycle.

(e)

Find the average power (approximate value) dissipated in the resistor. Since only a finite number o f the infinite

50 kO

series is included in the calculation, the answer is only approximate. What is the error when only harmonics up to the third are considered? To deter mine this error, compute the exact average power using the result o f part (a).

Figure P20.21 A.\S\X ER,S: (a)

H (s) =

-•S

i()()0.v +

: ( b ) ± 0 .2 4 5 V


R

R.

'. W ©

R.

v jt )

'» '« 6

O '

Figure P20.23

(a)

Avjt)

0

pT

T

■> t T+pT

(b)

Figure P20.22 23. Consider the second-order circuit of Figure P20.23, in which = Ifl, /?2 = 8 Q, /?3 = 6 £2, Cj = 0.25 F, and = 0.125 E (a) Derive the transfer function 16 3 r + 7J + 10 If the input voltage is a square wave shown in Figure 20.23 with peak-topeak voltage 18 V and T = \ sec, find the Fourier series for and list the first four components. (c) Find the approximate average power delivered to R2 considering only the first four components. (d) Assuming the input has been applied for a long time so that the circuit is in steady state, find the expression for during the time interval (0, 7). Use MATLAB to plot the waveform. Hint: Expand H{s) into par tial fractions. Represent the system as the paral lel connection of two first-order subsystems. Then use the result of Problem 17 to solve each subsystem. Finally, combine the solutions in time domain to obtain the expression for (b)

Vw/

^

w

------ o

I

N

D

E

X

//(double slash) notation, 75, 90 ABCD parameters. Transmission (/-) parameters Abscissa of (absolute) convergence, 556, 591 AC (alternating current) adaptors, 883 defined, 40 Active ban-pass circuit, 827-828, 861 Active element, 82, 177 Active networks, Thevenin and Norton equivalent circuits for, 241-245 Active realization, 1045-1051, 1071 Active region, 171 Admittance, 481 of capacitor, 608, 658 defined, 606, 607, 658 of inductor, 658 of the inductor, 608 Laplace transform analysis circuit appli cations, 605-609, 659-662 Laplace transform analysis circuit appli cations, manipulation, 609-617 phasor admittance, 454, 455^61 ^'-parameters, 1010 Admittance (y-) parameters, 961, 968-973 Algebra, 781, 797 Ammeter, 10 .\mpere, 3—4, 29, 40 Amplifier. See Operational amplifier (op amp) Amplifier circuit, 117-118, 120, 121, 136 small-signal low-frcquenc)' equivalent, 117 Amplifiers, 959-960 Analytic continuation, 557, 591 Angular frequency, 384, 417, 433 Apparent power, 510-515, 531 Approximation problem, 1035, 1037-1042, 1071 Armature winding, 736 Associativit)', 766, 797 As)-mptotes, 732, 740 Attenuation (dB), 1033, 1072 Automobile batter)', 14-15, 16, 18, 27 Auxiliar)-voltage variables, 136 Available power, 528

Average pow-er, 531 complex power and SSS, 510-515 SSS power calculation and, 500-505, 533, 534-536 Average value, 10, 1089 Average value of p e r i o d i c 1089, 1108 Balanced terminal, 157 Band-pass circuit, 470, 481, 811-882 defined, 861 general structure of, with one pair of complex poles, 851-860 ideal band-pass filter, 814-816 with practical components, 830-838, 868-872 resonance phenomenon and resonant circuits, 839-851 resonant circuits with applications, 872-876 touch-tone phones and, 811-812 transfer function problems, 863-868 transfer functions with no zeros/single zero off the origin, 876-879 Band-pass filter design, 1062-1069, 1072 Band-reject circuit, 471, 481 Bandwidth with respect 1062, 1072 3 dB bandwidth, 815-816, 861 Batter)', 12, 14-15, 16, 40 automobile, 14-15, 16, 18, 27 ideal, 26, 30, 84 Bell Systems TechnicalJournal, 1069 Bode plots, 730-736, 760-761 bode magnitude plot, 731-733, 740 bode phase plot, 731, 734-735, 740 Bounded current/voltage, 278, 306, 555, 556, 690, 740 BP to LP frequency transformation (pr?), 1072 Branch, 53-54, 89 Branch current, 107, 110, 128 Branch voltage, 107 Breakdown voltage, 376 Breakpoint, 733, 740

120

Index

Buffer, 164, 177 Bulbs, 19, 30, 34-35, 37 Butterworth, Stephen, 1034 Buctenvorth filter, 634-635, 856-858, 1045-1051. See also Filtering active reaIi/;uion ol low-pass Butterworth filters, 1045-1051 Butterworth LP design algorithm based on S&K circuit, 1046-1051 loss function properties, 1052-1053 normalized Buttenvorth loss functions, 1072 “On the Theory of Filter Amplifiers” (Butterworth), 1034 passive realization, 1042-1044 singly terminated low-pass networks, 1069-1071 solution to approximation problem, 1036-1042 Capacitance, 284 ol a pair of conductors, 306 Capacitors, 269-319. See also Inductors admittance of, 608, 658 capacitive smoothing in power supplies, 269-270, 303-305 definitions and examples, 274-283, 284-293 discharging, through an inductor, 382-385 impedance of, 452, 608, 658 inductor and, 271-273 linear, 306 in parallel, 301 phasor relationships for resistors, inductors, and, 449-455 in series, 298 series and parallel inductors and capacitors, 293-302 v-i relationship and Laplace transform method, 580-582 Car heater fan, 51-52 Cascaded connections, two-ports and, 1002-1009 Cathode ray tube, 2 Characteristic equation, 388, 417, 591 Characteristic roots, 417 Charge, 3-6, 11-12, 16, 22, 27, 38, 40 Circuit model, 34, 36 Circuit response computation graphical convolution and, 774-781 Circuits, 2. Sec also Magnetically coupled circuits; RL/RC circuits connected, 62

critically damped, 390, 392, 417 difierentiator, 358, 364 elements, 2, 14-15, 53, 58, 60 integrator, 359, 364 model, 34, 36 open, 30, 37 oscillator, 385 overdamped, 389, 392, 417 parallel, 53 passive, 339 planar, 128 series, 53. 67 series-parallel, 73-78 short, 27, 30, 64 stable, 433, 482 tuned, 844 undamped, 385, 417 underdamped, 390, 392, 417 Classical difference amplifier, 164 Closed node sequence, 62, 89 Closed path, 61, 62, 63, 89 Coefficient-matching technique, 1072 Coefficient of coupling, 913, 937 Coils, 306. See also Inductors coupled inductors, 937 mutual inductance (M), 885, 886-893 primary/secondary, 898, 917, 938 qualit)' factor of, 861 Common-ground rwo-port, 1002-1009 Commutative propert)’, of convolution, 766 Comparator, 174-175, 177 Complementary function, 352 Complete response, 352, 364, 693, 740 Complex exponential forcing functions, 481 capacitor, 452 complex sinusoid, 441 inductor, 451 in SSS computation, 444-447 Complex numbers, 435-441 Complex power, 510-515, 531 Complex power calculations, 536-537 Computation of response for linear timeinvariant systems, 795-797 Computing equivalent inductances, 940-943 Conductance, 30, 40, 454, 482 equivalent, 70 in parallel, 70, 72 transfer, 79 Conductors, 3, 40 ideal, 3, 4, 41 Connected circuit, 62, 63, 89, 140 Conservation of charge, 288 Laplace transform analysis circuit

121

Index

applications, 6 4 5 - 6 5 2 principle of, 3 0 6 , 6 4 5 Conservation o f complex power in AC

Current, 2 , 4 0 alternating, 5, 8 - 1 0 branch, 107, 110, 128

circuits, 515 Conservation o f power (energy), 19, 2 0 , 3 6 , 4 0 ,5 1 5 . 5 3 1 ,5 3 7 Constant inputs, second-order linear networks with, 4 0 0 —411 Constant-resistance network, 6 4 0 Continuity property o f the capacitor, 2 8 7 , 306

direct, 8 - 9 , 3 0 , 33 loop, 110, 128, 141 mesh, 128, 141 Current division, 8 9 , 4 8 2 , 6 0 4 , 6 5 8 formula, 6 8 - 7 3 , 4 5 8 Current source, 4 0 , 5 6 nonideal, 8 5 - 8 8

Continuity property o f the inductor, 2 7 8 ,

v-i characteristics, 56 Current transformation property, 9 1 5

306

's ^

Critically damped circuit, 3 9 0 , 3 9 2 , 4 1 7

Controlled source, 2 8 , 2 9 , 7 8 , 80

C u to ff frequency, 1033, 1072

current-controlled current (C C C S), 2 8 , 39 current-controlled voltage (C C V S), 2 8 , 39 voltage-controlled current (V C C S ), 2 8 , 7 8 , 80

W

Damped oscillation frequency, 4 1 7 , 8 5 2 , 861 Dam ping, 3 9 2

voltage-controlled voltage (V C V S ), 28, 7 8 , 80

D C (d irea current) defined, 41

Convolution, 7 6 3 - 8 0 9 averaging by finite time integrator

first order R L and R C circuits, 3 3 6 - 3 4 7 motors, transfer function analysis of,

circuit, 7 6 3

7 3 6 -7 3 9

basic properties, examples, 7 6 6 - 7 7 2 , 7 9 0 -7 9 3

257

circuit response com putation and graphical convolution, 7 7 4 -7 8 1

O '

circuit response com putation using, 7 8 6 -7 9 0

defined, 7 6 6 , 7 9 7

Dependent (controlled) current source, 4 0 , 89

graphical convolution, 7 7 4 —7 8 1 ,

Dependent (controlled) voltage source, 4 1 , 89 Dependent source, 78

by integral, 7 9 9 - 8 0 2

Derivative/integration property, 1100, 1108

interconnection o f active circuits and,

Dielectric, 2 8 4 , 3 0 6 Difference amplifier, 1 6 3 -1 6 4 , 167, 177

by inverse Laplace transform, 8 0 6 - 8 0 7

Differential equations, 3 6 4

Laplace transformation and, 7 7 2 —7 7 4

com puting SSS and, 4 4 2 —444

tim e domain derivation of, for linear

general second-order circuit, 3 8 7 - 3 8 8

time-invariant circuits, 7 9 3 - 7 9 7

ideal transformers and, 914

Convolution algebra, 7 8 1 , 7 9 7

ordinary, 591

Copper wire resistance, 3 3

series-parallel R L C circuit, 3 8 6 - 3 8 7

C o m er (break) frequency, 7 4 0

SSS analysis and, 4 8 3 Differentiator circuit, 3 5 8 , 3 6 4

Coulom b, 3 , 3 0 6 Coupled circuits.

See Magnetically coupled

circuits Coupled inductors, models for pair of, 9 3 0 ,

937 Coupled inductors (coils), 8 8 6 - 8 8 7 , 9 3 0 -9 3 6 , 937 Coupling network, SSS calculation and, 52 9 , 530 Cram ers rule, 140

W

gain in dB, 1033 D elta function, 5 5 2 , 591

8 0 7 -8 0 9

'Sw/

257 Decade, 7 3 3 , 7 4 0

convolution theorem, 7 7 3 , 7 9 8

8 0 2 -8 0 4

O

Deactivating an independent voltage source,

Decibel (dB ), 7 3 3 ,7 4 0

convolution algebra, 7 8 1 - 7 8 6 , 8 0 4 - 8 0 6

W

Deactivating an independent current source,

Diodes, 3 0 3 Dirac delta function, 591 Dirichlet conditions, 1 0 9 8 -1 0 9 9 Distributivity, 7 6 6 , 7 9 8 D ot notation, 5 4 6 , 5 9 1 , 9 0 5 , 9 3 7 mutual inductance and, 8 8 6 - 8 9 3 Driving point admittance, 6 2 6 Driving point impedance, 6 2 6 D uty cycle, 1095, 1108

\

122

Index

Effective value, 1 0 ,4 1 , 505, 5 3 2 ,1 0 9 7 ,1 1 0 8 o f nonsinusoidal signals, 5 3 3 - 5 3 4

Butterworth solution to approximation problem, 1 0 3 6 -1 0 4 2

n

o f signal and average power, 5 0 6 - 5 1 0

filter, defined, 1033, 1072

single-frequency analysis widi, 5 0 9 - 5 1 0

high-pass active realization, 1 0 8 3 -1 0 8 4

Elapsed tim e formula, 3 5 5 E le a ric field, 11 Electric m otor analysis, Laplace transform analysis and, 6 8 3 Electron, 2 , 3 Elements active, 8 2 , 177

high-pass filter design w ith passive realization, 1 0 5 3 - 1 0 5 9 high-pass passive design, 1 0 8 1 -1 0 8 3 input attenuation for active circuit design, 1 0 5 1 -1 0 5 2 loudspeakers and crossover network, 1 0 3 1 -1 0 3 2

circuit, 2 , 1 4 -1 5

low-pass approximation, 1 0 7 4 - 1 0 7 6

linear resistive element, 215

low-pass filter basics, 1 0 3 4 -1 0 3 6 ,

passive elements, 178 two-terminal, 16, 2 0 , 4 0 , 58 Energy, 16, 2 3 , 32 energy storage in capacitors, 2 9 0 - 2 9 1 kinetic, 12, 32

1 0 7 3 -1 0 7 4 passive low-pass realization, 1 0 7 6 -1 0 7 9 singly terminated Bunerw orth low-pass networks, 1 0 6 9 -1 0 7 1 types of, 1 0 3 2 -1 0 3 4

net e n e i^ , 2 7 9

Final-value theorem, 7 2 6 - 7 2 9 , 7 4 0

n o n -D C power and, 2 2 - 2 5

Finite pole, 6 8 5

potential, 16

Finite tim e integrator circuit, 7 6 3

stored, 9 1 0 - 9 1 4

Finite zero, 6 8 5

stored in pair o f coupled inductors, 9 3 7

First order circuits, 3 2 2 - 3 7 8 .

Equilibrium state, 3 3 9 , 3 6 4 Equilibrium value, 3 5 4

Sef also RL/RC

D C or step response of, 3 3 6 - 3 4 7 first-order differential equaiton o f a circuit, 3 6 4 first-order R C op amp circuits, 3 5 7 - 3 6 3

Equivalent n-terminal networks, 2 5 7

m ath used for, 3 2 4 - 3 2 8

Equivalent resistance, 6 5 , 6 9 , 70

oscillator responses in second-order linear

Equivalent two-terminal networks, 2 1 5

r>

n

o

networks and, 4 1 2

Euler identity, 4 3 6 , 4 8 2

response classifications, 3 5 2

Eulers formula, 3 8 9

sawtooth, 3 2 2 - 3 2 3 , 3 6 5

Even function, 1108

source-fiee or zero-input response, 3 2 8 -3 3 6

Exponential (complex) Fourier series, 1108

r>

analysis and theory, 3 5 2 - 3 5 7

Equivalent circuits, 6 1 8 - 6 2 5 Equivalent networks, 2 1 2 - 2 1 3

r\

circuits

Equivalent, defined, 2 5 7 Equivalent conductance, 6 9

n

superposition and linearity, 3 4 7 - 3 5 2

Exponential decay, 3 9 2

First-order tim e differentiation formula,

Fan speed control, 51

Floating source, 1 2 3 -1 2 8 , 140

Faradays law o f induction, 3 0 6 - 3 0 7

Fluorescence, 5 1 8 , 6 0 3 - 6 0 4 , 6 2 3 - 6 2 5

Farad (F), 2 8 4 , 3 0 6

Forced response, 3 5 2 , 3 6 4 , 7 0 0 , 7 4 0

Femto, 2 9

Forcing function, 3 3 8

Fields, 11, 12

Fourier series, 1 0 8 5 -1 1 1 7

5 7 6 -5 7 9

Filtering, 1 0 3 1 -1 0 8 4 active low-pass design, 1 0 7 9 -1 0 8 1 active realization o f high-pass filters, 1 0 5 9 -1 0 6 2 a a iv e realization o f low-pass Butterworth filters, 1 0 4 5 -1 0 5 1 band-pass filter design, 1 0 6 2 -1 0 6 9 Butterworth loss function properties, 1 0 5 2 -1 0 5 3 Butterworth passive realization, 1 0 4 2 -1 0 4 4

n

additional properties and computational shortcuts for, 1 0 9 9 -1 1 0 7 coefficients by integration, 1110 complex exponential Fourier series, 1090

n

SSS analysis with multi-fi%quency inputs, 1 1 1 2 -1 1 1 7 trigonom etric and exponential forms, 1 0 8 8 -1 0 9 9 trigpnometric Fourier series, 1089 using properties w ithout integration,

1110-1112

n

1123

Index

Four-terminal device, 80

Ideal constant current source, 86

Frequenqr, 4 3 3 ,4 8 2

Ideal constant voltage, 86

dom ain analysis, 5 4 7 , 5 5 4 , 591 response, 4 6 7 , 4 8 2 Frequency response, 7 0 7 - 7 0 8 , 7 4 0

Ideal current source, 2 7 , 3 8 Ideal delay o f

T seconds, 7 8 9

Laplace transform analysis and, 7 0 7 - 7 1 4

Ideal insulator, 41

magnitude, 1 0 3 3 , 1072

Ideal integrator, 3 6 0

SSS analysis and, 4 9 4 —495

Ideal (linear) resistor, 15, 3 7

Frequency s ^ i n g , 5 8 2 - 5 8 3 , 7 4 0 Laplace transform m ethod and, 5 8 2 - 5 8 3 m agnitude scaling and, 7 1 4 - 7 2 6 Frequency shift property, 5 7 5 - 5 7 6 , 591 Fundamental com ponent (first harm onic), 10 8 9 , 1108 Fundamental frequency, 1088, 1108

O '

Ideal constant voltage source, 8 6

Ideal operational amplifier (op amp), 1 5 8 ,1 7 7 Ideal transformer, 9 3 7 defined, 9 1 7 Ideal transformers, 9 1 4 - 9 2 4 , 9 3 0 - 9 3 6 , 9 5 0 -9 5 6 Imaginary part, 4 3 5 , 4 8 2 Impedance, 4 8 2

Fundamental period, 2 8 0 , 6 9 5 , 1 0 8 8 , 1108

o f the capacitor, 4 5 2 , 6 0 8 , 6 5 8

Fundamental period o f p e r i o d i c 7 4 0

defined, 6 0 6 , 6 5 8

Fuse, 3 2

impedance transformation property, 9 2 0 o f inductor, 6 5 8

Gain

w

Laplace transform analysis circuit

in d B , 10 3 3 . 1072 power, 8 0 voltage, 8 0 G ain magnitude, 10 3 3 , 1072

phasor impedance, 4 5 4 , 4 5 5 - 4 6 1

Generalized conductance, 6 0 6

o f resistor, 6 0 7 - 6 0 8 , 6 5 8

General summ ing amplifier, 1 6 7 - 1 7 0 , 177

two-port impedance parameters, 9 7 6 -9 8 1

General voltage division formula, 6 6 Generator, 12, 2 6 Giga, 2 9 Graphical convolution, 7 7 4 - 7 8 1 , 8 0 2 - 8 0 4 Ground, 6 0 virtual, 160, 178

o

Grounded voltage sources examples, 1 1 1 -1 2 2

w w o

Laplace transform analysis circuit applications, 6 3 1 - 6 3 4 Impulse response theorem, 7 9 7 Independent (ideal) current source, 2 7 ,3 8 ,4 1 Independent (ideal) voltage source, 2 5 - 2 7 , 3 7 ,4 1 Induced voltage, 8 8 4 - 8 8 5 Inductance, 2 7 4

Half-wave symmetric, 1100, 1109

H enry (H ), 2 7 3 , 2 7 4 , 3 0 7

Half-wave symmetry property, 1100, 1109

linear inductor and, 2 7 4

H enry (H ), 2 7 3 , 2 7 4 , 3 0 7

Inductor impedance, 451

H i^ -p a s s circuit, 4 6 9 , 4 8 2

Inductors, 15, 2 6 9 - 3 1 9 .

also Filtering H P (high-pass) to LP (low-pass) f l u e n c y transform ation, 1 0 5 3 , 1072 Hybrid (A-) parameters, 9 6 1 , 9 8 5 - 9 9 1 ,

o

1010 Impulse response, 6 2 5 , 6 3 1 - 6 3 4 , 7 1 9 , 7 4 0

Half-power frequencies, 8 1 5

Hom ogenous differential equation, 4 0 0 , 4 1 7

o

two-ports and, 9 7 6 - 9 8 1 , 9 8 1 - 9 8 4 Impedance (z-) parameters, 9 6 1 , 9 7 6 - 9 8 1 ,

Half-power points, 1 0 3 3 ,1 0 7 2

High-pass filters, 10 3 3 , 1 0 5 9 -1 0 6 2 , 1072.

o

applications, manipulation, 6 0 9 - 6 1 7 Laplace transform m ethod and, 5 4 4

Gaussian curves (surfaces), 5 8 , 8 9 , 123, 141

General summ ing integrators, 6 5 2 - 6 5 7

O '

applications, 6 0 5 - 6 0 9 , 6 5 9 - 6 6 2 Laplace transform analysis circuit

1 0 1 0 , 1 0 2 0 -1 0 2 3

See also Capacitors

adm ittance of, 6 0 8 , 6 5 8 capacitive sm oothing in power supplies and, 2 6 9 - 2 7 0 , 3 0 3 - 3 0 5 capacitor, dischaiging through, 3 8 2 - 3 8 5 capacitor and, 2 8 4 - 2 9 3 defined, 2 7 1 - 2 7 3 examples, 2 7 4 - 2 8 3 impedance of, 6 5 8

Ideal band-pass filter, 8 1 4 - 8 1 6

o

■^

W

w

linear inductor, defined, 2 7 4 , 3 0 7

Ideal banery, 84

in parallel, 2 9 5 - 2 9 7

Ideal co nd u aor, 3 , 4 , 41

passivity principle for ind uaors, 9 0 9

24

Index

n phasor relationships for resistors, capacitors, and, 4 4 9 - 4 5 5 series and parallel inductors and

nodal and loop analysis in i^-domain, 6 3 4 SSS analysis with phasors, 4 8 3 - 4 8 8

n

statements of, 6 1 , 6 3

capacitors, 2 9 3 - 3 0 2 Initial-value theorem, 7 2 6 - 7 2 9 , 7 4 0

Ladder network, 2 0 6 - 2 0 8

Input admittance, 9 7 4 , 10 1 0

Lagging, 4 5 1 , 5 1 8 , 5 2 6

Input attenuation, 1 0 5 1 - 1 0 5 2

Laplace, Pierre Sim on, 543

Input impedance, 9 6 2 , 9 9 0 , 1010

Laplace transform analysis

Instability, 3 6 2

basic signals, 5 4 8 - 5 5 4

5 0 0 - 5 0 5 , 5 3 2 , 533

basic signals and signal

Insulators, 3 , 41 In t^ ra tin g factor m ethod, 3 2 7 , 3 6 4 Integration, Laplace transform m ethod and, 5 7 9 -5 8 0 , 5 7 9 -5 8 2 Integrator, 3 5 9 , 3 6 4 , 6 2 9 , 6 5 8

n

basics, 5 4 3 - 6 0 2

Instantaneous power, 2 2 - 2 5 , 4 1 , 2 7 9 , 3 0 7 , Instantaneous stored e n e ^ , 2 8 0 , 291

/n

representation, 5 9 2 - 5 9 5 circuit responses, applied to

r> o

differential equations, 6 0 0 - 6 0 2 frequency shift property, 5 7 5 - 5 7 6

n

integration, 5 7 9 - 5 8 0 inverse Laplace transform, 5 6 5 - 5 7 5

ideal, 3 6 0

Laplace transform pairs, 564

leaky, 3 5 9 , 3 6 4

one-sided, 5 5 4 - 5 6 3

Int^ro-difFerential equations, 5 8 5 - 5 9 0

overview, 5 4 7 - 5 4 8

Internal resistance, 6 4

by partial fraction expansion, 5 9 8 -6 0 0

Inverse Laplace transform, 5 6 5 - 5 7 5 , 591

properties, 5 8 4

Inverting amplifier, 159, 160, 173, 177

properties, finding, 5 9 5 - 5 9 8

Isolation amplifier, 164

second order tim e domain methods,

Joule, 12

5 4 4 -5 4 7 solution o f integro-difierential

KUo, 2 9

time dif&rendadon formula, 5 7 6 -5 7 8

Kilowatt-hour (kW h) meter, 5 2 6

time-/frequency-scaling property,

o

equations, 5 8 5 - 5 9 0

KirchhofTs current law (K C L ), 5 1 - 1 0 6 , 53, 5 5 - 6 0 , 89 definitions, 5 3 - 5 4 explained, 5 5 - 6 0 first order R L and R C circuits, 3 4 8 for Gaussian curves/surfaces, 59 Laplace transform analysis circuit applications, 6 0 9 Laplace transform m ethod and, 5 5 7 , 591 linearity, superposition, source transformation and, 191 nodal and loop analysis and, 1 0 7 -1 0 8

5 8 2 -5 8 3

V‘i relationship o f capacitor, 5 8 0 - 5 8 2 circuit applications, 6 0 3 - 6 8 1 design o f general summ ing in t^ rators,

652-^57

inductors and capacitors, 6 1 8 - 6 2 5 impedance, admittance, voltage division, source transformations, 6 5 9 -6 6 2 impedance and admittance,

capacitors and, 2 8 7

manipulation, 6 0 9 - 6 1 7 nodal and loop analysis in x-domain, 6 3 4 -6 4 0 op amp integrator design and, 6 8 0

explained, 6 0 - 6 4

response calculation with initial

Laplace transform analysis circuit applications, 6 0 9 Laplace transform m ethod and, 5 5 7 , 591 linearity, superposition, source

o

impulse and step responses, 6 3 1 - 6 3 4

defm itions, 5 3 - 5 4 first order R L and R C circuits, 3 4 8

r\

Thevenin, N orton equivalents,

SSS analysis with phasors, 4 8 3 - 4 8 8

5 3 - 5 5 , 6 0 - 6 4 . 90

o

fluorescence, 6 0 3 - 6 0 4 , 6 2 3 - 6 2 5

impedance and adm ittance, 6 0 5 - 6 0 9

KirchhoflPs voltage law (K V L ), 5 1 - 1 0 6 ,

n

equivalent circuits for initialized

nodal and loop analysis in ^-domain, 6 3 4 statements of, 55

r\

n

conditions, 6 6 8 - 6 7 6 sawtooth waveform generadon and, 6 8 0 -6 8 1

o

switched capacitor circuits and conservation o f charge, 6 4 5 - 6 5 2

transform ation and, 191

r

112^

Index

switched capacitor networks, 6 7 8 - 6 8 0

Linear resistive circuit/networks, 2 1 5

switching problems, 6 7 6 - 6 7 8

Linear resistive element, 2 1 5

transfer functions, 6 2 5 -6 3 0 , 6 6 2 - 6 6 8

Linear voltage sweep, 3 2 2 - 3 2 3

convolution and, 7 7 2 - 7 7 4

Loading effect, 7 6

inverse, and convolution, 8 0 6 - 8 0 7

Loop analysis, 1 0 7 -1 5 4

Laplace transform, defmed, 5 5 4 , 591

concepts of, 1 1 0 -1 1 1

Pierre Laplace and, 543

explained, 1 2 8 -1 3 9

second-order linear networks with

floating voltage sources, 1 2 3 -1 2 8 ,

constant inputs, 4 0 0 transfer function applications, 6 8 3 -7 6 1 applications and bode techniques. 7 6 0 -7 6 1 classification o f responses, 6 9 3 -7 0 1 D C m otors and, 7 3 6 - 7 3 9 electric m otor analysis and, 6 8 3

Laplace transform analysis circuit applications, 6 3 4 - 6 4 0 nodal analysis and, 110, 141 terminology, 1 0 9 - 1 1 0 Loop (closed path), 110, 141

frequency scaling and magnitude

Loop current, 110, 128, 141

initial- and Bnal-value theorems. 7 2 6 -7 2 9 . 7 5 9 -7 6 0

Loop equations, 128, 136, 140 Lossless device, 2 8 0 , 3 0 7 Loss magnitude, 1033, 1072

poles, zeros, 6 8 5 - 6 9 3 , 7 4 2 - 7 4 4

Lossy device, 3 0 7

responses and classifications, 7 4 6 - 7 4 9

Lower half-power frequency, 8 1 6 , 861

SSS for stable networks and systems,

Low-pass (brickwall) filter specification, 1072

7 0 1 -7 0 7

Low-pass Butterworth filter, 6 3 4 .

stability problems, 7 4 5 - 7 4 6 steady-state calculation, 7 4 9 - 7 5 3 L C circuit, 3 7 9 , 3 8 0

See also

Bunerw orth filter Low-pass filters, 10 0 3 , 1072.

See also

Filtering

angular frequency o f oscillation and.

V J

grounded voltage sources, 1 1 1 -1 2 2

frequency response, 7 0 7 - 7 1 4 ,7 5 3 - 7 5 9 scaling, 7 1 4 - 7 2 6

j

1 4 6 -1 4 8 general SSS analysis, 4 9 2 - 4 9 4 history of, 1 0 7 -1 0 8

bode plots, 7 3 0 - 7 3 6

L

Linear m atrix equation, 140, 141

switching in R L C circuits, 6 4 0 - 6 4 5

3 8 4 ,4 1 7

basics of, 1 0 3 4 -1 0 3 6 low-pass filter (brickwall) specification.

differential equation of, 3 8 2

1034

parallel, 385

LP to B P frequency transformation, 1072

undriven, 4 1 8

LP to H P frequency transformation, 1056,

L C resonance frequency, 861

1072

Leading, 451 Leaky integrator circuit, 3 5 9 , 3 6 4 , 5 8 8 - 5 9 0

A C adaptors and, 8 8 3

Linear active r ^ o n , 177

analysis of, with open-circuited

Linear circuit, 7 4 0 .

See also First order

circuits; Second order linear circuits Linear differential equations, 3 2 6 Linearity, 1 9 1 -2 2 5 .

See also Source

transformation; Superposition

coefficient o f coupling and energy

linearity property, 2 1 5 , 4 4 0 , 1099, 1109 linearity theorem, 194 source transformations and, 2 0 8 - 2 1 2 superposition and, 3 4 7 - 3 5 2 , 3 6 5 superposition and proportionality, 2 0 1 -2 0 8 W

w

transformers, 9 5 0 - 9 5 6

equivalent networks and, 2 1 2 - 2 1 3

Laplace transform analysis and, 5 5 7 , 6 9 3

»

9 0 1 -9 0 9 analysis o f circuits containing ideal applications, 9 5 6 - 9 5 7

inverse Laplace transform and, 5 6 5

W

secondary, 8 9 5 - 9 0 0 analysis of, with terminated secondary.

convolution method and, 7 9 3 , 7 9 8 explained, 1 9 3 - 2 0 0

I

M agnetically coupled circuits, 8 8 3 - 9 5 7

Light-em itting diode (L E D ), 189

calculations, 9 0 9 - 9 1 4 com puting equivalent inductances.

Z-Js), responses for simple circuits, 9 4 0 -9 4 3 coupled inductors modeled with ideal transformer, 9 3 0 - 9 3 6 coupling coefficient problems and energy calculations, 9 4 7 - 9 4 9

1126

Index

r\ diflferential equation, Laplace transform, phasor models of, 8 9 3 - 8 9 5 dot placement,

M, and basic equations,

9 3 9 -9 4 0

Modulus, 4 3 5 , 4 8 2 M ultiplication-by-r property, 5 5 9 - 5 6 0 Mutual inductance

r^

(M), 8 8 5 , 9 3 7 . See also

M agnetically coupled circuits

general analysis o f circuits with coupled inductors, 9 4 3 - 9 4 7

dot convention and, 8 8 6 - 8 9 3 rule for induced voltage drop due to, 8 8 7

ideal transformers, 9 1 4 - 9 2 4 magnetically coupled, defined, 883

Nano, 2 9

modek for practical transformers, 9 2 4 -9 3 0

Natural fi-equency o f a circuit, 3 6 5 , 3 8 8 ,

mutual inductance and dot convention,

o

417, 687 Natural modes o f vibration, 3 2 6

8 8 6 -8 9 3 M agnitude (modulus), 4 3 3 , 4 8 2

Natural response, 3 5 2 , 3 6 5 , 7 0 0 , 741

M agnitude response, 7 0 8 - 7 0 9 , 7 4 0

Network equivalency, 3 7 3

M agnitude scale factor, 6 3 0 , 7 1 4 - 7 2 6 ,

Network fim ction, 6 2 6

7 4 0 -7 4 1

Networks

r s

M atching networks, 8 4 8 , 861

equivalent networks, 2 1 2 - 2 1 3

M A TLA B

two-terminal networks, 2 0 9 , 2 1 2 , 2 1 5 ,

first order R L and R C circuits, 3 2 4 , 3 3 2 , 341

2 2 8 ,2 5 7

Laplace transform and partial fraction expansion, 572

Nodal analysis, 1 0 7 -1 5 4 , 110, 141 concepts of, 1 1 0 -1 1 1

n

defined, 110, l 4 l

linearity, superposition, source transformation and, 192

floating voltage sources, 1 2 3 -1 2 8 ,1 4 6 -1 4 8

residue comm and, 671

grounded voltage sources, 1 1 1 -1 2 2 history of, 1 0 7 -1 0 8

M atrix inverse, 141

o

general SSS analysis, 4 9 2 - 4 9 4

series-parallel resistive circuits and, 1 0 4 -1 0 6

r>


M atrix notation, 6 3 6

applications, 6 3 4 - 6 4 0

M atrix partitioning, 9 6 6 , 101 0

linearity and, 195

M axim dly flat, 1 0 5 2 -1 0 5 3 , 1072

loop analysis and, 1 2 8 - 1 3 9

M axim um Power Transfer theorem,

o f pressure-sensing device, 4 7 7 —481

See also N orton

theorem; Thevenin theorem for A C circuits, 5 2 6 defined, 5 3 2

and, 461

r>

terminology, 1 0 9 -1 1 0 Node, 54, 90 Node voltage, 6 0 , 9 0 , 109, I 4 l

Maxwell’s equations, 2 7 3 , 3 0 7

Non-ideal constant current source, 87

Mega, 2 9

Non-ideal constant voltage source, 87

M em ory

Non-ideal dependent current source, 8 7

capacitors and, 2 8 6 inductors and, 2 7 5 Memoryless device, 41

Non-inverting amplifier, 162, 163, 177 Nonlinear controlled source, 80 Normalized Butterworth loss functions, 1072

Mesh current, 128, 141

Normalized circuits, 7 1 4

Metastable, 6 9 0

Normalized fi^quency, 1 0 3 5 - 1 0 3 6

m ho, 3 0 , 4 0 , 41

N orton, E . L , 2 2 7 , 1069

M icro, 29

N orton theorem, 2 2 7 - 2 6 8 .

MilU, 2 9 Modified loop equations, 136 Modified n o ^ analysis (M D A ), 110, 123,

n

Nonlinear, 8 0

Mesh analysis, 128, 141, 461

3 7 9 -3 8 0

r\

Non-ideal dependent voltage source, 87

Mesh, I 4 l

Microwaves, second order linear circuits and,

o

steady-state circuit analysis using phasors

Maxwell, Jam es Clerk, 107

140, 141

r\

with V/I division, SSS analysis, 4 9 0 - 4 9 1

for fi«quency response plots, 4 7 2 —4 7 6

2 4 9 -2 5 6 , 257, 5 3 9 -5 4 2 .

r>

o

See also

M axim um Power Transfer theorem; Thevenin theorem

o

corollary to Thevenin and N ortons theorem’s for passive networks, 2 3 6 equivalent o f one-port, 1010

o

1127

Index

general approach to finding Thcvenin

O utput impedance, 9 8 1 , 9 9 3 , 1010

and N orton equivalents, 2 3 6 -2 4 1

Overdamped circuit, 3 8 9 , 3 9 2 , 4 1 7

Laplace transform analysis circuit applications, 6 5 9 - 6 6 2 maximum power transfer theorem and, 2 4 9 -2 5 6 N orton equivalent circuit, 2 2 9 , 2 4 2 , 2 5 7 for passive networks, 231 Thevenin theorem and active networks, 2 4 1 - 2 4 5

Parallel, 3 5 , 37 Parallel circuit, 90 Parallel connection, 53 two-ports and, 1 0 0 2 -1 0 0 9 Parallel resistance, 6 8 - 7 3 Parallel resonant circuit, 8 4 2 - 8 4 6 Parallel R L C circuit, 8 3 3 - 8 3 8

linear passive networks, 2 2 9 - 2 3 6

Parsevals theorem, 1098

op amp circuits, 2 4 6 - 2 4 9

Partial fi^ction expansions, 5 6 5 , 591

SSS analysis, 4 9 1 - 4 9 2 two-ports and, 9 6 4 - 9 6 8

distinct complex poles, 5 7 3 - 5 7 5 repeated poles, 5 6 9 - 5 7 2 Particular integral, 3 5 2

Octave, 7 3 3 , 741

Panitioned matrix, 9 6 6 , 1010

Odd (u n a io n , 1094, 1109

Passive analog filters, 1072

OfF-set terminal, 157

Passive circuit, 3 3 9

O hm , 2 9 , 41

Passive elements, 178

O hm ’s law, 3 0 - 3 7 , 41 linearity, superposition, source transformation and, 191 phasor relationships for resistors, inductors, capacitors, 4 4 9 - 4 5 5 O ne-port networks, 9 6 1 , 9 6 2 - 9 6 8 ,

1011-1012 “O n the T heory o f Filter Amplifiers” (Butterw orth), 1034

theorems for, 2 3 6 N ortons theorem for, 231 Thevenin’s theorem for, 231 Passive R L C circuit, 3 6 5 Passive sign convention, 1 5 - 1 7 , 2 2 , 2 3 , 27, 3 0 ,4 1 , 54 Passivity principle for inductors, 9 0 9

Open-circuited secondary, magnetically

Peak frequency (center frequency), 8 1 5 , 861 Peak-to-peak value, 10, 41

O pen-circuit impedance parameters, 1010

Peak value, 10, 41

O pen-circuit output admittance, 9 8 7 , 1010

Periodic current waveform, 2 7 9 - 2 8 0

O pen-circuit voltage, 231

Periodic signal, 6 9 5 , 7 4 1 , 1088, 1109

O pen-loop gain, 172, 177

Period o f waveform, 4 3 3

Operational amplifier (op amp), 121, 1 5 5 -1 8 9

Phase response, 741

design o f general summ ing amplifiers, 1 6 7 -1 7 0

Phasors.

See abo Sinusoidal steady state (SSS)

defined, 4 8 2

first-order R C op amp circuit, 3 5 7 - 3 6 3

phasor analysis, 4 3 5

idealized, 1 5 7 -1 6 6

relationships for resistors, ind uaors,

op amp circuits and SSS analysis, 4 9 5 -4 9 7 op amp open-loop gain, 741 phasor m ethod for analyzing, 4 6 0 - 4 6 1 , 4 7 5 -4 7 6 saturation and active region of, 1 7 1 -1 7 6 Thevenin and N orton equivalent circuits for op amp circuits, 2 4 6 - 2 4 9

capacitors, 4 4 9 —455 representations o f sinusoidal signals, 4 4 7 -4 4 9 steady-state circuit analysis using, 4 6 1 -4 6 6 Photo timer, 3 7 7 Pico, 29

Ordinary differential equation model, 591

Piecewise linear curve, 7 3 4 , 741

Oscillation

Piecewise linear relationship, 1 7 1 -1 7 2

natural frequency o f a circuit, 3 6 5

Jt-equivalent circuit, 9 0 8 , 9 3 8 , 1010

oscillator circuit, 3 8 5 , 4 1 7

Planar circuit, 128

sinusoidal, 378

Polar coordinates, 4 3 6 , 4 8 2

W ien bridge, 4 1 2 , 4 1 4 , 4 7 5 - 4 7 6

Poles, 5 5 7

O utput admittance, 9 7 4 , 1010

w

corollary to Thevenin and N ortons

O pen circuit, 3 0 , 3 7 , 41 coupled circuits and, 8 9 5 - 9 0 0

'< J

op amp and, 156 Passive networks

band-pass transfer funcdon vwth, 8 5 1 -8 6 0

1128

Index

n

n

See RL/RC circuits

distinct complex poles, 5 7 3 - 5 7 5

R C circuit.

Laplace transform analysis transfer

R C to C R transformation, 1062, 1072

function applications, 6 8 5 - 6 9 3 ,

Reactance, 4 5 6 , 4 8 2 , 8 4 3

7 4 2 -7 4 4

Reactive power, 5 1 0 - 5 1 5 , 5 3 2

pole frequency, 8 1 9

Real part, 4 3 5 , 4 8 2

pole Q. 8 5 2 , 861

Real power, 5 3 2

pole-zero, 8 1 6

Reciprocity

o f rational fiinction fmite, 5 9 1 ,6 8 5 ,7 4 1 simple, 5 9 1 , 741 repeated, 5 6 9 - 5 7 2 , 6 8 5 Potential difference, 12, 16 Power, 1 6 - 2 5 , 4 1 .

See abo Sinusoidal steady

state (SSS)

reciprocal networks, 9 9 8 , 1010 two-ports and, 9 9 7 - 1 0 0 2 , 1 0 2 9 -1 0 3 0

Region o f convergence (R O C ), 5 56, 591 Residue theorem, 5 65, 5 6 6

average power and SSS power

Resistance, 3, 3 0 , 4 1 , 4 8 2 converter, 2 4 7

complex, 5 1 0 - 5 1 5 , 5 3 1 , 5 3 6 - 5 3 7

equivalent, 6 5 , 6 9 , 70

conservation o f power, 19, 2 0 , 3 6 , 4 0 ,

internal, 64 negative, 6 8 , 2 4 6

half-power frequencies, 815

output, 2 5 0

half-power points, 1033, 1072

parallel, 6 8 - 7 3

instantaneous, 2 2 - 2 5 , 4 1 , 2 7 9 , 3 0 7 ,

phasor admittance/impedance, 4 5 6 Resistivity, 3 3 , 41

M axim um Power Transfer theorem,

Resistors. 15, 16, 2 2 , 3 0 , 3 1 , 4 1 ideal (linear), 15, 37

power gain, 80

impedance of, 6 0 7 - 6 0 8 , 6 5 8

reactive, 5 1 0 - 5 1 5 , 5 3 2

nonlinear, 41

real, 5 3 2

phasor relationships for inductors,

upper half-power frequency, 8 1 6 , 8 6 2 Power factor (pO. 5 1 7 - 5 2 6 , 5 3 2 . 5 3 7 - 5 3 9

Resonance.

See abo Band-pass circuit

phenomenon, 8 3 9 - 8 4 2

Practical inductors, 8 3 0

resonance frequency, 8 6 2

Practical source, 80

series and parallel resonant circuits,

Primary coils, 8 9 8 , 9 1 7 , 9 3 8 Product over sum rule, 71 Proportionality, 2 0 1 - 2 0 8

8 4 2 -8 4 6

r\ r\ r\ n

classifications, 3 5 2

natural frequency o f a circuit and, 215

complete, 3 5 2 , 3 6 4 forced, 3 5 2 , 3 6 4

Pulse function, 549

r\

series-parallel resonant circuits, 8 4 6 -8 5 1 Response

property, 215 proportionality property, 2 0 1 - 2 0 8

r\

capacitors, and, 4 4 9 - 4 5 5

Power gain, 80

Practical transformers, 9 2 4 - 9 3 0

r>

transfer, 79

lower half-power frequency, 8 1 6 , 861 2 4 9 -2 5 6 , 257, 526, 532, 5 3 9 -5 4 2

o

7 9 4 -7 9 5 Rectangular coordinates, 4 3 6 , 4 8 2

available, 528

5 0 0 - 5 0 5 , 53 2 , 533

r>

Rectangular approximations to signals,

Repeated pole, 685

5 1 5 . 5 3 1 ,5 3 7

r\

reciprocal two-port, 1010

apparent, 5 1 0 - 5 1 5 , 531

calculation, 5 0 0 - 5 0 5 , 5 33, 5 3 4 - 5 3 6

n

natural, 3 5 2 , 3 6 5 source-free, 3 2 8 - 3 3 6 , 3 6 5 step, 3 3 6 - 3 4 7 , 365

Quality factor, 8 1 6 o f a band-pass circuit, 861 o f a capacitor, 861 o f a coil, 861 o f L and C components, 8 3 0 - 8 3 3 o f a reactive com ponent, 861

unstable, 3 3 5 , 3 6 5 zero-input, 3 2 8 - 3 3 6 , 3 5 2 , 3 6 5 zero-state, 3 5 2 , 365 Reverse open-circuit voltage gain, 9 8 6 - 9 8 7 ,

1010 Reverse voltage gain, 9 9 2 R L C circuit.

Ramp function, 55 0 , 5 6 0 , 5 9 1 , 741 Rational function, 5 6 5 , 6 8 5 , 741

r\

See abo Second order linear

circuits computing SSS response and, 4 4 2 —444

n

1129

Index

damping and, 3 9 2

Series connections, 1 0 0 2 -1 0 0 9


Series inductors, 2 9 3 - 2 9 5 Series-parallel circuit, 73

applications, 6 4 0 - 6 4 5

Series-parallel com binations

parallel, 8 3 3 - 8 3 8 passive R C or R L circuits

vs., 3 9 7 - 4 0 0

source-free, 3 8 3 - 4 0 0

Series-parallel impedance

steady-state analysis of, 6 0 4 switching in, 6 4 0 - 6 4 5 RL/RC circuits, 3 2 2 - 3 7 8 .

with V/I division, SSS analysis, 4 8 8 —490 Series-parallel resistive circuit

See also First order

circuits

O '

capacitors, 301 inductors, 2 9 7 - 2 9 8

resonance and, 8 3 9

definitions, 5 1 - 1 0 6 , 5 3 - 5 4 dependent sources and, 7 8 - 8 4 , 1 0 3 -1 0 4

analysis and theory, 3 5 2 - 3 5 7

explained, 6 4 - 6 8

com puting SSS response and, 4 4 2 - 4 4 4

M A TLA B and, 1 0 4 - 1 0 6

D C or step response of, 3 3 6 - 3 4 7

model for non-ideal battery and, 8 4 - 8 5

first-order R C op amp circuits, 3 5 7 - 3 6 3

non-ideal sources and, 8 5 - 8 8

m ath used for, 3 2 4 - 3 2 8

parallel resistance and current division,

passive R L C circuit, 3 6 5 sawtooth, 3 2 2 - 3 2 3 , 3 6 5 source-free or zero-input response, 3 2 8 -3 3 6 steady-state circuit analysis using phasors, 4 6 1 -4 6 6

6 8 -7 3 , 9 6 -9 8 RgO and related calculations of, 9 9 - 1 0 3 series-parallel interconnections, 7 3 - 7 8 Series-parallel resonant circuit, 8 4 6 -8 5 1 Series-parallel R LC circuit, 3 8 6 - 3 8 7

superposition and linearity, 3 4 7 - 3 5 2 Root-mean-square (rms), 10, 4 1 , 50 6 , 53 2 .

See also Effective value (Thevenin’s equivalent resistance), 2 5 7 .

See also Thevenin theorem Running average, 7 6 3 , 7 8 8 , 7 9 0 , 7 9 8

W

Series resonant circuit, 8 4 2 - 8 4 6 Shifted step, 5 4 9 , 591 Short circuit, 2 7 , 3 0 , 4 1 , 64 Short-circuit admittance parameters, 9 69, 1010 Short-circuit forward current gain, 9 86, 1010 Short-circuit input impedance, 9 8 6 , 1010 Siemens, 2 9 , 4 0 , 41

Sallen and Key circuit, 1046

Sifting property, 55 3 , 5 9 1 , 7 6 7 , 7 9 8

Sallen and Key normalized low-pass

Simultaneous equations, 9 3 0

Butterworth filter, 6 3 4 - 6 3 5 , 8 5 6 - 8 5 8 , 1 0 4 5 -1 0 5 1 .

See also Butterworth filter

Saraga design, 1 0 4 7 -1 0 5 0 Saturation, 164, 171 Saturation regions, 178 Sawtoodi waveform, 3 2 2 - 3 2 3 , 3 65, 6 8 0 -6 8 1

Single-frequency analysis, 5 0 9 - 5 1 0 Sinusoidal oscillation, 3 7 8 , 4 1 2 - 4 1 5 Sinusoidal steady state (SSS) analysis, 4 3 1 —497 complex exponential forcing functions in, 4 4 4 - 4 4 7

Scaled sum o f waveforms, 4 1 7

complex numbers and, 4 3 5 -4 4 1

5-domain

defined. 4 3 3 , 4 8 2

Laplace transform analysis circuit applications, 6 3 4 - 6 4 0

elementary impedance concepts, 4 4 9 -4 5 5

nodal and loop analysis in, 6 3 4 - 6 4 0

frequency response and, 4 6 7 - 4 7 6

poles, zeros, and, 6 8 5 - 6 9 3 , 7 4 2 - 7 4 4

as high-accuracy pressure sensor

Secondary coils, 8 9 8 , 9 1 7 , 9 3 8 Second order linear circuits, 3 7 9 —430 with constant inputs, 4 0 0 -4 1 1 defined, 4 1 7 discharging a capacitor through an inductor, 3 8 2 - 3 8 5 oscillator application, 4 1 2 - 4 1 5 source-free, 3 8 5 - 4 0 0 Second order time domain methods, 5 4 4 -5 4 7 Seleaivity, 8 1 6 , 8 6 2 Semiconductors, 4 , 5 Series circuit, 90

application, 4 3 1 —4 3 2 Laplace transform analysis and, 7 0 1 -7 0 7 naive technique for computing, 442^44 nodal analysis o f pressure-sensing device, 4 7 7 -4 8 1 phasor impedance and admittance, 4 5 5 -4 6 1 phasor representations o f sinusoidal signals, 4 4 7 —449 o f R L C circuit, 6 0 4

1130

Index

r> using phasors, 4 6 1 —4 6 6

Square wave, 3 4 5 , 3 4 6

power calculation, 4 9 9 - 5 4 2

Stable circuit, 4 3 3

complex power and, 5 1 0 - 5 1 5 complex power conservation, 5 1 3 -5 1 7 effective value o f signal and average

defined, 4 8 2 stability and boimdedness, 6 9 0 Stable transfer function, 741 Steady state, 4 3 3 .

power, 5 0 6 - 5 0 9 instantaneous and average powers,

n

See also Fourier series

Steady-state analysis, 6 0 4 , 6 5 8 Steady-state calculation, 7 4 9 - 7 5 3

5 0 0 -5 0 5 m aximum power transfer in SSS, 5 2 6 -5 3 0

Steady-state circuit output response, 701 Steady-state response, 6 9 6 , 741

power factor/power (actor correction, 5 1 7 -5 2 6

Step function, 4 1 7 Step response, 3 6 5 , 4 1 7 , 6 3 1 - 6 3 4 , 741

single-frequency analysis with effective values, 5 0 9 - 5 1 0

Stored energy, 3 7 6 , 9 1 0 - 9 1 4 Stray capacitance, 3 6 1 , 3 6 2 , 3 6 5

Sm oothing, 2 6 9 - 2 7 0 , 3 0 3 - 3 0 5

Supermesh, 136

Solar cell, 3 7 6

Supernode, 123

Source, 16

Superposition, 1 9 1 -2 2 5 , 3 6 5 .

n See also

Linearity; Source transformation

controlled, 2 8 , 2 9 , 7 8 , 80 current, 56

o f average power, 5 0 3 - 5 0 5

current-controlled ciurent (C C C S), 28, 39

linearity and, 1 9 3 -2 0 0 , 3 4 7 - 3 5 2

current-controlled voltagp (C C Y S), 28, 39

proportionality and, 2 0 1 - 2 0 8

floating, 1 2 3 -1 2 8 , 140, 1 8 8 -1 8 9

source transformations and, 2 0 8 - 2 1 2

ideal current, 2 7 , 38

source transformations and equivalent

n n

networks, 2 1 2 - 2 1 3

independent current, 2 7 , 38 independent (ideal) current, 2 7 , 38

superposition property, 2 0 1 , 215

r\

independent (ideal) voltage, 2 5 - 2 7 , 3 7

Susceptance, 4 8 2 , 8 6 2

nonideal, 8 5 - 8 8

Switched capacitor circuit, 6 4 5 - 6 5 2 , 6 5 8

nonlinear, 80

Switching first order R L and R C circuits, 3 2 3 , 3 4 5

nonlinear controlled, 80

in R L C circuit, 6 4 0 - 6 4 5

practical, 2 6 , 7 9 , 8 0 , 81

v-i characteristics, 8 0 , 86

Sym m etric matrbc, 141

voltage, 6 4

Sym m etry properties o f Fourier series, 1094,

n

1109

Source free, 4 1 7 Source-free response, 3 2 8 - 3 3 6 , 365

System identification, 7 6 5

r>

Tank circuit, 8 4 3 , 8 6 2

r\

Source-free second order linear circuits, 3 8 5 -4 0 0 Source transformation, 1 9 1 -2 2 5 .

See also

Linearity; Superposition defmed, 2 1 5

Tank frequency, 8 4 3 , 8 6 2 Temperature measurement, nodal and loop

r\

analysis, 143

explained, 2 0 8 - 2 1 2

T-equivalent circuit, 9 0 8 , 9 3 8 , 1010


Tera, 29

applications, 6 5 9 - 6 6 2 linearity and, 1 9 3 -2 0 0

Term inated secondary, magnetically coupled circuits and, 9 0 1 - 9 0 9

linearity and superposition, 2 0 1 - 2 0 8

Term inated two-ports, 9 7 3 , 1010

source transformation theorem and

Thevenin, M . L ., 2 2 7

equivalent networks, 2 1 2 - 2 1 3 source transformation theorem for independent sources, 2 0 9 Source transformation property, 6 1 5 , 6 5 8 S P IC E (Simulation Program with Integrated Circuit Emphasis), 178 first order R L and R C circuits, 3 4 6 for frequency response plots, 4 7 2 - 4 7 6 W ien bridge oscillator and, 4 1 4

r\

Thevenin theorem, 2 2 7 - 2 6 8 .

See also

M axim iun Power Transfer theorem; N orton theorem corollary to Thevenin and N ortons theorem s for passive networks, 2 3 6 equivalent o f a one-port, 1010 general approach to finding Thevenin and N orton equivalents, 2 3 6 -2 4 1

n

1131

Index

O Laplace transform analysis circuit applications, 6 5 9 - 6 6 2 M axim um Power Transfer theorem and, 2 4 9 -2 5 6 N orton theorem and

Transfer resistance, 7 9 Transformers.

(Thevenin’s equivalent resistance), 2 5 7 SSS power calculations and, 4 9 9 steady-state circuit analysis using phasors and, 4 6 4 - 4 6 6

Transient analysis, 5 4 4 , 6 5 8 Transient response, 4 3 3 , 6 9 6 , 741 Translation o f a plot, 1 0 9 2 - 1 0 9 3 , 1109

Thevenin equivalent resistance, 231

Translation property o f Fourier series,

3 dB bandwidth, 8 1 5 - 8 1 6 , 861

1 0 9 3 -1 0 9 4 , 1109 Transmission

{t-) parameters, 9 6 1 , 9 9 1 - 9 9 4 ,

1010, 1 0 2 3 -1 0 2 5 Triangular waveform, 3 5 5 - 3 5 7 Trigonom etric Fourier series, 1089, 1109

o f B P filter, 1071

Tuned circuit, 8 4 4 , 8 6 2

o f LP filter, 1071

Two-dependent source equivalent circuit,

3 dB down point, 1033 3 dB firequency, 7 3 6 , 7 4 0

9 7 2 - 9 7 3 , 1010 Two-ports, 9 5 9 - 1 0 3 0

Three-term ind device, 8 0 , 9 0 8

admittance parameters, 9 6 8 - 9 7 3

T im e constant, 3 2 9 , 3 6 5

amplifiers, 9 5 9 - 9 6 0

T im e differentiadon formula, 5 7 6 - 5 7 8

general relations am ong two-port

T im e dom ain circuit response com putation.

See Convolution T im e invariance, 6 5 8 , 7 9 3 , 7 9 8

parameters, 9 9 4 - 9 9 7 A-parameters, 9 6 1 , 9 8 5 - 9 9 1 , 1 0 2 0 -1 0 2 3 impedance and gain calculations o f

Tim e-scaling property, 5 8 2 - 5 8 3

terminated two-ports modeled by

T im e shift property, 5 5 8 - 5 5 9 , 5 9 1 , 7 6 7

z-parameters, 9 8 1 - 9 8 4

Touch-tone phones, 8 1 1 - 8 1 2

impedance parameters, 9 7 6 -9 8 1

Transconductance, 7 9

one-port networks, 9 6 1 , 9 6 2 - 9 6 8 ,

Transfer admittance, 6 2 6

1 0 1 1 -1 0 1 2

Transfer conductance, 7 9

parallel, series, cascaded co n n eaio n s of,

Transfer function, 6 2 5 - 6 3 0 , 6 5 8 , 6 8 3 - 7 6 1

1 0 0 2 -1 0 0 9 parameter conversion and inter

applicadons and bode techniques, 7 6 0 -7 6 1 band-pass circuits, 8 6 3 - 8 6 8 , 8 7 6 - 8 7 9

connection o f two-ports, 1 0 2 5 -1 0 2 9

bode plots, 7 3 0 - 7 3 6

reciprocity, 9 9 7 - 1 0 0 2 , 1 0 2 9 -1 0 3 0

classification o f responses, 6 9 3 -7 0 1

^-parameters, 9 6 1 , 9 9 1 - 9 9 4 , 1 0 2 3 -1 0 2 5

D C m otors and, 7 3 6 - 7 3 9

transmission parameters, 9 9 1 - 9 9 4

electric m otor analysis and, 6 8 3

^ p aram eter analysis o f terminated

frequency response, 7 0 7 - 7 1 4 , 7 5 3 - 7 5 9 frequency scaling and magnitude scaling, 7 1 4 -7 2 6 initial- and finai-value theorems, 7 2 6 - 7 2 9 ,7 5 9 - 7 6 0 Laplace transform analysis circuit applications, 6 2 5 - 6 3 0

-V

ideal, 9 1 4 - 9 2 4 , 9 3 0 - 9 3 6 , 9 5 0 - 9 5 6 practical, 9 2 4 - 9 3 0

Transistor photo timer, 3 7 7

231 two-ports and, 9 6 4 - 9 6 8

w

defined, 885

Thevenin equivalent, deBned, 2 2 9 , 2 4 2

Thevenin theorem for passive networks,

u

See also M agnetically coupled

circuits

superposition and, 3 4 9

Thevenins equivalent circuit, 2 5 7

Vw^

steady-state calculation, 7 4 9 - 7 5 3 Transfer impedance, 6 2 6

linear passive networks, 2 2 9 - 2 3 6 SSS analysis, 4 9 1 - 4 9 2

'O

7 0 1 -7 0 7 stability problems, 7 4 5 - 7 4 6

active networks, 2 4 1 - 2 4 5 op amp circuits, 2 4 6 - 2 4 9

L >

SSS for stable networks and systems,

two-ports, 9 7 3 - 9 7 6 , 1 0 1 2 -1 0 1 6 z-parameters, 1 0 1 6 - 1 0 2 0 Two-terminal circuit clem ent, 58 Two-terminal device, 15, 3 0 Tw o-terminal networks, 2 0 9 , 2 1 2 , 2 1 5 Thevenin and Nortons theorems, 228, 257

V 257

Laplace transform m ethod and, 5 4 4 poles, zeros, 6 8 5 - 6 9 3 , 7 4 2 - 7 4 4

Unbounded voltage or current, 3 0 7

responses and classifications, 7 4 6 - 7 4 9

Undamped circuit, 3 8 5 , 4 1 7 , 8 6 2

1132

n

Index

n Undamped natural frequency, 8 4 3 , 8 6 2

Wattage, 19, 41

Underdamped circuit, 3 9 0 , 3 9 2 , 4 1 7

Waveform

U nit impulse function, 5 5 2 , 591

effective value of, 505

U nit step fu n a io n , 3 2 4 - 3 2 5 , 3 6 5 , 5 4 8 , 591

periodic current, 2 7 9 - 2 8 0

r\ n

U nity coupling, 9 3 1 , 9 3 8

period of, 4 3 3

Universal resonance curves, 8 2 6 , 8 6 2

sawtooth, 3 2 2 - 3 2 3 , 3 6 5

Unstable response, 3 3 5 , 3 6 5

scaled sum o f waveforms, 4 1 7

Upper half-power frequency, 8 1 6 , 8 6 2

second order linear circuit and, 3 8 3 - 3 8 5

n

sinusoidal voltage, 4 1 2

v-i characteristic, 2 6 , 3 7 - 3 8 , 41 v-i relationship o f capacitor Laplace transform m ethod, 5 8 0 - 5 8 2 Virtual ground, 160, 178

triangular, 3 5 5 - 3 5 7 , 6 8 0 -6 8 1 W ien bridge oscillator, 4 1 2 , 4 1 4 , 4 7 5 —476 W ire, resistance o f copper wire, 33.

See also

Resistivity

V inually grounded, 160 Virtual short circuit model, 1 5 8 ,1 6 0 , 1 7 7 ,1 7 8

V 257

j^-parameters 9 7 3 - 9 7 6 ,1 0 1 2 - 1 0 1 6

Voltage, 2, 16 branch, 107

n

analysis o f terminated two-ports, z-parameters and, 9 7 6 - 9 8 1

gain, 117 node, 6 0 regulators, 3 0 3 Voltagp division, 9 0 ,2 9 5 ,2 9 9 ,3 0 0 ,4 8 2 ,6 5 8

Zero finite zero, 6 8 5 Laplace transform analysis transfer

formula, 6 6 , 4 5 6

function applicadons, 6 8 5 - 6 9 3 ,


7 4 2 -7 4 4

applications, 6 5 9 - 6 6 2 Voltage drop, 10 Voltage follower, 1 64, 178

Zero-input response, 3 2 8 - 3 3 6 , 3 5 2 , 3 6 5 ,

Voltage gain, 80

Zero-state response, 3 5 2 , 3 6 5 , 6 9 3 , 7 4 1 , 7 9 8

Voltage (potential difference), 41

Z Js). 9 4 0 - 9 4 3 2-parameters, 9 7 6 - 9 8 1 , 1 0 1 6 -1 0 2 0

Voltage regulator, 3 0 7

n

4 8 2 , 6 9 3 , 741 Zeros o f rational function, 5 6 6 , 5 91, 741

Voltage source, 4 1 , 6 4 nonideal, 8 6

v-i charaaeristics, 8 0 , 8 6

n

Voltage-to-current converter, 188 Voltage transformation property, 9 1 4 , 9 3 1 , 933, 937, 938

n

Voltmeter, 14, 7 6 sensitivity, 7 7

n o n

r s

n

o

o

Linear Circuit Analysis- DeCarlo-Lin 3e

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