AP2212 – Lecture AP2212 – Lecture 9 AC analysis 2
Course organization TSANG, Stephen E-mail: Tel: Office:
[email protected] 3442-4618 P6706
2
Lecture 08 - Review
Managed several examples of circuits with Resistive (R ), ), Capacitive (C (C ) and inductive elements (L (L). C
v
v R
d I V L dt
Ri
I P R sin sin(w t )
Q
v L
w LI
P
cos(w t )
v C
-
V
i
I p C
dV C dt
cos(w t )
w
Reactance (denoted X, Reactance (denoted X, units of ohms ohms)) is the ratio (magnitude only) voltage to current: measures how a component opposes the flow of electricity X R = R X L = w L X C = 1/w C Impedance, Z (magnitude Impedance, (magnitude and phase) of the current Z R = R Z L = j = jw L Z C = - j X X C 1 = 1/jw C
2
Lecture 08 - Review
Managed several examples of circuits with Resistive (R ), ), Capacitive (C (C ) and inductive elements (L (L). C
v
v R
d I V L dt
Ri
I P R sin sin(w t )
Q
v L
w LI
P
cos(w t )
v C
-
V
i
I p C
dV C dt
cos(w t )
w
Reactance (denoted X, Reactance (denoted X, units of ohms ohms)) is the ratio (magnitude only) voltage to current: measures how a component opposes the flow of electricity X R = R X L = w L X C = 1/w C Impedance, Z (magnitude Impedance, (magnitude and phase) of the current Z R = R Z L = j = jw L Z C = - j X X C 1 = 1/jw C
3
Lecture 08 - Review
Graphical representation of complex impedance Impedances combine in the same way as resistors complex impedances can be Impedance are added , subtracted , multiplied and and divided in the same way as other complex quantities They can also be expressed in a range of forms such as the rectangular , polar and and exponential exponential forms forms
4
Power, voltage, and current gains
Power gain is defined as: Gain
=
10log
P in
dB
÷
Voltage gain is defined as: (why?) Gain = 20log
P out
V out V in
dB
÷
Current gain is defined as: (why?) Gain = 20log
I out I in
dB
÷
5
Power, voltage, and current gains
Pin,
Vin
X Rin
Proof
Pout,
Vout
Rout
that power gain = voltage gain if R in equal to R out
6
Lecture 09 - Outline Application and engineering manipulation of RCL circuit
High-pass RC filter
Filter representation
Bode diagram, phasor diagram
Transfer function
Low-pass RL filter
Resonance RLC circuit
Charging and discharging in RCL circuit
7
Application of RCL circuit Still remember how does a radio works?
1.
Convert sound to electrical signal
4.
Received by an antenna
3.
2.
Modulate with carrier frequency
Transmit in air AM
FM How
to extract the sound? Source: National Radio and Astronomy
Observatory
8
Application of RCL circuit
FM receiver In human, the audible range of frequency: 20Hz to 20kHz FM carrier frequency: 88-108MHz
There
are multiple channels in air: e.g. 88.1MHz, 99.7MHz We have to find a way to select the frequency We
concern: The frequency response The resonance(center) frequency The width of the spectrum (Bandwidth) The amplification (Gain) A
resonance circuit
Frequency
response
9
A High-Pass RC Filter
Consider the following circuit
We want to know what is its function?
If we know : v input
find V p sinw t v output
This circuit is commonly drawn in the following way:
?
10
RC Filter 1/2 1) Use complex numbers:
vi
=
V p sin t ® vi
=
j t
V p e
~ Z R - j / w C R 2 (1/ w C )2 e - j
2) Find impendence:
where : tan-1
~ ~ ~ i v i / Z
3) Calculate current:
1 w RC
(w )
V pe j (w t ) R 2 (1/ w 2C 2 )
i I p sin(w t )
(w I p )
11
RC Filter 2/2 v i
~ V p sinw t v i
Find output:
v o
Define:
v o v i
iR
V pe j
V pR 2
R
2 2
1 w 2
2
2
sin(w t )
2
(1/ w C )
= RC , then:
w
t
w
(w
v o
v i
2
w
V pw
2 2
w
v o
V pRC w
)
2
2
R C
sin(w t ) 1
sin(w t ) 1
This is a high pass filter!
12
High-pass Filter (as function of
w )
For High-pass RC filter , output is taken from resistor Since :
v o v i
2
If :
w 1
If :
w 1
If : w 1 The
2
w
2
1 w v o v i v o v i
v o v i
2
, Then high f signals can pass
1
0,
Then low f signals can' t pass
1
2
0.707
frequency times the time constant is equal to 1
13
High-pass Filter (as function of f ) = RC has units of (1/angular frequency)
(Remember definition of w :
w
= 2p f)
then we can write: =1/( 2p f C ), and w
= (2p f )/(2p f C ) = f /f C )
the gain response has two asymptotes that meet at the cutoff frequency figures of this form are called Bode diagrams(plots)
14
High-pass Filter (as phasor diagrams)
The behaviour in these three regions can be illustrated using phasor diagrams
15
Differentiating Circuit v i
As we saw before : v R
iR
V p sinw t
V p sin(w t )
1 (w RC )
-
2
Therefore it w 1/RC tan 1/w CR p /2
v R
V p sin(w t )
1 (w RC )
v o
-
2
v R
w RCV p
RC
cos w t
dv i dt
This is a differentiating circuit!
16
Transfer function
Consider the potential divider shown here
From considerations on this circuit: v o
v i
Z2 Z1
Z2
rearranging, the gain of the circuit is
v o v i
Z2 Z1
Z2
this is also called the transfer function of the circuit
17
A Low-Pass RL Network
Low-pass networks can also be produced using RL circuits
they behave similarly to the corresponding CR circuit the voltage gain is v o v i
Z R Z R
ZL
1
R
the cut-off frequency is
R jw L
w c
R L
1 jw
1
L R
rad/s
f c
w c
2p
R
2p L
Hz
18
A High-Pass RL Network
High-pass networks can also be produced using RL circuits
these behave similarly to the corresponding CR circuit the voltage gain is v o v i
ZL Z R
the cut-off frequency is
ZL
w
c
jw L R jw L
R L
1
1
1
R
jw L
rad/s
1
1 - j
f c
R w L
w c
2p
R
2p L
Hz
19
A Comparison of RC and RL Networks
Circuits using RC and RL techniques have similar characteristics
20
Make filters to condition a signal High-pass
filter
Low-pass
filter
21
Noise spectrum (From L5)
22
Series RLC Circuits and Resonance
the impedance is given by Z
R jw L
1 jw C
R j(w L -
1
)
w C
if the magnitude of the reactance of the inductor and capacitor are equal, the imaginary part is zero, and the impedance is simply R this occurs when w L
1 w C
2
w
1
LC
w
1
LC
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Resonance in RLC circuits The condition
w
1
is known as resonance
LC
We define
w
LC
as the resonant angular frequency
The resonant frequency is f o
1
1 2p LC
in the series resonant circuit, the impedance is at a minimum at resonance the current is at a maximum at resonance
24
Quality factor The resonant effect can be quantified by the quality factor, Q
Q is the ratio of the maximum energy stored to the energy dissipated in each cycle Q
max. =
energy dissipated in one cycle
it can be shown that: and:
energy stored
L Q R C 1
Quality factor Q
WHY?
X L R
X C R
What happens when R 0?
25
RLC Resonance
Current resonance occurs for RLC series circuit. w o is called resonance frequency. w
o
1
LC
f o
1 2p LC
The resonance peak at w w o is prominent and sharp for lower resistance values. Why?
Q
L R C 1
26
Key points
A combination of resistor (R), capacitor (C), and inductor (L) can be used to construct filters and resonance for electrical signals. In such frequency dependent system, we concern about the cutoff frequency, phase changes, quality(Q) factor. Bode plot and phasor diagram are effectively ways to represent the frequency response of the circuit. In a resonance circuit, the Q fact is maximum when the reactance of C and L are equal.
27
Transient response
We have looked at the behavior of systems in response to:
Fixed DC signals (L06)
“Constant” AC signals (L07-L09)
What happens before these circuits reach “steady-state”?
this is referred to as the transient response
What happens to the circuit on the right when at t=0 the switch is closed? 1. Was the capacitor charged or discharged at t=0-? 2. What is the value (phase) of V at t=0?
28
Charging Capacitors (1/2)
Kirchhoff’s voltage law:
iR v
In a capacitor we have:
i
Therefore : CR
C
V
dv dt
dv v V dt
(First-order differential equation with constant coefficients) Assuming V C = 0 at t = 0, this can be solved to give: Also since i = C( dv /dt ) (still assuming V C = 0 at t = 0), then
-
v V (1 - e
-
i I e
t CR
-
) V (1 - e )
t CR
t
-
I e
t
(I=V/R)
29
Charging Capacitors(2/2)
Thus both the voltage and current have an exponential form
-
v V (1 - e
t CR
-
t
) V (1 - e )
-
i I e
t CR
-
I e
t
30
Energizing Inductors
A similar analysis of an RL circuit gives -
v V e
Rt L
-
V e
t
where I = V/R
-
i I (1 - e
Rt L
-
)I (1 - e
t
)
31
Discharging Capacitors
Consider this circuit for discharging a capacitor (At t = 0, V C = V )
Kirchhoff’s voltage law:
Then:
CR
dv dt
iR v
0
v 0
Solving this equation as before gives: -
v V e -
i -I e
t CR
-
CR
V e
t
-
-I e
(I = V/R )
t
t
32
De-energizing Inductors
A similar analysis of this circuit gives -
v -V e
Rt L
(I = V/R)
-
-V e
t
-
i I e
Rt L
-
I e
t
33
A comparison of the four circuits
34
Response of First-Order Systems
Initial and final value formulae
Increasing or decreasing exponential waveforms (for either voltage or current) are given by: v V f (V i - V f )e
- t /
i I f (I i - I f )e
- t /
V i and I i are the initial values of the voltage and current
V f and I f are the final values of the voltage and current
1. The first term in each case is the steady-state response 2. The second term represents the transient response 3. The combination gives the total response of the arrangement
35
Tutorial The input voltage of this CR circuit changes from 5 V to 10 V at t = 0. What is the output voltage?
36
About exponential curves
37
Output of first-order systems to a square waves [for different time response (T)] see
38
Output of first-order systems to a square waves [for different frequencies ( f )] See
39
Second-Order Systems Circuits with capacitance and inductance result in second-order differential equations.
for example, the circuit:
is described by the equation: 2
LC
d v C dt
2
RC
dv C dt
v C
V
Second order systems also have transients.
They will be more complex than first order systems
Transient solutions depend on the equation’s coefficients useful to find an equation’s “general form”
40
Second order differential equation When a step input is applied to a second-order system, the form of the resultant transient depends on the relative magnitudes of the coefficients of its differential equation. The general form of the response is 1
w n
2
2
d
y 2
dt
2 dy
w n
dt
y x
is the u n d a m p e d n a t u r a l f r eq u e n c y (rad/s)
n
(Greek Zeta) is the d a m p i n g f ac t o r
41
Response of second-order systems =0 undamped <1 under damped =1 critically damped >1 over damped
Will all responses reach “steady-state”?
Which one reaches “steady-state” faster?
Which one oscillates?
42
Key Points
The charging or discharging of a capacitor are each associated with exponential voltage and current waveforms (Same for the energizing and de-energizing of an inductor) Circuits that contain resistance, and either capacitance or inductance, are termed first-order systems The increasing or decreasing exponential waveforms of firstorder systems can be described by the initial and final value formulae Circuits that contain both capacitance and inductance are usually second-order systems. These are characterized by their undamped natural frequency and their damping factor
43
Tutorial As
we increase R, the frequency range over which the dissipative characteristics dominate the behavior of the circuit increases. In order to quantify this behavior we define a parameter called the Q u a l i t y F a c t o r Q which is related to the sharpness of the peak and it is given by Q
=
2
max. energy stored
total energy lost per cycle at resonance
=
2
E S E D
which
represents the ratio of the energy stored to the energy dissipated in a circuit. The energy stored in the circuit is
E S
For
=
1 2
2
LI
+
1
2
CV
2
Vc = Asin(ωt ) the current flowing in the circuit is I = C dVc/dt = ωCAcos(ωt ) . The total energy stored in the reactive elements is
44
E S
1 =
2
L
2
2
2
C A cos
2
1
( t ) +
2
CA sin
2
At
the resonance frequency where ω = ω0 the energy stored in the circuit becomes E S
=
1
CA
2
( t ) 1
0
=
LC
2
2
The
energy dissipated per period is equal to the average resistive power dissipated times the oscillation period. 2 2
E D = R I
And
0
= R
2
C A 2
2
2 =
÷
2
0
so the ratio Q becomes Q
=
L
0
R
1
L
RC R
C
1 =
=
0
1 RC 2
L
0
A
2
