Kanodia EC Solved Paper

GATE

ELECTRONICS & COMMUNICATION

Solved Paper ( 2013-1996 )

RK Kanodia Ashish Murolia

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GATE Electronics & Communication Solved Paper (2013 - 1996 ) RK Kanodia & Ashish Murolia

Copyright © By NODIA & COMPANY Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that NODIA & COMPANY and its author are supplying information but are not attempting to render engineering or other professional services.

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SYLLABUS GENERAL ABILITY Verbal Ability : English grammar, sentence completion, verbal analogies, word groups, instructions, critical reasoning and verbal deduction. Numerical Ability : Numerical computation, numerical estimation, numerical reasoning and data interpretation.

ENGINEERING MATHEMATICS Linear Algebra: Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors. Calculus : Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems. Differential equations : First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method. Complex variables : Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s and Laurent’ series, Residue theorem, solution integrals. Probability and Statistics : Sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Discrete and continuous distributions, Poisson, Normal and Binomial distribution, Correlation and regression analysis. Numerical Methods : Solutions of non-linear algebraic equations, single and multi-step methods for differential equations. Transform Theory : Fourier transform, Laplace transform, Z-transform.

Electronics and Communication Engineering Networks : Network graphs: matrices associated with graphs; incidence, fundamental cut set and fundamental circuit matrices. Solution methods: nodal and mesh analysis. Network theorems: superposition, Thevenin and Norton’s maximum power transfer, Wye-Delta transformation. Steady state sinusoidal analysis using phasors. Linear constant coefficient differential equations; time domain analysis of simple RLC circuits, Solution of network equations using Laplace transform: frequency domain analysis of RLC circuits. 2-port network parameters: driving point and transfer functions. State equations for networks. Electronic Devices : Energy bands in silicon, intrinsic and extrinsic silicon. Carrier transport in silicon: diffusion current, drift current, mobility, and resistivity. Generation and recombination of carriers. p-n junction diode, Zener diode, tunnel diode, BJT, JFET, MOS capacitor, MOSFET, LED, p-I-n and avalanche photo diode, Basics of LASERs. Device technology: integrated circuits fabrication process, oxidation, diffusion, ion implantation, photolithography,

n-tub, p-tub and twin-tub CMOS process. Analog Circuits : Small Signal Equivalent circuits of diodes, BJTs, MOSFETs and analog CMOS. Simple diode circuits, clipping, clamping, rectifier. Biasing and bias stability of transistor and FET amplifiers. Amplifiers: single-and multi-stage, differential and operational, feedback, and power. Frequency response of amplifiers. Simple op-amp circuits. Filters. Sinusoidal oscillators; criterion for oscillation; single-transistor and op-amp configurations. Function generators and wave-shaping circuits, 555 Timers. Power supplies. Digital circuits : Boolean algebra, minimization of Boolean functions; logic gates; digital IC families (DTL, TTL, ECL, MOS, CMOS). Combinatorial circuits: arithmetic circuits, code converters, multiplexers, decoders, PROMs and PLAs. Sequential circuits: latches and flipflops, counters and shift-registers. Sample and hold circuits, ADCs, DACs. Semiconductor memories. Microprocessor(8085): architecture, programming, memory and I/O interfacing. Signals and Systems : Definitions and properties of Laplace transform, continuous-time and discrete-time Fourier series, continuous-time and discrete-time Fourier Transform, DFT and FFT, z-transform. Sampling theorem. Linear Time-Invariant (LTI) Systems: definitions and properties; causality, stability, impulse response, convolution, poles and zeros, parallel and cascade structure, frequency response, group delay, phase delay. Signal transmission through LTI systems. Control Systems : Basic control system components; block diagrammatic description, reduction of block diagrams. Open loop and closed loop (feedback) systems and stability analysis of these systems. Signal flow graphs and their use in determining transfer functions of systems; transient and steady state analysis of LTI control systems and frequency response. Tools and techniques for LTI control system analysis: root loci, Routh-Hurwitz criterion, Bode and Nyquist plots. Control system compensators: elements of lead and lag compensation, elements of Proportional-Integral-Derivative (PID) control. State variable representation and solution of state equation of LTI control systems. Communications : Random signals and noise: probability, random variables, probability density function, autocorrelation, power spectral density. Analog communication systems: amplitude and angle modulation and demodulation systems, spectral analysis of these operations, superheterodyne receivers; elements of hardware, realizations of analog communication systems; signal-to-noise ratio (SNR) calculations for amplitude modulation (AM) and frequency modulation (FM) for low noise conditions. Fundamentals of information theory and channel capacity theorem. Digital communication systems: pulse code modulation (PCM), differential pulse code modulation (DPCM), digital modulation schemes: amplitude, phase and frequency shift keying schemes (ASK, PSK, FSK), matched filter receivers, bandwidth consideration and probability of error calculations for these schemes. Basics of TDMA, FDMA and CDMA and GSM. Electromagnetics : Elements of vector calculus: divergence and curl; Gauss’ and Stokes’ theorems, Maxwell’s equations: differential and integral forms. Wave equation, Poynting vector. Plane waves: propagation through various media; reflection and refraction; phase and group velocity; skin depth. Transmission lines: characteristic impedance; impedance transformation; Smith chart; impedance matching; S parameters, pulse excitation. Waveguides: modes in rectangular waveguides; boundary conditions; cut-off frequencies; dispersion relations. Basics of propagation in dielectric waveguide and optical fibers. Basics of Antennas: Dipole antennas; radiation pattern; antenna gain.

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PREFACE This book doesn’t make promise but provides complete satisfaction to the readers. The market scenario is confusing and readers don’t find the optimum quality books. This book provides complete set of problems appeared in competition exams as well as fresh set of problems. The book is categorized into units which are then sub-divided into chapters and the concepts of the problems are addressed in the relevant chapters. The aim of the book is to avoid the unnecessary elaboration and highlights only those concepts and techniques which are absolutely necessary. Again time is a critical factor both from the point of view of preparation duration and time taken for solving each problem in the examination. So the problems solving methods is the books are those which take the least distance to the solution. But however to make a comment that this book is absolute for GATE preparation will be an inappropriate one. The theory for the preparation of the examination should be followed from the standard books. But for a wide collection of problems, for a variety of problems and the efficient way of solving them, what one needs to go needs to go through is there in there in the book. Each unit (e.g. Networks) is subdivided into average seven number of chapters on an average each of which contains 40 problems which are selected so as to avoid unnecessary redundancy and highly needed completeness. I shall appreciate and greatly acknowledge the comments and suggestion from the users of this book.

R. K. Kanodia Ashish Murolia

CONTENTS CHAP 1

Engineering Mathematics

1 - 31

CHAP 2

Networks

32 - 100

CHAP 3

Electronics Devices

100 - 136

CHAP 4

Analog Circuits

137 - 213

CHAP 5

Digital Circuits

214 - 281

CHAP 6

Signals and Systems

282 - 328

CHAP 7

Control Systems

329 - 385

CHAP 8

Electromagnetic

386 - 435

CHAP 8

Communication Systems

436 - 502

***********

CHAPTER 1 ENGINEERING MATHEMATICS

2013

ONE MARK

MCQ 1.1

The maximum value of q until which the approximation sin q . q holds to within 10% error is (A) 10c (B) 18c (C) 50c (D) 90c

MCQ 1.2

The minimum eigen value of the following matrix is R3 5 2V S W S5 12 7W SS2 7 5WW T X (A) 0 (B) 1 (C) 2 (D) 3

MCQ 1.3

A polynomial f (x) = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 with all coefficients positive has (A) no real roots (B) no negative real root (C) odd number of real roots

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(D) at least one positive and one negative real root 2013 MCQ 1.4

Let A be an m # n matrix and B an n # m matrix. It is given that determinant Îm + AB h = determinant În + BAh, where Ik is the k # k identity matrix. Using the above property, the determinant of the matrix given below is R V S2 1 1 1W S1 2 1 1W S1 1 2 1W S W S1 1 1 2W T X (A) 2 (B) 5 (C) 8 (D) 16 2012

MCQ 1.5

ONE MARK

With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (B) x = t 2 - 1 (A) x = t - 1 2 2 2

t (D) x = 2

(C) x = t 2 MCQ 1.6

TWO MARKS

Given f (z) =

1 - 2 . z+1 z+3

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Chapter 1

If C is a counter clockwise path in the z -plane such that z + 1 = 1, the value of 1 # f (z) dz is 2p j C (B) - 1 (A) - 2 (C) 1 (D) 2 MCQ 1.7

If x = - 1, then the value of xx is (A) e- p/2 (C) x

(B) e p/2 (D) 1

2012 MCQ 1.8

TWO MARKS

Consider the differential equation d 2 y (t) dy (t) dy +2 + y (t)= d (t) with y (t) t = 0 =- 2 and 2 dt dt dt dy The numerical value of is dt t = 0 (B) - 1 (A) - 2 (C) 0 (D) 1 -

=0 t = 0-

+

nodia

MCQ 1.9

The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) - 2 (B) 2 (C) 1 (D) 0

MCQ 1.10

A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (A) 1/3 (B) 1/2 (C) 2/3 (D) 3/4

MCQ 1.11

The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41 (D) 46

MCQ 1.12

Given that

-5 -3 1 0 , the value of A3 is A=> and I = > 2 0H 0 1H (B) 19A + 30I (A) 15A + 12I (C) 17A + 15I (D) 17A + 21I 2011 MCQ 1.13

Consider a closed surface S surrounding volume V . If rv is the position vector of a point inside S , with nt the unit normal on S , the value of the integral ## 5rv $ nt dS S is (A) 3V (C) 10V

MCQ 1.14

ONE MARK

The solution of the differential equation (A) x = ce-ky

(B) 5V (D) 15V dy = ky, y (0) = c is dx (B) x = kecy

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(D) y = ce-kx

(C) y = cekx MCQ 1.15

The value of the integral by (A) 0 (C) 4/5

Page 3

# c

- 3z + 4 dz where c is the circle z = 1 is given (z 2 + 4z + 5) (B) 1/10 (D) 1

2011

TWO MARKS

MCQ 1.16

A numerical solution of the equation f (x) + x - 3 = 0 can be obtained using Newton- Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (A) 0.306 (B) 0.739 (C) 1.694 (D) 2.306

MCQ 1.17

The system of equations x+y+z = 6 x + 4y + 6y = 20 x + 4y + lz = m

nodia

has NO solution for values of l and μ given by (B) l = 6, m = (A) l = 6, m = 20 Y 20 (C) l = (D) l = Y 6, m = 20 Y 6, m = 20 MCQ 1.18

A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (A) 2/36 (B) 2/6 (C) 5/12

(D) 1/2

2010

ONE MARKS

MCQ 1.19

The eigen values of a skew-symmetric matrix are (A) always zero (B) always pure imaginary (C) either zero or pure imaginary (D) always real

MCQ 1.20

The trigonometric Fourier series for the waveform f (t) shown below contains

(A) only cosine terms and zero values for the dc components (B) only cosine terms and a positive value for the dc components (C) only cosine terms and a negative value for the dc components (D) only sine terms and a negative value for the dc components



MCQ 1.21


A function n (x) satisfied the differential equation d 2 n (x) n (x) - 2 =0 dx 2 L where L is a constant. The boundary conditions are : n (0) = K and n (3) = 0 . The solution to this equation is (B) n (x) = K exp (- x/ L ) (A) n (x) = K exp (x/L) 2 (C) n (x) = K exp (- x/L) (D) n (x) = K exp (- x/L) 2010

MCQ 1.22

MCQ 1.23

TWO MARKS

If ey = x1/x , then y has a (A) maximum at x = e (C) maximum at x = e-1

(B) minimum at x = e (D) minimum at x = e-1

A fair coin is tossed independently four times. The probability of the event “the number of time heads shown up is more than the number of times tail shown up” (A) 1/16 (B) 1/3

nodia

(C) 1/4 MCQ 1.24

Chapter 1

v = xyatx + x 2 aty , then If A

(D) 5/16

# Av $ dlv over the path shown in the figure is

C

(A) 0 (C) 1 MCQ 1.25

MCQ 1.26

MCQ 1.27

(B) 2 3 (D) 2 3

The residues of a complex function 1 - 2z x (z) = z (z - 1) (z - 2) at its poles are (B) 1 , - 1 and - 1 (A) 1 , - 1 and 1 2 2 2 2 (C) 1 , 1 and - 3 (D) 1 , - 1 and 3 2 2 2 2 dy (x) Consider differential equation - y (x) = x , with the initial dx condition y (0) = 0 . Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is (A) 0.01 (B) 0.031 (C) 0.0631 (D) 0.1 3s + 1 Given f (t) = L-1 ; 3 . If lim f (t) = 1, then the value t"3 s + 4s2 + (k - 3) s E of k is




(A) 1 (C) 3

Page 5

(B) 2 (D) 4

2009 MCQ 1.28

ONE MARK

The order of the differential equation (A) 1 (C) 3

d2y dy 3 -t 4 is 2 + c dt m + y = e dt (B) 2 (D) 4

MCQ 1.29

A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield heads? 2 2 (B) 10C2 b 1 l (A) c 1 m 2 2 10 10 (C) c 1 m (D) 10C2 b 1 l 2 2

MCQ 1.30

If f (z) = c 0 + c1 z-1 , then

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(A) 2pc1 (C) 2pjc1 2009 MCQ 1.31

unit circle

(B) 2p (1 + c0) (D) 2p (1 + c0)

TWO MARKS

The Taylor series expansion of sin x at x = p is given by x-p (x - p) 2 (x - p) 2 (A) 1 + (B) - 1 + ... + ... 3! 3! (C) 1 -

MCQ 1.32

# 1 +zf (z) dz is given by

(x - p) 2 + ... 3!

Match each differential equation Group II Group I dy y A. = dx x dy y B. =dx x dy x C. = dx y dy D. =- x dx y (A) A - 2, B - 3, C - 3, D - 1 (B) A - 1, B - 3, C - 2, D - 1

(D) - 1 +

(x - p) 2 + ... 3!

in Group I to its family of solution curves from Group II 1. Circles 2. Straight lines 3. Hyperbolas

(C) A - 2, B - 1, C - 3, D - 3 (D) A - 3, B - 2, C - 1, D - 2 MCQ 1.33

The Eigen values of following matrix are V R S- 1 3 5 W S- 3 - 1 6 W SS 0 0 3 WW X T




(A) 3, 3 + 5j, 6 - j (C) 3 + j, 3 - j, 5 + j

Chapter 1

(B) - 6 + 5j, 3 + j, 3 - j (D) 3, - 1 + 3j, - 1 - 3j

2008 MCQ 1.34

MCQ 1.35

ONE MARKS

p11 p12 All the four entries of the 2 # 2 matrix P = = are nonzero, p21 p22 G and one of its eigenvalue is zero. Which of the following statements is true? (A) p11 p12 - p12 p21 = 1 (B) p11 p22 - p12 p21 =- 1 (C) p11 p22 - p12 p21 = 0 (D) p11 p22 + p12 p21 = 0 The system of linear equations 4 x + 2y = 7 2x + y = 6 has (A) a unique solution (B) no solution (C) an infinite number of solutions (D) exactly two distinct solutions

nodia

MCQ 1.36

The equation sin (z) = 10 has (A) no real or complex solution (B) exactly two distinct complex solutions (C) a unique solution (D) an infinite number of complex solutions

MCQ 1.37

For real values of x , the minimum value of the function f (x) = exp (x) + exp (- x) is (A) 2 (B) 1 (C) 0.5 (D) 0

MCQ 1.38

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0 ? (A) sin (x3) (B) sin (x2) (C) cos (x3) (D) cos (x2)

MCQ 1.39

Which of the following is a solution to the differential equation dx (t) + 3x (t) = 0 ? dt (B) x (t) = 2e - 3t (A) x (t) = 3e - t (C) x (t) =- 23 t2

(D) x (t) = 3t2

2008 MCQ 1.40

TWO MARKS

The recursion relation to solve x = e (A) xn + 1 = e-x n

-x (C) xn + 1 = (1 + xn) e -x 1+e

using Newton - Raphson method is (B) xn + 1 = xn - e-x x 2 - e-x (1 - xn) - 1 (D) xn + 1 = n xn - e-x n

n

n

MCQ 1.41

-x

n

The residue of the function f (z) =

n

1 at z = 2 is (z + 2) 2 (z - 2) 2




(A) - 1 32 (C) 1 16 MCQ 1.42

Page 7

(B) - 1 16 (D) 1 32

0 1 Consider the matrix P = = . The value of e p is - 2 - 3G 2e-2 - 3e-1 e-1 - e-2 e-1 + e-1 2e-2 - e-1 (B) > -1 (A) > -2 H -1 -2 -1H 2e - 2e 5e - e 2e - 4e2 3e-1 + 2e-2 5e-2 - e-1 3e-1 - e-2 (C) > -2 H 2e - 6e-1 4e-2 + 6-1

2e-1 - e-2 e-1 - e-2 (D) > H -1 -2 - 2e + 2e - e-1 + 2e-2

MCQ 1.43

In the Taylor series expansion of exp (x) + sin (x) about the point x = p , the coefficient of (x - p) 2 is (A) exp (p) (B) 0.5 exp (p) (C) exp (p) + 1 (D) exp (p) - 1

MCQ 1.44

The value of the integral of the function g (x, y) = 4x3 + 10y 4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x - y plane is (A) 33 (B) 35 (C) 40 (D) 56

MCQ 1.45

Consider points P and Q in the x - y plane, with P = (1, 0) and Q = (0, 1). The

nodia

line integral 2 its diameter (A) is - 1 (B) is 0

Q

#P

(xdx + ydy) along the semicircle with the line segment PQ as

(C) is 1 (D) depends on the direction (clockwise or anit-clockwise) of the semicircle 2007 MCQ 1.46

The following plot shows a function which varies linearly with x . The value of the integral I =

MCQ 1.47

ONE MARK 2

#1 ydx

is

(A) 1.0

(B) 2.5

(C) 4.0

(D) 5.0

For x << 1, coth (x) can be approximated as (B) x2 (A) x (C) 1 (D) 12 x x



MCQ 1.48

MCQ 1.49


sin b q l 2 is lim q"0 q (A) 0.5 (C) 2

(B) 1 (D) not defined

Which one of following functions is strictly bounded? (A) 1/x2 (B) ex (D) e - x

(C) x2 MCQ 1.50

Chapter 1

2

For the function e - x , the linear approximation around x = 2 is (A) (3 - x) e - 2 (B) 1 - x (C) 63 + 3 2 - (1 -

2 ) x @e - 2

(D) e - 2

2007

TWO MARKS 2

MCQ 1.51

d y The solution of the differential equation k2 2 = y - y2 under the boundary dx conditions (i) y = y1 at x = 0 and (ii) y = y2 at x = 3 , where k, y1 and y2 are constants, is

nodia

(A) y = (y1 - y2) exp a- x2 k + y2 k (C) y = ^y1 - y2h sinh a x k + y1 k MCQ 1.52

The equation x3 - x2 + 4x - 4 = 0 is to be solved using the Newton - Raphson method. If x = 2 is taken as the initial approximation of the solution, then next approximation using this method will be (A) 2/3 (B) 4/3 (C) 1

MCQ 1.53

(B) y = (y2 - y1) exp a- x k + y1 k (D) y = ^y1 - y2h exp a- x k + y2 k

(D) 3/2

Three functions f1 (t), f2 (t) and f3 (t) which are zero outside the interval [0, T] are shown in the figure. Which of the following statements is correct?




(A) f1 (t) and f2 (t) are orthogonal (C) f2 (t) and f3 (t) are orthogonal MCQ 1.54

Page 9

(B) f1 (t) and f3 (t) are orthogonal D) f1 (t) and f2 (t) are orthonormal

If the semi-circular control D of radius 2 is as shown in the figure, then the value of the integral # 2 1 ds is (s - 1) D

(B) - jp (D) p

(A) jp (C) - p MCQ 1.55

It is given that X1, X2 ...XM at M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X1, X2,... XM , - X1, - X2,... - XM is (B) M + 1 (A) 2M (C) M (D) dependent on the choice of X1, X2,... XM

MCQ 1.56

Consider the function f (x) = x2 - x - 2 . The maximum value of f (x) in the closed interval [- 4, 4] is (B) 10 (A) 18 (C) - 225 (D) indeterminate

MCQ 1.57

An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is (A) 0.5 (B) 0.18

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(C) 0.12

(D) 0.06

2006 MCQ 1.58

MCQ 1.59

MCQ 1.60

V R S1 1 1 W The rank of the matrix S1 - 1 0 W is SS1 1 1 WW X T (A) 0 (C) 2

ONE MARK

(B) 1 (D) 3

4#4# P , where P is a vector, is equal to (B) 4 2 P + 4 (4# P) (A) P #4# P - 4 2 P (C) 4 2 P + 4# P (D) 4 (4$ P) - 4 2 P

## (4# P) $ ds , where P is a vector, is equal to (B) # 4#4# P $ dl (A) # P $ dl (C) # 4# P $ dl (D) ### 4$ Pdv




Chapter 1

MCQ 1.61

A probability density function is of the form p (x) = Ke- a x , x ! (- 3, 3). The value of K is (A) 0.5 (B) 1 (C) 0.5a (D) a

MCQ 1.62

A solution for the differential equation xo (t) + 2x (t) = d (t) with initial condition x (0-) = 0 is (A) e - 2t u (t) (B) e2t u (t) (C) e - t u (t)

(D) et u (t)

2006

TWO MARKS

MCQ 1.63

The eigenvalue and the corresponding eigenvector of 2 # 2 matrix are given by Eigenvalue Eigenvector 1 l1 = 8 v1 = = G 1 1 l2 = 4 v2 = = G -1 The matrix is 6 2 4 6 (B) = (A) = G 2 6 6 4G 2 4 4 8 (C) = (D) = 4 2G 8 4G

MCQ 1.64

For the function of a complex variable W = ln Z (where, W = u + jv and Z = x + jy , the u = constant lines get mapped in Z -plane as (A) set of radial straight lines (B) set of concentric circles (C) set of confocal hyperbolas (D) set of confocal ellipses

MCQ 1.65

The value of the constant integral

nodia z-j

jp 2 jp (C) 2 (A)

MCQ 1.66

MCQ 1.67

The integral (A) 1 2 (C) 4 3

#0

p

1 dz is positive sense is z2 + 4 =2 (B) - p 2 (D) p 2

#

sin3 qdq is given by

(B) 2 3 (D) 8 3

Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below Company

% of Computer Supplied

Probability of being supplied defective

X

60%

0.01

Y

30%

0.02

Z

10%

0.03




Page 11

Given that a computer is defective, the probability that was supplied by Y is (A) 0.1 (B) 0.2 (C) 0.3 (D) 0.4 MCQ 1.68

MCQ 1.69

4 2 101 For the matrix = the eigenvalue corresponding to the eigenvector = G is 2 4G 101 (A) 2 (B) 4 (C) 6 (D) 8 For the differential equation

d2 y + k2 y = 0 the boundary conditions are dx2

(ii) y = 0 for x = a (i) y = 0 for x = 0 and The form of non-zero solutions of y (where m varies over all integers) are (A) y = / Am sin mpx (B) y = / Am cos mpx a a m m (C) y = / Am x

mp a

(D) y = / Am e -

m

MCQ 1.70

mpx a

m

nodia

As x increased from - 3 to 3, the function f (x) = (A) monotonically increases (B) monotonically decreases

ex 1 + ex

(C) increases to a maximum value and then decreases (D) decreases to a minimum value and then increases 2005 MCQ 1.71

The following differential equation has d2 y dy 3 3 c 2 m + 4 c m + y2 + 2 = x dt dt (A) degree = 2 , order = 1 (C) degree = 4 , order = 3

MCQ 1.72

(B) degree = 1, order = 2 (D) degree = 2 , order = 3

A fair dice is rolled twice. The probability that an odd number will follow an even number is (A) 1/2 (B) 1/6 (C) 1/3

MCQ 1.73

(D) 1/4

A solution of the following differential equation is given by (A) y = e2x + e-3x (C) y = e-2x + 33x

(B) y = e2x + e3x (D) y = e-2x + e-3x

2005 MCQ 1.74

ONE MARK

d2 y dy -5 + 6y = 0 2 dx dx

TWO MARKS

In what range should Re (s) remain so that the Laplace transform of the function e(a + 2) t + 5 exits. (A) Re (s) > a + 2 (B) Re (s) > a + 7 (C) Re (s) < 2

(D) Re (s) > a + 5




Chapter 1

MCQ 1.75

The derivative of the symmetric function drawn in given figure will look like

MCQ 1.76

Match the following and choose the correct combination: Group I Group 2 E. Newton-Raphson method 1. Solving nonlinear equations F. Runge-kutta method 2. Solving linear simultaneous equations G. Simpson’s Rule 3. Solving ordinary differential equations H. Gauss elimination 4. Numerical integration 5. Interpolation 6. Calculation of Eigenvalues (A) E - 6, F - 1, G - 5, H - 3 (B) E - 1, F - 6, G - 4, H - 3 (C) E - 1, F - 3, G - 4, H - 2 (D) E - 5, F - 3, G - 4, H - 1

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MCQ 1.77

MCQ 1.78

MCQ 1.79

-4 2 Given the matrix = , the eigenvector is 4 3G 3 4 (B) = G (A) = G 2 3 2 -1 (C) = G (D) = G -1 2

1 2 - 0.1 a Let, A = = and A - 1 = = 2 G. Then (a + b) = G 0 3 0 b (A) 7/20 (B) 3/20 (C) 19/60 (D) 11/20

The value of the integral I = (A) 1 (C) 2

MCQ 1.80

1 2p

#0

3

2

exp c- x m dx is 8 (B) p (D) 2p

Given an orthogonal matrix R1 1 1 1 V W S S1 1 - 1 - 1W A =S 1 - 1 0 0W W S S0 0 1 1 W X T T 6AA @- 1 is




R1 S4 S0 (A) S S0 S0 T R1 S S0 (C) S 0 S S0 T

V 0 0 0W 1 W 4 0 0 W 1 0 2 0W 0 0 12 W X 0 0 0 VW 1 0 0W 0 1 0W W 0 0 1W X

R1 S2 S0 (B) S S0 S0 TR 1 S4 S0 (D) S S0 S0 T

Page 13

V 0 0 0W 1 W 2 0 0 W 1 0 2 0W 0 0 12 W XV 0 0 0W 1 W 4 0 0 W 0 14 0 W 0 0 14 W X

***********

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Chapter 1

SOLUTIONS SOL 1.1

Option (B) is correct. Here, as we know Lim sin q . 0 q"0

but for 10% error, we can check option (B) first, q = 18c = 18c # p = 0.314 180c sin q = sin 18c = 0.309 % error = 0.314 - 0.309 # 100% = 0.49% 0.309 Now, we check it for q = 50c q = 50c = 50c # p = 0.873 180c sin q = sin 50c = 0.77 % error = 0.77 - 0.873 =- 12.25% 0.873 so, the error is more than 10% . Hence, for error less than 10%, q = 18c can have the approximation sin q . q SOL 1.2

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Option (A) is correct. For, a given matrix 6A@ the eigen value is calculated as A - lI = 0 where l gives the eigen values of matrix. Here, the minimum eigen value among the given options is l =0 We check the characteristic equation of matrix for this eigen value (for l = 0 ) A - lI = A 3 5 2 = 5 12 7 2 7 5 = 3 ^60 - 49h - 5 ^25 - 14h + 2 ^35 - 24h = 33 - 55 + 22 =0 Hence, it satisfied the characteristic equation and so, the minimum eigen value is l =0

SOL 1.3

Option (D) is correct. Given, the polynomial

f ^x h = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 Since, all the coefficients are positive so, the roots of equation is given by f ^x h = 0 It will have at least one pole in right hand plane as there will be least one sign change from â1h to â 0h in the Routh matrix 1 st column. Also, there will be a




Page 15

corresponding pole in left hand plane i.e.; at least one positive root (in R.H.P) and at least one negative root (in L.H.P) Rest of the roots will be either on imaginary axis or in L.H.P SOL 1.4

Option (B) is correct. Consider the given matrix be R V S2 1 1 1W S1 2 1 1W Im + AB = S W S1 1 2 1W S1 1 1 2W T X where m = 4 so, we obtain R V R V S2 1 1 1W S1 0 0 0W S1 2 1 1W S0 1 0 0W AB = S W-S W S1 1 2 1W S0 0 1 0W S1 1 1 2W S0 0 0 1W TR X VX T S1 1 1 1W S1 1 1 1W =S W S1 1 1 1W S1 1 1 1W X RT V S1W S1W = S W 61 1 1 1@ S1W S1W T X Hence, we get R V S1W S1W A = S W, B = 81 1 1 1B S1W S1W T X R V Therefore, BA = 81 1 1 1B S1W S1W S1W S W S1W T X =4 From the given property Det Îm + AB h = Det Îm + BAh R V V ZR _ S2 1 1 1W ]S1 0 0 0W b ]S0 1 0 0W b S1 2 1 1W = Det [S + 4` Det S & W W 0 0 1 0W S1 1 2 1W ]S b ]S0 0 0 1W b S1 1 1 2W \T a T X X = 1+4 =5 Note : Determinant of identity matrix is always 1.

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SOL 1.5

Option (D) is correct. t dx + x = t dt




Chapter 1

dx + x = 1 t dt dx + Px = Q (General form) dt IF = e # = e = e lnt = t

Integrating factor,

1 # dt t

Pdt

Solution has the form,

# ^Q # IF hdt + C x # t = # (1) (t) dt + C

x # IF =

2

xt = t + C 2 Taking the initial condition, x (1) = 0.5 0.5 = 1 + C & C = 0 2 So, SOL 1.6

2 xt = t & x = t 2 2

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Option (C) is correct.

f (z) =

1 2p j

# f (z) dz C

1 - 2 z+1 z+3

= sum of the residues of the poles which lie inside the given closed region.

C & z+1 = 1 Only pole z =- 1 inside the circle, so residue at z =- 1 is. (z + 1) (- z + 1) 2 -z + 1 f (z) = = lim = =1 2 z "- 1 (z + 1) (z + 3) (z + 1) (z + 3) 1 So f (z) dz = 1 2pj #C SOL 1.7

Option (A) is correct. x=

- 1 = i = cos p + i sin p 2 2 p

x = ei 2

So,

xx = êi 2 h & êi 2 h = e- 2 p i

p x

SOL 1.8

p

Option (D) is correct. d 2 y (t) 2dy (t) + + y (t) = d (t) dt dt 2 By taking Laplace transform with initial conditions dy 2 ;s Y (s) - sy (0) - dt E + 2 [sy (s) - y (0)] + Y (s) = 1 t=0 &

2 6s Y (s) + 2s - 0@ + 2 6sY (s) + 2@ + Y (s) = 1

Y (s) [s2 + 2s + 1] = 1 - 2s - 4 Y (s) = 2- 2s - 3 s + 2s + 1 We know that, If,

y (t)

L

Y (s)




Page 17

dy (t) dt

then,

L

sY (s) - y (0)

(- 2s - 3) s +2 (s2 + 2s + 1) 2 2 = - 2s - 32 s + 2s + 4s + 2 (s + 2s + 1) + + s 2 s 1 1 sY (s) - y (0) = = + (s + 1) 2 (s + 1) 2 (s + 1) 2 1 = 1 + s + 1 (s + 1) 2 Taking inverse Laplace transform dy (t) = e-t u (t) + te-t u (t) dt dy At t = 0+ , = e0 + 0 = 1 dt t = 0 sY (s) - y (0) =

So,

+

SOL 1.9

Option (A) is correct. Divergence of A in spherical coordinates is given as d:A = 12 2 (r 2 Ar ) = 12 2 (krn + 2) r 2r r 2r = k2 (n + 2) rn + 1 r = k (n + 2) rn - 1 = 0 (given)

nodia n+2 = 0

SOL 1.10

Option (C) is correct. Probability of appearing a head is 1/2. If the number of required tosses is odd, we have following sequence of events. H, TTH, TTTTH, ........... 3 5 P = 1 + b 1 l + b 1 l + ..... 2 2 2

Probability

P =

SOL 1.11

n =- 2

&

1 2

1 - 14

=2 3

Option (B) is correct.

&

f (x) = x3 - 9x2 + 24x + 5 df (x) = 3x2 - 18x + 24 = 0 dx df (x) = x2 - 6x + 8 = 0 dx

x = 4, x = 2

d 2 f (x) = 6x - 18 dx 2 d 2 f (x) For x = 2, = 12 - 18 =- 6 < 0 dx2 So at x = 2, f (x) will be maximum f (x)

max

= (2) 3 - 9 (2) 2 + 24 (2) + 5 = 8 - 36 + 48 + 5 = 25



SOL 1.12


Chapter 1

Option (B) is correct. Characteristic equation. A - lI = 0 -5 - l -3 =0 2 -l 5l + l2 + 6 = 0 l2 + 5l + 6 = 0 Since characteristic equation satisfies its own matrix, so A2 + 5A + 6 = 0 & A2 =- 5A - 6I Multiplying with A A3 + 5A2 + 6A = 0 A3 + 5 (- 5A - 6I) + 6A = 0 A3 = 19A + 30I

SOL 1.13

Option (D) is correct. From Divergence theorem, we have v = #A v $ nt ds v Adv ### 4$ s The position vector rv = ûtx x + uty y + utz z h v v Here, A = 5r , thus v = utx 2 + uty 2 + utz 2 : utx x + uty y + utz z 4$ A h c 2x 2y 2z m ^ dy dz 5 = 3 # 5 = 15 = c dx + + dx dy dz m So, ## 5rv $ nt ds = ### 15 dv = 15V

nodia s

SOL 1.14

Option (C) is correct. We have

#

Integrating or Since y (0) = c thus So, we get, or

ln y = kx + A ln c = A ln y = kx + ln c ln y = ln ekx + ln c y = cekx

or SOL 1.15

dy = ky dx dy = # k dx + A y

Option (A) is correct. 3z + 4 dz where C is circle z = 1 C R Integrals is # 2 + z 4z + 5 C # f (z) dz = 0 if poles are outside C. C

2

z + 4z + 5 = 0 (z + 2) 2 + 1 = 0 Thus z1, 2 =- 2 ! j & z1, 2 > 1 So poles are outside the unit circle. Now

SOL 1.16





f (x) = x + x - 3 = 0 f l (x) = 1 + 1 2 x Substituting x 0 = 2 we get f l (x 0) = 1.35355 and f (x 0) = 2 +

Page 19

We have

2 - 3 = 0.414

Newton Raphson Method x1 = x 0 -

f (x 0) f l (x 0)

Substituting all values we have x 1 = 2 - 0.414 = 1.694 1.3535 SOL 1.17

Option (B) is correct. Writing A: B we have R V S1 1 1 : 6 W S1 4 6 : 20W S W S1 4 l : m W T X Apply R 3 " R 3 - R2 R V 6 W S1 1 1 : S1 4 6 : 20 W S W S0 0 l - 6 : m - 20W T X For equation to have solution, rank of A and A: B must be same. Thus for no solution; l = 6, m ! 20

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SOL 1.18

Option (C) is correct. Total outcome are 36 out of which favorable outcomes are : (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6); (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15. Thus P (E) = No. of favourable outcomes = 15 = 5 36 12 No. of total outcomes

SOL 1.19

Option (C) is correct. Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in conjugate pairs.

SOL 1.20

Option (C) is correct. For a function x (t) trigonometric fourier series is x (t) = Ao + Where,

3

/ [An cos nwt + Bn sin nwt]

n=1

Ao = 1 # x (t) dt T0 T

T0 " fundamental period

0

An = 2 # x (t) cos nwt dt T0 T 0

Bn = 2 # x (t) sin nwt dt T0 T For an even function x (t), Bn = 0 Since given function is even function so coefficient Bn = 0 , only cosine and constant terms are present in its fourier series representation. Constant term : 0




Chapter 1

3T/4 A0 = 1 # x (t) dt T -T/4 T/4 3T/4 = 1 : # Adt + # - 2AdtD T -T/4 T/4 = 1 :TA - 2AT D =- A 2 2 T 2 Constant term is negative.

SOL 1.21

Option (D) is correct. Given differential equation d 2 n (x) n (x) - 2 =0 dx 2 L Let n (x) = Aelx lx So, Al2 elx - Ae2 = 0 L l2 - 12 = 0 & l = ! 1 L L Boundary condition, n (3) = 0 so take l =- 1 L

nodia x

So, SOL 1.22

n (x) = Ae- L n (0) = Ae0 = K & A = K n (x) = Ke- (x/L)

Option (A) is correct. Given that

1

ey = x x

1

ln ey = ln x x or y = 1 ln x x 1 dy Now = 1 1 + ln x ^- x- x h = 12 - ln2 xx dx x x For maxima and minima : dy = 12 (1 - ln x) = 0 dx x ln x = 1 " x = e 1 d 2y Now =- 23 - ln x b- 23 l - 12 b 1 l dx 2 x x x x =- 22 + 2 ln3 x - 13 x x x 2 2 1 d 2x = 2 + 3- 3<0 e dy 2 at x = e e e 1 So, y has a maximum at x = e or

2

1

SOL 1.23

Option (D) is correct. According to given condition head should comes 3 times or 4 times 4 3 P (Heads comes 3 times or 4 times) = 4C 4 b 1 l + 4C 3 b 1 l b 1 l 2 2 2 = 1: 1 +4:1 :1 = 5 16 8 2 16

SOL 1.24





Page 21

v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) # Av : dl C

C

# (xydx + x 2 dy)

=

C

=

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] 2 3 3 2 3 3 3 3 =1 SOL 1.25

Option (C) is correct. Given function 1 - 2z z (z - 1) (z - 2) Poles are located at z = 0, z = 1, and z = 2 At Z = 0 residues is R 0 = z : X (z) Z = 0 = 1 - 2 # 0 = 1 2 (0 - 1) (0 - 2) X (z ) =

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at z = 1,

R1 = (Z - 1) : X (Z ) Z = 1 = 1-2#1 = 1 1 (1 - 2)

R2 = (z - 2) : X (z) z = 2

At z = 2 ,

= 1 - 2 # 2 =- 3 2 2 (2 - 1)

SOL 1.26

Option (B) is correct. Taking step size

h = 0.1, y (0) = 0

x

y

dy = x+y dx

0

0

0

y1 = 0 + 0.1 (0) = 0

0.1

0

0.1

y2 = 0 + 0.1 (0.1) = 0.01

0.2

0.01

0.21

y 3 = 0.01 + 0.21 # 0.1 = 0.031

0.3

0.031

yi + 1 = yi + h

dy dx

From table, at x = 0.3, y (x = 0.3) = 0.031 SOL 1.27

Option (D) is correct. Given that 3s + 1 f (t) = L - 1 ; 3 s + 4s 2 + (K - 3) s E lim f (t) = 1

t"3

By final value theorem lim f (t) = lim sF (s) = 1

t"3

or

lim s"0

s"0

s : (3s + 1) =1 s3 + 4s2 + (K - 3) s




or

lim s"0

Chapter 1

s (3s + 1) =1 s [s + 4s + (K - 3)] 1 =1 K-3 2

K =4

or SOL 1.28

Option (B) is correct. The highest derivative terms present in DE is of 2nd order.

SOL 1.29

Option (C) is correct. Number of elements

SOL 1.30

210 .

Only 1 "H, H, T, T, T, T, T, T, T, T , is event. Thus probability is 10 2 Option (C) is correct. We have in

sample

space

is

one

element

f (z) = c0 + c1 z - 1 1 + f (z) 1 + c0 + c1 z - 1 z (1 + c0) + c1 = f1 (z) = = z z z2 Since f1 (z) has double pole at z = 0 , the residue at z = 0 is z (1 + c0) + c1 Res f1 (z) z = 0 = lim z2 .f1 (z) = lim z2 . c m = c1 z"0 z"0 z2 Hence # f1(z) dz = # [1 +zf (z)] dz = 2pj [Residue at z = 0]

nodia unit circle

unit circle

= 2pjc1

SOL 1.31

Option (D) is correct.

f (x) = sin x x-p

We have

Substituting x - p = y ,we get sin (y + p) sin y f (y + p) = = - 1 (sin y) =y y y 3 5 y y = - 1 cy - + - ...m y 3! 5! y2 y 4 - + ... 3! 5! Substituting x - p = y we get (x - p) 2 (x - p) 4 f (x) =- 1 + + ... 3! 5! or

SOL 1.32

f (y + p) =- 1 +

Option (A) is correct. (A) or

#

dy y = dx x dy = # dx x y

log y = log x + log c or y = cx Thus option (A) and (C) may be correct. dy y (B) =dx x or

Straight Line




# dyy

or

Page 23

=- # dx x

log y =- log x + log c log y = log 1 + log c x y =c x

or or or

Hyperbola

SOL 1.33

Option (D) is correct. Sum of the principal diagonal element of matrix is equal to the sum of Eigen values. Sum of the diagonal element is - 1 - 1 + 3 = 1.In only option (D), the sum of Eigen values is 1.

SOL 1.34

Option (C) is correct. The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero. Thus p11 p22 - p12 p21 = 0

SOL 1.35

Option (B) is correct. The given system is 4 2 x 7 =2 1G=y G = =6 G 4 2 We have A == G 2 1 4 2 and =0 A = 2 1 4 2 7 Now C == G 2 1 6

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Rank of matrix r (A) < 2 Rank of matrix r (C) = 2

Since r (A) ! r (C) there is no solution. SOL 1.36

Option (A) is correct. sin z can have value between - 1 to + 1. Thus no solution.

SOL 1.37

Option (A) is correct. We have f (x) = ex + e-x For x > 0 , ex > 1 and 0 < e-x < 1 For x < 0 , 0 < ex < 1 and e-x > 1 Thus f (x) have minimum values at x = 0 and that is e0 + e-0 = 2 .

SOL 1.38

Option (A) is correct. 3 5 sin x = x + x + x + ... 3! 5! 2 4 cos x = 1 + x + x + ... 2! 4!

Thus only sin (x3) will have odd power of x . SOL 1.39

Option (B) is correct. dx (t) We have + 3x (t) = 0 dt or Since m =- 3 ,

(D + 3) x (t) = 0 x (t) = Ce - 3t

Thus only (B) may be solution.



SOL 1.40


Chapter 1

Option (C) is correct. x = e-x

We have

f (x) = x - e - x

or

f'( x) = 1 + e - x The Newton-Raphson iterative formula is f (xn) xn + 1 = xn f'( xn) f (xn) = xn - e - x

Now

f'( xn) = 1 + e - x

n

n

- xn

xn + 1 = xn - xn - e- x = 1+e

Thus

n

SOL 1.41

(1 + xn) e - x 1 + e-x

n

n

Option (A) is correct. Res f (z) z = a =

1 dn - 1 6(z - a) n f (z)@ z=a (n - 1)! dzn - 1

Here we have n = 2 and a = 2

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Thus Res f (z) z = 2 =

1 d (z - 2) 2 1 (2 - 1)! dz ; (z - 2) 2 (z + 2) 2 Ez = a

-2 = d ; 1 2E = dz (z + 2) z = a ; (z + 2) 3 Ez = a =- 2 =- 1 32 64

SOL 1.42


eP = L- 1 6(sI - A) - 1@ s 0 0 1 -1 = L- 1 e= G - = 0 s - 2 - 3Go s - 1 -1 o = L- 1 e= 2 s + 3G = L f> -1

s+3 (s + 1)( s + 2) -2 (s + 1)( s + 2)

1 (s + 1)( s + 2) s (s + 1)( s + 2)

Hp

2e - 1 - e - 2 e-1 - e-2 == G - 2e - 1 + 2e - 2 - e - 1 + 2e - 2 SOL 1.43

Option (B) is correct. Taylor series is given as f (x) = f (a) + x - a f'( a) + 1!

(x - a) 2 f"( a) + ... 2!

f (x) = f (p) + x - p f'( p) + 1!

(x - p) 2 f"( x)... 2!

For x = p we have Thus Now

f (x) = ex + sin x f'( x) = ex + cos x f"( x) = ex - sin x f"( p) = e p - sin p = e p




f"( p) 2!

Thus the coefficient of (x - p) 2 is SOL 1.44

Page 25

Option (A) is correct. The equation of straight line from (0, 0) to (1, 2) is y = 2x . Now g (x, y) = 4x3 + 10y 4 g (x, 2x) = 4x3 + 160x 4

or,

1

1

#0 g (x, 2x) = #0 (4x3 + 160x4) dx

Now

= [x 4 + 32x5] 10 = 33 SOL 1.45

SOL 1.46


#P

=2

#P

=2

#1

Q

0

(xdx + ydy) Q

xdx + 2

#P ydy

xdx + 2

#0 ydy = 0

1

nodia

Option (B) is correct. The given plot is straight line whose equation is x +y =1 -1 1 or Now

y = x+1 I =

2

#1 ydx

=;

SOL 1.47

Q

I =2

=

2

#1 (x + 1) dx

(x + 1) 2 2 9 4 E = - = 2.5 2 2 2

Option (C) is correct. coth x = cosh x sinh x as x << 1, cosh x . 1 and sinh x . x Thus coth x . 1 x

SOL 1.48

Option (A) is correct. sin ^ q2 h sin ^ q h sin ^ q2 h = lim = 1 lim q 2 = 1 = 0.5 lim q q " 0 2^ h q"0 2 2 q"0 ^ 2h q 2

SOL 1.49

Option (D) is correct. We have,

lim 12 = 3 x"0 x lim x2 = 3 x"3

lim e - x = 3

x"3

lim e - x = 0 2

x"3

lim e - x = 1 2

x"0

SOL 1.50

Thus e - x is strictly bounded. 2

Option (A) is correct. We have

f (x) = e - x = e - (x - 2) - 2 = e - (x - 2) e - 2 (x - 2) 2 = ;1 - (x - 2) + ...Ee - 2 2!




= 61 - (x - 2)@e - 2 = (3 - x) e SOL 1.51

Chapter 1

Neglecting higher powers

-2

Option (D) is correct. d2 y = y - y2 dx2 d2 y y y =- 22 dx2 k2 k k2

We have or

D2 - 12 = 0 k

A.E.

D =! 1 k

or

C.F. = C1 e - + C2 e x k

P.I. =

1 D 2

Thus solution is

1 k2

c

x k

- y22 m = y2 k2

nodia y = C1 e - + C2 e + y2 x k

x k

From y (0) = y1 we get C1 + C2 = y1 - y2 From y (3) = y2 we get that C1 must be zero. Thus C2 = y1 - y2

y = (y1 - y2) e - + y2 x k

SOL 1.52

Option (B) is correct. We have

f (x) = x3 - x2 + 4x - 4 f'( x) = 3x2 - 2x + 4 Taking x0 = 2 in Newton-Raphosn method 23 - 22 + 4 (2) - 4 f (x0) =4 x1 = x0 = 23 f'( x0) 3 (2) 2 - 2 (2) + 4 SOL 1.53

Option (C) is correct. For two orthogonal signal f (x) and g (x) +3

#- 3

f (x) g (x) dx = 0

i.e. common area between f (x) and g (x) is zero. SOL 1.54

Option (A) is correct. We know that [sum of residues] # s2 -1 1 ds = 2pj D Singular points are at s = ! 1 but only s =+ 1 lies inside the given contour, Thus Residue at s =+ 1 is lim (s - 1) f (s) = lim (s - 1) 2 1 = 1 s"1 s"1 s -1 2 # s2 -1 1 ds = 2pj` 12 j = pj D




Page 27

SOL 1.55

Option (C) is correct. For two orthogonal vectors, we require two dimensions to define them and similarly for three orthogonal vector we require three dimensions to define them. 2M vectors are basically M orthogonal vector and we require M dimensions to define them.

SOL 1.56

Option (A) is correct. We have f (x) = x2 - x + 2 f'( x) = 2x - 1 = 0 " x = 1 2 f"( x) = 2 Since f"( x) = 2 > 0 , thus x = 1 is minimum point. The maximum value in 2 closed interval 6- 4, 4@ will be at x =- 4 or x = 4 Now maximum value = max [f (- 4), f (4)]

nodia = max (18, 10) = 18

SOL 1.57

Option (C) is correct. Probability of failing in paper 1 is P (A) = 0.3 Possibility of failing in Paper 2 is P (B) = 0.2 Probability of failing in paper 1, when student has failed in paper 2 is P ^ BA h = 0.6 We know that (P + B) Pb A l = B P (B) or P (A + B) = P (B) P b A l = 0.6 # 0.2 = 0.12 B

SOL 1.58


V R V R S1 1 1 W S1 1 1 W A = S1 - 1 0 W + S1 - 1 0 W SS1 1 1 WW SS0 0 0 WW X T X T Since one full row is zero, r (A) < 3 1 1 Now =- 2 ! 0 , thus r (A) = 2 1 -1 SOL 1.59

R3 - R1

Option (D) is correct. The vector Triple Product is A # (B # C) = B (A $ C) - C (A $ B) Thus

SOL 1.60

4#4# P = 4 (4$ P) - P (4$4) = 4 (4$ P) - 4 2 P

Option (A) is correct. The Stokes theorem is

## (4# F) $ ds = # A $ dl SOL 1.61





# p (x) dx 3

We know

-3

#

Thus

3

-3

#

or

0

-3

Keax dx +

Chapter 1

=1

Ke- a x dx = 1

# Ke 3

- ax

0

dx = 1

K eax 0 + k e- ax 3 = 1 @0 a 6 @- 3 (- a) 6 K +K =1 or a a or K =a 2 Option (A) is correct. We have xo (t) + 2x (t) = s (t) or

SOL 1.62

Taking Laplace transform both sides sX (s) - x (0) + 2X (s) = 1 or

sX (s) + 2X (s) = 1

nodia X (s) =

Since x (0 -) = 0

1 s+2

Now taking inverse Laplace transform we have x (t) = e - 2t u (t)

SOL 1.63

Option (A) is correct. Sum of the Eigen values must be equal to the sum of element of principal diagonal of matrix. 6 2 Only matrix = G satisfy this condition. 2 6

SOL 1.64


W = ln z u + jv = ln (x + jy)

eu + jv = x + jy eu e jv = x + jy eu (cos v + j sin v) = x + jy Now x = eu cos v and y = eu sin v Thus x 2 + y 2 = e 2u or or

SOL 1.65

Option (D) is correct. We have # z2 +1 4 dz = z-j = 2

# z-j = 2

Equation of circle

1 dz (z + 2i) (z - 2i)

P (0, 2) lies inside the circle z - j = 2 and P (0, - 2) does not lie. Thus By cauchy’s integral formula 1 = # 2pi = p I = 2pi lim (z - 2i) z " 2i 2i + 2i 2 (z + 2i)( z - 2i) C

SOL 1.66

Option (C) is correct. I =

#0

p

sin3 qdq

=

#0

p

`

3 sin q - sin 3q dq j 4

sin 3q = 3 sin q - 4 sin3 q




Page 29

p p = :- 3 cos qD = : ws3q D = 8 3 + 3 B - 8 1 + 1 B = 4 12 4 4 4 12 12 3 0 0

SOL 1.67

Option (D) is correct. Let d " defective and y " supply by Y P (y + d) y pa k = d P (d) P (y + d) = 0.3 # 0.02 = 0.006 P (d) = 0.6 # 0.1 + 0.3 # 0.02 + 0.1 # 0.03 = 0.015 y P a k = 0.006 = 0.4 d 0.015

SOL 1.68

Option (C) is correct. 4 2 A == 2 4G

We have

6A - lI @[X] = 0

Now or or or SOL 1.69

4 - l 2 101 0 = 2 4 - l G=101G = =0 G

nodia (101)( 4 - l) + 2 (101) = 0

l =6

Option (A) is correct. We have or

d2 y + k2 y = 0 dx2 D2 y + k2 y = 0

The AE is m2 + k2 = 0 The solution of AE is m = ! ik Thus y = A sin kx + B cos kx From x = 0 , y = 0 we get B = 0 and x = a, y = 0 we get A sin ka = 0 or sin ka = 0 k = mpx a Thus y = / Am sin ` mpx j a m SOL 1.70

Option (A) is correct. ex 1 + ex For x " 3, the value of f (x) monotonically increases. We have

f (x) =

SOL 1.71

Option (B) is correct. Order is the highest derivative term present in the equation and degree is the power of highest derivative term. Order = 2 , degree = 1

SOL 1.72

Option (D) is correct. Probability of coming odd number is 12 and the probability of coming even number is 12 . Both the events are independent to each other, thus probability of coming odd number after an even number is 12 # 12 = 14 .



SOL 1.73


Chapter 1

Option (B) is correct. d2 y dy -5 + 6y = 0 2 dx dx m2 - 5m + 6 = 0

We have The A.E. is

m = 3, 2 The CF is

yc = C1 e3x + C2 e2x

Since Q = 0 , thus

y = C1 e3x + C2 e2x

Thus only (B) may be correct. SOL 1.74

Option (A) is correct. We have f (t) = e(a + 2) t + 5 = e5 .e(a + 2) t Taking Laplace transform we get 1 F (s) = e5 ; s - (a + 2) E

Thus Re (s) > (a + 2)

SOL 1.75

Option (C) is correct. For x > 0 the slope of given curve is negative. Only (C) satisfy this condition.

SOL 1.76

Option (C) is correct. Newton - Raphson Runge - kutta Method

nodia

Simpson’s Rule

Gauss elimination SOL 1.77

" Method-Solving nonlinear eq. " Solving ordinary differential eq. " Numerical Integration " Solving linear simultaneous eq.


-4 2 A == 4 3G

We have

Characteristic equation is A - lI = 0 or

4-l 2 =0 4 3-l

or

(- 4 - l)(3 - l) - 8 = 0

or

- 12 + l + l2 - 8 = 0 l2 + l - 20 = 0 l =- 5, 4

or or Eigen vector for l =- 5 (A - lI) Xi = 0 1 - (- 5) 2 x1 0 = = 4 8 - 4 G=x2 G =0 G 1 2 x1 0 =0 0G=x G = =0 G 2

Eigen values

R2 - 4R1

x1 + 2x2 = 0 Let - x1 = 2 & x2 =- 1, Thus SOL 1.78

X ==

2 -1G

Eigen vector

Option (A) is correct.




Page 31

We have 1 2 - 0.1 a and A - 1 = = 2 G A == G 0 3 0 b

AA - 1 = I

Now or

1 0 2 - 0. 1 1 a =0 3 G= 2 G = =0 1G 0 b

or

1 2a - 0.1b 1 0 == G G =0 3b 0 1

2a - 0.1 = 0 and 3b = 1 Thus solving above we have b = 1 and a = 1 3 60 1 1 7 Therefore a+b = + = 3 60 20 or

SOL 1.79

Option (A) is correct. Gaussian PDF is

nodia 1 2p s

f (x) =

and

# f (x) dx 3

-3

#

3 - (x - m)2 2s2

-3

e

dx

for - 3 # x # 3

=1

Substituting m = 0 and s = 2 in above we get 3 1 e dx = 1 # 2p 2 - 3 x2 8

or or SOL 1.80

1 2 3e- dx = 1 # 2p 2 0 x2 8

1 2p

# 0

3 - x2 8

e

dx = 1

Option (C) is correct. From orthogonal matrix [AAT ] = I Since the inverse of I is I , thus [AAT ] -1 = I-1 = I *********


CHAPTER 2 NETWORKS

2013 MCQ 2.1

MCQ 2.2

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k , k > 0 , the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2

(B) k

(C) 1/k

(D)

The transfer function

k

V2 ^s h of the circuit shown below is V1 ^s h

nodia

(A) 0.5s + 1 s+1 (C) s + 2 s+1 MCQ 2.3

ONE MARK

(B) 3s + 6 s+2 (D) s + 1 s+2

A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + j3h W . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be (A) 3 (B) 4 (C) 5 (D) 7 2013

MCQ 2.4

TWO MARKS

In the circuit shown below, if the source voltage VS = 100+53.13c V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

(A) 100+90c

(B) 800+0c


Networks

(C) 800+90c MCQ 2.5

Page 33

(D) 100+60c

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100 V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 /VWX1 , VWX2 /VYZ2 are respectively,

(A) 125/100 and 80/100 (C) 100/100 and 100/100

(B) 100/100 and 80/100 (D) 80/100 and 80/100

MCQ 2.6

Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2 . When connected in series, their effective Q factor at the same operating frequency is (A) q1 + q2 (B) ^1/q1h + ^1/q2h (C) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h

MCQ 2.7

Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and 2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in mC stored in the effective capacitance across the terminals are respectively,

nodia

(A) 2.8 and 36

(B) 7 and 119

(C) 2.8 and 32

(D) 7 and 80

Common Data For Q. 8 and 9: Consider the following figure



Networks

Chapter 2

MCQ 2.8

The current IS in Amps in the voltage source, and voltage VS in Volts across the current source respectively, are (B) 8, - 10 (A) 13, - 20 (C) - 8, 20 (D) - 13, 20

MCQ 2.9

The current in the 1W resistor in Amps is (A) 2 (B) 3.33 (C) 10 (D) 12 2012

MCQ 2.10

ONE MARK

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is

nodia

(A) zero (B) a step function (C) an exponentially decaying function (D) an impulse function MCQ 2.11

The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W

MCQ 2.12

In the circuit shown below, the current through the inductor is

2 A 1+j (C) 1 A 1+j (A)

2012 MCQ 2.13

(B) - 1 A 1+j (D) 0 A TWO MARKS

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is



Networks

(A) 0.8 W (C) 2 W MCQ 2.14

Page 35

(B) 1.4 W (D) 2.8 W

If VA - VB = 6 V then VC - VD is

nodia

(A) - 5 V (C) 3 V

(B) 2 V (D) 6 V

Common Data For Q. 15 and 16 :

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed : (i) 1 W connected at port B draws a current of 3 A (ii) 2.5 W connected at port B draws a current of 2 A

MCQ 2.15

With 10 V dc connected at port A, the current drawn by 7 W connected at port B is (B) 5/7 A (A) 3/7 A (C) 1 A (D) 9/7 A

MCQ 2.16

For the same network, with 6 V dc connected at port A, 1 W connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is (B) 7 V (A) 6 V (C) 8 V (D) 9 V 2011

MCQ 2.17

ONE MARK

In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is



Networks

(A) 6.4 - j 4.8 (C) 10 + j 0 MCQ 2.18

(B) 6.56 - j 7.87 (D) 16 + j 0

In the circuit shown below, the value of RL such that the power transferred to RL is maximum is

nodia

(A) 5 W (C) 15 W MCQ 2.19

Chapter 2

(B) 10 W (D) 20 W

The circuit shown below is driven by a sinusoidal input vi = Vp cos (t/RC ). The steady state output vo is

(A) (Vp /3) cos (t/RC ) (C) (Vp /2) cos (t/RC )

(B) (Vp /3) sin (t/RC ) (D) (Vp /2) sin (t/RC )

2011 MCQ 2.20

In the circuit shown below, the current I is equal to

(A) 1.4+0c A (C) 2.8+0c A MCQ 2.21

TWO MARKS

(B) 2.0+0c A (D) 3.2+0c A

In the circuit shown below, the network N is described by the following Y matrix: 0.1 S - 0.01 S . the voltage gain V2 is Y=> 0.01 S 0.1 SH V1



Networks

(A) 1/90 (C) –1/99 MCQ 2.22

Page 37

(B) –1/90 (D) –1/11

In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0 . The current i (t) at a time t after the switch is closed is

nodia

(A) i (t) = 15 exp (- 2 # 103 t) A (B) i (t) = 5 exp (- 2 # 103 t) A (C) i (t) = 10 exp (- 2 # 103 t) A (D) i (t) =- 5 exp (- 2 # 103 t) A 2010 MCQ 2.23

For the two-port network shown below, the short-circuit admittance parameter matrix is

4 (A) > -2 1 (C) > 0.5 MCQ 2.24

ONE MARK

-2 S 4H 0.5 S 1H

1 - 0. 5 (B) > S - 0.5 1H 4 2 (D) > S 2 4H

For parallel RLC circuit, which one of the following statements is NOT correct ? (A) The bandwidth of the circuit decreases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value.



Networks

Chapter 2

2010 MCQ 2.25

TWO MARKS

In the circuit shown, the switch S is open for a long time and is closed at t = 0 . The current i (t) for t $ 0+ is

(A) i (t) = 0.5 - 0.125e-1000t A (C) i (t) = 0.5 - 0.5e-1000t A MCQ 2.26

The current I in the circuit shown is

nodia

(A) - j1 A (C) 0 A MCQ 2.27

(B) i (t) = 1.5 - 0.125e-1000t A (D) i (t) = 0.375e-1000t A

(B) j1 A (D) 20 A

In the circuit shown, the power supplied by the voltage source is

(A) 0 W

(B) 5 W

(C) 10 W

(D) 100 W

GATE 2009 MCQ 2.28

ONE MARK

In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power.

Which of the following can be the value of the current source I ?



Networks

(A) 10 A (C) 15 A MCQ 2.29

MCQ 2.30

Page 39

(B) 13 A (D) 18 A

If the transfer function of the following network is Vo (s) 1 = 2 + sCR Vi (s)

The value of the load resistance RL is (A) R 4

(B) R 2

(C) R

(D) 2R

A fully charged mobile phone with a 12 V battery is good for a 10 minute talktime. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

nodia

(A) 220 J (C) 13.2 kJ

(B) 12 kJ (D) 14.4 J

GATE 2009 MCQ 2.31

An AC source of RMS voltage 20 V with internal impedance Zs = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W in the figure below. The reactive power consumed by the load is

(A) 8 VAR (C) 28 VAR MCQ 2.32

TWO MARK

(B) 16 VAR (D) 32 VAR

The switch in the circuit shown was on position a for a long time, and is move to position b at time t = 0 . The current i (t) for t > 0 is given by



Networks

(A) 0.2e-125t u (t) mA (C) 0.2e-1250t u (t) mA MCQ 2.33

MCQ 2.34

Chapter 2

(B) 20e-1250t u (t) mA (D) 20e-1000t u (t) mA

In the circuit shown, what value of RL maximizes the power delivered to RL ?

nodia

(A) 2.4 W

(B) 8 W 3

(C) 4 W

(D) 6 W

The time domain behavior of an RL circuit is represented by L di + Ri = V0 (1 + Be-Rt/L sin t) u (t). dt For an initial current of i (0) = V0 , the steady state value of the current is given R by (B) i (t) " 2V0 (A) i (t) " V0 R R V 2 (C) i (t) " 0 (1 + B) (D) i (t) " V0 (1 + B) R R GATE 2008

MCQ 2.35

In the following graph, the number of trees (P) and the number of cut-set (Q) are

(A) P = 2, Q = 2 (C) P = 4, Q = 6 MCQ 2.36

ONE MARK

(B) P = 2, Q = 6 (D) P = 4, Q = 10

In the following circuit, the switch S is closed at t = 0 . The rate of change of current di (0+) is given by dt



Networks

(A) 0 (R + Rs) Is (C) L

Page 41

(B) Rs Is L (D) 3

GATE 2008 MCQ 2.37

The Thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is

(A) 1

MCQ 2.38

TWO MARKS

nodia (B) 1 + s + 1 s

2 (C) 2 + s + 1 (D) s2 + s + 1 s s + 2s + 1 The driving point impedance of the following network is given by Z (s) = 2 0.2s s + 0.1s + 2

The component values are (A) L = 5 H, R = 0.5 W, C = 0.1 F (B) L = 0.1 H, R = 0.5 W, C = 5 F (C) L = 5 H, R = 2 W, C = 0.1 F (D) L = 0.1 H, R = 2 W, C = 5 F MCQ 2.39

The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows: For 2nT # t # (2n + 1) T , (n = 0, 1, 2,..) S1 to P1 and S2 to P2 For (2n + 1) T # t # (2n + 2) T, (n = 0, 1, 2,...) S1 to Q1 and S2 to Q2



Networks

Chapter 2

Assume that the capacitor has zero initial charge. Given that u (t) is a unit step function , the voltage vc (t) across the capacitor is given by (A)

3

/ (- 1) n tu (t - nT) n=1 3

(B) u (t) + 2 / (- 1) n u (t - nT) n=1 3

(C) tu (t) + 2 / (- 1) n u (t - nT) (t - nT) n=1

(D) / 60.5 - e- (t - 2nT) + 0.5e- (t - 2nT) - T @ 3

n=1

nodia

Common Data For Q.40 and 41 :

The following series RLC circuit with zero conditions is excited by a unit impulse functions d (t).

MCQ 2.40

For t > 0 , the output voltage vC ^ t h is

MCQ 2.41

-1 (B) 2 te 2 t (A) 2 ê t - e t h 3 3 1 -1 (C) 2 e 2 t cos c 3 t m (D) 2 e 2 t sin c 3 t m 2 2 3 3 For t > 0 , the voltage across the resistor is -1 2

(A) 1 _e 3 (B) e (C)

3t 2

3 2

- e- 2 t i 1

3 1 sin 3 t c 2 mG =cos c 2 t m 3 2 e -21 t sin 3 t c 2 m 3

-1 t 2

-1 (D) 2 e 2 t cos c 3 t m 2 3

Statement for linked Answers Questions 42 and 43: A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1, V2 and ammeters. A1, A2 (all assumed to be ideal), as indicated



Networks

Page 43

Under following conditions, the readings obtained are: (1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V, A2 = 1 A (2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V, A2 = 0 MCQ 2.42

The z -parameter matrix for this network is 1. 5 (A) = 4. 5 1.5 (C) = 1.5

MCQ 2.43

1.5 1.5G 4.5 1.5 G

1.5 (B) = 1.5 4.5 (D) = 1. 5

4.5 4.5G 1. 5 4.5G

The h -parameter matrix for this network is -3 3 (A) = - 1 0.67 G 3 3 (C) = 1 0.67 G

-3 -1 (B) = 3 0.67 G 3 1 (D) = - 3 - 0.67 G

nodia

GATE 2007

ONE MARK

MCQ 2.44

An independent voltage source in series with an impedance Zs = Rs + jXs delivers a maximum average power to a load impedance ZL when (B) ZL = Rs (A) ZL = Rs + jXs (C) ZL = jXs (D) ZL = Rs - jXs

MCQ 2.45

The RC circuit shown in the figure is

(A) a low-pass filter (C) a band-pass filter GATE 2007 MCQ 2.46

(B) a high-pass filter (D) a band-reject filter TWO MARKS

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2 . the value B1 is B2



Networks

(A) 4 (C) 1/2 MCQ 2.47

(B) 1 (D) 1/4

For the circuit shown in the figure, the Thevenin voltage and resistance looking into X - Y are

(A) (C) MCQ 2.48

Chapter 2

4 3 4 3

V, 2 W V, 23 W

(B) 4 V, 23 W (D) 4 V, 2 W

In the circuit shown, vC is 0 volts at t = 0 sec. For t > 0 , the capacitor current iC (t), where t is in seconds is given by

nodia

(A) 0.50 exp (- 25t) mA (B) 0.25 exp (- 25t) mA (C) 0.50 exp (- 12.5t) mA (D) 0.25 exp (- 6.25t) mA MCQ 2.49

In the ac network shown in the figure, the phasor voltage VAB (in Volts) is

(A) 0 (C) 12.5+30c

(B) 5+30c (D) 17+30c

GATE 2006 MCQ 2.50

TWO MARKS

A two-port network is represented by ABCD parameters given by V1 A B V2 =I G = =C D G=- I G 1 2 If port-2 is terminated by RL , the input impedance seen at port-1 is given by (B) ARL + C (A) A + BRL C + DRL BRL + D (C) DRL + A BRL + C

(D) B + ARL D + CRL



MCQ 2.51

Networks

Page 45

In the two port network shown in the figure below, Z12 and Z21 and respectively

(A) re and br0 (C) 0 and bro

(B) 0 and - br0 (D) re and - br0

MCQ 2.52

The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by (A) RL network only (B) RC network only (C) LC network only (D) RC as well as RL networks

MCQ 2.53

A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

nodia

(A) 0.5 A (C) 1.0 A MCQ 2.54

In the figure shown below, assume that all the capacitors are initially uncharged. If vi (t) = 10u (t) Volts, vo (t) is given by

(A) 8e -t/0.004 Volts (C) 8u (t) Volts MCQ 2.55

(B) 2.0 A (D) 0.0 A

(B) 8 (1 - e -t/0.004) Volts (D) 8 Volts

A negative resistance Rneg is connected to a passive network N having driving point impedance as shown below. For Z2 (s) to be positive real,



Networks

(A) Rneg # Re Z1 (jw), 6w (C) Rneg # Im Z1 (jw), 6w

Chapter 2

(B) Rneg # Z1 (jw) , 6w (D) Rneg # +Z1 (jw), 6w

GATE 2005 MCQ 2.56

The condition on R, L and C such that the step response y (t) in the figure has no oscillations, is

L C (C) R $ 2 (D) R = 1 LC The ABCD parameters of an ideal n: 1 transformer shown in the figure are n 0 >0 x H (A) R $ 1 2

MCQ 2.57

ONE MARK

L C L C

nodia

The value of x will be (A) n (C) n2 MCQ 2.58

(B) 1 n (D) 12 n

In a series RLC circuit, R = 2 kW , L = 1 H, and C = 1 mF The resonant 400 frequency is (A) 2 # 10 4 Hz (B) 1 # 10 4 Hz p (C) 10 4 Hz

MCQ 2.59

(B) R $

(D) 2p # 10 4 Hz

The maximum power that can be transferred to the load resistor RL from the voltage source in the figure is

(A) 1 W (C) 0.25 W

(B) 10 W (D) 0.5 W



MCQ 2.60

Networks

Page 47

The first and the last critical frequency of an RC -driving point impedance function must respectively be (A) a zero and a pole (B) a zero and a zero (C) a pole and a pole (D) a pole and a zero

GATE 2005 MCQ 2.61

MCQ 2.62

For the circuit shown in the figure, the instantaneous current i1 (t) is

(A) 10 3 90c A 2

(B) 10 3 - 90c A 2

(C) 5 60c A

(D) 5 - 60c A

nodia

Impedance Z as shown in the given figure is

(A) j29 W (C) j19 W MCQ 2.63

(B) j9 W (D) j39 W

For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a - b is

(A) 5 V and 2 W (C) 4 V and 2 W MCQ 2.64

TWO MARKS

(B) 7.5 V and 2.5 W (D) 3 V and 2.5 W

If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is



Networks

(A) 0.238 V (C) - 0.238 V MCQ 2.65

(B) 0.138 V (D) 1 V

The h parameters of the circuit shown in the figure are

0. 1 0. 1 (A) = - 0. 1 0. 3 G 30 20 (C) = 20 20G MCQ 2.66

Chapter 2

10 - 1 (B) = 1 0.05G 10 1 (D) = - 1 0.05G

A square pulse of 3 volts amplitude is applied to C - R circuit shown in the figure. The capacitor is initially uncharged. The output voltage V2 at time t = 2 sec is

nodia

(A) 3 V (C) 4 V

(B) - 3 V (D) - 4 V

GATE 2004 MCQ 2.67

MCQ 2.68

ONE MARK

Consider the network graph shown in the figure. Which one of the following is NOT a ‘tree’ of this graph ?

(A) a

(B) b

(C) c

(D) d

The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is



Networks

(A) L1 + L2 + M (C) L1 + L2 + 2M MCQ 2.69

nodia (B) 5 sin (2t - 53.1c) (D) 25 sin (2t - 53.1c)

For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is vi (t) = 2 sin 103 t . The output voltage vo (t) is equal to

(A) sin (103 t - 45c) (C) sin (103 t - 53c) MCQ 2.71

(B) L1 + L2 - M (D)L1 + L2 - 2M

The circuit shown in the figure, with R = 1 W, L = 1 H and C = 3 F 3 4 has input voltage v (t) = sin 2t . The resulting current i (t) is

(A) 5 sin (2t + 53.1c) (C) 25 sin (2t + 53.1c) MCQ 2.70

Page 49

(B) sin (103 t + 45c) (D) sin (103 t + 53c)

For the R - L circuit shown in the figure, the input voltage vi (t) = u (t). The current i (t) is



Networks

Chapter 2

GATE 2004 MCQ 2.72

For the lattice shown in the figure, Za = j2 W and Zb = 2 W . The values of the open z11 z12 are circuit impedance parameters 6z @ = = z21 z22 G

1-j (A) = 1+j 1+j (C) = 1-j MCQ 2.73

TWO MARKS

1+j 1 + jG 1+j 1 - jG

1-j 1+j (B) = -1 + j 1 - j G 1 + j -1 + j (D) = -1 + j 1 + j G

The circuit shown in the figure has initial current iL (0-) = 1 A through the inductor and an initial voltage vC (0-) =- 1 V across the capacitor. For input v (t) = u (t), the Laplace transform of the current i (t) for t $ 0 is

nodia s+2 s2 + s + 1 (D) 2 1 s +s+1

s s2 + s + 1 (C) 2 s - 2 s +s+1 (A)

MCQ 2.74

(B)

The transfer function H (s) =

Vo (s) of an RLC circuit is given by Vi (s)

106 s + 20s + 106 The Quality factor (Q-factor) of this circuit is (A) 25 (B) 50 (C) 100 (D) 5000 H (s) =

MCQ 2.75

2

For the circuit shown in the figure, the initial conditions are zero. Its transfer V (s) is function H (s) = c Vi (s)

(A)

1 s + 106 s + 106

(B)

106 s + 103 s + 106

(C)

103 s + 103 s + 106

(D)

106 s + 106 s + 106

2

2

2

2



MCQ 2.76

Networks

Page 51

Consider the following statements S1 and S2 S1 : At the resonant frequency the impedance of a series RLC circuit is zero. S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor. Which one of the following is correct? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) S1 is TRUE and S2 is FALSE (D) Both S1 and S2 are FALSE GATE 2003

MCQ 2.77

ONE MARK

The minimum number of equations required to analyze the circuit shown in the figure is

(A) 3 (C) 6

nodia (B) 4 (D) 7

MCQ 2.78

A source of angular frequency 1 rad/sec has a source impedance consisting of 1 W resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is (A) 1 W resistance (B) 1 W resistance in parallel with 1 H inductance (C) 1 W resistance in series with 1 F capacitor (D) 1 W resistance in parallel with 1 F capacitor

MCQ 2.79

A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100 . If each of R, L and C is doubled from its original value, the new Q of the circuit is (A) 25 (B) 50 (C) 100 (D) 200

MCQ 2.80

The differential equation for the current i (t) in the circuit of the figure is

2 (A) 2 d 2i + 2 di + i (t) = sin t dt dt 2 (C) 2 d 2i + 2 di + i (t) = cos t dt dt

2 (B) d 2i + 2 di + 2i (t) = cos t dt dt 2 (D) d 2i + 2 di + 2i (t) = sin t dt dt



Networks

Chapter 2

GATE 2003

TWO MARKS

MCQ 2.81

Twelve 1 W resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is (B) 1 W (A) 5 W 6 (C) 6 W (D) 3 W 5 2

MCQ 2.82

The current flowing through the resistance R in the circuit in the figure has the form P cos 4t where P is

(A) (0.18 + j0.72) (C) - (0.18 + j1.90)

(B) (0.46 + j1.90) (D) - (0.192 + j0.144)

nodia


Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0 .

MCQ 2.83

MCQ 2.84

At t = 0+ , the current i1 is (A) - V (B) - V 2R R (C) - V (D) zero 4R I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively. The equations for the loop currents I1 (s) and I2 (s) for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t = 0 , are V R + Ls + Cs1 - Ls I1 (s) s (A) > == G G 1 H= R + Cs I2 (s) - Ls 0 R + Ls + Cs1 - Ls I1 (s) - Vs (B) > == G G 1 H= R + Cs I2 (s) - Ls 0 R + Ls + Cs1 - Ls I1 (s) - Vs (C) > == G G 1 H= R + Ls + Cs I2 (s) - Ls 0 V R + Ls + Cs1 - Cs I1 (s) s (D) > == G G 1 H= R + Ls + Cs I2 (s) - Ls 0



MCQ 2.85

Networks

The driving point impedance Z (s) of a network has the pole-zero locations as shown in the figure. If Z (0) = 3 , then Z (s) is

3 (s + 3) 2 (s + 3) (B) 2 s + 2s + 3 s + 2s + 2 3 (s + 3) 2 (s - 3) (C) 2 (D) 2 s + 2s + 2 s - 2s - 3 An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) V is applied to a series combination of resistance R = 1 W and an inductance L = 1 H. The resulting steady-state current i (t) in ampere is (A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan-1 2) (A)

MCQ 2.86

Page 53

2

(B) 10 cos (t + 55c) + 10 23 cos (2t + 55c) (C) 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)

nodia

(D) 10 cos (t - 35c) + MCQ 2.87

3 2

cos (2t - 35c)

The impedance parameters z11 and z12 of the two-port network in the figure are

(A) z11 = 2.75 W and z12 = 0.25 W (C) z11 = 3 W and z12 = 0.25 W

(B) z11 = 3 W and z12 = 0.5 W (D) z11 = 2.25 W and z12 = 0.5 W

GATE 2002 MCQ 2.88

The dependent current source shown in the figure

(A) delivers 80 W (C) delivers 40 W MCQ 2.89

ONE MARK

(B) absorbs 80 W (D) absorbs 40 W

In the figure, the switch was closed for a long time before opening at t = 0 . The voltage vx at t = 0+ is



Networks

(A) 25 V (C) - 50 V

Chapter 2

(B) 50 V (D) 0 V

GATE 2002 MCQ 2.90

MCQ 2.91

TWO MARKS

In the network of the fig, the maximum power is delivered to RL if its value is

nodia

(A) 16 W

(B) 40 W 3

(C) 60 W

(D) 20 W

If the 3-phase balanced source in the figure delivers 1500 W at a leading power factor 0.844 then the value of ZL (in ohm) is approximately

(A) 90+32.44c (C) 80+ - 32.44c

(B) 80+32.44c (D) 90+ - 32.44c

GATE 2001 MCQ 2.92

The Voltage e0 in the figure is

(A) 2 V (C) 4 V MCQ 2.93

ONE MARK

(B) 4/3 V (D) 8 V

If each branch of Delta circuit has impedance equivalent Wye circuit has impedance (A) Z (B) 3Z 3

3 Z , then each branch of the



Networks

Page 55

(D) Z 3 The admittance parameter Y12 in the 2-port network in Figure is (C) 3 3 Z

MCQ 2.94

(A) - 0.02 mho (C) - 0.05 mho

(B) 0.1 mho (D) 0.05 mho

GATE 2001 MCQ 2.95

TWO MARKS

The voltage e0 in the figure is

nodia

(A) 48 V (C) 36 V

(B) 24 V (D) 28 V

MCQ 2.96

When the angular frequency w in the figure is varied 0 to 3, the locus of the current phasor I2 is given by

MCQ 2.97

In the figure, the value of the load resistor RL which maximizes the power delivered



Networks

Chapter 2

to it is

(A) 14.14 W (C) 200 W MCQ 2.98

(B) 10 W (D) 28.28 W

The z parameters z11 and z21 for the 2-port network in the figure are

nodia

(A) z11 = 6 W; z21 = 16 W 11 11

(B) z11 = 6 W; z21 = 4 W 11 11

(C) z11 = 6 W; z21 =- 16 W 11 11

(D) z11 = 4 W; z21 = 4 W 11 11

GATE 2000 MCQ 2.99

The circuit of the figure represents a

(A) Low pass filter (C) band pass filter MCQ 2.100

(B) High pass filter (D) band reject filter

In the circuit of the figure, the voltage v (t) is

(A) eat - ebt (C) aeat - bebt MCQ 2.101

ONE MARK

(B) eat + ebt (D) aeat + bebt

In the circuit of the figure, the value of the voltage source E is



Networks

(A) - 16 V (C) - 6 V

Page 57

(B) 4 V (D) 16 V

GATE 2000 MCQ 2.102

TWO MARKS

Use the data of the figure (a). The current i in the circuit of the figure (b)

nodia

(A) - 2 A (C) - 4 A

(B) 2 A

(D) 4 A

GATE 1999 MCQ 2.103

Identify which of the following is NOT a tree of the graph shown in the given figure is

(A) begh (C) abfg MCQ 2.104

ONE MARK

(B) defg (D) aegh

A 2-port network is shown in the given figure. The parameter h21 for this network can be given by

(A) - 1/2

(B) + 1/2



Networks

(C) - 3/2

Chapter 2

(D) + 3/2

GATE 1999 MCQ 2.105

The Thevenin equivalent voltage VTH appearing between the terminals A and B of the network shown in the given figure is given by

(A) j16 (3 - j4) (C) 16 (3 + j4) MCQ 2.106

(B) j16 (3 + j4) (D) 16 (3 - j4)

The value of R (in ohms) required for maximum power transfer in the network shown in the given figure is

(A) 2 (C) 8 MCQ 2.107

TWO MARK

nodia (B) 4 (D) 16

A Delta-connected network with its Wye-equivalent is shown in the given figure. The resistance R1, R2 and R3 (in ohms) are respectively

(A) 1.5, 3 and 9 (C) 9, 3 and 1.5

(B) 3, 9 and 1.5 (D) 3, 1.5 and 9

GATE 1998

ONE MARK

MCQ 2.108

A network has 7 nodes and 5 independent loops. The number of branches in the network is (A) 13 (B) 12 (C) 11 (D) 10

MCQ 2.109

The nodal method of circuit analysis is based on (A) KVL and Ohm’s law (B) KCL and Ohm’s law (C) KCL and KVL (D) KCL, KVL and Ohm’s law

MCQ 2.110

Superposition theorem is NOT applicable to networks containing



Networks

(A) nonlinear elements (C) dependent current sources MCQ 2.111

MCQ 2.113

(B) IR + IL > 1 mA (D) IR + IC > 1 mA

0 - 1/2 The short-circuit admittance matrix a two-port network is > 1/2 0 H The two-port network is (A) non-reciprocal and passive (B) non-reciprocal and active (C) reciprocal and passive (D) reciprocal and active

nodia

The voltage across the terminals a and b in the figure is

(A) 0.5 V (C) 3.5 V MCQ 2.114

(B) dependent voltage sources (D) transformers

The parallel RLC circuit shown in the figure is in resonance. In this circuit

(A) IR < 1 mA (C) IR + IC < 1 mA MCQ 2.112

Page 59

(B) 3.0 V (D) 4.0 V

A high-Q quartz crystal exhibits series resonance at the frequency ws and parallel resonance at the frequency wp . Then (A) ws is very close to, but less than wp (B) ws << wp (C) ws is very close to, but greater than wp (D) ws >> wp GATE 1997

MCQ 2.115

The current i4 in the circuit of the figure is equal to

(A) 12 A (C) 4 A MCQ 2.116

ONE MARK

(B) - 12 A (D) None or these

The voltage V in the figure equal to



Networks

(A) 3 V (C) 5 V

MCQ 2.117

nodia

MCQ 2.120

(B) 5 V (D) None of the above

The voltage V in the figure is

(A) 10 V (C) 5 V MCQ 2.119

(B) - 3 V (D) None of these

The voltage V in the figure is always equal to

(A) 9 V (C) 1 V MCQ 2.118

Chapter 2

(B) 15 V (D) None of the above

In the circuit of the figure is the energy absorbed by the 4 W resistor in the time interval (0, 3) is

(A) 36 Joules

(B) 16 Joules

(C) 256 Joules

(D) None of the above

In the circuit of the figure the equivalent impedance seen across terminals a, b, is



Networks

(A) b 16 l W 3 (C) b 8 + 12j l W 3

Page 61

(B) b 8 l W 3 (D) None of the above

GATE 1996 MCQ 2.121

nodia

In the given figure, A1, A2 and A3 are ideal ammeters. If A2 and A3 read 3 A and 4 A respectively, then A1 should read

(A) 1 A (C) 7 A MCQ 2.122

ONE MARK

(B) 5 A (D) None of these

The number of independent loops for a network with n nodes and b branches is (A) n - 1 (B) b - n (C) b - n + 1 (D) independent of the number of nodes GATE 1996

MCQ 2.123

TWO MARKS

The voltages VC1, VC2, and VC3 across the capacitors in the circuit in the given figure, under steady state, are respectively.

(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V

(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V



Networks

Chapter 2

***********

nodia



Networks

Page 63

SOLUTIONS SOL 2.1

Option (B) is correct. In the equivalent star connection, the resistance can be given as Rb Ra RC = Ra + Rb + Rc Ra Rc RB = Ra + Rb + Rc Rb Rc RA = Ra + Rb + Rc So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then 2 ^k Rb h^k Rc h Rb Rc RAl = = k RA =k kRa + kRb + kRc k Ra + Rb + Rc

nodia

Hence, it is also scaled by a factor k SOL 2.2

SOL 2.3

SOL 2.4

Option (D) is correct. For the given capacitance, C = 100mF in the circuit, we have the reactance. 4 1 = 10 XC = 1 = 6 sc s s # 100 # 10 So, 10 4 + 10 4 V2 ^s h = 4 s = s+1 s+2 V1 ^s h 10 + 10 4 + 10 4 s s Option (C) is correct. For the purely resistive load, maximum average power is transferred when 2 2 RL = RTh + XTh where RTh + jXTh is the equivalent thevenin (input) impedance of the circuit. Hence, we obtain RL = 42 + 32 5 W Option (C) is correct. For evaluating the equivalent thevenin voltage seen by the load RL , we open the circuit across it (also if it consist dependent source). The equivalent circuit is shown below

As the circuit open across RL so or,

I2 = 0 j40I2 = 0



Networks

Chapter 2

i.e., the dependent source in loop 1 is short circuited. Therefore, ^ j4h Vs VL1 = j4 + 3 40 90c j40 VTh = 10 VL1 = 100 53.13c = 100 53.13c j4 + 3 5 53.13c = 800 90c SOL 2.5

Option (C) is correct. For the given transformer, we have V = 1.25 1 VWX

Since, So, or, at

VYZ = 0.8 (attenuation factor) V VYZ = 0.8 1.25 = 1 ^ h^ h VWX VYZ = VWX V VWX = 100 V ; YZ = 100 100 VWX VWX VWZ = 100 V ; = 100 100 VYZ

nodia 1

1

1

at

2

2

2

SOL 2.6

SOL 2.7

Option (C) is correct. The quality factor of the inductances are given by q 1 = wL 1 R1 w and q2 = L2 R2

So, in series circuit, the effective quality factor is given by XLeq = wL 1 + wL 2 Q = Req R1 + R 2 q1 q wL 1 + wL 2 + 2 q R + q2 R2 R R R R 2 2 1 R2 1 R2 = 1 1 = = 1 + 1 1 + 1 R1 + R 2 R 2 R1 R 2 R1 Option (C) is correct.

Consider that the voltage across the three capacitors C1 , C2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 ....(1) V3 C2 Since, Voltage is inversely proportional to capacitance



SOL 2.8

Networks

Page 65

Now, given that C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for ^V3hmax = 2 We obtain, V2 = 2 # 2 = 0.8 volt < 5 5 i.e., V2 < ^V2hmax Hence, this is the voltage at C2 . Therefore, V3 = 2 volt V2 = 0.8 volt and V1 = V2 + V3 = 2.8 volt Now, equivalent capacitance across the terminal is Ceq = C2 C 3 + C1 = 5 # 2 + 10 = 80 mF 5+2 7 C2 + C3 Equivalent voltage is (max. value) Vmax = V1 = 2.8 So, charge stored in the effective capacitance is Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC 7 Option (D) is correct.

nodia

At the node 1, voltage is given as V1 = 10 volt Applying KCL at node 1 IS + V1 + V1 - 2 = 0 2 1 10 10 IS + + - 2 = 0 2 1 IS =- 13 A Also, from the circuit, VS - 5 # 2 = V1 & VS = 10 + V1 = 20 volt SOL 2.9

SOL 2.10

Option (C) is correct. Again from the shown circuit, the current in 1 W resistor is I = V1 = 10 = 10 A 1 1 Option (D) is correct. The s -domain equivalent circuit is shown as below.



Networks

I (s) =

Chapter 2

vc (0) /s v (0) = c 1 + 1 1 + 1 C1 s C 2 s C1 C 2

I (s) = b C1 C2 l (12 V) = 12Ceq vC (0) = 12 V C1 + C 2 Taking inverse Laplace transform for the current in time domain, i (t) = 12Ceq d (t) SOL 2.11

Option (B) is correct. In phasor form,

(Impulse)

nodia Z = 4 - j 3 = 5 - 36.86cW I = 5 100c A

Average power delivered. Pavg. = 1 I 2 Z cos q = 1 # 25 # 5 cos 36.86c = 50 W 2 2

Alternate Method: Pavg SOL 2.12

Z = (4 - j3) W , I = 5 cos (100pt + 100) A 2 1 1 = Re $ I Z . = # Re "(5) 2 # (4 - j3), = 1 # 100 = 50 W 2 2 2

Option (C) is correct

Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1

Current

V1 (j 1 + 1) + j 1 + 1 0c = j1 V1 = - 1 1 + j1 1 V1 + 1 0c - 1 + j + 1 j I1 = = = = 1 A j1 j1 (1 + j) j 1 + j



SOL 2.13

Networks

Page 67

Option (A) is correct. We obtain Thevenin equivalent of circuit B .

Thevenin Impedance :

ZTh = R Thevenin Voltage : VTh = 3 0c V

nodia

Now, circuit becomes as

I1 = 10 - 3 2+R Power transfer from circuit A to B Current in the circuit,

or

SOL 2.14

P = (I 12) 2 R + 3I1 2 = :10 - 3D R + 3 :10 - 3D 2+R 2+R P = 42 + 70R2 (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0 & R = 0.8 W

Option (A) is correct. In the given circuit VA - VB = 6 V So current in the branch will be IAB = 6 = 3 A 2 We can see, that the circuit is a one port circuit looking from terminal BD as shown below



Networks

Chapter 2

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e.

IDC = IAB = 3 A

nodia

The total current in the resistor 1 W will be

So, SOL 2.15

I1 = 2 + IDC = 2+3 = 5A VCD = 1 # (- I1) =- 5 V

(By writing KCL at node D )

Option (C) is correct. When 10 V is connected at port A the network is

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh .

IL =

VTh,10 V RTh + RL

For RL = 1 W , IL = 3 A 3=

VTh,10 V RTh + 1

...(i)



Networks

Page 69

For RL = 2.5 W , IL = 2 A 2=

VTh,10 V RTh + 2.5

...(ii)

Dividing above two 3 = RTh + 2.5 2 RTh + 1 3RTh + 3 = 2RTh + 5 RTh = 2 W Substituting RTh into equation (i) VTh,10 V = 3 (2 + 1) = 9 V Note that it is a non reciprocal two port network. Thevenin voltage seen at port B depends on the voltage connected at port A. Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is as shown below :

nodia

For RL = 7 W , SOL 2.16

IL =

VTh,10 V = 9 = 1A 2 + RL 2 + 7

Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3

VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3 This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V 9 = 10a + b ` When V1 = 6 V , VTh, 6 V = 9 V ` 7 = 6a + b Solving (i) and (ii)

...(i) ...(ii)

a = 0.5 , b = 4 Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be



Networks

So, For SOL 2.17

VTh, V = 0.5V1 + 4 V1 = 8 V VTh,8 V = 0.5 # 8 + 4 = 8 = Voc

Chapter 2

1

(open circuit voltage)

Option (A) is correct. Replacing P - Q by short circuit as shown below we have

Using current divider rule the current Isc is 25 ISC = (16 0 ) = (6.4 - j4.8) A 25 + 15 + j30 SOL 2.18

Option (C) is correct. Power transferred to RL will be maximum when RL is equal to the Thevenin resistance. We determine Thevenin resistance by killing all source as follows :

nodia RTH = 10 # 10 + 10 = 15 W 10 + 10

SOL 2.19

Option (A) is correct. The given circuit is shown below

For parallel combination of R and C equivalent impedance is R$ 1 jwC R Zp = = 1 + jwRC R+ 1 jw C Transfer function can be written as R 1 + jwRC Vout = Z p = Vin R Zs + Zp R+ 1 + jwC 1 + jwRC jwRC jwRC + (1 + jwRC) 2 j = j + (1 + j) 2 =

Here w = 1 RC



Networks

j Vout = =1 3 Vin (1 + j) 2 + j Vp v out = b l cos (t/RC) 3

Thus SOL 2.20

Page 71

Option (B) is correct. From star delta conversion we have

Thus

R1 =

Ra Rb 6.6 = = 2W Ra + Rb + Rc 6 + 6 + 6

R1 = R 2 = R 3 = 2 W Replacing in circuit we have the circuit shown below : Here

nodia

Now the total impedance of circuit is (2 + j4) (2 - j4) Z = +2 = 7W (2 + j4) (2 - j4) Current I = 14+0c = 2+0c 7 SOL 2.21

Option (D) is correct. From given admittance matrix we get I1 = 0.1V1 - 0.01V2 and I2 = 0.01V1 + 0.1V2 Now, applying KVL in outer loop;

...(1) ...(2)

V2 =- 100I2 I2 =- 0.01V2 From eq (2) and eq (3) we have

or

...(3)

- 0.01V2 = 0.01V1 + 0.1V2 - 0.11V2 = 0.01V1 V2 = - 1 11 V1 SOL 2.22

Option (A) is correct. Here we take the current flow direction as positive. At t = 0- voltage across capacitor is



Networks

Chapter 2

-3 Q =- 2.5 # 10-6 =- 50 V C 50 # 10 Thus VC (0+) =- 50 V In steady state capacitor behave as open circuit thus

VC (0-) =-

V (3) = 100 V VC (t) = VC (3) + (VC (0+) - VC (3)) e-t/RC

Now,

-t

= 100 + (- 50 - 100) e 10 # 50 # 10

-6

3

= 100 - 150e- (2 # 10 t) ic (t) = C dV dt

Now

3

= 50 # 10-6 # 150 # 2 # 103 e-2 # 10 t A 3

= 15e-2 # 10 t ic (t) = 15 exp (- 2 # 103 t) A SOL 2.23

Option (A) is correct. Given circuit is as shown below

nodia

Writing node equation at input port I1 = V1 + V1 - V2 = 4V1 - 2V2 0. 5 0. 5 Writing node equation at output port I2 = V2 + V2 - V1 =- 2V1 + 4V2 0.5 0.5 From (1) and (2), we have admittance matrix 4 -2 Y => - 2 4H SOL 2.24

...(1)

...(2)

Option (D) is correct. A parallel RLC circuit is shown below :

1 1 + 1 + jw C R jw L

Input impedance

Z in =

At resonance

1 = wC wL Z in = 1 = R 1/R

So,

(maximum at resonance)

Thus (D) is not true.



Networks

Page 73

Furthermore bandwidth is wB i.e wB \ 1 and is independent of L, R Hence statements A, B, C, are true. SOL 2.25

Option (A) is correct. Let the current i (t) = A + Be-t/t t " Time constant When the switch S is open for a long time before t < 0 , the circuit is

At t = 0 , inductor current does not change simultaneously, So the circuit is

nodia

Current is resistor (AB) i (0) = 0.75 = 0.375 A 2

Similarly for steady state the circuit is as shown below

i (3) = 15 = 0.5 A 3 -3 = 10-3 sec t = L = 15 # 10 Req 10 + (10 || 10) t

i (t) = A + Be- 1 # 10 = A + Be-100t i (0) = A + B = 0.375 -3

Now and So, Hence SOL 2.26

i (3) = A = 0.5 B = 0.375 - 0.5 =- 0.125 i (t) = 0.5 - 0.125e-1000 t A

Option (A) is correct. Circuit is redrawn as shown below

Where,

Z1 = jwL = j # 103 # 20 # 10-3 = 20j Z2 = R || XC 1 XC = 1 = =- 20j jwC j # 103 # 50 # 10-6



Networks

Z2 =

Chapter 2

1 (- 20j) 1 - 20j

R = 1W

Voltage across Z2 VZ = 2

Z2 : 20 0 = Z1 + Z 2

=c

- 20j c 1 - 20j m 20j c 20j - 1 - 20j m

: 20

(- 20j) : 20 =- j 20j + 400 - 20j m

Current in resistor R is j V I = Z =- =- j A 1 R 2

SOL 2.27

Option (A) is correct. The circuit can be redrawn as

nodia

Applying nodal analysis VA - 10 + 1 + VA - 0 = 0 2 2 Current,

2VA - 10 + 2 = 0 = V4 = 4 V I1 = 10 - 4 = 3 A 2

Current from voltage source is I 2 = I1 - 3 = 0 Since current through voltage source is zero, therefore power delivered is zero. SOL 2.28

Option (A) is correct. Circuit is as shown below

Since 60 V source is absorbing power. So, in 60 V source current flows from + to - ve direction So, I + I1 = 12 I = 12 - I1



Networks

Page 75

I is always less then 12 A So, only option (A) satisfies this conditions. SOL 2.29

Option (C) is correct. For given network we have (RL XC ) Vi V0 = R + (RL XC ) RL V0 (s) RL + 1 sRL C = = R Vi (s) R RR + L L sC + RL R+ 1 + sRL C =

RL 1 = R R + RRL sC + RL 1+ + RsC RL

But we have been given V (s) 1 = T .F. = 0 2 + sCR Vi (s) Comparing, we get 1 + R = 2 & RL = R RL SOL 2.30

nodia

Option (C) is correct. The energy delivered in 10 minutes is E =

SOL 2.31

t

t

#0 VIdt = I #0Vdt

= I # Area

= 2 # 1 (10 + 12) # 600 = 13.2 kJ 2 Option (B) is correct. From given circuit the load current is 20+0c = = 20+0c IL = V Zs + ZL (1 + 2j) + (7 + 4j) 8 + 6j = 1 (8 - 6j) = 20+0c = 2+ - f 10+f 5

where f = tan - 1 3 4

The voltage across load is VL = IL ZL The reactive power consumed by load is Pr = VL IL* = IL ZL # IL* = ZL IL 2 2 = (7 # 4j) 20+0c = (7 + 4j) = 28 + 16j 8 + 6j Thus average power is 28 and reactive power is 16. SOL 2.32

Option (B) is correct. At t = 0- , the circuit is as shown in fig below :

V (0-) = 100 V Thus V (0+) = 100 V At t = 0+ , the circuit is as shown below



Networks

Chapter 2

I (0+) = 100 = 20 mA 5k At steady state i.e. at t = 3 is I (3)= 0 i (t) = I (0+) e-

Now

t RCeq

u (t)

(0.5m + 0.3m) 0.2m = 0.16 m F 0.5m + 0.3m + 0.2m 1 1 = = 1250 RCeq 5 # 103 # 0.16 # 10-6 i (t) = 20e-1250t u (t) mA Ceq =

SOL 2.33

Option (C) is correct. For Pmax the load resistance RL must be equal to thevenin resistance Req i.e. RL = Req . The open circuit and short circuit is as shown below

nodia

The open circuit voltage is From fig

Voc = 100 V I1 = 100 = 12.5 A 8 Vx =- 4 # 12.5 =- 50 V I2 = 100 + Vx = 100 - 50 = 12.5 A 4 4 Isc = I1 + I2 = 25 A Rth = Voc = 100 = 4 W Isc 25

Thus for maximum power transfer RL = Req = 4 W SOL 2.34

Option (A) is correct. Steady state all transient effect die out and inductor act as short circuits and forced response acts only. It doesn’t depend on initial current state. From the given time domain behavior we get that circuit has only R and L in series with V0 . Thus at steady state i (t) " i (3) = V0 R

SOL 2.35

Option (C) is correct. The given graph is



Networks

Page 77

There can be four possible tree of this graph which are as follows:

There can be 6 different possible cut-set.

SOL 2.36

Option (B) is correct. Initially i (0-) = 0 therefore due to inductor i (0+) = 0 . Thus all current Is will flow in resistor R and voltage across resistor will be Is Rs . The voltage across inductor will be equal to voltage across Rs as no current flow through R.

nodia

Thus but Thus SOL 2.37

vL (0+) = Is Rs di (0+) vL (0+) = L dt di (0+) v (0+) Is Rs = L = L L dt

Option (A) is correct. Killing all current source and voltage sources we have,

or Alternative :

Zth = (1 + s) ( s1 + 1) (1 + s)( s1 + 1) [ 1 + 1 + 1 + s] = s 1 = 1 (1 + s) + ( s + 1) s+ s +1+1 Zth = 1

Here at DC source capacitor act as open circuit and inductor act as short circuit. Thus we can directly calculate thevenin Impedance as 1 W



SOL 2.38

Networks

Option (D) is correct. Z (s) = R 1 sL = 2 sC s + We have been given Z (s) = 2 0.2s s + 0.1s + 2 Comparing with given we get 1 = 0.2 or C = 5 F C 1 = 0.1 or R = 2 W RC 1 = 2 or L = 0.1 H LC

SOL 2.39

Chapter 2

s C s RC

+

1 LC

Option (C) is correct. Voltage across capacitor is t Vc = 1 # idt C 0

nodia

Here C = 1 F and i = 1 A. Therefore Vc =

t

#0 dt

For 0 < t < T , capacitor will be charged from 0 V Vc =

t

#0 dt = t

At t = T, Vc = T Volts For T < t < 2T , capacitor will be discharged from T volts as Vc = T -

t

#T dt = 2T - t

At t = 2T, Vc = 0 volts For 2T < t < 3T , capacitor will be charged from 0 V Vc =

t

#2Tdt = t - 2T

At t = 3T, Vc = T Volts For 3T < t < 4T , capacitor will be discharged from T Volts Vc = T -

t

#3Tdt = 4T - t

At t = 4T, Vc = 0 Volts For 4T < t < 5T , capacitor will be charged from 0 V Vc =

t

#4Tdt = t - 4T

At t = 5T, Vc = T Volts Thus the output waveform is



Networks

Page 79

Only option C satisfy this waveform. SOL 2.40

Option (D) is correct. Writing in transform domain we have 1 Vc (s) = 2 1 = 1 s Vs (s) ^ s + s + 1h (s + s + 1) Since Vs (t) = d (t) " Vs (s) = 1 and Vc (s) = 2 1 (s + s + 1) 3

2 Vc (s) = 2 = G 3 (s + 12 ) 2 + 43 Taking inverse Laplace transform we have

or

Vt = 2 e- sin c 3 t m 2 3 t 2

SOL 2.41

Option (B) is correct. Let voltage across resistor be vR VR (s) s = 1 1 = VS (s) ( s + s + 1) (s2 + s + 1) Since vs = d (t) " Vs (s) = 1 we get s VR (s) = 2 s = (s + s + 1) (s + 12 ) 2 + 43

nodia =

or

(s + 12 ) (s + 12 ) 2 +

vR (t) = e- cos 1 2

(s +

1 2 3 1 2 2) + 4

3 t-1 2 e- sin 3 t 2 2# 3 2

t

= e- 2 =cos SOL 2.42

3 4

-

1 2

3 t - 1 sin 3 t 2 2 G 3

Option (C) is correct. From the problem statement we have = 6 = 1. 5W z11 = v1 i1 i = 0 4 = 4.5 = 4.5W z12 = v1 i2 i = 0 1 v z21 = 2 = 6 = 1.5W i1 i = 0 4 = 1.5 = 1.5W z22 = v2 i2 i = 0 1 2

1

2

2

Thus z -parameter matrix is z11 z12 1.5 4.5 =z z G = =1.5 1.5 G 21 22 SOL 2.43

Option (A) is correct. From the problem statement we h12 = v1 = v2 i = 0 = h22 = i2 v2 i = 0 1

1

have 4. 5 = 3 1. 5 1 = 0.67 1. 5

From z matrix, we have



Networks

Chapter 2

v1 = z11 i1 + z12 i2 v2 = z21 i1 + z22 i2 i2 = - z21 = - 1.5 =- 1 = h 21 i1 z22 1.5

If v2 = 0 then

or i2 =- i1 Putting in equation for v1, we get v1 i1

v1 = (z11 - z12) i1 v2 = 0

= h11 = z11 - z12 = 1.5 - 4.5 =- 3

Hence h -parameter will be h11 h12 -3 3 =h h G = =- 1 0.67 G 21 22 SOL 2.44

Option (D) is correct. According to maximum Power Transform Theorem ZL = Zs* = (Rs - jXs)

SOL 2.45

Option (C) is correct. At w " 3 , capacitor acts as short circuited and circuit acts as shown in fig below

nodia Here we get V0 = 0 Vi

At w " 0 , capacitor acts as open circuited and circuit look like as shown in fig below

Here we get also V0 = 0 Vi So frequency response of the circuit is as shown in fig and circuit is a Band pass filter.

SOL 2.46

Option (D) is correct. We know that bandwidth of series RLC circuit is R . Therefore L Bandwidth of filter 1 is B1 = R L1 Bandwidth of filter 2 is B2 = R = R = 4R L2 L1 L1 /4 B 1 Dividing above equation 1 = B2 4



SOL 2.47

Networks

Page 81

Option (D) is correct. Here Vth is voltage across node also. Applying nodal analysis we get

Vth + Vth + Vth - 2i = 2 2 1 1 But from circuit i = Vth = Vth 1 Therefore Vth + Vth + Vth - 2Vth = 2 2 1 1 or Vth = 4 volt From the figure shown below it may be easily seen that the short circuit current at terminal XY is isc = 2 A because i = 0 due to short circuit of 1 W resistor and all current will pass through short circuit.

nodia

Therefore SOL 2.48

Rth = Vth = 4 = 2 W isc 2

Option (A) is correct. The voltage across capacitor is At t = 0+ , Vc (0+) = 0 At t = 3 , VC (3) = 5 V The equivalent resistance seen by capacitor as shown in fig is Req = 20 20 = 10kW

Time constant of the circuit is t = Req C = 10k # 4m = 0.04 s Using direct formula Vc (t) = VC (3) - [Vc (3) - Vc (0)] e-t/t or

= VC (3) (1 - e-t/t) + VC (0) e-t/t = 5 (1 - e-t/0.04) Vc (t) = 5 (1 - e-25t)



Networks

IC (t) = C

Now

Chapter 2

dVC (t) dt

= 4 # 10-6 # (- 5 # 25e-25t) = 0.5e-25t mA SOL 2.49

Option (D) is correct. Impedance

= (5 - 3j) (5 + 3j) = =

(5 - 3j) # (5 + 3j) 5 - 3j + 5 + 3j

(5) 2 - (3j) 2 = 25 + 9 = 3.4 10 10

VAB = Current # Impedance = 5+30c # 34 = 17+30c SOL 2.50

Option (D) is correct. The network is shown in figure below.

Now and

nodia V1 = AV2 - BI2

...(1)

I1 = CV2 - DI2

...(2) ...(3)

also V2 =- I2 RL From (1) and (2) we get V1 = AV2 - BI2 Thus I1 CV2 - DI2

Substituting value of V2 from (3) we get Input Impedance Zin = - A # I2 RL - BI2 - C # I2 RL - DI2 or Zin = ARL + B CRL + D SOL 2.51

Option (B) is correct. The circuit is as shown below.

V1 = re I1 At output port V2 = r0 (I2 - bI1) =- r0 bI1 + r0 I2 Comparing standard equation At input port

V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 z12 = 0 and z21 =- r0 b SOL 2.52

Option (B) is correct. For series RC network input impedance is Zins = 1 + R = 1 + sRC sC sC



Networks

Page 83

Thus pole is at origin and zero is at - 1 RC For parallel RC network input impedance is 1 R sC sC = Zin = 1 +R 1 + sRC sC Thus pole is at - 1 and zero is at infinity. RC SOL 2.53

Option (A) is correct. We know

v = Ldi dt

Taking Laplace transform we get V (s) = sLI (s) - Li (0+) As per given in question - Li (0+) =- 1 mV Thus i (0+) = 1 mV = 0.5 A 2 mH SOL 2.54

nodia

Option (B) is correct. At initial all voltage are zero. So output is also zero. Thus v0 (0+) = 0 At steady state capacitor act as open circuit.

Thus,

v0 (3) = 4 # vi = 4 # 10 = 8 5 5

The equivalent resistance and capacitance can be calculate after killing all source

Req = 1 4 = 0.8 kW Ceq = 4 1 = 5 mF t = Req Ceq = 0.8kW # 5mF = 4 ms v0 (t) = v 0 (3) - [v 0 (3) - v 0 (0+)] e-t/t = 8 - (8 - 0) e-t/0.004



Networks

Chapter 2

v0 (t) = 8 (1 - e-t/0.004) Volts SOL 2.55

Option (A) is correct. Here Z2 (s) = Rneg + Z1 (s) or Z2 (s) = Rneg + Re Z1 (s) + j Im Z1 (s) For Z2 (s) to be positive real, Re Z2 (s) $ 0 Thus or But Rneg

Rneg + Re Z1 (s) $ 0 Re Z1 (s) $- Rneg is negative quantity and - Rneg is positive quantity. Therefore Re Z1 (s) $ Rneg Rneg # Re Z1 (jw)

or SOL 2.56

For all w.

Option (C) is correct. Transfer function is 1 1 Y (s) 1 sC LC = = 2 = U (s) s LC + scR + 1 R + sL + 1 s2 + R s + 1 sC L LC

nodia

Comparing with s2 + 2xwn s + wn2 = 0 we have Here 2xwn = R , L and wn = 1 LC R Thus LC = R C x = 2L 2 L For no oscillations, x $ 1 R C $1 Thus 2 L or SOL 2.57

R $2

L C

Option (B) is correct. For given transformer I2 = V1 = n I1 V2 1 or I1 = I2 and V1 = nV2 n Comparing with standard equation V1 = AV2 + BI2

Thus SOL 2.58

I1 = CV2 + DI2 A B n 0 =C D G = =0 1 G n 1 x = n

Option (B) is correct. We have L = 1H and C = 1 # 10-6 400 Resonant frequency f0 =

1 == 2p LC 2p

1 1 # 1 # 10 - 6 400



Networks

Page 85

3 4 = 10 # 20 = 10 Hz 2p p

SOL 2.59

Option (C) is correct. Maximum power will be transferred when RL = Rs = 100W In this case voltage across RL is 5 V, therefore 2 Pmax = V = 5 # 5 = 0.25 W R 100

SOL 2.60

Option (C) is correct. For stability poles and zero interlace on real axis. In RC series network the driving point impedance is Zins = R + 1 = 1 + sRC Cs sC Here pole is at origin and zero is at s =- 1/RC , therefore first critical frequency is a pole and last critical frequency is a zero. For RC parallel network the driving point impedance is R 1 R Cs = Zinp = 1 1 + sRC R+ Cs

nodia

Here pole is s =- 1/RC and zero is at 3, therefore first critical frequency is a pole and last critical frequency is a zero. SOL 2.61

Option (A) is correct. Applying KCL we get

i1 (t) + 5+0c = 10+60c

or or SOL 2.62

i1 (t) = 10+60c - 5+0c = 5 + 5 3j - 5 i1 (t) = 5 3 +90c = 10 3 +90c 2

Option (B) is correct. If L1 = j5W and L3 = j2W the mutual induction is subtractive because current enters from dotted terminal of j2W coil and exit from dotted terminal of j5W. If L2 = j2W and L3 = j2W the mutual induction is additive because current enters from dotted terminal of both coil. Thus Z = L1 - M13 + L2 + M23 + L3 - M31 + M32 = j5 + j10 + j2 + j10 + j2 - j10 + j10 = j9

SOL 2.63

Option (B) is correct. Open circuit at terminal ab is shown below

Applying KCL at node we get Vab + Vab - 10 = 1 5 5 or

Vab = 7.5 = Vth



Networks

Chapter 2

Short circuit at terminal ab is shown below

Short circuit current from terminal ab is Isc = 1 + 10 = 3 A 5 V th Thus Rth = = 7.5 = 2.5 W Isc 3 Here current source being in series with dependent voltage source make it ineffective. SOL 2.64

Option (C) is correct. Here Va = 5 V because R1 = R2 and total voltage drop is 10 V. Now Vb = R3 # 10 = 1.1 # 10 = 5.238 V R3 + R4 2.1

nodia V = Va - Vb = 5 - 5.238 =- 0.238 V

SOL 2.65

Option (D) is correct. For h parameters we have to write V1 and I2 in terms of I1 and V2 . V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 Applying KVL at input port V1 = 10I1 + V2 Applying KCL at output port V2 = I + I 1 2 20 or I2 =- I1 + V2 20 Thus from above equation we get h11 h12 10 1 =h h G = =- 1 0.05G 12 22

SOL 2.66

Option (B) is correct. RC = 0.1 # 10 - 6 # 103 = 10 - 4 sec Since time constant RC is very small, so steady state will be reached in 2 sec. At t = 2 sec the circuit is as shown in fig.

Time constant

Vc = 3 V V2 =- Vc =- 3 V



Networks

Page 87

SOL 2.67

Option (B) is correct. For a tree there must not be any loop. So a, c, and d don’t have any loop. Only b has loop.

SOL 2.68

Option (D) is correct. The sign of M is as per sign of L If current enters or exit the dotted terminals of both coil. The sign of M is opposite of L If current enters in dotted terminal of a coil and exit from the dotted terminal of other coil. Thus Leq = L1 + L2 - 2M

SOL 2.69

Option (A) is correct. Here w = 2 and V = 1+0c Y = 1 + jw C + 1 R jw L = 3 + j2 # 3 + 1 1 = 3 + j4 j2 # 4 -1 4 = 5+ tan = 5+53.11c 3

nodia I = V * Y = (1+0c)( 5+53.1c) = 5+53.1c

Thus SOL 2.70

i (t) = 5 sin (2t + 53.1c)


vi (t) =

2 sin 103 t

Here w = 103 rad and Vi = Now

2 +0c

1 jw C 1 V0 = .Vt = V 1 + jwCR i R+ 1 jw C 1 1 + j # 103 # 10 - 3 = 1 - 45c

=

2 + 0c

v0 (t) = sin (103 t - 45c) SOL 2.71

Option (C) is correct. Input voltage

vi (t) = u (t)

Taking Laplace transform Impedance

or

Vi (s) = 1 s

Z (s) = s + 2 V (s) 1 I (s) = i = s + 2 s (s + 2) I (s) = 1 ; 1 - 1 E 2 s s+2

Taking inverse Laplace transform i (t) = 1 (1 - e-2t) u (t) 2 At t = 0 , i (t) = 0 1 At t = 2 , i (t) = 0.31 At t = 3 , i (t) = 0.5 Graph (C) satisfies all these conditions. SOL 2.72




Networks

Chapter 2

We know that

where

V1 = z11 I1 + z12 I2 V2 = z11 I1 + z22 I2 z11 = V1 I1 I = 0 z21 = V2 I1 I = 0 2

1

Consider the given lattice network, when I2 = 0 . There is two similar path in the circuit for the current I1. So I = 1 I1 2

nodia

For z11 applying KVL at input port we get Thus

V1 = I (Za + Zb) V1 = 1 I1 (Za + Zb) 2 1 z11 = (Za + Zb) 2

For Z21 applying KVL at output port we get V2 = Za I1 - Zb I1 2 2 Thus V2 = 1 I1 (Za - Zb) 2 z21 = 1 (Za - Zb) 2 For this circuit z11 = z22 and z12 = z21. Thus V R SZa + Zb Za - Zb W z11 z12 2 2 =z z G = SSZa - Zb Za + Zb WW 21 22 S 2 2 W X T Here Za = 2j and Zb = 2W z11 z12 1+j j-1 Thus =z z G = =j - 1 1 + j G 21 22 SOL 2.73

Option (B) is correct. Applying KVL, v (t) = Ri (t) +

Ldi (t) 1 + dt C

#0

3

i (t) dt

Taking L.T. on both sides, V (s) = RI (s) + LsI (s) - Li (0+) +

Hence

I (s) vc (0+) + sC sC

v (t) = u (t) thus V (s) = 1 s 1 = I (s) + sI (s) - 1 + I (s) - 1 s s s



Networks

or SOL 2.74

Page 89

2 + 1 = I (s) 6s2 + s + 1@ s s I (s) = 2 s + 2 s +s+1

Option (B) is correct. Characteristics equation is s2 + 20s + 106 = 0 Comparing with s2 + 2xwn s + wn2 = 0 we have wn = 106 = 103 2xw = 20 2x = 203 = 0.02 10 Q = 1 = 1 = 50 2x 0.02

Thus Now SOL 2.75

Option (D) is correct. 1 V0 (s) 1 sC = = 2 1 Vi (s) s LC sCR + 1 + R + sL + sC 1 = 2 -2 s (10 # 10-4) + s (10-4 # 10 4) + 1 106 = -6 2 1 = 2 10 s + s + 1 s + 106 s + 106

nodia H (s) =

SOL 2.76

Option (D) is correct. Impedance of series RLC circuit at resonant frequency is minimum, not zero. Actually imaginary part is zero. Z = R + j ` wL - 1 j wC At resonance wL - 1 = 0 and Z = R that is purely resistive. Thus S1 is false wC Now quality factor Q =R C L Since G = 1 , Q = 1 C G L R If G - then Q . provided C and L are constant. Thus S2 is also false.

SOL 2.77

Option (B) is correct. Number of loops = b - n + 1 = minimum number of equation Number of branches = b = 8 Number of nodes = n = 5 Minimum number of equation = 8 - 5 + 1 = 4

SOL 2.78

Option (C) is correct. For maximum power transfer Thus

SOL 2.79

ZL = ZS* = Rs - jXs ZL = 1 - 1j

Option (B) is correct. Q = 1 R

L C



Networks

When R, L and C are doubled, 2L = 1 Q' = 1 2R 2C 2R 100 Thus = 50 Q' = 2 SOL 2.80

Chapter 2

L =Q C 2

Option (C) is correct. Applying KVL we get, di (t) 1 + # i (t) dt dt C di (t) sin t = 2i (t) + 2 + # i (t) dt dt sin t = Ri (t) + L or

Differentiating with respect to t , we get 2di (t) 2d2 i (t) cos t = + i (t) + dt dt2 SOL 2.81

Option (A) is correct. For current i there is 3 similar path. So current will be divide in three path

nodia

so, we get Vab - b i # 1l - b i # 1l - b 1 # 1l = 0 3 6 3 Vab = R = 1 + 1 + 1 = 5 W eq i 6 3 6 3 SOL 2.82

Option ( ) is correct. Data are missing in question as L1 &L2 are not given.

SOL 2.83

Option (A) is correct. At t = 0 - circuit is in steady state. So inductor act as short circuit and capacitor act as open circuit.

At t = 0 - ,

i1 (0 -) = i2 (0 -) = 0

vc (0 -) = V At t = 0+ the circuit is as shown in fig. The voltage across capacitor and



Networks

Page 91

current in inductor can’t be changed instantaneously. Thus

i1 = i2 =- V 2R

At t = 0+ , SOL 2.84

Option (C) is correct. When switch is in position 2, as shown in fig in question, applying KVL in loop (1), RI1 (s) + V + 1 I1 (s) + sL [I1 (s) - I2 (s)] = 0 s sC or I1 (s) 8R + 1 + sL B - I2 (s) sL = - V s sc

nodia z11 I1 + z12 I2 = V1

Applying KVL in loop 2,

sL [I2 (s) - I1 (s)] + RI2 (s) + 1 I2 (s) = 0 sC or

Z12 I1 + Z22 I2 = V2 - sLI1 (s) + 8R + sL + 1 BI2 (s) = 0 sc

Now comparing with

Z11 Z12 I1 V1 =Z Z G=I G = =V G 21 22 2 2 we get

SOL 2.85

R - sL SR + sL + 1 sC S R + sL + 1 - sL SS sC T Option (B) is correct. Zeros =- 3 Pole1 =- 1 + j

V W I1 (s) -V W= => s H G WW I2 (s) 0 X

Pole 2 =- 1 - j K (s + 3) K (s + 3) K (s + 3) Z (s) = = = 2 2 (s + 1 + j)( s + 1 - j) (s + 1) - j (s + 1) 2 + 1 From problem statement Z (0) w = 0 = 3 Thus 3K = 3 and we get K = 2 2 2 (s + 3) Z (s) = 2 s + 2s + 2 SOL 2.86

Option (C) is correct. v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) 1 4444 2 4 4 44 3 1 4 4 4 4 2 4 4 4 4 3 v1

v2

Thus we get w1 = 1 and w2 = 2 Now Z1 = R + jw1 L = 1 + j1



Networks

Chapter 2

Z2 = R + jw2 L = 1 + j2 i (t) = =

v1 (t) v2 (t) 10 2 cos (t + 10c) 10 5 cos (2t + 10c) + = + 1+j 1 + j2 Z1 Z2 10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 12 + 22 + tan-1 1 12 + 22 tan-1 2

10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 2 + tan-1 45c 5 tan-1 2 i (t) = 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2) =

SOL 2.87

Option (A) is correct. Using 3- Y conversion

nodia 2 # 1 = 2 = 0. 5 2+1+1 4 R2 = 1 # 1 = 1 = 0.25 2+1+1 4 R3 = 2 # 1 = 0.5 2+1+1 R1 =

Now the circuit is as shown in figure below.

Now

z11 = V1 I1

I2 = 0

= 2 + 0.5 + 0.25 = 2.75

z12 = R3 = 0.25 SOL 2.88

Option (A) is correct. Applying KCL at for node 2,

V2 + V2 - V1 = V1 5 5 5 or V2 = V1 = 20 V Voltage across dependent current source is 20 thus power delivered by it is PV2 # V1 = 20 # 20 = 80 W 5 5 It deliver power because current flows from its +ive terminals.



SOL 2.89

Networks

Page 93

Option (C) is correct. When switch was closed, in steady state, iL (0 -) = 2.5 A

At t = 0+ , iL (0+) = iL (0 -) = 2.5 A and all this current of will pass through 2 W resistor. Thus Vx =- 2.5 # 20 =- 50 V SOL 2.90

Option (A) is correct. For maximum power delivered, RL must be equal to Rth across same terminal.

nodia

Applying KCL at Node, we get 0.5I1 = Vth + I1 20 or but Thus

Vth + 10I1 = 0 I1 = Vth - 50 40 Vth + Vth - 50 = 0 4

or Vth = 10 V For Isc the circuit is shown in figure below.

but

Isc = 0.5I1 - I1 =- 0.5I1 I1 =- 50 =- 1.25 A 40 Isc =- 0.5 # - 12.5 = 0.625 A Rth = Vth = 10 = 16 W Isc 0.625

SOL 2.91

Option (D) is correct. IP , VP " Phase current and Phase voltage IL, VL " Line current and line voltage



Networks

Now So,

also

Chapter 2

VP = c VL m and IP = IL 3 Power = 3VP IL cos q 1500 = 3 c VL m (IL) cos q 3 V L IL = c 3 ZL m 1500 = 3 c VL mc VL m cos q 3 3 ZL ZL =

(400) 2 (.844) = 90 W 1500

As power factor is leading So, cos q = 0.844 " q = 32.44 As phase current leads phase voltage ZL = 90+ - q = 90+ - 32.44c SOL 2.92

or SOL 2.93

e0 = 4 V

Option (A) is correct. The star delta circuit is shown as below

Here and

Now SOL 2.94

nodia

Option (C) is correct. Applying KCL, we get e0 - 12 + e0 + e0 = 0 4 4 2+2

ZAB = ZBC = ZCA = 3 Z ZAB ZCA ZA = ZAB + ZBC + ZCA ZAB ZBC ZB = ZAB + ZBC + ZCA ZBC ZCA ZC = ZAB + ZBC + ZCA ZA = ZB = ZC =

3Z 3Z = Z 3Z+ 3Z+ 3Z 3

Option (C) is correct. y11 y12 y1 + y3 - y3 =y y G = = - y y + y G 21 22 3 2 3 y12 =- y3 y12 =- 1 =- 0.05 mho 20



SOL 2.95

Networks

Page 95

Option (D) is correct. We apply source conversion the circuit as shown in fig below.

Now applying nodal analysis we have e0 - 80 + e0 + e0 - 16 = 0 10 + 2 12 6 4e0 = 112 e0 = 112 = 28 V 4

or

SOL 2.96

Option (A) is correct. jwC I2 = Em +01c = Em +0c + 1 jwCR2 R2 + jwC

nodia

+90c + tan-1 wCR2 E m wC I2 = + (90c - tan-1 wCR2) 1 + w2 C2 R 22 At w = 0 I2 = 0 and at w = 3, I2 = Em R2 Only figure given in option (A) satisfies both conditions. +I2 =

SOL 2.97

Option (A) is correct. Xs = wL = 10 W For maximum power transfer RL =

SOL 2.98

Rs2 + Xs2 =

102 + 102 = 14.14 W

Option (C) is correct. Applying KVL in LHS loop or

E1 = 2I1 + 4 (I1 + I2) - 10E1 E1 = 6I1 + 4I2 11 11

Thus z11 = 6 11 Applying KVL in RHS loop E2 = 4 (I1 + I2) - 10E1 = 4 (I1 + I2) - 10 c 6I1 + 4I2 m =- 16I1 + 4I2 11 11 11 11 Thus z21 =- 16 11 SOL 2.99

Option (D) is correct. At w = 0 , circuit act as shown in figure below.



Networks

V0 = RL Vs RL + Rs

Chapter 2

(finite value)

At w = 3 , circuit act as shown in figure below:

nodia V0 = RL Vs RL + Rs

At resonant frequency w =

(finite value)

1 circuit acts as shown in fig and V = 0 . 0 LC

Thus it is a band reject filter. SOL 2.100

Option (D) is correct. Applying KCL we get Now

SOL 2.101

iL = eat + ebt V (t) = vL = L diL = L d [eat + ebt] = aeat + bebt dt dt

Option (A) is correct. Going from 10 V to 0 V

10 + 5 + E + 1 = 0 or SOL 2.102

E =- 16 V

Option (C) is correct. This is a reciprocal and linear network. So we can apply reciprocity theorem



Networks

Page 97

which states “Two loops A & B of a network N and if an ideal voltage source E in loop A produces a current I in loop B , then interchanging positions an identical source in loop B produces the same current in loop A. Since network is linear, principle of homogeneity may be applied and when volt source is doubled, current also doubles. Now applying reciprocity theorem i = 2 A for 10 V V = 10 V, i = 2 A V =- 20 V, i =- 4 A SOL 2.103

Option (C) is correct. Tree is the set of those branch which does not make any loop and connects all the nodes. abfg is not a tree because it contains a loop l node (4) is not connected

SOL 2.104

Option (A) is correct. For a 2-port network the parameter h21 is defined as h21 = I2 I1 V = 0 (short circuit)

nodia 2

Applying node equation at node a we get Va - V1 + Va - 0 + Va - 0 = 0 R R R 3Va = V1 & Va = V1 3 V1 - V1 3 = 2V1 Now I1 = V1 - Va = R R 3R 0 - V1 - V 3 = 1 and I2 = 0 - Va = R R 3R - V1 /3R - 1 I2 Thus = h21 = = 2 I1 V = 0 2V1 /3R 2

SOL 2.105

Option (A) is correct. Applying node equation at node A Vth - 100 (1 + j0) Vth - 0 =0 + 3 4j or or

4jVth - 4j100 + 3Vth = 0 Vth (3 + 4j) = 4j100



Networks

Vth = By simplifying Vth =

Chapter 2

4j100 3 + 4j

4j100 3 - 4j 3 + 4j # 3 - 4j

Vth = 16j (3 - j4) SOL 2.106

Option (C) is correct. For maximum power transfer RL should be equal to RTh at same terminal. so, equivalent Resistor of the circuit is

Req = 5W 20W + 4W

nodia Req = 5.20 + 4 = 4 + 4 = 8 W 5 + 20

SOL 2.107

Option (D) is correct. Delta to star conversion Rab Rac = 5 # 30 = 150 = 3 W R1 = 50 Rab + Rac + Rbc 5 + 30 + 15 R R 5 15 # ab bc = = 1. 5 W R2 = Rab + Rac + Rbc 5 + 30 + 15 Rac Rbc = 15 # 30 = 9 W R3 = Rab + Rac + Rbc 5 + 30 + 15

SOL 2.108

Option (C) is correct. No. of branches = n + l - 1 = 7 + 5 - 1 = 11

SOL 2.109

Option (B) is correct. In nodal method we sum up all the currents coming & going at the node So it is based on KCL. Furthermore we use ohms law to determine current in individual branch. Thus it is also based on ohms law.

SOL 2.110

Option (A) is correct. Superposition theorem is applicable to only linear circuits.

SOL 2.111


SOL 2.112

Option (B) is correct. For reciprocal network y12 = y21 but here y12 =- 12 ! y21 = 12 . Thus circuit is non reciprocal. Furthermore only reciprocal circuit are passive circuit.

SOL 2.113

Option (C) is correct. Taking b as reference node and applying KCL at a we get Vab - 1 + Vab = 3 2 2 or or

SOL 2.114

Vab - 1 + Vab = 6 Vab = 6 + 1 = 3.5 V 2




SOL 2.115

Networks

Page 99

Option (B) is correct. The given figure is shown below.

Applying KCL at node a we have I = i 0 + i1 = 7 + 5 = 12 A Applying KCL at node f so SOL 2.116

I =- i 4 i 4 =- 12 amp


so

nodia V = 3 - 0 = 3 volt

SOL 2.117

Option (D) is correct. Can not determined V without knowing the elements in box.

SOL 2.118

Option (A) is correct. The voltage V is the voltage across voltage source and that is 10 V.

SOL 2.119

Option (B) is correct. Voltage across capacitor -t

VC (t) = VC (3) + (VC (0) - VC (3)) e RC Here VC (3) = 10 V and (VC (0) = 6 V. Thus -t

-t

-t

VC (t) = 10 + (6 - 10) e RC = 10 - 4e RC = 10 - 4e 8 Now

VR (t) = 10 - VC (t) -t

-t

= 10 - 10 + 4e RC = 4e RC Energy absorbed by resistor E SOL 2.120

#0

3

V R2 (t) = R

#0

3

-t

16e 4 = 4

-t

#0 3 4e 4

= 16 J

Option (B) is correct. It is a balanced whetstone bridge R1 R 3 b R2 = R 4 l so equivalent circuit is



Networks

Chapter 2

Zeq = (4W 8W) = 4 # 8 = 8 3 4+8 SOL 2.121

Option (B) is correct. Current in A2 , I2 = 3 amp Inductor current can be defined as I2 =- 3j Current in A 3 , Total current

I3 = 4 I1 = I2 + I 3 = 4 - 3j I =

(4) 2 + (3) 2 = 5 amp

SOL 2.122

Option (C) is correct. For a tree we have (n - 1) branches. Links are the branches which from a loop, when connect two nodes of tree. so if total no. of branches = b No. of links = b - (n - 1) = b - n + 1 Total no. of links in equal to total no. of independent loops.

SOL 2.123

Option (B) is correct. In the steady state condition all capacitors behaves as open circuit & Inductors behaves as short circuits as shown below :

nodia

Thus voltage across capacitor C1 is VC = 100 # 40 = 80 V 10 + 40 1

Now the circuit faced by capacitor C2 and C 3 can be drawn as below :

Voltage across capacitor C2 and C 3 are VC = 80 C 3 = 80 # 3 = 48 volt 5 C2 + C3 VC = 80 C2 = 80 # 2 = 32 volt 5 C2 + C3 2

3

***********


CHAPTER 3 ELECTRONICS DEVICES

2013

ONE MARK

MCQ 3.1

In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is (A) injection, and subsequent diffusion and recombination of minority carriers (B) injection, and subsequent drift and generation of minority carriers (C) extraction, and subsequent diffusion and generation of minority carriers (D) extraction, and subsequent drift and recombination of minority carriers

MCQ 3.2

In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces (A) superior quality oxide with a higher growth rate (B) inferior quality oxide with a higher growth rate (C) inferior quality oxide with a lower growth rate (D) superior quality oxide with a lower growth rate

MCQ 3.3

nodia

In a MOSFET operating in the saturation region, the channel length modulation effect causes (A) an increase in the gate-source capacitance (B) a decrease in the transconductance (C) a decrease in the unity-gain cutoff frequency (D) a decrease in the output resistance 2013

MCQ 3.4

TWO MARKS

The small-signal resistance (i.e., dVB /dID ) in kW offered by the n-channel MOSFET M shown in the figure below, at a bias point of VB = 2 V is (device data for M: device transconductance parameter kN = mn C 0' x ^W/L h = 40 mA/V2 , threshold voltage VTN = 1 V , and neglect body effect and channel length modulation effects)

(A) 12.5 (B) 25 (C) 50 (D) 100


Electronics Devices

2012

Chapter 3

TWO MARKS 10

3

MCQ 3.5

The source of a silicon (ni = 10 per cm ) n -channel MOS transistor has an area of 1 sq mm and a depth of 1 mm . If the dopant density in the source is 1019 /cm3 , the number of holes in the source region with the above volume is approximately (A) 107 (B) 100 (C) 10 (D) 0

MCQ 3.6

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

(A) Vin < 1.875 V (B) 1.875 V < Vin < 3.125 V

nodia

(C) Vin > 3.125 V (D) 0 < Vin < 5 V

Common Data For Q. 7 and 8

In the three dimensional view of a silicon n -channel MOS transistor shown below, d = 20 nm . The transistor is of width 1 mm . The depletion width formed at every p-n junction is 10 nm. The relative permittivity of Si and SiO 2 , respectively, are 11.7 and 3.9, and e0 = 8.9 # 10-12 F/m .

MCQ 3.7

The gate source overlap capacitance is approximately (A) 0.7 fF (B) 0.7 pF (C) 0.35 fF (D) 0.24 pF

MCQ 3.8

The source-body junction capacitance is approximately (B) 7 fF (A) 2 fF (C) 2 pF (D) 7 pF



Electronics Devices

2011

Page 103

ONE MARK

MCQ 3.9

Drift current in the semiconductors depends upon (A) only the electric field (B) only the carrier concentration gradient (C) both the electric field and the carrier concentration (D) both the electric field and the carrier concentration gradient

MCQ 3.10

A Zener diode, when used in voltage stabilization circuits, is biased in (A) reverse bias region below the breakdown voltage (B) reverse breakdown region (C) forward bias region (D) forward bias constant current mode

MCQ 3.11

A silicon PN junction is forward biased with a constant current at room temperature. When the temperature is increased by 10ºC, the forward bias voltage across the PN junction (A) increases by 60 mV (B) decreases by 60 mV

nodia

(C) increases by 25 mV (D) decreases by 25 mV 2011

TWO MARKS


The channel resistance of an N-channel JFET shown in the figure below is 600 W when the full channel thickness (tch ) of 10 μm is available for conduction. The built-in voltage of the gate P+ N junction (Vbi ) is - 1 V . When the gate to source voltage (VGS ) is 0 V, the channel is depleted by 1 μm on each side due to the built in voltage and hence the thickness available for conduction is only 8 μm

MCQ 3.12

The channel resistance when VGS =- 3 V is (B) 917 W (A) 360 W (C) 1000 W (D) 3000 W

MCQ 3.13

The channel resistance when VGS = 0 V is (A) 480 W (B) 600 W



Electronics Devices

(C) 750 W

Chapter 3

(D) 1000 W

2010

ONE MARK

MCQ 3.14

At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n -channel MOSFET is (A) 450 cm2 / V-s (B) 1350 cm2 / V-s (C) 1800 cm2 / V-s (D) 3600 cm2 / V-s

MCQ 3.15

Thin gate oxide in a CMOS process in preferably grown using (A) wet oxidation (B) dry oxidation (C) epitaxial oxidation (D) ion implantation 2010

MCQ 3.16

TWO MARKS

In a uniformly doped BJT, assume that NE , NB and NC are the emitter, base and collector doping in atoms/cm3 , respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following condition is TRUE (A) NE = NB = NC (B) NE >> NB and NB > NC

nodia

(C) NE = NB and NB < NC (D) NE < NB < NC MCQ 3.17

Compared to a p-n junction with NA = ND = 1014 /cm3 , which one of the following statements is TRUE for a p-n junction with NA = ND = 1020 /cm3 ? (A) Reverse breakdown voltage is lower and depletion capacitance is lower (B) Reverse breakdown voltage is higher and depletion capacitance is lower (C) Reverse breakdown voltage is lower and depletion capacitance is higher (D) Reverse breakdown voltage is higher and depletion capacitance is higher

Statements for Linked Answer Question : 18 and 19 The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T = 300 K electronic charge = 1.6 # 10-19 C , thermal voltage = 26 mV and electron mobility = 1350 cm2 / V-s

MCQ 3.18

The magnitude of the electric field at x = 0.5 mm is (A) 1 kV/cm (B) 5 kV/cm (C) 10 kV/cm

MCQ 3.19

(D) 26 kV/cm

The magnitude of the electron of the electron drift current density at x = 0.5 mm



Electronics Devices

is (A) 2.16 # 10 4 A/cm2 (C) 4.32 # 103 A/cm2

Page 105

(B) 1.08 # 10 4 A/m2 (D) 6.48 # 102 A/cm2

2009

ONE MARK

MCQ 3.20

In an n-type silicon crystal at room temperature, which of the following can have a concentration of 4 # 1019 cm - 3 ? (A) Silicon atoms (B) Holes (C) Dopant atoms (D) Valence electrons

MCQ 3.21

The ratio of the mobility to the diffusion coefficient in a semiconductor has the units (A) V - 1 (B) cm.V1 (C) V.cm - 1 (D) V.s 2009

MCQ 3.22

nodia

TWO MARKS

Consider the following two statements about the internal conditions in a n channel MOSFET operating in the active region. S1 : The inversion charge decreases from source to drain S2 : The channel potential increases from source to drain. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, but S2 is not a reason for S1 (D) Both S1 and S2 are true, and S2 is a reason for S1

Common Data For Q. 23 and 24 Consider a silicon p - n junction at room temperature having the following parameters: Doping on the n -side = 1 # 1017 cm - 3 Depletion width on the n -side = 0.1mm Depletion width on the p -side = 1.0mm Intrinsic carrier concentration = 1.4 # 1010 cm - 3 Thermal voltage = 26 mV Permittivity of free space = 8.85 # 10 - 14 F.cm - 1 Dielectric constant of silicon = 12 MCQ 3.23

The built-in potential of the junction (A) is 0.70 V (B) is 0.76 V (C) is 0.82 V (D) Cannot be estimated from the data given

MCQ 3.24

The peak electric field in the device is (A) 0.15 MV . cm - 1, directed from p -region to n -region



Electronics Devices

Chapter 3

(B) 0.15 MV . cm - 1, directed from n -region to p -region (C) 1.80 MV . cm - 1, directed from p-region to n -region (D) 1.80 MV . cm - 1, directed from n -region to p -region 2008

ONE MARK

MCQ 3.25

Which of the following is NOT associated with a p - n junction ? (A) Junction Capacitance (B) Charge Storage Capacitance (C) Depletion Capacitance (D) Channel Length Modulations

MCQ 3.26

Which of the following is true? (A) A silicon wafer heavily doped with boron is a p+ substrate (B) A silicon wafer lightly doped with boron is a p+ substrate (C) A silicon wafer heavily doped with arsenic is a p+ substrate (D) A silicon wafer lightly doped with arsenic is a p+ substrate

MCQ 3.27

A silicon wafer has 100 nm of oxide on it and is furnace at a temperature above 1000c C for further oxidation in dry oxygen. The oxidation rate (A) is independent of current oxide thickness and temperature (B) is independent of current oxide thickness but depends on temperature (C) slows down as the oxide grows (D) is zero as the existing oxide prevents further oxidation

MCQ 3.28

The drain current of MOSFET in saturation is given by ID = K (VGS - VT ) 2 where K is a constant. The magnitude of the transconductance gm is

nodia

(A)

K (VGS - VT ) 2 VDS

(B) 2K (VGS - VT )

(C)

Id VGS - VDS

(D)

2008

K (VGS - VT ) 2 VGS TWO MARKS

MCQ 3.29

The measured trans conductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD . Which of the following figures represents the expected dependence of gm on VG ?

MCQ 3.30

Silicon is doped with boron to a concentration of 4 # 1017 atoms cm3 . Assume



Electronics Devices

Page 107

the intrinsic carrier concentration of silicon to be 1.5 # 1010 / cm 3 and the value of kT/q to be 25 mV at 300 K. Compared to undopped silicon, the fermi level of doped silicon (A) goes down by 0.31 eV (B) goes up by 0.13 eV (C) goes down by 0.427 eV (D) goes up by 0.427 eV MCQ 3.31

The cross section of a JFET is shown in the following figure. Let Vc be - 2 V and let VP be the initial pinch -off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual trans conductances of the initial and the modified JFET is

(A) 4 (C) e MCQ 3.32

nodia

1 - 2/Vp o 1 - 1/2Vp

1 - 2/Vp (B) 1 e 2 1 - 1/2Vp o 1 - (2 - Vp ) (D) 1 - [1 (2 Vp )]

Consider the following assertions. S1 : For Zener effect to occur, a very abrupt junction is required. S2 : For quantum tunneling to occur, a very narrow energy barrier is required. Which of the following is correct ? (A) Only S2 is true (B) S1 and S2 are both true but S2 is not a reason for S1 (C) S1 and S2 and are both true but S2 is not a reason for S1 (D) Both S1 and S2 are false 2007

MCQ 3.33

MCQ 3.34

ONE MARK

The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration of NA per cm3 (where NA >> ni , the electron concentration per cm3 at 300 K will be) (A) ni (B) ni + NA 2 (C) NA - ni (D) ni NA + In a p n junction diode under reverse biased the magnitude of electric field is maximum at (A) the edge of the depletion region on the p-side (B) the edge of the depletion region on the n -side (C) the p+ n junction (D) the centre of the depletion region on the n -side



Electronics Devices

2007 MCQ 3.35

Chapter 3

TWO MARKS

Group I lists four types of p - n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of the device in its normal mode of operation. Group - I Group-II (P) Zener Diode (1) Forward bias (Q) Solar cell (2) Reverse bias (R) LASER diode (S) Avalanche Photodiode (A) P - 1, Q - 2, R - 1, S - 2 (B) P - 2, Q - 1, R - 1, S - 2 (C) P - 2, Q - 2, R - 1, S- -2 (D) P - 2, Q - 1, R - 2, S - 2

MCQ 3.36

Group I lists four different semiconductor devices. match each device in Group I with its charactecteristic property in Group II Group-I Group-II (P) BJT (1) Population iniversion (Q) MOS capacitor (2) Pinch-off voltage (R) LASER diode (3) Early effect (S) JFET (4) Flat-band voltage

nodia

(A) P - 3, Q - 1, R - 4, S - 2 (B) P - 1, Q - 4, R - 3, S - 2 (C) P - 3, Q - 4, R - 1, S - 2

(D) P - 3, Q - 2, R - 1, S - 4 MCQ 3.37

A p+ n junction has a built-in potential of 0.8 V. The depletion layer width a reverse bias of 1.2 V is 2 mm. For a reverse bias of 7.2 V, the depletion layer width will be (A) 4 mm (B) 4.9 mm (C) 8 mm (D) 12 mm

MCQ 3.38

The DC current gain (b) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is (A) 0.980 (B) 0.985 (C) 0.990 (D) 0.995

Common Data For Q. 39 to 41 The figure shows the high-frequency capacitance - voltage characteristics of Metal/Sio 2 /silicon (MOS) capacitor having an area of 1 # 10 - 4 cm 2 . Assume that the permittivities (e0 er ) of silicon and Sio2 are 1 # 10 - 12 F/cm and 3.5 # 10 - 13 F/cm respectively.



Electronics Devices

Page 109

MCQ 3.39

The gate oxide thickness in the MOS capacitor is (A) 50 nm (B) 143 nm (C) 350 nm (D) 1 mm

MCQ 3.40

The maximum depletion layer width in silicon is (A) 0.143 mm (B) 0.857 mm (C) 1 mm (D) 1.143 mm

MCQ 3.41

Consider the following statements about the C - V characteristics plot : S1 : The MOS capacitor has as n -type substrate

nodia

S2 : If positive charges are introduced in the oxide, the C - V polt will shift to the left. Then which of the following is true? (A) Both S1 and S2 are true (B) S1 is true and S2 is false (C) S1 is false and S2 is true (D) Both S1 and S2 are false 2006

ONE MARK

MCQ 3.42

The values of voltage (VD) across a tunnel-diode corresponding to peak and valley currents are Vp, VD respectively. The range of tunnel-diode voltage for VD which the slope of its I - VD characteristics is negative would be (B) 0 # VD < Vp (A) VD < 0 (C) Vp # VD < Vv (D) VD $ Vv

MCQ 3.43

The concentration of minority carriers in an extrinsic semiconductor under equilibrium is (A) Directly proportional to doping concentration (B) Inversely proportional to the doping concentration (C) Directly proportional to the intrinsic concentration (D) Inversely proportional to the intrinsic concentration

MCQ 3.44

Under low level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the (A) Diffusion current (B) Drift current (C) Recombination current (D) Induced current

MCQ 3.45

The phenomenon known as “Early Effect” in a bipolar transistor refers to a reduction of the effective base-width caused by (A) Electron - hole recombination at the base



Electronics Devices

Chapter 3

(B) The reverse biasing of the base - collector junction (C) The forward biasing of emitter-base junction (D) The early removal of stored base charge during saturation-to-cut off switching 2006 MCQ 3.46

TWO MARKS

In the circuit shown below, the switch was connected to position 1 at t < 0 and at t = 0 , it is changed to position 2. Assume that the diode has zero voltage drop and a storage time ts . For 0 < t # ts, vR is given by (all in Volts)

(A) vR =- 5 (C) 0 # vR < 5

(B) vR =+ 5 (D) - 5 # vR < 0

nodia

MCQ 3.47

The majority carriers in an n-type semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B . The electric field E induced due to Hall effect acts in the direction (A) v # B (B) B # v (C) along v (D) opposite to v

MCQ 3.48

Find the correct match between Group 1 and Group 2 Group 1 Group 2 E - Varactor diode 1. Voltage reference F - PIN diode 2. High frequency switch G - Zener diode 3. Tuned circuits H - Schottky diode 4. Current controlled attenuator (A) E - 4, F - 2, G - 1, H - 3 (B) E - 3, F - 4, G - 1, H - 3 (C) E - 2, F - 4, G - 1, H - 2 (D) E - 1, F - 3, G - 2, H - 4

MCQ 3.49

A heavily doped n - type semiconductor has the following data: Hole-electron ratio : 0.4 Doping concentration : 4.2 # 108 atoms/m3 Intrinsic concentration : 1.5 # 10 4 atoms/m 3 The ratio of conductance of the n -type semiconductor to that of the intrinsic semiconductor of same material and ate same temperature is given by (A) 0.00005 (B) 2000 (C) 10000 (D) 20000 2005

MCQ 3.50

ONE MARK

The bandgap of Silicon at room temperature is



Electronics Devices

(A) 1.3 eV (C) 1.1 eV

Page 111

(B) 0.7 eV (D) 1.4 eV

MCQ 3.51

A Silicon PN junction at a temperature of 20c C has a reverse saturation current of 10 pico - Ameres (pA). The reserve saturation current at 40cC for the same bias is approximately (A) 30 pA (B) 40 pA (C) 50 pA (D) 60 pA

MCQ 3.52

The primary reason for the widespread use of Silicon in semiconductor device technology is (A) abundance of Silicon on the surface of the Earth. (B) larger bandgap of Silicon in comparison to Germanium. (C) favorable properties of Silicon - dioxide (SiO2) (D) lower melting point 2005

MCQ 3.53

MCQ 3.54

nodia

TWO MARKS

A Silicon sample A is doped with 1018 atoms/cm 3 of boron. Another sample B of identical dimension is doped with 1018 atoms/cm 3 phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is (A) 3 (B) 1 3 2 (C) (D) 3 3 2 A Silicon PN junction diode under reverse bias has depletion region of width 10 mm. The relative permittivity of Silicon, er = 11.7 and the permittivity of free space e0 = 8.85 # 10 - 12 F/m. The depletion capacitance of the diode per square meter is (A) 100 mF (B) 10 mF (C) 1 mF

(D) 20 mF

MCQ 3.55

A MOS capacitor made using p type substrate is in the accumulation mode. The dominant charge in the channel is due to the presence of (A) holes (B) electrons (C) positively charged icons (D) negatively charged ions

MCQ 3.56

For an n -channel MOSFET and its transfer curve shown in the figure, the threshold voltage is

(A) 1 V and the device is in active region (B) - 1 V and the device is in saturation region (C) 1 V and the device is in saturation region (D) - 1 V and the device is an active region



Electronics Devices

Chapter 3

2004

ONE MARK

MCQ 3.57

The impurity commonly used for realizing the base region of a silicon n - p - n transistor is (A) Gallium (B) Indium (C) Boron (D) Phosphorus

MCQ 3.58

If for a silicon npn transistor, the base-to-emitter voltage (VBE ) is 0.7 V and the collector-to-base voltage (VCB) is 0.2 V, then the transistor is operating in the (A) normal active mode (B) saturation mode (C) inverse active mode (D) cutoff mode

MCQ 3.59

Consider the following statements S1 and S2. S1 : The b of a bipolar transistor reduces if the base width is increased. S2 : The b of a bipolar transistor increases if the dopoing concentration in the base is increased. Which remarks of the following is correct ? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

MCQ 3.60

Given figure is the voltage transfer characteristic of

nodia

(A) an NOMS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load (C) a CMOS inverter (D) a BJT inverter MCQ 3.61

Assuming VCEsat = 0.2 V and b = 50 , the minimum base current (IB) required to drive the transistor in the figure to saturation is

(A) 56 mA (C) 60 mA

(B) 140 mA (D) 3 mA



Electronics Devices

2004

Page 113

TWO MARKS

MCQ 3.62

In an abrupt p - n junction, the doping concentrations on the p -side and n -side are NA = 9 # 1016 /cm 3 respectively. The p - n junction is reverse biased and the total depletion width is 3 mm. The depletion width on the p -side is (A) 2.7 mm (B) 0.3 mm (C) 2.25 mm (D) 0.75 mm

MCQ 3.63

The resistivity of a uniformly doped n -type silicon sample is 0.5W - mc. If the electron mobility (mn) is 1250 cm 2 /V-sec and the charge of an electron is 1.6 # 10 - 19 Coulomb, the donor impurity concentration (ND) in the sample is (A) 2 # 1016 /cm 3 (B) 1 # 1016 /cm 3 (C) 2.5 # 1015 /cm 3 (D) 5 # 1015 /cm 3

MCQ 3.64

Consider an abrupt p - n junction. Let Vbi be the built-in potential of this junction and VR be the applied reverse bias. If the junction capacitance (Cj ) is 1 pF for Vbi + VR = 1 V, then for Vbi + VR = 4 V, Cj will be (A) 4 pF (B) 2 pF (C) 0.25 pF (D) 0.5 pF

MCQ 3.65

Consider the following statements Sq and S2. S1 : The threshold voltage (VT ) of MOS capacitor decreases with increase in gate oxide thickness. S2 : The threshold voltage (VT ) of a MOS capacitor decreases with increase in substrate doping concentration. Which Marks of the following is correct ? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

MCQ 3.66

The drain of an n-channel MOSFET is shorted to the gate so that VGS = VDS . The threshold voltage (VT ) of the MOSFET is 1 V. If the drain current (ID) is 1 mA for VGS = 2 V, then for VGS = 3 V, ID is (A) 2 mA (B) 3 mA (C) 9 mA (D) 4 mA

MCQ 3.67

The longest wavelength that can be absorbed by silicon, which has the bandgap of 1.12 eV, is 1.1 mm. If the longest wavelength that can be absorbed by another material is 0.87 mm, then bandgap of this material is (A) 1.416 A/cm 2 (B) 0.886 eV (C) 0.854 eV

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(D) 0.706 eV MCQ 3.68

The neutral base width of a bipolar transistor, biased in the active region, is 0.5 mm. The maximum electron concentration and the diffusion constant in the base are 1014 / cm 3 and Dn = 25 cm 2 /sec respectively. Assuming negligible recombination in the base, the collector current density is (the electron charge is 1.6 # 10 - 19 Coulomb)



Electronics Devices

(A) 800 A/cm 2 (C) 200 A/cm 2

Chapter 3

(B) 8 A/cm 2 (D) 2 A/cm 2

2003

ONE MARK

MCQ 3.69

n -type silicon is obtained by doping silicon with (A) Germanium (B) Aluminium (C) Boron (D) Phosphorus

MCQ 3.70

The Bandgap of silicon at 300 K is (A) 1.36 eV (C) 0.80 eV

MCQ 3.71

(B) 1.10 eV (D) 0.67 eV

The intrinsic carrier concentration of silicon sample at 300 K is 1.5 # 1016 /m 3 . If after doping, the number of majority carriers is 5 # 1020 /m 3 , the minority carrier density is (A) 4.50 # 1011/m 3 (B) 3.333 # 10 4 /m 3

nodia

(C) 5.00 # 1020 /m 3

(D) 3.00 # 10 - 5 /m 3

MCQ 3.72

Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photo diode are operated in X bias ad Y bias respectively (A) X: reverse, Y: reverse (B) X: reverse, Y: forward (C) X: forward, Y: reverse (D) X: forward, Y: forward

MCQ 3.73

For an n - channel enhancement type MOSFET, if the source is connected at a higher potential than that of the bulk (i.e. VSB > 0 ), the threshold voltage VT of the MOSFET will (A) remain unchanged (B) decrease (C) change polarity

(D) increase

2003 MCQ 3.74

An n -type silicon bar 0.1 cm long and 100 mm i cross-sectional area has a majority carrier concentration of 5 # 1020 /m 2 and the carrier mobility is 0.13 m2 /V-s at 300 K. If the charge of an electron is 1.5 # 10 - 19 coulomb, then the resistance of the bar is (A) 106 Ohm (B) 10 4 Ohm (C) 10 - 1 Ohm

MCQ 3.75

TWO MARKS 2

(D) 10 - 4 Ohm

The electron concentration in a sample of uniformly doped n -type silicon at 300 K varies linearly from 1017 /cm 3 at x = 0 to 6 # 1016 /cm 3 at x = 2mm . Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 # 10 - 19 coulomb and the diffusion constant Dn = 35 cm 2 /s, the current density in the silicon, if no electric field is present, is (A) zero (B) -112 A/cm 2



Electronics Devices

Page 115

(C) +1120 A/cm 2 (D) -1120 A/cm 2 MCQ 3.76

Match items in Group 1 with items in Group 2, most suitably. Group 1 Group 2 P. LED 1. Heavy doping Q. Avalanche photo diode 2. Coherent radiation R. Tunnel diode 3. Spontaneous emission S. LASER 4. Current gain (A) P - 1, Q - 2, R - 4, S - 3 (B) P - 2, Q - 3, R - 1, S - 4 (C) P - 3 Q - 4, R - 1, S - 2 (D) P - 2, Q - 1, R - 4, S - 3

MCQ 3.77

At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions state above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (B) 5

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(C) 4 # 103

(D) 8 # 103

MCQ 3.78

A particular green LED emits light of wavelength 5490 Ac. The energy bandgap of the semiconductor material used there is (Plank’s constant = 6.626 # 10 - 34 J - s ) (A) 2.26 eV (B) 1.98 eV (C) 1.17 eV (D) 0.74 eV

MCQ 3.79

When the gate-to-source voltage (VGs) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation, the drain current for an applied VGS of 1400 mV is (A) 0.5 mA (B) 2.0 mA (C) 3.5 mA (D) 4.0 mA

MCQ 3.80

If P is Passivation, Q is n -well implant, R is metallization and S is source/ drain diffusion, then the order in which they are carried out in a standard n -well CMOS fabrication process, is (A) P - Q - R - S (B) Q - S - R - P (C) R - P - S - Q (D) S - R - Q - P

MCQ 3.81

The action of JFET in its equivalent circuit can best be represented as a (A) Current controlled current source (B) Current controlled voltage source (C) Voltage controlled voltage source (D) Voltage controlled current source



Electronics Devices

Chapter 3

2002 MCQ 3.82

ONE MARK

In the figure, silicon diode is carrying a constant current of 1 mA. When the temperature of the diode is 20cC, VD is found to be 700 mV. If the temperature rises to 40cC, VD becomes approximately equal to

(A) 740 mV (C) 680 mV MCQ 3.83

(B) 660 mV (D) 700 mV

If the transistor in the figure is in saturation, then

nodia

(A) IC is always equal to bdc IB (B) IC is always equal to - bde IB

(C) IC is greater than or equal to bdc IB (D) IC is less than or equal to bdc IB 2001

ONE MARK

MCQ 3.84

MOSFET can be used as a (A) current controlled capacitor (B) voltage controlled capacitor (C) current controlled inductor (D) voltage controlled inductor

MCQ 3.85

The effective channel length of MOSFET in saturation decreases with increase in (A) gate voltage (B) drain voltage (C) source voltage (D) body voltage 1999

MCQ 3.86

ONE MARK

The early effect in a bipolar junction transistor is caused by (A) fast turn-on (B) fast turn-off (C) large collector-base reverse bias (D) large emitter-base forward bias



Electronics Devices

Page 117

1999

TWO MARKS

MCQ 3.87

An n -channel JEFT has IDSS = 2 mA and Vp =- 4 V . Its transconductance gm (in milliohm) for an applied gate-to-source voltage VGS of - 2 V is (A) 0.25 (B) 0.5 (C) 0.75 (D) 1.0

MCQ 3.88

An npn transistor (with C = 0.3 pF ) has a unity-gain cutoff frequency fT of 400 MHz at a dc bias current Ic = 1 mA . The value of its Cm (in pF) is approximately (VT = 26 mV) (A) 15 (B) 30 (C) 50 (D) 96 1998

ONE MARK

MCQ 3.89

The electron and hole concentrations in a intrinsic semiconductor are ni and pi respectively. When doped with a p-type material, these change to n and p, respectively, Then (A) n + p = ni + pi (B) n + ni = p + pi (C) npi = ni p (D) np = ni pi

MCQ 3.90

The fT of a BJT is related to its gm, C p and C m as follows

nodia

Cp + Cm gm gm (C) fT = Cp + Cm (A) fT =

2p (C p + C m) gm gm (D) fT = 2p (C p + C m) (B) fT =

MCQ 3.91

The static characteristic of an adequately forward biased p-n junction is a straight line, if the plot is of (A) log I vs log V (B) log I vs V (C) I vs log V (D) I vs V

MCQ 3.92

A long specimen of p-type semiconductor material (A) is positively charged (B) is electrically neutral (C) has an electric field directed along its length (D) acts as a dipole

MCQ 3.93

Two identical FETs, each characterized by the parameters gm and rd are connected in parallel. The composite FET is then characterized by the parameters g g (A) m and 2rd (B) m and rd 2 2 2 r (C) 2gm and d (D) 2gm and 2rd 2 q The units of are kT (A) V (B) V-1

MCQ 3.94



Electronics Devices

(C) J

Chapter 3

(D) J/K

1997

ONE MARK

MCQ 3.95

For a MOS capacitor fabricated on a p-type semiconductor, strong inversion occurs when (A) surface potential is equal to Fermi potential (B) surface potential is zero (C) surface potential is negative and equal to Fermi potential in magnitude (D) surface potential is positive and equal to twice the Fermi potential

MCQ 3.96

The intrinsic carrier density at 300 K is 1.5 # 1010 /cm3 , in silicon. For n -type silicon doped to 2.25 # 1015 atoms/cm3 , the equilibrium electron and hole densities are (A) n = 1.5 # 1015 /cm3, p = 1.5 # 1010 /cm3 (B) n = 1.5 # 1010 /cm3, p = 2.25 # 1015 /cm3 (C) n = 2.25 # 1015 /cm3, p = 1.0 # 1015 /cm3 (D) n = 1.5 # 1010 /cm3, p = 1.5 # 1010 /cm3 1996

MCQ 3.97

nodia

ONE MARK

The p-type substrate in a conventional pn -junction isolated integrated circuit should be connected to (A) nowhere, i.e. left floating (B) a DC ground potential (C) the most positive potential available in the circuit (D) the most negative potential available in the circuit

MCQ 3.98

If a transistor is operating with both of its junctions forward biased, but with the collector base forward bias greater than the emitter base forward bias, then it is operating in the (A) forward active mode (B) reverse saturation mode (C) reverse active mode (D) forward saturation mode

MCQ 3.99

The common-emitter short-circuit current gain b of a transistor (A) is a monotonically increasing function of the collector current IC (B) is a monotonically decreasing function of IC (C) increase with IC , for low IC , reaches a maximum and then decreases with further increase in IC (D) is not a function of IC

MCQ 3.100

A n -channel silicon (Eg = 1.1 eV) MOSFET was fabricated using n +poly-silicon gate and the threshold voltage was found to be 1 V. Now, if the gate is changed to p+ poly-silicon, other things remaining the same, the new threshold voltage should be (A) - 0.1 V (B) 0 V



Electronics Devices

(C) 1.0 V

Page 119

(D) 2.1 V

1996

TWO MARKS

MCQ 3.101

In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base-emitter junction (A) doubles (B) halves (C) increases by about 20 mV (D) decreases by about 20 mV

MCQ 3.102

An npn transistor has a beta cut-off frequency fb of 1 MHz and common emitter short circuit low-frequency current gain bo of 200 it unity gain frequency fT and the alpha cut-off frequency fa respectively are (A) 200 MHz, 201 MHz (B) 200 MHz, 199 MHz (C) 199 MHz, 200 MHz (D) 201 MHz, 200 MHz

MCQ 3.103

A silicon n MOSFET has a threshold voltage of 1 V and oxide thickness of Ao.

nodia

[er (SiO 2) = 3.9, e0 = 8.854 # 10-14 F/cm, q = 1.6 # 10-19 C] The region under the gate is ion implanted for threshold voltage tailoring. The dose and type of the implant (assumed to be a sheet charge at the interface) required to shift the threshold voltage to - 1 V are (A) 1.08 # 1012 /cm2 , p-type (B) 1.08 # 1012 /cm2 , n-type (C) 5.4 # 1011 /cm2 , p-type (D) 5.4 # 1011 /cm2 , n-type

***********



Electronics Devices

Chapter 3

SOLUTIONS SOL 3.1

Option (A) is correct. The potential barrier of the pn junction is lowered when a forward bias voltage is applied, allowing electrons and holes to flow across the space charge region (Injection) when holes flow from the p region across the space charge region into the n region, they become excess minority carrier holes and are subject to diffuse, drift and recombination processes.

SOL 3.2

Option (D) is correct. In IC technology, dry oxidation as compared to wet oxidation produces superior quality oxide with a lower growth rate

SOL 3.3

Option (D) is correct. In a MOSFET operating in the saturation region, the channel length modulation effect causes a decrease in output resistance.

SOL 3.4

Option (A) is correct. Given,

nodia VB = 2V VTN = 1V

So, we have Drain voltage

VD = 2 volt VG = 2 volt VS = 0 (Ground) Therefore, VGS = 2 > VTN and VDS = 2 > VGS - VTN So, the MOSFET is in the saturation region. Therefore, drain current is ID = kN ^VGS - VTN h2

or, ID = kN ^VB - 1h Differentiating both side with respect to ID 1 = kN 2 ^VB - 1hdVB dID Since, VBQ = 2 volt (at D.C. Voltage) Hence, we obtain dVB = 1 1 = dID 2kN ^VB - 1h 2 # 40 # 10-6 # ^2 - 1h = 12.5 # 103 W = 12.5 kW 2

SOL 3.5

Option (D) is correct. For the semiconductor,

Volume of given device,

n 0 p 0 = n i2 2 20 p 0 = n i = 1019 = 10 per cm3 n 0 10 V = Area # depth = 1 mm2 # 1 mm = 10-8 cm2 # 10-4 cm = 10-12 cm3

So total no. of holes is,



Electronics Devices

Page 121

p = p 0 # V = 10 # 10-12 = 10-11 Which is approximately equal to zero. SOL 3.6

Option (A) is correct. Given the circuit as below :

Since all the parameters of PMOS and NMOS are equal. So, mn = mp W COX b l = COX bW l = COX bW l L M1 L M2 L Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0

nodia

Where I1 is drain current in M1 and I2 is drain current in M2 . m C 2 Since, I1 = p OX bW l82VSD ^VSG - VTp h - V SD B 2 L m C 2 0 = p OX bW l [2VSD ^VSG - VTp h - V SD & ] 2 L Solving it we get, 2 ^VSG - VTp h = VSD

2 ^5 - Vin - 1h = 5 - VD Vin = VD + 3 2

& & For So, So for the NMOS

I1 = 0 , VD = 5 V Vin = 5 + 3 = 4 V 2

VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn So it can’t be in cutoff region. Case 2 : M2 must be in saturation region. So,

I1 = I 2 mp COX W mn COX W 2 2 2 (VSG - VTp) VSD - V SD @ = 2 L (VGS - VTn) 2 L6

&

2 2 (VSG - VTp) VSD - V SD = (VGS - VTn) 2 2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2

& & 2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2 Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1 &

2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2



Electronics Devices

Chapter 3

48 - 36 - 8Vin =- 2Vin + 1 6Vin = 11 & Vin = 11 = 1.833 V 6 So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V

& &

SOL 3.7

Option (B) is correct. Gate source overlap capacitance. Co = dWeox e0 (medium Sio 2 ) tox -9 -6 -12 = 20 # 10 # 1 # 10 #-93.9 # 8.9 # 10 = 0.69 # 10-15 F 1 # 10

SOL 3.8

Option (B) is correct. Source body junction capacitance. Cs = Aer e0 d A = (0.2 mm + 0.2 mm + 0.2 mm) # 1 mm + 2 (0.2 mm # 0.2 mm)

nodia

= 0.68 mm2 d = 10 nm (depletion width of all junction) -12 8.9 # 10-12 = 7 10-15 F Cs = 0.68 # 10 # 11.7 -# # 9 10 # 10

SOL 3.9

Option (C) is correct. Drift current Id = qnmn E It depends upon Electric field E and carrier concentration n

SOL 3.10

Option (B) is correct. Zener diode operates in reverse breakdown region.

SOL 3.11

Option (D) is correct. For every 1c C increase in temperature, forward bias voltage across diode decreases by 2.5 mV. Thus for 10c C increase, there us 25 mV decreases.

SOL 3.12

Option (B) is correct. Full channel resistance is r L r # = 600 W W#a If VGS is applied, Channel resistance is r L rl = # W#b Pinch off voltage, qN Vp = D a2 2e

...(1) where b = a c1 -

VGS Vp m ...(2)



Electronics Devices

Page 123

If depletion on each side is d = 1 μm at VGS = 0 . qN Vj = D d2 2e qN qND or 1 = D (1 # 10-6) 2 & = 1012 2e 2e Now from equation (2), we have or At VGS =- 3 V ;

Vp = 1012 # (5 # 10-6) 2 Vp =- 25 V

- 3 mm = 3.26 mm - 25 l rL rL a = 600 5 rl = = # 3.26 = 917 W W # b Wa # b b = 5 b1 -

SOL 3.13

Option (C) is correct. At VGS = 0 V , Thus

b = 4 mm rL a = 600 5 = 750 W rl = #4 Wa # b

since 2b = 8 mm

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SOL 3.14

Option (A) is correct. At room temperature mobility of electrons for Si sample is given mn = 1350 cm2 /Vs . For an n -channel MOSFET to create an inversion layer of electrons, a large positive gate voltage is to be applied. Therefore, induced electric field increases and mobility decreases. So, Mobility mn < 1350 cm2 /Vs for n -channel MOSFET

SOL 3.15

Option (B) is correct. Dry oxidation is used to achieve high quality oxide growth.

SOL 3.16

Option (B) is correct. Emitter injection efficiency is given as 1 g = 1 + NB NE To achieve g = 1, NE >> NB

SOL 3.17

Option (C) is correct. Reverse bias breakdown or Zener effect occurs in highly doped PN junction through tunneling mechanism. In a highly doped PN junction, the conduction and valence bands on opposite sides of the junction are sufficiently close during reverse bias that electron may tunnel directly from the valence band on the p-side into the conduction band on n -side. Breakdown voltage VB \ 1 NA ND So, breakdown voltage decreases as concentration increases Depletion capacitance 1/2 ees NA ND C =' 1 2 (Vbi + VR) (NA + ND) Thus C \ NA ND Depletion capacitance increases as concentration increases

SOL 3.18

Option (C) is correct. Sample is in thermal equilibrium so, electric field



Electronics Devices

Chapter 3

1 = 10 kV/cm 1 mm Option (A) is correct. Electron drift current density E =

SOL 3.19

Jd = ND mn eE = 1016 # 1350 # 1.6 # 10-19 # 10 # 1013 = 2.16 # 10 4 A/cm2 SOL 3.20

Option (C) is correct. Only dopant atoms can have concentration of 4 # 1019 cm - 3 in n -type silicon at room temperature.

SOL 3.21

Option (A) is correct. 2 Unit of mobility mn is = cm V. sec 2 Unit of diffusion current Dn is = cm sec 2 2 mn Thus unit of is = cm / cm = 1 = V-1 V $ sec sec V Dn

nodia

SOL 3.22

Option (D) is correct. Both S1 and S2 are true and S2 is a reason for S1.

SOL 3.23

Option (B) is correct. We know that or

NA WP = ND WN 17 -6 NA = ND WN = 1 # 10 # 0.1-6# 10 = 1 # 1016 WP 1 # 10

The built-in potential is

D Vbi = VT 1nc NA N n i2 m 17 16 # 10 = 0.760 = 26 # 10-3 ln e 1 # 10 # 1 10 o 2 (1.4 # 10 )

SOL 3.24

Option (B) is correct. The peak electric field in device is directed from p to n and is from p to n E =- eND xn es from n to p = eND xn es -19 17 -5 # 1 # 10 = 0.15 MV/cm = 1.6 # 10 # 1 #-10 14 8.85 # 10 # 12

SOL 3.25

Option (D) is correct. Channel length modulation is not associated with a p - n junction. It is being associated with MOSFET in which effective channel length decreases, producing the phenomenon called channel length modulation.

SOL 3.26

Option (A) is correct. Trivalent impurities are used for making p - type semiconductors. So, Silicon wafer heavily doped with boron is a p+ substrate.

SOL 3.27

Option (D) is correct. Oxidation rate is zero because the existing oxide prevent the further oxidation.



Electronics Devices

Page 125

SOL 3.28

Option (B) is correct. gm = 2ID = 2 K (VGS - VT ) 2 = 2K (VGS - VT ) 2VGS 2VGS

SOL 3.29

Option (C) is correct. As VD = constant Thus gm \ (VGS - VT )

SOL 3.30

Which is straight line.

Option (C) is correct. E2 - E1 = kT ln NA ni NA = 4 # 1017 ni = 1.5 # 1010 17 E2 - E1 = 25 # 10-3 e ln 4 # 10 10 = 0.427 eV 1.5 # 10 Hence fermi level goes down by 0.427 eV as silicon is doped with boron.

SOL 3.31


nodia 2 VP = eW ND es

Pinch off voltage

VP = VP1 VP1 = W12 = W2 Now VP2 W22 (2W) 2 or 4VP1 = VP2 Initial transconductance gm = Kn ;1 - Vbi - VGS E Vp Let

For first condition

gm1 = Kn =1 -

0 - (- 2) = Kn ;1 VP1 G

2 VP1 E

For second condition gm2 = Kn =1 Dividing Hence SOL 3.32


SOL 3.33

Option (D) is correct. As per mass action law

0 - (- 2) = K2 ;1 VP2 G

2 4VP1 E

1 - 2/VP1 gm1 =f p gm2 1 - 1/ (2VP1) VP = VP1

np = ni2 If acceptor impurities are introduces Thus or

p = NA nNA = ni2 2 n = ni NA

SOL 3.34

Option (C) is correct. The electric field has the maximum value at the junction of p+ n .

SOL 3.35




Electronics Devices

Chapter 3

Zener diode and Avalanche diode works in the reverse bias and laser diode works in forward bias. In solar cell diode works in forward bias but photo current is in reverse direction. Thus Zener diode : Reverse Bias Solar Cell : Forward Bias Laser Diode : Forward Bias Avalanche Photo diode : Reverse Bias SOL 3.36

Option (C) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A reduction in base width will causes the gradient in minority carrier concentration to increase, which in turn causes an increased in the diffusion current. This effect si known as base modulation as early effect. In JFET the gate to source voltage that must be applied to achieve pinch off voltage is described as pinch off voltage and is also called as turn voltage or threshold voltage. In LASER population inversion occurs on the condition when concentration of electrons in one energy state is greater than that in lower energy state, i.e. a non equilibrium condition. In MOS capacitor, flat band voltage is the gate voltage that must be applied to create flat ban condition in which there is no space charge region in semiconductor under oxide. Therefore BJT : Early effect MOS capacitor : Flat-band voltage

nodia LASER diode : Population inversion JFET : Pinch-off voltage

SOL 3.37

Option (A) is correct. W = K V + VR Now 2m = K 0.8 + 1.2 From above two equation we get 0.8 + 7.2 = 0.8 + 1.2 W2 = 4 m m

W = 2m or SOL 3.38

8 =2 2

Option (B) is correct. a=

b = 50 = 50 b + 1 50 + 1 51

Current Gain = Base Transport Factor # Emitter injection Efficiency a = b1 # b2 50 or b1 = a = = 0.985 51 # 0.995 b2 SOL 3.39

Option (A) is correct. At low voltage when there is no depletion region and capacitance is decide by



Electronics Devices

Page 127

SiO2 thickness only,

or SOL 3.40

C = e0 er1 A D -13 10-4 = 50 nm D = e0 er1 A = 3.5 # 10 -# 12 C 7 # 10

Option (B) is correct. The construction of given capacitor is shown in fig below

When applied voltage is 0 volts, there will be no depletion region and we get C1 = 7 pF When applied voltage is V , a depletion region will be formed as shown in fig an

nodia

total capacitance is 1 pF. Thus or or

CT = 1 pF CT = C1 C2 = 1 pF C1 + C2 1 = 1 + 1 CT C1 C2

Substituting values of CT and C1 we get C2 = 7 pF 6 Now D2 = e0 er2 A C2

- 12 -4 = 1 # 710 #- 1210 = 6 # 10 - 4 cm 7 6 # 10 = 0.857 mm

SOL 3.41

Option (C) is correct. Depletion region will not be formed if the MOS capacitor has n type substrate but from C-V characteristics, C reduces if V is increased. Thus depletion region must be formed. Hence S1 is false If positive charges is introduced in the oxide layer, then to equalize the effect the applied voltage V must be reduced. Thus the C - V plot moves to the left. Hence S2 is true.

SOL 3.42

Option (C) is correct. For the case of negative slope it is the negative resistance region



SOL 3.43

Electronics Devices

Chapter 3

Option (A) is correct. For n -type p is minority carrier concentration np = ni2 np = Constant p \ 1 n

Since ni is constant

Thus p is inversely proportional to n . SOL 3.44

Option (A) is correct. Diffusion current, since the drift current is negligible for minority carrier.

SOL 3.45

Option (B) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A reduction in base width will causes the gradient in minority carrier concentration to increases, which in turn causes an increases in the diffusion current. This effect si known as base modulation as early effect.

SOL 3.46

Option (A) is correct. For t < 0 diode forward biased and VR = 5 . At t = 0 diode abruptly changes to reverse biased and current across resistor must be 0. But in storage time 0 < t < ts diode retain its resistance of forward biased. Thus for 0 < t < ts it will be ON and

nodia VR =- 5 V

SOL 3.47

Option (B) is correct. According to Hall effect the direction of electric field is same as that of direction of force exerted. or

SOL 3.48

E =- v # B E = B#v

Option (B) is correct. The varacter diode is used in tuned circuit as it can provide frequently stability. PIN diode is used as a current controlled attenuator. Zener diode is used in regulated voltage supply or fixed voltage reference. Schottkey diode has metal-semiconductor function so it has fast switching action so it is used as high frequency switch Varactor diode : Tuned circuits PIN Diode : Current controlled attenuator Zener diode : Voltage reference Schottky diode : High frequency switch

SOL 3.49

Option (D) is correct. mP We have = 0.4 mn Conductance of n type semiconductor sn = nqmn Conductance of intrinsic semiconductor Ratio is

si = ni q (mn + mp) nmn sn = n = si ni (mn + mp) ni ^1 +

mp mn h



Electronics Devices

= SOL 3.50

Page 129

4.2 # 108 = 2 # 10 4 1.5 # 10 4 (1 + 0.4)

Option (C) is correct. For silicon at 0 K, Eg0 = 1.21 eV At any temperature EgT = Eg0 - 3.6 # 10 - 4 T At T = 300 K, Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be remembered.

SOL 3.51

Option (B) is correct. The reverse saturation current doubles for every 10cC rise in temperature as follows : I0 (T) = I 01 # 2(T - T )/10 Thus at 40c C, I0 = 40 pA 1

nodia

SOL 3.52

Option (A) is correct. Silicon is abundant on the surface of earth in the from of SiO2 .

SOL 3.53


sn = nqmn sp = pqmp sp m = p =1 sn mn 3

SOL 3.54

(n = p)


or

C = e0 er A d C = e0 er = 8.85 # 10-12 # 11.7 = 10.35 m F d A 10 # 10-6

SOL 3.55

Option (B) is correct. In accumulation mode for NMOS having p -substrate, when positive voltage is applied at the gate, this will induce negative charge near p - type surface beneath the gate. When VGS is made sufficiently large, an inversion of electrons is formed and this in effect forms and n - channel.

SOL 3.56

Option (C) is correct. From the graph it can be easily seen that Vth = 1 V Now VGS = 3 - 1 = 2 V and VDS = 5 - 1 = 4 V VDS > VGS $ VDS > VGS - Vth Thus MOSFET is in saturation region.

Since SOL 3.57

Option (C) is correct. Trivalent impurities are used for making p type semiconductor. Boron is trivalent.

SOL 3.58

Option (A) is correct. Here emitter base junction is forward biased and base collector junction is reversed biased. Thus transistor is operating in normal active region.



SOL 3.59

Electronics Devices

Chapter 3

Option (D) is correct. b =

We have

a 1-a

a -" b -

Thus

a ." b . If the base width increases, recombination of carrier in base region increases and a decreases & hence b decreases. If doping in base region increases, recombination of carrier in base increases and a decreases thereby decreasing b . Thus S1 is true and S2 is false. SOL 3.60


SOL 3.61

Option (A) is correct. Applying KVL we get or Now

SOL 3.62

VCC - IC RC - VCE = 0 IC = VCC - VCE = 3 - 0.2 = 2.8 mA RC 1k I 2.8 m C = = 56 mA IB = b 50

nodia

Option (B) is correct. We know that or

Wp NA = Wn ND Wp = Wn # ND NA =

SOL 3.63

Option (B) is correct. Conductivity or resistivity Thus

3 m # 1016 = 0.3 m m 9 # 1016

s = nqun r = 1 = 1 s nqmn 1 n = 1 = = 1016 /cm 3 qrmn 1.6 # 10 - 19 # 0.5 # 1250

For n type semiconductor n = ND SOL 3.64

Option (D) is correct. We know that eeS NA ND Cj = ; 2 (Vbi + VR)( NA + ND) E 1 Cj \ (Vbi + VR) C j2 (Vbi + VR) 1 1 =1 = = (Vbi + VR) 2 4 2 C j1 C Cj2 = j1 = 1 = 0.5 pF 2 2 1 2

Thus Now or SOL 3.65

Option (C) is correct. Increase in gate oxide thickness makes difficult to induce charges in channel. Thus VT increases if we increases gate oxide thickness. Hence S1 is false. Increase in substrate doping concentration require more gate voltage because initially induce charges will get combine in substrate. Thus VT increases if we



Electronics Devices

Page 131

increase substrate doping concentration. Hence S2 is false. SOL 3.66

Option (D) is correct. We know that ID = K (VGS - VT ) 2 2 IDS = (VGS2 - VT ) IDI (VGS1 - VT ) 2 Substituting the values we have

Thus

2 ID2 = (3 - 1) = 4 ID1 (2 - 1) 2 ID2 = 4IDI = 4 mA

or SOL 3.67

Option (A) is correct. Eg \ 1 l Eg2 = l1 = 1.1 l2 0.87 Eg1 Eg2 = 1.1 # 1.12 = 1.416 eV 0.87

Thus or SOL 3.68

nodia

Option (B) is correct. Concentration gradient

1014 dn = = 2 # 1018 dx 0.5 # 10 - 4 q = 1.6 # 10 - 19 C Dn = 25 1014 dn = dx 0.5 # 10 - 4 JC = qDn dn dx

= 1.6 # 10 - 19 # 25 # 2 # 1018 = 8 A/cm 2 SOL 3.69

Option (D) is correct. Pentavalent make n -type semiconductor and phosphorous is pentavalent.

SOL 3.70

Option (C) is correct. For silicon at 0 K Eg0 = 1.21 eV At any temperature EgT = Eg0 - 3.6 # 10 - 4 T At T = 300 K, Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be remembered.

SOL 3.71

Option (A) is correct. By Mass action law np = ni2 2 16 .5 # 1016 = 4.5 # 1011 p = ni = 1.5 # 10 # 120 n 5 # 10

SOL 3.72

Option (C) is correct. Tunnel diode shows the negative characteristics in forward bias. It is used in



Electronics Devices

Chapter 3

forward bias. Avalanche photo diode is used in reverse bias. SOL 3.73


SOL 3.74

Option (A) is correct. R =

We that

rl , r = 1 and a = nqun A s

From above relation we have 1 R = nqmn A 0.1 # 10 - 2 = 106 W 5 # 10 # 1.6 # 10 - 19 # 0.13 # 100 # 10 - 12 Option (D) is correct. =

SOL 3.75

Now

20

dn = 6 # 1016 - 1017 =- 2 # 1020 dx 2 # 10 - 4 - 0 Jn = nqme E + Dn q dn dx

nodia

Since no electric field is present, E = 0 and we get So, Jn = qDn dn dx

= 1.6 # 10 - 19 # 35 # (- 2 # 1020) =- 1120 A/cm 2

SOL 3.76

Option (C) is correct. LED works on the principal of spontaneous emission. In the avalanche photo diode due to the avalanche effect there is large current gain. Tunnel diode has very large doping. LASER diode are used for coherent radiation.

SOL 3.77


I = Io è h V - 1j where h = 1 for germanium and h = 2 silicon. As per question VD1

We know that

si

T

Io è e - 1j = Io è hV - 1j VDsi

VDGe

hVT

n

or

Ge

Io = Io si si

SOL 3.78

VDsi e hVT VDGe e hVT

T

0.718

- 1 = e 2 # 26 # 10 - 1 = 4 103 # 0.1435 e 26 # 10 - 1 -1 -3

-3


In eV

Eg = hc l -34 8 = 6.626 # 10 # -310# 10 = 3.62 J 54900 # 10 -19 Eg (J) Eg (eV) = = 3.62 # 10-19 = 2.26 eV e 1.6 # 10

Alternatively 1.24 Eg = 1.24 eV = = 2.26 eV l (mm) 5490 # 10-4 mm SOL 3.79

Option (D) is correct. We know that



Electronics Devices

Page 133

ID = K (VGS - VT ) 2 2 ID2 = (VGS2 - VT ) Thus 2 ID1 (VGS1 - VT ) Substituting the values we have 2 ID2 = (1.4 - 0.4) = 4 ID1 (0.9 - 0.4) 2 ID2 = 4IDI = 4 mA

or SOL 3.80

Option (B) is correct. In n -well CMOS fabrication following are the steps : (A) n - well implant (B) Source drain diffusion (C) Metalization (D) Passivation

SOL 3.81

Option (D) is correct. For a JFET in active region we have 2 IDS = IDSS c1 - VGS m VP

nodia

From above equation it is clear that the action of a JFET is voltage controlled current source. SOL 3.82

Option (B) is correct. At constant current the rate of change of voltage with respect to temperature is dV =- 2.5 mV per degree centigrade dT Here

Thus Therefore,

3 T = T2 - T1 = 40 - 20 = 20cC 3 VD =- 2.5 # 20 = 50 mV VD = 700 - 50 = 650 mV

SOL 3.83

Option (D) is correct. Condition for saturation is IC < bIB

SOL 3.84

Option (B) is correct. The metal area of the gate in conjunction with the insulating dielectric oxide layer and semiconductor channel, form a parallel plate capacitor. It is voltage controlled capacitor because in active region the current voltage relationship is given by IDS = K (VGS - VT ) 2

SOL 3.85

Option (D) is correct. In MOSFET the body (substrate) is connected to power supply in such a way to maintain the body (substrate) to channel junction in cutoff condition. The resulting reverse bias voltage between source and body will have an effect on device function. The reverse bias will widen the depletion region resulting the reduction in channel length.

SOL 3.86

Option (C) is correct. At a given value of vBE , increasing the reverse-bias voltage on the collector-base junction and thus increases the width of the depletion region of this junction. This in turn results in a decrease in the effective base width W . Since IS is inversely proportional to W , IS increases and that iC increases proportionally.



Electronics Devices

Chapter 3

This is early effect. SOL 3.87

Option (B) is correct. For an n -channel JEFT trans-conductance is -3 (- 2) gm = - 2IDSS b1 - VGS l = - 2 # 2 # 10 =1 -4 VP VP (- 4)G = 10-3 # 1 = 0.5 mho 2

SOL 3.88

Option (A) is correct. We have Now or or or

SOL 3.89

gm = IC = 1 26 VT gm fT = 2p (C p + C m) 1/26 400 = 2p (0.3 # 10-12 + C m) 1 = 15.3 # 10-12 (0.3 # 10-12 + C m) = 2p # 26 # 400 C m 15.3 # 10-12 - 0.3 # 10-12 = 15 # 10-12 15 pF

nodia

Option (D) is correct. For any semiconductor (Intrinsic or extrinsic) the product n p remains constant at a given temperature so here np = ni pi

SOL 3.90


fT =

SOL 3.91

gm 2p (C p + C m)

Option (B) is correct. For a Forward Bias p-n junction, current equation I = I 0 (eV/kT - 1) or or

I + 1 = eV/kT I0 kT log b I + 1l = V I0

So if we plot log I vs V we get a straight line. SOL 3.92

Option (B) is correct. A specimen of p - type or n - type is always electrical neutral.

SOL 3.93


SOL 3.94

Option (B) is correct. The unit of q is e and unit of kT is eV. Thus unit of e/kT is e/eV = V-1 .

SOL 3.95


SOL 3.96

Option (C) is correct. ni = 1.5 # 1010 /cm3 Nd = 2.25 # 1015 atoms/cm3 For n type doping we have electron concentration n - Nd = 2.25 # 1015 atom/cm3 For a given temperature

We have

np = n i2



Electronics Devices

Hole concentration

Page 135

2 (1.5 # 1010) 2 p = ni = = 1.0 # 105 /cm3 n 2.25 # 1015

SOL 3.97

Option (D) is correct. In p n -junction isolated circuit we should have high impedance, so that p n junction should be kept in reverse bias. (So connect p to negative potential in the circuit)

SOL 3.98


If both junction are forward biased and collector base junction is more forward biased then IC will be flowing out wards (opposite direction to normal mode) the collector and it will be in reverse saturation mode. SOL 3.99

Option (C) is correct. For normal active mode we have b = IC IB

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For small values of IC , if we increases IC , b also increases until we reach (IC ) saturation. Further increases in IC (since transistor is in saturation mode know) will increases IB and b decreases. SOL 3.100

Option (C) is correct. For a n -channel mosfet thresholds voltage is given by VTN = VGS - VDS (sat) for p-channel [p+ polysilicon used in gate] VTP = VSD (sat) - VGS so VTP =- VDS (sat) + VGS so threshold voltage will be same.

SOL 3.101

Option (C) is correct. Emitter current is given by IE = I 0 (eV /kT - 1) IE = I 0 eV /kT VBE = kT ln b IE l I0 BE

or or Now

or

BE

eV

BE

/kT

>> 1

(VBE ) 1 = kT ln b IE 1 l I0 (VBE ) 2 = kT ln b IE 2 l I0 (VBE ) 2 - (VBE ) 1 = kT ;ln b IE 2 lE = kT ln b 2IE 1 l IE 1 IE 1

Now if emitter current is double i.e. IE 2 = 2IE1 (VBE ) 2 = (VBE ) 1 + (25 # 0.60) m volt = (VBE ) 1 + 15 m volt Thus if emitter current is doubled the base emitter junction voltage is increased



Electronics Devices

Chapter 3

by 15 mV. SOL 3.102

Option (A) is correct. Unity gain frequency is given by fT = fB # b = 106 # 200 = 200 MHz a-cutoff frequency is given by f fb fa = b = = fb (b + 1) 1-a 1 - b +b 1 = 106 # (200 + 1) = 201 MHz

SOL 3.103

Option (A) is correct. ***********

nodia


CHAPTER 4 ANALOG CIRCUITS

2013 MCQ 4.1

ONE MARK

In the circuit shown below what is the output voltage ^Vouth if a silicon transistor Q and an ideal op-amp are used?

nodia

(A) - 15 V (C) + 0.7 V MCQ 4.2

In a voltage-voltage feedback as shown below, which one of the following statements is TRUE if the gain k is increased?

(A) (B) (C) (D)

The The The The

2013 MCQ 4.3

(B) - 0.7 V (D) + 15 V

input input input input

impedance impedance impedance impedance

increases and output impedance decreases increases and output impedance also increases decreases and output impedance also decreases decreases and output impedance increases TWO MARKS

The ac schematic of an NMOS common-source state is shown in the figure below, where part of the biasing circuits has been omitted for simplicity. For the n -channel MOSFET M, the transconductance gm = 1 mA/V , and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in HZ of the circuit is approximately at


Analog Circuits

(A) 8 (C) 50 MCQ 4.4

Chapter 4

(B) 32 (D) 200

In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA . To maintain 5 V across RL , the minimum value of RL in W and the minimum power rating of the Zener diode in mW, respectively, are

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(A) 125 and 125 (C) 250 and 125 MCQ 4.5

In the circuit shown below the op-amps are ideal. Then, Vout in Volts is

(A) 4 (C) 8 MCQ 4.6

(B) 125 and 250 (D) 250 and 250

(B) 6 (D) 10

In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is + 5 V , X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is



Analog Circuits

(A) XY (C) XY MCQ 4.7

Page 139

(B) XY (D) XY

A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the voltage measured across WX in Volts, is

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(A) sin wt (C) ^sin wt - sin wt h /2 MCQ 4.8

(B) _sin wt + sin wt i /2 (D) 0 for all t

In the circuit shown below, the silicon npn transistor Q has a very high value of b . The required value of R2 in kW to produce IC = 1 mA is

(A) 20 (C) 40

(B) 30 (D) 50

2012 MCQ 4.9

ONE MARK

The i -v characteristics of the diode in the circuit given below are v - 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V



Analog Circuits

Chapter 4

The current in the circuit is (A) 10 mA (B) 9.3 mA (C) 6.67 mA (D) 6.2 mA MCQ 4.10

The current ib through the base of a silicon npn transistor is 1 + 0.1 cos (10000pt) mA At 300 K, the rp in the small signal model of the transistor is

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(A) 250 W (C) 25 W MCQ 4.11

The diodes and capacitors in the circuit shown are ideal. The voltage v (t) across the diode D1 is

(A) cos (wt) - 1 (C) 1 - cos (wt) MCQ 4.12

(B) 27.5 W (D) 22.5 W

(B) sin (wt) (D) 1 - sin (wt)

The impedance looking into nodes 1 and 2 in the given circuit is

(A) 50 W (C) 5 kW

(B) 100 W (D) 10.1 kW



Analog Circuits

Page 141

2012 MCQ 4.13

TWO MARKS

The circuit shown is a

1 rad/s (R1 + R2) C (B) high pass filter with f3dB = 1 rad/s R1 C (C) low pass filter with f3dB = 1 rad/s R1 C 1 (D) high pass filter with f3dB = rad/s (R1 + R2) C (A) low pass filter with f3dB =

MCQ 4.14

nodia

The voltage gain Av of the circuit shown below is

(A) Av . 200 (C) Av . 20

(B) Av . 100 (D) Av . 10

2011 MCQ 4.15

ONE MARK

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude vo at 10 M vi rad/s is

(A) maximum (C) unity

(B) minimum (D) zero



MCQ 4.16

Analog Circuits

Chapter 4

The circuit below implements a filter between the input current ii and the output voltage vo . Assume that the op-amp is ideal. The filter implemented is a

(A) low pass filter (C) band stop filter

(B) band pass filter (D) high pass filter

2011 MCQ 4.17

TWO MARKS

In the circuit shown below, for the MOS transistors, mn Cox = 100 mA/V 2 and the threshold voltage VT = 1 V . The voltage Vx at the source of the upper transistor is

nodia

(A) 1 V (C) 3 V

(B) 2 V (D) 3.67 V

MCQ 4.18

For a BJT, the common base current gain a = 0.98 and the collector base junction reverse bias saturation current ICO = 0.6 mA . This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20 mA . The collector current IC for this mode of operation is (A) 0.98 mA (B) 0.99 mA (C) 1.0 mA (D) 1.01 mA

MCQ 4.19

For the BJT, Q1 in the circuit shown below, b = 3, VBEon = 0.7 V, VCEsat = 0.7 V . The switch is initially closed. At time t = 0 , the switch is opened. The time t at which Q1 leaves the active region is



Analog Circuits

(A) 10 ms (C) 50 ms

Page 143

(B) 25 ms (D) 100 ms

Statement for Linked Answer Questions: 20 and 21 In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV . The small signal input vi = Vp cos ^wt h where Vp = 100 mV.

nodia

MCQ 4.20

The bias current IDC through the diodes is (A) 1 mA (B) 1.28 mA (C) 1.5 mA (D) 2 mA

MCQ 4.21

The ac output voltage vac is (A) 0.25 cos ^wt h mV (C) 2 cos (wt) mV 2010

MCQ 4.22

(B) 1 cos (wt) mV (D) 22 cos (wt) mV

ONE MARK

The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is true

(A) The input resistance Ri increases and magnitude of voltage gainAV decreases (B) The input resistance Ri decreases and magnitude of voltage gain AV increases (C) Both input resistance Ri and magnitude of voltage gain AV decreases



Analog Circuits

Chapter 4

(D) Both input resistance Ri and the magnitude of voltage gain AV increases MCQ 4.23

In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2

The value of current Io is approximately (A) 0.5 mA (B) 2 mA (C) 9.3 mA (D) 15 mA MCQ 4.24

Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

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(A) - R2 R1 R || R 3 (C) - 2 R1 2010

(B) - R 3 R1 (D) -b R2 + R 3 l R1 TWO MARKS

Common Data For Q. 25 and 26 Consider the common emitter amplifier shown below with the following circuit parameters: b = 100, gm = 0.3861 A/V, r0 = 259 W, RS = 1 kW, RB = 93 kW, RC = 250 kW, RL = 1 kW, C1 = 3 and C2 = 4.7 mF



MCQ 4.25

Analog Circuits

The resistance seen by the source vS is (A) 258 W (C) 93 kW

Page 145

(B) 1258 W (D) 3

MCQ 4.26

The lower cut-off frequency due to C2 is (A) 33.9 Hz (B) 27.1 Hz (C) 13.6 Hz (D) 16.9 Hz

MCQ 4.27

The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)

nodia 2009 MCQ 4.28

TWO MARKS

In the circuit below, the diode is ideal. The voltage V is given by

(A) min (Vi, 1) (C) min (- Vi, 1)

(B) max (Vi, 1) (D) max (- Vi, 1)



MCQ 4.29

Analog Circuits

Chapter 4

In the following a stable multivibrator circuit, which properties of v0 (t) depend on R2 ?

(A) Only the frequency (B) Only the amplitude (C) Both the amplitude and the frequency (D) Neither the amplitude nor the frequency

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Statement for Linked Answer Question 30 and 31

Consider for CMOS circuit shown, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p -MOSFET is kept constant at 3 V. Assume, that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1mA. V - 2

MCQ 4.30

For small increase in VG beyond 1V, which of the following gives the correct description of the region of operation of each MOSFET (A) Both the MOSFETs are in saturation region (B) Both the MOSFETs are in triode region (C) n-MOSFETs is in triode and p -MOSFET is in saturation region (D) n- MOSFET is in saturation and p -MOSFET is in triode region

MCQ 4.31

Estimate the output voltage V0 for VG = 1.5 V. [Hints : Use the appropriate current-voltage equation for each MOSFET, based on the answer to Q.4.16] (B) 4 + 1 (A) 4 - 1 2 2 3 (C) 4 (D) 4 + 3 2 2

MCQ 4.32

In the circuit shown below, the op-amp is ideal, the transistor has VBE = 0.6 V and b = 150 . Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp.



Analog Circuits

Page 147

(A) Positive feedback, V = 10 V (B) Positive feedback, V = 0 V (C) Negative feedback, V = 5 V (D) Negative feedback, V = 2 V MCQ 4.33

A small signal source Vi (t) = A cos 20t + B sin 106 t is applied to a transistor amplifier as shown below. The transistor has b = 150 and hie = 3W . Which expression best approximate V0 (t)

nodia (A) V0 (t) =- 1500 (A cos 20t + B sin 106 t) (B) V0 (t) = - 1500( A cos 20t + B sin 106 t) (C) V0 (t) =- 1500B sin 106 t (D) V0 (t) =- 150B sin 106 t 2008 MCQ 4.34

ONE MARK

In the following limiter circuit, an input voltage Vi = 10 sin 100pt is applied. Assume that the diode drop is 0.7 V when it is forward biased. When it is forward biased. The zener breakdown voltage is 6.8 V The maximum and minimum values of the output voltage respectively are

(A) 6.1 V, - 0.7 V (C) 7.5 V, - 0.7 V

(B) 0.7 V, - 7.5 V (D) 7.5 V, - 7.5 V



Analog Circuits

Chapter 4

2008 MCQ 4.35

MCQ 4.36

TWO MARSK

For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded.

The current Ix is related to Ibias as (B) Ix = Ibias (A) Ix = Ibias + Is V out (C) Ix = Ibias - cVDD (D) Ix = Ibias - Is RE m Consider the following circuit using an ideal OPAMP. The I-V characteristic of V the diode is described by the relation I = I 0 _eV - 1i where VT = 25 mV, I0 = 1m A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage Vi =- 1 V , the output voltage V0 is

nodia t

(A) 0 V (C) 0.7 V MCQ 4.37

(B) 0.1 V (D) 1.1 V

The OPAMP circuit shown above represents a

(A) high pass filter (B) low pass filter (C) band pass filter (D) band reject filter MCQ 4.38

Two identical NMOS transistors M1 and M2 are connected as shown below. Vbias is chosen so that both transistors are in saturation. The equivalent gm of the pair is defied to be 2Iout at constant Vout 2Vi The equivalent gm of the pair is



Analog Circuits

Page 149

(A) the sum of individual gm ' s of the transistors (B) the product of individual gm ’s of the transistors

MCQ 4.39

(C) nearly equal to the gm of M1 g (D) nearly equal to m of M2 g0 Consider the Schmidt trigger circuit shown below A triangular wave which goes from -12 to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from +15 V to -15 V. The voltage at the non-inverting input switches between

nodia

(A) - 12V to +12 V (C) -5 V to +5 V

(B) -7.5 V to 7.5 V (D) 0 V and 5 V

Statement for Linked Answer Question 40 and 41 In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV/IE , and b and all the capacitances are very large

MCQ 4.40

MCQ 4.41

The value of DC current IE is (A) 1 mA (C) 5 mA

(B) 2 mA (D) 10 mA

The mid-band voltage gain of the amplifier is approximately (A) -180 (B) -120 (C) -90 (D) -60



Analog Circuits

Chapter 4

2007

ONE MARK

MCQ 4.42

The correct full wave rectifier circuit is

MCQ 4.43

In a transconductance amplifier, it is desirable to have (A) a large input resistance and a large output resistance (B) a large input resistance and a small output resistance (C) a small input resistance and a large output resistance (D) a small input resistance and a small output resistance 2007

MCQ 4.44

nodia

For the Op-Amp circuit shown in the figure, V0 is

(A) -2 V (C) -0.5 V MCQ 4.45

MCQ 4.46

TWO MARKS

(B) -1 V (D) 0.5 V

For the BJT circuit shown, assume that the b of the transistor is very large and VBE = 0.7 V. The mode of operation of the BJT is

(A) cut-off

(B) saturation

(C) normal active

(D) reverse active

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT ). For Vi = 2V, V0 = V01, and for Vi = 4V, V0 = V02 . The relationship between V01 and V02 is



Analog Circuits

(A) V02 =

2 Vo1 (C) Vo2 = Vo1 1n2 MCQ 4.47

(B) Vo2 = e2 Vo1 (D) Vo1 - Vo2 = VT 1n2

In the CMOS inverter circuit shown, if the trans conductance parameters of the NMOS and PMOS transistors are W kn = kp = mn Cox Wn = mCox p = 40mA/V2 Ln Lp and their threshold voltages ae VTHn = VTHp = 1 V the current I is

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(A) 0 A (C) 45 mA MCQ 4.48

Page 151

(B) 25 mA (D) 90 mA

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 W. If the input voltage (Vi) range is from 10 to 16 V, the output voltage (V0) ranges from

(A) 7.00 to 7.29 V (C) 7.14 to 7.43 V

(B) 7.14 to 7.29 V (D) 7.29 to 7.43 V

Statement for Linked Answer Questions 49 and 50: Consider the Op-Amp circuit shown in the figure.

MCQ 4.49

The transfer function V0 (s)/ Vi (s) is (A) 1 - sRC 1 + sRC

(B) 1 + sRC 1 - sRC



Analog Circuits

(C) MCQ 4.50

1 1 - sRC

Chapter 4

(D)

1 1 + sRC

If Vi = V1 sin (wt) and V0 = V2 sin (wt + f), then the minimum and maximum values of f (in radians) are respectively (A) - p and p (B) 0 and p 2 2 2 (C) - p and 0 (D) - p and 0 2 2006

ONE MARK

MCQ 4.51

The input impedance (Zi) and the output impedance (Z0) of an ideal transconductance (voltage controlled current source) amplifier are (A) Zi = 0, Z0 = 0 (B) Zi = 0, Z0 = 3 (C) Zi = 3, Z0 = 0 (D) Zi = 3, Z0 = 3

MCQ 4.52

An n-channel depletion MOSFET has following two points on its ID - VGs curve: (i) VGS = 0 at ID = 12 mA and (ii) VGS =- 6 Volts at ID = 0 mA Which of the following Q point will given the highest trans conductance gain for small signals? (A) VGS =- 6 Volts (B) VGS =- 3 Volts (C) VGS = 0 Volts (D) VGS = 3 Volts

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2006 MCQ 4.53

MCQ 4.54

TWO MARKS

For the circuit shown in the following figure, the capacitor C is initially uncharged. At t = 0 the switch S is closed. The Vc across the capacitor at t = 1 millisecond is In the figure shown above, the OP-AMP is supplied with !15V .

(A) 0 Volt

(B) 6.3 Volt

(C) 9.45 Volts

(D) 10 Volts

For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is



Analog Circuits

Page 153

Common Data For Q. 55 to 57 In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters:

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bDC = 60 , VBE = 0.7V, hie " 3 The capacitance CC can be assumed to be infinite. In the figure above, the ground has been shown by the symbol 4

MCQ 4.55

Under the DC conditions, the collector-or-emitter voltage drop is (A) 4.8 Volts (B) 5.3 Volts (C) 6.0 Volts (D) 6.6 Volts

MCQ 4.56

If bDC is increased by 10%, the collector-to-emitter voltage drop (A) increases by less than or equal to 10% (B) decreases by less than or equal to 10% (C) increase by more than 10% (D) decreases by more than 10%

MCQ 4.57

The small-signal gain of the amplifier vc is vs (A) -10 (B) -5.3 (C) 5.3 (D) 10

Common Data For Q. 58 and 59 A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output Vout . Use the component values shown in the figure.



Analog Circuits

Chapter 4

MCQ 4.58

The power dissipation across the transistor Q1 shown in the figure is (A) 4.8 Watts (B) 5.0 Watts (C) 5.4 Watts (D) 6.0 Watts

MCQ 4.59

If the unregulated voltage increases by 20%, the power dissipation across the transistor Q1 (A) increases by 20% (B) increases by 50% (C) remains unchanged (D) decreases by 20% 2005

MCQ 4.60

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ONE MARK

The input resistance Ri of the amplifier shown in the figure is

(A) 30 kW 4

(B) 10 kW

(C) 40 kW

(D) infinite

MCQ 4.61

The effect of current shunt feedback in an amplifier is to (A) increase the input resistance and decrease the output resistance (B) increases both input and output resistance (C) decrease both input and output resistance (D) decrease the input resistance and increase the output resistance

MCQ 4.62

The cascade amplifier is a multistage configuration of (A) CC - CB (B) CE - CB (C) CB - CC (D) CE - CC 2005

MCQ 4.63

TWO MARKS

In an ideal differential amplifier shown in the figure, a large value of (RE ). (A) increase both the differential and common - mode gains. (B) increases the common mode gain only. (C) decreases the differential mode gain only. (D) decreases the common mode gain only.



MCQ 4.64

Analog Circuits

For an npn transistor connected as shown in figure VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature 300 K is 10 - 13 A, the emitter current is

(A) 30 mA (C) 49 mA MCQ 4.65

Page 155

(B) 39 mA (D) 20 mA

The voltage e0 is indicated in the figure has been measured by an ideal voltmeter. Which of the following can be calculated ?

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(A) Bias current of the inverting input only (B) Bias current of the inverting and non-inverting inputs only (C) Input offset current only (D) Both the bias currents and the input offset current MCQ 4.66

MCQ 4.67

The Op-amp circuit shown in the figure is filter. The type of filter and its cut. Off frequency are respectively

(A) high pass, 1000 rad/sec.

(B) Low pass, 1000 rad/sec

(C) high pass, 1000 rad/sec

(D) low pass, 10000 rad/sec

The circuit using a BJT with b = 50 and VBE = 0.7V is shown in the figure. The base current IB and collector voltage by VC and respectively



Analog Circuits

(A) 43 mA and 11.4 Volts (C) 45 mA and 11 Volts MCQ 4.68

(B) 40 mA and 16 Volts (D) 50 mA and 10 Volts

The Zener diode in the regulator circuit shown in the figure has a Zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this current ensuring proper functioning over the input voltage range between 20 and 30 volts, is

(A) 23.7 mA (C) 13.7 mA MCQ 4.69

Chapter 4

(B) 14.2 mA (D) 24.2 mA

Both transistors T1 and T2 show in the figure, have a b = 100 , threshold voltage of 1 Volts. The device parameters K1 and K2 of T1 and T2 are, respectively, 36 mA/V2 and 9 mA/V 2 . The output voltage Vo i s

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(A) 1 V (C) 3 V

(B) 2 V (D) 4 V

Common Data For Q. 70 to 72 Given, rd = 20kW , IDSS = 10 mA, Vp =- 8 V

MCQ 4.70

MCQ 4.71

Zi and Z0 of the circuit are respectively (A) 2 MW and 2 kW (B) 2 MW and 20 kW 11 (C) infinity and 2 MW (D) infinity and 20 kW 11 ID and VDS under DC conditions are respectively (A) 5.625 mA and 8.75 V (B) 1.875 mA and 5.00 V (C) 4.500 mA and 11.00 V (D) 6.250 mA and 7.50 V



Analog Circuits

Page 157

MCQ 4.72

Transconductance in milli-Siemens (mS) and voltage gain of the amplifier are respectively (A) 1.875 mS and 3.41 (B) 1.875 ms and -3.41 (C) 3.3 mS and -6 (D) 3.3 mS and 6

MCQ 4.73

Given the ideal operational amplifier circuit shown in the figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage.

nodia 2004 MCQ 4.74

An ideal op-amp is an ideal (A) voltage controlled current source (B) voltage controlled voltage source (C) current controlled current source (D) current controlled voltage source

MCQ 4.75

Voltage series feedback (also called series-shunt feedback) results in (A) increase in both input and output impedances (B) decrease in both input and output impedances

ONE MARK

(C) increase in input impedance and decrease in output impedance (D) decrease in input impedance and increase in output impedance



MCQ 4.76

Analog Circuits

Chapter 4

The circuit in the figure is a

(A) low-pass filter (C) band-pass filter

(B) high-pass filter (D) band-reject filter

2004

TWO MARKS

MCQ 4.77

A bipolar transistor is operating in the active region with a collector current of 1 mA. Assuming that the b of the transistor is 100 and the thermal voltage (VT ) is 25 mV, the transconductance (gm) and the input resistance (rp) of the transistor in the common emitter configuration, are (A) gm = 25 mA/V and rp = 15.625 kW (B) gm = 40 mA/V and rp = 4.0 kW (C) gm = 25 mA/V and rp = 2.5 k W (D) gm = 40 mA/V and rp = 2.5 kW

MCQ 4.78

The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is

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(A) 1 mF 2p 1 mF (C) 2p 6 MCQ 4.79

(B) 2p mF (D) 2p 6 mF

In the op-amp circuit given in the figure, the load current iL is

(A) - Vs R2

(B) Vs R2



Analog Circuits

(C) - Vs RL MCQ 4.80

MCQ 4.81

MCQ 4.82

(D) Vs R1

In the voltage regulator shown in the figure, the load current can vary from 100 mA to 500 mA. Assuming that the Zener diode is ideal (i.e., the Zener knee current is negligibly small and Zener resistance is zero in the breakdown region), the value of R is

(A) 7 W (B) 70 W 70 (C) (D) 14 W W 3 In a full-wave rectifier using two ideal diodes, Vdc and Vm are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are (B) Idc = 2 Vm , PIV = 2Vm (A) Vdc = Vm , PIV = 2Vm p p (C) Vdc = 2 Vm , PIV = Vm (D) Vdc Vm , PIV = Vm p p Assume that the b of transistor is extremely large and VBE = 0.7V, IC and VCE in the circuit shown in the figure

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(A) IC = 1 mA, VCE = 4.7 V (C) IC = 1 mA, VCE = 2.5 V 2003 MCQ 4.83

Page 159

(B) IC = 0.5 mA, VCE = 3.75 V (D) IC = 0.5 mA, VCE = 3.9 V ONE MARK

Choose the correct match for input resistance of various amplifier configurations shown below : Configuration Input resistance CB : Common Base LO : Low CC : Common Collector MO : Moderate CE : Common Emitter HI : High (A) CB - LO, CC - MO, CE - HI (B) CB - LO, CC - HI, CE - MO (C) CB - MO, CC - HI, CE - LO (D) CB - HI, CC - LO, CE - MO



MCQ 4.84

Analog Circuits

The circuit shown in the figure is best described as a

(A) bridge rectifier (C) frequency discriminator MCQ 4.85

Chapter 4

(B) ring modulator (D) voltage double

If the input to the ideal comparators shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparators has a duty cycle of

(A) 1/2 (C) 1/6

(B) 1/3 (D) 1/2

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MCQ 4.86

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is (A) 23 dB (B) 25 dB (C) 46 dB (D) 50 dB

MCQ 4.87

Generally, the gain of a transistor amplifier falls at high frequencies due to the (A) internal capacitances of the device (B) coupling capacitor at the input (C) skin effect (D) coupling capacitor at the output 2003

MCQ 4.88

An amplifier without feedback has a voltage gain of 50, input resistance of 1 k W and output resistance of 2.5 kW. The input resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is (A) 1 kW (B) 1 kW 11 5 (C) 5 kW

MCQ 4.89

TWO MARKS

(D) 11 kW

In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3 V and IC = 1.5 mA when its b is 150. For a transistor with b of 200, the operating point (VCE , IC ) is



Analog Circuits

(A) (2 V, 2 mA) (C) (4 V, 2 mA) MCQ 4.90

1 (2p 6 RC) 1 (C) ( 6 RC)

1 (2pRC)

(D)

6 (2pRC)

nodia (B) 6 V (D) 12 V

If the op-amp in the figure is ideal, the output voltage Vout will be equal to

(A) 1 V (C) 14 V MCQ 4.93

(B)

The output voltage of the regulated power supply shown in the figure is

(A) 3 V (C) 9 V MCQ 4.92

(B) (3 V, 2 mA) (D) (4 V, 1 mA)

The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is

(A)

MCQ 4.91

Page 161

(B) 6 V (D) 17 V

Three identical amplifiers with each one having a voltage gain of 50, input resistance of 1 kW and output resistance of 250 W are cascaded. The opened circuit voltages gain of the combined amplifier is (A) 49 dB (B) 51 dB (C) 98 dB (D) 102 dB



MCQ 4.94

Analog Circuits

Chapter 4

An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 mF in every cycle. The charging requires (A) Constant voltage source of 3 V for 1 ms (B) Constant voltage source of 3 V for 2 ms (C) Constant voltage source of 1 mA for 1 ms (D) Constant voltage source of 3 mA for 2 ms 2002

ONE MARK

MCQ 4.95

In a negative feedback amplifier using voltage-series (i.e. voltage-sampling, series mixing) feedback. (A) Ri decreases and R0 decreases (B) Ri decreases and R0 increases (D) Ri increases and R0 increases (C) Ri increases and R0 decreases (Ri and R0 denote the input and output resistance respectively)

MCQ 4.96

A 741-type opamp has a gain-bandwidth product of 1 MHz. A non-inverting amplifier suing this opamp and having a voltage gain of 20 dB will exhibit a -3 dB bandwidth of (A) 50 kHz (B) 100 kHz 1000 kHz (D) 1000 kHz (C) 17 7.07 Three identical RC-coupled transistor amplifiers are cascaded. If each of the amplifiers has a frequency response as shown in the figure, the overall frequency response is as given in

MCQ 4.97

nodia



Analog Circuits

Page 163

2002 MCQ 4.98

TWO MARKS

The circuit in the figure employs positive feedback and is intended to generate V (f) 1 sinusoidal oscillation. If at a frequency f0, B (f) = 3 f = +0c, then to sustain V0 (f) 6 oscillation at this frequency

(A) R2 = 5R1 (C) R2 = R1 6 MCQ 4.99

An amplifier using an opamp with a slew-rate SR = 1 V/m sec has a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal signals from dc to 20 kHz without introducing any slew-rate induced distortion, then the input signal level must not exceed. (A) 795 mV (B) 395 mV

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(C) 79.5 mV MCQ 4.100

(D) 39.5 mV

A zener diode regulator in the figure is to be designed to meet the specifications: IL = 10 mA V0 = 10 V and Vin varies from 30 V to 50 V. The zener diode has Vz = 10 V and Izk (knee current) =1 mA. For satisfactory operation

(A) R # 1800W (C) 3700W # R # 4000W MCQ 4.101

(B) R2 = 6R1 (D) R2 = R1 5

(B) 2000W # R # 2200W (D) R $ 4000W

The voltage gain Av = v0 of the JFET amplifier shown in the figure is IDSS = 10 vt mA Vp =- 5 V(Assume C1, C2 and Cs to be very large

(A) +16 (C) +8

(B) -16 (D) -6



Analog Circuits

Chapter 4

2001 MCQ 4.102

ONE MARK

The current gain of a BJT is (A) gm r0 (C) gm rp

MCQ 4.103

gm r g (D) m rp (B)

Thee ideal OP-AMP has the following characteristics. (A) Ri = 3, A = 3, R0 = 0 (B) Ri = 0, A = 3, R0 = 0 (C) Ri = 3, A = 3, R0 = 3 (D) Ri = 0, A = 3, R0 = 3

MCQ 4.104

Consider the following two statements : Statement 1 : A stable multi vibrator can be used for generating square wave. Statement 2: Bistable multi vibrator can be used for storing binary information. (A) Only statement 1 is correct (B) Only statement 2 is correct (C) Both the statements 1 and 2 are correct

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(D) Both the statements 1 and 2 are incorrect 2001

-14

-13

TWO MARKS

MCQ 4.105

An npn BJT has gm = 38 mA/V, C m = 10 F, C p = 4 # 10 gain b0 = 90 . For this transistor fT and fb are (A) fT = 1.64 # 108 Hz and fb = 1.47 # 1010 Hz (B) fT = 1.47 # 1010 Hz and fb = 1.64 # 108 Hz (C) fT = 1.33 # 1012 Hz and fb = 1.47 # 1010 Hz (D) fT = 1.47 # 1010 Hz and fb = 1.33 # 1012 Hz

MCQ 4.106

The transistor shunt regulator shown in the figure has a regulated output voltage of 10 V, when the input varies from 20 V to 30 V. The relevant parameters for the zener diode and the transistor are : Vz = 9.5 , VBE = 0.3 V, b = 99 , Neglect the current through RB . Then the maximum power dissipated in the zener diode (Pz ) and the transistor (PT ) are

F, and DC current

(A) Pz = 75 mW, PT = 7.9 W (B) Pz = 85 mW, PT = 8.9 W (C) Pz = 95 mW, PT = 9.9 W (D) Pz = 115 mW, PT = 11.9 W



MCQ 4.107

Analog Circuits

Page 165

The oscillator circuit shown in the figure is

4 (A) Hartely oscillator with foscillation = 79.6 MHz (B) Colpitts oscillator with foscillation = 50.3 MHz (C) Hartley oscillator with foscillation = 159.2 MHz (D) Colpitts oscillator with foscillation = 159.3 MHz MCQ 4.108

The inverting OP-AMP shown in the figure has an open-loop gain of 100.

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The closed-loop gain V0 is Vs (A) - 8 (C) - 10 MCQ 4.109

(B) - 9 (D) - 11

In the figure assume the OP-AMPs to be ideal. The output v0 of the circuit is

(A) 10 cos (100t) (C) 10 - 4 2000

t

#0 cos (100t) dt

(B) 10

t

#0 cos (100t) dt

(D) 10 - 4 d cos (100t) dt ONE MARK

MCQ 4.110

Introducing a resistor in the emitter of a common amplifier stabilizes the dc operating point against variations in (A) only the temperature (B) only the b of the transistor (C) both temperature and b (D) none of the above

MCQ 4.111

In the differential amplifier of the figure, if the source resistance of the current



Analog Circuits

Chapter 4

source IEE is infinite, then the common-mode gain is

(A) zero (C) indeterminate MCQ 4.112

(B) infinite (D) Vin1 + Vin2 2VT

In the circuit of the figure, V0 is

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(A) -1 V (C) +1 V

(B) 2 V (D) +15 V

MCQ 4.113

The current gain of a bipolar transistor drops at high frequencies because of (A) transistor capacitances (B) high current effects in the base (C) parasitic inductive elements (D) the Early effect

MCQ 4.114

If the op-amp in the figure, is ideal, then v0 is

(A) zero (C) - (V1 + V2) sin wt MCQ 4.115

(B) (V1 - V2) sin wt (D) (V1 + V2) sin wt

The configuration of the figure is a



Analog Circuits

(A) precision integrator (C) Butterworth high pass filter MCQ 4.116

(B) triangular wave (D) sine wave

The most commonly used amplifier is sample and hold circuits is (A) a unity gain inverting amplifier (B) a unity gain non-inverting amplifier (C) an inverting amplifier with a gain of 10 (D) an inverting amplifier with a gain of 100 2000

MCQ 4.118

(B) Hartely oscillator (D) Wien-bridge oscillator

Assume that the op-amp of the figure is ideal. If vi is a triangular wave, then v0 will be

(A) square wave (C) parabolic wave MCQ 4.117

Page 167

nodia

TWO MARKS

In the circuit of figure, assume that the transistor is in the active region. It has a large b and its base-emitter voltage is 0.7 V. The value of Ic is

(A) Indeterminate since Rc is not given (B) 1 mA (C) 5 mA (D) 10 mA MCQ 4.119

If the op-amp in the figure has an input offset voltage of 5 mV and an open-loop voltage gain of 10000, then v0 will be

(A) 0 V (C) + 15 V or -15 V

(B) 5 mV (D) +50 V or -50 V



Analog Circuits

Chapter 4

1999

ONE MARK

MCQ 4.120

The first dominant pole encountered in the frequency response of a compensated op-amp is approximately at (A) 5 Hz (B) 10 kHz (C) 1 MHz (D) 100 MHz

MCQ 4.121

Negative feedback in an amplifier (A) reduces gain (B) increases frequency and phase distortions (C) reduces bandwidth (D) increases noise

MCQ 4.122

In the cascade amplifier shown in the given figure, if the common-emitter stage (Q1) has a transconductance gm1 , and the common base stage (Q2) has a transconductance gm2 , then the overall transconductance g (= i 0 /vi) of the cascade amplifier is

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(A) gm1 g (C) m1 2 MCQ 4.123

(B) gm2 g (D) m2 2

Crossover distortion behavior is characteristic of (A) Class A output stage (B) Class B output stage (C) Class AB output stage (D) Common-base output stage 1999

TWO MARK

MCQ 4.124

An amplifier has an open-loop gain of 100, an input impedance of 1 kW,and an output impedance of 100 W. A feedback network with a feedback factor of 0.99 is connected to the amplifier in a voltage series feedback mode. The new input and output impedances, respectively, are (A) 10 W and 1W (B) 10 W and 10 kW (C) 100 kW and 1 W (D) 100 kW and 1 kW

MCQ 4.125

A dc power supply has a no-load voltage of 30 V, and a full-load voltage of 25 V at a full-load current of 1 A. Its output resistance and load regulation, respectively, are (A) 5 W and 20% (B) 25 W and 20% (C) 5 W and 16.7% (D) 25 W and 16.7% 1998

MCQ 4.126

ONE MARK

The circuit of the figure is an example of feedback of the following type



Analog Circuits

(A) current series (C) voltage series

Page 169

(B) current shunt (D) voltage shunt

MCQ 4.127

In a differential amplifier, CMRR can be improved by using an increased (A) emitter resistance (B) collector resistance (C) power supply voltages (D) source resistance

MCQ 4.128

From a measurement of the rise time of the output pulse of an amplifier whose is a small amplitude square wave, one can estimate the following parameter of the amplifier (A) gain-bandwidth product (B) slow rate (C) upper 3–dB frequency (D) lower 3–dB frequency

MCQ 4.129

The emitter coupled pair of BJT’s given a linear transfer relation between the differential output voltage and the differential output voltage and the differential input voltage Vid is less a times the thermal voltage, where a is (A) 4 (B) 3 (C) 2 (D) 1

MCQ 4.130

In a shunt-shunt negative feedback amplifier, as compared to the basic amplifier (A) both, input and output impedances,decrease (B) input impedance decreases but output impedance increases (C) input impedance increase but output

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(D) both input and output impedances increases. 1998 MCQ 4.131

TWO MARKS

A multistage amplifier has a low-pass response with three real poles at s =- w1 - w2 and w3 . The approximate overall bandwidth B of the amplifier will be given by (A) B = w1 + w2 + w3 (B) 1 = 1 + 1 + 1 w1 w2 w3 B (C) B = (w1 + w2 + w3) 1/3

(D) B =

w12 + w22 + w23

MCQ 4.132

One input terminal of high gain comparator circuit is connected to ground and a sinusoidal voltage is applied to the other input. The output of comparator will be (A) a sinusoid (B) a full rectified sinusoid (C) a half rectified sinusoid (D) a square wave

MCQ 4.133

In a series regulated power supply circuit, the voltage gain Av of the ‘pass’ transistor satisfies the condition (A) Av " 3 (B) 1 << Av < 3 (C) Av . 1 (D) Av << 1



Analog Circuits

Chapter 4

MCQ 4.134

For full wave rectification, a four diode bridge rectifier is claimed to have the following advantages over a two diode circuit : (A) less expensive transformer, (B) smaller size transformer, and (C) suitability for higher voltage application. Of these, (A) only (1) and (2) are true (B) only (1) and (3) are true (C) only (2) and (3) are true (D) (1), (2) as well as (3) are true

MCQ 4.135

In the MOSFET amplifier of the figure is the signal output V1 and V2 obey the relationship

MCQ 4.136

nodia

(A) V1 = V2 2

(B) V1 =-V2 2

(C) V1 = 2V2

(D) V1 =- 2V2

For small signal ac operation, a practical forward biased diode can be modelled as (A) a resistance and a capacitance in series (B) an ideal diode and resistance in parallel (C) a resistance and an ideal diode in series (D) a resistance 1997

MCQ 4.137

ONE MARK

In the BJT amplifier shown in the figure is the transistor is based in the forward active region. Putting a capacitor across RE will

(A) decrease the voltage gain and decrease the input impedance (B) increase the voltage gain and decrease the input impedance (C) decrease the voltage gain and increase the input impedance



Analog Circuits

Page 171

(D) increase the voltage gain and increase the input impedance MCQ 4.138

A cascade amplifier stags is equivalent to (A) a common emitter stage followed by a common base stage (B) a common base stage followed by an emitter follower (C) an emitter follower stage followed by a common base stage (D) a common base stage followed by a common emitter stage

MCQ 4.139

In a common emitter BJT amplifier, the maximum usable supply voltage is limited by (A) Avalanche breakdown of Base-Emitter junction (B) Collector-Base breakdown voltage with emitter open (BVCBO) (C) Collector-Emitter breakdown voltage with base open (BVCBO) (D) Zener breakdown voltage of the Emitter-Base junction 1997

MCQ 4.140

nodia

In the circuit of in the figure is the current iD through the ideal diode (zero cut in voltage and forward resistance) equals

(A) 0 A (C) 1 A MCQ 4.141

(B) 4 A (D) None of the above

The output voltage V0 of the circuit shown in the figure is

(A) - 4 V (C) 5 V MCQ 4.142

TWO MARKS

(B) 6 V (D) - 5.5 V

A half wave rectifier uses a diode with a forward resistance Rf . The voltage is Vm sin wt and the load resistance is RL . The DC current is given by Vm (B) (A) Vm p (R f + RL) 2 RL (C) 2Vm (D) Vm RL p



Analog Circuits

Chapter 4

1996 MCQ 4.143

ONE MARK

In the circuit of the given figure, assume that the diodes are ideal and the meter is an average indicating ammeter. The ammeter will read

(A) 0.4 2 A (C) 0.8 A p MCQ 4.144

The circuit shown in the figure is that of

nodia

(A) a non-inverting amplifier (C) an oscillator 1996 MCQ 4.145

(B) an inverting amplifier (D) a Schmitt trigger TWO MARKS

In the circuit shown in the given figure N is a finite gain amplifier with a gain of k , a very large input impedance, and a very low output impedance. The input impedance of the feedback amplifier with the feedback impedance Z connected as shown will be

(A) Z b1 - 1 l k (C) Z (k - 1) MCQ 4.146

(B) 0.4 A (D) 0.4 mamp p

(B) Z (1 - k) (D) Z (1 - k)

A Darlington stage is shown in the figure. If the transconductance of Q1 is gm1 and c Q2 is gm2 , then the overall transconductance gmc ;T i cc E is given by vbe



Analog Circuits

(A) gm1 (C) gm2 MCQ 4.147

(B) 0.5 gm1 (D) 0.5 gm2

Value of R in the oscillator circuit shown in the given figure, so chosen that it just oscillates at an angular frequency of w. The value of w and the required value of R will respectively be

nodia

(A) 105 rad/ sec, 2 # 10 4 W (C) 2 # 10 4 rad/ sec, 105 W MCQ 4.148

Page 173

(B) 2 # 10 4 rad/ sec, 2 # 10 4 W (D) 105 rad/ sec, 105 W

A zener diode in the circuit shown in the figure is has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage V0 constant at 6 V?

(A) 0 mA, 180 mA (C) 10 mA, 55 mA

(B) 5 mA, 110 mA (D) 60 mA, 180 mA ***********



Analog Circuits

Chapter 4

SOLUTIONS SOL 4.1

Option (B) is correct. For the given ideal op-amp, negative terminal will be also ground (at zero voltage) and so, the collector terminal of the BJT will be at zero voltage. i.e., VC = 0 volt The current in 1 kW resistor is given by I = 5 - 0 = 5 mA 1 kW This current will flow completely through the BJT since, no current will flow into the ideal op-amp ( I/P resistance of ideal op-amp is infinity). So, for BJT we have

nodia

VC = 0 VB = 0 IC = 5 mA i.e.,the base collector junction is reverse biased (zero voltage) therefore, the collector current (IC ) can have a value only if base-emitter is forward biased. Hence, VBE = 0.7 volts & VB - VE = 0.7 & 0 - Vout = 0.7 or, Vout =- 0.7 volt SOL 4.2

Option (A) is correct. The i/p voltage of the system is given as Vin = V1 + Vf = V1 + k Vout = V1 + k A 0 V1 ^Vout = A 0 V1h = V1 ^1 + k A 0h Therefore, if k is increased then input voltage is also increased so, the input impedance increases. Now, we have Vin Vout = A 0 V1 = A 0 = A 0 Vin ^1 + k A 0h ^1 + k A 0h Since, Vin is independent of k when seen from output mode, the output voltage decreases with increase in k that leads to the decrease of output impedance. Thus, input impedance increases and output impedance decreases.

SOL 4.3

Option (A) is correct. For the given circuit, we obtain the small signal model as shown in figure below :



Analog Circuits

Page 175

We obtain the node voltage at V1 as V1 + V1 + gm Vi = 0 RD R + 1 L sC - gm Vi 1 + 1 RD R + 1 L sC Therefore, the output voltage V0 is obtained as - gm Vi N RL J V0 = V1 RL = K 1 O 1 1 1 RL + RL + + sC sC KK RD R + 1 OO L sC P L so, the transfer function is V0 = - RD RL sCgm Vi 1 + sC ^RD + RL h 1 Then, we have the pole at w = C ^RD + RL h It gives the lower cutoff frequency of transfer function. 1 i.e., w0 = C ^RD + RL h 1 1 or, f0 = = 2pC ^RD + RL h 2p # 10-6 # 20 # 103 = 7.97 . 8 Hz V1 =

&

SOL 4.4

nodia


From the circuit, we have Is = I Z + I L or, (1) I Z = Is - I L Since, voltage across zener diode is 5 V so, current through 100 W resistor is obtained as Is = 10 - 5 = 0.05 A 100 Therefore, the load current is given by IL = 5 RL Since, for proper operation, we must have IZ $ Iknes So, from Eq. (1), we write 0.05 A - 5 $ 10 mA RL 50 mA - 5 $ 10 mA RL



Analog Circuits

Chapter 4

40 mA $ 5 RL -3 40 # 10 $ 5 RL 1 # RL 5 40 # 10-3 5 # RL 40 # 10-3 or, 125 W # RL Therefore, minimum value of RL = 125 W Now, we know that power rating of Zener diode is given by PR = VZ IZ^maxh IZ^maxh is maximum current through zener diode in reverse bias. Maximum currrent through zener diode flows when load current is zero. i.e., IZ^maxh = Is = 10 - 5 = 0.05 100 Therefore, PR = 5 # 0.05 W = 250 mW SOL 4.5


nodia

For the given ideal op-Amps we can assume V 2- = V 2+ = V2 (ideal) V 1+ = V 1- = V1 (ideal) So, by voltage division V1 = Vout # 1 2 Vout = 2V1 and, as the I/P current in Op-amp is always zero therefore, there will be no voltage drop across 1 KW in II op-amp i.e., V2 = 1 V Therefore, V1 - V2 = V2 - ^- 2h 1 1 & V1 - 1 = 1 + 2 or, V1 = 4 Hence, Vout = 2V1 = 8 volt SOL 4.6

Option (B) is correct. For the given circuit, we can make the truth table as below



Analog Circuits

X 0 0 1 1

Y 0 1 0 1

Page 177

Z 0 1 0 0

Logic 0 means voltage is v = 0 volt and logic 1 means voltage is 5 volt For x = 0 , y = 0 , Transistor is at cut off mode and diode is forward biased. Since, there is no drop across forward biased diode. So, Z =Y=0 For x = 0 , y = 1, Again Transistor is in cutoff mode, and diode is forward biased. with no current flowing through resistor. So, Z =Y=1 For x = 1, y = 0 , Transistor is in saturation mode and so, z directly connected to ground irrespective of any value of Y . i.e., Z = 0 (ground) Similarly for X = Y = 1 Z = 0 (ground) Hence, from the obtained truth table, we get Z =XY SOL 4.7

nodia

Option (D) is correct. Given, the input voltage VYZ = 100 sin wt

For + ve half cycle VYZ > 0 i.e., VY is a higher voltage than VZ So, the diode will be in cutoff region. Therefore, there will no voltage difference between X and W node. i.e., VWX = 0 Now, for - ve half cycle all the four diodes will active and so, X and W terminal is short circuited i.e., VWX = 0 Hence, VWX = 0 for all t SOL 4.8

Option (C) is correct. The equivalent circuit can be shown as



Analog Circuits

Chapter 4

R2 = 3R2 R1 + R 2 R1 + R 2 R 2 R1 and RTh = R 2 + R1 Since, IC = bIB has b . 3 (very high) so, IB is negative in comparison to IC . Therefore, we can write the base voltage VTh = VCC

VB = VTh VTh - 0.7 - IC RE = 0

So, or, or, or, or,

nodia

Hence, SOL 4.9

3R2 = ^60 kWh^1.2h + 1.2R2

1.8R2 = ^60 kWh # ^1.2h

R2 = 60 # 1.2 = 40 kW 1.8

Option (D) is correct. Let v > 0.7 V and diode is forward biased. By applying Kirchoff’s voltage law 10 - i # 1k - v = 0 10 - :v - 0.7 D (1000) - v = 0 500

So, SOL 4.10

3R2 - 0.7 - 10-3 500 = 0 ^ h^ h R1 + R 2 3R 2 = 0.7 + 0.5 60 kW + R2

10 - (v - 0.7) # 2 - v = 0 10 - 3v + 1.4 = 0 v = 11.4 = 3.8 V > 0.7 3 v 0 . 7 3 . i= = 8 - 0.7 = 6.2 mA 500 500

Option (C) is correct. Given ib = 1 + 0.1 cos (1000pt) mA So, IB = DC component of ib = 1 mA In small signal model of the transistor bVT rp = IC = VT = VT = VT IB IB IC /b So, rp = 25 mV = 25 W 1 mA

SOL 4.11

(Assumption is true)

VT " Thermal voltage IC = I B b VT = 25 mV, IB = 1 mA

Option (A) is correct. The circuit composed of a clamper and a peak rectifier as shown.



Analog Circuits

Page 179

Clamper clamps the voltage to zero voltage, as shown

The peak rectifier adds + 1 V to peak voltage, so overall peak voltage lowers down by - 1 volt. So, vo = cos wt - 1 SOL 4.12

nodia

Option (A) is correct. We put a test source between terminal 1, 2 to obtain equivalent impedance

ZTh = Vtest Itest Applying KCL at top right node Vtest + Vtest - 99I = I b test 9 k + 1k 100 Vtest + Vtest - 99I = I b test 10 k 100 But Ib =- Vtest =-Vtest 9k + 1k 10k

SOL 4.13

...(i)

Substituting Ib into equation (i), we have Vtest + Vtest + 99Vtest = I test 10 k 100 10 k 100Vtest + Vtest = I test 10 # 103 100 2Vtest = I test 100 ZTh = Vtest = 50 W Itest Option (B) is correct. First we obtain the transfer function.



Analog Circuits

Chapter 4

0 - Vi (jw) 0 - Vo (jw) =0 + 1 +R R2 1 jw C - Vi (jw) Vo (jw) = 1 +R R2 1 jw C Vi (jw) R2 R1 - j 1 wC 1 " 3, so V = 0 o wC

Vo (jw) =At w " 0 (Low frequencies),

nodia

At w " 3 (higher frequencies)

1 " 0, so V (jw) =- R2 V (jw) o R1 i wC The filter passes high frequencies so it is a high pass filter. H (jw) = Vo = - R2 Vi R1 - j 1 wC R R 2 2 H (3) = = R1 R1

2 times of maximum gain 6H (3)@ H ^ jw0h = 1 H (3) 2 R2 R2 1 = b R1 l 1 2 2 R1 + 2 2 w0 C

At 3 dB frequency, gain will be

So,

2R 12 = R 12 +

1 w02 C 2

1 w 2C 2 w0 = 1 R1 C

R 12 =

SOL 4.14

Option (D) is correct. DC Analysis :

Using KVL in input loop,



Analog Circuits

VC - 100IB - 0.7 = 0 VC = 100IB + 0.7 IC - IE = 13.7 - VC = (b + 1) IB 12k 13.7 - VC = 100I B 12 # 103 Solving equation (i) and (ii),

Page 181

...(i)

...(ii)

IB = 0.01 mA Small Signal Analysis : Transforming given input voltage source into equivalent current source.

nodia

This is a shunt-shunt feedback amplifier. Given parameters, rp = VT = 25 mV = 2.5 kW IB 0.01 mA b 100 gm = = = 0.04 s rp 2.5 # 1000 Writing KCL at output node v0 + g v + v0 - vp = 0 m p RC RF v 0 : 1 + 1 D + v p :gm - 1 D = 0 RC RF RF Substituting RC = 12 kW, RF = 100 kW, gm = 0.04 s

v 0 (9.33 # 10-5) + v p (0.04) = 0 v 0 =- 428.72Vp Writing KCL at input node vi = v p + v p + v p - vo Rs Rs rp RF vi = v 1 + 1 + 1 - v 0 p: Rs Rs rp RF D RF vi = v (5.1 10-4) - v 0 # p Rs RF Substituting Vp from equation (i)

...(i)

vi = - 5.1 # 10-4 v - v 0 0 428.72 Rs RF vi Rs = 10 kW (source resistance) =- 1.16 # 10-6 v 0 - 1 # 10-5 v 0 10 # 103 vi =- 1.116 # 10-5 10 # 103 1 Av = v 0 = - 8.96 vi 10 # 103 # 1.116 # 10-5 SOL 4.15




Analog Circuits

Chapter 4

For the parallel RLC circuit resonance frequency is, 1 wr = 1 = = 10 M rad/s LC 10 # 10-6 # 1 # 10-9 Thus given frequency is resonance frequency and parallel RLC circuit has maximum impedance at resonance frequency Gain of the amplifier is gm # (ZC RL) where ZC is impedance of parallel RLC circuit. At w = wr , ZC = R = 2 kW = ZC max . Hence at this frequency (wr ), gain is Gain w = w = gm (ZC RL) = gm (2k 2k) = gm # 103 which is maximum. Therefore gain is maximum at wr = 10 M rad/ sec . r

SOL 4.16

Option (D) is correct. The given circuit is shown below :

nodia

From diagram we can write Ii = Vo + Vo R1 sL1 Transfer function

or At w = 0 At w = 3 SOL 4.17

H (s) = Vo = sR1 L1 I1 R1 + sL1 jw R 1 L 1 H (jw) = R 1 + jw L 1 H (jw) = 0 H (jw) = R1 = constant .

Hence HPF.

Option (C) is correct. Given circuit is shown below.

For transistor M2 , VGS = VG - VS = Vx - 0 = Vx VDS = VD - VS = Vx - 0 = Vx



Analog Circuits

Page 183

Since VGS - VT = Vx - 1 < VDS , thus M2 is in saturation. By assuming M1 to be in saturation we have IDS (M ) = IDS (M ) mn C 0x m C (4) (5 - Vx - 1) 2 = n 0x 1 (Vx - 1) 2 2 2 1

2

4 (4 - Vx ) 2 = (Vx - 1) 2 or 2 (4 - Vx ) = ! (Vx - 1) Taking positive root, 8 - 2Vx = Vx - 1 Vx = 3 V At Vx = 3 V for M1,VGS = 5 - 3 = 2 V < VDS . Thus our assumption is true and Vx = 3 V . SOL 4.18

Option (D) is correct. We have

a = 0.98 Now b = a = 4.9 1-a In active region, for common emitter amplifier,

nodia

Substituting ICO

IC = bIB + (1 + b) ICO = 0.6 mA and IB = 20 mA in above eq we have,

...(1)

IC = 1.01 mA

SOL 4.19

Option (C) is correct. In active region VBEon = 0.7 V Emitter voltage VE = VB - VBEon =- 5.7 V V - (- 10) - 5.7 - (- 10) Emitter Current IE = E = = 1 mA 4.3k 4.3k Now IC . IE = 1 mA Applying KCL at collector Since or

i1 = 0.5 mA i1 = C dVC dt VC = 1 # i1 dt = i1 t C C

...(1)

with time, the capacitor charges and voltage across collector changes from 0 towards negative. When saturation starts,

VCE = 0.7 & VC =+ 5 V (across capacitor)



Analog Circuits

Chapter 4

+ 5 = 0.5 mA T 5 mA

Thus from (1) we get,

-6 T = 5 # 5 # 10 = 50 m sec -3 0.5 # 10

or SOL 4.20

Option (A) is correct. The current flows in the circuit if all the diodes are forward biased. In forward biased there will be 0.7 V drop across each diode. 12.7 - 4 (0.7) Thus IDC = = 1 mA 9900

SOL 4.21

Option (B) is correct. The forward resistance of each diode is r = VT = 25 mV = 25 W IC 1 mA 4 (r) Thus Vac = Vi # e 4 (r) + 9900 o = 100 mV cos (wt) 0.01 = 1 cos (wt) mV

SOL 4.22

nodia

Option (A) is correct. The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)

Input impedance Ri = RB || r p Voltage gain AV = gm RC Now, if CE is disconnected, resistance RE appears in the circuit

R in = RB || [rp + (b + 1)] RE Input impedance increases gm RC Voltage gain AV = 1 + gm R E Input impedance

SOL 4.23

Voltage gain decreases.

Option (B) is correct. Since, emitter area of transistor Q1 is half of transistor Q2 , so current IE = 1 IE and IB = 1 IB 2 2 1

2

1

2

The circuit is as shown below :



Analog Circuits

Collector current

Page 185

VB =- 10 - (- 0.7) =- 9.3 V 0 - (- 9.3) I1 = = 1 mA (9.3 kW)

b 1 = 700 (high), So IC . IE Applying KCL at base we have 1 - IE = IB + IB 1 - (b 1 + 1) IB = IB + IB 1

1

2

1

2

1

nodia 1 = (700 + 1 + 1)

IB + IB 2 2

2

IB . 2 702 2

I 0 = IC = b 2 : IB = 715 # 2 . 2 mA 702 Option (A) is correct. The circuit is as shown below : 2

SOL 4.24

So, or SOL 4.25

2

0 - Vi + 0 - Vo = 0 R1 R2 Vo =- R2 Vi R1

Option (B) is correct. By small signal equivalent circuit analysis

Input resistance seen by source vs R in = vs = Rs + Rs || rs = (1000 W) + (93 kW || 259 W) = 1258 W is



SOL 4.26

Analog Circuits

Chapter 4

Option (B) is correct. Cut-off frequency due to C2 1 2p (RC + RL) C2 1 = = 271 Hz 2 # 3.14 # 1250 # 4.7 # 10-6 f Lower cut-off frequency fL . o = 271 = 27.1 Hz 10 10 fo =

SOL 4.27

Option (B) is correct. The circuit is as shown below

nodia

I = 20 - 0 + Vi - 0 = 5 + Vi 4R R R If I > 0, diode D2 conducts So, for 5 + VI > 0 & VI > - 5, D2 conducts 2 Equivalent circuit is shown below Current

Output is Vo = 0 . If I < 0 , diode D2 will be off 5 + VI < 0 & V < - 5, D is off I 2 R The circuit is shown below

0 - Vi + 0 - 20 + 0 - Vo = 0 R 4R R or

Vo =- Vi - 5

At Vi =- 5 V,

Vo = 0 Vo = 5 V

At Vi =- 10 V,



Analog Circuits

Page 187

SOL 4.28

Option (A) is correct. Let diode be OFF. In this case 1 A current will flow in resistor and voltage across resistor will be V = 1.V Diode is off, it must be in reverse biased, therefore Vi - 1 > 0 " Vi > 1 Thus for Vi > 1 diode is off and V = 1V Option (B) and (C) doesn’t satisfy this condition. Let Vi < 1. In this case diode will be on and voltage across diode will be zero and V = Vi Thus V = min (Vi, 1)

SOL 4.29

Option (A) is correct. The R2 decide only the frequency.

SOL 4.30

Option (D) is correct. For small increase in VG beyond 1 V the n - channel MOSFET goes into saturation as VGS "+ ive and p - MOSFET is always in active region or triode region.

SOL 4.31


SOL 4.32

Option (D) is correct. The circuit is shown in fig below

nodia

The voltage at non inverting terminal is 5 V because OP AMP is ideal and inverting terminal is at 5 V. Thus IC = 10 - 5 = 1 mA 5k VE = IE RE = 1m # 1.4k = 1.4V = 0.6 + 1.4 = 2V Thus the feedback is negative and output voltage is V = 2V . SOL 4.33

IE = IC

Option (D) is correct. The output voltage is V0 = Ar Vi .-

hfe RC Vi hie

Here RC = 3 W and hie = 3 kW Thus

V0 . - 150 # 3k Vi 3k

.- 150 (A cos 20t + B sin 106 t) Since coupling capacitor is large so low frequency signal will be filtered out, and best approximation is



Analog Circuits

Chapter 4

V0 .- 150B sin 106 t SOL 4.34

Option (C) is correct. For the positive half of Vi , the diode D1 is forward bias, D2 is reverse bias and the zener diode is in breakdown state because Vi > 6.8 . Thus output voltage is V0 = 0.7 + 6.8 = 7.5 V For the negative half of Vi, D2 is forward bias thus V0 =- 0.7 V

Then SOL 4.35

Option (B) is correct. By Current mirror,

^ L h2 Ibias W ^ L h1 W

Ix =

Since MOSFETs are identical, W W Thus b L l =b L l 2 2 Hence SOL 4.36

nodia Ix = Ibias

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground.

Thus current will flow from -ive terminal (0 Volt) to -1 Volt source. Thus the current I is 0 - (- 1) I = = 1 100k 100k The current through diode is

I = I 0 _eV - 1i Now VT = 25 mV and I0 = 1 mA V

t

Thus or Now SOL 4.37

V I = 10-6 8e 25 # 10 - 1B = 1 5 10 V = 0.06 V V0 = I # 4k + V = 1 # 4k + 0.06 = 0.1 V 100k -3

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground. Thus we can write vi = -Rv R1 + sL sR C + 1 2

2

or

2

v0 =R2 vi (R1 + sL)( sR2 C2 + 1)



Analog Circuits

Page 189

and from this equation it may be easily seen that this is the standard form of T.F. of low pass filter K H (s) = (R1 + sL)( sR2 C2 + 1) and form this equation it may be easily seen that this is the standard form of T.F. of low pass filter H (s) = 2 K as + bs + b

nodia

SOL 4.38

Option (C ) is correct. The current in both transistor are equal. Thus gm is decide by M1.

SOL 4.39

Option (C) is correct. Let the voltage at non inverting terminal be V1, then after applying KCL at non inverting terminal side we have 15 - V1 + V0 - V1 = V1 - (- 15) 10 10 10 or V1 = V0 3 If V0 swings from -15 to +15 V then V1 swings between -5 V to +5 V.

SOL 4.40

Option (A) is correct. For the given DC values the Thevenin equivalent circuit is as follows

The Thevenin resistance and voltage are VTH = 10 # 9 = 3 V 10 + 20 and total RTH = 10k # 20k = 6.67 kW 10k + 20k Since b is very large, therefore IB is small and can be ignored Thus IE = VTH - VBE = 3 - 0.7 = 1 mA RE 2 . 3k SOL 4.41

Option (D) is correct. The small signal model is shown in fig below



Analog Circuits

gm =

IC = 1m = 1 A/V VT 25m 25

Vo =- gm Vp # (3k 3k ) =- 1 Vin (1.5k) 25

IC . IE

Vp = Vin

=- 60Vin Am = Vo =- 60 Vin

or SOL 4.42

Chapter 4

Option (C) is correct. The circuit shown in (C) is correct full wave rectifier circuit.

nodia

SOL 4.43

Option (A) is correct. In the transconductance amplifier it is desirable to have large input resistance and large output resistance.

SOL 4.44

Option (C) is correct. We redraw the circuit as shown in fig.

Applying voltage division rule We know that Thus Now and or SOL 4.45

v+ = 0.5 V v+ = vv- = 0.5 V i = 1 - 0.5 = 0.5 mA 1k - v0 = 0.5 mA 0 . 5 i = 2k v0 = 0.5 - 1 =- 0.5 V

Option (B) is correct. If we assume b very large, then IB = 0 and IE = IC ; VBE = 0.7 V. We assume that



Analog Circuits

Page 191

BJT is in active, so applying KVL in Base-emitter loop IE = 2 - VBE = 2 - 0.7 = 1.3 mA 1k RE Since b is very large, we have IE = IC , thus IC = 1.3 mA Now applying KVL in collector-emitter loop 10 - 10IC - VCE - IC = 0 VCE =- 4.3 V

or Now Since VBC SOL 4.46

VBC = VBE - VCE = 0.7 - (- 4.3) = 5 V > 0.7 V, thus transistor in saturation.

Option (D) is correct. Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e. IR = ID

or or

nodia Vi = I eV /V s R VD = VT 1n Vi Is R D

T

For the first condition

VD = 0 - Vo1 = VT 1n 2 Is R

For the first condition

VD = 0 - Vo1 = VT 1n 4 Is R

Subtracting above equation

or SOL 4.47

Vo1 - Vo2 = VT 1n 4 - VT 1n 2 Is R Is R Vo1 - Vo2 = VT 1n 4 = VT 1n2 2

Option (D) is correct. We have Vthp = Vthp = 1 V W W P and = N = 40mA/V2 LP LN From figure it may be easily seen that Vas for each NMOS and PMOS is 2.5 V mA Thus ID = K (Vas - VT ) 2 = 40 2 (2.5 - 1) 2 = 90 m A V

SOL 4.48

Option (C) is correct. We have VZ = 7 volt, VK = 0, RZ = 10W Circuit can be modeled as shown in fig below



Analog Circuits

Chapter 4

Since Vi is lies between 10 to 16 V, the range of voltage across 200 kW V200 = Vi - VZ = 3 to 9 volt The range of current through 200 kW is 3 = 15 mA to 9 = 45 mA 200k 200k The range of variation in output voltage 15m # RZ = 0.15 V to 45m # RZ = 0.45 Thus the range of output voltage is 7.15 Volt to 7.45 Volt SOL 4.49

Option (A) is correct. The voltage at non-inverting terminal is V+ =

1 sC

R+

1 sC

V- = V+ =

Now

Vi =

1 V 1 + sCR i

1 V 1 + sCR i

Applying voltage division rule (V + Vi) V+ = R1 (V0 + Vi) = o R1 + R1 2 (V + Vi) 1 or Vi = o 1 + sCR 2 V 2 o or =- 1 + 1 + sRC Vi V0 = 1 - sRC Vi 1 + sRC SOL 4.50

nodia


V0 = H (s) = 1 - sRC 1 + sRC Vi 1 - jwRC H (jw) = 1 + jwRC

+H (jw) = f =- tan - 1 wRC - tan - 1 wRC =- 2 tan - 2 wRC = - p (at w " 3)

Minimum value,

fmin

Maximum value,

fmax = 0( at w = 0)

SOL 4.51

Option (D) is correct. In the transconductance amplifier it is desirable to have large input impedance and large output impedance.

SOL 4.52


SOL 4.53

Option (D) is correct. The voltage at inverting terminal is V- = V+ = 10 V Here note that current through the capacitor is constant and that is I = V- = 10 = 10 mA 1k 1k Thus the voltage across capacitor at t = 1 msec is 1m 1m VC = 1 # Idt = 1 # 10mdt = 10 4 C 0 1m 0

Im

#0 dt = 10 V



SOL 4.54

Analog Circuits

Page 193

Option (A) is correct. In forward bias Zener diode works as normal diode. Thus for negative cycle of input Zener diode is forward biased and it conducts giving VR = Vin . For positive cycle of input Zener diode is reversed biased when 0 < Vin < 6 , Diode is OFF and VR = 0 when Vin > 6 Diode conducts and voltage across diode is 6 V. Thus voltage across is resistor is VR = Vin - 6 Only option (B) satisfy this condition.

SOL 4.55

Option (C) is correct. The circuit under DC condition is shown in fig below

nodia

Applying KVL we have

SOL 4.56

VCC - RC (IC + IB) - VCE = 0

...(1)

and VCC - RB IB - VBE = 0 Substituting IC = bIB in (1) we have

...(2)

VCC - RC (bIB + IB) - VCE = 0 Solving (2) and (3) we get VCE = VCC - VCC - VBE RB 1+ RC (1 + b) Now substituting values we get 12 - 0.7 VCE = 12 = 5.95 V 53 1+ 1 + (1 + 60)

...(3) ...(4)


b' = 110 # 60 = 66 100

Substituting b' = 66 with other values in (iv) in previous solutions 12 - 0.7 VCE = 12 = 5.29 V 53 1+ 1 + (1 + 66) Thus change is

= 5.29 - 59.5 # 100 =- 4.3% 5.95

SOL 4.57


SOL 4.58

Option (C) is correct. The Zener diode is in breakdown region, thus V+ = VZ = 6 V = Vin



Analog Circuits

Chapter 4

Rf R1 m = Vo = 6`1 + 12k j = 9 V 24k

Vo = Vin c1 +

We know that or

Vout

The current in 12 kW branch is negligible as comparison to 10 W. Thus Current IC . IE . = Vout = 9 = 0.9 A RL 10 VCE = 15 - 9 = 6 V The power dissipated in transistor is Now

P = VCE IC = 6 # 0.9 = 5.4 W SOL 4.59

Option (B) is correct. If the unregulated voltage increase by 20%, them the unregulated voltage is 18 V, but the VZ = Vin = 6 remain same and hence Vout and IC remain same. There will be change in VCE Thus, VCE - 18 - 9 = 9 V IC = 0.9 A Power dissipation P = VCE IC = 9 # 0.9 = 8.1 W Thus % increase in power is 8.1 - 5.4 # 100 = 50% 5.4

nodia

SOL 4.60

Option (B) is correct. Since the inverting terminal is at virtual ground, the current flowing through the voltage source is Is = Vs 10k Vs = 10 kW = R or in Is

SOL 4.61

Option (D) is correct. The effect of current shunt feedback in an amplifier is to decrease the input resistance and increase the output resistance as : Rif = Ri 1 + Ab where

Rof = R0 (1 + Ab) Ri " Input resistance without feedback Rif " Input resistance with feedback.

SOL 4.62

Option (B) is correct. The CE configuration has high voltage gain as well as high current gain. It performs basic function of amplifications. The CB configuration has lowest Ri and highest Ro . It is used as last step to match a very low impedance source and to drain a high impedance load Thus cascade amplifier is a multistage configuration of CE-CB

SOL 4.63

Option (D) is correct. Common mode gain ACM =- RC 2RE And differential mode gain



Analog Circuits

Page 195

ADM =- gm RC Thus only common mode gain depends on RE and for large value of RE it decreases. SOL 4.64


IE = Is è nV - 1j = 10 - 13 c VBE

T

SOL 4.65

0. 7 - 1m = 49 mA e1 # 26 # 10 -3

Option (C) is correct. The circuit is as shown below

nodia

Writing equation for I- have e 0 - V- = I 1M

or e0 = I- (1M) + VWriting equation for I+ we have 0 - V+ = I+ 1M

...(1)

or V+ = - I+ (1M) Since for ideal OPAMP V+ = V- , from (1) and (2) we have

...(2)

e0 = I- (1M) - I + (1M) = (I- - I+) (1M) = IOS (1M) Thus if e0 has been measured, we can calculate input offset current IOS only. SOL 4.66

SOL 4.67

Option (C) is correct. At low frequency capacitor is open circuit and voltage acr s non-inverting terminal is zero. At high frequency capacitor act as short circuit and all input voltage appear at non-inverting terminal. Thus, this is high pass circuit. The frequency is given by 1 = 1000 rad/sec w = 1 = 3 RC 1 # 10 # 1 # 10 - 6 Option (B) is correct. The circuit under DC condition is shown in fig below



Analog Circuits

Chapter 4

Applying KVL we have or or

VCC - RB IB - VBE - RE IE = 0 VCC - RB IB - VBE - RE (b + 1) IB = 0 IB = VCC - VBE RB + (b + 1) RE 20 - 0.7 = = 40m A 430k + (50 + 1)1 k

IC = bIB = 50 # 40m = 2 mA VC = VCC - RC IC = 20 - 2m # 2k = 16 V

Now SOL 4.68

Since IE = IB + bIB

Option (A) is correct. The maximum load current will be at maximum input voltage i.e.

or or

Vmax = 30 V i.e. Vmax - VZ = I + I L Z 1k 30 - 5.8 = I = 0.5 m L 1k

nodia IL = 24.2 - 0.5 = 23.7 mA

SOL 4.69


SOL 4.70

Option (B) is correct. The small signal model is as shown below

From the figure we have Zin = 2 MW and SOL 4.71

Z0 = rd RD = 20k 2k = 20 kW 11

Option (A) is correct. The circuit in DC condition is shown below

Since the FET has high input resistance, gate current can be neglect and we get VGS =- 2 V Since VP < VGS < 0 , FET is operating in active region



Analog Circuits

2 (- 2) 2 ID = IDSS c1 - VGS m = 10 c1 = 5.625 mA (- 8) m VP

Now

VDS = VDD - ID RD = 20 - 5.625 m # 2 k = 8.75 V

Now SOL 4.72

Option (B) is correct. The transconductance is gm =

VP

2 ID IDSS

= 2 5.625mA # 10mA = 1.875 mS 8

or, The gain is So, SOL 4.73

Page 197

A =- gm (rd RD) = 1.875ms # 20 K =- 3.41 11

Option (B) is correct. Only one diode will be in ON conditions When lower diode is in ON condition, then Vu = 2k Vsat = 2 10 = 8 V 2.5k 2. 5

nodia

when upper diode is in ON condition Vu = 2k Vsat = 2 (- 10) =- 5 V 2.5k 4 SOL 4.74

Option (B) is correct. An ideal OPAMP is an ideal voltage controlled voltage source.

SOL 4.75

Option (C) is correct. In voltage series feed back amplifier, input impedance increases by factor (1 + Ab) and output impedance decreases by the factor (1 + Ab). Rif = Ri (1 + Ab) Ro Rof = (1 + Ab)

SOL 4.76

Option (A) is correct. This is a Low pass filter, because V0 = 0 At w = 3 Vin V0 = 1 and at w = 0 Vin

SOL 4.77

Option (D) is correct. When IC >> ICO IC = 1mA = 0.04 = 40 mA/V VT 25mV b rp = = 100 - 3 = 2.5 kW gm 40 # 10

gm =

SOL 4.78

Option (A) is correct. The given circuit is wein bridge oscillator. The frequency of oscillation is 2pf = 1 RC 1 or = 1 m C = 1 = 2pRf 2p 2p # 103 # 103



SOL 4.79

Analog Circuits

Chapter 4

Option (A) is correct. The circuit is as shown below

We know that for ideal OPAMP V- = V+ Applying KCL at inverting terminal V- - Vs + V- - V0 = 0 R1 R1

nodia

or 2V- - Vo = Vs Applying KCL at non-inverting terminal V+ V - Vo =0 + IL + + R2 R2

...(1)

or 2V+ - Vo + IL R2 = 0 Since V- = V+ , from (1) and (2) we have

...(2)

Vs + IL R2 = 0

or SOL 4.80

IL =- Vs R2

Option (D) is correct. If IZ is negligible the load current is 12 - Vz = I L R as per given condition 100 mA # 12 - VZ # 500 mA R 12 5 At IL = 100 mA = 100 mA R or

VZ = 5 V

R = 70W

At IL = 500 mA 12 - 5 = 500 mA R

VZ = 5 V

R = 14 W Thus taking minimum we get or

R = 14 W SOL 4.81


SOL 4.82

Option (C) is correct. The Thevenin equivalent is shown below



Analog Circuits

VT =

Page 199

R1 V = 1 #5 = 1 V R1 + R2 C 4+1

Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE = 1 - 0.7 = 3 mA RE 300 VCE = 5 - 2.2kIC - 300IE = 5 - 2.2k # 1m - 300 # 1m = 2.5 V

Now SOL 4.83

Option (B) is correct. For the different combinations the table is as follows

nodia

CE

CE

CC

CB

Ai

High

High

Unity

Av

High

Unity

High

Ri

Medium

High

Low

Ro

Medium

Low

High

SOL 4.84

Option (D) is correct. This circuit having two diode and capacitor pair in parallel, works as voltage doubler.

SOL 4.85

Option (B) is correct. If the input is sinusoidal signal of 8 V (peak to peak) then Vi = 4 sin wt The output of comparator will be high when input is higher than Vref = 2 V and will be low when input is lower than Vref = 2 V. Thus the waveform for input is shown below

From fig, first crossover is at wt1 and second crossover is at wt2 where 4 sin wt1 = 2V



Analog Circuits

wt1 = sin - 1 1 = p 2 6 wt2 = p - p = 5p 6 6 5p p -6 Duty Cycle = 6 =1 2p 3

Thus

SOL 4.86

Chapter 4

Thus the output of comparators has a duty cycle of 1 . 3 Option (C) is correct. CMMR = Ad Ac or 20 log CMMR = 20 log Ad - 20 log Ac = 48 - 2 = 46 dB Where Ad "Differential Voltage Gain and AC " Common Mode Voltage Gain

SOL 4.87

Option (B) is correct. The gain of amplifier is Ai =

- gm gb + jwC

nodia

Thus the gain of a transistor amplifier falls at high frequencies due to the internal capacitance that are diffusion capacitance and transition capacitance. SOL 4.88

Option (A) is correct. We have Ri = 1kW, b = 0.2, A = 50 Ri Thus, Rif = = 1 kW (1 + Ab) 11

SOL 4.89

Option (A) is correct. The DC equivalent circuit is shown as below. This is fixed bias circuit operating in active region.

In first case or or

VCC - IC1 R2 - VCE1 = 0 6 - 1.5mR2 - 3 = 0

In second case IB2 Thus

R2 = 2kW I C1 = 1.5m = 0.01 mA IB1 = 150 b1 will we equal to IB1 as there is no in R1. IC2 = b2 IB2 = 200 # 0.01 = 2 mA VCE2 = VCC - IC2 R2 = 6 - 2m # 2 kW = 2 V

SOL 4.90

Option (A) is correct. The given circuit is a R - C phase shift oscillator and frequency of its oscillation is 1 f = 2p 6 RC



SOL 4.91

Analog Circuits

Option (C) is correct. If we see th figure we find that the voltage at non-inverting terminal is 3 V by the zener diode and voltage at inverting terminal will be 3 V. Thus Vo can be get by applying voltage division rule, i.e. 20 V = 3 20 + 40 o V0 = 9 V

or SOL 4.92

Page 201


nodia 8 (3) = 8 kW 1+8 3 8 V+ = V- = V 3 V+ =

Now applying KCL at inverting terminal we get V- - 2 + V- - Vo = 0 1 5 or

SOL 4.93

Vo = 6V- - 10 = 6 # 8 - 10 = 6 V 3

Option (C) is correct. The equivalent circuit of 3 cascade stage is as shown in fig.

1k 50V1 = 40V1 1k + 0.25k 1k 50V2 = 40V2 V3 = 1k + 0.25k V2 =

Similarly or or or SOL 4.94

V3 = 40 # 40V1 Vo = 50V3 = 50 # 40 # 40V1 AV = Vo = 50 # 40 # 40 = 8000 V1 20 log AV = 20 log 8000 = 98 dB

Option (D) is correct. If a constant current is made to flow in a capacitor, the output voltage is integration of input current and that is sawtooth waveform as below :



Analog Circuits

Chapter 4

t VC = 1 # idt C 0 The time period of wave form is T = 1 = 1 = 2 m sec f 500 20 # 10 1 idt 6 # 2 # 10 0 or i (2 # 10 - 3 - 0) = 6 # 10 - 6 or i = 3 mA Thus the charging require 3 mA current source for 2 msec. -3

Thus

SOL 4.95

3=

Option (C) is correct. In voltage-amplifier or voltage-series amplifier, the Ri increase and Ro decrease because Rif = Ri (1 + Ab) Ro Rof = (1 + Ab)

SOL 4.96

nodia

Option (B) is correct. Let x be the gain and it is 20 db, therefore

20 log x = 20 or x = 10 Since Gain band width product is 106 Hz, thus So, bandwidth is

6 6 BW = 10 = 10 = 105 Hz = 100 kHz 10 Gain

SOL 4.97

Option (A) is correct. In multistage amplifier bandwidth decrease and overall gain increase. From bandwidth point of view only options (A) may be correct because lower cutoff frequency must be increases and higher must be decreases. From following calculation we have We have fL = 20 Hz and fH = 1 kHz For n stage amplifier the lower cutoff frequency is fL 20 f = = = 39.2 . 40 Hz Ln

1

2n - 1

1

23 - 1

The higher cutoff frequency is fHn = fH SOL 4.98

1

2 2 - 1 = 0.5 kHz

Option (A) is correct. As per Barkhousen criterion for sustained oscillations Ab $ 1 and phase shift must be or 2pn . V (f) Now from circuit A= O = 1 + R2 Vf (f) R1 V ( f ) b (f) = 1 +0 = f 6 VO (f) Thus from above equation for sustained oscillation 6 = 1 + R2 R1 or

R2 = 5R1



SOL 4.99

Analog Circuits

Option (C) is correct. Let the gain of OPAMP be AV then we have 20 log AV = 40 dB or AV = 100 Let input be Vi = Vm sin wt then we have VO = VV Vi = Vm sin wt dVO = A V w cos wt V m dt

Now Slew Rate or or SOL 4.100

Page 203

dVO = AV Vm w = AV Vm 2pf c dt m max 1 = -6 Vm = SR AV V2pf 10 # 100 # 2p # 20 # 103 VM = 79.5 mV

Option (A) is correct. The circuit is shown as below

nodia

I For satisfactory operations Vin - V0 R When Vin = 30 V, 30 - 10 R 20 or R or when Vin = 50 V

> IZ + IL

[IZ + IL = I]

$ (10 + 1) mA $ 11 mA

R # 1818 W 50 - 10 $ (10 + 1) mA R 40 $ 11 # 10 - 3 R

or SOL 4.101

= IZ + IL

R # 3636W

Thus R # 1818W

Option (D) is correct. We have Now and Thus Now

So,

IDSS = 10 mA and VP =- 5 V VG = 0 VS = ID RS = 1 # 2.5W = 2.5 V VGS = VG - VS = 0 - 2.5 =- 2.5 V gm = 2IDSS 81 - ` - 2.5 jB = 2 mS VP -5 V AV = 0 =- gm RD Vi =- 2ms # 3k =- 6



SOL 4.102

Analog Circuits

Chapter 4

Option (C) is correct. The current gain of a BJT is hfe = gm rp

SOL 4.103

Option (A) is correct. The ideal op-amp has following characteristic : Ri " 3 and

R0 " 0 A"3

SOL 4.104

Option (C) is correct. Both statements are correct because (1) A stable multivibrator can be used for generating square wave, because of its characteristic (2) Bi-stable multivibrator can store binary information, and this multivibrator also give help in all digital kind of storing.

SOL 4.105

Option (B) is correct. If fT is the frequency at which the short circuit common emitter gain attains unity magnitude then gm 38 # 10 - 3 = fT = = 1.47 # 1010 Hz 2p (Cm + Cp) 2p # (10 - 14 + 4 # 10 - 13) If fB is bandwidth then we have 10 f fB = T = 1.47 # 10 = 1.64 # 108 Hz b 90

SOL 4.106

nodia

Option (C) is correct. If we neglect current through RB then it can be open circuit as shown in fig.

Maximum power will dissipate in Zener diode when current through it is maximum and it will occur at Vin = 30 V I = Vin - Vo = 30 - 10 = 1 A 20 20 I IC + IZ = bIB + IZ or

= bIZ + IZ = (b + 1) IZ IZ = I = 1 = 0.01 A b + 1 99 + 1

Since IC = bIB since IB = IZ

Power dissipated in zener diode is PZ = VZ IZ = 9.5 # 0.01 = 95 mW IC = bIZ = 99 # 0.1 = 0.99 A VCE = Vo = 10 V Power dissipated in transistor is PT = VC IC = 10 # 0.99 = 9.9 W



Analog Circuits

Page 205

SOL 4.107

Option (B) is correct. From the it may be easily seen that the tank circuit is having 2-capacitors and one-inductor, so it is colpits oscillator and frequency is 1 f = 2p LCeq Ceq = C1 C2 = 2 # 2 = 1 pF 4 C1 + C2 9 1 = 1 # 10 = 50.3 MHz f = -6 - 12 2p 10 2p 10 # 10 # 10

SOL 4.108

Option (D) is correct. The circuit is as shown below

nodia

Let V- be the voltage of inverting terminal, since non inverting terminal a at ground, the output voltage is Vo = AOL VNow applying KCL at inverting terminal we have V- - Vs + V- - V0 = 0 R1 R2

SOL 4.109

...(1) ...(2)

From (1) and (2) we have VO = A = - R2 CL Vs R - R2 + R1 ROL Substituting the values we have - 10k ACL = =- 1000 . - 11 89 10 k 1 k + 1k 100k Option (A) is correct. The first OPAMP stage is the differentiator and second OPAMP stage is integrator. Thus if input is cosine term, output will be also cosine term. Only option (A) is cosine term. Other are sine term. However we can calculate as follows. The circuit is shown in fig

Applying KCL at inverting terminal of first OP AMP we have V1 = - wjL = - 100 # 10 # 10 - 3 = - 1 10 R 10 VS - jVS or V1 = = j cos 100t 10



Analog Circuits

Chapter 4

Applying KCL at inverting terminal of second OP AMP we have VO = - 1/jwC V1 100 1 == j10 j100 # 10 # 10 - 6 # 100 or V0 = j10V2 = j10 (- j cos 100t) V0 = 10 cos 100t SOL 4.110

Option (C) is correct. With the addition of RE the DC abis currents and voltages remain closer to the point where they were set by the circuit when the outside condition such as temperature and transistor parameter b change.

SOL 4.111

Option (A) is correct. Common mode gain is AC = aRC REE Since source resistance of the current source is infinite REE = 3 , common mode gain AC = 0

nodia

SOL 4.112

Option (D) is correct. In positive feed back it is working as OP-AMP in saturation region, and the input applied voltage is +ve. So, V0 =+ Vsat = 15 V

SOL 4.113

Option (A) is correct. At high frequency

gm ' gbc + jw (C) 1 Ai \ Capacitance 1 Ai a frequency Ai =-

or, and

Thus due to the transistor capacitance current gain of a bipolar transistor drops. SOL 4.114

Option (C) is correct. As OP-AMP is ideal, the inverting terminal at virtual ground due to ground at non-inverting terminal. Applying KCL at inverting terminal sC (v1 sin wt - 0) + sC (V2 sin wt - 0) + sC (Vo - 0) = 0 or Vo =- (V1 + V2) sin wt

SOL 4.115

Option (D) is correct. There is R - C , series connection in parallel with parallel R - C combination. So, it is a wein bridge oscillator because two resistors R1 and R2 is also in parallel with them.

SOL 4.116

Option (A) is correct. The given circuit is a differentiator, so the output of triangular wave will be square wave.

SOL 4.117

Option (B) is correct. In sampling and hold circuit the unity gain non-inverting amplifier is used.



SOL 4.118

Analog Circuits

Page 207

Option (D) is correct. The Thevenin equivalent is shown below

R1 V = 5 # 15 = 5 V 10 + 5 R1 + R2 C Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE RE 4.3 = 5 - 0.7 = = 10 mA 0.430kW 0.430KW VT =

SOL 4.119

nodia

Option (C) is correct. The output voltage will be input offset voltage multiplied by open by open loop gain. Thus So V0 = 5mV # 10, 000 = 50 V But V0 = ! 15 V in saturation condition So, it can never be exceeds !15 V So,

V0 = ! Vset = ! 15V

SOL 4.120


SOL 4.121

Option (A) is correct. Negative feedback in amplifier reduces the gain of the system.

SOL 4.122

Option (A) is correct. By drawing small signal equivalent circuit

by applying KCL at E2 gm1 Vp 1

Vp = gm2 Vp rp 2

2

2

at C2 from eq (1) and (2) gm1 Vp + 1

i 0 =- gm2 Vp

2

i 0 =- i 0 gm2 rp 2

gm1 Vp =- i 0 :1 + 1

1 gm2 rp D 2



Analog Circuits

Chapter 4

gm2 rp = b >> 1 so gm1 Vp =- i 0 i 0 =- g m1 Vp i0 = g a Vp = Vi m1 Vi Option (B) is correct. Crossover behavior is characteristic of calss B output stage. Here 2 transistor are operated one for amplifying +ve going portion and other for -ve going portion. 2

1

1

1

SOL 4.123

SOL 4.124

Option (C) is correct. In Voltage series feedback mode input impedance is given by R in = Ri (1 + bv Av) bv = feedback factor , Av = openloop gain

where and

Ri = Input impedance

nodia

So, R in = 1 # 103 (1 + 0.99 # 100) = 100 kW Similarly output impedance is given by R0 ROUT = R 0 = output impedance (1 + bv Av) 100 Thus ROUT = = 1W (1 + 0.99 # 100) SOL 4.125


Regulation = Vno - load - Vfuel - load Vfull - load 30 25 100 = 20% = 25 # Output resistance = 25 = 25 W 1

SOL 4.126

Option (D) is correct. This is a voltage shunt feedback as the feedback samples a portion of output voltage and convert it to current (shunt).

SOL 4.127

Option (A) is correct. In a differential amplifier CMRR is given by (1 + b) IQ R 0 CMRR = 1 ;1 + E 2 VT b So where R 0 is the emitter resistance. So CMRR can be improved by increasing emitter resistance.

SOL 4.128

Option (C) is correct. We know that rise time (tr ) is tr = 0.35 fH where fH is upper 3 dB frequency. Thus we can obtain upper 3 dB frequency it rise time is known.

SOL 4.129

Option (D) is correct. In a BJT differential amplifier for a linear response Vid < VT .



SOL 4.130

Analog Circuits

Page 209

Option (D) is correct. In a shunt negative feedback amplifier. Input impedance Ri R in = (1 + bA) where

Ri = input impedance of basic amplifier b = feedback factor A = open loop gain

So, R in < Ri Similarly ROUT =

R0 (1 + bA)

ROUT < R 0 Thus input & output impedances decreases. SOL 4.131


SOL 4.132

Option (D) is correct. Comparator will give an output either equal to + Vsupply or - Vsupply . So output is a square wave.

SOL 4.133

Option (C) is correct. In series voltage regulator the pass transistor is in common collector

nodia

configuration having voltage gain close to unity. SOL 4.134

Option (D) is correct. In bridge rectifier we do not need central tap transformer, so its less expensive and smaller in size and its PIV (Peak inverse voltage) is also greater than the two diode circuit, so it is also suitable for higher voltage application.

SOL 4.135

Option (C) is correct. In the circuit we have V2 = IS # RD 2 and

V1 = IS # RD V2 = 1 2 V1 V1 = 2V2

SOL 4.136


SOL 4.137

Option (C) is correct. The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)

Input impedance Voltage gain

Ri = RB || r p AV = gm RC



Analog Circuits

Chapter 4

Now, if CE is disconnected, resistance RE appears in the circuit

Input impedance R in = RB || [rp + (b + 1)] RE Input impedance increases gm RC Voltage gain AV = 1 + gm R E

Voltage gain decreases.

SOL 4.138

Option (A) is correct. In common emitter stage input impedance is high, so in cascaded amplifier common emitter stage is followed by common base stage.

SOL 4.139

Option (C) is correct. We know that collect-emitter break down voltage is less than compare to collector base breakdown voltage.

nodia

BVCEO < BVCBO both avalanche and zener break down. Voltage are higher than BVCEO .So BVCEO limits the power supply. SOL 4.140


If we assume consider the diode in reverse bias then Vn should be greater than VP . VP < Vn by calculating VP = 10 # 4 = 5 Volt 4+4 Vn = 2 # 1 = 2 Volt here VP > Vn (so diode cannot be in reverse bias mode).

apply node equation at node a Va - 10 + Va + Va = 2 1 4 4 6Va - 10 = 8 Va = 3 Volt



Analog Circuits

Ib = 0 - 3 + 10 - 3 4 4 Ib = 10 - 6 = 1 amp 4

so current

SOL 4.141

Page 211

Option (D) is correct. Applying node equation at terminal (2) and (3) of OP -amp

Va - Q Va - V0 + =0 5 10

nodia 2Va - 4 + Va - V0 = 0

V0 = 3Va - 4 Va - V0 + Va - 0 = 0 100 10

So

Va - V0 + 10Va = 0 11Va = V0 Va = V0 11 V0 = 3V0 - 4 11 8V0 =- 4 11

V0 =- 5.5 Volts SOL 4.142

Option (B) is correct. Circuit with diode forward resistance looks

So the DC current will IDC = SOL 4.143

Vm p (R f + RL)

Option (D) is correct. For the positive half cycle of input diode D1 will conduct & D2 will be off. In negative half cycle of input D1 will be off & D2 conduct so output voltage wave from across resistor (10 kW) is –



Analog Circuits

Chapter 4

Ammeter will read rms value of current so I rms = Vm (half wave rectifier) pR 4 = 0.4 mA = p (10 kW) p SOL 4.144

Option (D) is correct. In given circuit positive feedback is applied in the op-amp., so it works as a Schmitt trigger.

SOL 4.145

Option (D) is correct. Gain with out feedback factor is given by V0 = kVi after connecting feedback impedance Z

nodia

given input impedance is very large, so after connecting Z we have Ii = Vi - V0 Z V Ii = i kVi Z input impedance Zin = Vi = Z Ii (1 - k)

V0 = kVi

SOL 4.146


SOL 4.147

Option (A) is correct. For the circuit, In balanced condition It will oscillated at a frequency 1 w= 1 = = 105 rad/ sec LC 10 # 10-3 # .01 # 10-6 In this condition R1 = R 3 R2 R4 5 =R 100 1 R = 20 kW = 2 # 10 4 W

SOL 4.148

Option (C) is correct. V0 kept constant at so current in 50 W resistor

V0 = 6 volt I = 9-6 50 W I = 60 m amp



Analog Circuits

Page 213

Maximum allowed power dissipation in zener PZ = 300 mW Maximum current allowed in zener PZ = VZ (IZ ) max = 300 # 10-3 & = 6 (IZ ) max = 300 # 10-3 & = (IZ ) max = 50 m amp Given knee current or minimum current in zener In given circuit

(IZ ) min = 5 m amp I = IZ + I L I L = I - IZ (IL) min = I - (IZ ) max = (60 - 50) m amp = 10 m amp (IL) max = I - (IZ ) min = (60 - 5) = 55 m amp

nodia ***********


CHAPTER 5 DIGITAL CIRCUITS

2013

ONE MARK

MCQ 5.1

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) and AND gate (B) an OR gate (C) an XOR gate (D) a NAND gate

MCQ 5.2

For 8085 microprocessor, the following program is executed. MVI A, 05H; MVI B, 05H; PTR: ADD B; DCR B; JNZ PTR; ADI 03H; HLT; At the end of program, accumulator contains (A) 17H (B) 20H (C) 23H (D) 05H

nodia

2013 MCQ 5.3

TWO MARKS

There are four chips each of 1024 bytes connected to a 16 bit address bus as shown in the figure below, RAMs 1, 2, 3 and 4 respectively are mappped to addresses


Digital Circuits

Page 215

(A) 0C00H-0FFFH, 1C00H-1FFFH, 2C00H-2FFFH, 3C00H-3FFFH (B) 1800H-1FFFH, 2800H-2FFFH, 3800H-3FFFH, 4800H-4FFFH (C) 0500H-08FFH, 1500H-18FFH, 3500H-38FFH, 5500H-58FFH (D) 0800H-0BFFH, 1800H-1BFFH, 2800H-2BFFH, 3800H-3BFFH 2012 MCQ 5.4

ONE MARK

Consider the given circuit

In this circuit, the race around (A) does not occur (B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0 MCQ 5.5

(C) 8 MCQ 5.6

nodia

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is (A) 4 (B) 6

In the circuit shown

(A) Y = A B + C (C) Y = (A + B ) C MCQ 5.7

(D) 10

(B) Y = (A + B) C (D) Y = AB + C

In the sum of products function f (X, Y, Z) = / (2, 3, 4, 5), the prime implicants are (A) XY, XY (B) XY, X Y Z , XY Z (C) XY Z , XYZ, XY (D) XY Z , XYZ, XY Z , XY Z



Digital Circuits

2012 MCQ 5.8

TWO MARKS

The state transition diagram for the logic circuit shown is

2011 MCQ 5.9

Chapter 5

nodia

ONE MARK

The output Y in the circuit below is always ‘1’ when

(A) two or more of the inputs P, Q, R are ‘0’ (B) two or more of the inputs P, Q, R are ‘1’ (C) any odd number of the inputs P, Q, R is ‘0’ (D) any odd number of the inputs P, Q, R is ‘1’ MCQ 5.10

When the output Y in the circuit below is “1”, it implies that data has

(A) changed from “0” to “1” (B) changed from “1” to “0” (C) changed in either direction (D) not changed



MCQ 5.11

Digital Circuits

Page 217

The logic function implemented by the circuit below is (ground implies a logic “0”)

(A) F = AND ^P, Q h (C) F = XNOR ^P, Q h

(B) F = OR ^P, Q h (D) F = XOR ^P, Q h

2011

TWO MARKS

nodia

MCQ 5.12

The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is

MCQ 5.13

Two D flip-flops are connected as a synchronous counter that goes through the following QB QA sequence 00 " 11 " 01 " 10 " 00 " .... The connections to the inputs DA and DB are (A) DA = QB, DB = QA (B) DA = Q A, DB = Q B (C) DA = (QA Q B + Q A QB), DB = QA (D) DA = (QA QB + Q A Q B), DB = Q B



MCQ 5.14

Digital Circuits

Chapter 5

An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is

(A) 8CH (C) 23H

(B) 64H (D) 15H

2010 MCQ 5.15

ONE MARK

Match the logic gates in Column A with their equivalents in Column B

nodia

(A) P-2, Q-4, R-1, S-3 (C) P-2, Q-4, R-3, S-1 MCQ 5.16

(B) P-4, Q-2, R-1, S-3 (D) P-4, Q-2, R-3, S-1

In the circuit shown, the device connected Y5 can have address in the range

(A) 2000 - 20FF (C) 2E00 - 2EFF

(B) 2D00 - 2DFF (D) FD00 - FDFF



MCQ 5.17

Digital Circuits

Page 219

For the output F to be 1 in the logic circuit shown, the input combination should be

(A) A = 1, B = 1, C = 0 (C) A = 0, B = 1, C = 0

(B) A = 1, B = 0, C = 0 (D) A = 0, B = 0, C = 1

2010 MCQ 5.18

TWO MARKS

Assuming that the flip-flop are in reset condition initially, the count sequence observed at QA , in the circuit shown is

nodia

(A) 0010111... (C) 0101111... MCQ 5.19

The Boolean function realized by the logic circuit shown is

(A) F = Sm (0, 1, 3, 5, 9, 10, 14) (C) F = Sm (1, 2, 4, 5, 11, 14, 15) MCQ 5.20

(B) 0001011... (D) 0110100....

(B) F = Sm (2, 3, 5, 7, 8, 12, 13) (D) F = Sm (2, 3, 5, 7, 8, 9, 12)

For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

(A) 00H (C) 67H

(B) 45H (D) E7H



Digital Circuits

Chapter 5

2009

ONE MARK

MCQ 5.21

The full form of the abbreviations TTL and CMOS in reference to logic families are (A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor (D) Tristate Transistor Logic and Complementary Metal Oxide Silicon

MCQ 5.22

In a microprocessor, the service routine for a certain interrupt starts from a fixed location of memory which cannot be externally set, but the interrupt can be delayed or rejected Such an interrupt is (A) non-maskable and non-vectored (B) maskable and non-vectored (C) non-maskable and vectored (D) maskable and vectored 2009

MCQ 5.23

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TWO MARKS

If X = 1 in logic equation 6X + Z {Y + (Z + XY )}@{X + X (X + Y)} = 1, then (B) Y = Z (A) Y = Z (C) Z = 1 (D) Z = 0

MCQ 5.24

What are the minimum number of 2- to -1 multiplexers required to generate a 2- input AND gate and a 2- input Ex-OR gate (A) 1 and 2 (B) 1 and 3 (C) 1 and 1 (D) 2 and 2

MCQ 5.25

What are the counting states (Q1, Q2) for the counter shown in the figure below

(A) 11, 10, 00, 11, 10,... (C) 00, 11, 01, 10, 00...

(B) 01, 10, 11, 00, 01... (D) 01, 10, 00, 01, 10...

Statement for Linked Answer Question 26 & 27 : Two products are sold from a vending machine, which has two push buttons P1 and P2 . When a buttons is pressed, the price of the corresponding product is displayed in a 7 - segment display. If no buttons are pressed, '0' is displayed signifying ‘Rs 0’. If only P1 is pressed, ‘2’ is displayed, signifying ‘Rs. 2’ If only P2 is pressed ‘5’ is displayed, signifying ‘Rs. 5’



Digital Circuits

Page 221

If both P1 and P2 are pressed, 'E' is displayed, signifying ‘Error’ The names of the segments in the 7 - segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’ are shown below.

Consider (1) push buttons pressed/not pressed in equivalent to logic 1/0 respectively. (2) a segment glowing/not glowing in the display is equivalent to logic 1/0 respectively. MCQ 5.26

If segments a to g are considered as functions of P1 and P2 , then which of the following is correct (A) g = P 1 + P2, d = c + e (B) g = P1 + P2, d = c + e (C) g = P1 + P2, e = b + c (D) g = P1 + P2, e = b + c

MCQ 5.27

What are the minimum numbers of NOT gates and 2 - input OR gates required to design the logic of the driver for this 7 - Segment display (A) 3 NOT and 4 OR (B) 2 NOT and 4 OR

nodia

(C) 1 NOT and 3 OR MCQ 5.28

(D) 2 NOT and 3 OR

Refer to the NAND and NOR latches shown in the figure. The inputs (P1, P2) for both latches are first made (0, 1) and then, after a few seconds, made (1, 1). The corresponding stable outputs (Q1, Q2) are

(A) NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0) (B) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (1, 0) (C) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (0, 0) (D) NAND : first (1, 0) then (1, 1) NOR : first (0, 1) then (0, 1) 2008 MCQ 5.29

TWO MARKS

The logic function implemented by the following circuit at the terminal OUT is

(A) P NOR Q (C) P OR Q

(B) P NAND Q (D) P AND Q



Digital Circuits

Chapter 5

MCQ 5.30

The two numbers represented in signed 2’s complement form are P + 11101101 and Q = 11100110 . If Q is subtracted from P , the value obtained in signed 2’s complement is (B) 00000111 (A) 1000001111 (C) 11111001 (D) 111111001

MCQ 5.31

Which of the following Boolean Expressions correctly represents the relation between P, Q, R and M1

(A) M1 = (P OR Q) XOR R (B) M1 = (P AND Q) X OR R (C) M1 = (P NOR Q) X OR R (D) M1 = (P XOR Q) XOR R MCQ 5.32

nodia

For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible Which of the following statements is true

(A) Q (B) Q (C) Q (D) Q MCQ 5.33

goes to 1 at the CLK transition and stays at 1 goes to 0 at the CLK transition and stays 0 goes to 1 at the CLK tradition and goes to 0 when D goes to 1 goes to 0 at the CLK transition and goes to 1 when D goes to 1

For each of the positive edge-triggered J - K flip flop used in the following figure, the propagation delay is 3 t .

Which of the following wave forms correctly represents the output at Q1 ?



Digital Circuits

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Statement For Linked Answer Question 34 & 35 : In the following circuit, the comparators output is logic “1” if V1 > V2 and is logic "0" otherwise. The D/A conversion is done as per the relation

nodia 3

VDAC = / 2n - 1 bn Volts, where b3 (MSB), b1, b2 and b0 (LSB) are the counter 0 outputs.n =The counter starts from the clear state.

MCQ 5.34

The stable reading of the LED displays is (A) 06 (B) 07 (C) 12 (D) 13

MCQ 5.35

The magnitude of the error between VDAC and Vin at steady state in volts is (A) 0.2 (B) 0.3 (C) 0.5 (D) 1.0

MCQ 5.36

For the circuit shown in the following, I0 - I3 are inputs to the 4:1 multiplexers, R(MSB) and S are control bits. The output Z can be represented by



Digital Circuits

(A) (B) (C) (D) MCQ 5.37

Chapter 5

PQ + PQS + QRS PQ + PQR + PQS PQR + PQR + PARS + QRS PQR + PQRS + PQRS + QRS

An 8085 executes the following instructions 2710 LXI H, 30A0 H 2713 DAD H 2414 PCHL All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct ? PC = 2715H PC = 30A0H (A) (B) HL = 30A0H HL = 2715H (C)

PC = 6140H HL = 6140H

2007

(D)

PC = 6140H HL = 2715H

nodia

ONE MARK

MCQ 5.38

X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is (A) 100111 (B) 0010000 (C) 000111 (D) 101001

MCQ 5.39

The Boolean function Y = AB + CD is to be realized using only 2 - input NAND gates. The minimum number of gates required is (A) 2 (B) 3 (C) 4 2007

MCQ 5.40

TWO MARKS

In the following circuit, X is given by

(A) X = ABC + ABC + ABC + ABC (C) X = AB + BC + AC MCQ 5.41

(D) 5

(B) X = ABC + ABC + ABC + ABC (D) X = AB + BC + AC

The Boolean expression Y = ABC D + ABCD + ABC D + ABC D can minimized to (A) Y = ABC D + ABC + AC D (B) Y = ABC D + BCD + ABC D (C) Y = ABCD + BC D + ABC D (D) Y = ABCD + BC D + ABC D

be



MCQ 5.42

Digital Circuits

Page 225

The circuit diagram of a standard TTL NOT gate is shown in the figure. Vi = 25 V, the modes of operation of the transistors will be

(A) Q1: revere active; Q2: normal active; Q3: saturation; Q4: cut-off (B) Q1: revere active; Q2: saturation; Q3: saturation; Q4: cut-off (C) Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation (D) Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active MCQ 5.43

nodia

The following binary values were applied to the X and Y inputs of NAND latch shown in the figure in the sequence indicated below : X = 0,Y = 1; X = 0, Y = 0; X = 1; Y = 1 The corresponding stable P, Q output will be.

(A) P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 (B) P = 1, Q = 0; P = 0, Q = 1; or P = 0, Q = 1; P = 0, Q = 1 (C) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 (D) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1 MCQ 5.44

An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped I/O as show in the figure. The address lines A0 and A1 of the 8085 are used by the 8255 chip to decode internally its thee ports and the Control register. The address lines A3 to A7 as well as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is

(A) F8H - FBH

(B) F8GH - FCH

(C) F8H - FFH

(D) F0H - F7H



Digital Circuits

Chapter 5

Statement for Linked Answer Question 45 and 46 : In the Digital-to-Analog converter circuit shown in the figure below, VR = 10V and R = 10kW

MCQ 5.45

MCQ 5.46

The current is (A) 31.25mA (C) 125mA The voltage V0 is (A) - 0.781 V

(B) 62.5mA (D) 250mA

nodia

(C) - 3.125 V

(B) - 1.562 V (D) - 6.250 V

Statement for Linked Answer Questions 47 & 48 : An 8085 assembly language program is given below. Line 1: MVI A, B5H 2: MVI B, OEH 3: XRI 69H 4: ADD B 5: ANI 9BH 6: CPI 9FH 7: STA 3010H 8: HLT MCQ 5.47

The contents of the accumulator just execution of the ADD instruction in line 4 will be (A) C3H (B) EAH (C) DCH (D) 69H

MCQ 5.48

After execution of line 7 of the program, the status of the CY and Z flags will be (A) CY = 0, Z = 0 (B) CY = 0, Z = 1 (C) CY = 1, Z = 0 (D) CY = 1, Z = 1

MCQ 5.49

For the circuit shown, the counter state (Q1 Q0) follows the sequence



Digital Circuits

(A) 00, 01, 10, 11, 00 (C) 00, 01, 11, 00, 01

Page 227

(B) 00, 01, 10, 00, 01 (D) 00, 10, 11, 00, 10

2006 MCQ 5.50

ONE MARK

The number of product terms in the minimized sum-of-product expression obtained through the following K - map is (where, "d" denotes don’t care states)

(A) 2 (C) 4 2006 MCQ 5.51

(B) 3 (D) 5

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An I/O peripheral device shown in Fig. (b) below is to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H - D7 H, its chip-select (CS ) should be connected to the output of the decoder shown in as below :

(A) output 7 (C) output 2 MCQ 5.52

TWO MARKS

(B) output 5 (D) output 0

For the circuit shown in figures below, two 4 - bit parallel - in serial - out shift registers loaded with the data shown are used to feed the data to a full adder. Initially, all the flip - flops are in clear state. After applying two clock pulse, the output of the full-adder should be

(A) S = 0, C0 = 0 (C) S = 1, C0 = 0

(B) S = 0, C0 = 1 (D) S = 1, C0 = 1



Digital Circuits

Chapter 5

MCQ 5.53

A new Binary Coded Pentary (BCP) number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 10001001101 corresponds of the following number is base-5 system (A) 423 (B) 1324 (C) 2201 (D) 4231

MCQ 5.54

A 4 - bit DAC is connected to a free - running 3 - big UP counter, as shown in the following figure. Which of the following waveforms will be observed at V0 ?

nodia

In the figure shown above, the ground has been shown by the symbol 4

MCQ 5.55

Following is the segment of a 8085 assembly language program LXI SP, EFFF H CALL 3000 H : : : 3000 H LXI H, 3CF4 PUSH PSW SPHL POP PSW RET On completion of RET execution, the contents of SP is (A) 3CF0 H (B) 3CF8 H (C) EFFD H (D) EFFF H

MCQ 5.56

Two D - flip - flops, as shown below, are to be connected as a synchronous counter that goes through the sequence 00 " 01 " 11 " 10 " 00 " ... The inputs D0 and D1 respectively should be connected as,



Digital Circuits

(A) Q 1 and Q0 (C) Q1 Q0 and Q 1 Q0 MCQ 5.57

Page 229

(B) Q 0 and Q1 (D) Q 1 Q 0 and Q1 Q0

The point P in the following figure is stuck at 1. The output f will be

(A) ABC (C) ABC

(B) A (D) A

2005

ONE MARK

MCQ 5.58

Decimal 43 in Hexadecimal and BCD number system is respectively (A) B2, 0100 011 (B) 2B, 0100 0011 (C) 2B, 0011 0100 (D) B2, 0100 0100

MCQ 5.59

The Boolean function f implemented in the figure using two input multiplexes is

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(A) ABC + ABC (C) ABC + ABC

(B) ABC + ABC (D) ABC + ABC

2005 MCQ 5.60

TWO MARKS

The transistors used in a portion of the TTL gate show in the figure have b = 100 . The base emitter voltage of is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1 A and the output is at logic 0, then the current IR will be equal to

(A) 0.65 mA (C) 0.75 mA

(B) 0.70 mA (D) 1.00 mA



MCQ 5.61

Digital Circuits

Chapter 5

The Boolean expression for the truth table shown is

(A) B (A + C)( A + C ) (C) B (A + C )( A + C)

(B) B (A + C )( A + C) (D) B (A + C)( A + C )

MCQ 5.62

The present output Qn of an edge triggered JK flip-flop is logic 0. If J = 1, then Qn + 1 (A) Cannot be determined (B) Will be logic 0 (C) will be logic 1 (D) will rave around

MCQ 5.63

The given figure shows a ripple counter using positive edge triggered flip-flops. If the present state of the counter is Q2 Q1 Q0 = 001 then is next state Q2 Q1 Q will be

nodia

(A) 010 (C) 100 MCQ 5.64

(B) 111 (D) 101

What memory address range is NOT represents by chip # 1 and chip # 2 in the figure A0 to A15 in this figure are the address lines and CS means chip select.

(A) 0100 - 02FF (C) F900 - FAFF

(B) 1500 - 16FF (D) F800 - F9FF



Digital Circuits

Page 231

Statement For Linked Answer Questions 65 & 66 : Consider an 8085 microprocessor system. MCQ 5.65

The following program starts at location 0100H. LXI SP, OOFF LXI H, 0701 MVI A, 20H SUB M The content of accumulator when the program counter reaches 0109 H is (A) 20 H (B) 02 H (C) 00 H (D) FF H

MCQ 5.66

If in addition following code exists from 019H onwards, ORI 40 H ADD M What will be the result in the accumulator after the last instruction is executed ? (A) 40 H (B) 20 H (C) 60 H (D) 42 H 2004

MCQ 5.67

nodia

ONE MARK

A master - slave flip flop has the characteristic that (A) change in the output immediately reflected in the output (B) change in the output occurs when the state of the master is affected (C) change in the output occurs when the state of the slave is affected (D) both the master and the slave states are affected at the same time

MCQ 5.68

The range of signed decimal numbers that can be represented by 6-bits 1’s complement number is (A) -31 to +31 (B) -63 to +63 (C) -64 to +63

(D) -32 to +31

MCQ 5.69

A digital system is required to amplify a binary-encoded audio signal. The user should be able to control the gain of the amplifier from minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is (A) 8 (B) 6 (C) 5 (D) 7

MCQ 5.70

Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 most appropriate item in Group 2. Group 1 Group 2 P. Shift register 1. Frequency division Q. Counter 2. Addressing in memory chips R. Decoder 3. Serial to parallel data conversion (A) P - 3, Q - 2, R - 1 (B) P - 3, Q - 1, R - 2 (C) P - 2, Q - 1, R - 3 (D) P - 1, Q - 2, R - 2

MCQ 5.71

The figure the internal schematic of a TTL AND-OR-OR-Invert (AOI) gate. For



Digital Circuits

Chapter 5

the inputs shown in the figure, the output Y is

(A) 0 (C) AB

(B) 1 (D) AB

2004 MCQ 5.72

TWO MARKS

11001, 1001, 111001 correspond to the 2’s complement representation of which one of the following sets of number (A) 25,9, and 57 respectively (B) -6, -6, and -6 respectively (C) -7, -7 and -7 respectively

MCQ 5.73

(D) -25, -9 and -57 respectively

In the modulo-6 ripple counter shown in figure, the output of the 2- input gate is used to clear the J-K flip-flop The 2-input gate is

nodia

(A) a NAND gate (C) an OR gate

(B) a NOR gate (D) a AND gare

MCQ 5.74

The minimum number of 2- to -1 multiplexers required to realize a 4- to -1 multiplexers is (A) 1 (B) 2 (C) 3 (D) 4

MCQ 5.75

The Boolean expression AC + BC is equivalent to (A) AC + BC + AC (B) BC + AC + BC + ACB (C) AC + BC + BC + ABC (D) ABC + ABC + ABC + ABC

MCQ 5.76

A Boolean function f of two variables x and y is defined as follows : f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0 Assuming complements of x and y are not available, a minimum cost solution for realizing f using only 2-input NOR gates and 2- input OR gates (each having unit cost) would have a total cost of (A) 1 unit (B) 4 unit (C) 3 unit (D) 2 unit

MCQ 5.77

The 8255 Programmable Peripheral Interface is used as described below. (i) An A/D converter is interface to a microprocessor through an 8255. The conversion is initiated by a signal from the 8255 on Port C. A signal on



Digital Circuits

Page 233

Port C causes data to be stobed into Port A. (ii) Two computers exchange data using a pair of 8255s. Port A works as a bidirectional data port supported by appropriate handshaking signals. The appropriate modes of operation of the 8255 for (i) and (ii) would be (A) Mode 0 for (i) and Mode 1 for (ii) (B) Mode 1 for (i) and Mode 2 for (ii) (C) Mode for (i) and Mode 0 for (ii) (D) Mode 2 for (i) and Mode 1 for (ii) MCQ 5.78

The number of memory cycles required to execute the following 8085 instructions (i) LDA 3000 H (ii) LXI D, FOF1H would be (A) 2 for (i) and 2 for (ii) (B) 4 for (i) and 3 for (ii) (C) 3 for (i) and 3 for (ii) (D) 3 for (i) and 4 for (ii)

MCQ 5.79

Consider the sequence of 8085 instructions given below LXI H, 9258 MOV A, M CMA MOV M, A Which one of the following is performed by this sequence ? (A) Contents of location 9258 are moved to the accumulator

nodia

(B) Contents of location 9258 are compared with the contents of the accumulator (C) Contents of location 8529 are complemented and stored in location 8529 (D) Contents of location 5892 are complemented and stored in location 5892 MCQ 5.80

It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below : MVI A, 00H LOOP --------------HLT END The sequence of instructions to complete the program would be (A) JNX LOOP, ADD B, DCR C (B) ADD B, JNZ LOOP, DCR C (C) DCR C, JNZ LOOP, ADD B (D) ADD B, DCR C, JNZ LOOP

2003 MCQ 5.81

ONE MARK

The number of distinct Boolean expressions of 4 variables is (A) 16 (B) 256 (C) 1023 (D) 65536



Digital Circuits

Chapter 5

MCQ 5.82

The minimum number of comparators required to build an 8-bits flash ADC is (A) 8 (B) 63 (C) 255 (D) 256

MCQ 5.83

The output of the 74 series of GATE of TTL gates is taken from a BJT in (A) totem pole and common collector configuration (B) either totem pole or open collector configuration (C) common base configuration (D) common collector configuration

MCQ 5.84

Without any additional circuitry, an 8:1 MUX can be used to obtain (A) some but not all Boolean functions of 3 variables (B) all functions of 3 variables but non of 4 variables (C) all functions of 3 variables and some but not all of 4 variables (D) all functions of 4 variables

MCQ 5.85

A 0 to 6 counter consists of 3 flip flops and a combination circuit of 2 input gate (s). The common circuit consists of (A) one AND gate (B) one OR gate

nodia

(C) one AND gate and one OR gate (D) two AND gates 2003 MCQ 5.86

TWO MARKS

The circuit in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P 5 Q 5 R and Z = RQ + PR + QP The circuit acts as a

(A) 4 bit adder giving P + Q (B) 4 bit subtractor giving P - Q (C) 4 bit subtractor giving Q-P (D) 4 bit adder giving P + Q + R MCQ 5.87

If the function W, X, Y and Z are as follows W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + P .Q Z = R + S + PQ + P .Q .R + PQ .S Then, (A) W = Z, X = Z (B) W = Z, X = Y (C) W = Y (D) W = Y = Z



Digital Circuits

Page 235

MCQ 5.88

A 4 bit ripple counter and a bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then (A) R = 10 ns, S = 40 ns (B) R = 40 ns, S = 10 ns (C) R = 10 ns S = 30 ns (D) R = 30 ns, S = 10 ns

MCQ 5.89

In the circuit shown in the figure, A is parallel-in, parallel-out 4 bit register, which loads at the rising edge of the clock C . The input lines are connected to a 4 bit bus, W . Its output acts at input to a 16 # 4 ROM whose output is floating when the input to a partial table of the contents of the ROM is as follows Data

0011

1111

0100

1010

1011

1000

0010

1000

Address

0

2

4

6

8

10

11

14

The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is

nodia (A) 1111 (C) 1000 MCQ 5.90

(B) 1011 (D) 0010

The DTL, TTL, ECL and CMOS famil GATE of digital ICs are compared in the following 4 columns (P)

(Q)

(R)

(S)

Fanout is minimum

DTL

DTL

TTL

CMOS

Power consumption is minimum

TTL

CMOS

ECL

DTL

Propagation delay is minimum

CMOS

ECL

TTL

TTL



MCQ 5.91

Digital Circuits

Chapter 5

The correct column is (A) P

(B) Q

(C) R

(D) S

The circuit shown in figure converts

(A) BCD to binary code (C) Excess -3 to gray code

(B) Binary to excess - 3 code (D) Gray to Binary code

nodia

MCQ 5.92

In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator is less than that of register B . As a result (A) Carry flag will be set but Zero flag will be reset (B) Carry flag will be rest but Zero flag will be set (C) Both Carry flag and Zero flag will be rest (D) Both Carry flag and Zero flag will be set

MCQ 5.93

The circuit shown in the figure is a 4 bit DAC

The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistance and the 5 v inputs have a tolerance of !10%. The specification (rounded to nearest multiple of 5%) for the tolerance of the DAC is (A) !35% (B) !20% (C) !10% (D) !5% 2002

ONE MARK

MCQ 5.94

4 - bit 2’s complement representation of a decimal number is 1000. The number is (A) +8 (B) 0 (C) -7 (D) -8

MCQ 5.95

The number of comparators required in a 3-bit comparators type ADC (A) 2 (B) 3



Digital Circuits

(C) 7 MCQ 5.96

Page 237

(D) 8

If the input to the digital circuit (in the figure) consisting of a cascade of 20 XOR - gates is X , then the output Y is equal to

(A) 0 (C) X

(B) 1 (D) X

2002 MCQ 5.97

TWO MARKS

The gates G1 and G2 in the figure have propagation delays of 10 ns and 20 ns respectively. If the input V1, makes an output change from logic 0 to 1 at time t = t0 , then the output waveform V0 is

nodia

MCQ 5.98

MCQ 5.99

If the input X3, X2, X1, X0 to the ROM in the figure are 8 4 2 1 BCD numbers, then the outputs Y3, Y2, Y1, Y0 are

(A) gray code numbers

(B) 2 4 2 1 BCD numbers

(C) excess - 3 code numbers

(D) none of the above

Consider the following assembly language program MVI B, 87H MOV A, B START : JMP NEXT



Digital Circuits

Chapter 5

MVI B, 00H XRA B OUT PORT1 HLT NEXT : XRA B JP START OUT PORT2 HTL The execution of above program in an 8085 microprocessor will result in (A) an output of 87H at PORT1 (B) an output of 87H at PORT2 (C) infinite looping of the program execution with accumulator data remaining at 00H (D) infinite looping of the program execution with accumulator data alternating between 00H and 87H MCQ 5.100

The circuit in the figure has two CMOS NOR gates. This circuit functions as a:

nodia

(A) flip-flop (C) Monostable multivibrator

(B) Schmitt trigger (D) astable multivibrator

2001

ONE MARKS

MCQ 5.101

The 2’s complement representation of -17 is (A) 101110 (B) 101111 (C) 111110 (D) 110001

MCQ 5.102

For the ring oscillator shown in the figure, the propagation delay of each inverter is 100 pico sec. What is the fundamental frequency of the oscillator output

MCQ 5.103

(A) 10 MHz

(B) 100 MHz

(C) 1 GHz

(D) 2 GHz

Ab 8085 microprocessor based system uses a 4K # 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is (A) OFFFH (B) 1000H (C) B9FFH (D) BA00H



Digital Circuits

Page 239

2001 MCQ 5.104

TWO MARKS

In the TTL circuit in the figure, S2 and S0 are select lines and X7 and X0 are input lines. S0 and X0 are LSBs. The output Y is

(A) indeterminate (C) A 5 B MCQ 5.105

(B) A 5 B (D) C (A 5 B ) + C (A 5 B)

The digital block in the figure is realized using two positive edge triggered D-flipflop. Assume that for t < t0, Q1 = Q2 = 0 . The circuit in the digital block is given by

nodia MCQ 5.106

In the DRAM cell in the figure, the Vt of the NMOSFET is 1 V. For the following three combinations of WL and BL voltages.

(A) 5 V; 3 V; 7 V



Digital Circuits

Chapter 5

(B) 4 V; 3 V; 4 V (C) 5 V; 5 V; 5 V (D) 4 V; 4 V; 4 V MCQ 5.107

In the figure, the LED

(A) emits light when both S1 and S2 are closed (B) emits light when both S1 and S2 are open (C) emits light when only of S1 and S2 is closed

nodia

(D) does not emit light, irrespective of the switch positions. 2000

ONE MARKS

MCQ 5.108

An 8 bit successive approximation analog to digital communication has full scale reading of 2.55 V and its conversion time for an analog input of 1 V is 20 ms. The conversion time for a 2 V input will be (A) 10 ms (B) 20 ms (C) 40 ms (D) 50 ms

MCQ 5.109

The number of comparator in a 4-bit flash ADC is (A) 4 (B) 5 (C) 15 (D) 16

MCQ 5.110

For the logic circuit shown in the figure, the required input condition (A, B, C) to make the output (X) = 1 is

(A) 1,0,1 (B) 0,0,1 (C) 1,1,1 (D) 0,1,1 MCQ 5.111

The number of hardware interrupts (which require an external signal to interrupt) present in an 8085 microprocessor are (A) 1 (B) 4 (C) 5 (D) 13

MCQ 5.112

In the microprocessor, the RST6 instruction transfer the program execution to



Digital Circuits

the following location : (A)30 H (C) 48 H

Page 241

(B) 24 H (D) 60 H

2000

TWO MARKS

MCQ 5.113

The contents of register (B) and accumulator (A) of 8085 microprocessor are 49J are 3AH respectively. The contents of A and status of carry (CY) and sign (S) after execution SUB B instructions are (A) A = F1, CY = 1, S = 1 (B) A = 0F, CY = 1, S = 1 (C) A = F0, CY = 0, S = 0 (D) A = 1F, CY = 1, S = 1

MCQ 5.114

For the logic circuit shown in the figure, the simplified Boolean expression for the output Y is

nodia

(A) A + B + C (C) B MCQ 5.115

For the 4 bit DAC shown in the figure, the output voltage V0 is

(A) 10 V (C) 4 V MCQ 5.116

(B) A (D) C

(B) 5 V (D) 8 V

A sequential circuit using D flip-flop and logic gates is shown in the figure, where X and Y are the inputs and Z is the inputs. The circuit is

(A) S - R Flip-Flop with inputs X = R and Y = S (B) S - R Flip-Flop with inputs X = S and Y = R



Digital Circuits

Chapter 5

(C) J - K Flip-Flop with inputs X = J and Y = K (D) J - K Flip-Flop with input X = K and Y = J MCQ 5.117

In the figure, the J and K inputs of all the four Flip-Flips are made high. The frequency of the signal at output Y is

(A) 0.833 kHz (C) 0.91 kHz 1999

(B) 1.0 kHz (D) 0.77 kHz

nodia

ONE MARK

MCQ 5.118

The logical expression y = A + AB is equivalent to (A) y = AB (B) y = AB (C) y = A + B (D) y = A + B

MCQ 5.119

A Darlington emitter follower circuit is sometimes used in the output stage of a TTL gate in order to (A) increase its IOL (B) reduce its IOH (C) increase its speed of operation (D) reduce power dissipation

MCQ 5.120

Commercially available ECL gears use two ground lines and one negative supply in order to (A) reduce power dissipation (B) increase fan-out (C) reduce loading effect (D) eliminate the effect of power line glitches or the biasing circuit

MCQ 5.121

The resolution of a 4-bit counting ADC is 0.5 volts. For an analog input of 6.6 volts, the digital output of the ADC will be (A) 1011 (B) 1101 (C) 1100 (D) 1110 1999

TWO MARKS

MCQ 5.122

The minimized form of the logical expression (ABC + ABC + ABC + ABC ) is (A) AC + BC + AB (B) AC + BC + AB (C) AC + BC + AB (D) AC + BC + AB

MCQ 5.123

For a binary half-subtractor having two inputs A and B, the correct set of logical



Digital Circuits

Page 243

expressions for the outputs D (= A minus B) and X (= borrow) are (A) D = AB + AB, X = AB (B) D = AB + AB + AB , X = AB (C) D = AB + AB , X = AB (D) D = AB + AB , X = AB MCQ 5.124

If CS = A15 A14 A13 is used as the chip select logic of a 4 K RAM in an 8085 system, then its memory range will be (A) 3000 H - 3 FFF H (B) 7000 H - 7 FFF H (C) 5000 H - 5 FFF H and 6000 H - 6 FFF H (D) 6000 H - 6 FFF H and 7000 H - 7 FFF H

MCQ 5.125

The ripple counter shown in the given figure is works as a

nodia

(A) mod-3 up counter (C) mod-3 down counter

(B) mod-5 up counter (D) mod-5 down counter

1998

ONE MARK

MCQ 5.126

The minimum number of 2-input NAND gates required to implement of Boolean function Z = ABC , assuming that A, B and C are available, is (A) two (B) three (C) five (D) six

MCQ 5.127

The noise margin of a TTL gate is about (B) 0.4 V (A) 0.2 V (C) 0.6 V (D) 0.8 V

MCQ 5.128

In the figure is A = 1 and B = 1, the input B is now replaced by a sequence 101010....., the output x and y will be

(A) fixed at 0 and 1, respectively (B) x = 1010.....while y = 0101...... (C) x = 1010.....and y = 1010......



Digital Circuits

Chapter 5

(D) fixed at 1 and 0, respectively MCQ 5.129

An equivalent 2’s complement representation of the 2’s complement number 1101 is (A) 110100 (B) 01101 (C) 110111 (D) 111101

MCQ 5.130

The threshold voltage for each transistor in the figure is 2 V. For this circuit to work as an inverter, Vi must take the values

(A) - 5 V and 0 V (C) - 0 V and 3 V

(B) - 5 V and 5 V (D) 3 V and 5 V

nodia

MCQ 5.131

An I/O processor control the flow of information between (A) cache memory and I/O devices (B) main memory and I/O devices (C) two I/O devices (D) cache and main memories

MCQ 5.132

Two 2’s complement number having sign bits x and y are added and the sign bit of the result is z . Then, the occurrence of overflow is indicated by the Boolean function (A) xyz (B) x y z (C) x yz + xyz (D) xy + yz + zx

MCQ 5.133

The advantage of using a dual slope ADC in a digital voltmeter is that (A) its conversion time is small (B) its accuracy is high (C) it gives output in BCD format (D) it does not require a

MCQ 5.134

For the identity AB + AC + BC = AB + AC , the dual form is (A) (A + B) (A + C) (B + C) = (A + B) (A + C) (B) (A + B ) (A + C ) (B + C ) = (A + B ) (A + C ) (C) (A + B) (A + C) (B + C) = (A + B ) (A + C ) (D) AB + AC + BC = AB + AC

MCQ 5.135

An instruction used to set the carry Flag in a computer can be classified as (A) data transfer (B) arithmetic (C) logical (D) program control

MCQ 5.136

The figure is shows a mod-K counter, here K is equal to



Digital Circuits

(A) 1 (C) 3 MCQ 5.137

(B) 2 (D) 4

The current I through resistance r in the circuit shown in the figure is

nodia

(A) - V 12R (C) V 6R MCQ 5.138

Page 245

(B) V 12R (D) V 3T

The K -map for a Boolean function is shown in the figure is the number of essential prime implicates for this function is

(A) 4 (C) 6

(B) 5 (D) 8

1997 MCQ 5.139

Each cell of a static Random Access Memory contains (A) 6 MOS transistors (B) 4 MOS transistors and 2 capacitors (C) 2 MOS transistors and 4 capacitors (D) 1 MOS transistors and 1 capacitors

MCQ 5.140

A 2 bit binary multiplier can be implemented using (A) 2 inputs ANSs only

ONE MARK

(B) 2 input XORs and 4 input AND gates only (C) Two 2 inputs NORs and one XNO gate



Digital Circuits

Chapter 5

(D) XOR gates and shift registers MCQ 5.141

In standard TTL, the ‘totem pole’ stage refers to (A) the multi-emitter input stage (B) the phase splitter (C) the output buffer (D) open collector output stage

MCQ 5.142

The inverter 74 ALSO4 has the following specifications IOH max =- 0.4 A, IOL max = 8 mA, IIH max = 20 mA, IIL max =- 0.1 mA The fan out based on the above will be (A) 10 (B) 20 (C) 60 (D) 100

MCQ 5.143

The output of the logic gate in the figure is

(A) 0 (C) A

nodia (B) 1 (D) F

MCQ 5.144

In an 8085 mP system, the RST instruction will cause an interrupt (A) only if an interrupt service routine is not being executed (B) only if a bit in the interrupt mask is made 0 (C) only if interrupts have been enabled by an EI instruction (D) None of the above

MCQ 5.145

The decoding circuit shown in the figure is has been used to generate the active low chip select signal for a microprocessor peripheral. (The address lines are designated as AO to A7 for I/O address)

The peripheral will correspond to I/O address in the range (A) 60 H to 63 H (B) A4 to A 7H (C) 30 H to 33 H MCQ 5.146

(D) 70 H to 73 H

The following instructions have been executed by an 8085 mP ADDRESS (HEX)

INSTRUCTION

6010

LXI H, 8 A 79 H

6013

MOV A, L

6015

ADDH



Digital Circuits

6016

DAA

6017

MOV H, A

6018

PCHL

Page 247

From which address will the next instruction be fetched ? (A) 6019 (B) 6379 (C) 6979 (D) None of the above MCQ 5.147

A signed integer has been stored in a byte using the 2’s complement format. We wish to store the same integer in a 16 bit word. We should (A) copy the original byte to the less significant byte of the word and fill the more significant with zeros (B) copy the original byte to the more significant byte of the word and fill the less significant byte with zeros (C) copy the original byte to the less significant byte of the word and make each fit of the more significant byte equal to the most significant bit of the original byte (D) copy the original byte to the less significant byte as well as the more significant byte of the word 1997

MCQ 5.148

nodia

For the NMOS logic gate shown in the figure is the logic function implemented is

(A) ABCDE (C) A : (B + C) + D : E MCQ 5.149

TWO MARKS

(B) (AB + C ) : (D + E ) (D) (A + B ) : C + D : E

In a J–K flip-flop we have J = Q and K = 1. Assuming the flip flop was initially cleared and then clocked for 6 pulses, the sequence at the Q output will be

(A) 010000 (C) 010010

(B) 011001 (D) 010101



Digital Circuits

Chapter 5

MCQ 5.150

The gate delay of an NMOS inverter is dominated by charge time rather than discharge time because (A) the driver transistor has larger threshold voltage than the load transistor (B) the driver transistor has larger leakage currents compared to the load transistor (C) the load transistor has a smaller W/L ratio compared to the driver transistor (D) none of the above

MCQ 5.151

The boolean function A + BC is a reduced form of (A) AB + BC (B) (A + B) : (A + C) (C) AB + ABC (D) (A + C) : B 1996

ONE MARK

MCQ 5.152

Schottky clamping is resorted in TTl gates (A) to reduce propagation delay (B) to increase noise margins (C) to increase packing density (D) to increase fan-out

MCQ 5.153

A pulse train can be delayed by a finite number of clock periods using (A) a serial-in serial-out shift register (B) a serial-in parallel-out shift register (C) a parallel-in serial-out shift register (D) a parallel-in parallel-out shift register

MCQ 5.154

A 12-bit ADC is operating with a 1 m sec clock period and the total conversion time is seen to be 14 m sec . The ADC must be of the (A) flash type (B) counting type (C) intergrating type (D) successive approximation type

MCQ 5.155

The total number of memory accesses involved (inclusive of the op-code fetch) when an 8085 processor executes the instruction LDA 2003 is (A) 1 (B) 2 (C) 3 (D) 4

nodia

1996

TWO MARKS

MCQ 5.156

A dynamic RAM cell which hold 5 V has to be refreshed every 20 m sec, so that the stored voltage does not fall by more than 0.5 V. If the cell has a constant discharge current of 1 pA, the storage capacitance of the cell is (A) 4 # 10-6 F (B) 4 # 10-9 F (C) 4 # 10-12 F (D) 4 # 10-15 F

MCQ 5.157

A 10-bit ADC with a full scale output voltage of 10.24 V is designed to have a ! LSB/2 accuracy. If the ADC is calibrated at 25c C and the operating temperature ranges from 0c C to 25c C , then the maximum net temperature coefficient of the ADC should not exceed (A) ! 200 mV/cC (B) ! 400 mV/cC



Digital Circuits

(C) ! 600 mV/cC MCQ 5.158

(D) ! 800 mV/cC

A memory system of size 26 K bytes is required to be designed using memory chips which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is (A) 2 (B) 4 (C) 8

MCQ 5.159

Page 249

(D) 13

The following sequence of instructions are executed by an 8085 microprocessor: 1000 LXI SP, 27 FF 1003 CALL 1006 1006 POP H The contents of the stack pointer (SP) and the HL, register pair on completion of execution of these instruction are (A) SP = 27 FF, HL = 1003 (B) SP = 27 FD, HL = 1003 (C) SP = 27 FF, HL = 1006 (D) SP = 27 FD, HL = 1006

nodia ***********



Digital Circuits

Chapter 5

SOLUTIONS SOL 5.1

Option (C) is correct. Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position. Following are the truth table given for different combinations of A and B A

B

Y(Bulb)

up(1)

up(1)

OFF(0)

Down(0)

Down(0)

OFF(0)

up(1)

Down(0)

ON(1)

Down(0) up(1) ON(1) When the switches A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the switch changes its position leads to the ON state of bulb. Hence, from the truth table, we get

nodia Y = A5B

i.e., the XOR gate SOL 5.2

Option (A) is correct. The program is being executed as follows MVI A, 0.5H; A = 05H MVI B, 0.5H; B = 05H At the next instruction, a loop is being introduced in which for the instruction “DCR B” if the result is zero then it exits from loop so, the loop is executed five times as follows : Content in B

Output of ADD B (Stored value at A)

05

05 + 05

04

05 + 05 + 04

03

05 + 05 + 04 + 03

02

05 + 05 + 04 + 03 + 02

01

05 + 05 + 04 + 03 + 02 + 01

00

System is out of loop

i.e., A = 05 + 05 + 04 + 03 + 02 + 01 = 144 At this stage, the 8085 microprocessor exits from the loop and reads the next instruction. i.e., the accumulator is being added to 03 H. Hence, we obtain A = A + 03 H = 14 + 03 = 17 H SOL 5.3

Option (D) is correct. For chip-1, we have the following conclusions: it is enable when (i) and (ii) For S1 S 0 = 0 0 We have

S1 S 0 = 0 0 Input = 1

A13 = A12 = 0



Digital Circuits

Page 251

and for I/p = 1we obtain A10 = 1 or A10 = 0 A11 = 1 A14 = 1 or A14 = 0 A15 = 1 or A15 = 0 Since, A 0 - A 9 can have any value 0 or 1 Therefore, we have the address range as A15 A14 A13 A12 A11 A10 A 9 A 8 A7 A 6 A5 A 4 A 3 A2 A1 A 0 From

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

to

0

0

0

0

1

0

1

1

1

1

1

1

1

1

1

1

In Hexadecimal & 0800 H to 0BFFH Similarly, for chip 2, we obtain the range as follows E = 1 for S1 S 0 = 0 1 so, A13 = 0 and A12 = 1 and also the I/P = 1 for

nodia

A10 = 0 , A11 = 1, A14 = 0 , A15 = 0 so, the fixed I/ps are A15

A14

A13

A12

A11

A10

0

0

0

1

1

0

Therefore, the address range is

A15 A14 A13 A12 A11 A10 A 9 A 8 A7 A 6 A5 A 4 A 3 A2 A1 A 0

From to

0

0

0

1

1

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1

1

0

1

1

1

1

1

1

1

1

1

1

In hexadecimal it is from 1800 H to 1BFFH . There is no need to obtain rest of address ranged as only (D) is matching to two results. SOL 5.4

Option (A) is correct. The given circuit is

Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 .



Digital Circuits

Chapter 5

2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. SOL 5.5

Option (B ) is correct.

Y = 1, when A > B A = a1 a 0, B = b1 b 0

a1 0 1 1 1 1 1

nodia a0

b1

b0

Y

1

0

0

1

0

0

0

1

0

0

1

1

1

0

0

1

1

0

1

1

1

1

0

1

Total combination = 6 SOL 5.6

Option (A) is correct. Parallel connection of MOS & OR operation Series connection of MOS & AND operation The pull-up network acts as an inverter. From pull down network we write Y = (A + B) C = (A + B) + C = A B + C

SOL 5.7

Option (A) is correct. Prime implicants are the terms that we get by solving K-map

F = XY + XY 1prime 44 2 44 3 implicants SOL 5.8

Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure. D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn

X 0 = Q , X1 = Q



Digital Circuits

If A = 0, If A = 1, So state diagram is

SOL 5.9

Qn + 1 = Qn Qn + 1 = Qn

Page 253

(toggle of previous state)

Option (B) is correct. The given circuit is shown below:

nodia

(PQ QR ) PR = (PQ + QR PR ) = PQ + QR + PR = PQ + QR + PR If any two or more inputs are ‘1’ then output y will be 1. SOL 5.10

Option (A) is correct. For the output to be high, both inputs to AND gate should be high. The D-Flip Flop output is the same, after a delay. Let initial input be 0; (Consider Option A) then Q = 1 (For 1st D-Flip Flop). This is given as input to 2nd FF. Let the second input be 1. Now, considering after 1 time interval; The output of 1st Flip Flop is 1 and 2nd FF is also 1. Thus Output = 1.

SOL 5.11

Option (D) is correct. F = S1 S 0 I 0 + S1 S 0 I1 + S1 S 0 I 2 + S1 S 0 I 3 I0 = I3 = 0 F = PQ + PQ = XOR (P, Q)

SOL 5.12

( S1 = P, S 0 = Q )

Option (A) is correct. All the states of the counter are initially unset.

State Initially are shown below in table : Q2

Q1

Q0

0

0

0

0

1

0

0

4

1

1

0

6

1

1

1

7



SOL 5.13

Digital Circuits

0

1

1

3

0

0

1

1

0

0

0

0

Chapter 5

Option (D) is correct. The sequence is QB QA 00 " 11 " 01 " 10 " 00 " ... QB

QA

QB (t + 1)

QA (t + 1)

0

0

1

1

1

1

0

1

0

1

1

0

1 QB ^t + 1h

0

0

0

QB ^t + 1h = Q A

nodia DA = Q A Q B + QA QB

SOL 5.14

Option (C) is correct. Initially Carry Flag, C = 0 MVI A, 07 H ; A = 0000 0111 RLC ; Rotate left without carry. A = 0000 1110 MVO B, A ; B = A = 0000 1110 RLC ; A = 0001 1100 RLC ; A = 0011 1000 ADD B ; A = 0011 1000 + 0000 1110 ; 0100 0110 ; RRC Thus A = 23 H

SOL 5.15

; Rotate Right with out carry, A = 0010 0011

Option ( ) is correct.



SOL 5.16

Digital Circuits

Page 255

Option (B) is correct. Since G2 is active low input, output of NAND gate must be 0 G2 = A15 : A14 A13 A12 A11 = 0 So, A15 A14 A13 A12 A11 = 00101 To select Y5 Decoder input ABC = A 8 A 9 A10 = 101 Address range A15 A14 A13 A12 A11 A10 A 9 A 8 ...............A 0 0011101........A 0 S S 2 D ^2D00 - 2DFF h

SOL 5.17

Option (A) (B) (C) are correct. In the circuit F = (A 5 B) 9 (A 9 B) 9 C For two variables A5B = A9B So, (A 5 B) 9 (A 9 B) = 0 (always) F = 09C = 0$C+1$C = C

nodia

So, F = 1 when C = 1 or C = 0 SOL 5.18

Option (D) is correct. Let QA (n), QB (n), QC (n) are present states and QA (n + 1), QB (n + 1), QC (n + 1) are next states of flop-flops. In the circuit QA (n + 1) = QB (n) 9 QC (n) QB (n + 1) QA (n) QC (n + 1) QB (n) Initially all flip-flops are reset 1st clock pulse QA = 0 9 0 = 1 QB = 0 QC = 0 2 nd clock pulse QA = 0 9 0 = 1 QB = 1 QC = 0 3 rd clock pulse QA = 1 9 0 = 0 QB = 1 QC = 1 4 th clock pulse QA = 1 9 1 = 1 QB = 0 So, sequence

SOL 5.19

QC = 1 QA = 01101.......




Digital Circuits

Chapter 5

Output of the MUX can be written as F = I 0 S 0 S1 + I1 S 0 S1 + I 2 S 0 S1 + I 3 S 0 S1 Here, I 0 = C, I1 = D, I2 = C , I 3 = CD and S 0 = A, S1 = B So, F = C A B + D A B + C A B + C DA B Writing all SOP terms F = A B C D + A B C D + A BCD + A B C D + A B C D + A B C D + ABC D 1 44 2 44 3 1 44 2 44 3 S 1 44 2 4 4 3 1 44 2 4 4 3 1 44 2 44 3 S m m m m m m m 3

SOL 5.20

7

2

5

F = / m (2, 3, 5, 7, 8, 9, 12)

9

8

12

Option (C) is correct. By executing instruction one by one MVI A, 45 H & MOV 45 H into accumulator, A = 45 H STC & Set carry, C = 1 CMC & Complement carry flag, C = 0 RAR & Rotate accumulator right through carry

nodia A = 00100010 XRA B & XOR A and B

A = A 5 B = 00100010 5 01000101 = 01100111 = 674

SOL 5.21

Option (C) is correct. TTL " Transistor - Transistor logic CMOS " Complementary Metal Oxide Semi-conductor

SOL 5.22

Option (D) is correct. Vectored interrupts : Vectored interrupts are those interrupts in which program control transferred to a fixed memory location. Maskable interrupts : Maskable interrupts are those interrupts which can be rejected or delayed by microprocessor if it is performing some critical task.

SOL 5.23

Option (D) is correct. We have 6X + Z {Y + (Z + XY )}@[X + Z (X + Y)] = 1 Substituting X = 1 and X = 0 we get [1 + Z {Y + (Z + 1Y )}][ 0 + Z (1 + Y)] = 1 or [1][ Z (1)] = 1 1 + A = 1 and 0 + A = A or Z =1)Z=0

SOL 5.24

Option (A) is correct. The AND gate implementation by 2:1 mux is as follows

Y = AI 0 + AI1 = AB



Digital Circuits

Page 257

The EX - OR gate implementation by 2:1 mux is as follows

Y = BI0 + BI1 = AB + BA SOL 5.25

Option (A) is correct. The given circuit is as follows.

nodia

The truth table is as shown below. Sequence is 00, 11, 10, 00 ... CLK 1 2 3 4 SOL 5.26

J1

K1

Q1

J2

K2

Q2

1

1

0

1

1

0

1

1

1

1

1

1

0

0

1

0

1

0

1

1

0

1

1

0

Option (B) is correct. The given situation is as follows

The truth table is as shown below P1

P2

a

b

c

d

e

f

g

0

0

1

1

1

1

1

1

0

0

1

1

0

1

1

0

1

1

1

0

1

1

0

1

1

0

1

1

1

1

0

0

1

1

1

1

From truth table we can write a =1 b = P 1 P 2 + P1 P 2 = P 2 c = P1 P2 + P1 P2 = P1 d = 1 = c+e

1 NOT Gate 1 NOT Gate



Digital Circuits

and

c = P1 P2 = P1 + P2

Chapter 5

1 OR GATE

1 OR GATE f = P1 P2 = P1 + P2 1 OR GATE g = P1 P2 = P1 + P2 Thus we have g = P1 + P2 and d = 1 = c + e . It may be observed easily from figure that Led g does not glow only when both P1 and P2 are 0. Thus g = P1 + P2 LED d is 1 all condition and also it depends on d = c+e SOL 5.27

Option (D) is correct. As shown in previous solution 2 NOT gates and 3-OR gates are required.

SOL 5.28

Option (C) is correct. For the NAND latche the stable states are as follows

nodia

For the NOR latche the stable states are as follows

SOL 5.29

Option (D) is correct. From the figure shown below it may be easily seen upper MOSFET are shorted and connected to Vdd thus OUT is 1 only when the node S is 0,

Since the lower MOSFETs are shorted to ground, node S is 0 only when input P and Q are 1. This is the function of AND gate. SOL 5.30

Option (B) is correct. MSB of both number are 1, thus both are negative number. Now we get 11101101 = (- 19) 10 and 11100110 = (- 26) 10 P - Q = (- 19) - (- 26) = 7 Thus 7 signed two’s complements form is (7) 10 = 00000111



SOL 5.31

Digital Circuits

Page 259

Option (D) is correct. The circuit is as shown below

X = PQ Y = (P + Q) So and SOL 5.32

Z = PQ (P + Q) = (P + Q )( P + Q) = PQ + PQ = P 5 Q M1 = Z 5 R = (P 5 Q) 5 R


nodia

The truth table is shown below. When CLK make transition Q goes to 1 and when D goes to 1, Q goes to 0 SOL 5.33

Option (B) is correct. Since the input to both JK flip-flop is 11, the output will change every time with clock pulse. The input to clock is

The output Q0 of first FF occurs after time 3 T and it is as shown below

The output Q1 of second FF occurs after time 3 T when it gets input (i.e. after 3 T from t1) and it is as shown below

SOL 5.34




Digital Circuits

VDAC =

Chapter 5

3

/ 2n - 1bn = 2- 1b0 + 20 b1 + 21b2 + 22 b3 n=0

or VDAC = 0.5b0 + b1 + 2b2 + 4b3 The counter outputs will increase by 1 from 0000 till Vth > VDAC . The output of counter and VDAC is as shown below Clock

b3 b3 b2 b0

VDAC

1

0001

0

2

0010

0.5

3

0011

1

4

0100

1.5

5

0101

2

6

0110

2.5

7

0111

3

8

1000

3.5

9

1001

4

10

1010

4.5

11

1011

5

12

1100

5.5

13

1101

6

14

1110

6.5

nodia

and when VADC = 6.5 V (at 1101), the output of AND is zero and the counter stops. The stable output of LED display is 13. SOL 5.35

Option (B) is correct. The VADC - Vin at steady state is = 6.5 - 6.2 = 0.3V

SOL 5.36

Option (A) is correct. Z = I0 RS + I1 RS + I2 RS + I3 RS = (P + Q ) RS + PRS + PQRS + PRS = PRS + QRS + PRS + PQRS + PRS The k - Map is as shown below

Z = PQ + PQS + QRS SOL 5.37

Option (C) is correct. 2710H LXI H, 30A0H 2713H DAD H

; Load 16 bit data 30A0 in HL pair ; 6140H " HL



Digital Circuits

Page 261

2714H PCHL ; Copy the contents 6140H of HL in PC Thus after execution above instruction contests of PC and HL are same and that is 6140H SOL 5.38

Option (C) is correct. MSB of Y is 1, thus it is negative number and X is positive number Now we have X = 01110 = (14) 10 and Y = 11001 = (- 7) 10 X + Y = (14) + (- 7) = 7 In signed two’s complements from 7 is (7) 10 = 000111

SOL 5.39

Option (B) is correct. Y = AB + CD = AB .CD This is SOP form and we require only 3 NAND gate

SOL 5.40


nodia Y = AB + AB

and

X = YC + YC = (AB + AB ) C + (AB + AB ) C = (AB + AB) C + (AB + AB ) C = ABC + ABC + ABC + ABC

SOL 5.41

Option (D) is correct. Y = ABCD + ABCD + ABC D + ABC D = ABCD + ABC D + ABC D + ABC D = ABCD + ABC D + BC D (A + A) = ABCD + ABC D + BC D

SOL 5.42

A+A = 1

Option (B) is correct. In given TTL NOT gate when Vi = 2.5 (HIGH), then Q1 " Reverse active Q2 " Saturation Q3 " Saturation Q4 " cut - off region

SOL 5.43

Option (C) is correct. For X = 0, Y = 1 For X = 0, Y = 0 For X = 1, Y = 1

P = 1, Q = 0 P = 1, Q = 1 P = 1, Q = 0 or P = 0, Q = 1



Digital Circuits

Chapter 5

SOL 5.44

Option (C) is correct. Chip 8255 will be selected if bits A3 to A7 are 1. Bit A0 to A2 can be 0 or. 1. Thus address range is 11111000 F8H 11111111 FFH

SOL 5.45

Option (B) is correct. Since the inverting terminal is at virtual ground the resistor network can be reduced as follows

nodia

The current from voltage source is I = VR = 10 = 1 mA R 10k

This current will be divide as shown below

Now

-3 i = I = 1 # 10 = 62.5 m A 16 16

SOL 5.46

Option (C) is correct. The net current in inverting terminal of OP - amp is I - = 1 + 1 = 5I 4 16 16 So that V0 =- R # 5I =- 3.125 16

SOL 5.47

Option (B) is correct. Line 1 : MVI A, B5H 2 : MVI B, 0EH 3 : XRI 69H

; ; ; ;

Move B5H to A Move 0EH to B [A] XOR 69H and store in A Contents of A is CDH



Digital Circuits

Page 263

4 : ADDB

; Add the contents of A to contents of B and ; store in A, contents of A is EAH 5 : ANI 9BH ; [a] AND 9BH, and store in A, ; Contents of A is 8 AH 6 : CPI 9FH ; Compare 9FH with the contents of A ; Since 8 AH < 9BH, CY = 1 7 : STA 3010 H ; Store the contents of A to location 3010 H 8 : HLT ; Stop Thus the contents of accumulator after execution of ADD instruction is EAH. SOL 5.48

Option (C) is correct. The CY = 1 and Z = 0

SOL 5.49

Option (A) is correct. For this circuit the counter state (Q1, Q0) follows the sequence 00, 01, 10, 00 ... as shown below Clock 1st 2nd 3rd

SOL 5.50

D1 D0

Q1 NOR Q0

Q1 Q0

nodia 00

1

01

10

0

10

01

0

00

00

0

Option (A) is correct. As shown below there are 2 terms in the minimized sum of product expression. 1

0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

SOL 5.51

Option (B) is correct. The output is taken from the 5th line.

SOL 5.52

Option (D) is correct. After applying two clock poles, the outputs of the full adder is S = 1, C0 = 1 A B Ci S Co 1st 1 0 0 0 1 2nd 1 1 1 1 1

SOL 5.53

Option (D) is correct. 100010011001 SSSS 4 2 3 1

SOL 5.54




Digital Circuits

Chapter 5

In this the diode D2 is connected to the ground. The following table shows the state of counter and D/A converter Q2 Q1 Q0

D3 = Q2

D2 = 0

D1 = Q1

D0 = Q0

Vo

000

0

0

0

0

0

001

0

0

0

1

1

010

0

0

1

0

2

011

0

0

1

1

3

100

1

0

0

0

8

101

1

0

0

1

9

110

1

0

1

0

10

111

1

0

1

1

11

000

0

0

0

0

0

001

0

0

0

1

1

Thus option (B) is correct

nodia

SOL 5.55

Option (B) is correct. LXI, EFFF H ; Load SP with data EFFH CALL 3000 H ; Jump to location 3000 H : : : 3000H LXI H, 3CF4 ; Load HL with data 3CF4H PUSH PSW ; Store contnets of PSW to Stack POP PSW ; Restore contents of PSW from stack PRE ; stop Before instruction SPHL the contents of SP is 3CF4H. After execution of POP PSW, SP + 2 " SP After execution of RET, SP + 2 " SP Thus the contents of SP will be 3CF4H + 4 = 3CF8H

SOL 5.56

Option (A) is correct. The inputs D0 and D1 respectively should be connected as Q1 and Q0 where Q0 " D1 and Q1 " D0

SOL 5.57

Option (D) is correct. If the point P is stuck at 1, then output f is equal to A

SOL 5.58

Option (B) is correct. Dividing 43 by 16 we get 2 16 43 32 11

g



Digital Circuits

Page 265

11 in decimal is equivalent is B in hexamal. Thus 4310 * 2B16 Now 410 * 01002 310 * 00112 Thus 4310 * 01000011BCD SOL 5.59

Option (A) is correct. The diagram is as shown in fig

nodia f' = BC + BC

f = f' A + f ' 0 = f'A = ABC + ABC

SOL 5.60


If output is at logic 0, the we have V0 = 0 which signifies BJT Q3 is in saturation and applying KVL we have or or

VBE3 = IR # 1k 0.75 = IR # 1k IR = 0.75 mA

SOL 5.61

Option (A) is correct. We have f = ABC + ABC = B (AC + AC ) = B (A + C)( A + C )

SOL 5.62

Option (C) is correct. Characteristic equation for a jk flip-flop is written as Where So,

Qn + 1 = JQ n + K Qn Qn is the present output Qn + 1 is next output Qn + 1 = 10 + K : 0 Qn + 1 = 1

Qn = 0



Digital Circuits

Chapter 5

SOL 5.63

Option (C) is correct. Since T2 T1 T0 is at 111, at every clock Q2 Q1 Q0 will be changes. Ir present state is 011, the next state will be 100.

SOL 5.64


SOL 5.65

Option (C) is correct. 0100H LXI SP, 00FF 0103H LXI H, 0701 0106H MVI A, 20H 0108 H SUB M

; Load SP with 00FFG ; Load HL with 0107H ; Move A with 20 H ; Subtract the contents of memory ; location whose address is stored in HL ; from the A and store in A 0109H ORI 40H ; 40H OR [A] and store in A 010BH ADD M ; Add the contents of memeory location ; whose address is stored in HL to A ; and store in A HL contains 0107H and contents of 0107H is 20H Thus after execution of SUB the data of A is 20H - 20H = 00 SOL 5.66

nodia

Option (C) is correct. Before ORI instruction the contents of A is 00H. On execution the ORI 40H the contents of A will be 40H 00H = 00000000 40H = 01000000 ORI 01000000 After ADD instruction the contents of memory location whose address is stored in HL will be added to and will be stored in A 40H + 20 H = 60 H

SOL 5.67

Option (C) is correct. A master slave D-flip flop is shown in the figure.

In the circuit we can see that output of flip-flop call be triggered only by transition of clock from 1 to 0 or when state of slave latch is affected. SOL 5.68

Option (A) is correct. The range of signed decimal numbers that can be represented by n - bits 1’s complement number is - (2n - 1 - 1) to + (2n - 1 - 1). Thus for n = 6 we have Range =- (26 - 1 - 1) to + (26 - 1 - 1) =- 31 to + 31

SOL 5.69

Option (D) is correct. The minimum number of bit require to encode 100 increment is 2n $ 100



Digital Circuits

or

Page 267

n $7

SOL 5.70

Option (B) is correct. Shift Register " Serial to parallel data conversion Counter " Frequency division Decoder " Addressing in memory chips.

SOL 5.71

Option (A) is correct. For the TTL family if terminal is floating, then it is at logic 1. Thus Y = (AB + 1) = AB .0 = 0

SOL 5.72

Option (C) is correct. 11001 1001 111001 00110 0110 000110 +1 +1 +1 00111 0111 000111 7 7 7 Thus 2’s complement of 11001, 1001 and 111001 is 7. So the number given in the question are 2’s complement correspond to -7.

SOL 5.73

Option (C) is correct. In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after 101 or at 110) all states must be cleared. Thus when CB is 11 the all states must be cleared. The input to 2-input gate is C and B and the desired output should be low since the CLEAR is active low Thus when C and B are 0, 0, then output must be 0. In all other case the output must be 1. OR gate can implement this functions.

SOL 5.74

Option (C) is correct. Number of MUX is 4 = 2 and 2 = 1. Thus the total number 3 multiplexers is 3 2 required.

SOL 5.75


nodia AC + BC = AC1 + BC 1 = AC (B + B ) + BC (A + A) = ACB + ACB + BC A + BC A

SOL 5.76

Option (D) is correct. We have f (x, y) = xy + xy + xy = x (y + y) + xy = x + xy or f (x, y) = x + y Here compliments are not available, so to get x we use NOR gate. Thus desired circuit require 1 unit OR and 1 unit NOR gate giving total cost 2 unit.

SOL 5.77

Option (D) is correct. For 8255, various modes are described as following. Mode 1 : Input or output with hand shake In this mode following actions are executed 1. Two port (A & B) function as 8 - bit input output ports. 2. Each port uses three lines from C as a hand shake signal 3.

Input & output data are latched.

Form (ii) the mode is 1. Mode 2 : Bi-directional data transfer This mode is used to transfer data between two computer. In this mode port A



Digital Circuits

Chapter 5

can be configured as bidirectional port. Port A uses five signal from port C as hand shake signal. For (1), mode is 2 SOL 5.78

Option (B) is correct. LDA 16 bit & Load accumulator directly this instruction copies data byte from memory location (specified within the instruction) the accumulator. It takes 4 memory cycle-as following. 1. in instruction fetch 2. in reading 16 bit address 1. in copying data from memory to accumulator LXI D, (F0F1) 4 & It copies 16 bit data into register pair D and E. It takes 3 memory cycles.

SOL 5.79

Option (A) is correct. LXI H, 9258H MOV A, M CMa MOV M, A This program complement

; 9258H " HL ; (9258H) " A ; A"A ; A"M the data of memory location 9258H.

nodia

SOL 5.80

Option (D) is correct. MVI A, 00H ; Clear accumulator LOOP ADD B ; Add the contents of B to A DCR C ; Decrement C JNZ LOOP ; If C is not zero jump to loop HLT END This instruction set add the contents of B to accumulator to contents of C times.

SOL 5.81

Option (D) is correct. The number of distinct boolean expression of n variable is 22n . Thus 22 = 216 = 65536 4

SOL 5.82

Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bit.s So,

2n - 1 = 28 - 1 = 255

SOL 5.83

Option (B) is correct. When output of the 74 series gate of TTL gates is taken from BJT then the configuration is either totem pole or open collector configuration .

SOL 5.84

Option (D) is correct. A 2n: 1 MUX can implement all logic functions of (n + 1) variable without andy additional circuitry. Here n = 3 . Thus a 8 : 1 MUX can implement all logic functions of 4 variable.

SOL 5.85

Option (D) is correct. Counter must be reset when it count 111. This can be implemented by following circuitry



SOL 5.86

Digital Circuits

Page 269

Option (B) is correct. We have Y = P5Q5R Z = RQ + PR + QP Here every block is a full subtractor giving P - Q - R where R is borrow. Thus circuit acts as a 4 bit subtractor giving P - Q .

SOL 5.87

Option (A) is correct. W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + PQ = RS + PR $ PQ $ PQ = RS + (P + R )( P + Q)( P + Q) = RS + (P + PQ + PR + QR )( P + Q) = RS + PQ + QR (P + P ) + QR = RS + PQ + QR

nodia

Z = R + S + PQ + PQR + PQS = R + S + PQ $ PQR $ PQS = R + S + (P + Q )( P + Q + R)( P + Q + S) = R + S + PQ + PQ + PQS + PR + PQR + PRS + PQ + PQS + PQR + QRS

= R + S + PQ + PQS + PR + PQR + PRS + PQS + PQR + QRS

= R + S + PQ (1 + S) + PR (1 + P ) + PRS + PQS + PQR + QRS = R + S + PQ + PR + PRS + PQS + PQR + QRS

= R + S + PQ + PR (1 + Q ) + PQS + QRS = R + S + PQ + PR + PQS + QRS Thus W = Z and X = Z SOL 5.88

Option (B) is correct. Propagation delay of flip flop is tpd = 10 nsec Propagation delay of 4 bit ripple counter R = 4tpd = 40 ns and in synchronous counter all flip-flop are given clock simultaneously, so S = tpd = 10 ns

SOL 5.89

Option (C) is correct. After t = t1, at first rising edge of clock, the output of shift register is 0110, which in input to address line of ROM. At 0110 is applied to register. So at this time data stroed in ROM at 1010 (10), 1000 will be on bus. When W has the data 0110 and it is 6 in decimal, and it’s data value at that add is 1010 then 1010 i.e. 10 is acting as odd, at time t2 and data at that movement is 1000.

SOL 5.90

Option (B) is correct. The DTL has minimum fan out and CMOS has minimum power consumption. Propagation delay is minimum in ECL.



Digital Circuits

Chapter 5

SOL 5.91

Option (D) is correct. Let input be 1010; output will be 1101 Let input be 0110; output will be 0100 Thus it convert gray to Binary code.

SOL 5.92

Option (A) is correct. CMP B & Compare the accumulator content with context of Register B If A < R CY is set and zero flag will be reset.

SOL 5.93

Option (A) is correct. Vo =- V1 :R bo + R b1 + R b2 + R b 3D R 2R 4R 4R Exact value when V1 = 5 , for maximum output VoExact =- 5 :1 + 1 + 1 + 1 D =- 9.375 2 4 8 Maximum Vout due to tolerance Vo max =- 5.5 :110 + 110 + 110 + 110 D 90 2 # 90 4 # 90 8 # 90

nodia =- 12.604

Tolerance

= 34.44% = 35%

SOL 5.94

Option (D) is correct. If the 4- bit 2’s complement representation of a decimal number is 1000, then the number is -8

SOL 5.95

Option (C) is correct. In the comparator type ADC, the no. of comparators is equal to 2n - 1, where n is no. of bit.s So, 23 - 1 = 7

SOL 5.96

Option (B) is correct. Output of 1 st XOR = = X $ 1 + X $ 1 = X Output of 2 nd XOR = X X + XX = 1 So after 4,6,8,...20 XOR output will be 1.

SOL 5.97

Option (B) is correct. They have prorogation delay as respectively, G1 " 10 nsec G2 " 20 nsec For abrupt change in Vi from 0 to 1 at time t = t0 we have to assume the output of NOR then we can say that option (B) is correct waveform.

SOL 5.98

Option (B) is correct. Let X3 X2 X1 X0 be 1001 then Y3 Y2 Y1 Y0 will be 1111.



Digital Circuits

Page 271

Let X3 X2 X1 X0 be 1000 then Y3 Y2 Y1 Y0 will be 1110 Let X3 X2 X1 X0 be 0110 then Y3 Y2 Y1 Y0 will be 1100 So this converts 2-4-2-1 BCD numbers. SOL 5.99

Option (B) is correct. MVI B, 87H MOV A, B START : JMP NEXT XRA B JP START JMP NEXT XRA

NEXT :

JP START OUT PORT2 SOL 5.100

; B = 87 ; A = B = 87 ; Jump to next ; A 5 B " A, ; A = 00, B = 87 ; Since A = 00 is positive ; so jump to START ;Jump to NEXT ; unconditionally ; B ; A 5 B " A, A = 87 , ; B = 87 H ; will not jump as D7 , of A is 1 ; A = 87 " PORT2

nodia


The circuit shown is monostable multivibrator as it requires an external triggering and it has one stable and one quasistable state. SOL 5.101

Option (B) is correct. The two’s compliment representation of 17 is 17 = 010001 Its 1’s complement is 101110 So 2’s compliment is +

101110 1 101111

SOL 5.102

Option (C) is correct. The propagation delay of each inverter is tpd then The fundamental frequency of oscillator output is 1 = 1 GHz f = 1 = 2ntpd 2 # 5 # 100 # 10 - 12

SOL 5.103

Option (C) is correct. 4K # 8 bit means 102410 location of byte are present Now 102410 * 1000H It starting address is AA00H then address of last byte is AA00H + 1000H - 0001H = B9FFH



SOL 5.104

Digital Circuits

Chapter 5

Option (D) is correct. Y = I0 + I3 + I5 + I6 = C BA + C AB + CBA + CBA = C (BA + AB) + C (AB + BA) or

Y = C (A 5 B ) + C (A 5 B)

SOL 5.105

Option (C) is correct. The output of options (C) satisfy the given conditions

SOL 5.106


SOL 5.107

Option (D) is correct. For the LED to glow it must be forward biased. Thus output of NAND must be LOW for LED to emit light. So both input to NAND must be HIGH. If any one or both switch are closed, output of AND will be LOW. If both switch are open, output of XOR will be LOW. So there can’t be both input HIGH to NAND. So LED doesn’t emit light.

SOL 5.108

Option (B) is correct. Conversion time of successive approximate analog to digital converters is independent of input voltage. It depends upon the number of bits only. Thus it remains unchanged.

SOL 5.109

Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bits. So, 2 4 - 1 = 15

SOL 5.110

Option (D) is correct. As the output of AND is X = 1, the all input of this AND must be 1. Thus ...(1) AB + AB = 1 ...(2) BC + BC = 1

nodia

C =1 From (2) and (3), if C = 1, then B = 1 If B = 1, then from (1) A = 0 . Thus A = 0, B = 1 and C = 1 SOL 5.111

...(3)

Option (C) is correct. Interrupt is a process of data transfer by which an external device can inform the processor that it is ready for communication. 8085 microprocessor have five interrupts namely TRAP, INTR, RST 7.5, RST 6.5 and RST 5.5



Digital Circuits

Page 273

SOL 5.112

Option (A) is correct. For any RST instruction, location of program transfer is obtained in following way. RST x & (x ) 8) 10 " convert in hexadecimal So for RST 6 & (6 ) 8) 10 = (48) 10 = (30) H

SOL 5.113

Option (A) is correct. Accumulator contains A = 49 H Register B = 3 AH SUB B = A minus B A = 49 H = 01001001 B = 3 AH = 00111010 2’s complement of (- B) = 11000110 A - B = A + (- B) 010 010 01 & +1 1 0 0 0 1 1 0 0 0 0 0 1111

nodia Carry = 1 so here outputA = 0 F Carry CY = 1 Sign flag S = 1

SOL 5.114

Option (C) is correct. The circuit is as shown below :

Y = B + (B + C ) = B (B + C ) = B SOL 5.115


The voltage at non-inverting terminal is V+ = 1 + 1 = 5 8 2 8



Digital Circuits

V- = V+ = 5 8 Now applying voltage divider rule V- = 1k V% = 1 Vo 1k + 7k 8

Chapter 5

...(1)

...(2)

From (1) and (2) we have Vo = 8 # 5 = 5V 8 SOL 5.116

Option (D) is correct. The truth table is shown below Z = XQ + YQ Comparing from the truth table of J - K FF Y = J, X =K X 0 0 1 1

Y

Z

nodia 0

Q

1

0

0

1

1

Q1

SOL 5.117

Option (B) is correct. In the figure the given counter is mod-10 counter, so frequency of output is 10k = 1k 10

SOL 5.118

Option (D) is correct. We have y = A + AB we know from Distributive property Thus

x + yz = (x + y) (x + z) y = (A + A) (A + B) = A + B

` A+A = 1

SOL 5.119

Option (C) is correct. Darligton emitter follower provides a low output impedance in both logical state (1 or 0). Due to this low output impedance, any stray capacitance is rapidly charged and discharged, so the output state changes quickly. It improves speed of operation.

SOL 5.120


SOL 5.121

Option (B) is correct. For ADC we can write Analog input = (decimal eq of digital output) # resol 6.6 = (decimal eq. of digital output) # 0.5 6.6 = decimal eq of digital. output 0.5 13.2 = decimal equivalent of digital output so output of ADC is 1101.

SOL 5.122

Option (A) is correct. We use the K -map as below.



Digital Circuits

Page 275

So given expression equal to = AC + BC + AB SOL 5.123

Option (C) is correct. For a binary half-subtractor truth table si given below.

nodia

from truth table we can find expressions of D & X D = A 5 B = AB + AB X = AB

SOL 5.124

Option (B) is correct. We have 4 K RAM (12 address lines)

so here chip select logic CS = A15 A14 A13 address range (111) S

initial address final address

A15 A14 A13 A12 A11 A10 A 9 A 8 A7 A6 A5 A 4 A 3 A2 A1 A 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 &7000H 1 1

1 1 1 1 1 1 1 1 1 1 1 1 11 &7FFFH so address range is (7 0 0 0 H – 7 F F F H)



SOL 5.125

Digital Circuits

Chapter 5

Option (D) is correct. From the given figure we can write the output

For the state 010 all preset = 1 and output QA QB QC = 111 so here total no. of states = 5 (down counter) SOL 5.126

Option (C) is correct. Given boolean function is

nodia Z = ABC

Now

Z = ABC = ACB = AC + B

Z = AC + B we have Z = X + Y (1 NOR gate) where X = AC (1 NAND gate) To implement a NOR gate we required 4 NAND gates as shown below in figure. Thus

here total no. of NAND gates required = 4+1 = 5 SOL 5.127

Option (B) is correct. For TTL worst cases low voltages are VOL (max) = 0.4 V VIL (max) = 0.8 V Worst case high voltages are VOH (min) = 2.4 V VIH (min) = 2 V The difference between maximum input low voltage and maximum output low voltage is called noise margin. It is 0.4 V in case of TTL.

SOL 5.128

Option (D) is correct. From the figure we can see If A =1 then y =1

B=0

x=0 If A =1 B=1 then also y =1 x=0 so for sequence B = 101010....output x and y will be fixed at 0 and 1 respectively.



Digital Circuits

Page 277

SOL 5.129

Option (D) is correct. Given 2’s complement no. 1101; the no. is 0011 for 6 digit output we can write the no. is – 000011 2’s complement representation of above no. is 111101

SOL 5.130


SOL 5.131

Option (B) is correct. An I/O Microprocessor controls data flow between main memory and the I/O device which wants to communicate.

SOL 5.132


SOL 5.133

Option (B) is correct. Dual slope ADC is more accurate.

SOL 5.134

Option (A) is correct. Dual form of any identity can be find by replacing all AND function to OR and vice-versa. so here dual form will be (A + B) (A + C) (B + C) = (A + B) (A + C)

nodia

SOL 5.135

Option (B) is correct. Carry flag will be affected by arithmetic instructions only.

SOL 5.136

Option (C) is correct. This is a synchronous counter. we can find output as QA QB 0 0 1 0 0 1 0 0 h So It counts only three states. It is a mod-3 counter. K =3

SOL 5.137


SOL 5.138

Option (A) is correct. Essential prime implicates for a function is no. of terms that we get by solving K -map. Here we get 4 terms when solve the K -map.

y = B D + A C D + C AB + CA B so no of prime implicates is 4



Digital Circuits

SOL 5.139


SOL 5.140

Option (B) is correct. For a 2 bit multiplier

C3

B1 # A1 A 0 B1 # A1 B1 A1 B 0 C2 C1

Chapter 5

B0 A0 A0 B0 C0

This multiplication is identical to AND operation and then addition. SOL 5.141

Option (C) is correct. In totem pole stage output resistance will be small so it acts like a output buffer.

SOL 5.142

Option (B) is correct. Consider high output state fan out = IOH max = 400 mA = 20 IIH max 20 mA Consider low output state fan out = IOL max = 8 mA = 80 IIL max 0.1 mA

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Thus fan out is 20 SOL 5.143

Option (A) is correct. The given gate is ex-OR so output

F = AB + AB B = 0 so,

Here input

F = A1 + A0 = A

SOL 5.144

Option (C) is correct. EI = Enabled Interput flag ,RST will cause an Interrupt only it we enable EI .

SOL 5.145

Option (A) is correct. Here only for the range 60 to 63 H chipselect will be 0, so peripheral will correspond in this range only chipselect = 1 for rest of the given address ranges.

SOL 5.146

Option (B) is correct. By executing instructions one by one LXI H, 8A79 H (Load HL pair by value 8A79) H = 8AH L = 79 H MOV A, L (copy contain of L to accumulator) A = 79 H ADDH (add contain of H to accumulator) A = 79 H = 0 1111 0 0 1 H = 8AH = add 1 0 0 0 1 0 1 0 = A = 0 0 0 0 0 0 11 Carry = 1 DAA (Carry Flag is set, so DAA adds 6 to high order four bits) 0 1111 0 0 1 DAA add 1 0 0 0 1 0 1 0



Digital Circuits

Page 279

A = 0 0 0 0 0 0 1 1 = 63 H MOV H, A (copy contain of A to H) H = 63 H PCHL (Load program counter by HL pair) PC = 6379 H SOL 5.147


SOL 5.148

Option (C) is correct. NMOS In parallel makes OR Gate & in series makes AND so here we can have F = A (B + C) + DE we took complement because there is another NMOS given above (works as an inverter)

SOL 5.149

Option (D) is correct. For a J -K flip flop we have characteristic equation as Q (t + 1) = JQ (t) + KQ (t) Q (t) & Q (t + 1) are present & next states. In given figure J = Q (t), K = 1 so

nodia

Q (t + 1) = Q (t) Q (t) + 0Q (t) Q (t + 1) = Q (t)[complement of previous state] we have initial input Q (t) = 0

so for 6 clock pulses sequence at output Q will be 010101 SOL 5.150


SOL 5.151

Option (B) is correct. By distributive property in boolean algebra we have (A + BC) = (A + B) (A + C) (A + B) (A + C) = AA + AC + AB + BC

= A (1 + C) + AB + BC = A + AB + BC = A (1 + B) + BC = A + BC SOL 5.152

Option (A) is correct. The current in a p n junction diode is controlled by diffusion of majority carriers while current in schottky diode dominated by the flow of majority carrier over the potential barrier at metallurgical junction. So there is no minority carrier storage in schottky diode, so switching time from forward bias to reverse bias is very short compared to p n junction diode. Hence the propagation delay will reduces.

SOL 5.153


SOL 5.154

Option (D) is correct. The total conversion time for different type of ADC are given as– t is clock period For flash type & 1t Counter type & (2n - t) = 4095 m sec n = no.of bits Integrating type conver time > 4095 m sec successive approximation type nt = 12 m sec



Digital Circuits

here n = 12 nt = 12

Chapter 5

so

12t = 12 so this is succ. app. type ADC. SOL 5.155

Option (D) is correct. LDA 2003 (Load accumulator by a value 2003 H) so here total no. of memory access will be 4. 1 = Fetching instruction 2 = Read the value from memory 1 = write value to accumulator

SOL 5.156

Option (D) is correct. Storage capacitance -12 C = i = 1 # 10 5 - 0. 5 dv b dt l b 20 10-3 l #

nodia -12

= 1 # 10

SOL 5.157

= 4.4 # 10-15 F


or or SOL 5.158

-3

# 20 # 10 4.5

Accuracy ! 1 LSB = Tcoff # DT 2 1 10.24 = T coff # DT 2 # 210 10.24 Tcoff = = 200 mV/cC 2 # 1024 # (50 - 25) cC

Option (D) is correct. 210 # 8 = 13 No. of chips = 26 # 12 2 #4

SOL 5.159

Option (C) is correct. Given instruction set 1000 LXI SP 27FF 1003 CALL 1006 1006 POP H First Instruction will initialize the SP by a value 27FF SP ! 27FF CALL 1006 will “Push PC” and Load PC by value 1006 PUSH PC will store value of PC in stack PC = 1006

now POP H will be executed



Digital Circuits

Page 281

which load HL pair by stack values HL = 1006 and SP = SPl + 2 SP = SPl + 2 = SP - 2 + 2 = SP SP = 27FF ***********

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CHAPTER 6 SIGNALS AND SYSTEMS

2013 MCQ 6.1

ONE MARK

Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected in cascade. Then the overall impulse response of the cascaded system is given by (A) product of h1 ^ t h and h2 ^ t h (B) sum of h1 ^ t h and h2 ^ t h (C) convolution of h1 ^ t h and h2 ^ t h (D) subtraction of h2 ^ t h from h1 ^ t h

MCQ 6.2

The impulse response of a system is h ^ t h = tu ^ t h. For an input u ^t - 1h, the output is 2 t ^t - 1h (A) t u ^ t h (B) u ^t - 1h 2 2

MCQ 6.3

^t - 1h2 u ^t - 1h 2

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(C)

2 (D) t - 1 u ^t - 1h 2

For a periodic signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the fundamental frequency in rad/s (A) 100 (B) 300 (C) 500 (D) 1500

MCQ 6.4

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is (A) 5 kHz (B) 12 kHz (C) 15 kHz (D) 20 kHz

MCQ 6.5

Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system? (A) All the poles of the system must lie on the left side of the jw axis (B) Zeros of the system can lie anywhere in the s-plane (C) All the poles must lie within s = 1 (D) All the roots of the characteristic equation must be located on the left side of the jw axis.

MCQ 6.6

Assuming zero initial condition, the response y ^ t h of the system given below to a unit step input u ^ t h is

(A) u ^ t h 2 (C) t u ^ t h 2

(B) tu ^ t h (D) e-t u ^ t h


MCQ 6.7

Signals and Systems

Page 283

Let g ^ t h = e- pt , and h ^ t h is a filter matched to g ^ t h. If g ^ t h is applied as input to h ^ t h, then the Fourier transform of the output is (B) e- pf /2 (A) e- pf 2

2

(C) e- p f

2

(D) e-2pf

2

2013 MCQ 6.8

MCQ 6.9

MCQ 6.10

TWO MARKS

The impulse response of a continuous time system is given by h ^ t h = d ^t - 1h + d ^t - 3h . The value of the step response at t = 2 is (A) 0 (B) 1 (C) 2 (D) 3 d2 y dy A system described by the differential equation + 5 + 6y ^ t h = x ^ t h. Let dt dt2 x ^ t h be a rectangular pulse given by 1 0
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A system described by a linear, constant coefficient, ordinary, first order differential equation has an exact solution given by y ^ t h for t > 0 , when the forcing function is x ^ t h and the initial condition is y ^0 h. If one wishes to modify the system so that the solution becomes - 2y ^ t h for t > 0 , we need to (A) change the initial condition to - y ^0 h and the forcing function to 2x ^ t h (B) change the initial condition to 2y ^0 h and the forcing function to - x ^ t h (C) change the initial condition to j 2 y ^0 h and the forcing function to j 2 x^t h (D) change the initial condition to - 2y ^0 h and the forcing function to - 2x ^ t h

MCQ 6.11

The DFT of a vector 8a b c dB is the vector 8a b g dB . Consider the product R V Sa b c d W Sd a b c W 8p q r sB = 8a b c dBSc d a b W S W Sb c d aW T X The DFT of the vector 8p q r sB is a scaled version of (A) 9a2 b2 g2 d2C (B) 9 a b g dC (C) 8a + b b + d d + g g + aB (D) 8a b g dB 2012

MCQ 6.12

ONE MARK

The unilateral Laplace transform of f (t) is 2 1 . The unilateral Laplace s +s+1 transform of tf (t) is (B) - 2 2s + 1 2 (A) - 2 s (s + s + 1) 2 (s + s + 1)



Signals and Systems

(C) MCQ 6.13

s (s2 + s + 1) 2

(D)

Chapter 6

2s + 1 (s2 + s + 1) 2

If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (B) 1 < z < 1 (A) 1 < z < 3 3 3 2 (C) 1 < z < 3 (D) 1 < z 2 3 2012

MCQ 6.14

TWO MARKS

The input x (t) and output y (t) of a system are related as y (t) = The system is (A) time-invariant and stable (C) time-invariant and not stable

t

# x (t) cos (3t) dt . -3

(B) stable and not time-invariant (D) not time-invariant and not stable

MCQ 6.15

The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1 (D) 2

MCQ 6.16

Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (C) 1

nodia (B) 1/2 (D) 3/2

2011 MCQ 6.17

ONE MARK d2 y dt2

dy dt

The differential equation 100 - 20 + y = x (t) describes a system with an input x (t) and an output y (t). The system, which is initially relaxed, is excited by a unit step input. The output y ^ t h can be represented by the waveform



Signals and Systems

Page 285

MCQ 6.18

The trigonometric Fourier series of an even function does not have the (A) dc term (B) cosine terms (C) sine terms (D) odd harmonic terms

MCQ 6.19

A system is defined by its impulse response h (n) = 2n u (n - 2). The system is (A) stable and causal (B) causal but not stable (C) stable but not causal (D) unstable and non-causal

MCQ 6.20

If the unit step response of a network is (1 - e- at), then its unit impulse response is (B) a-1 e- at (A) aeat (C) (1 - a-1) e- at (D) (1 - a) e- at 2011

MCQ 6.21

TWO MARKS

An input x (t) = exp (- 2t) u (t) + d (t - 6) is applied to an LTI system with impulse response h (t) = u (t) . The output is (A) [1 - exp (- 2t)] u (t) + u (t + 6) (B) [1 - exp (- 2t)] u (t) + u (t - 6)

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(C) 0.5 [1 - exp (- 2t)] u (t) + u (t + 6) (D) 0.5 [1 - exp (- 2t)] u (t) + u (t - 6) MCQ 6.22

Two systems H1 (Z ) and H2 (Z ) are connected in cascade as shown below. The overall output y (n) is the same as the input x (n) with a one unit delay. The transfer function of the second system H2 (Z ) is

1 - 0.6z-1 z (1 - 0.4z-1) z-1 (1 - 0.4z-1) (C) (1 - 0.6z-1) (A)

MCQ 6.23

-1

(B)

z-1 (1 - 0.6z-1) (1 - 0.4z-1)

(D)

1 - 0.4 z-1 z (1 - 0.6z-1)

The first six points of the 8-point DFT of a real valued sequence are 5, 1 - j 3, 0, 3 - j 4, 0 and 3 + j 4 . The last two points of the DFT are respectively (B) 0, 1 + j 3 (A) 0, 1 - j 3 (C) 1 + j3, 5 (D) 1 - j 3, 5 2010

MCQ 6.24

-1

ONE MARK 2

-1

Consider the z -transform x (z) = 5z + 4z + 3; 0 < z < 3. The inverse z transform x [n] is (A) 5d [n + 2] + 3d [n] + 4d [n - 1] (B) 5d [n - 2] + 3d [n] + 4d [n + 1] (C) 5u [n + 2] + 3u [n] + 4u [n - 1] (D) 5u [n - 2] + 3u [n] + 4u [n + 1]



MCQ 6.25

Signals and Systems

Chapter 6

The trigonometric Fourier series for the waveform f (t) shown below contains

(A) (B) (C) (D)

only only only only

cosine terms and zero values for the dc components cosine terms and a positive value for the dc components cosine terms and a negative value for the dc components sine terms and a negative value for the dc components

MCQ 6.26

Two discrete time system with impulse response h1 [n] = d [n - 1] and h2 [n] = d [n - 2] are connected in cascade. The overall impulse response of the cascaded system is (A) d [n - 1] + d [n - 2] (B) d [n - 4] (C) d [n - 3] (D) d [n - 1] d [n - 2]

MCQ 6.27

For a N -point FET algorithm N = 2m which one of the following statements is TRUE ? (A) It is not possible to construct a signal flow graph with both input and output in normal order (B) The number of butterflies in the m th stage in N/m (C) In-place computation requires storage of only 2N data (D) Computation of a butterfly requires only one complex multiplication.

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2010 MCQ 6.28

TWO MARKS

3s + 1 Given f (t) = L-1 ; 3 . If lim f (t) = 1, then the value of k is t"3 s + 4s2 + (k - 3) s E (A) 1 (B) 2 (C) 3 (D) 4

MCQ 6.29

A continuous time LTI system is described by dx (t) d 2 y (t) dy (t) + 4x (t) +4 + 3y (t) = 2 dt dt dt 2 Assuming zero initial conditions, the response y (t) of the above system for the input x (t) = e-2t u (t) is given by (B) (e-t - e-3t) u (t) (A) (et - e3t) u (t) (C) (e-t + e-3t) u (t) (D) (et + e3t) u (t)

MCQ 6.30

The transfer function of a discrete time LTI system is given by 2 - 34 z-1 H (z) = 1 - 34 z-1 + 18 z-2 Consider the following statements: S1: The system is stable and causal for ROC: z > 1/2 S2: The system is stable but not causal for ROC: z < 1/4



Signals and Systems

Page 287

S3: The system is neither stable nor causal for ROC: 1/4 < z < 1/2 Which one of the following statements is valid ? (A) Both S1 and S2 are true (B) Both S2 and S3 are true (C) Both S1 and S3 are true (D) S1, S2 and S3 are all true 2009

ONE MARK

MCQ 6.31

The Fourier series of a real periodic function has only (P) cosine terms if it is even (Q) sine terms if it is even (R) cosine terms if it is odd (S) sine terms if it is odd Which of the above statements are correct ? (A) P and S (B) P and R (C) Q and S (D) Q and R

MCQ 6.32

A function is given by f (t) = sin2 t + cos 2t . Which of the following is true ? (A) f has frequency components at 0 and 21p Hz (B) f has frequency components at 0 and p1 Hz (C) f has frequency components at 21p and p1 Hz

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(D) f has frequency components at MCQ 6.33

0. 1 2p

and

1 p

Hz

The ROC of z -transform of the discrete time sequence n n x (n) = b 1 l u (n) - b 1 l u (- n - 1) is 3 2 1 1 (A) < z < (B) z > 1 3 2 2 1 (C) z < (D) 2 < z < 3 3 2009

MCQ 6.34

TWO MARKS

Given that F (s) is the one-side Laplace transform of f (t), the Laplace transform of

t

#0 f (t) dt is

(A) sF (s) - f (0) (C) MCQ 6.35

#0

s

F (t) dt

(B) 1 F (s) s (D) 1 [F (s) - f (0)] s

A system with transfer function H (z) has impulse response h (.) defined as h (2) = 1, h (3) =- 1 and h (k) = 0 otherwise. Consider the following statements. S1 : H (z) is a low-pass filter. S2 : H (z) is an FIR filter. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, and S2 is a reason for S1 (D) Both S1 and S2 are true, but S2 is not a reason for S1



MCQ 6.36

Signals and Systems

Chapter 6

Consider a system whose input x and output y are related by the equation y (t) =

# x (t - t) g (2t) dt where h (t) is shown in the graph. 3

-3

Which of the following four properties are possessed by the system ? BIBO : Bounded input gives a bounded output. Causal : The system is causal, LP : The system is low pass. LTI : The system is linear and time-invariant. (A) Causal, LP (B) BIBO, LTI (C) BIBO, Causal, LTI (D) LP, LTI MCQ 6.37

The 4-point Discrete Fourier Transform (DFT) of a discrete time sequence {1,0,2,3} is (A) [0, - 2 + 2j , 2, - 2 - 2j ] (B) [2, 2 + 2j , 6, 2 - 2j ] (C) [6, 1 - 3j , 2, 1 + 3j ] (D) [6, - 1 + 3j , 0, - 1 - 3j ]

MCQ 6.38

An LTI system having transfer function s +s 2+s 1+ 1 and input x (t) = sin (t + 1) is in steady state. The output is sampled at a rate ws rad/s to obtain the final output {x (k)}. Which of the following is true ? (A) y (.) is zero for all sampling frequencies ws (B) y (.) is nonzero for all sampling frequencies ws

nodia 2

2

(C) y (.) is nonzero for ws > 2 , but zero for ws < 2 (D) y (.) is zero for ws > 2 , but nonzero for w2 < 2 2008

ONE MARK

MCQ 6.39

The input and output of a continuous time system are respectively denoted by x (t) and y (t). Which of the following descriptions corresponds to a causal system ? (A) y (t) = x (t - 2) + x (t + 4) (B) y (t) = (t - 4) x (t + 1) (C) y (t) = (t + 4) x (t - 1) (D) y (t) = (t + 5) x (t + 5)

MCQ 6.40

The impulse response h (t) of a linear time invariant continuous time system is described by h (t) = exp (at) u (t) + exp (bt) u (- t) where u (- t) denotes the unit step function, and a and b are real constants. This system is stable if (A) a is positive and b is positive (B) a is negative and b is negative (C) a is negative and b is negative (D) a is negative and b is positive 2008

MCQ 6.41

TWO MARKS

A linear, time - invariant, causal continuous time system has a rational transfer function with simple poles at s =- 2 and s =- 4 and one simple zero at s =- 1.



Signals and Systems

Page 289

A unit step u (t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is (A) [exp (- 2t) + exp (- 4t)] u (t) (B) [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u (t) (C) [- 4 exp (- 2t) + 12 exp (- 4t)] u (t) (D) [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u (t) MCQ 6.42

The signal x (t) is described by 1 for - 1 # t # + 1 x (t) = ) 0 otherwise Two of the angular frequencies at which its Fourier transform becomes zero are (A) p, 2p (B) 0.5p, 1.5p (C) 0, p (D) 2p, 2.5p

MCQ 6.43

A discrete time linear shift - invariant system has an impulse response h [n] with h [0] = 1, h [1] =- 1, h [2] = 2, and zero otherwise The system is given an input sequence x [n] with x [0] = x [2] = 1, and zero otherwise. The number of nonzero samples in the output sequence y [n], and the value of y [2] are respectively (A) 5, 2 (B) 6, 2 (C) 6, 1 (D) 5, 3

MCQ 6.44

Let x (t) be the input and y (t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4 Properties Relations P1 : Linear but NOT time - invariant R1 : y (t) = t2 x (t) P2 : Time - invariant but NOT linear R2 : y (t) = t x (t) P3 : Linear and time - invariant R3 : y (t) = x (t) R4 : y (t) = x (t - 5) (A) (P1, R1), (P2, R3), (P3, R4) (B) (P1, R2), (P2, R3), (P3, R4) (C) (P1, R3), (P2, R1), (P3, R2) (D) (P1, R1), (P2, R2), (P3, R3)

MCQ 6.45

{x (n)} is a real - valued periodic sequence with a period N . x (n) and X (k) form N-point Discrete Fourier Transform (DFT) pairs. The DFT Y (k) of the sequence N-1 y (n) = 1 / x (r) x (n + r) is N r=0 N-1 (B) 1 / X (r) X (k + r) (A) X (k) 2 N r=0

nodia

N-1 (C) 1 / X (r) X (k + r) N r=0

(D) 0

Statement for Linked Answer Question 46 and 47: In the following network, the switch is closed at t = 0- and the sampling starts from t = 0 . The sampling frequency is 10 Hz.



Signals and Systems

Chapter 6

MCQ 6.46

The samples x (n), n = (0, 1, 2, ...) are given by (A) 5 (1 - e-0.05n) (B) 5e-0.05n (C) 5 (1 - e-5n) (D) 5e-5n

MCQ 6.47

The expression and the region of convergence of the z -transform of the sampled signal are 5z , z < e-0.05 (B) (A) 5z 5 , z < e-5 z - e-0.05 z-e 5z , z > e-0.05 (C) (D) 5z -5 , z > e-5 z-e z - e-0.05

Statement for Linked Answer Question 48 & 49: The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t), where u (t) denotes the unit step function. MCQ 6.48

The frequency response H (w) of this system in terms of angular frequency w, is given by H (w) 1 (A) (B) sin w 1 + j2w w jw 1 (C) (D) 2 + jw 2 + jw

MCQ 6.49

The output of this system, to the sinusoidal input x (t) = 2 cos 2t for all time t , is (A) 0 (B) 2-0.25 cos (2t - 0.125p) (C) 2-0.5 cos (2t - 0.125p) (D) 2-0.5 cos (2t - 0.25p) 2007

MCQ 6.50

nodia

ONE MARK

1 If the Laplace transform of a signal Y (s) = , then its final value is s (s - 1) (A) - 1 (B) 0 (C) 1

(D) Unbounded

2007 MCQ 6.51

TWO MARKS -t

The 3-dB bandwidth of the low-pass signal e u (t), where u (t) is the unit step function, is given by (B) 1 (A) 1 Hz 2 - 1 Hz 2p 2p (C) 3

MCQ 6.52

A 5-point sequence x [n] is given as x [- 3] = 1, x [- 2] = 1, x [- 1] = 0, x [0] = 5 and x [1] = 1. Let X (eiw) denoted the discrete-time Fourier transform of x [n]. The value of (A) 5 (C) 16p

MCQ 6.53

(D) 1 Hz

#

p

-p

X (e jw) dw is (B) 10p (D) 5 + j10p

The z -transform X (z) of a sequence x [n] is given by X [z] = 1 -0.25z . It is given that the region of convergence of X (z) includes the unit circle. The value of x [0] is (B) 0 (A) - 0.5 (C) 0.25 (D) 05 -1



MCQ 6.54

MCQ 6.55

Signals and Systems

A Hilbert transformer is a (A) non-linear system (C) time-varying system

Page 291

(B) non-causal system (D) low-pass system

The frequency response of a linear, time-invariant system is given by H (f) = 1 + j510pf . The step response of the system is t (A) 5 (1 - e-5t) u (t) (B) 5 61 - e- 5@ u (t) t (C) 1 (1 - e-5t) u (t) (D) 1 ^1 - e- 5 h u (t) 2 5 2006

ONE MARK

MCQ 6.56

Let x (t) * X (jw) be Fourier Transform pair. The Fourier Transform of the signal x (5t - 3) in terms of X (jw) is given as j3w j3w jw jw (A) 1 e- 5 X b l (B) 1 e 5 X b l 5 5 5 5 jw jw (C) 1 e-j3w X b l (D) 1 e j3w X b l 5 5 5 5

MCQ 6.57

The Dirac delta function d (t) is defined as 1 t=0 (A) d (t) = ) 0 otherwise 3 t=0 (B) d (t) = ) 0 otherwise 3 1 t=0 (C) d (t) = ) and # d (t) dt = 1 -3 0 otherwise 3 3 t=0 (D) d (t) = ) and # d (t) dt = 1 -3 0 otherwise If the region of convergence of x1 [n] + x2 [n] convergence of x1 [n] - x2 [n] includes (A) 1 < z < 3 (B) 3 (C) 3 < z < 3 (D) 2

MCQ 6.58

MCQ 6.59

nodia

is 1 < z < 2 then the region of 3 3 2 < z <3 3 1< z <2 3 3

In the system shown below, x (t) = (sin t) u (t) In steady-state, the response y (t) will be

1 sin t - p ` 4j 2 (C) 1 e-t sin t 2 (A)

(B)

1 sin t + p ` 4j 2

(D) sin t - cos t

2006 MCQ 6.60

TWO MARKS

Consider the function f (t) having Laplace transform F (s) = 2 w0 2 Re [s] > 0 s + w0



Signals and Systems

The final value of f (t) would be (A) 0 (C) - 1 # f (3) # 1

Chapter 6

(B) 1 (D) 3

MCQ 6.61

A system with input x [n] and output y [n] is given as y [n] = (sin 56 pn) x [n]. The system is (A) linear, stable and invertible (B) non-linear, stable and non-invertible (C) linear, stable and non-invertible (D) linear, unstable and invertible

MCQ 6.62

The unit step response of a system starting from rest is given by c (t) = 1 - e-2t for t $ 0 . The transfer function of the system is 1 (A) (B) 2 1 + 2s 2+s 1 (C) (D) 2s 2+s 1 + 2s

MCQ 6.63

The unit impulse response of a system is f (t) = e-t, t $ 0 . For this system the steady-state value of the output for unit step input is equal to (A) - 1 (B) 0 (C) 1 (D) 3 2005

nodia

ONE MARK

MCQ 6.64

Choose the function f (t); - 3 < t < 3 for which a Fourier series cannot be defined. (A) 3 sin (25t) (B) 4 cos (20t + 3) + 2 sin (710t) (C) exp (- t ) sin (25t) (D) 1

MCQ 6.65

The function x (t) is shown in the figure. Even and odd parts of a unit step function u (t) are respectively,

MCQ 6.66

MCQ 6.67

(A) 1 , 1 x (t) (B) - 1 , 1 x (t) 2 2 2 2 1 1 (C) , - x (t) (D) - 1 , - 1 x (t) 2 2 2 2 The region of convergence of z - transform of the sequence 5 n 6 n b 6 l u (n) - b 5 l u (- n - 1) must be (A) z < 5 (B) z > 5 6 6 5 6 6 (C) < z < (D) < z < 3 6 5 5 Which of the following can be impulse response of a causal system ?



Signals and Systems

Page 293

MCQ 6.68

Let x (n) = ( 12 ) n u (n), y (n) = x2 (n) and Y (e jw) be the Fourier transform of y (n) then Y (e j0) (A) 1 (B) 2 4 (C) 4 (D) 4 3

MCQ 6.69

The power in the signal s (t) = 8 cos (20p - p2 ) + 4 sin (15pt) is (A) 40 (B) 41 (C) 42 (D) 82 2005

MCQ 6.70

nodia

TWO MARKS

The output y (t) of a linear time invariant system is related to its input x (t) by the following equations y (t)= 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td + T)

The filter transfer function H (w) of such a system is given by (B) (1 + 0.5 cos wT) e-jwt (A) (1 + cos wT) e-jwt (C) (1 - cos wT) e-jwt (D) (1 - 0.5 cos wT) e-jwt MCQ 6.71

d

d

d

Match the following and choose the correct combination. Group 1 E. Continuous and aperiodic signal F. Continuous and periodic signal G. Discrete and aperiodic signal H. Discrete and periodic signal Group 2 1. Fourier representation is continuous and aperiodic 2. Fourier representation is discrete and aperiodic 3. Fourier representation is continuous and periodic 4. Fourier representation is discrete and periodic (A) E - 3, (B) E - 1, (C) E - 1, (D) E - 2,

MCQ 6.72

d

F - 2, G - 4, H - 1 F - 3, G - 2, H - 4 F - 2, G - 3, H - 4 F - 1, G - 4, H - 3

A signal x (n) = sin (w0 n + f) is the input to a linear time- invariant system having a frequency response H (e jw). If the output of the system Ax (n - n0) then the most general form of +H (e jw) will be



Signals and Systems

Chapter 6

(A) - n0 w0 + b for any arbitrary real (B) - n0 w0 + 2pk for any arbitrary integer k (C) n0 w0 + 2pk for any arbitrary integer k (D) - n0 w0 f

Statement of linked answer question 73 and 74 : A sequence x (n) has non-zero values as shown in the figure.

x ( n2 - 1),

MCQ 6.73

The sequence y (n) = *

will be

nodia 0,

MCQ 6.74

For n even For n odd

The Fourier transform of y (2n) will be (A) e-j2w [cos 4w + 2 cos 2w + 2] (C) e-jw [cos 2w + 2 cos w + 2]

(B) cos 2w + 2 cos w + 2 (D) e-j2w [cos 2w + 2 cos + 2]



MCQ 6.75

Signals and Systems

Page 295

For a signal x (t) the Fourier transform is X (f). Then the inverse Fourier transform of X (3f + 2) is given by j4pt (A) 1 x` t j e j3pt (B) 1 x` t j e - 3 2 2 3 3 (C) 3x (3t) e-j4pt

(D) x (3t + 2)

2004 MCQ 6.76

ONE MARK

The impulse response h [n] of a linear time-invariant system is given by h [n] = u [n + 3] + u [n - 2) - 2n [n - 7] where u [n] is the unit step sequence. The above system is (A) stable but not causal (B) stable and causal (C) causal but unstable

(D) unstable and not causal

MCQ 6.77

The z -transform of a system is H (z) = z -z0.2 . If the ROC is z < 0.2 , then the impulse response of the system is (A) (0.2) n u [n] (B) (0.2) n u [- n - 1] n (C) - (0.2) u [n] (D) - (0.2) n u [- n - 1]

MCQ 6.78

The Fourier transform of a conjugate symmetric function is always (A) imaginary (B) conjugate anti-symmetric (C) real (D) conjugate symmetric 2004

nodia

TWO MARKS

MCQ 6.79

Consider the sequence x [n] = [- 4 - j51 + j25]. The conjugate anti-symmetric part of the sequence is (A) [- 4 - j2.5, j2, 4 - j2.5] (B) [- j2.5, 1, j2.5] (C) [- j2.5, j2, 0] (D) [- 4, 1, 4]

MCQ 6.80

A causal LTI system is described by the difference equation 2y [n] = ay [n - 2] - 2x [n] + bx [n - 1] The system is stable only if (B) a > 2, b > 2 (A) a = 2 , b < 2 (C) a < 2 , any value of b (D) b < 2 , any value of a

MCQ 6.81

The impulse response h [n] of a linear time invariant system is given as -2 2

n = 1, - 1

h [ n] = * 4 2 n = 2, - 2 0 otherwise If the input to the above system is the sequence e jpn/4 , then the output is (B) 4 2 e-jpn/4 (A) 4 2 e jpn/4 j p n /4 (C) 4e (D) - 4e jpn/4 MCQ 6.82

Let x (t) and y (t) with Fourier transforms F (f) and Y (f) respectively be related as shown in Fig. Then Y (f) is



Signals and Systems

(A) - 1 X (f/2) e-jpf 2

(B) - 1 X (f/2) e j2pf 2

(C) - X (f/2) e j2pf

(D) - X (f/2) e-j2pf

2003 MCQ 6.83

MCQ 6.84

Chapter 6

ONE MARK

2 The Laplace transform of i (t) is given by I (s) = At t " 3, The value of s (1 + s) i (t) tends to (A) 0 (B) 1 (C) 2 (D) 3

nodia

The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = / cn e j2pf t . It is given that c3 = 3 + j5 . Then c-3 is n =- 3 (A) 5 + j3 (B) - 3 - j5 (C) - 5 + j3 (D) 3 - j5 0

MCQ 6.85

Let x (t) be the input to a linear, time-invariant system. The required output is 4p (t - 2). The transfer function of the system should be (A) 4e j4pf (B) 2e-j8pf (C) 4e-j4pf (D) 2e j8pf

MCQ 6.86

A sequence x (n) with the z -transform X (z) = z 4 + z2 - 2z + 2 - 3z-4 is applied as an input to a linear, time-invariant system with the impulse response h (n) = 2d (n - 3) where 1, n = 0 d (n) = ) 0, otherwise The output at n = 4 is (B) zero (A) - 6 (C) 2 (D) - 4 2003

MCQ 6.87

TWO MARKS

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship, x (n) n$1 y (n) = *0, n= 0 x (n + 1) n # - 1 where x (n) is the input and y (n) is the output. The above system has the properties (A) P, S but not Q, R (B) P, Q, S but not R (C) P, Q, R, S (D) Q, R, S but not P



Signals and Systems

Page 297

Common Data For Q. 88 & 89 : The system under consideration is an RC low-pass filter (RC-LPF) with R = 1 kW and C = 1.0 mF. MCQ 6.88

Let H (f) denote the frequency response of the RC-LPF. Let f1 be the highest H (f1) frequency such that 0 # f # f1 $ 0.95 . Then f1 (in Hz) is H (0) (A) 324.8 (B) 163.9 (C) 52.2 (D) 104.4

MCQ 6.89

Let tg (f) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg (f2) in ms, is (A) 0.717 (B) 7.17 (C) 71.7 (D) 4.505 2002

ONE MARK

MCQ 6.90

Convolution of x (t + 5) with impulse function d (t - 7) is equal to (A) x (t - 12) (B) x (t + 12) (C) x (t - 2) (D) x (t + 2)

MCQ 6.91

Which of the following cannot be the Fourier series expansion of a periodic signal? (A) x (t) = 2 cos t + 3 cos 3t (B) x (t) = 2 cos pt + 7 cos t (C) x (t) = cos t + 0.5 (D) x (t) = 2 cos 1.5pt + sin 3.5pt

MCQ 6.92

The Fourier transform F {e-1 u (t)} is equal to

nodia

(A) e f u (f) (C) e f u (- f) MCQ 6.93

1 1 . Therefore, F ' is 1 + j 2p f 1 + j 2p t 1 -f (B) e u (f) (D) e-f u (- f)

A linear phase channel with phase delay Tp and group delay Tg must have (A) Tp = Tg = constant (B) Tp \ f and Tg \ f (C) Tp = constant and Tg \ f ( f denote frequency) (D) Tp \ f and Tp = constant 2002

TWO MARKS 5-s s2 - s - 2

MCQ 6.94

The Laplace transform of continuous - time signal x (t) is X (s) = Fourier transform of this signal exists, the x (t) is (B) - e2t u (- t) + 2e-t u (t) (A) e2t u (t) - 2e-t u (t) (C) - e2t u (- t) - 2e-t u (t) (D) e2t u (- t) - 2e-t u (t)

MCQ 6.95

If the impulse response of discrete - time system is h [n] =- 5n u [- n - 1], then the system function H (z) is equal to (A) - z and the system is stable (B) z and the system is stable z-5 z-5 z (C) and the system is unstable (D) z and the system is unstable z-5 z-5

. If the



Signals and Systems

Chapter 6

2001 MCQ 6.96

ONE MARK

The transfer function of a system is given by H (s) = 2 1 . The impulse s (s - 2) response of the system is (A) (t2 * e-2t) u (t) (B) (t * e2t) u (t) -2 (C) (te t) u (t) (D) (te-2t) u (t)

MCQ 6.97

The region of convergence of the z - transform of a unit step function is (B) z < 1 (A) z > 1 (C) (Real part of z ) > 0 (D) (Real part of z ) < 0

MCQ 6.98

Let d (t) denote the delta function. The value of the integral

d (t) cos b 3t l dt is 2

If a signal f (t) has energy E , the energy of the signal f (2t) is equal to (A) 1 (B) E/2 (C) 2E (D) 4E 2001

MCQ 6.100

3

-3

(B) - 1 (D) p2

(A) 1 (C) 0 MCQ 6.99

#

nodia

TWO MARKS

The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by u (t) and h 4 (t) = e-3t u (t) h1 (t) = 1, h2 (t) = u (t), h3 (t) = t+1 where u (t) is the unit step function. Which of these systems is time invariant, causal, and stable? (A) S1 (B) S2 (C) S3 (D) S4 2000

MCQ 6.101

s2 + 1 Given that L [f (t)] = s2+ 2 , L [g (t)] = and h (t) = ( s + 3) (s + 2) s +1 L [h (t)] is 2 (A) s + 1 s+3

(C) MCQ 6.102

ONE MARK

s2 + 1 + s+2 (s + 3)( s + 2) s2 + 1

(B)

t

#0 f (t) g (t - t) dt .

1 s+3

(D) None of the above 2

The Fourier Transform of the signal x (t) = e-3t is of the following form, where A and B are constants : (B) Ae-Bf (A) Ae-B f (C) A + B f 2 (D) Ae-Bf 2

MCQ 6.103

A system with an input x (t) and output y (t) is described by the relations : y (t) = tx (t). This system is (A) linear and time - invariant (B) linear and time varying (C) non - linear and time - invariant (D) non - linear and time - varying



MCQ 6.104

Signals and Systems

Page 299

A linear time invariant system has an impulse response e2t, t > 0 . If the initial conditions are zero and the input is e3t , the output for t > 0 is (B) e5t (A) e3t - e2t (C) e3t + e2t (D) None of these 2000

MCQ 6.105

TWO MARKS

One period (0, T) each of two periodic waveforms W1 and W2 are shown in the figure. The magnitudes of the nth Fourier series coefficients of W1 and W2 , for n $ 1, n odd, are respectively proportional to

(A) n-3 and n-2 (C) n-1 and n-2

(B) n-2 and n-3 (D) n-4 and n-2

nodia

MCQ 6.106

Let u (t) be the step function. Which of the waveforms in the figure corresponds to the convolution of u (t) - u (t - 1) with u (t) - u (t - 2) ?

MCQ 6.107

A system has a phase response given by f (w), where w is the angular frequency. The phase delay and group delay at w = w0 are respectively given by df (w) f (w0) d2 f (w0) (B) f (wo), (A) , dw w = w w0 dw2 w = w w df (w) (C) wo , (D) wo f (wo), # f (l) f (wo) d (w) w = w -3 0

o

o

o

1999 MCQ 6.108

ONE MARK

The z -transform F (z) of the function f (nT) = anT is (B) z T (A) z T z-a z+a z z (C) (D) z - a-T z + a-T



Signals and Systems

Chapter 6

MCQ 6.109

If [f (t)] = F (s), then [f (t - T)] is equal to (A) esT F (s) (B) e-sT F (s) F (s) F (s) (C) (D) sT 1-e 1 - e-sT

MCQ 6.110

A signal x (t) has a Fourier transform X (w). If x (t) is a real and odd function of t , then X (w) is (A) a real and even function of w (B) a imaginary and odd function of w (C) an imaginary and even function of w (D) a real and odd function of w 1999

MCQ 6.111

TWO MARKS

The Fourier series representation of an impulse train denoted by s (t) =

3

/ d (t - nT0) is given by

n =- 3

nodia

3 j2pnt (A) 1 / exp T0 n =- 3 T0 3 j nt p (C) 1 / exp T0 n =- 3 T0

MCQ 6.112

3 jpnt (B) 1 / exp T0 n =- 3 T0 3 j 2 nt p (D) 1 / exp T0 n =- 3 T0

(A) 1/4

1z-1 (1 - z-4) . Its final value is 4 (1 - z-1) 2 (B) zero

(C) 1.0

(D) infinity

The z -transform of a signal is given by C (z) =

1998

ONE MARK

w , then the value of Limf (t) t"3 s2 + w2 (A) cannot be determined (B) is zero (C) is unity (D) is infinite

MCQ 6.113

If F (s) =

MCQ 6.114

The trigonometric Fourier series of a even time function can have only (A) cosine terms (B) sine terms (C) cosine and sine terms (D) d.c and cosine terms

MCQ 6.115

1, t < T1 A periodic signal x (t) of period T0 is given by x (t) = * 0, T1 < t < T0 2 The dc component of x (t) is (A) T1 T0 (C) 2T1 T0

MCQ 6.116

(B) T1 2T0 (D) T0 T1

The unit impulse response of a linear time invariant system is the unit step function u (t). For t > 0 , the response of the system to an excitation e-at u (t), a > 0 will be (B) (1/a) (1 - e-at) (A) ae-at (C) a (1 - e-at) (D) 1 - e-at



MCQ 6.117

Signals and Systems

The z-transform of the time function

3

/ d (n - k) is k=0

(A) z - 1 z

Page 301

(B)

z z-1

(z - 1) 2 z (D) 2 z (z - 1) A distorted sinusoid has the amplitudes A1, A2, A 3, .... of the fundamental, second harmonic, third harmonic,..... respectively. The total harmonic distortion is 2 2 (A) A2 + A 3 + .... (B) A 2 + A 3 + ..... A1 A1 (C)

MCQ 6.118

(C) MCQ 6.119

2 2 (D) c A 2 + A 3 + ..... m A

A 22 + A 32 + ..... A 12 + A 22 + A 32 + ....

1

The Fourier transform of a function x (t) is X (f). The Fourier transform of will be dX (f) df (C) jfX (f) (A)

1997 MCQ 6.120

(B) j2pfX (f) X (f) (D) jf

nodia

ONE MARK

The function f (t) has the Fourier Transform g (w). The Fourier Transform ff (t) g (t) e =

(A) 1 f (w) 2p

3

# g (t) e-jwt dt o is

-3

(C) 2pf (- w) MCQ 6.121

dX (t) df

(B) 1 f (- w) 2p


The Laplace Transform of eat cos (at) is equal to (s - a) (s + a) (A) (B) (s - a) 2 + a2 (s - a) 2 + a2 1 (C) (D) None of the above (s - a) 2 1996

ONE MARK

MCQ 6.122

The trigonometric Fourier series of an even function of time does not have the (A) dc term (B) cosine terms (C) sine terms (D) odd harmonic terms

MCQ 6.123

The Fourier transform of a real valued time signal has (A) odd symmetry (B) even symmetry (C) conjugate symmetry

(D) no symmetry

***********



Signals and Systems

Chapter 6

SOLUTIONS SOL 6.1

Option (C) is correct. If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

SOL 6.2

Option (C) is correct. Given, the input x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s

nodia

The impulse response of system is given

h^t h = t u^t h Its Laplace transform is H ^s h = 12 s Hence, the overall response at the output is -s Y ^s h = X ^s h H ^s h = e 3 s Its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 SOL 6.3

Option (A) is correct. Given, the signal

v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h

So we have w1 = 100 rad/s , w2 = 300 rad/s and w3 = 500 rad/s Therefore, the respective time periods are T1 = 2p = 2p sec , T2 = 2p = 2p sec and T3 = 2p sec 300 w1 100 w2 500 So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h or, T0 = 2p 100 Hence, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10



SOL 6.4

Signals and Systems

Page 303

Option (A) is correct. Given, the maximum frequency of the band-limited signal fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy fs $ fN fs $ 10 kHz only the option (A) doesn’t satisfy the condition therefore, 5 kHz is not a valid sampling frequency.

SOL 6.5

Option (C) is correct. For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies an LTI system stability and causality both. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also.

SOL 6.6

Option (B) is correct. The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l = 12 s s s For zero initial condition, we check dy ^ t h u^t h = dt U ^s h = SY ^s h - y ^0 h & & U ^s h = s c 12 m - y ^0 h s 1 or, U ^s h = s Hence, the O/P is correct which is Y ^s h = 12 s its inverse Laplace transform is given by y ^ t h = tu ^ t h

SOL 6.7

nodia ^y ^0 h = 0h

No Option is correct. The matched filter is characterized by a frequency response that is given as H ^ f h = G * ^ f h exp ^- j2pfT h f where g^t h G^f h Now, consider a filter matched to a known signal g ^ t h. The fourier transform of the resulting matched filter output g 0 ^ t h will be G 0 ^ f h = H ^ f h G ^ f h = G * ^ f h G ^ f h exp ^- j2pfT h = G ^ f h 2 exp ^- j2pfT h T is duration of g ^ t h



Signals and Systems

Chapter 6

Assume exp ^- j2pfT h = 1 So, G0 ^ f h = G_ f i 2 Since, the given Gaussian function is g ^ t h = e- pt Fourier transform of this signal will be f g ^ t h = e- pt e- pf = G ^ f h Therefore, output of the matched filter is 2 G 0 ^ f h = e- pf 2

2

2

2

SOL 6.8

Option (B) is correct. Given, the impulse response of continuous time system h ^ t h = d ^t - 1h + d ^t - 3h From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as

y ^ t h = u ^ t h * h ^ t h = u ^ t h * 6d ^t - 1h + d ^t - 3h@

nodia = u ^t - 1h + u ^t - 3h

At t = 2 SOL 6.9

y ^2 h = u ^2 - 1h + u ^2 - 3h = 1

Option (B) is correct. Given, the differential equation d2y dy + 5 + 6y ^ t h = x ^ t h dt dt2

Taking its Laplace transform with zero initial conditions, we have

s2 Y ^s h + 5sY ^s h + 6Y ^s h = X ^s h Now, the input signal is 1 0
....(1)

Option (D) is correct. The solution of a system described by a linear, constant coefficient, ordinary, first order differential equation with forcing function x ^ t h is y ^ t h so, we can define a function relating x ^ t h and y ^ t h as below dy P + Qy + K = x ^ t h dt where P , Q , K are constant. Taking the Laplace transform both the sides, we get ....(1) P sY ^s h - Py ^0 h + Q Y ^s h = X ^s h Now, the solutions becomes y1 ^ t h =- 2y ^ t h



Signals and Systems

Page 305

or, Y1 ^s h =- 2Y ^s h So, Eq. (1) changes to

P sY1 ^s h - P y1 ^0 h + Q Y1 ^s h = X1 ^s h or, ....(2) - 2PSY ^s h - P y1 ^0 h - 2QY1 ^s h = X1 ^s h Comparing Eq. (1) and (2), we conclude that X1 ^s h =- 2X ^s h y1 ^0 h =- 2y ^0 h Which makes the two equations to be same. Hence, we require to change the initial condition to - 2y ^0 h and the forcing equation to - 2x ^ t h SOL 6.11

Option (A) is correct. Given, the DFT of vector 8a b c dB as D.F.T. %8a b c dB/ = 8a b g dB Also, we have R V Sa b c d W Sd a b c W ...(1) 8p q r sB = 8a b c dBSc d a b W S W Sb c d aW T X For matrix circular convolution, we know Rh h h VRx V S 0 2 1WS 0W x 6n@ * h 6n@ = Sh1 h 0 h2WSx1W SSh h h WWSSx WW 1 0 1 2 XT X T where "x 0, x1, x2, are three point signals for x 6n@ and similarly for h 6n@, h 0 , h1 and h2 are three point signals. Comparing this transformation to Eq(1), we get R VT Sa d c W Sb a d W 6p q r s@ = Sc b aW 8a b c dB S W Sd c b W T X = 6a b c d @T * 6a b c d @T R V R V Sa W Sa W Sb W Sb W =S W * S W c S W Sc W Sd W Sd W T X T X Now, we know that

nodia

So,

SOL 6.12

x1 6n@ * x2 6n@ = X1DFT 6k @ X2, DFT 6k @ R V R V R V R V Sa W Sa W SaW SaW Sb W Sb W SbW SbW Sc W * Sc W = SgW * SgW S W S W S W S W Sd W Sd W Sd W Sd W T X T X T X2 2 2T X2 = 9a b g d C

Option (D) is correct. Using s -domain differentiation property of Laplace transform. If

f (t)

L

tf (t)

L

F (s) dF (s) ds



Signals and Systems

2s + 1 L [tf (t)] = - d ; 2 1 = ds s + s + 1E (s2 + s + 1) 2

So, SOL 6.13

Chapter 6

Option (C) is correct. n n x [n] = b 1 l - b 1 l u [n] 3 2 n -n n 1 x [n] = b l u [n] + b 1 l u [- n - 1] - b 1 l u (n) 3 3 2 Taking z -transform X 6z @ =

3

/ n =- 3

1 n -n b 3 l z u [ n] + = =

3

1 -n -n b 3 l z u [ - n - 1] -

n =- 3 -1 -n

/ b 13 l z

n=0 3

3

/

n

+

/

n =- 3 3

1 -n b3l z -n

/ b 31z l + / b 13 z l

3

3

/ n =- 3

1 n -n b 2 l z u [ n]

/ b 12 l z n

-n

n=0

3

- /b 1 l 2z m=1 n=0 1 44 2 4 4 3 1 4 4 2 4 43 II III

n

n=0

14 42 4 43 I

m

n

Taking m =- n

1 < 1 or z > 1 3z 3 1 Series II converges if z < 1 or z < 3 3 Series III converges if 1 < 1 or z > 1 2z 2 Region of convergence of X (z) will be intersection of above three So, ROC : 1 < z < 3 2 Option (D) is correct. Series I converges if

SOL 6.14

nodia y (t) =

t

# x (t) cos (3t) dt -3

Time Invariance : Let, x (t) = d (t) y (t) =

t

# d (t) cos (3t) dt -3

= u (t) cos (0) = u (t)

For a delayed input (t - t 0) output is y (t, t 0) = Delayed output,

t

# d (t - t ) cos (3t) dt -3

0

= u (t) cos (3t 0)

y (t - t 0) = u (t - t 0) y (t, t 0) ! y (t - t 0) Stability : Consider a bounded input x (t) = cos 3t t t y (t) = # cos2 3t = # 1 - cos 6t = 1 2 2 -3 -3 As t " 3, y (t) " 3 (unbounded) SOL 6.15

System is not time invariant.

# 1dt - 12 # cos 6t dt t

t

-3

-3

System is not stable.

Option (C) is correct. (2 cos w) (sin 2w) = sin 3w + sin w H (jw) = w w w We know that inverse Fourier transform of sin c function is a rectangular function.



Signals and Systems

Page 307

So, inverse Fourier transform of H (jw) h (t) = h1 (t) + h2 (t) h (0) = h1 (0) + h2 (0) = 1 + 1 = 1 2 2

SOL 6.16

nodia

Option (A) is correct. Convolution sum is defined as

y [n] = h [n] * g [n] =

For causal sequence,

y [n] =

3

/ h [n] g [n - k]

k =- 3

3

/ h [n] g [n - k]

k=0

y [n] = h [n] g [n] + h [n] g [n - 1] + h [n] g [n - 2] + .....

For n = 0 ,

y [0] = h [0] g [0] + h [1] g [- 1] + ........... y [0] = h [0] g [0] y [0] = h [0] g [0]

For n = 1,

y [1] = h [1] g [1] + h [1] g [0] + h [1] g [- 1] + .... y [1] = h [1] g [1] + h [1] g [0] 1 = 1 g [1] + 1 g [0] h [1] = 1 1 = 1 b2l 2 2 2 2

From equation (i), So, SOL 6.17

g [- 1] = g [- 2] = ....0 ...(i)

1 = g [ 1] + g [ 0 ] g [ 1] = 1 - g [ 0] y [ 0] 1 g [ 0] = = =1 h [ 0] 1 g [ 1] = 1 - 1 = 0

Option (A) is correct. d2 y dy We have 100 2 - 20 + y = x (t) dt dt Applying Laplace transform we get or

100s2 Y (s) - 20sY (s) + Y (s) = X (s) Y (s) 1 H (s) = = X (s) 100s2 - 20s + 1 1/100 A = 2 = s - (1/5) s + 1/100 s2 + 2xwn s + w2



Signals and Systems

Chapter 6

Here wn = 1/10 and 2xwn =- 1/5 giving x =- 1 Roots are s = 1/10, 1/10 which lie on Right side of s plane thus unstable. SOL 6.18

Option (C) is correct. For an even function Fourier series contains dc term and cosine term (even and odd harmonics).

SOL 6.19

Option (B) is correct. Function h (n) = an u (n) stable if a < 1 and Unstable if a H 1We We have h (n) = 2n u (n - 2); Here a = 2 therefore h (n) is unstable and since h (n) = 0 for n < 0 Therefore h (n) will be causal. So h (n) is causal and not stable.

SOL 6.20


SOL 6.21

Impulse response = d (step response) dt = d (1 - e- at) = 0 + aeat = aeat dt Option (D) is correct. We have x (t) = exp (- 2t) m (t) + s (t - 6) and h (t) = u (t) Taking Laplace Transform we get X (s) = b 1 + e-6s l and H (s) = 1 s+2 s Now

or

nodia Y (s) = H (s) X (s) -6s 1 = 1 : 1 + e-6sD = +e s s+2 s s (s + 2) -6s 1 Y (s) = 1 +e 2s 2 (s + 2) s

y (t) = 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)

Thus SOL 6.22

Option (B) is correct. y (n) = x (n - 1) Y (z) = z-1 X (z)

or

Y (z) = H (z) = z-1 X (z)

or Now

H1 (z) H2 (z) = z-1 1 - 0.4z-1 -1 c 1 - 0.6z-1 m H2 (z) = z z-1 (1 - 0.6z-1) H2 (z) = (1 - 0.4z-1)

SOL 6.23

Option (B) is correct. For 8 point DFT, x* [1] = x [7]; x* [2] = x [6]; x* [3] = x [5] and it is conjugate symmetric about x [4], x [6] = 0 ; x [7] = 1 + j3

SOL 6.24

Option (A) is correct. We know that Given that Inverse z-transform

SOL 6.25

Inverse Z - transform

ad [n ! a] aZ ! a 2 -1 X (z) = 5z + 4z + 3 x [n] = 5d [n + 2] + 4d [n - 1] + 3d [n]

Option (C) is correct. For a function x (t) trigonometric fourier series is



Signals and Systems

x (t) = Ao +

3


n=1

Ao 1 # x (t) dt T0 T

Where,

Page 309

T0 "fundamental period

0

An = 2 # x (t) cos nwt dt T0 T Bn = 2 # x (t) sin nwt dt T0 T For an even function x (t), Bn = 0 Since given function is even function so coefficient Bn = 0 , only cosine and constant terms are present in its fourier series representation 3T/4 T/4 3T/4 Constant term A0 = 1 # - 2AdtD x (t) dt = 1 : # Adt + # T -T/4 T -T/4 T/4 = 1 :TA - 2AT D =- A 2 2 T 2 Constant term is negative. and

0

0

SOL 6.26


nodia

and Response of cascaded system or,

h1 [n] = d [n - 1] or H1 [Z ] = Z - 1 h 2 [n] = d [n - 2] or H2 (Z ) = Z - 2

H (z ) = H1 (z ) : H2 (z ) = z-1 : z-2 = z-3 h [n] = d [n - 3]

SOL 6.27

Option (D) is correct. For an N-point FET algorithm butterfly operates on one pair of samples and involves two complex addition and one complex multiplication.

SOL 6.28

Option (D) is correct. We have and

3s + 1 f (t) = L - 1 ; 3 s + 4s 2 + (k - 3) s E lim f (t) = 1

t"3

By final value theorem lim f (t) = lim sF (s) = 1

t"3

or

or

k =4

or

SOL 6.29

s"0

s. (3s + 1) =1 lim 3 s " 0 s + 4s 2 + (k - 3) s s (3s + 1) =1 lim 2 s " 0 s [s + 4s + (k - 3)] 1 =1 k-3

Option (B) is correct. System is described as dx (t) d 2 y (t) dt (t) + 4x (t) +4 + 3y (t) = 2 dt dt dt 2 Taking Laplace transform on both side of given equation s 2 Y (s) + 4sY (s) + 3Y (s) = 2sX (s) + 4X (s) (s 2 + 4s + 3) Y (s) = 2 (s + 2) X (s) s



Signals and Systems

Chapter 6

Transfer function of the system Y (s) 2 (s + 2) 2 (s + 2) H (s) = = = X (s) s 2 + 4s + 3 (s + 3) (s + 1) x (t) = e-2t u (t) X (s) = 1 (s + 2)

Input or, Output

Y (s) = H (s) : X (s) =

2 (s + 2) : 1 (s + 3) (s + 1) (s + 2)

By Partial fraction 1 - 1 s+1 s+3 Taking inverse Laplace transform Y (s) =

y (t) = (e-t - e-3t) u (t) SOL 6.30

Option (C) is correct. 2 - 34 z - 1 1 - 34 z - 1 + 18 z - 2 By partial fraction H (z ) can be written as 1 1 H (z ) = 1 -1 + 1 -1 1 z 1 ^ h ^ h 2 4z For ROC : z > 1/2 We have

H (z) =

nodia

n n h [n] = b 1 l u [n] + b 1 l u [n], n > 0 2 4

1 = an u [n], z > a 1 - z -1 Thus system is causal. Since ROC of H (z ) includes unit circle, so it is stable also. Hence S1 is True For ROC : z < 1 4 n n h [n] =-b 1 l u [- n - 1] + b 1 l u (n), z > 1 , z < 1 2 2 4 4 System is not causal. ROC of H (z ) does not include unity circle, so it is not stable and S 3 is True SOL 6.31

Option (A) is correct. The Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd.

SOL 6.32

Option (B) is correct. Given function is f (t) = sin2 t + cos 2t = 1 - cos 2t + cos 2t = 1 + 1 cos 2t 2 2 2

SOL 6.33

The function has a DC term and a cosine function. The frequency of cosine terms is w = 2 = 2pf " f = 1 Hz p The given function has frequency component at 0 and 1 Hz. p Option (A) is correct. n n x [n] = b 1 l u (n) - b 1 l u (- n - 1) 3 2 Taking z transform we have



Signals and Systems

X (z) =

Page 311

n=3

n

n =- 1

/ b 13 l z-n - /

n =- 3

n=0 n=3

1 n -n b2l z

n =- 1 n n = / b 1 z-1 l - / b 1 z-1 l 3 2 n =- 3 n=0

1 z-1 < 1 " 3 1 z-1 > 1 " 2

First term gives Second term gives

1< z 3 1> z 2

Thus its ROC is the common ROC of both terms. that is 1< z <1 3 2 SOL 6.34

Option (B) is correct. By property of unilateral Laplace transform t 0 #- 3 f (t) dt L Fs(s) + s1 #- 3 f (t) dt -

Here function is defined for 0 < t < t , Thus t #0 f (t) L Fs(s) SOL 6.35

nodia

Option (A) is correct. We have h (2) = 1, h (3) =- 1 otherwise h (k) = 0 . The diagram of response is as follows :

It has the finite magnitude values. So it is a finite impulse response filter. Thus S2 is true but it is not a low pass filter. So S1 is false. SOL 6.36

Option (B) is correct. Here h (t) ! 0 for t < 0 . Thus system is non causal. Again any bounded input x (t) gives bounded output y (t). Thus it is BIBO stable. Here we can conclude that option (B) is correct.

SOL 6.37

Option (D) is correct. We have x [n] = {1, 0, 2, 3) and N = 4 X [k ] = For N = 4 ,

X [k ] =

N-1

/ x [n] e

n=0 3

/ x [ n] e

-j2pnk/N

-j2pnk/4

k = 0, 1...N - 1 k = 0, 1,... 3

n=0

Now

X [ 0] =

3

/ x [ n ] = x [ 0] + x [ 1] + x [ 2] + x [ 3] = 1 + 0 + 2 + 3 = 6 n=0

x [1] =

3

/ x [ n] e

-jpn/2

= x [0] + x [1] e-jp/2 + x [2] e-jp + x [3] e-jp3/2

n=0

= 1 + 0 - 2 + j3 =- 1 + j3 X [ 2] =

3

/ x [ n] e

-jpn

= x [0] + x [1] e-jp + x [2] e-j2p + x [3] e-jp3

n=0



Signals and Systems

Chapter 6

= 1+0+2-3 = 0 X [ 3] =

3

/ x [n] e

-j3pn/2

= x [0] + x [1] e-j3p/2 + x [2] e-j3p + x [3] e-j9p/2

n=0

= 1 + 0 - 2 - j3 =- 1 - j3 Thus [6, - 1 + j3, 0, - 1 - j3] SOL 6.38


SOL 6.39

Option (C) is correct. The output of causal system depends only on present and past states only. In option (A) y (0) depends on x (- 2) and x (4). In option (B) y (0) depends on x (1). In option (C) y (0) depends on x (- 1). In option (D) y (0) depends on x (5). Thus only in option (C) the value of y (t) at t = 0 depends on x (- 1) past value. In all other option present value depends on future value.

SOL 6.40

Option (D) is correct. We have h (t) = eat u (t) + e bt u (- t) This system is stable only when bounded input has bounded output For stability at < 0 for t > 0 that implies a < 0 and bt > 0 for t > 0 that implies

nodia

b > 0 . Thus, a is negative and b is positive. SOL 6.41


K (s + 1) , and R (s) = 1 s (s + 2)( s + 4) K (s + 1) C (s) = G (s) R (s) = s (s + 2)( s + 4) K = K + - 3K 8s 4 (s + 2) 8 (s + 4) c (t) = K :1 + 1 e-2t - 3 e-4tD u (t) 8 4 8

G (s) =

Thus

At steady-state , c (3) = 1 K = 1 or K = 8 Thus 8 8 (s + 1) Then, G (s) = = 12 - 4 (s + 2)( s + 4) (s + 4) (s + 2) h (t) = L-1 G (s) = (- 4e-2t + 12e-4t) u (t) SOL 6.42

Option (A) is correct. 1 x (t) = ) 0 Fourier transform is We have

#- 33e-jwt x (t) dt

for - 1 # t # + 1 otherwise

= 1 [e-jwt]-11 - jw 1 -jw jw (e - e ) = 1 (- 2j sin w) = 2 sin w = - jw - jw w

=

1

#-1 e-jwt 1dt

This is zero at w = p and w = 2p SOL 6.43

Option (D) is correct. Given h (n) = [1, - 1, 2]



Signals and Systems

Page 313

x (n) = [1, 0, 1] y (n) = x (n)* h (n) The length of y [n] is = L1 + L2 - 1 = 3 + 3 - 1 = 5 y (n) = x (n) * h (n) =

3

/ x (k) h (n - k) k =- 3

y (2) =

3

/ x (k) h (2 - k) k =- 3

= x (0) h (2 - 0) + x (1) h (2 - 1) + x (2) h (2 - 2) = h (2) + 0 + h (0) = 1 + 2 = 3 There are 5 non zero sample in output sequence and the value of y [2] is 3. SOL 6.44

Option (B) is correct. Mode function are not linear. Thus y (t) = x (t) is not linear but this functions is time invariant. Option (A) and (B) may be correct. The y (t) = t x (t) is not linear, thus option (B) is wrong and (a) is correct. We can see that R1: y (t) = t2 x (t) Linear and time variant. R2: y (t) = t x (t) Non linear and time variant. R3: y (t) = x (t) Non linear and time invariant R4: y (t) = x (t - 5) Linear and time invariant

SOL 6.45

Option (A) is correct. Given :

nodia y (n) = 1 N

N-1

/ x (r) x (n + r)

r=0

It is Auto correlation. Hence SOL 6.46

y (n) = rxx (n)

DFT

X (k) 2

Option (B) is correct. Current through resistor (i.e. capacitor) is Here,

I = I (0+) e-t/RC I (0+) = V = 5 = 25mA R 200k RC = 200k # 10m = 2 sec

I = 25e- m A = VR # R = 5e- V Here the voltages across the resistor is input to sampler at frequency of 10 Hz. Thus t 2

x (n) = 5e

-n 2 # 10

t 2

= 5e-0.05n For t > 0

SOL 6.47

Option (C) is correct. Since x (n) = 5e-0.05n u (n) is a causal signal Its z transform is 1 5z X (z) = 5 : = 1 - e-0.05 z-1 D z - e-0.05 Its ROC is e-0.05 z-1 > 1 " z > e-0.05

SOL 6.48

Option (C) is correct. h (t) = e-2t u (t) H (jw) =

#- 33 h (t) e-jwt dt



Signals and Systems

= SOL 6.49

#0 3e-2t e-jwt dt = #0 3e-(2 + jw)t dt

Chapter 6

=

1 (2 + jw)

Option (D) is correct. H (jw) =

1 (2 + jw)

The phase response at w = 2 rad/sec is +H (jw) =- tan-1 w =- tan-1 2 =- p =- 0.25p 2 2 4 Magnitude response at w = 2 rad/sec is 1 H (jw) = = 1 22 + w2 2 2 Input is x (t) = 2 cos (2t) Output is = 1 # 2 cos (2t - 0.25p) 2 2 = 1 cos [2t - 0.25p] 2 SOL 6.50

Option (D) is correct. Y (s) =

nodia

1 s (s - 1) Final value theorem is applicable only when all poles of system lies in left half of S -plane. Here s = 1 is right s -plane pole. Thus it is unbounded. SOL 6.51


x (t) = e-t u (t) Taking Fourier transform X (jw) = 1 1 + jw X (jw) = 1 2 1+w

Magnitude at 3dB frequency is 1 = 1 2 1 + w2 w = 1 rad f = 1 Hz 2p

Thus or or SOL 6.52

1 2

Option (B) is correct. For discrete time Fourier transform (DTFT) when N " 3 p x [n] = 1 # X (e jw) e jwn dw 2p - p Putting n = 0 we get x [0] = 1 2p or

SOL 6.53

#

p

-p

X (e jw) e jw0 dw = 1 2p -p

#

p

#

p

-p

X (e jw) dw

jw

X (e ) dw = 2px [0] = 2p # 5 = 10p

Option (B) is correct. 0. 5 1 - 2z-1 Since ROC includes unit circle, it is left handed system X (z) =



Signals and Systems

Page 315

x (n) =- (0.5) (2) -n u (- n - 1) x (0) = 0 If we apply initial value theorem x (0) = lim X (z) = lim 0.5 -1 = 0.5 z"3 z " 31 - 2z That is wrong because here initial value theorem is not applicable because signal x (n) is defined for n < 0 . SOL 6.54

Option (A) is correct. A Hilbert transformer is a non-linear system.

SOL 6.55

Option (B) is correct. 5 1 + j10pf 5 = 1 H (s) = 5 = 1 + 5s 5^s + 15 h s + 15 Y (s) = 1 a 1 = 1 1 1 = 5 - 5 s ^s + 5 h s ^s + 5 h s s+ -t/5 y (t) = 5 (1 - e ) u (t) H (f) =

Step response or

1 5

nodia

SOL 6.56

Option (A) is correct. F x (t) X (jw) Using scaling we have F 1 X jw x (5t) 5 c 5 m Using shifting property we get F 1 X jw e- j35w x ;5 bt - 3 lE 5 5 b5l

SOL 6.57

Option (D) is correct. Dirac delta function d (t) is defined at t = 0 and it has infinite value a t = 0 . The area of dirac delta function is unity.

SOL 6.58

Option (D) is correct. The ROC of addition or subtraction of two functions x1 (n) and x2 (n) is R1 + R2 . We have been given ROC of addition of two function and has been asked ROC of subtraction of two function. It will be same.

SOL 6.59

Option (A) is correct. As we have x (t) = sin t , Now H (s) = 1 s+1 or H (jw) = 1 = 1 jw + 1 j+1 or H (jw) = 1 + - 45c 2 1 Thus y (t) = sin (t - p4 ) 2

SOL 6.60

thus w = 1

Option (C) is correct. w0 s + w2 L-1 F (s) = sin wo t f (t) = sin wo t F (s) =

2



Signals and Systems

Chapter 6

Thus the final value is - 1 # f (3) # 1 SOL 6.61

Option (C) is correct. y (n) = b sin 5 pn l x (n) 6 Let Now

SOL 6.62

x (n) = d (n) y (n) = sin 0 = 0 (bounded)

BIBO stable

Option (B) is correct. c (t) = 1 - e-2t Taking Laplace transform C (s) 2 = C (s) = #s = 2 s (s + 2) s+2 U (s)

SOL 6.63

Option (C) is correct. 1 s+1 L x (t) = u (t) X (s) = 1 s 1 1 Y (s) = H (s) X (s) = # =1- 1 s+1 s s s+1 h (t) = e-t

L

H (s) =

nodia

y (t) = u (t) - e-t In steady state i.e. t " 3, y (3) = 1 SOL 6.64

Option (C) is correct. Fourier series is defined for periodic function and constant. 3 sin (25t) is a periodic function. 4 cos (20t + 3) + 2 sin (710t) is sum of two periodic function and also a periodic function. e- t sin (25t) is not a periodic function, so FS can’t be defined for it. 1 is constant

SOL 6.65

Option (A) is correct. g (t) + g (- t) 2 g (t) - g (- t) odd{g (t)} = 2 Ev{g (t)} =

Here Thus

SOL 6.66

g (t) = u (t) u (t) + u (- t) ue (t) = = 2 u (t) - u (- t) uo (t) = = 2

1 2 x (t) 2

Option (C) is correct. Here

x1 (n) = ` 5 jn u (n) 6 1 X1 (z) = 1 - ^ 65 z-1h

ROC : R1 " z > 5 6

x2 (n) =-` 6 jn u (- n - 1) 5 1 X1 (z) = 1 1 - ^ 65 z-1h

ROC : R2 " z < 6 5



Signals and Systems

Page 317

Thus ROC of x1 (n) + x2 (n) is R1 + R2 which is 5 < z < 6 6 5 SOL 6.67

Option (D) is correct. For causal system h (t) = 0 for t # 0 . Only (D) satisfy this condition.

SOL 6.68

Option (D) is correct. x (n) = b 1 l u (n) 2 n

y (n) = x2 (n) = b 1 l u2 (n) 2 2n

or

2 n n y (n) = ;b 1 l E u (n) = b 1 l u (n) 2 4

Y (e jw) =

n=3

/ y (n) e-jwn

=

n=3

/ b 14 l e-jwn

n =- 3

or

Y (e j0) =

/ ` 14 j

n=0

n

n=0

n

n=3

...(1)

1 3 4 = 1 +b1l +b1l+b1l +b1l = 1 4 4 4 4 1-

=4 3

1 4

Alternative : Taking z transform of (1) we get 1 Y (z) = 1 - 14 z-1 Substituting z = e jw we have 1 Y (e jw) = 1 - 14 e-jw Y (e j0) = 1 1 = 4 3 1- 4 SOL 6.69

nodia


s (t) = 8 cos ` p - 20pt j + 4 sin 15pt 2

= 8 sin 20pt + 4 sin 15pt Here A1 = 8 and A2 = 4 . Thus power is 2 2 2 2 P = A1 + A2 = 8 + 4 = 40 2 2 2 2 SOL 6.70

Option (A) is correct. y (t) = 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td - T) Taking Fourier transform we have Y (w) = 0.5e-jw (- t + T) X (w) + e-jwt X (w) + 0.5e-jw (- t - T) X (w) Y (w) = e-jwt [0.5e jwT + 1 + 0.5e-jwT ] X (w) d

or

d

d

d

= e-jwt [0.5 (e jwT + e-jwT ) + 1] = e-jwt [cos wT + 1] d

or SOL 6.71

H (w) =

d

Y (w) = e-jwt (cos wT + 1) X (w) d

Option (C) is correct. For continuous and aperiodic signal Fourier representation is continuous and aperiodic. For continuous and periodic signal Fourier representation is discrete and aperiodic. For discrete and aperiodic signal Fourier representation is continuous and periodic. For discrete and periodic signal Fourier representation is discrete and periodic.



SOL 6.72

Signals and Systems

Chapter 6

Option (B) is correct. y (n) = Ax (n - no) Taking Fourier transform Y (e jw) = Ae-jw n X (e jw) Y (e jw) or = Ae-jw n H (e jw) = X (e jw) Thus +H (e jw) =- wo no For LTI discrete time system phase and frequency of H (e jw) are periodic with period 2p. So in general form o

o

o

o

q (w) =- no wo + 2pk SOL 6.73

Option (A) is correct. From x (n) = [ 12 , 1, 2, 1, 1, 12 ] y (n) = x ^ n2 - 1h, n even = 0 , for n odd n =- 2 , n =- 1, n = 0,

y (- 2) = x ( -22 - 1) = x (- 2) = 12 y (- 1) = 0 y (0) = x ( 20 - 1) = x (- 1) = 1

nodia

n = 1, n=2 n = 3, n=4

y (1) = 0 y (2) = x ( 22 - 1) = x (0) = 2

y (3) = 0 y (4) = x ( 24 - 1) = x (1) = 1

n = 5, n=6

SOL 6.74

y (5) = 0

y (6) = x ( 26 - 1) = x (2) = 12 y (n) = 1 d (n + 2) + d (n) + 2d (n - 2) + d (n - 4) 2

Hence + 1 d (n - 6) 2 Option (C) is correct. Here y (n) is scaled and shifted version of x (n) and again y (2n) is scaled version of y (n) giving z (n) = y (2n) = x (n - 1) = 1 d (n + 1) + d (n) + 2d (n - 1) + d (n - 2) + 1 d (n - 3) 2 2 Taking Fourier transform. Z (e jw) = 1 e jw + 1 + 2e-jw + e-2jw + 1 e-3jw 2 2 = e-jw b 1 e2jw + e jw + 2 + e-jw + 1 e-2jw l 2 2 = e-jw b e or

SOL 6.75

2jw

+ e-2jw + e jw + 2 + e-jw l 2

Z (e jw) = e-jw [cos 2w + 2 cos w + 2]

Option (B) is correct. x (t)

F

x (at)

F

X (f)

Using scaling we have 1 X f a ca m



Signals and Systems

xb 1 f l 3

Thus

F

Page 319

3X (3f)

Using shifting property we get e-j2pf t x (t) = X (f + f0) F 1 e-j 43 pt x 1 t X (3f + 2) b3 l 3 F 3X (3 (f + 23 )) e-j2p t x b 1 t l 3 F 1 e-jp t x 1 t X [3 (f + 23 )] b3 l 3 0

Thus

2 3

4 3

SOL 6.76

Option (A) is correct. 3

/

A system is stable if

h (n) < 3. The plot of given h (n) is

n =- 3

Thus

nodia 3

/

6

/

h (n) =

h (n)

n =- 3

n =- 3

= 1+1+1+1+2+2+2+2+2

= 15 < 3 Hence system is stable but h (n) ! 0 for n < 0 . Thus it is not causal. SOL 6.77

Option (D) is correct. z z - 0.2

H (z) = We know that

z < 0.2

1 1 - az-1 h [n] =- (0.2) n u [- n - 1]

- an u [- n - 1] * Thus

z
SOL 6.78

Option (C) is correct. The Fourier transform of a conjugate symmetrical function is always real.

SOL 6.79


x (n) = [- 4 - j5, 1 + 2j,

4]

-

x *( n) = [- 4 + j5, 1 - 2j, 4] -

x *( - n) = [4,

1 - 2j, - 4 + j5] -

x (n) - x* (- n) xcas (n) = = [- 4 - j 25 , 2 SOL 6.80

2j -

4 - j 25 ]

Option (C) is correct. We have 2y (n) = ay (n - 2) - 2x (n) + bx (n - 1) Taking z transform we get 2Y (z) = aY (z) z-2 - 2X (z) + bX (z) z-1



Signals and Systems

Chapter 6

Y (z) bz-1 - 2 ...(i) =c m X (z) 2 - az-2 z ( b - z) or H (z) = 22 a (z - 2 ) It has poles at ! a/2 and zero at 0 and b/2 . For a stable system poles must lie inside the unit circle of z plane. Thus a <1 2 or

a <2 But zero can lie anywhere in plane. Thus, b can be of any value. or SOL 6.81


x (n) = e jpn/4

and

h (n) = 4 2 d (n + 2) - 2 2 d (n + 1) - 2 2 d (n - 1) + 4 2 d (n - 2)

Now

y (n) = x (n)* h (n)

nodia 3

2

k =- 3

k =- 2

/ x (n - k) h (k) = / x (n - k) h (k)

=

or

y (n) = x (n + 2) h (- 2) + x (n + 1) h (- 1) + x (n - 1) h (1) + x (n - 2) h (2) = 4 2 ej

p 4

(n + 2)

= 4 2 6e j

- 2 2 ej

p 4

(n + 1)

- 2 2 ej

p 4

(n - 1)

+ 4 2 ej

@ - 2 2 6e j (n + 1) + e j = 4 2 e j n 6e j + e-j @ - 2 2 e j n 6e j + e-j @ p 4

(n + 2)

p 4

+ ej

p 2

p 4

(n - 2) p 2

p 4

or SOL 6.82

p 4

p 2

p 4

p 4

p 4

(n - 2)

@

(n - 1)

p 4

p 4

= 4 2 e j n [0] - 2 2 e j n [2 cos p4 ] y (n) =- 4e j n r 4

Option (B) is correct. From given graph the relation in x (t) and y (t) is y (t) =- x [2 (t + 1)] x (t)

F

x (at)

F

x (2t)

F

X (f)

Using scaling we have

Thus

1 X f a ca m 1X f 2 c2m

Using shifting property we get x (t - t0) = e-j2pft X (f) 0

x [2 (t + 1)]

F

- x [2 (t + 1)]

F

Thus

SOL 6.83

j2pf f f e-j2pf (- 1) 1 X b l = e X b l 2 2 2 2 j 2p f f Xc m -e 2 2

Option (C) is correct. From the Final value theorem we have lim i (t) = lim sI (s) = lim s

t"3

SOL 6.84

s"0

s"0

2 = lim 2 =2 s (1 + s) s " 0 (1 + s)

Option (D) is correct. Here C3 = 3 + j5



Signals and Systems

Page 321

For real periodic signal Thus SOL 6.85

C-k = Ck* C-3 = Ck = 3 - j5

Option (C) is correct. y (t) = 4x (t - 2) Taking Fourier transform we get

or Thus SOL 6.86

Y (e j2pf ) = 4e-j2pf2 X (e j2pf ) Y (e j2pf ) = 4e-4jpf X (e j2pf ) H (e j2pf ) = 4e-4jpf

Time Shifting property

Option (B) is correct. We have h (n) = 3d (n - 3) or Taking z transform H (z) = 2z-3 4 2 -4 X (z) = z + z - 2z + 2 - 3z Now Y (z) = H (z) X (z) = 2z-3 (z 4 + z2 - 2z + 2 - 3z-4)

nodia

= 2 (z + z-1 - 2z-2 + 2z-3 - 3z-7) Taking inverse z transform we have y (n) = 2[ d (n + 1) + d (n - 1) - 2d (n - 2) + 2d (n - 3) - 3d (n - 7)] At n = 4 , y (4) = 0 SOL 6.87

Option (A) is correct. System is non causal because output depends on future value For n # 1 y (- 1) = x (- 1 + 1) = x (0) y (n - n0) = x (n - n0 + 1) y (n) = x (n + 1)

Time varying

Depends on Future i.e. None causal y (1) = x (2) For bounded input, system has bounded output. So it is stable. y (n) = x (n) for n $ 1 = 0 for n = 0 = x (x + 1) for n #- 1 So system is linear. SOL 6.88

Option (C) is correct. The frequency response of RC-LPF is 1 H (f) = 1 + j2pfRC Now

or or or or

H (0) = 1 H (f1) 1 = $ 0.95 H (0) 1 + 4p2 f12 R2 C2 1 + 4p2 f12 R2 C2 # 1.108 4p2 f12 R2 C2 # 0.108 2pf1 RC # 0.329 f1 # 0.329 2pRC



Signals and Systems

f1 # 0.329 2pRC f1 # 0.329 2p1k # 1m

or or or Thus SOL 6.89

Chapter 6

f1 # 52.2 Hz f1 max = 52.2 Hz

Option (A) is correct. 1 H (w) = 1 + jwRC q (w) =- tan-1 wRC dq (w) 10-3 RC tg == = 0.717 ms = 2 2 2 2 dw 1 + 4p # 10 4 # 10-6 1+w R C

SOL 6.90

Option (C) is correct. If Then Thus

SOL 6.91

x (t)* h (t) = g (t)

x (t - t1)* h (t - t2) = y (t - t1 - t2) x (t + 5)* d (t - 7) = x (t + 5 - 7) = x (t - 2)

nodia

Option (B) is correct. In option (B) the given function is not periodic and does not satisfy Dirichlet condition. So it cant be expansion in Fourier series. x (t) = 2 cos pt + 7 cos t T1 = 2p = 2 w 2 T2 = p = 2p 1 T1 = 1 = irrational p T2

SOL 6.92

Option (C) is correct. From the duality property of fourier transform we have x (t)

FT

X (f)

X (t)

FT

x (- f) 1 1 + j 2p f

If Then Therefore if Then SOL 6.93

e u (t) 1 1 + j2pt

FT

FT

e f u (- f)


and Thus SOL 6.94

-t

q (w) =- wt0 - q (w) tp = = t0 w dq (w) tg == t0 dw tp = tg = t0 = constant

Option (*) is correct. 5-s = -2 + 1 X (s) = 2 5 - s = s+1 s-2 ( s 1 )( s 2 ) + s -s-2 Here three ROC may be possible.



Signals and Systems

Page 323

Re (s) < - 1 Re (s) > 2 - 1 < Re (s) < 2 Since its Fourier transform exits, only - 1 < Re (s) < 2 include imaginary axis. so this ROC is possible. For this ROC the inverse Laplace transform is x (t) = [- 2e-t u (t) - 2e2t u (- t)] SOL 6.95

Option (B) is correct. For left sided sequence we have - an u (- n - 1)

z

Thus

- 5n u (- n - 1)

z

or

- 5n u (- n - 1)

z

1 1 - az-1 1 1 - 5z-1 z z-5

where z < a where z < 5 where z < 5

Since ROC is z < 5 and it include unit circle, system is stable. Alternative :

nodia h (n) =- 5n u (- n - 1) 3

/ h (n) z

H (z) =

-n

-1

/-5 z

=

n =- 3

n -n

-1

/ (5z

=-

n =- 3

-1 n

)

n =- 3

Let n =- m, then

H (z) =-

-3

/ (5z

-1 -m

)

= 1-

n =- 1

3

/ (5

-1

z) -m

m=0

1 = 1, 1 - 5-1 z = 1- 5 = z 5-z z-5

SOL 6.96

5-1 z < 1 or z < 5

Option (B) is correct. 1 = 12 # 1 2 s-2 s (s - 2) s L 1 # 1 (t * e2t) u (t) s2 s - 2 Here we have used property that convolution in time domain is multiplication in s - domain LT

X1 (s) X2 (s) SOL 6.97

x1 (t)* x2 (t)

Option (A) is correct. We have h (n) = u (n) 3

/ x (n) .z

H (z) =

-n

n =- 3

=

3

/ 1. z

-n

n=0

=

3

/ (z

-1 n

)

n=0

H (z) is convergent if 3

/ (z-1) n < 3 n=0

and this is possible when z-1 < 1. Thus ROC is z-1 < 1 or z > 1 SOL 6.98

Option (A) is correct. We know that d (t) x (t) = x (0) d (t) and

# d (t) = 1 3

-3



Signals and Systems

Chapter 6

Let x (t) = cos ( 23 t), then x (0) = 1

# d (t) x (t) = # x (0) d (t) dt

Now SOL 6.99

3

3

-3

-3

=

# d (t) dt = 1 3

-3

Option (B) is correct. Let E be the energy of f (t) and E1 be the energy of f (2t), then

# E1 = # E =

and

3

-3 3

-3

[f (t)] 2 dt [f (2t)] 2 dt

Substituting 2t = p we get E1 =

#

3

-3

dp 1 = 2 2

[f (p)] 2

#

3

-3

[f (p)] 2 dp = E 2

SOL 6.100

Option (B) is correct. Since h1 (t) ! 0 for t < 0 , thus h1 (t) is not causal h2 (t) = u (t) which is always time invariant, causal and stable. u (t) is time variant. h3 (t) = 1+t h 4 (t) = e-3t u (t) is time variant.

SOL 6.101


nodia

h (t) = f (t)* g (t) We know that convolution in time domain is multiplication in s - domain. L

Thus SOL 6.102

f (t)* g (t) = h (t) H (s) = F (s) # G (s) s2 + 1 H (s) = s2+ 2 # = 1 s + 1 (s + 2)( s + 3) s + 3

Option (B) is correct. Since normalized Gaussion function have Gaussion FT p eThus e-at a 2

SOL 6.103

FT

p2 f2 a

Option (B) is correct. Let x (t) = ax1 (t) + bx2 (t) ay1 (t) = atx1 (t) by2 (t) = btx2 (t) Adding above both equation we have ay1 (t) + by2 (t) = atx1 (t) + btx2 (t) = t [ax1 (t) + bx2 (t)] = tx (t) or Thus system is linear ay1 (t) + by2 (t) = y (t) If input is delayed then we have yd (d) = tx (t - t0) If output is delayed then we have y (t - t0) = (t - t0) x (t - t0) which is not equal. Thus system is time varying.

SOL 6.104


h (t) = e2t

LS

and

x (t) = e3t

LS

1 s-2 X (s) = 1 s-3

H (s) =



Signals and Systems

Y (s) = H (s) X (s) =

Now output is Thus SOL 6.105

Page 325

1 # 1 = 1 - 1 s-2 s-3 s-3 s-2

y (t) = e3t - e2t

Option (C) is correct. We know that for a square wave the Fourier series coefficient sin nw t Cnsq = At nw t2 T 2 1 Cnsq \ n 0

...(i)

0

Thus

If we integrate square wave, triangular wave will be obtained, Hence Cntri \ 12 n SOL 6.106

Option (B) is correct. u (t) - u (t - 1) = f (t)

L

u (t) - u (t - 2) = g (t)

L

F (s) = 1 [1 - e-s] s G (s) = 1 [1 - e-2s] s

nodia f (t)* g (t)

L

F (s) G (s)

= 12 [1 - e-s] [1 - e-2s] s = 12 [1 - e-2s - e-s + e-3s] s -2s -s -3s L or = 12 - e 2 - e 2 + e 2 f (t)* g (t) s s s s Taking inverse Laplace transform we have

f (t)* g (t) = t - (t - 2) u (t - 2) - (t - 1) u (t - 1) + (t - 3) u (t - 3) The graph of option (B) satisfy this equation. SOL 6.107


SOL 6.108

Option (A) is correct. We have f (nT) = anT Taking z -transform we get F (z) =

3

/ anT z-n n =- 3

SOL 6.109

3

/ (aT ) n z-n

=

n =- 3

=

T n

/ b az 3

n=0

l =

z z - aT

Option (B) is correct. If L [f (t)] = F (s) Applying time shifting property we can write L [f (t - T)] = e-sT F (s)

SOL 6.110


SOL 6.111


SOL 6.112

Option (C) is correct. Given z transform C (z) =

z-1 (1 - z-4) 4 (1 - z-1) 2

Applying final value theorem lim f (n) = lim (z - 1) f (z)

n"3

z"1



Signals and Systems

Chapter 6

z-1 (1 - z-4) z-1 (1 - z-4) (z - 1) = lim 1 2 z"1 z"1 4 (1 - z ) 4 (1 - z-1) 2 z-1 z-4 (z 4 - 1) (z - 1) = lim z"1 4z-2 (z - 1) 2 -3 (z - 1) (z + 1) (z2 + 1) (z - 1) = lim z z"1 4 (z - 1) 2 -3 = lim z (z + 1) (z2 + 1) = 1 z"1 4

lim (z - 1) F (z) = lim (z - 1) z"1

SOL 6.113

Option (A) is correct. w s2 + w2 lim f (t) final value theorem states that: t"3 lim f (t) = lim sF (s) F (s) =

We have

t"3

s"0

It must be noted that final value theorem can be applied only if poles lies in –ve half of s -plane. Here poles are on imaginary axis (s1, s2 = ! jw) so can not apply final value theorem. so lim f (t) cannot be determined.

nodia t"3

SOL 6.114

Option (D) is correct. Trigonometric Fourier series of a function x (t) is expressed as : x (t) = A 0 +

3


n=1

For even function x (t), Bn = 0 So

x (t) = A 0 +

3

/ An cos nwt

n=1

Series will contain only DC & cosine terms. SOL 6.115

Option (C) is correct. Given periodic signal t < T1 1, x (t) = * 0, T1 < t < T0 2 The figure is as shown below.

For x (t) fourier series expression can be written as x (t) = A 0 +

3


n=1

where dc term T /2 A 0 = 1 # x (t) dt = 1 # x (t) dt T0 T T0 -T /2 T -T 1 = : # x (t) dt + # x (t) dt + T0 -T /2 -T 0

0

0

1

0

1

1

#T

T0 /2

1

x (t) dtD = 1 60 + 2T1 + 0@ T0



Signals and Systems

Page 327

A 0 = 2T1 T0 SOL 6.116

Option (B) is correct. The unit impulse response of a LTI system is u (t) Let h (t) = u (t) Taking LT we have H (s) = 1 s If the system excited with an input x (t) = e-at u (t), a > 0 , the response Y (s) = X (s) H (s) X (s) = L [x (t)] = 1 (s + a) so Y (s) = 1 1 = 1 :1 - 1 D a s s+a (s + a) s Taking inverse Laplace, the response will be y (t) = 1 61 - e-at@ a

SOL 6.117

nodia


x [n] =

We have

X (z) =

3

/ d (n - k)

k=0 3

/ x [n] z-n

=

k=0

3

3

/ ; / d (n - k) z-nE n =- 3 k = 0

Since d (n - k) defined only for n = k so 3 1 X (z) = / z-k = = z (z - 1) (1 - 1/z) k=0 SOL 6.118


SOL 6.119

Option (B) is correct. F

X (f) x (t) by differentiation property; dx (t) = jwX (w) F; dt E dx (t) or = j2pfX (f) F; dt E SOL 6.120

Option (C) is correct. We have g (w) f (t) by duality property of fourier transform we can write F

g (t) so

2pf (- w)

F 3

# g (t) e-jwt dt = 2pf (- w)

F [g (t)] =

-3

SOL 6.121

Option (B) is correct. Given function Now

x (t) = eat cos (at) L s cos (at) s2 + a2 x (t)

L

X (s)

e x (t)

L

X (s - s 0)

If then

s0 t

shifting in s-domain



Signals and Systems

eat cos (at)

so SOL 6.122

Chapter 6

(s - a) (s - a) 2 + a2

L

Option (C) is correct. For a function x (t), trigonometric fourier series is : x (t) = A 0 +

3

/ [An cos nwt + Bn sin nwt] n=1

where A 0 = 1 # x (t) dt T0 T An = 2 # x (t) cos nwtdt T0 T Bn = 2 # x (t) sin nwtdt T0 T For an even function x (t), coefficient Bn = 0

T0 =Fundamental period

0

0

0

A0 = 0 An = 0 so if x (t) is even function its fourier series will not contain sine terms. for an odd function x (t),

SOL 6.123

nodia

Option (C) is correct. The conjugation property allows us to show if x (t) is real, then X (jw) has conjugate symmetry, that is X (- jw) = X)(jw)

[ x (t) real]

Proof :

3

# x (t) e-jwt dt

X (jw) =

replace w by - w then

-3

3

# x (t) e jwt dt

X (- jw) =

-3 3

X)(jw) = = if x (t) real x)(t) = x (t) then

X)(jw) =

)

# x (t) e-jwt dtG

-3

=

3

# x)(t) e jwt dt -3

3

# x (t) e jwt dt = X (- jw) -3

***********


CHAPTER 7 CONTROL SYSTEMS

2013 MCQ 7.1

ONE MARK

The Bode plot of a transfer function G ^s h is shown in the figure below.

nodia

The gain _20 log G ^s h i is 32 dB and - 8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all w. Then G ^s h is (B) 392.8 (A) 39.8 s s 32 32 (C) (D) 2 s s 2013 MCQ 7.2

TWO MARKS

Y ^s h The signal flow graph for a system is given below. The transfer function for U ^s h this system is

s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2 (A)

s+1 s 2 + 6s + 2 (D) 2 1 5s + 6s + 2 (B)

Statement for Linked Answer Questions 4 and 5: The state diagram of a system is shown below. A system is described by the o = AX + Bu ; y = CX + Du state-variable equations X


Control Systems

Chapter 7

MCQ 7.3

The state-variable equations of the system shown in the figure above are o = >- 1 0 H X + >- 1H u o = >- 1 0 H X + >- 1H u X X (A) (B) -1 -1 1 -1 1 1 y = 61 - 1@ X + u y = 6- 1 - 1@ X + u 1 0 1 o => o = >- 1 - 1H X + >- 1H u X X + > Hu X (C) (D) - 1 - 1H 1 0 -1 1 y = 6- 1 - 1@ X - u y = 61 - 1@ X - u

MCQ 7.4

The state transition matrix eAt of the system shown in the figure above is e-t 0 e-t 0 (A) > -t -tH (B) > H te e - te-t e-t

nodia

e-t 0 (C) > -t -tH e e MCQ 7.5

e-t - te-t (D) > H 0 e-t

The open-loop transfer function of a dc motor is given as V ^^shh = 1 +1010s . When connected in feedback as shown below, the approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is ws a

(A) 1 (C) 10

(B) 5 (D) 100

2012 MCQ 7.6

ONE MARK 2

(s + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at (A) w = 1 rad/s (B) w = 2 rad/s (C) w = 3 rad/s (D) w = 4 rad/s A system with transfer function G (s) =

2012 MCQ 7.7

TWO MARKS

The feedback system shown below oscillates at 2 rad/s when



Control Systems

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5 MCQ 7.8

Page 331

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N Jx1N K O K OK O K O K O y = _1 0 0iKx2O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka Kx 3O 0 0OKx 3O K 1O 3 3 L P L PL P L P L P where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 2011

MCQ 7.9

ONE MARK

The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

nodia s (s + 1) (s + 2) (s + 3)

(s + 1) s (s + 2) (s + 3) 2 (s + 1) 1 (C) G ^s h H ^s h = k (D) G ^s h H ^s h = k s (s + 2) (s + 3) s (s - 1) (s + 2) (s + 3) (A) G ^s h H ^s h = k

MCQ 7.10

(B) G ^s h H ^s h = k

For the transfer function G (jw) = 5 + jw , the corresponding Nyquist plot for positive frequency has the form



Control Systems

2011 MCQ 7.11

Chapter 7

TWO MARKS

The block diagram of a system with one input u and two outputs y1 and y2 is given below.

A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]T is (A) xo = [2] x + [1] u ; y = [1 2] x 1 (B) xo = [- 2] x + [1] u; y = > H x 2 -2 0 1 (C) xo = > x + > H u ; y = 81 2B x H 0 -2 1 2 0 1 1 (D) xo = > H x + > H u ; y = > H x 0 2 1 2

nodia


100 . s (s + 10) 2 The plant is placed in a unity negative feedback configuration as shown in the figure below.

The input-output transfer function of a plant H (s) =

MCQ 7.12

The gain margin of the system under closed loop unity negative feedback is (A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

MCQ 7.13

The signal flow graph that DOES NOT model the plant transfer function H (s) is



Control Systems

Page 333

2010 MCQ 7.14

ONE MARK

The transfer function Y (s) /R (s) of the system shown is

(A) 0 (C) MCQ 7.15

2 s+1

Y (s) = s has an output y (t) = cos a2t - p k 3 X (s) s + p p for the input signal x (t) = p cos a2t - k. Then, the system parameter p is 2 (B) 2/ 3 (A) 3

A system with transfer function

(C) 1 MCQ 7.16

1 s+1 (D) 2 s+3

(B)

nodia (D)

3 /2

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

(A) 10s + 1 0.1s + 1 (C) 100s 10s + 1

(B) 100s + 1 0.1s + 1 (D) 0.1s + 1 10s + 1

2010

TWO MARKS

Common Data For Q. 18 and 19 The signal flow graph of a system is shown below:

MCQ 7.17

The state variable representation of the system can be -1 1 0 1 1 0 o xo = > x +> Hu (B) (A) x = >- 1 0H x + >2H u - 1 0H 2 yo = 80 0.5B x yo = [0 0.5] x



MCQ 7.18

MCQ 7.19

Control Systems

1 1 0 xo = > x +> Hu H (C) -1 0 2 o y = 80.5 0.5B x The transfer function of the system is (A) s2+ 1 s +1 (C) 2 s + 1 s +s+1

Chapter 7

-1 xo = > (D) -1 o y = 80.5

1 0 x +> Hu H 0 2 0.5B x

(B) s2- 1 s +1 (D) 2 s - 1 s +s+1

A unity negative feedback closed loop system has a plant with the transfer function G (s) = s + 21s + 2 and a controller Gc (s) in the feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is (A) Gc (s) = s + 1 (B) Gc (s) = s + 2 s+2 s+1 (s + 1) (s + 4) (C) Gc (s) = (D) Gc (s) = 1 + 2 + 3s s (s + 2) (s + 3) 2

2009 MCQ 7.20

nodia

The magnitude plot of a rational transfer function G (s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?

(A) Lead compensator (C) PID compensator MCQ 7.21

ONE MARK

(B) Lag compensator (D) Lead-lag compensator

Consider the system dx = Ax + Bu with A = =1 0G and B = =p G q 0 1 dt where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true ? (A) The system is completely state controllable for any nonzero values of p and q (B) Only p = 0 and q = 0 result in controllability (C) The system is uncontrollable for all values of p and q (D) We cannot conclude about controllability from the given data 2009

MCQ 7.22

TWO MARKS

The feedback configuration and the pole-zero locations of 2 G (s) = s2 - 2s + 2 s + 2s + 2 are shown below. The root locus for negative values of k , i.e. for - 3 < k < 0



Control Systems

Page 335

, has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

MCQ 7.23

(A) ! 2 and 0c

(B) ! 2 and 45c

(C) ! 3 and 0c

(D) ! 3 and 45c

The unit step response of an under-damped second order system has steady state value of -2. Which one of the following transfer functions has theses properties ? (A) 2 - 2.24 (B) 2 - 3.82 s + 2.59s + 1.12 s + 1.91s + 1.91 (C)

- 2.24 s - 2.59s + 1.12 2

(D)

- 382 s - 1.91s + 1.91 2

nodia


The Nyquist plot of a stable transfer function G (s) is shown in the figure are interested in the stability of the closed loop system in the feedback configuration shown.

MCQ 7.24

Which of the following statements is true ? (A) G (s) is an all-pass filter (B) G (s) has a zero in the right-half plane (C) G (s) is the impedance of a passive network (D) G (s) is marginally stable

MCQ 7.25

The gain and phase margins of G (s) for closed loop stability are (B) 3 dB and 180c (A) 6 dB and 180c (C) 6 dB and 90c (D) 3 dB and 90c 2008

MCQ 7.26

ONE MARKS

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems ?



MCQ 7.27

Control Systems

The pole-zero given below correspond to a

nodia

(A) Law pass filter (C) Band filter 2008 MCQ 7.28

Chapter 7

(B) High pass filter (D) Notch filter

TWO MARKS

A signal flow graph of a system is given below

The set of equalities that corresponds to this signal flow graph is Jx1N R b - g 0 VJx1N R0 0 V WK O S W u1 K O S (A) d Kx2O = S g a 0 WKx2O + S0 1 We o u2 dt K O S x3 S- a b 0 WWKx3O SS1 0 WW L P L P Jx1N RT0 a g XVJx1N TR1 0 XV W S K O K O S W u1 (B) d Kx2O = S0 - a - g WKx2O + S0 1 We o u2 dt K O S x3 S0 b - b WWKx3O SS0 0 WW L P TR L P Jx1N - a b 0 VXJx1N RT1 0 VX W S K O K O S W u1 (C) d Kx2O = S- b - g 0 WKx2O + S0 1 We o u2 dt K O S x3 S a g 0 WWKx3O SS0 0 WW L P TR L P Jx1N - a 0 b XVJx1N TR1 0XV W S K O K O S W u1 (D) d Kx2O = S g 0 a WKx2O + S0 1 We o u2 dt K O S x3 S- b 0 - a WWKx3O SS0 0 WW L P T XL P T X



MCQ 7.29

Control Systems

Page 337

Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

nodia

(A) P - 3, Q - 1, R - 4, S - 2 (B) P - 3, Q - 2, R - 4, S - 1 (C) P - 2, Q - 1, R - 4, S - 2 (D) P - 3, Q - 4, R - 1, S - 2 MCQ 7.30

A certain system has transfer function G (s) = 2 s + 8 s + as - 4 where a is a parameter. Consider the standard negative unity feedback configuration as shown below

Which of the following statements is true? (A) The closed loop systems is never stable for any value of a (B) For some positive value of a, the closed loop system is stable, but not for all positive values. (C) For all positive values of a, the closed loop system is stable. (D) The closed loop system stable for all values of a, both positive and negative. MCQ 7.31

The number of open right half plane of 10 is G (s) = 5 4 3 s + 2s + 3s + 6s2 + 5s + 3 (A) 0 (B) 1 (C) 2 (D) 3



MCQ 7.32

MCQ 7.33

Control Systems

Chapter 7

The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec. The transfer function of the 3 system is 500 (A) 2 (B) 2 375 s + 10s + 100 s + 5s + 75 720 (C) 2 (D) 2 1125 s + 12s + 144 s + 25s + 225 Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R2 C2 > R1 C1. The transfer functions V0 represents a kind of controller. Vi

Match the impedances in Group I with the type of controllers in Group II

nodia

(A) Q - 1, R - 2

(B) Q - 1, R - 3

(C) Q - 2, R - 3

(D) Q - 3, R - 2

2007 MCQ 7.34

ONE MARK

s-5 If the closed-loop transfer function of a control system is given as T (s) (s + 2)( s + 3) , then It is (A) an unstable system (B) an uncontrollable system (C) a minimum phase system (D) a non-minimum phase system 2007

MCQ 7.35

TWO MARKS

A control system with PD controller is shown in the figure. If the velocity error constant KV = 1000 and the damping ratio z = 0.5 , then the value of KP and KD are

(A) KP = 100, KD = 0.09 (C) KP = 10, KD = 0.09

(B) KP = 100, KD = 0.9 (D) KP = 10, KD = 0.9



Control Systems

Page 339

5 (s + 5)( s2 + s + 1) The second-order approximation of T (s) using dominant pole concept is 1 5 (A) (B) (s + 5)( s + 1) (s + 5)( s + 1) (C) 2 5 (D) 2 1 s +s+1 s +s+1 T (s) =

MCQ 7.36

The transfer function of a plant is

MCQ 7.37

The open-loop transfer function of a plant is given as G (s) = s 1- 1 . If the plant is operated in a unity feedback configuration, then the lead compensator that an stabilize this control system is 10 (s - 1) 10 (s + 4) (B) (A) s+2 s+2 10 (s + 2) 2 (s + 2) (C) (D) s + 10 s + 10

MCQ 7.38

A unity feedback control system has an open-loop transfer function K G (s) = s (s2 + 7s + 12)

2

nodia

The gain K for which s = 1 + j1 will lie on the root locus of this system is (A) 4 (B) 5.5 (C) 6.5 (D) 10 MCQ 7.39

The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G (s) corresponding to this Bode plot is

1 (s + 1)( s + 20) 100 (C) s (s + 1)( s + 20)

1 s (s + 1)( s + 20) 100 (D) s (s + 1)( 1 + 0.05s)

(A)

MCQ 7.40

(B)

The state space representation of a separately excited DC servo motor dynamics is given as dw dt

-1

1 w

0

> di H = =- 1 - 10G=ia G + =10Gu dt o

where w is the speed of the motor, ia is the armature current and u is the w (s) armature voltage. The transfer function of the motor is U (s) 1 (A) 2 10 (B) 2 s + 11s + 11 s + 11s + 11 (C) 2 10s + 10 (D) 2 1 s + s + 11 s + 11s + 11



Control Systems

Chapter 7

Statement for linked Answer Question 41 and 42 :

MCQ 7.41

MCQ 7.42

Consider a linear system whose state space representation is x (t) = Ax (t). If 1 the initial state vector of the system is x (0) = = G, then the system response is 2 1 e-2x , then x (t) = > -2tH . If the itial state vector of the system changes to x (0) = = - 2G - 2e -t e the system response becomes x (t) = > -tH -e The eigenvalue and eigenvector pairs (li vi) for the system are 1 1 1 1 (A) e- 1 = Go and e- 2 = Go (B) e- 1, = Go and e2, = Go -1 -2 -1 -2 1 1 1 1 (C) e- 1, = Go and e- 2, = Go (D) e- 2 = Go and e1, = Go -1 -2 -1 -2 The system matrix A is 0 1 (A) = - 1 1G 2 1 (C) = - 1 - 1G 2006

MCQ 7.43

1 1 (B) = - 1 - 2G 0 1 (D) = - 2 - 3G

nodia

ONE MARK

The open-loop function of a unity-gain feedback control system is given by K G (s) = (s + 1)( s + 2) The gain margin of the system in dB is given by (A) 0 (B) 1 (C) 20 (D) 3 2006

MCQ 7.44

TWO MARKS

1 and G2 (s) = 2 s . s2 + as + b s + as + b The 3-dB bandwidths of their frequency responses are, respectively (B) a2 + 4b , a2 - 4b (A) a2 - 4b , a2 + 4b

Consider two transfer functions G1 (s) =

(C)

a2 - 4b , a2 - 4b

(D)

a 2 + 4b , a 2 + 4b

MCQ 7.45

The Nyquist plot of G (jw) H (jw)for a closed loop control system, passes through (- 1, j0) point in the GH plane. The gain margin of the system in dB is equal to (A) infinite (B) greater than zero (C) less than zero (D) zero

MCQ 7.46

The positive values of K and a so that the system shown in the figures below oscillates at a frequency of 2 rad/sec respectively are

(A) 1, 0.75 (C) 1, 1

(B) 2, 0.75 (D) 2, 2



MCQ 7.47

MCQ 7.48

Control Systems

Page 341

The transfer function of a phase lead compensator is given by Gc (s) = 1 + 3Ts 1 + Ts where T > 0 The maximum phase shift provide by such a compensator is (B) p (A) p 2 3 (C) p (D) p 4 6 A linear system is described by the following state equation 0 1 Xo (t) = AX (t) + BU (t), A = = - 1 0G The state transition matrix of the system is cos t sin t - cos t sin t (B) = (A) = - sin t cos t G - sin t - cos t G - cos t - sin t cos t - sin t (C) = (D) = - sin t cos t G cos t sin t G

nodia

Statement for Linked Answer Questions 49 and 50:

MCQ 7.49

Consider a unity - gain feedback control system whose open - loop transfer 1 function is : G (s) = as + s2 The value of a so that the system has a phase - margin equal to p is approximately 4 equal to (A) 2.40 (B) 1.40 (C) 0.84

MCQ 7.50

(D) 0.74

With the value of a set for a phase - margin of p , the value of unit - impulse 4 response of the open - loop system at t = 1 second is equal to (A) 3.40 (B) 2.40 (C) 1.84 (D) 1.74 2005

ONE MARK

MCQ 7.51

Which one of the following polar diagrams corresponds to a lag network ?

MCQ 7.52

A linear system is equivalently represented by two sets of state equations : Xo = AX + BU and Wo = CW + DU



Control Systems

Chapter 7

The eigenvalues of the representations are also computed as [l] and [m]. Which one of the following statements is true ? (B) [l] = [m] and X ! W (A) [l] = [m] and X = W (C) [l] ! [m] and X = W (D) [l] = [m] and X ! W MCQ 7.53

Despite the presence of negative feedback, control systems still have problems of instability because the (A) Components used have non- linearities (B) Dynamic equations of the subsystem are not known exactly. (C) Mathematical analysis involves approximations. (D) System has large negative phase angle at high frequencies. 2005

MCQ 7.54

TWO MARKS

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

nodia MCQ 7.55

(A) K < 5 and 1 < K < 1 (B) K < 1 and 1 < K < 5 2 8 8 2 1 1 (C) K < and 5 < K (D) K > and 5 > K 8 8 In the derivation of expression for peak percent overshoot - px Mp = exp e o # 100% 1 - x2 Which one of the following conditions is NOT required ? (A) System is linear and time invariant (B) The system transfer function has a pair of complex conjugate poles and no zeroes. (C) There is no transportation delay in the system. (D) The system has zero initial conditions.

MCQ 7.56

MCQ 7.57

A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively (A) 1 and 20 (B) 0 and 20 (D) 1 and 1 (C) 0 and 1 20 20 A double integrator plant G (s) = K/s2, H (s) = 1 is to be compensated to achieve the damping ratio z = 0.5 and an undamped natural frequency, wn = 5 rad/sec



MCQ 7.58

Control Systems

Page 343

which one of the following compensator Ge (s) will be suitable ? (A) s + 3 (B) s + 99 s + 99 s+3 (C) s - 6 (D) s - 6 s + 8.33 s K (1 - s) An unity feedback system is given as G (s) = . Indicate the correct root s (s + 3) locus diagram.

nodia

Statement for Linked Answer Question 59 and 60

MCQ 7.59

The open loop transfer function of a unity feedback system is given by -2s G (s) = 3e s (s + 2) The gain and phase crossover frequencies in rad/sec are, respectively (A) 0.632 and 1.26 (B) 0.632 and 0.485 (C) 0.485 and 0.632 (D) 1.26 and 0.632

MCQ 7.60

Based on the above results, the gain and phase margins of the system will be (A) -7.09 dB and 87.5c (B) 7.09 dB and 87.5c (C) 7.09 dB and - 87.5c (D) - 7.09 and - 87.5c 2004

ONE MARK

MCQ 7.61

The gain margin for the system with open-loop transfer function 2 (1 + s) , is G (s) H (s) = s2 (B) 0 (A) 3 (C) 1 (D) - 3

MCQ 7.62

K Given G (s) H (s) = s (s + 1)( .The point of intersection of the asymptotes of the s + 3) root loci with the real axis is (A) - 4 (B) 1.33 (C) - 1.33 (D) 4



Control Systems

Chapter 7

2004 MCQ 7.63

TWO MARKS

Consider the Bode magnitude plot shown in the fig. The transfer function H (s) is

(A)

(s + 10) (s + 1)( s + 100)

(B)

10 (s + 1) (s + 10)( s + 100)

102 (s + 1) 103 (s + 100) (D) (s + 10)( s + 100) (s + 1)( s + 10) A causal system having the transfer function H (s) = 1/ (s + 2) is excited with 10u (t). The time at which the output reaches 99% of its steady state value is (A) 2.7 sec (B) 2.5 sec (C) 2.3 sec (D) 2.1 sec (C)

MCQ 7.64

MCQ 7.65

MCQ 7.66

nodia

A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is (A) - 90c (B) 0c (C) 90c (D) - 180c Consider the signal flow graph shown in Fig. The gain x5 is x1

(A)

abcd 1 - (be + cf + dg) + bedg -2 2 If A = = , then sin At is 1 - 3G (C)

MCQ 7.67

1 - (be + cf + dg) abcd

bedg 1 - (be + cf + dg) 1 - (be + cf + dg) + bedg (D) abcd (B)

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) (A) 1 = G 3 - sin (- 4t) + sin (- t) 2 sin (- 4t) + sin (- t) sin (- 2t) sin (2t) (B) = sin (t) sin (- 3t)G sin (4t) + 2 sin (t) 2 sin (- 4t) - 2 sin (- t) (C) 1 = 2 sin (4t) + sin (t) G 3 - sin (- 4t) + sin (t) cos (- t) + 2 cos (t) 2 cos (- 4t) + 2 cos (- t) (D) 1 = G 3 - cos (- 4t) + cos (- t) - 2 cos (- 4t) + cos (t) MCQ 7.68

The open-loop transfer function of a unity feedback system is K G (s) = s (s2 + s + 2)( s + 3) The range of K for which the system is stable is (A) 21 > K > 0 (B) 13 > K > 0 4



MCQ 7.69

MCQ 7.70

MCQ 7.71

Control Systems

(C) 21 < K < 3 (D) - 6 < K < 3 4 For the polynomial P (s) = s2 + s 4 + 2s3 + 2s2 + 3s + 15 the number of roots which lie in the right half of the s -plane is (A) 4 (B) 2 (C) 3 (D) 1 The state variable equations of a system are : xo1 =- 3x1 - x2 = u, xo2 = 2x1 and y = x1 + u . The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable (D) controllable and observable 1 0 Given A = = G, the state transition matrix eAt is given by 0 1 0 e-t (A) > -t H e 0

et 0 (B) = t G 0 e

e-t 0 (C) > H 0 e-t

0 et (D) = t G e 0

2003 MCQ 7.72

Page 345

nodia

ONE MARK

Fig. shows the Nyquist plot of the open-loop transfer function G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the closed-loop system is

(A) always stable (B) unstable with one closed-loop right hand pole (C) unstable with two closed-loop right hand poles (D) unstable with three closed-loop right hand poles MCQ 7.73

A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot 2003

MCQ 7.74

TWO MARKS

The signal flow graph of a system is shown in Fig. below. The transfer function C (s)/ R (s) of the system is



Control Systems

(A)

MCQ 7.76

6 s2 + 29s + 6

(B)

s (s + 2) s + 29s + 6

(D)

6s s2 + 29s + 6

s (s + 27) s + 29s + 6 K The root locus of system G (s) H (s) = has the break-away point s (s + 2)( s + 3) located at (B) (- 2.548, 0) (A) (- 0.5, 0) (C) (- 4, 0) (D) (- 0.784, 0) (C)

MCQ 7.75

Chapter 7

2

2

The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is

nodia

(s + 0.1) 3 (s + 0.1) 3 (B) 107 2 (s + 10)( s + 100) (s + 10) (s + 100) 2 (s + 0.1) (s + 0.1) 3 (C) (D) (s + 10) 2 (s + 100) (s + 10)( s + 100) 2 A second-order system has the transfer function C (s) = 2 4 R (s) s + 4s + 4 With r (t) as the unit-step function, the response c (t) of the system is represented by (A) 108

MCQ 7.77

MCQ 7.78

The gain margin and the phase margin of feedback system with 8 are G (s) H (s) = (s + 100) 3



Control Systems

(A) dB, 0c (C) 3, 0c MCQ 7.79

Page 347

(B) 3, 3 (D) 88.5 dB, 3

The zero-input response of a system given by the state-space equation x1 (0) 1 1 0 x1 xo1 =xo G = =1 1G=x G and =x (0)G = =0 G is 2 2 2 tet (A) = G t et (C) = t G te

et (B) = G t t (D) = t G te

2002 MCQ 7.80

MCQ 7.81

ONE MARK

Consider a system with transfer function G (s) = 2s + 6 . Its damping ratio ks + s + 6 will be 0.5 when the value of k is (A) 2 (B) 3 6 (C) 1 (D) 6 6 Which of the following points is NOT on the root locus of a system with the openk loop transfer function G (s) H (s) = s (s + 1)( s + 3) (A) s =- j 3 (B) s =- 1.5 (C) s =- 3 (D) s =- 3

nodia

MCQ 7.82

The phase margin of a system with the open - loop transfer function (1 - s) G (s) H (s) = (1 + s)( 2 + s) (A) 0c (B) 63.4c (C) 90c (D) 3

MCQ 7.83

The transfer function Y (s)/ U (s) of system described by the state equation xo (t) =- 2x (t) + 2u (t) and y (t) = 0.5x (t) is 1 (A) 0.5 (B) (s - 2) (s - 2) 1 (C) 0.5 (D) (s + 2) (s + 2) 2002

TWO MARKS

MCQ 7.84

The system shown in the figure remains stable when (A) k < - 1 (B) - 1 < k < 3 (C) 1 < k < 3 (D) k > 3

MCQ 7.85

The transfer function of a system is G (s) = (s + 1)(100 . For a unit - step input to s + 100) the system the approximate settling time for 2% criterion is

(A)100 sec (C) 1 sec

(B) 4 sec (D) 0.01 sec



MCQ 7.86

Control Systems

Chapter 7

The characteristic polynomial of a system is q (s) = 2s5 + s 4 + 4s3 + 2s2 + 2s + 1 The system is (A) stable (C) unstable

MCQ 7.87

(B) marginally stable (D) oscillatory

1 The system with the open loop transfer function G (s) H (s) = has a 2 ( + s s s + 1) gain margin of (A) - 6 db (B) 0 db (C) 35 db (D) 6 db 2001

ONE MARK

MCQ 7.88

The Nyquist plot for the open-loop transfer function G (s) of a unity negative feedback system is shown in the figure, if G (s) has no pole in the right-half of s plane, the number of roots of the system characteristic equation in the right-half of s -plane is (A) 0 (B) 1 (C) 2 (D) 3

MCQ 7.89

The equivalent of the block diagram in the figure is given is

MCQ 7.90

The root-locus diagram for a closed-loop feedback system is shown in the figure. The system is overdamped.

nodia

(A) only if 0 # k # 1

(B) only if 1 < k < 5



Control Systems

(C) only if k > 5 MCQ 7.91

Page 349

(D) if 0 # k < 1 or k > 5

If the characteristic equation of a closed - loop system is s2 + 2s + 2 = 0 , then the system is (A) overdamped (B) critically damped (C) underdamped (D) undamped 2001

MCQ 7.92

An electrical system and its signal-flow graph representations are shown the figure (A) and (B) respectively. The values of G2 and H , respectively are

nodia

Z3 (s) - Z3 (s) - Z3 (s) - Z3 (s) (B) , , Z1 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) Z3 (s) Z3 (s) - Z3 (s) Z3 (s) (C) (D) , , Z2 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) The open-loop DC gain of a unity negative feedback system with closed-loop transfer function 2 s + 4 is s + 7s + 13 (A) 4 (B) 4 13 9 (A)

MCQ 7.93

TWO MARK

(C) 4 MCQ 7.94

The feedback control system in the figure is stable

(A) for all K $ 0 (C) only if 0 # K < 1 2000 MCQ 7.95

(D) 13

(B) only if K $ 0 (D) only if 0 # K # 1 ONE MARK

An amplifier with resistive negative feedback has tow left half plane poles in its open-loop transfer function. The amplifier (A) will always be unstable at high frequency (B) will be stable for all frequency (C) may be unstable, depending on the feedback factor (D) will oscillate at low frequency.



Control Systems

Chapter 7

2000 MCQ 7.96

TWO MARKS

1 A system described by the transfer function H (s) = 3 is stable. 2 s + a s + ks + 3 The constraints on a and k are. (A) a > 0, ak < 3 (B) a > 0, ak > 3 (C) a < 0, ak > 3 (D) a > 0, ak < 3 1999

ONE MARK

MCQ 7.97

For a second order system with the closed-loop transfer function T (s) = 2 9 s + 4s + 9 the settling time for 2-percent band, in seconds, is (A) 1.5 (B) 2.0 (C) 3.0 (D) 4.0

MCQ 7.98

The gain margin (in dB) of a system a having the loop transfer function

(A) 0 (C) 6 MCQ 7.99

2 is s (s + 1)

nodia G (s) H (s) =

(B) 3

(D) 3

The system modeled described by the state equations is 0 1 0 x + > Hu X => H 2 -3 1 Y = 81 1B x

(A) controllable and observable (C) observable, but not controllable MCQ 7.100

(B) controllable, but not observable (D) neither controllable nor observable

The phase margin (in degrees) of a system having the loop transfer function G (s) H (s) = 2 3 is s (s + 1) (B) - 30c (A) 45c (C) 60c (D) 30c 1999

TWO MARKS

MCQ 7.101

An amplifier is assumed to have a single-pole high-frequency transfer function. The rise time of its output response to a step function input is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to as sinusoidal input is approximately at (A) 4.55 (B) 10 (C) 20 (D) 28.6

MCQ 7.102

If the closed - loop transfer function T (s) of a unity negative feedback system is given by an - 1 s + an T (s) = n s + a1 sn - 1 + .... + an - 1 s + an then the steady state error for a unit ramp input is



Control Systems

(A) an an - 1 (C) an - 2 an

Page 351

(B) an an - 2 (D) zero

MCQ 7.103

Consider the points s1 =- 3 + j4 and s2 =- 3 - j2 in the s-plane. Then, for a system with the open-loop transfer function G (s) H (s) = K 4 (s + 1) (A) s1 is on the root locus, but not s2 (B) s2 is on the root locus, but not s1 (C) both s1 and s2 are on the root locus (D) neither s1 nor s2 is on the root locus

MCQ 7.104

For the system described by the state equation R0V R 0 1 0V W S W S xo = S 0 0 1W x + S0W u SS1WW SS0.5 1 2WW X T X T If the control signal u is given by u = [- 0.5 - 3 - 5] x + v , then the eigen values of the closed-loop system will be (A) 0, - 1, - 2 (B) 0, - 1, - 3 (C) - 1, - 1, - 2 (D) 0, - 1, - 1 1998

nodia 3

ONE MARK

2

MCQ 7.105

The number of roots of s + 5s + 7s + 3 = 0 in the left half of the s -plane is (A) zero (B) one (C) two (D) three

MCQ 7.106

The transfer function of a tachometer is of the form (B) K (A) Ks s K (C) K (D) (s + 1) s (s + 1)

MCQ 7.107

Consider a unity feedback control system with open-loop transfer function K . G (s) = s (s + 1) The steady state error of the system due to unit step input is (A) zero (B) K (C) 1/K (D) infinite

MCQ 7.108

The transfer function of a zero-order-hold system is (B) (1/s) (1 - e-sT ) (A) (1/s) (1 + e-sT ) (C) 1 - (1/s) e-sT (D) 1 + (1/s) e-sT

MCQ 7.109

In the Bode-plot of a unity feedback control system, the value of phase of G (jw) at the gain cross over frequency is - 125c. The phase margin of the system is (A) - 125c (B) - 55c (C) 55c

(D) 125c



MCQ 7.110

MCQ 7.111

MCQ 7.112

MCQ 7.113

MCQ 7.114

Control Systems

Chapter 7

Consider a feedback control system with loop transfer function K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) The type of the closed loop system is (A) zero (B) one (C) two (D) three The transfer function of a phase lead controller is 1 + 3Ts . The maximum value 1 + Ts of phase provided by this controller is (A) 90c (B) 60c (C) 45c (D) 30c The Nyquist plot of a phase transfer function g (jw) H (jw) of a system encloses the (–1, 0) point. The gain margin of the system is (A) less than zero (B) zero (C) greater than zero (D) infinity 2 The transfer function of a system is 2s + 26s + 5 . The characteristic equation (s + 1) (s + 2) of the system is 2 (A) 2s + 6s + 5 = 0 (B) (s + 1) 2 (s + 2) = 0 (C) 2s2 + 6s + 5 + (s + 1) 2 (s + 2) = 0 (D) 2s2 + 6s + 5 - (s + 1) 2 (s + 2) = 0

nodia

In a synchro error detector, the output voltage is proportional to [w (t)] n, where w (t) is the rotor velocity and n equals (A) –2 (B) –1 (D) 2 (C) 1 1997

MCQ 7.115

In the signal flow graph of the figure is y/x equals

(A) 3 (C) 2 MCQ 7.116

ONE MARK

(B) 5 2 (D) None of the above

A certain linear time invariant system has the state and the output equations given below 1 - 1 X1 0 Xo1 > o H = >0 1 H>X H + >1H u 2 X2

(A) 1 (C) 0

dy y = 81 1B: X1 D, If X1 (0) = 1, X2 (0) =- 1, u (0) = 0, then X2 dt (B) –1 (D) None of the above

is t=0

***********



Control Systems

Page 353

SOLUTIONS SOL 7.1

Option (B) is correct. From the given plot, we obtain the slope as 20 log G2 - 20 log G1 Slope = log w2 - log w1 From the figure 20 log G2 20 log G1 and w1 w2 So, the slope is

=- 8 dB = 32 dB = 1 rad/s = 10 rad/s

Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w In decibel, 20 log G ^ jwh = 20 log k = 32

nodia 32

SOL 7.2

or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s Option (A) is correct. For the given SFG, we have two forward paths 20

Pk1 = ^1 h^s-1h^s-1h^1 h = s-2 Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to the paths Pk1 and Pk2 so, Dk 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are

L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so, D = 1 - ^L1 + L2 + L 3 + L 4h

= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h



Control Systems

Chapter 7

= 5 + 6s1 + 2s2 From Mason’s gain formulae Y ^s h s-2 + s-1 = SPk Dk = = 2s+1 D U ^s h 5s + 6s + 2 5 + 6s-1 + 2s-2 SOL 7.3

Option (A) is correct. For the shown state diagram we can denote the states x1 , x2 as below

So, from the state diagram, we obtain xo1 =- x1 - u

and

xo2 =- x2 + ^1 h^- 1h^1 h^- 1h u + ^- 1h^1 h^- 1h x1 xo2 =- x2 + x1 + u y = ^- 1h^1 h x2 + ^- 1h^1 h^- 1h x1 + ^1 h^- 1h^1 h^- 1h^1 h u

nodia

= x1 - x 2 + u Hence, in matrix form we can write the state variable equations - 1 0 x1 -1 xo1 > o H = > 1 - 1H >x H + > 1 H u x2 2 x1 and y = 81 - 1B > H + u x2 which can be written in more general form as -1 0 -1 X +> H Xo = > 1 - 1H 1 y = 81 - 1B X + u SOL 7.4

Option (A) is correct. From the obtained state-variable equations We have -1 0 A => 1 - 1H So,

S+1 0 SI - A = > - 1 S + 1H

1 >S + 1 0 H ^S + 1h2 1 S + 1 R 1 V S 0 W S+1 W =S 1 1 W S 2 S^S + 1h S + 1W T X Hence, the state transition matrix is obtained as eAt = L-1 ^SI - Ah-1 V_ ZR 1 ]]S 0 Wbb -1 0 S+1 W` = >e = L-1 [S 1 H -t 1 S W te e-t ]S^S + 1h2 S + 1Wb \T Xa and

^SI - Ah-1 =



SOL 7.5

Control Systems

Page 355

Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t 1 10

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka Ka H ^s h = = = 1 + G ^s h 1 + 10s + 10Ka s + ^Ka + 101 h Taking inverse Laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 or, (approximately) tcl = 1 ka Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 or, = 10 100 ka Hence, ka = 10 a

SOL 7.6

1 10

nodia


(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if G (s) =

G (jw) = 0 -w 2 + 9 = 0 SOL 7.7

&

w = 3 rad/s

Option (A) is correct. K (s + 1) [R (s) - Y (s)] s3 + as2 + 2s + 1 K (s + 1) K (s + 1) Y (s) ;1 + 3 R (s) = 3 E 2 s + as + 2s + 1 s + as2 + 2s + 1 Y (s) [s3 + as2 + s (2 + k) + (1 + k)] = K (s + 1) R (s) Y (s) K (s + 1) Transfer Function, H (s) = = R (s) s3 + as2 + s (2 + k) + (1 + k) Routh Table : Y (s) =



Control Systems

For oscillation,

Chapter 7

a (2 + K) - (1 + K) =0 a a = K+1 K+2 A (s) = as2 + (k + 1) = 0 s2 =- k + 1 = - k + 1 (k + 2) =- (k + 2) a (k + 1)

Auxiliary equation

s = j k+2 jw = j k + 2 w = k+2 = 2 k =2 a = 2 + 1 = 3 = 0.75 2+2 4

and SOL 7.8

(Oscillation frequency)

Option (D) is correct. General form of state equations are given as xo = Ax + Bu

nodia

yo = Cx + Du For the given problem R0V R 0 a 0V 1 S W W S A = S 0 0 a2W, B = S0W SS1WW SSa 0 0WW 3 RT 0 a 0VXR0V R 0VT X 1 WS W S W S AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW S 2 A B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 X XT X T T For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 X T a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not. SOL 7.9

Option (B) is correct. For given plot root locus exists from - 3 to 3, So there must be odd number of poles and zeros. There is a double pole at s =- 3 Now

poles = 0, - 2, - 3, - 3 zeros =- 1

Thus transfer function

G (s) H (s) =

k (s + 1) s (s + 2) (s + 3) 2

SOL 7.10

Option (A) is correct. We have G (jw) = 5 + jw Here s = 5 . Thus G (jw) is a straight line parallel to jw axis.

SOL 7.11

Option (B) is correct. Here

dy x = y1 and xo = 1 dx



Control Systems

Page 357

y1 x 1 y = > H = > H = > Hx y2 2x 2 y1 = 1 u s+2

Now

y1 (s + 2) yo1 + 2y1 xo + 2x xo xo

=u =u =u =- 2x + u

= [- 2] x + [1] u Drawing SFG as shown below

Thus

xo1 = [- 2] x1 + [1] u y1 = x1 ; y2 = 2x1

nodia y1 1 y = > H = > H x1 y2 2

Here SOL 7.12

x1 = x


100 s (s + 10) 2 100 Now G (jw) H (jw) = jw (jw + 10) 2 If wp is phase cross over frequency +G (jw) H (jw) = 180c We have

G (s) H (s) =

Thus

- 180c = 100 tan-1 0 - tan-1 3 - 2 tan-1 a

or or or or

- 180c =- 90 - 2 tan-1 (0.1wp) 45c = tan-1 (0.1wp)

Now

wp 10 k

tan 45c 0.1wp = 1 wp = 10 rad/se 100 G (jw) H (jw) = w (w2 + 100)

At w = wp G (jw) H (jw) =

100 = 1 10 (100 + 100) 20

Gain Margin =- 20 log 10 G (jw) H (jw) =- 20 log 10 b 1 l 20 = 26 dB SOL 7.13

Option (D) is correct. From option (D) TF = H (s) 100 100 = ! s (s2 + 100) s (s + 10) 2



SOL 7.14

Control Systems

Chapter 7

Option (B) is correct. From the given block diagram

H (s) = Y (s) - E (s) $

1 s+1

E (s) = R (s) - H (s) = R (s) - Y (s) +

E (s) (s + 1)

1 = R (s) - Y (s) s + 1D sE (s) = R (s) - Y (s) (s + 1) E (s) Y (s) = s+1

E (s) :1 -

nodia

...(1) ...(2)

sY (s) = R (s) - Y (s)

From (1) and (2)

(s + 1) Y (s) = R (s)

Transfer function

Y (s) = 1 R (s) s + 1

SOL 7.15

Option (B) is correct. Transfer function is given as Y (s) = s X (s) s + p jw H (jw) = jw + p H (s) =

Amplitude Response H (jw) = Phase Response Input Output

or or

w w2+ p 2

qh (w) = 90c - tan-1 a w k p x (t) = p cos a2t - p k 2 y (t) = H (jw) x (t - qh) = cos a2t - p k 3 w H (jw) = p = w2+ p 2 2 1 = , (w = 2 rad/ sec) p 4+p2 4p 2 = 4 + p 2 & 3p 2 = 4 p = 2/ 3

Alternative : qh = 9- p - a- p kC = p 3 2 6



Control Systems

So,

p 6 tan-1 a w k p w p 2 p

SOL 7.17

= p - tan-1 a w k 2 p =p-p =p 2 6 3 p = tan a k = 3 3 =

3,

(w = 2 rad/ sec)

p = 2/ 3

or SOL 7.16

Page 359

Option (A) is correct. Initial slope is zero, so K = 1 At corner frequency w 1 = 0.5 rad/ sec , slope increases by + 20 dB/decade, so there is a zero in the transfer function at w 1 At corner frequency w 2 = 10 rad/ sec , slope decreases by - 20 dB/decade and becomes zero, so there is a pole in transfer function at w 2 K a1 + s k w1 Transfer function H (s) = s a1 + w 2 k 1 a1 + s k (1 + 10s) 0. 1 = = s (1 + 0.1s) a1 + 0.1 k Option (D) is correct. Assign output of each integrator by a state variable

nodia xo1 =- x1 + x2 xo2 =- x1 + 2u

y = 0.5x1 + 0.5x2 State variable representation -1 1 0 x + > Hu xo = > H -1 0 2 yo = [0.5 0.5] x SOL 7.18

Option (C) is correct. By masson’s gain formula

Transfer function Y (s) H (s) = = U (s)

/ PK DK D



Control Systems

Chapter 7

Forward path given P1 (abcdef ) = 2 # 1 # 1 # 0.5 = 12 s s s 1 P2 (abcdef ) = 2 # # 1 # 0.5 3 Loop gain L1 (cdc) =- 1 s L2 (bcdb) = 1 # 1 # - 1 = -21 s s s D = 1 - [L1 + L2] = 1 - :- 1 - 12 D = 1 + 1 + 12 s s s s So,

SOL 7.19

D1 = 1, D2 = 2 Y (s) H (s) = = P1 D 1 + P2 D 2 D U (s) 1 :1+1:1 2 (1 + s) s =s = 2 1 1 ( + s + 1) s 1+ + 2 s s

nodia

Option (D) is correct. Steady state error is given as eSS = lim s"0

sR (s) 1 + G (s) GC (s)

R (s) = 1 s

(unit step unit)

1 1 + G (s) GC (s) 1 = lim s"0 GC (s) 1+ 2 s + 2s + 2 eSS will be minimum if lim GC (s) is maximum s"0 In option (D) lim GC (s) = lim 1 + 2 + 3s = 3 s s"0 s"0 1 So, eSS = lim = 0 (minimum) s"0 3 eSS = lim s"0

SOL 7.20

Option (C) is correct. This compensator is roughly equivalent to combining lead and lad compensators in the same design and it is referred also as PID compensator.

SOL 7.21

Option (C) is correct. Here

p 0 and B = = G q 1G 0 p p = 1G=q G =q G p q S = 8B AB B = = q pG

1 A == 0 1 AB = = 0

S = pq - pq = 0 Since S is singular, system is completely uncontrollable for all values of p and q .



SOL 7.22

Control Systems

Page 361

Option (B) is correct. The characteristic equation is 1 + G (s) H (s) = 0 K (s2 - 2s + 2) or =0 1+ s2 + 2s + 2 or s2 + 2s + 2 + K (s2 - 2s + 2) = 0 2 or K =- s2 + 2s + 2 s - 2s + 2 For break away & break in point differentiating above w.r.t. s we have 2 2 dK =- (s - 2s + 2)( 2s + 2) - (s + 2s + 2)( 2s - 2) = 0 2 ds (s - 2s + 2) 2 Thus (s2 - 2s + 2)( 2s + 2) - (s2 + 2s + 2)( 2s - 2) = 0 or s =! 2 Let qd be the angle of departure at pole P , then

nodia - qd - qp1 + qz1 + qz2 = 180c

- qd = 180c - (- qp1 + qz1 + q2) = 180c - (90c + 180 - 45c) =- 45c

SOL 7.23

Option (B) is correct. For under-damped second order response

kwn2 s + 2xwn s + wn2 Thus (A) or (B) may be correct For option (A) wn = 1.12 and 2xwn = 2.59 " x = 1.12 For option (B) wn = 1.91 and 2xwn = 1.51 " x = 0.69 T (s) =

2

where x < 1

SOL 7.24

Option (B) is correct. The plot has one encirclement of origin in clockwise direction. Thus G (s) has a zero is in RHP.

SOL 7.25

Option (C) is correct. The Nyzuist plot intersect the real axis ate - 0.5. Thus G. M. =- 20 log x =- 20 log 0.5 = 6.020 dB And its phase margin is 90c.

SOL 7.26

Option (C) is correct. Transfer function for the given pole zero plot is: (s + Z1)( s + Z2) (s + P1)( s + P2) From the plot Re (P1 and P2 )>(Z1 and Z2 ) So, these are two lead compensator. Hence both high pass filters and the system is high pass filter.



SOL 7.27

Control Systems

Chapter 7

Option (C) is correct. Percent overshoot depends only on damping ratio, x . 2

Mp = e- xp 1 - x If Mp is same then x is also same and we get x = cos q Thus q = constant The option (C) only have same angle. SOL 7.28

Option (C) is correct. We labeled the given SFG as below :

nodia

From this SFG we have xo1 =- gx1 + bx3 + m1 xo2 = gx1 + ax3

SOL 7.29

SOL 7.30

xo3 =- bx1 - ax3 + u2 R V R VR V R V Sx1 W S- g 0 b WSx1 W S0 1 W u1 Sx2 W = S g 0 a WSx2 W + S0 0 We o Thus SSx WW SS- b 0 - a WWSSx WW SS1 0 WW u2 3 3 T X T XT X T X Option (D) is correct. P = 2 25 2xwn = 0, x = 0 " Undamped s + 25

Graph 3

Q=

62 s + 20s + 62

2xwn = 20, x > 1 " Overdamped

Graph 4

R=

62 s + 12s + 62

2xwn = 12, x = 1 " Critically

Graph 1

S=

72 s + 7s + 7 2

2xwn = 7, x < 1 " underdamped

Graph 2

2

2

2

Option (C) is correct. The characteristic equation of closed lop transfer function is 1 + G (s) H (s) = 0 =0 1+ 2 s+8 s + as - 4 or s 2 + as - 4 + s + 8 = 0 or s2 + (a + 1) s + 4 = 0 This will be stable if (a + 1) > 0 " a > - 1. Thus system is stable for all positive value of a.

SOL 7.31

Option (C) is correct. The characteristic equation is 1 + G (s) = 0



Control Systems

Page 363

s5 + 2s 4 + 3s3 + 6s2 + 5s + 3 = 0 Substituting s = z1 we have or

3z5 + 5z 4 + 6z3 + 3z2 + 2z + 1 = 0 The routh table is shown below. As there are two sign change in first column, there are two RHS poles. z5

3

6

2

z4

5

3

1

3

21 5

7 5

z2

4 3

3

z1

- 74

z0

1

z

SOL 7.32

Option (A) is correct. For underdamped second order system the transfer function is

nodia

Kwn2 s2 + 2xwn s + wn2 It peaks at resonant frequency. Therefore T (s) =

wr = wn 1 - 2x2

Resonant frequency and peak at this frequency

mr =

5 2x 1 - x2

We have wr = 5 2 , and mr = 10 . Only options (A) satisfy these values. 3 1 wn = 10, x = 2 where wr = 10 1 - 2` 1 j = 5 2 4 5 and Hence satisfied = 10 mr = 1 2 2 1 - 14 3 SOL 7.33

Option (B) is correct. The given circuit is a inverting amplifier and transfer function is Vo = - Z = - Z (sC1 R1 + 1) R Vi R1 sC R + 1 1

1

For Q ,

(sC2 R2 + 1) sC2 (sC2 R2 + 1) (sC1 R1 + 1) =# sC2 R1 R2 = (sC2 R2 + 1) (sC1 R1 + 1) R2 =# (sC2 R2 + 1) R1

Z = Vo Vi

For R,

1

Z Vo Vi

PID Controller

Since R2 C2 > R1 C1, it is lag compensator. SOL 7.34

Option (D) is correct. In a minimum phase system, all the poles as well as zeros are on the left half of



Control Systems

Chapter 7

the s -plane. In given system as there is right half zero (s = 5), the system is a non-minimum phase system. SOL 7.35

Option (B) is correct. We have Kv = lim sG (s) H (s) s"0

1000 = lim s

or

s"0

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equations is 1 + G (s) H (s) = 0 1000 = lims " 0 s

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equation is 1 + G (s) H (s) = 0 (100 + KD s) 100 =0 1+ s (s + 10)

or

Kp = 100

nodia

or s2 + (10 + 100KD) s + 10 4 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get or SOL 7.36

2xwn = 10 + 100KD KD = 0.9


5 (s + 5)( s2 + s + 1) 5 = = 2 1 5`1 + s j (s2 + s + 1) s +s+1 5 In given transfer function denominator is (s + 5)[( s + 0.5) 2 + 43 ]. We can see We have

T (s) =

easily that pole at s =- 0.5 ! j have approximated it. SOL 7.37

3 2

is dominant then pole at s =- 5 . Thus we

Option (A) is correct. G (s) =

1 = 1 s2 - 1 (s + 1)( s - 1)

1 The lead compensator C (s) should first stabilize the plant i.e. remove (s - 1) term. From only options (A), C (s) can remove this term Thus

SOL 7.38

10 (s - 1) 1 # (s + 1)( s - 1) (s + 2) 10 Only option (A) satisfies. = (s + 1)( s + 2)

G (s) C (s) =

Option (D) is correct. For ufb system the characteristics equation is 1 + G (s) = 0 K or 1+ =0 s (s2 + 7s + 12) or s (s2 + 7s + 12) + K = 0 Point s =- 1 + j lie on root locus if it satisfy above equation i.e (- 1 + j)[( - 1 + j) 2 + 7 (- 1 + j) + 12) + K] = 0



Control Systems

K =+ 10

or SOL 7.39

Page 365

Option (D) is correct. At every corner frequency there is change of -20 db/decade in slope which indicate pole at every corner frequency. Thus K G (s) = s (1 + s)`1 + s j 20 Bode plot is in (1 + sT) form 20 log K = 60 dB = 1000 w w = 0. 1 K =5

Thus

G (s) =

Hence SOL 7.40

100 s (s + 1)( 1 + .05s)

Option (A) is correct. We have or and

dw dt

-1

1 w

0

> di H = =- 1 - 10G=in G + =10Gu dt a

dw =- w + i n dt dia =- w - 10i + 10u a dt

nodia

...(1) ...(2)

Taking Laplace transform (i) we get

sw (s) =- w (s) = Ia (s) or (s + 1) w (s) = Ia (s) Taking Laplace transform (ii) we get or or or or SOL 7.41

sIa (s) =- w (s) - 10Ia (s) + 10U (s) w (s) = (- 10 - s) Ia (s) + 10U (s) = (- 10 - s)( s + 1) w (s) + 10U (s) w (s) =- [s2 + 11s + 10] w (s) + 10U (s) (s2 + 11s + 11) w (s) = 10U (s) w (s) = 2 10 U (s) (s + 11s + 11)

...(3)

From (3)

Option (A) is correct. We have xo (t) = Ax (t) p q Let A == r sG 1 e-2t For initial state vector x (0) = = G the system response is x (t) = > H -2 - 2e-2t Thus

e-2t > d (- 2e-2t)H dt d dt

t=0

or

- 2e-2 (0)

p q

p q 1 == r s G=- 2G 1

> 4e-2 (0) H = =r s G=- 2G -2 p - 2q = 4 G = =r - 2s G

We get

p - 2q =- 2 and r - 2s = 4

...(i)



Control Systems

Chapter 7

1 e-t For initial state vector x (0) = = G the system response is x (t) = > -tH -1 -e Thus

d dt

e-t

> d (- e-t)H dt

t=0

- e- (0)

p q 1 == r s G=- 1G

p q

1

> e- (0) H = =r s G=- 1G -1 p-q = 1 G = =r - s G We get p - q =- 1 and r - s = 1 Solving (1) and (2) set of equations we get p q 0 1 =r s G = =- 2 - 3G

...(2)

The characteristic equation lI - A = 0

nodia l -1 =0 2 l+3

or

l (l + 3) + 2 = 0

or l =- 1, - 2 Thus Eigen values are - 1 and - 2 Eigen vectors for l1 =- 1

(l1 I - A) X1 = 0

or

or or

l1 - 1 x11 = 2 l + 3G=x G = 0 1 21 - 1 - 1 x11 = 2 2 G=x G = 0 21 - x11 - x21 = 0

x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K , then x21 =- K , the Eigen vector will be x11 K 1 =x G = =- K G = K =- 1G 21 Now Eigen vector for l2 =- 2 or or or

(l2 I - A) X2 = 0 l2 - 1 x12 = 2 l + 3G=x G = 0 2 22 - 2 - 1 x11 = 2 1 G=x G = 0 21 - x11 - x21 = 0

x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K, then x21 =- K , the Eigen vector will be x12 K 1 =x G = =- 2K G = K =- 2G 22 or



Control Systems

Page 367

SOL 7.42

Option (D) is correct. As shown in previous solution the system matrix is 0 1 A == - 2 - 3G

SOL 7.43

Option (D) is correct. Given system is 2nd order and for 2nd order system G.M. is infinite.

SOL 7.44


SOL 7.45

Option (D) is correct. If the Nyquist polt of G (jw) H (jw) for a closed loop system pass through (- 1, j0) point, the gain margin is 1 and in dB GM =- 20 log 1 = 0 dB

SOL 7.46

Option (B) is correct. The characteristics equation is

nodia 1 + G (s) H (s) = 0

1+

K (s + 1) =0 s3 + as2 + 2s + 1

s3 + as2 + (2 + K) s + K + 1 = 0 The Routh Table is shown below. For system to be oscillatory stable a (2 + K) - (K + 1) =0 a or a = K+1 K+2

...(1)

Then we have as2 + K + 1 = 0 At 2 rad/sec we have s = jw " s2 =- w2 =- 4 , - 4a + K + 1 = 0 Solving (i) and (ii) we get K = 2 and a = 0.75 . Thus

s3 s

SOL 7.47

2

1

2+K

a

1+K

s1

(1 + K) a - (1 + K) a

s0

1+K

Option (D) is correct. The transfer function of given compensator is Gc (s) = 1 + 3Ts 1 + Ts Comparing with Gc (s) = 1 + aTs we get a = 3 1 + Ts

...(2)

T>0

The maximum phase sift is



Control Systems

fmax = tan-1 a - 1 = tan-1 3 - 1 = tan-1 1 2 a 2 3 3 p fmax = 6

or SOL 7.48

Chapter 7

Option (A) is correct. s 0 0 1 s -1 == (sI - A) = = G - = G 0 s -1 0 1 sG (sI - A) -1 =

s 1 =s - 1G = >s + 1 1 s2 + 1 1 s s +1 2

2

1 s2 + 1 s s2 + 1

H

cos t sin t f (t) = eAt = L-1 [(sI - A)] -1 = = - sin t cos t G SOL 7.49


1 G (s) = as + s2 +G (jw) = tan-1 (wa) - p Since PM is p i.e. 45c, thus 4 p = p + +G (jw ) w " Gain cross over Frequency g g 4 p = p + tan-1 (w a) - p or g 4 p = tan-1 (w a) or g 4 We have

nodia

or awg = 1 At gain crossover frequency G (jwg) = 1 Thus or or SOL 7.50

1 + a2 wg2 =1 wg2

1 + 1 = wg2

(as awg = 1)

wg = (2)

1 4

Option (C) is correct. For a = 0.84 we have G (s) = 0.84s2 + 1 s Due to ufb system H (s) = 1 and due to unit impulse response R (s) = 1, thus C (s) = G (s) R (s) = G (s) = 0.84s2 + 1 = 12 + 0.84 s s s Taking inverse Laplace transform At t = 1,

SOL 7.51

c (t) = (t + 0.84) u (t) c (1 sec) = 1 + 0.84 = 1.84

Option (D) is correct. The transfer function of a lag network is T (s) = 1 + sT 1 + sbT T (jw) =

b > 1; T > 0

1 + w2 T2 1 + w2 b2 T2



Control Systems

and At w = 0 , At w = 0 , At w = 3 ,

+T (jw) = tan-1 (wT) - tan-1 (wbT) T (jw) = 1 +T (jw) =- tan-1 0 = 0 T (jw) = 1 b

At w = 3 , SOL 7.52

Page 369

+T (jw) = 0

Option (C) is correct. We have where l is set of Eigen values Xo = AX + BU where m is set of Eigen values and Wo = CW + DU If a liner system is equivalently represented by two sets of state equations, then for both sets, states will be same but their sets of Eigne values will not be same i.e. X = W but l ! m

SOL 7.53

Option (A) is correct. Despite the presence of negative feedback, control systems still have problems of instability because components used have nonlinearity. There are always some variation as compared to ideal characteristics.

SOL 7.54


SOL 7.55

Option (C) is correct. The peak percent overshoot is determined for LTI second order closed loop system with zero initial condition. It’s transfer function is

nodia

wn2 s + 2xwn s + wn2 Transfer function has a pair of complex conjugate poles and zeroes. T (s) =

SOL 7.56

2

Option (A) is correct. For ramp input we have R (s) = 12 s Now ess = lim sE (s) s"0

R (s) 1 = lim 1 + G (s) s " 0 s + sG (s) ess = lim 1 = 5% = 1 s " 0 sG (s) 20 kv = 1 = lim sG (s) = 20 s"0 ess = lim s s"0

or But

Finite

kv is finite for type 1 system having ramp input. SOL 7.57


SOL 7.58

Option (C) is correct. Any point on real axis of s - is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option. The characteristics equation is 1 + G (s) H (s) = 0 or

1+

K (1 - s) =0 s (s + 3)



Control Systems

or

Chapter 7

2 K = s + 3s 1-s

For break away & break in point dK = (1 - s)( 2s + 3) + s2 + 3s = 0 ds - s2 + 2s + 3 = 0 which gives s = 3 , - 1 Here - 1 must be the break away point and 3 must be the break in point. or

SOL 7.59

Option (D) is correct. -2s G (s) = 3e s (s + 2) -2jw or G (jw) = 3e jw (jw + 2) 3 G (jw) = w w2 + 4 Let at frequency wg the gain is 1. Thus 3 =1 wg (wg2 + 4)

or or or Now

nodia wg4 + 4wg2 - 9 = 0 wg2 = 1.606 wg = 1.26 rad/sec +G (jw) =- 2w - p - tan-1 w 2 2

Let at frequency wf we have +GH =- 180c w - p =- 2wf - p - tan-1 f 2 2 w or 2wf + tan-1 f = p 2 2 wf 1 wf 3 or - ` jm = p 2wf + c 2 2 3 2 or

or SOL 7.60

5wf wf3 =p 2 2 24 5wf .p 2 2 wf = 0.63 rad

Option (D) is correct. The gain at phase crossover frequency wf is 3 3 = G (jwg) = wf (wf2 + 4) 0.63 (0.632 + 4) or G (jwg) = 2.27 G.M. =- 20 log G (jwg) - 20 log 2.26 =- 7.08 dB Since G.M. is negative system is unstable. The phase at gain cross over frequency is w +G (jwg) =- 2wg - p - tan-1 g 2 2

1 2



Control Systems

Page 371

=- 2 # 1.26 - p - tan-1 1.26 2 2 or

=- 4.65 rad or - 266.5c PM = 180c + +G (jwg) = 180c - 266.5c =- 86.5c

SOL 7.61

Option (D) is correct. The open loop transfer function is 2 (1 + s) G (s) H (s) = s2 Substituting s = jw we have 2 (1 + jw) G (jw) H (jw) = - w2 +G (jw) H (jw) =- 180c + tan-1 w The frequency at which phase becomes - 180c, is called phase crossover frequency. Thus - 180 =- 180c + tan-1 wf

...(1)

tan-1 wf = 0 wf = 0 The gain at wf = 0 is or or

SOL 7.62

SOL 7.63

nodia

w2 = 3 G (jw) H (jw) = 2 1 + 2 w 1 Thus gain margin is = = 0 and in dB this is - 3 . 3 Option (C) is correct. Centroid is the point where all asymptotes intersects. SReal of Open Loop Pole - SReal Part of Open Loop Pole s = SNo.of Open Loop Pole - SNo.of Open Loop zero = - 1 - 3 =- 1.33 3 Option (C) is correct. The given bode plot is shown below

At w = 1 change in slope is +20 dB " 1 zero at w = 1 At w = 10 change in slope is - 20 dB " 1 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K (s + 1) Thus T (s) = s s + 1) ( 10 + 1)( 100 Now 20 log10 K =- 20 " K = 0.1 0.1 (s + 1) 100 (s + 1) Thus = T (s) = s s + ( s 10)( s + 100) ( 10 + 1)( 100 + 1) SOL 7.64




Control Systems

We have or Now

Chapter 7

r (t) = 10u (t) R (s) = 10 s H (s) = 1 s+2 C (s) = H (s) $ R (s) =

1 $ 10 10 s + 2 s s (s + 2)

C (s) = 5 - 5 s s+2

or

c (t) = 5 [1 - e-2t] The steady state value of c (t) is 5. It will reach 99% of steady state value reaches at t , where 5 [1 - e-2t] = 0.99 # 5 1 - e-2t = 0.99

or or or

e-2t = 0.1 - 2t = ln 0.1 t = 2.3 sec

nodia

SOL 7.65

Option (A) is correct. Approximate (comparable to 90c) phase shift are Due to pole at 0.01 Hz " - 90c Due to pole at 80 Hz " - 90c Due to pole at 80 Hz " 0 Due to zero at 5 Hz " 90c Due to zero at 100 Hz " 0 Due to zero at 200 Hz " 0 Thus approximate total - 90c phase shift is provided.

SOL 7.66

Option (C) is correct. Mason Gain Formula T (s) =

Spk 3 k 3

In given SFG there is only one forward path and 3 possible loop. p1 = abcd 31 = 1 3= 1 - (sum of indivudual loops) - (Sum of two non touching loops) = 1 - (L1 + L2 + L3) + (L1 L3) Non touching loop are L1 and L3 where L1 L2 = bedg Thus

SOL 7.67

C (s) p1 3 1 = 1 - (be + cf + dg) + bedg R (s) abcd = 1 - (be + cf + dg) + bedg


-2 2 A == 1 - 3G

Characteristic equation is



Control Systems

Page 373

[lI - A] = 0 l + 2 -2 =0 -1 l + 3

or

(l + 2)( l + 3) - 2 = 0 or l2 + 5l + 4 = 0 Thus l1 =- 4 and l2 =- 1 Eigen values are - 4 and - 1. Eigen vectors for l1 =- 4 or

(l1 I - A) X1 = 0 l1 + 2 - 2 x11 = 1 l + 3G=x G = 0 1 21 - 2 - 2 x11 =- 1 - 1G=x G = 0 21

or

- 2x11 - 2x21 = 0

or

x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x21 = K , then x11 =- K , the Eigen vector will be x11 -K -1 =x G = = K G = K = 1 G 21 or

nodia

Now Eigen vector for l2 =- 1 or or

(l2 I - A) X2 = 0 l2 + 2 - 2 x12 = - 1 l + 3G=x G = 0 2 22 1 - 2 x12 =- 1 2 G=x G = 0 22

We have only one independent equation x12 = 2x22 Let x22 = K , then x12 = 2K . Thus Eigen vector will be 2 x12 2K =x G = = K G = K =1 G 22 Digonalizing matrix

Now

-1 2 x11 x12 = M == x21 x22 G = 1 1G 1 -2 M-1 = ` - 1 j= 1 - 1G 3

Now Diagonal matrix of sin At is D where sin (l1 t) 0 sin (- 4t) 0 = D == 0 sin (l2 t)G = 0 sin (l2 t)G Now matrix

B = sin At = MDM-1 - 1 2 sin (- 4t) 0 1 -2 =-` 1 j= 1 1G= 0 sin (- t)G=- 1 - 1G 3 - sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) + 2 sin (t) - 2 sin (- 4t) - sin (- t)G 3 - sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) - sin (- t) - 2 sin (- 4t) + 2 sin (- t)G 3



Control Systems

Chapter 7

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) = ` 1 j= Gs 3 - sin (- 4t + sin (- t) 2 sin (- 4t) + sin (- t) SOL 7.68

Option (A) is correct. For ufb system the characteristic equation is 1 + G (s) = 0 1 + G (s)

K =0 s (s + 2s + 2)( s + 3) s 4 + 4s3 + 5s2 + 6s + K = 0 The routh table is shown below. For system to be stable, (21 - 4K) 0 < K and 0 < 2/7 21 This gives 0
s4 s

3

s

2

s1 s SOL 7.69

0

2

1

5

K

nodia 4

6

7 2

K

21 - 4K 7/2

0

0

K

Option (B) is correct. We have P (s) = s5 + s 4 + 2s3 + 3s + 15 The routh table is shown below. If e " 0+ then 2e +e 12 is positive and -15e2-e +2412e - 144 is negative. Thus there are two sign change in first column. Hence system has 2 root on RHS of plane. 2

s5

1

2

3

4

1

2

15

s3

e

- 12

0

s2

2e + 12 e

15

0

s

1

SOL 7.70

s

-15e2 - 24e - 144 2e + 12

s0

0

Option (D) is correct. - 3 - 1 x1 x1 1 =x G = = 2 0 G=x G + =0 Gu 2 2 x1 1 and Y = [1 0]= G + = Gu x2 2 -3 -1 1 Here , B = = G and C = [1 0] A == G 2 0 0 The controllability matrix is QC = [B AB ] 1 -3 == 0 2G We have



Control Systems

Page 375

Thus controllable

det QC ! 0 The observability matrix is Q0 = [CT AT CT ] 1 -3 == !0 0 - 1G

Thus observable

det Q0 ! 0 SOL 7.71

Option (B) is correct. s 0 1 0 s-1 0 (sI - A) = = G - = G = = 0 s 0 1 0 s - 1G (sI - A) -1 =

1 0 (s - 1) s-1 1 = G = > 0 (s - 1) 0 (s - 1) 2

0 1 s-1

H

eAt = L-1 [(sI - A)] -1 et 0 = = tG 0 e SOL 7.72


nodia

Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right hand side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 1 and P = 1

Thus Z =0 Hence there are no roots on RH of s -plane and system is always stable. SOL 7.73

Option (C) is correct. PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot.

SOL 7.74

Option (D) is correct. Mason Gain Formula T (s) =

Spk 3 k 3

In given SFG there is only forward path and 3 possible loop. p1 = 1 31 = 1 + 3 + 24 = s + 27 s s s L1 = - 2 , L2 = - 24 and L3 = - 3 s s s where L1 and L3 are non-touching C (s) p1 3 1 This = 1 - (loop gain) + pair of non - touching loops R (s)

^ s +s27 h = 1 - ^ - 24s - s2 h + -s2 . -s3 s (s + 27) = 2 s + 29s + 6 Option (D) is correct. We have =

SOL 7.75

-3 s

^

s + 27 s

h

1 + 29s + s62



Control Systems

or

Chapter 7

1 + G (s) H (s) = 0 K 1+ =0 s (s + 2)( s + 3) K =- s (s2 + 5s2 + 6s) dK =- (3s2 + 10s + 6) = 0 ds

or

which gives

s = - 10 ! 100 - 72 =- 0.784, - 2.548 6

The location of poles on s - plane is

Since breakpoint must lie on root locus so s =- 0.748 is possible. SOL 7.76

Option (A) is correct. The given bode plot is shown below

nodia

At w = 0.1 change in slope is + 60 dB " 3 zeroes at w = 0.1 At w = 10 change in slope is - 40 dB " 2 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 Thus Now or Thus SOL 7.77

K ( 0s.1 + 1) 3 s ( 10s + 1) 2 ( 100 + 1) 20 log10 K = 20 K = 10 10 ( 0s.1 + 1) 3 108 (s + 0.1) 3 T (s) = s = 2 s ( 10 + 1) ( 100 + 1) (s + 10) 2 (s + 100) T (s) =

Option (B) is correct. The characteristics equation is s2 + 4s + 4 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 4 and wn2 = 4 Thus x =1 ts = 4 = 4 = 2 1#2 xwn

SOL 7.78


SOL 7.79


Critically damped



Control Systems

1 xo1 =xo G = =1 2 1 A == 1 s (sI - A) = = 0

Page 377

0 x1 x1 (0) 1 and = = 1G=x2 G x2 (0)G =0 G 0 1G 0 1 0 s-1 0 = s G =1 1G = - 1 s - 1G

1 0 s-1 1 >(s - 1) = H > +1 (s - 1) 2 + 1 (s - 1) (s - 1) et 0 -1 -1 At L [(sI - A) ] = e = = t t G te e et 0 1 et x (t) = eAt # [x (t0)] = = t t G= G = = t G te e 0 te

0

(sI - A) -1 =

SOL 7.80

2

1 s-1

H

Option (C) is correct. The characteristics equation is or

ks2 + s + 6 = 0 s2 + 1 s + 6 = 0 K K

nodia

Comparing with s2 + 2xwn s + wn2 = 0 we have we get 2xwn = 1 and wn2 = 6 K K or 2 # 0.5 # 6 Kw = 1 K 6 = 1 & K =1 or K 6 K2

Given x = 0.5

SOL 7.81

Option (B) is correct. Any point on real axis lies on the root locus if total number of poles and zeros to the right of that point is odd. Here s =- 1.5 does not lie on real axis because there are total two poles and zeros (0 and - 1) to the right of s =- 1.5 .

SOL 7.82

Option (D) is correct. From the expression of OLTF it may be easily see that the maximum magnitude is 0.5 and does not become 1 at any frequency. Thus gain cross over frequency does not exist. When gain cross over frequency does not exist, the phase margin is infinite.

SOL 7.83

Option (D) is correct. We have xo (t) =- 2x (t) + 2u (t) Taking Laplace transform we get or or Now or or

...(i)

sX (s) =- 2X (s) + 2U (s) (s + 2) X (s) = 2U (s) 2U (s) X (s) = (s + 2) y (t) = 0.5x (t) Y (s) = 0.5X (s) 0.5 # 2U (s) Y (s) = s+2 Y (s) 1 = (s + 2) U (s)



Control Systems

SOL 7.84

Option (D) is correct. From Mason gain formula we can write transfer function as K Y (s) K s = = 3 R (s) 1 - ( s + -sK ) s - 3 (3 - K) For system to be stable (3 - K) < 0 i.e. K > 3

SOL 7.85

Option (B) is correct. The characteristics equation is (s + 1)( s + 100) = 0 s2 + 101s + 100 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 101 and wn2 = 100 Thus x = 101 20

Chapter 7

Overdamped

For overdamped system settling time can be determined by the dominant pole of the closed loop system. In given system dominant pole consideration is at s =- 1. Thus

nodia 1 =1 T

SOL 7.86

and Ts = 4 = 4 sec T

Option (B) is correct. Routh table is shown below. Here all element in 3rd row are zero, so system is marginal stable. s5 s

4

s3 s

2

4

2

1

2

1

0

0

0

2

s1 s0 SOL 7.87

Option (B) is correct. The open loop transfer function is 1 G (s) H (s) = s (s2 + s + 1) Substituting s = jw we have 1 G (jw) H (jw) = jw (- w2 + jw + 1) +G (jw) H (jw) =- p - tan-1 w 2 2 (1 - w ) The frequency at which phase becomes - 180c, is called phase crossover frequency. wf Thus - 180 =- 90 - tan-1 1 - wf2 wf or - 90 =- tan-1 1 - wf2 or 1 - w2f = 0



Control Systems

Page 379

wf = 1 rad/sec The gain margin at this frequency wf = 1 is GM =- 20 log10 G (jwf) H (jwf) = 20 log10 (wf (1 - w2f) 2 + w2f =- 20 log 1 = 0 SOL 7.88

Option (A) is correct. Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right had side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 0 (1 encirclement in CW direction and other in CCW) and P = 0 Thus Z = 0 Hence there are no roots on RH of s - plane.

SOL 7.89

Option (D) is correct. Take off point is moved after G2 as shown below

SOL 7.90

Option (D) is correct. If roots of characteristics equation lie on negative axis at different positions (i.e. unequal), then system response is over damped. From the root locus diagram we see that for 0 < K < 1, the roots are on imaginary axis and for 1 < K < 5 roots are on complex plain. For K > 5 roots are again on imaginary axis. Thus system is over damped for 0 # K < 1 and K > 5 .

SOL 7.91

Option (C) is correct. The characteristics equation is

nodia

s2 + 2s + 2 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 2 and wn2 = 2 wn =

2 1 and x = 2 Since x < 1 thus system is under damped SOL 7.92

Option (C) is correct. From SFG we have I1 (s) = G1 Vi (s) + HI2 (s) I2 (s) = G2 I1 (s) V0 (s) = G3 I2 (s) Now applying KVL in given block diagram we have Vi (s) = I1 (s) Z1 (s) + [I1 (s) - I2 (s)] Z3 (s)

...(1) ...(2) ...(3) ...(4)



Control Systems

0 = [I2 (s) - I1 (s)] Z3 (s) + I2 (s) Z2 (s) + I2 (s) Z4 (s) From (4) we have or Vi (s) = I1 (s)[ Z1 (s) + Z3 (S)] - I2 (s) Z3 (S) Z3 (s) 1 or I1 (s) = Vi + I2 Z1 (s) + Z3 (s) Z1 (s) + Z3 (s)

Chapter 7

...(5)

...(6)

From (5) we have or

I1 (s) Z3 (S) = I2 (s)[ Z2 (s) + Z3 (s) + Z4 (s)] I1 (s) Z3 (s) Is (s) = Z3 (s) + Z2 (s) + Z4 (s)

...(7)

Comparing (2) and (7) we have Z3 (s) G2 = Z3 (s) + Z2 (s) + Z4 (s) Comparing (1) and (6) we have Z3 (s) H = Z1 (s) + Z3 (s) SOL 7.93

Option (B) is correct. For unity negative feedback system the closed loop transfer function is G (s) s+4 CLTF = , G (s) " OL Gain = 1 + G (s) s2 + 7s + 13 2 1 + G (s) or = s + 7s + 13 G (s) s+4

nodia

1 = s2 + 7s + 13 - 1 = s2 + 6s + 9 G (s) s+4 s+4 or G (s) = 2 s + 4 s + 6s + 9 For DC gain s = 0 , thus Thus G (0) = 4 9 or

SOL 7.94

Option (C) is correct. From the Block diagram transfer function is G (s) T (s) = 1 + G (s) H (s) K (s - 2) Where G (s) = (s + 2) and H (s) = (s - 2) The Characteristic equation is 1 + G (s) H (s) = 0 K (s - 2) 1+ (s - 2) = 0 (s + 2) 2 or (s + 2) 2 + K (s - 2) 2 = 0 or (1 + K) s2 + 4 (1 - K) s + 4K + 4 = 0 Routh Table is shown below. For System to be stable 1 + k > 0 , and 4 + 4k > 0 and 4 - 4k > 0 . This gives - 1 < K < 1 As per question for 0 # K < 1



Control Systems

s2

1+k

4 + 4k

s1

4 - 4k

0

s0

4 + 4k

Page 381

SOL 7.95

Option (B) is correct. It is stable at all frequencies because for resistive network feedback factor is always less than unity. Hence overall gain decreases.

SOL 7.96

Option (B) is correct. The characteristics equation is s2 + as2 + ks + 3 = 0 The Routh Table is shown below For system to be stable a > 0 and aK - 3 > 0 a Thus a > 0 and aK > 3 s3 s2 s1 s0

SOL 7.97

1

K

a

3

aK - 3 a

0

nodia 3

Option (B) is correct. Closed loop transfer function is given as T (s) = 2 9 s + 4s + 9 by comparing with standard form we get natural freq. wA2 = 9 wn = 3 2xwn = 4

4 = 2/3 2#3 For second order system the setting time for 2-percent band is given by 4 ts = 4 = =4 =2 xwn 3 # 2/3 2 x =

Damping factor

SOL 7.98

Option (D) is correct. Given loop transfer function is G (s) H (s) =

2 s (s + 1)

2 jw (jw + 1) Phase cross over frequency can be calculated as G (jw) H (jw) =

So here

f (w) at w = w =- 180c f (w) =- 90c - tan-1 (w) - 90c - tan-1 (wp) =- 180c tan-1 (wp) = 90c p

wp = 3



Control Systems

Chapter 7

Gain margin 20 log 10 =

1 at w = wp G (jw) H (jw) G G.M. = 20 log 10 e

2 =0 w2p + 1 G.M. = 20 log 10 b 1 l = 3 0

G (jwp) H (jwp) = so SOL 7.99

1 G (jw) H (jwp) o

wp

Option (A) is correct. 0 1 0 , B = = G and C = [1 1] A == G 2 -3 1

Here

The controllability matrix is 0 1 QC = [B AB ] = = 1 - 3G Thus controllable

det QC ! 0 The observability matrix is

nodia 1 2 !0 Q0 = [CT AT CT ] = = 1 - 2G

det Q0 ! 0

SOL 7.100

Thus observable


G (s) H (s) = 2 3 s (s + 1)

we have or

G (jw) H (jw) =

2 3 jw (jw + 1)

Gain cross over frequency G (jw) H (jw) at w = w = 1 g

2 3 =1 w w2 + 1 12 = w2 (w2 + 1) w4 + w2 - 12 = 0 (w2 + 4) (w2 - 3) = 0 w2 = 3 and w2 =- 4

or

w1, w2 = ! 3

which gives wg =

3 f (w) at w = w =- 90 - tan-1 (wg) g

=- 90 - tan-1 3 =- 90 - 60 =- 150

Phase margin = 180 + f (w) at w = w = 180 - 150 = 30c g

SOL 7.101


SOL 7.102

Option (C) is correct. Closed-loop transfer function is given by an - 1 s + an T (s) = n s + a1 sn - 1 + ... + an - 1 s + an



Control Systems

Page 383

an - 1 s + an n n-1 s a + + ...an - 2 s2 1s = a n - 1 s + an 1+ n s + a1 sn - 1 + ...an - 2 s2 G (s) H (s) =

Thus

an - 1 s + an sn + a1 sn - 1 + ....an - 2 s2

For unity feed back H (s) = 1 an - 1 s + an sn + a1 sn - 1 + ....an - 2 s2 Steady state error is given by 1 E (s) = lim R (s) s"0 1 + G (s) H (s) for unity feed back H (s) = 1 Here input R (s) = 12 (unit Ramp) s 1 1 so E (s) = lim 2 s " 0 s 1 + G (s) n a1 sn - 1 + .... + an - 2 s2 = lim 12 s + n s"0 s s + a1 sn - 1 + .... + an = an - 2 an G (s) =

Thus

nodia

SOL 7.103


SOL 7.104


SOL 7.105

Option (A) is correct. Applying Routh’s criteria

s3 + 5s2 + 7s + 3 = 0 s3

1

s2

5

1

7#5-3 5

0

3

s s

7 3 =

32 5

0

There is no sign change in the first column. Thus there is no root lying in the left-half plane. SOL 7.106

Option (A) is correct. Techometer acts like a differentiator so its transfer function is of the form ks .

SOL 7.107

Option (A) is correct. Open loop transfer function is K G (s) = s (s + 1) Steady state error sR (s) E (s) = lim s " 0 1 + G (s) H (s) R (s) = input

Where

H (s) = 1 (unity feedback)

R (s) = 1 s



SOL 7.108

Control Systems

Chapter 7

s1 s (s + 1) s so E (s) = lim = lim 2 =0 s"0 s"0 s + s + K K 1+ s (s + 1) Option (B) is correct. Fig given below shows a unit impulse input given to a zero-order hold circuit which holds the input signal for a duration T & therefore, the output is a unit step function till duration T .

h (t) = u (t) - u (t - T) Taking Laplace transform we have H (s) = 1 - 1 e-sT = 1 61 - e-sT @ s s s SOL 7.109

Here so SOL 7.110

nodia

Option (C) is correct. Phase margin = 180c + qg where qg = value of phase at gain crossover frequency. qg =- 125c P.M = 180c - 125c = 55c

Option (B) is correct. Open loop transfer function is given by K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) Close looped system is of type 1.

It must be noted that type of the system is defined as no. of poles lies at origin lying in OLTF. SOL 7.111

Option (D) is correct. Transfer function of the phase lead controller is 1 + (3Tw) j T.F = 1 + 3Ts = 1+s 1 + (Tw) j Phase is f (w) = tan-1 (3Tw) - tan-1 (Tw) w f (w) = tan-1 ; 3Tw - T 1 + 3T 2 w2 E f (w) = tan-1 ; 2Tw2 2 E 1 + 3T w For maximum value of phase df (w) =0 dw or

1 = 3T 2 w2 Tw = 1 3



Control Systems

Page 385

So maximum phase is fmax = tan-1 ; 2Tw2 2 E at Tw = 1 1 + 3T w 3 1 2 3 1 = 30c -1 = tan-1 > H = tan ; 1 + 3 # 13 3E SOL 7.112

Option (A) is correct. G (jw) H (jw) enclose the (- 1, 0) point so here G (jwp) H (jwp) > 1 wp = Phase cross over frequency 1 Gain Margin = 20 log 10 G (jwp) H (jwp) so gain margin will be less than zero.

SOL 7.113

Option (B) is correct. The denominator of Transfer function is called the characteristic equation of the system. so here characteristic equation is (s + 1) 2 (s + 2) = 0

SOL 7.114

Option (C) is correct. In synchro error detector, output voltage is proportional to [w (t)], where w (t) is the rotor velocity so here n = 1

SOL 7.115

Option (C) is correct. By masson’s gain formulae / Dk Pk y = x D

nodia

Forward path gain

so gain SOL 7.116

P1 = 5 # 2 # 1 = 10 D = 1 - (2 # - 2) = 1 + 4 = 5

D1 = 1 y = 10 # 1 = 2 5 x

Option (C) is correct. By given matrix equations we can have Xo1 = dx1 = x1 - x2 + 0 dt Xo2 = dx2 = 0 + x2 + m dt x1 y = [1 1] > H = x1 + x2 x2 dy = dx1 + dx2 dt dt dt dy = x1 + m dt dy = x1 (0) + m (0) dt t = 0 = 1+0 = 0 ***********


CHAPTER 8 ELECTROMAGNETICS

2013 MCQ 8.1

ONE MARK

v ^rvh. The closed loop line integral # A v : dlv can be Consider a vector field A expressed as vh : dsv over the closed surface bounded by the loop (A) ## ^d # A (B) (C) (D)

### ^d : Avhdv over the closed volume bounded by the loop ### ^d : Avhdv over the open volume bounded by the loop ## ^d # Avh : dsv over the open surface bounded by the loop

MCQ 8.2

v = xatx + yaty + zatz is The divergence of the vector field A (A) 0 (B) 1/3 (C) 1 (D) 3

MCQ 8.3

The return loss of a device is found to be 20 dB. The voltage standing wave ratio (VSWR) and magnitude of reflection coefficient are respectively (A) 1.22 and 0.1 (B) 0.81 and 0.1 (C) – 1.22 and 0.1 (D) 2.44 and 0.2

nodia

2013

TWO MARKS

Statement for Linked Answer Questions 4 and 5: A monochromatic plane wave of wavelength l = 600 mm is propagating in the direction as shown in the figure below. Evi , Evr and Evt denote incident, reflected, and transmitted electric field vectors associated with the wave.

MCQ 8.4

The angle of incidence qi and the expression for Evi are p # 10 ^x + 2h (A) 60c and E 0 âtx - atz h e-j 3 2 V/m 2 4


Electromagnetics

Page 387

p # 10 z (B) 45c and E 0 âtx + atz h e-j 3 V/m 2 p # 10 ^x + z h E (C) 45c and 0 âtx - atz h e-j 3 2 V/m 2 p # 10 z (D) 60c and E 0 âtx - atz h e-j 3 V/m 2 4

4

4

MCQ 8.5

The expression for Evr is p # 10 ^x - z h (A) 0.23 E 0 âtx + atz h e-j 3 2 V/m 2 p # 10 z (B) - E 0 âtx + atz h e j 3 V/m 2 p # 10 ^x - z h (C) 0.44 E 0 âtx + atz h e-j 3 2 V/m 2 p # 10 ^x + z h 3 (D) E 0 âtx + atz h e-j V/m 2 4

4

4

4

2012 MCQ 8.6

nodia

ONE MARK

A plane wave propagating in air with E = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m is incident on a perfectly conducting slab positioned at x # 0 . The E field of the reflected wave is (A) (- 8ax - 6ay - 5az ) e j (wt + 3x + 4y) V/m -

(B) (- 8ax + 6ay - 5az ) e j (wt + 3x + 4y) V/m (C) (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m (D) (- 8ax + 6ay - 5az ) e j (wt - 3x - 4y) V/m MCQ 8.7

The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10 (ay + jaz ) e-j 25x . The frequency and polarization of the wave, respectively, are (A) 1.2 GHz and left circular (B) 4 Hz and left circular (C) 1.2 GHz and right circular (D) 4 Hz and right circular

MCQ 8.8

A coaxial-cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. Given m0 = 4p # 10-7 H/m, -9 e0 = 10 F/m , the characteristic impedance of the cable is 36p (B) 100 W (A) 330 W (C) 143.3 W (D) 43.4 W

MCQ 8.9

The radiation pattern of an antenna in spherical co-ordinates is given by F (q) = cos 4 q ; 0 # q # p/2 . The directivity of the antenna is (A) 10 dB (B) 12.6 dB (C) 11.5 dB (D) 18 dB 2012

MCQ 8.10

TWO MARKS

A transmission line with a characteristic impedance of 100 W is used to match a 50 W section to a 200 W section. If the matching is to be done both at 429 MHz and 1 GHz, the length of the transmission line can be approximately



Electromagnetics

(A) 82.5 cm (C) 1.58 cm MCQ 8.11

Chapter 8

(b) 1.05 m (D) 1.75 m

The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz)

The phase velocity v p of the wave inside the waveguide satisfies (A) v p > c (B) v p = c (C) 0 < v p < c

(D) v p = 0

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Statement for Linked Answer Question 12 and 13 :

An infinitely long uniform solid wire of radius a carries a uniform dc current of density J MCQ 8.12

The magnetic field at a distance r from the center of the wire is proportional to (B) 0 for r < a and 1/r for r > a (A) r for r < a and 1/r 2 for r > a (C) r for r < a and 1/r for r > a (D) 0 for r < a and 1/r 2 for r > a

MCQ 8.13

A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below.

The magnetic field inside the hole is (A) uniform and depends only on d (B) uniform and depends only on b (C) uniform and depends on both b and d (D) non uniform 2011 MCQ 8.14

ONE MARK

Consider the following statements regarding the complex Poynting vector Pv for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(Pv ) denotes the real part of Pv, S denotes a spherical surface whose centre is at the point source, and nt denotes the unit surface normal on S . Which of the following statements is TRUE?



Electromagnetics

Page 389

(A) Re(Pv ) remains constant at any radial distance from the source (B) Re(Pv ) increases with increasing radial distance from the source (C) ## Re (Pv) : nt dS remains constant at any radial distance from the source s

(D) MCQ 8.15

MCQ 8.16

##s Re (Pv) : nt dS

A transmission line of characteristic impedance 50 W is terminated by a 50 W load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be p/4 radians. The phase velocity of the wave along the line is (A) 0.8 # 108 m/s (B) 1.2 # 108 m/s (C) 1.6 # 108 m/s (D) 3 # 108 m/s The modes in a rectangular waveguide are denoted by TE mn where TM mn m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 10 mode of the waveguide does not exist (B) The TE 10 mode of the waveguide does not exist (C) The TM 10 and the TE 10 modes both exist and have the same cut-off frequencies (D) The TM 10 and the TM 01 modes both exist and have the same cut-off frequencies 2011

MCQ 8.17

decreases with increasing radial distance from the source

nodia

TWO MARKS

A current sheet Jv = 10uty A/m lies on the dielectric interface x = 0 between two dielectric media with er 1 = 5, mr 1 = 1 in Region-1 (x < 0) and er2 = 2, mr2 = 2 in Region-2 (x 2 0). If the magnetic field in Region-1 at x = 0- is Hv1 = 3utx + 30uty A/m the magnetic field in Region-2 at x = 0+ is

(A) Hv2 = 1.5utx + 30uty - 10utz A/m (B) Hv2 = 3utx + 30uty - 10utz A/m (C) Hv2 = 1.5utx + 40uty A/m (D) Hv2 = 3utx + 30uty + 10utz A/m MCQ 8.18

A transmission line of characteristic impedance 50 W is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is (A) 10 W (B) 250 W (C) (19.23 + j 46.15) W (D) (19.23 - j 46.15) W

MCQ 8.19

The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity er and relative permeability mr = 1



Electromagnetics

Chapter 8

are given by Ev = E p e j (wt - 280py) utz V/m and Hv = 3e j (wt - 280py) utx A/m . Assuming the speed of light in free space to be 3 # 108 m/s , the intrinsic impedance of free space to be 120p , the relative permittivity er of the medium and the electric field amplitude E p are (B) er = 3, E p = 360p (A) er = 3, E p = 120p (C) er = 9, E p = 360p (D) er = 9, E p = 120p 2010

ONE MARK

MCQ 8.20

If the scattering matrix [S ] of a two port network is 0.2+0c 0.9+90c , then the network is [S ] = > 0.9+90c 0.1+90cH (A) lossless and reciprocal (B) lossless but not reciprocal (C) not lossless but reciprocal (D) neither lossless nor reciprocal

MCQ 8.21

A transmission line has a characteristic impedance of 50 W and a resistance of 0.1 W/m . If the line is distortion less, the attenuation constant(in Np/m) is (A) 500 (B) 5 (C) 0.014 (D) 0.002

MCQ 8.22

The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2 ) is (B) 1 (A) 1 30p 60p (C) 1 (D) 1 120p 240p

nodia

2010 MCQ 8.23

v = xyatx + x 2 aty , then If A

(A) 0 (C) 1 MCQ 8.24

TWO MARKS

# Av $ dlv over the path shown in the figure is

o

C

(B) 2 3 (D) 2 3

A plane wave having the electric field components Evi = 24 cos ^3 # 108 - by h atx V/m and traveling in free space is incident normally on a lossless medium with m = m0 and e = 9e0 which occupies the region y $ 0 . The reflected magnetic field component is given by (A) 1 cos (3 # 108 t + y) atx A/m 10p



Electromagnetics

Page 391

1 cos (3 108 t + y) at A/m # x 20p (C) - 1 cos (3 # 108 t + y) atx A/m 20p (D) - 1 cos (3 # 108 t + y) atx A/m 10p (B)

MCQ 8.25

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 W line is

(A) 1.00 (C) 2.50 2009 MCQ 8.26

nodia

ONE MARK

Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y - z plane and parallel to the y - axis. The other wire is in the x - y plane and parallel to the x - axis. Which components of the resulting magnetic field are non-zero at the origin ?

(A) x, y, z components (C) y, z components MCQ 8.27

(B) 1.64 (D) 3.00

(B) x, y components (D) x, z components

Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?

(A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency (C) Only R has no cutoff-frequency



Electromagnetics

Chapter 8

(D) All three have cutoff-frequencies 2009 MCQ 8.28

TWO MARKS

If a vector field V is related to another vector field A through V = 4# A , which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . ) (B) # A $ dl = # # V $ d S (A) # V $ dl = # # A $ d S C

(C) MCQ 8.29

#C D # V $ dl = #S #C D # A $ d S

C

(D)

S C

#C D # V $ dl = #S #CV $ d S

A transmission line terminates in two branches, each of length l , as shown. 4 The branches are terminated by 50W loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.

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(A) 200W (C) 50W MCQ 8.30

S C

(B) 100W (D) 25W

A magnetic field in air is measured to be y B = B0 c 2 x 2 yt - 2 xt m x +y x + y2 What current distribution leads to this field ? [Hint : The algebra is trivial in cylindrical coordinates.] t t (A) J = B0 z c 2 1 2 m, r ! 0 (B) J =- B0 z c 2 2 2 m, r ! 0 m0 x + y m0 x + y t (C) J = 0, r ! 0 (D) J = B0 z c 2 1 2 m, r ! 0 m0 x + y 2008

ONE MARK

MCQ 8.31

For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is (B) 180c (A) 360c (C) 90c (D) 45c

MCQ 8.32

For static electric and magnetic fields in an inhomogeneous source-free medium, which of the following represents the correct form of Maxwell’s equations ? (A) 4$ E = 0 , 4# B = 0 (B) 4$ E = 0 , 4$ B = 0 (C) 4# E = 0 , 4# B = 0 (D) 4# E = 0 , 4$ B = 0 2008

MCQ 8.33

TWO MARKS

A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be



Electromagnetics

Page 393

operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz (C) 5.0 GHz (D) 3.75 GHz MCQ 8.34

One end of a loss-less transmission line having the characteristic impedance of 75W and length of 1 cm is short-circuited. At 3 GHz, the input impedance at the other end of transmission line is (A) 0 (B) Resistive (C) Capacitive (D) Inductive

MCQ 8.35

A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant e = 9 ). The magnitude of the reflection coefficient is (A) 0 (B) 0.3 (C) 0.5 (D) 0.8

MCQ 8.36

In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if (A) radius as well as operating wavelength are halved

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(B) radius as well as operating wavelength are doubled (C) radius is halved and operating wavelength is doubled (D) radius is doubled and operating wavelength is halved MCQ 8.37

At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70% efficiency is (A) 15 dB (B) 25 dB (C) 35 dB (D) 45 dB 2007

MCQ 8.38

ONE MARK

A plane wave of wavelength l is traveling in a direction making an angle 30c with positive x - axis and 90c with positive y - axis. The E field of the plane wave can be represented as (E0 is constant) 3p p p 3p t 0 e j c wt - l x - l z m t 0 e jc wt - l x - l z m (A) E = yE (B) E = yE t 0 e jc wt + (C) E = yE

MCQ 8.39

3 p x+ p z l l m

p

t 0 e jc wt - l x + (D) E = yE

3pz l m

If C is code curve enclosing a surface S , then magnetic field intensity H , the current density j and the electric flux density D are related by (A) ## H $ ds = ## c j + 2D m $ d t 2t S c (B)

#S H $ d l = ##S c j + 22Dt m $ dS

(C)

##S H $ dS = #C c j + 22Dt m $ d t

(D)

#C H $ d l # = ##S c j + 22Dt m $ ds c

2007 MCQ 8.40

TWO MARKS

The E field in a rectangular waveguide of inner dimension a # b is given by



Electromagnetics

Chapter 8

2 wm l H sin ` 2px j sin (wt - bz) yt 2 `2j 0 a h Where H0 is a constant, and a and b are the dimensions along the x - axis and the y - axis respectively. The mode of propagation in the waveguide is (A) TE20 (B) TM11 (C) TM20 (D) TE10

E =

MCQ 8.41

MCQ 8.42

A load of 50 W is connected in shunt in a 2-wire transmission line of Z0 = 50W as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

1 -1 2 (A) > 12 1H 2 2

0 1 (B) = G 1 0

2 - 13 3 (C) > 2 1H 3 -3

(D) > 43 -4

nodia 1

- 43 1 4

H

The parallel branches of a 2-wirw transmission line re terminated in 100W and 200W resistors as shown in the figure. The characteristic impedance of the line is Z0 = 50W and each section has a length of l . The voltage reflection coefficient G 4 at the input is

(A) - j 7 5 5 (C) j 7

(B) - 5 7 5 (D) 7

MCQ 8.43

The H field (in A/m) of a plane wave propagating in free space is given by H = xt 5 3 cos (wt - bz) + yt` wt - bz + p j . h0 2 The time average power flow density in Watts is h (A) 0 (B) 100 100 h0 (C) 50h20 (D) 50 h0

MCQ 8.44

An air-filled rectangular waveguide has inner dimensions of 3 cm # 2 cm. The



Electromagnetics

Page 395

wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance h0 = 377 W ) (B) 355 W (A) 308 W (C) 400 W (D) 461 W MCQ 8.45

A l2 dipole is kept horizontally at a height of l2 above a perfectly conducting infinite ground plane. The radiation pattern in the lane of the dipole (E plane) looks approximately as

MCQ 8.46

A right circularly polarized (RCP) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant xr2 is.

0

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(A)

2

(C) 2 2006 MCQ 8.47

(B)

3

(D) 3 ONE MARK

The electric field of an electromagnetic wave propagation in the positive direction is given by E = atx sin (wt - bz) + aty sin (wt - bz + p/2). The wave is (A) Linearly polarized in the z -direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized

MCQ 8.48

A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) 1 Watts (C) 0.1 Watts (D) 0.01 Watt 2006

MCQ 8.49

TWO MARKS

When a planes wave traveling in free-space is incident normally on a medium having the fraction of power transmitted into the medium is given by



MCQ 8.50

MCQ 8.51

Electromagnetics

(A) 8 (B) 1 9 2 (C) 1 (D) 5 3 6 A medium of relative permittivity er2 = 2 forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular cross-section over the interface. The area of the beam crosssection at the interface is given by (B) p2 m 2 (A) 2p m 2 (C) p m 2 (D) p m 2 2 A rectangular wave guide having TE10 mode as dominant mode is having a cut off frequency 18 GHz for the mode TE30 . The inner broad - wall dimension of the rectangular wave guide is (A) 5/3 cm (B) 5 cm

nodia

(C) 5/2 cm MCQ 8.52

Chapter 8

(D) 10 cm

A medium is divide into regions I and II about x = 0 plane, as shown in the figure below.

An electromagnetic wave with electric field E1 = 4atx + 3aty + 5atz is incident normally on the interface from region I . The electric file E2 in region II at the interface is (A) E2 = E1 (B) 4atx + 0.75aty - 1.25atz (C) 3atx + 3aty + 5atz (D) - 3atx + 3aty + 5atz MCQ 8.53

A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna is Ohms is 2 2 (B) p (A) 2p 5 5 2 (C) 4p 5

(D) 20p2

2005 MCQ 8.54

ONE MARK

The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) aty where aty , denotes the unit vector in y direction. The wave is propagating with a phase velocity. (A) 5 # 10 4 m/s (B) - 3 # 108 m/s 7 (C) - 1.25 # 10 m/s (D) 3 # 108 m/s



MCQ 8.55

Electromagnetics

Page 397

Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 1014 Hz in glass. Assume velocity of light is 3 # 108 m/s in vacuum (B) 3 mm (A) 3 mm (C) 2 mm (D) 1 mm 2005

TWO MARKS

MCQ 8.56

Which one of the following does represent the electric field lines for the mode in the cross-section of a hollow rectangular metallic waveguide ?

MCQ 8.57

Characteristic impedance of a transmission line is 50 W. Input impedance of the open-circuited line when the transmission line a short circuited, then value of the input impedance will be. (B) 100 + j150W (A) 50 W (C) 7.69 + j11.54W (D) 7.69 - j11.54W

MCQ 8.58

nodia

Two identical and parallel dipole antennas are kept apart by a distance of l in 4 the H - plane. They are fed with equal currents but the right most antenna has a phase shift of + 90c. The radiation pattern is given as.

Statement of Linked Answer Questions 59 & 60 : Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure.



MCQ 8.59

MCQ 8.60

MCQ 8.61

Electromagnetics

The value of the load resistance is (A) 50 W

Chapter 8

(C) 12.5 W

(B) 200 W (D) 0

The reflection coefficient is given by (A) - 0.6 (C) 0.6

(B) - 1 (D) 0

Many circles are drawn in a Smith Chart used for transmission line calculations. The circles shown in the figure represent

(A) Unit circles (B) Constant resistance circles

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(C) Constant reactance circles (D) Constant reflection coefficient circles. 2004

ONE MARK

MCQ 8.62

The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE10 mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space

MCQ 8.63

Consider a lossless antenna with a directive gain of + 6 dB. If 1 mW of power is fed to it the total power radiated by the antenna will be (A) 4 mW (B) 1 mW (C) 7 mW (D) 1/4 mW 2004

TWO MARKS

MCQ 8.64

A parallel plate air-filled capacitor has plate area of 10 - 4 m 2 and plate separation of 10 - 3 m. It is connect - ed to a 0.5 V, 3.6 GHz source. The magnitude of the displacement current is ( e = 361p 10 - 9 F/m) (A) 10 mA (B) 100 mA (C) 10 A (D) 1.59 mA

MCQ 8.65

Consider a 300 W, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 10 V, 50 W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is



Electromagnetics

(A) 10 V (C) 60 V

Page 399

(B) 5 V (D) 60/7 V

MCQ 8.66

In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz (A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies

MCQ 8.67

If E = (atx + jaty) e jkz - kwt Poynting vector is (A) null vector (C) (2k/wm) atz

MCQ 8.68

and H = (k/wm) (aty + katx ) e jkz - jwt , the time-averaged

nodia (B) (k/wm) atz (D) (k/2wm) atz

Consider an impedance Z = R + jX marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to

(A) adding an inductance in series with Z (B) adding a capacitance in series with Z (C) adding an inductance in shunt across Z (D) adding a capacitance in shunt across Z MCQ 8.69

A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with e > e0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 120p W (B) 60p W (C) 600p W (D) 24p W

MCQ 8.70

A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage of the power that is reflected back is (A) 57.73 (B) 33.33 (C) 0.11 (D) 11.11



Electromagnetics

Chapter 8

2003 MCQ 8.71

MCQ 8.72

ONE MARK

The unit of 4# H is (A) Ampere (C) Ampere/meter 2

(B) Ampere/meter (D) Ampere-meter

The depth of penetration of electromagnetic wave in a medium having conductivity s at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm (B) 12.50 cm (C) 50.00 cm (D) 100.00 cm 2003

MCQ 8.73

TWO MARKS

Medium 1 has the electrical permittivity e1 = 1.5e0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity e2 = 2.5e0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux - 3uy + 1uz ) volt/m, then E2 in medium 2 is (A) (2.0ux - 7.5uy + 2.5uz ) volt/m (B) (2.0ux - 2.0uy + 0.6uz ) volt/m (C) (2.0ux - 3.0uy + 1.0uz ) volt/m

nodia

(D) (2.0ux - 2.0uy + 0.6uz ) volt/m MCQ 8.74

If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m, the potential difference between X (2, 0, 0) and Y (1, 2, 3) is (A) + 1 volt (B) - 1 volt (C) + 5 volt (D) + 6 volt

MCQ 8.75

A uniform plane wave traveling in air is incident on the plane boundary between air and another dielectric medium with er = 4 . The reflection coefficient for the normal incidence, is (A) zero (B) 0.5+180c (B) 0.333+0c (D) 0.333+180c

MCQ 8.76

If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2p107 t - 0.1pz) V/m, then the velocity of the traveling wave is (A) 3.00 # 108 m/sec (B) 2.00 # 108 m/sec (C) 6.28 # 107 m/sec (D) 2.00 # 107 m/sec

MCQ 8.77

Two identical antennas are placed in the q = p/2 plane as shown in Fig. The elements have equal amplitude excitation with 180c polarity difference, operating at wavelength l. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for f = 0 , is



Electromagnetics

(A) 2 cos b 2ps l l (C) 2 cos a ps k l MCQ 8.78

(B) 2 sin b 2ps l l (D) 2 sin a ps k l

A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is

nodia

(A) (0.01 - j0.02) mho (C) (0.04 - j0.02) mho MCQ 8.79

Page 401

(B) (0.02 - j0.01) mho (D) (0.02 + j0) mho

A rectangular metal wave guide filled with a dielectric material of relative permittivity er = 4 has the inside dimensions 3.0 cm # 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz (B) 5.0 GHz (C) 10.0 GHz (D) 12.5 GHz 2002

ONE MARK

MCQ 8.80

The VSWR can have any value between (A) 0 and 1 (B) - 1 and + 1 (C) 0 and 3 (D) 1 and 3

MCQ 8.81

In in impedance Smith movement along a constant resistance circle gives rise to (A) a decrease in the value of reactance (B) an increase in the value of reactance (C) no change in the reactance value (D) no change in the impedance

MCQ 8.82

The phase velocity for the TE10 -mode in an air-filled rectangular waveguide is ( c is the velocity of plane waves in free space) (B) equal to c (A) less than c (C) greater than c (D) none of these 2002

MCQ 8.83

TWO MARKS

t jp/2) e jwt - jkz . This wave is A plane wave is characterized by E = (0.5xt + ye



Electromagnetics

(A) linearly polarized (C) elliptically polarized MCQ 8.84

Chapter 8

(B) circularly polarized (D) unpolarized

Distilled water at 25c C is characterized by s = 1.7 # 10 - 4 mho/m and e = 78eo at a frequency of 3 GHz. Its loss tangent tan d is ( e = 10 36p F/m) -9

(A) 1.3 # 10-5 (C) 1.3 # 10-4 /78 MCQ 8.85

(B) 1.3 # 10-3 (D) 1.3 # 10-5 /78e0

The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with e = 80eo . The surface charge density on the conductor is ( e = 10 36p F/m) (B) 2 C/m 2 (A) 0 C/m 2 -9

(C) 1.8 # 10 - 11 C/m 2 MCQ 8.86

A person with receiver is 5 Km away from the transmitter. What is the distance that this person must move further to detect a 3-dB decrease in signal strength (A) 942 m (B) 2070 m

nodia

(C) 4978 m 2001 MCQ 8.87

MCQ 8.88

(D) 1.41 # 10 - 9 C/m 2

(D) 5320 m

A transmission line is distortonless if (A) RL = 1 GC

(B) RL = GC

(C) LG = RC

(D) RG = LC

ONE MARK

2 2 If a plane electromagnetic wave satisfies the equal d E2x = c2 d E2x , the wave dZ dt propagates in the (A) x - direction (B) z - direction (C) y - direction

(D) xy plane at an angle of 45c between the x and z direction MCQ 8.89

The plane velocity of wave propagating in a hollow metal waveguide is (A) grater than the velocity of light in free space (B) less than the velocity of light in free space (C) equal to the velocity of light free space (D) equal to the velocity of light in free

MCQ 8.90

The dominant mode in a rectangular waveguide is TE10 , because this mode has (A) the highest cut-off wavelength (B) no cut-off (C) no magnetic field component (D) no attenuation



Electromagnetics

Page 403

2001 MCQ 8.91

TWO MARKS -2

A material has conductivity of 10 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is (A) 45 MHz (B) 90 MHz (C) 450 MHz

MCQ 8.92

A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ? (A) 10% (B) 25% (C) 50%

MCQ 8.93

MCQ 8.94

(D) 900 MHz

(D) 75%

A medium wave radio transmitter operating at a wavelength of 492 m has a tower antenna of height 124. What is the radiation resistance of the antenna? (A) 25 W (B) 36.5 W (C) 50 W (D) 73 W

nodia

In uniform linear array, four isotropic radiating elements are spaced l apart. The 4 progressive phase shift between required for forming the main beam at 60c off the end - fire is : (A) - p (B) - p2 radians (C) - p4 radians (D) - p8 radians 2000

ONE MARK

MCQ 8.95

The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100 W and 25 W respectively. The characteristic impedance of the line is, (A) 25 W (B) 50 W (C) 75 W (D) 100 W

MCQ 8.96

A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively, (A) minimum and minimum (B) maximum and maximum (C) minimum and maximum (D) maximum and minimum

MCQ 8.97

If the diameter of a l dipole antenna is increased from l to l , then its 2 100 50 (A) bandwidth increases (B) bandwidth decrease (C) gain increases (D) gain decreases 2000

MCQ 8.98

TWO MARKS

A uniform plane wave in air impings at 45c angle on a lossless dielectric material with dielectric constant dr . The transmitted wave propagates is a 30c direction with respect to the normal. The value of dr is (A) 1.5 (B) 1.5 (C) 2

(D)

2



Electromagnetics

Chapter 8

MCQ 8.99

A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is (A) 5 GHz (B) 10 GHz (C) 15 GHz (D) 12 GHz

MCQ 8.100

Two coaxial cable 1 and 2 are filled with different dielectric constants er1 and er2 respectively. The ratio of the wavelength in the cables (l1 /l2) is (B) er2 /er1 (A) er1 /er2 (C) er1 /er2 (D) er2 /er1

MCQ 8.101

For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the minimum distance required for far field measurement is closest to (A) 7.5 cm (B) 15 cm (C) 15 m (D) 150 m 1999

MCQ 8.102

ONE MARK

An electric field on a place is described by its potential

nodia

V = 20 (r-1 + r-2) where r is the distance from the source. The field is due to (A) a monopole (B) a dipole (C) both a monopole and a dipole (D) a quadruple MCQ 8.103

Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable (B) air-filled cylindrical waveguide (C) parallel twin-wire line in air (D) semi-infinite parallel plate wave guide

MCQ 8.104

Indicate which one of the following will NOT exist in a rectangular resonant cavity. (B) TE 011 (A) TE110 (C) TM110 (D) TM111

MCQ 8.105

Identify which one of the following will NOT satisfy the wave equation. (A) 50e j (wt - 3z) (B) sin [w (10z + 5t)] 2 (C) cos (y + 5t) (D) sin (x) cos (t) 1999

TWO MARKS

MCQ 8.106

In a twin-wire transmission line in air, the adjacent voltage maxima are at 12.5 cm and 27.5 cm . The operating frequency is (B) 1 GHz (A) 300 MHz (C) 2 GHz (D) 6.28 GHz

MCQ 8.107

A transmitting antenna radiates 251 W isotropically. A receiving antenna, located 100 m away from the transmitting antenna, has an effective aperture of 500 cm2 . The total received by the antenna is (A) 10 mW (B) 1 mW (C) 20 mW

(D) 100 mW



Electromagnetics

Page 405

MCQ 8.108

In air, a lossless transmission line of length 50 cm with L = 10 mH/m , C = 40 pF/m is operated at 25 MHz . Its electrical path length is (B) l meters (A) 0.5 meters (C) p/2 radians (D) 180 deg rees

MCQ 8.109

A plane wave propagating through a medium [er = 8, mr = 2, and s = 0] has its t - (z/3) sin (108 t - bz) V/m . The wave impedance, in electric field given by Ev = 0.5Xe ohms is (A) 377 (B) 198.5+180c (C) 182.9+14c

(D) 133.3

1998 MCQ 8.110

MCQ 8.111

ONE MARK

The intrinsic impedance of copper at high frequencies is (A) purely resistive (B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component

nodia

The Maxwell equation V # H = J + 2D is based on 2t (A) Ampere’s law (B) Gauss’ law (C) Faraday’s law

MCQ 8.112

(D) Coulomb’s law

All transmission line sections shown in the figure is have a characteristic impedance R 0 + j 0 . The input impedance Zin equals

(A) 2 R 0 3 (C) 3 R 0 2

(B) R 0 (D) 2R 0

1998 MCQ 8.113

TWO MARKS

The time averages Poynting vector, in W/m , for a wave with Ev = 24e j (wt + bz) avy V/m in free space is (B) 2.4 avz (A) - 2.4 avz p p (C) 4.8 avz (D) - 4.8 avz p p 2



Electromagnetics

Chapter 8

MCQ 8.114

The wavelength of a wave with propagation constant (0.1p + j0.2p) m-1 is 2 m (B) 10 m (A) 0.05 (C) 20 m (D) 30 m

MCQ 8.115

The depth of penetration of wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability (C) wavelength (D) permittivity

MCQ 8.116

MCQ 8.117

The polarization of wave with electric field vector Ev = E 0 e j^wt + bz h âvx + avy h is (A) linear (B) elliptical (C) left hand circular (D) right hand circular The vector H in the far field of an antenna satisfies (A) d $ Hv = 0 and d # Hv = 0 (B) d $ Hv ! 0 and d # Hv ! 0

nodia

(C) d $ Hv = 0 and d # Hv ! 0 (D) d $ Hv ! 0 and d # Hv = 0 MCQ 8.118

The radiation resistance of a circular loop of one turn is 0.01 W. The radiation resistance of five turns of such a loop will be (A) 0.002 W (B) 0.01 W (C) 0.05 W (D) 0.25 W

MCQ 8.119

An antenna in free space receives 2 mW is 20 mV/m rms. The effective aperture (A) 0.005 m2 (C) 1.885 m2

MCQ 8.120

The maximum usable frequency of an ionospheric layer at 60c incidence and with 8 MHz critical frequency is (A) 16 MHz (B) 16 MHz 3 (C) 8 MHz (D) 6.93 MHz

MCQ 8.121

A loop is rotating about they y -axis in a magnetic field Bv = B 0 cos (wt + f) avx T. The voltage in the loop is (A) zero (B) due to rotation only (C) due to transformer action only

of power when the incident electric field of the antenna is (B) 0.05 m2 (D) 3.77 m2

(D) due to both rotation and transformer action MCQ 8.122

The far field of an antenna varies with distance r as (B) 12 (A) 1 r r 1 (C) 3 (D) 1 r r



Electromagnetics

Page 407

1997

ONE MARK

MCQ 8.123

A transmission line of 50 W characteristic impedance is terminated with a 100 W resistance. The minimum impedance measured on the line is equal to (A) 0 W (B) 25 W (C) 50 W (D) 100 W

MCQ 8.124

A rectangular air filled waveguide has cross section of 4 cm #10 cm . The minimum frequency which can propagate in the waveguide is (A) 0.75 GHz (B) 2.0 GHz (C) 2.5 GHz (D) 3.0 GHz 1997

MCQ 8.125

A very lossy, l/4 long, 50 W transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately (A) 0 (B) 50 W (C) 3

MCQ 8.126

nodia

ONE MARK

A lossless transmission line having 50 W characteristic impedance and length l/4 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is (A) 0 (B) 0.02 A (C) 3

MCQ 8.128


The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity of an electromagnetic wave in the conductor at 1, 000 MHz is about (A) 6 # 106 m/ sec (B) 6 # 107 m/ sec 8 (C) 3 # 10 m/ sec (D) 6 # 108 m/ sec 1996

MCQ 8.127

TWO MARKS

(D) none of these

The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z 0 respectively. The velocity of a travelling wave on the transmission line is (B) 1 (A) Z 0 C Z0 C (C) Z 0 (D) C Z0 C 1996

TWO MARKS

MCQ 8.129

A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is (A) 0% (B) 4% (C) 20% (D) 100%

MCQ 8.130

The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer ?



Electromagnetics

(A) 0c (C) 45c MCQ 8.131

Chapter 8

(B) 30c (D) 90c

Some unknown material has a conductivity of 106 mho/m and a permeability of 4p # 10-7 H/m . The skin depth for the material at 1 GHz is (A) 15.9 mm (B) 20.9 mm (C) 25.9 mm (D) 30.9 mm ***********

nodia



Electromagnetics

Page 409

SOLUTION SOL 8.1

Option (D) is correct. v around a closed path Stoke’s theorem states that the circulation a vector field A v l is equal to the surface integral of the curl of A over the open surface S bounded by l . i.e., # Av : dlv = ## ^d # Avh : dsv Here, line integral is taken across a closed path which is denoted by a small vh is taken circle on the integral notation where as, the surface integral of ^d # A over open surface bounded by the loop.

SOL 8.2

Option (D) is correct. Given, the vector field

nodia v = xavx + yavy + zavz A

so,

v (Divergence of A v ) = 2Ax + 2Ay + 2Az d$A 2x 2y 2z = 1+1+1 = 3

SOL 8.3

Option (A) is correct. Given, the return loss of device as 20 dB i.e., G in dB =- 20 dB (loss) ^

or,

h

20 log G =- 20

& G = 10-1 = 0.1 Therefore, the standing wave ration is given by 1+ G VSWR = 1- G

SOL 8.4

= 1 + 0.1 = 1.1 = 1.22 1 - 0.1 0.9 Option (C) is correct. For the given incidence of plane wave, we have the transmitting angle qt = 19.2c From Snell’s law, we know n1 sin qi = n2 sin qt ...(1) c m1 e1 sin qi = c m2 e2 sin qt For the given interfaces, we have m1 = m2 = 1 e1 = 1, e2 = 4.5 So, from Eq. (1) sin qi = 4.5 sin 19.2 or, qi . 45c Now, the component of Evi can be obtained as Evi = _Eox avx - Eoz avz i e-jbk (observed from the shown figure) Since, the angle qi = 45c so,



Electromagnetics

Chapter 8

Eox = Eoz = Eo 2 Therefore, Evi = Eo _avx - avz i e-jbk 2 Now, the wavelength of EM wave is l = 600 mm

...(1)

b = 2p = p # 10 4 3 l Also, direction of propagation is v v avk = ax + az 2 So, k = x+z 2 Substituting it in equation (1), we get p # 10 ^x + z h Evi = Eo _avx - avz i e-j 3 2 2

So,

4

SOL 8.5

Option (A) is correct. We obtain the reflection coefficient for parallel polarized wave (since, electric field is in the plane of wave propagation) as h cos qt - h1 cos qi ...(1) Gz = 2 h2 cos qt + h1 cos qi

nodia

As we have already obtained

qi = 45c, qt = 19.2c m 1 = h0 Also, h2 = = h0 e 4. 5 4.5 m 1 and h1 = = h0 = h0 1 e Substituting these in eq. (1) we get G z = cos 19.2c - 4.5 cos 45c cos 19.2c + 4.5 cos 45c =- 0.227 .- 0.23 Therefore, the reflected field has the magnitude given by Ero = T 11' Eio or Ero = G z Eio =- 0.23 Eio Hence, the expression of reflected electric field is p # 10 Evr =- 0.23 Eo _- avx - avz i e-j 3 k 2 Again, we have the propagation vector of reflected wave as v v avk = ax - az 2 or, k = x-z 2 Substituting it in Eq. (2), we get p # 10 x - z Evr =- 0.23 Eo _- avx - avz i e-j 3 b 2 l 2 jp # 10 ^x - z h V m Evr = 0.23 Eo _avx + avz i e- 3 2 2 4

(2)

4

4

SOL 8.6

Option (C) is correct. Electric field of the propagating wave in free space is given as



Electromagnetics

Page 411

Ei = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m So, it is clear that wave is propagating in the direction (- 3ax + 4ay). Since, the wave is incident on a perfectly conducting slab at x = 0 . So, the reflection coefficient will be equal to - 1. i.e. Er = (- 1) Ei =- 8ax - 6ay - 5az Again, the reflected wave will be as shown in figure. 0

0

i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the electric field of the reflected wave will be. Ex = (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m

SOL 8.7

nodia

Option (A) is correct. The field in circular polarization is found to be

Es = E 0 (ay ! jaz ) e-jbx propagating in + ve x -direction. where, plus sign is used for left circular polarization and minus sign for right circular polarization. So, the given problem has left circular polarization. b = 25 = w c 8 2pf & f = 25 # c = 25 # 3 # 10 = 1.2 GHz 25 = c 2p 2 # 3.14 SOL 8.8

Option (B) is correct. Let b " outer diameter a " inner diameter Characteristic impedance, m0 4p # 10-7 # 36p ln 2.4 = 100 W Z0 = ln b b l = b 1 l a e0 er 10-9 # 10.89

SOL 8.9

Option (A) is correct. The directivity is defined as D = Fmax Favg Fmax = 1 Favg = 1 4p

# F (q, f) dW

= 1 ;# 4p 0

2p

# 0

p/2

2p = 1 ;# 4p 0

2p

# F (q, f) sin qdq dfE 0

cos 4 q sin qdqdfE

p/2 5 = 1 ;2p b- cos q lE = 1 # 2p :- 0 + 1 D 5 5 4p 4p 0 = 1 # 2p = 1 5 10 4p D = 1 = 10 10



Electromagnetics

D (in dB) = 10 log 10 = 10 dB

or, SOL 8.10

Chapter 8

Option (C) is correct. Z0 =

Since

Z1 Z 2

100 = 50 # 200 This is quarter wave matching. The length would be odd multiple of l/4 . l = (2m + 1) l 4 c = 3 # 108 = 0.174 m f1 # 4 429 # 106 # 4 8 f2 = 1 GHz , l2 = c = 3 # 10 = 0.075 m 9 f2 # 4 1 # 10 # 4 Only option (C) is odd multiple of both l1 and l2 . (2m + 1) = 1.58 = 9 l1 (2m + 1) = 1.58 - 21 l2 f1 = 429 MHz,

SOL 8.11

l1 =

nodia


Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz) bx = 2.094 # 102 by = 2.618 # 102 w = 6.283 # 1010 rad/s For the wave propagation, w 2 - (b 2 + b 2) x y c2

b =

Substituting above values,

6.283 # 1010 2 - (2.0942 + 2.6182) 10 4 - j261 # 3 # 108 m b is imaginary so mode of operation is non-propagating. vp = 0 b =

SOL 8.12

c

Option (C ) is correct. For r > a , Ienclosed = (pa2) J

# H : dl

= Ienclosed

H # 2pr = (pa2) J H = Io 2pr H \ 1 , for r > a r For r < a , So,

Ienclosed =

Io = (pa2) J

J (pr 2) Jr 2 = 2 a pa 2

# H : dl

= Ienclosed 2 H # 2pr = Jr2 a H = Jr 2 2p a H \ r , for r < a



SOL 8.13

Electromagnetics

Page 413

Option (A) is correct. Assuming the cross section of the wire on x -y plane as shown in figure.

nodia

Since, the hole is drilled along the length of wire. So, it can be assumed that the drilled portion carriers current density of - J . Now, for the wire without hole, magnetic field intensity at point P will be given as Hf1 (2pR) = J (pR2) Hf1 (2pR) = JR 2 Since, point o is at origin. So, in vector form H1 = J (xax + yay) 2 Again only due to the hole magnetic field intensity will be given as. (Hf2) (2pr) =- J (pr 2) Hf2 = - Jr 2 Again, if we take Ol at origin then in vector form H2 = - J (xlax + ylay) 2 where xl and yl denotes point ‘P ’ in the new co-ordinate system. Now the relation between two co-ordinate system will be. x = xl + d So,

y = yl H2 = - J [(x - d) ax + yay] 2

So, total magnetic field intensity = H1 + H2 = J dax 2 So, magnetic field inside the hole will depend only on ‘d ’. SOL 8.14

Option (C) is correct. Power radiated from any source is constant.

SOL 8.15

Option (C) is correct. We have d = 2 mm and f = 10 GHz



Electromagnetics

Chapter 8

Phase difference = 2p d = p ; 4 l = l = 8d = 8 # 2 mm = 16 mm

or

v = fl = 10 # 109 # 16 # 10-3 = 1.6 # 108 m/ sec SOL 8.16

Option (A) is correct. TM11 is the lowest order mode of all the TMmn modes.

SOL 8.17

Option (A) is correct. From boundary condition Bn1 = Bn2 m1 Hx1 = m2 Hx2 or Hx2 = Hx1 = 1.5 2 or Hx2 = 1.5utx Further if H z = 1.5utx + Auty + Buz Then from Boundary condition

nodia

10ut (3utx + 30uty) utx = (1.5utx + Auty + Butz ) xt + v y J =- 30utz =- Autz + Buty + 10uty Comparing we get A = 30 and B =- 10 So H 2 = 1.5utx + 30uty - 10utz A/m SOL 8.18

Option (A) is correct. Since voltage maxima is observed at a distance of l/4 from the load and we know that the separation between one maxima and minima equals to l/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive. Now G = s-1 s+1 R also RL = 0 (since voltage maxima is formed at the load) s RL = 50 = 10 W 5

SOL 8.19

Option (D) is correct. From the expressions of Ev & Hv , we can write, b = 280 p 2 p or = 280 p & l = 1 140 l v E Wave impedance, Zw = E = p = 120 p 3 v er H again, Now or or Now

f = 14 GHz 8 3 l = C = 3 # 10 9 = er f er 14 # 10 140 er 3 = 1 140 140 er er = 9 Ep = 120p = E p = 120p 3 9



SOL 8.20

Electromagnetics

Page 415

Option (C) is correct. For a lossless network S11 2 + S21 2 = 1 For the given scattering matrix S11 = 0.2 0c , S12 = 0.9 90c S21 = 0.9 90c , S22 = 0.1 90c Here, (0.2) 2 + (0.9) 2 ! 1 Reciprocity :

(not lossless)

S12 = S21 = 0.9 90c (Reciprocal) SOL 8.21

Option (D) is correct. For distortion less transmission line characteristics impedance Z0 = R G Attenuation constant a =

So,

RG a = R = 0.1 = 0.002 50 Z0

nodia

SOL 8.22

Option (C) is correct. Intrinsic impedance of EM wave m m0 h = = = 120p = 60p e 2 4e0 Time average power density 2 1 Pav = 1 EH = 1 E = = 1 2 # 60p 120p 2 2 h

SOL 8.23

Option (C) is correct. v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) = # (xydx + x 2 dy) # Av : dl

C

C

C

=

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] = 1 2 3 3 2 3 3 3 3 SOL 8.24

Option (A) is correct. In the given problem

Reflection coefficient h2 - h 1 = 400p - 120p =- 1 2 h2 + h 1 40p + 120p t is negative So magnetic field component does not change its direction Direction t =



Electromagnetics

Chapter 8

of incident magnetic field atE # atH = atK atZ # atH = aty atH = atx ( + x direction) So, reflection magnetic field component Hr = t # 24 cos (3 # 108 + by) atx , y $ 0 h 1 # 24 cos (3 # 108 + by) atx , y $ 0 = 2 # 120p

So, SOL 8.25

8 b = w = 3 # 108 = 1 vC 3 # 10 1 Hr = cos (3 # 108 t + y) atx , y $ 0 10p

Option (B) is correct. For length of l/4 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E ZL = 30 W , Zo = 30 W, b = 2p , l = l 4 l 2 p l So, tan bl = tan b : l=3 4 l R ZL V S tan bl + jZo W 2 W = Z 0 = 60 W Z in = Zo S S Zo + jZL W ZL S tan bl W T X For length of l/8 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E

nodia Zo = 30 W, ZL = 0 (short) tan bl = tan b 2p : l l = 1 8 l

Z in = jZo tan bl = 30j Circuit is shown below.

Reflection coefficient 60 + 3j - 60 = t = ZL - Zo = 60 + 3j + 60 ZL + Zo 1+ t VSWR = = 1 + 17 = 1.64 1- t 1 - 17 SOL 8.26

1 17




Electromagnetics

Page 417

Due to 1 A current wire in x - y plane, magnetic field be at origin will be in x direction. Due to 1 A current wire in y - z plane, magnetic field be at origin will be in z direction. Thus x and z component is non-zero at origin. SOL 8.27

Option (A) is correct. Rectangular and cylindrical waveguide doesn’t support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn’t have cut off frequency.

SOL 8.28

Option (B) is correct. We have V = 4# A By Stokes theorem

...(1)

# A $ dl

= ## (4# A) $ ds From (1) and (2) we get

# A $ dl

SOL 8.29

=

...(2)

## V $ ds

nodia


The transmission line are as shown below. Length of all line is l 4

2 2 Zi1 = Z01 = 100 = 200W ZL1 50 2 2 Zi2 = Z02 = 100 = 200W ZL2 50

ZL3 = Zi1 Zi2 = 200W 200W = 100W 2 2 Zi = Z0 = 50 = 25W ZL3 100

SOL 8.30

Option (C) is correct. We have Bv = B0

x a - y a y xm x2 + y2 x2 + y2 To convert in cylindrical substituting

and In (1) we have

Now

SOL 8.31

c

...(1)

x = r cos f and y = r sin f ax = cos far - sin faf ay = sin far + cos faf Bv = Bv0 af v Bv a Hv = B = 0 f m0 m0 Jv = 4# Hv = 0

constant since H is constant




Electromagnetics

Chapter 8

The beam-width of Hertizian dipole is 180c and its half power beam-width is 90c. SOL 8.32

Option (D) is correct. Maxwell equations 4- B = 0 4$ E = r/E 4# E =- B 4# Ht = D + J For static electric magnetic fields 4$ B = 0 4$ E = r/E 4# E = 0 4# Ht = J

SOL 8.33

Option (A) is correct. Cut-off Frequency is fc = c 2

m 2 n 2 ` a j +`b j

nodia

For TE11 mode,

3 # 1010 fc = 2

SOL 8.34

1 2 1 2 ` 4 j + ` 3 j = 6.25 GHz


Zin = Zo

ZL + iZo tan (bl) Zo + iZL tan (bl)

For ZL = 0 , Zin = iZo tan (bl) The wavelength is

SOL 8.35

8 l = c = 3 # 109 = 0.1 m or 10 cm f 3 # 10 2 p l = 2p # 1 = p bl = l 10 5 Thus Zin = iZo tan p 5 Thus Zin is inductive because Zo tan p is positive 5 Option (C) is correct. m We have h = e Reflection coefficient h - h1 G= 2 h2 + h1

Substituting values for h1 and h2 we have t =

mo eo er mo eo er

+

mo eo mo eo

= 11+

er = 1 er 1+

9 9

since er = 9

=- 0.5 SOL 8.36

Option (B) is correct. In single mode optical fibre, the frequency of limiting mode increases as radius decreases



Electromagnetics

Page 419

r \ 1 f

Hence

So. if radius is doubled, the frequency of propagating mode gets halved, and wavelength is doubled. SOL 8.37

Option (D) is correct. 8 l = c = 3 # 10 9 = 3 f 200 20 # 10 2 2 D 2 Gp = hp ` j = 0.7 # p2 c 13 m = 30705.4 l 100

Gain

= 44.87 dB SOL 8.38

Option (A) is correct. g = b cos 30cx ! b sin 30cy = 2p 3 x ! 2p 1 y l 2 l 2 = p 3 x! py l l

nodia E = ay E0 e j (wt - g) = ay E0 e j;wt - c

SOL 8.39

p 3 x! p y l l mE


4# H = J + 2D 2t

Maxwell Equations

## 4# H $ ds = ## `J + 22Dt j .ds

Integral form

## `J + 22Dt j .ds

Stokes Theorem

s

s

# H $ dl

=

s

SOL 8.40

Option (A) is correct. 2 wm p H sin ` 2px j sin (wt - bz) yt 2 `2j 0 a h This is TE mode and we know that mpy Ey \ sin ` mpx j cos ` a b j

E =

Thus m = 2 and n = 0 and mode is TE20 SOL 8.41

Option (C) is correct. The 2-port scattering parameter matrix is S11 S12 S == S21 S22 G (Z Z ) - Zo (50 50) - 50 S11 = L 0 = =- 1 (ZL Z0) + Zo (50 50) + 50 3 2 (ZL Zo) 2 (50 50) S12 = S21 = = =2 (ZL Zo) + Zo (50 50) + 50 3 (ZL Zo) - Zo (50 50) - 50 S22 = = =- 1 (ZL Zo) + Zo (50 50) + 50 3

SOL 8.42

Option (D) is correct. The input impedance is 2 Zin = Zo ; ZL

if l = l 4

2 2 Zin1 = Zo1 = 50 = 25 ZL1 100



Electromagnetics

Chapter 8

2 2 Zin2 = Zo2 = 50 = 12.5 ZL2 200

ZL = Zin1 Zin2 25 12.5 = 25 3

Now

(50) 2 = 300 25/3 G = ZS - Zo = 300 - 50 = 5 ZS + Zo 300 + 50 7

Zs =

SOL 8.43

Option (D) is correct. 2 2 2 = Hx2 + Hy2 = c 5 3 m + c 5 m = c 10 m ho ho ho ho H 2 E2 2 h P = = o c 10 m = 50 watts = ho 2 ho 2ho 2

We have

H

For free space SOL 8.44

Option (C) is correct. The cut-off frequency is

nodia fc = c 2

SOL 8.45

2

m 2 n 2 ` a j +`b j

Since the mode is TE20, m = 2 and n = 0 8 fc = c m = 3 # 10 # 2 = 10 GHz 2 2 2 # 0.03 ho 377 = h' = = 400W 10 2 fc 2 1-c m 1 - c 10 10 m f 3 # 10 Option (B) is correct. Using the method of images, the configuration is as shown below

Here d = l, a = p, thus bd = 2p bd cos y + a E 2 2p cos y + p = cos ; E = sin (p cos y) 2 = cos ;

Array factor is

SOL 8.46

Option (D) is correct. The Brewster angle is tan qn = tan 60c = or

SOL 8.47

er2 er1 er2 1

er2 = 3

Option (C) is correct. We have E = atxx sin (wt - bz) + aty sin (wt - bz + p/2) Here Ex = Ey and fx = 0, fy = p2



Electromagnetics

Phase difference is SOL 8.48

p 2

, thus wave is left hand circularly polarized.

Option (A) is correct. We have 10 log G = 10 dB or G = 10 Now gain G = Prad Pin or 10 = Prad 1W Prad = 10 Watts

or SOL 8.49

Option (A) is correct. G= = 1 + er = 1 - 4 =- 1 3 1 + er 1+ 4 The transmitted power is

or SOL 8.50

Page 421

h2 - h1 = h2 + h1

mo eo er mo eo er

+

mo eo mo eo

Pt = (1 - G2) Pi = 1 - 1 = 8 9 9 Pt = 8 9 Pi

nodia


or

sin q = 1 = 1 2 er p q = 45c = 4

The configuration is shown below. Here A is point source.

Now From geometry

AO = 1 m BO = 1 m = pr2 = p # OB = p m 2

Thus area SOL 8.51

Option (C) is correct. The cut-off frequency is fc = c 2

m 2 m 2 à j +`b j

Since the mode is TE30 , m = 3 and n = 0 fc = c m 2 a or or SOL 8.52

8 18 # 109 = 3 # 10 3 2 a 1 m = 5 cm a = 40 2

Option (C) is correct. We have E1 = 4ux + 3uy + 5uz



Electromagnetics

Chapter 8

Since for dielectric material at the boundary, tangential component of electric field are equal E21 = E1t = 3aty + 5atz at the boundary, normal component of displacement vector are equal i.e. Dn2 = Dn1 or e2 E2n = e1 E1n or 4eo E2n = 3eo 4atz E2n = 3atx E2 = E2t + E2a = 3atx + 3aty + 5atz

or Thus SOL 8.53

Option (C) is correct. Since antenna is installed at conducting ground, 2 2 2 50 Rrad = 80p2 ` dl j = 80p2 c = 4p W m 3 5 l 0.5 # 10

SOL 8.54

Option (C) is correct. w = 50, 000 and b =- 0.004 4 vP = w = 5 # 10 - 3 = 1.25 # 107 m/s b - 4 # 10

nodia

Phase Velocity is SOL 8.55

Option (C) is correct. Refractive index of glass m = 1.5 Frequency f = 1014 Hz

c = 3 # 108 m/sec

8 l = c = 3 # 10 = 3 # 10 - 6 14 f 10 wavelength in glass is

-6 lg = a = 3 # 10 = 2 # 10 - 6 m 1.5 m

SOL 8.56


SOL 8.57

Option (D) is correct. Zo2 = ZOC .ZSC 2 ZZC = Zo = 50 # 50 = 50 100 + j150 2 + 3j ZOC 50 (2 - 3j) = 7.69 - 11.54j = 13

SOL 8.58

Option (A) is correct. The array factor is A = cos b

bd sin q + a l 2

Here b = 2p , d = l and a = 90c l 4 2p l sin q + Thus A = cos c l 4 2

p 2

p p m = cos ` sin q + j 4 2

The option (A) satisfy this equation. SOL 8.59

Option (C) is correct. From the diagram, VSWR is



Electromagnetics

Page 423

s = Vmax = 4 = 4 Vmin 1 When minima is at load ZO = s.ZL or ZL = Zo = 50 = 12.5W s 4 SOL 8.60

Option (A) is correct. The reflection coefficient is G = ZL - ZO = 12.5 - 50 =- 0.6 ZL + ZO 125. + 50

SOL 8.61

Option (C) is correct. The given figure represent constant reactance circle.

SOL 8.62

Option (D) is correct. We know that vp > c > vg .

SOL 8.63


GD (q, f) =

4pU (q, f) Prad

nodia

For lossless antenna

Prad = Pin Here we have Prad = Pin = 1 mW and 10 log GD (q, f) = 6 dB or GD (q, f) = 3.98 Thus the total power radiated by antenna is

4pU (q, f) = Prad GD (q, f) = 1 m # 3.98 = 3.98 mW

SOL 8.64

Option (D) is correct. The capacitance is

- 12 -4 C = eo A = 8.85 # 10 - 3 # 10 = 8.85 # 10 - 13 d 10

The charge on capacitor is Q = CV = 8.85 # 10 - 13 = 4.427 # 10 - 13 Displacement current in one cycle Q I = = fQ = 4.427 # 10 - 13 # 3.6 # 109 = 1.59 mA T SOL 8.65

Option (C) is correct. VL = ZO Zin Vin or VL = ZO Vin = 10 # 300 = 60 V Zin 50

SOL 8.66


SOL 8.67

Option (A) is correct. Ravg = 1 Re [E # H*] 2 E # H* = (atx + jaty) e jkz - jwt # k (- jatx + aty) e-jkz + jwt wm = atz ; k - (- j) (j) k E = 0 wm wm



Electromagnetics

Ravg = 1 Re [E # H*] = 0 2

Thus SOL 8.68

Chapter 8

Option (A) is correct. Suppose at point P impedance is Z = r + j (- 1) If we move in constant resistance circle from point P in clockwise direction by an angle 45c, the reactance magnitude increase. Let us consider a point Q at 45c from point P in clockwise direction. It’s impedance is Z1 = r - 0.5j or Z1 = Z + 0.5j Thus movement on constant r - circle by an +45c in CW direction is the addition of inductance in series with Z .

SOL 8.69

Option (D) is correct. We have or Thus Now or or

SOL 8.70

1- G VSWR = Emax = 5 = Emin 1+ G G =2 3 G =- 2 3 h2 - h1 G= h2 + h1 h - 120p -2 = 2 h2 + 120p 3

nodia h2 = 24p

Option (D) is correct. The VSWR or Thus or

1- G 1+ G G =1 3 2=

Pref = G2= 1 9 Pinc Pref = Pinc 9

i.e. 11.11% of incident power is reflected. SOL 8.71

Option (C) is correct. By Maxwells equations 4# H = 2D + J 2t Thus 4# H has unit of current density J that is A/m2

SOL 8.72

Option (B) is correct. We know that Thus

d \ 1 f d2 = d1 d2 = 25

f1 f2 1 4



Electromagnetics

d2 =

or SOL 8.73

Page 425

1 # 25 = 12.5 cm 4

Option (C) is correct. We have E1 = 2ux - 3uy + 1uz E1t =- 3uy + uz and E1n = 2ux Since for dielectric material at the boundary, tangential component of electric field are equal E1t =- 3uy + uz = E2t E1n = 2ux At the boundary the for normal component of electric field are or or or Thus

SOL 8.74

(x = 0 plane)

D1n = D2n e1 E1n = e2 E2n 1.5eo 2ux = 2.5eo E2n E2n = 3 ux = 1.2ux 2.5

nodia E2 = E2t + E2n =- 3uy + uz + 1.2ux

Option (C) is correct. We have E = xux + yuy + zuz dl = utx dx + uty dy + utz dz Y

#1

VXY =- # E.dl = X

2

xdxutx +

0

0

#2 ydyutz + #3 zdzuzt

0

2 2 2 0 y2 =-= x + +z G 2 1 2 2 2 3 1 =- [22 - 12 + 02 - 22 + 02 - 32] = 5 2

SOL 8.75

Option (D) is correct. h =

m e

Reflection coefficient t =

h2 - h1 h2 + h1

Substituting values for h1 and h2 we have t =

mo eo er mo eo er

+

m0 eo mo eo

= 11+

er = 1 er 1+

4 4

since er = 4

= - 1 = 0.333+180c 3 SOL 8.76

SOL 8.77

Option (B) is correct. We have E (z, t) = 10 cos (2p # 107 t - 0.1pz) where w = 2p # 107 t b = 0.1p 7 Phase Velocity u = w = 2p # 10 = 2 # 108 m/s 0.1p b Option (D) is correct. y Normalized array factor = 2 cos 2



Electromagnetics

Chapter 8

y = bd sin q cos f + d q = 90c,

Now

d = 2 s, f = 45c, d = 180c bd sin q cos f + d y 2 cos = 2 cos ; E 2 2 = 2 cos 8 2p 2 s cos 45c + 180 B l. 2 2 s p p = 2 cos 8 + 90cB = 2 sin ` s j l l

SOL 8.78

Option (A) is correct. The fig of transmission line is as shown below. [Z + jZo tan bl] We know that Zin = Zo L [Zo + jZL tan bl] For line 1, l = l and b = 2p , ZL1 = 100W 2 l [ZL + jZo tan p] Thus Zin1 = Zo = ZL = 100W [Zo + jZL tan p] For line 2, l = l and b = 2p , ZL2 = 0 (short circuit) l 8 [0 + jZo tan p4 ] Thus Zin2 = Zo = jZo = j50W [Zo + 0] Y = 1 + 1 = 1 + 1 = 0.01 - j0.02 Zin1 Zin2 100 j50

nodia

SOL 8.79

Option (A) is correct. 8 u = c = 3 # 10 = 1.5 # 108 2 e0 In rectangular waveguide the dominant mode is TE10 and fC = v ` m j2 + ` n j2 2 a b 8 1 . 5 10 # 1 2 + 0 2 = 1.5 # 108 = 2.5 GHz = ` 2 0.03 j ` b j 0.06

SOL 8.80




Electromagnetics

s = 1+G 1-G

VSWR

Page 427

where G varies from 0 to 1

Thus s varies from 1 to 3. SOL 8.81

Option (B) is correct. Reactance increases if we move along clockwise direction in the constant resistance circle.

SOL 8.82

Option (C) is correct. Phase velocity VC

VP =

f 2 1-c c m f

When wave propagate in waveguide fc < f $ VP > VC SOL 8.83


p 2

t j ) e j (wt - kz) E = (0.5xt + ye Ex = 0.5e j (wt - kz)

nodia p

Ey = e j 2 e j (wt - kz)

Ex p = 0.5e- 2 Ey

Since SOL 8.84

Ex ! 1, it is elliptically polarized. Ey


1.7 # 10 - 4 tan a = s = we 2p # 3 # 109 # 78eo

Loss tangent

-4 9 = 1.7 # 10 9# 9 # 10 = 1.3 # 10 - 5 3 # 10 # 39

SOL 8.85

Option (D) is correct. The flux density is s = eE = e0 er E = 80 # 8.854 # 10 - 12 # 2 or

SOL 8.86

s = 1.41 # 10 - 9 C/m 2

Option (B) is correct. P \ 12 r P1 = r22 Thus P2 r12 3 dB decrease $ Strength is halved P1 = 2 Thus P2 Substituting values we have 2 2 = r22 5 or r2 = 5 2 kM = 7071 m Distance to move = 7071 - 5000 = 2071 m

SOL 8.87




Electromagnetics

Chapter 8

A transmission line is distortion less if LG = RC SOL 8.88

Option (B) is correct. d2 Ex = c2 d2 Ex dz2 dt2 This equation shows that x component of electric fields Ex is traveling in z direction because there is change in z direction. We have

SOL 8.89

Option (A) is correct. In wave guide vp > c > vg and in vacuum vp = c = vg where vp $ Phase velocity c $ Velocity of light vg $ Group velocity

SOL 8.90

Option (A) is correct. In a wave guide dominant gives lowest cut-off frequency and hence the highest cut-off wavelength.

SOL 8.91

Option (A) is correct. or or or or

SOL 8.92

nodia Ic = Id sE = jw d E s = 2pfeo er f =

w = 2pf and e = er e0

-2 s = 2s = 9 # 10 # 2 # 10 2p # eo er 4peo er 4 9

f = 45 # 106 = 45 MHz


or or Now

VSWR = 1 + G 1-G 3 = 1+G 1-G

G = 0.5 Pr = G2 = 0.25 Pi

Thus 25% of incident power is reflected. SOL 8.93

Option (A) is correct. l = 492 m

We have

and height of antenna = 124 m . l 4 It is a quarter wave monopole antenna and radiation resistance is 25 W. SOL 8.94

Option (C) is correct. The array factor is where

Thus or

y = bd cos q + d d =l 4

Distance between elements

y =0 q = 60c 0 = 2p # l cos 60c + d = p # 1 + d 2 2 l 4 d =- p 4

Because of end fire



SOL 8.95

Electromagnetics

Page 429

Option (B) is correct. ZOC .ZSC = 100 # 25 = 10 # 5 = 50W

Zo = SOL 8.96

Option (C) is correct. As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary.

SOL 8.97

Option (B) is correct. BW \

1 (Diameter)

As diameter increases Bandwidth decreases. SOL 8.98

Option (C) is correct. The fig is as shown below :

nodia

As per snell law sin qt = 1 sin qi er sin 30c = 1 or sin 45c er 1 1 2 = 1 er 2 or er = 2 SOL 8.99

Option (C) is correct. fc =

Cutoff frequency

vp 2

m 2 n 2 ` a j +`b j

For rectangular waveguide dominant mode is TE01 8 v Thus fc = p = 3 # 10- 2 = 15 # 109 2a 2 # 10 = 15 GHz SOL 8.100

Option (B) is correct. b = 2p = w me l l = 2p w me

Phase Velocity or

l \ 1 e

Thus we get SOL 8.101

For air vp = 3 # 108

l1 = l2

e2 e1

Option (D) is correct. 2 l ` 2 jd = l



Electromagnetics

Chapter 8

8 l = c = 3 # 10 9 = 3 m f 40 4 # 10 2 3 ` 40 # 2 j d = (2.4)

d =

or SOL 8.102

80 # (2.4) 2 . 150 m 3

Option (C) is correct. We know that for a monopole its electric field varies inversely with r 2 while its potential varies inversely with r . Similarly for a dipole its electric field varies inversely as r 3 and potential varies inversely as r 2 . In the given expression both the terms a _ r1 + r1 i are present, so this potential is due to both monopole & dipole. -1

-2

SOL 8.103

Option (D) is correct. In TE mode Ez = 0 , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide.

SOL 8.104


SOL 8.105

Option (C) is correct. A scalar wave equation must satisfy following relation 2 2 E - m 22 2 E = 0 ...(1) 2t 2 2z 2 Where m = w (Velocity) b Basically w is the multiply factor of t and b is multiply factor of z or x or y . In option (A) E = 50e j (wt - 3z) m =w=w 3 b

nodia

We can see that equations in option (C) does not satisfy equation (1) SOL 8.106

Option (B) is correct. We know that distance between two adjacent voltage maxima is equal to l/2 , where l is wavelength. l = 27.5 - 12.5 2 Frequency

l = 2 # 15 = 30 cm 10 u = C = 3 # 10 = 1 GHz 30 l

SOL 8.107

Option (D) is correct. Power received by antenna -4 PR = PT 2 # (apeture) = 251 # 500 # 102 = 100 mW 4p r 4 # p # (100)

SOL 8.108

Option (C) is correct. Electrical path length = bl Where b = 2p , l = 50 cm l We know that l =u =1# 1 f f LC

au=

1 LC



Electromagnetics

=

1 # 25 # 106

Page 431

1 10 # 10-6 # 40 # 10-12

7 = 5 # 10 6 = 2 m 25 # 10 Electric path length = 2p # 50 # 10-2 5 = p radian 2

SOL 8.109

Option (D) is correct. In a lossless dielectric (s = 0) median, impedance is given by m 0c h = e m0 mr = e0 er mr = 120p # er = 120p # 2 = 188.4 W 8

SOL 8.110

Option (D) is correct. Impedance is written as

nodia

jwm s + jwe Copper is good conductor i.e. s >> we jwm wm So h = 45c = s s Impedance will be complex with an inductive component. h =

SOL 8.111

Option (A) is correct. This equation is based on ampere’s law as we can see

or

#l H $ dl

= I enclosed

#l H $ dl

=

(ampere's law)

#s Jds

Applying curl theorem

#s (4 # H) $ ds

= # Jds s 4# H = J then it is modified to 4# H = J + 2D 2t SOL 8.112


SOL 8.113


SOL 8.114

Option (B) is correct. Propagation constant here

Based on continuity equation

r = a + ib = 0.1p + j0.2p b = 2p = 0.2p l l = 2 = 10 m 0. 2



SOL 8.115

Electromagnetics

Chapter 8

Option (C) is correct. The depth of penetration or skin depth is defined as – 1 d= pfms d\ 1 \ l f so depth increases with increasing in wavelength.

SOL 8.116

Option (A) is correct. Given j (wt + bz) v a x + e0 e j (wt + bz) avy E (z, t) = Eo e

...(1)

Generalizing ...(2) E (z) = avx E1 (z) + avy E2 (z) Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space quadrature but in time phase, their sum E will be linearly polarized along a line that makes an angle f with x -axis as shown below. SOL 8.117

nodia

Option (C) is correct. v v Hv = 1 4 A m # v is auxiliary potential function. where A So 4: H = 4: (4 # A) = 0

4# H = 4# (4 # A) = Y 0

SOL 8.118

Option (D) is correct. Radiation resistance of a circular loop is given as Rr = 8 hp3 :ND2 S D 3 l Rx \ N 2 N " no. of turns So, Rr 2 = N 2 # Rr 1 = (5) 2 # 0.01 = 0.25 W

SOL 8.119

Option (C) is correct. We have Aperture Area

Power Re ceived Polynting vector of incident wave A =W P =

2 P = E h0 = 120p is intrinsic impedance of space h0

So

-6 -6 A = 2 # 10 = 2 # 10 -3 2 # 120 # 3.14 2 E (20 # 10 ) c h0 m -6 3.14 = 1.884 m2 = 2 # 10 # 12 -# 400 # 10 6

SOL 8.120

Option (B) is correct. Maximum usable frequency fo fm = sin Ae



Electromagnetics

fm = 8MHz = sin 60c

Page 433

8 = 16 MHz 3 3 c 2 m

SOL 8.121

Option (D) is correct. When a moving circuit is put in a time varying magnetic field educed emf have two components. One for time variation of B and other turn motion of circuit in B.

SOL 8.122

Option (A) is correct. Far field \ 1 r

SOL 8.123

Option (B) is correct. = Z0 S where S = standing wave ratio 1 + GL S = 1 - GL Z in

min

nodia GL = reflection coefficient

GL = ZL - Z 0 = 100 - 50 = 50 = 1 ZL + Z 0 100 + 50 150 3 1+1 3 =2 1 13 50 = = 25 W 2

S =

Z in

SOL 8.124

min

Option (A) is correct. The cutoff frequency is given by ml m 2+ n 2 fc = 2 a a k a2k Here a < b , so minimum cut off frequency will be for mode TE 01 m = 0, n = 1 8 fc = 3 # 10 2#2

1 (10 # 10-12) 3 # 108 = 0.75 GHz = 2 # 2 # 10 # 10-2

SOL 8.125

a ml = c 2 *

c = 3 # 108

Option (A) is correct. For any transmission line we can write input impedance Z + jZ 0 tanh lg Zin = Z 0 ; L Z 0 + jZL tanh lg E Here given ZL = 3 (open circuited at load end) R jZ tanh lg V W S1 + 0 Z0 ZL W= so Zin = Z 0 lim S Z " 3S Z 0 j tanh lg W S ZL + j tanh lg W X T Option (A) is correct. We know that skin depth is given by L

SOL 8.126



Electromagnetics

s =

Chapter 8

1 = 1 # 10-2 m pf1 ms

1 = 10-2 p # 10 # 106 # m s -3 ms = 10 p

or or

f 1 = 10 MHz

Now phase velocity at another frequency f 2 = 1000 MHz is 4p f 2 V = ms -3

ms = 10 in above equation p

Put

V = SOL 8.127

4 # p # 1000 # 106 # p - 6 106 m/ sec # 10-3

Option (A) is correct. Input impedance of a lossless transmission line is given by Z + jZ 0 tan bl Zin = Z 0 ; L Z 0 + jZL tan bl E where

nodia Z 0 = Charateristic impedance of line

ZL = Load impedance b = 2p l = length l bl = 2p l = p 2 l 4

so here

ZL = 0 (Short circuited) Z 0 = 50 W 0 + j50 tan p/2 Zin = 50 = =3 50 + j0 tan p/2G

and so

Thus infinite impedance, and current will be zero. SOL 8.128

SOL 8.129

Option (B) is correct. For lossless transmission line, we have Velocity V =w= 1 b LC Characteristics impedance for a lossless transmission line Z0 = L C From eqn. (1) and (2) 1 V = = 1 C (Z 0 C ) Z 0 C

...(1)

...(2)

Option (C) is correct. Reflected power

So

Er = GEi G = Reflection coefficient h - h1 G = 2 = 1. 5 - 1 = 1 h 2 + h1 1. 5 + 1 5 Er = 1 # Ei 5

Ei " Incident power



Electromagnetics

Page 435

Er = 20% Ei SOL 8.130

Option (B) is correct. We have maximum usable frequency formulae as f0 fm = sin Ae 20 # 106 = 10 # 10 sin Ae sin Ae = 1 2

6

Ae = 30c SOL 8.131

Option (A) is correct. Skin depth

d=

1 pfms

Putting the given value d=

1 = 15.9 mm 3.14 # 1 # 109 # 4p # 10-7 # 106

nodia ***********


CHAPTER 9 COMMUNICATION SYSTEMS

2013 MCQ 9.1

ONE MARK

The bit rate of a digital communication system is R kbits/s . The modulation used is 32-QAM. The minimum bandwidth required for ISI free transmission is (A) R/10 Hz (B) R/10 kHz (C) R/5 Hz (D) R/5 kHz 2013

MCQ 9.2

TWO MARKS

Let U and V be two independent zero mean Gaussain random variables of variances 1 and 1 respectively. The probability P ^3V F 2U h is 9 4 (A) 4/9 (B) 1/2 (C) 2/3

(D) 5/9

nodia

MCQ 9.3

Consider two identically distributed zero-mean random variables U and V . Let the cumulative distribution functions of U and 2V be F ^x h and G ^x h respectively. Then, for all values of x (A) F ^x h - G ^x h # 0 (B) F ^x h - G ^x h $ 0 (C) ^F (x) - G (x)h .x # 0 (D) ^F (x) - G (x)h .x $ 0

MCQ 9.4

Let U and V be two independent and such that P Û =+ 1h = P Û =- 1h = 1 . 2 (A) 3/4 (C) 3/2

identically distributed random variables

The entropy H Û + V h in bits is (B) 1 (D) log 2 3

Common Data for Questions 5 and 6: Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the respective received signals for both bits are as shown below.

MCQ 9.5

If the detection threshold is 1, the BER will be (B) 1 (A) 1 2 4 1 (C) (D) 1 8 16


MCQ 9.6


The optimum threshold to achieve minimum bit error rate (BER) is (A) 1 (B) 4 2 5 (D) 3 2

(C) 1 2012 MCQ 9.7

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ONE MARK

The power spectral density of a real process X (t) for positive frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] , respectively, are

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(A) 6000/p, 0

(B) 6400/p, 0

(C) 6400/p, 20/ (p 2 )

(D) 6000/p, 20/ (p 2 )

MCQ 9.8

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maxi mum possible signaling rate in symbols per second is (A) 1750 (B) 2625 (C) 4000 (D) 5250

MCQ 9.9

A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount e and decreases that of the second by e. After encoding, the entropy of the source (A) increases (B) remains the same (C) increases only if N = 2 (D) decreases

MCQ 9.10

Two independent random variables X and Y are uniformly distributed in the interval 6- 1, 1@. The probability that max 6X, Y @ is less than 1/2 is (A) 3/4 (B) 9/16 (C) 1/4 (D) 2/3 2012

MCQ 9.11

TWO MARKS

A BPSK scheme operating over an AWGN channel with noise power spectral density of N 0 /2, uses equiprobable signals s1 (t) = 2E sin (wc t) and T s2 (t) =- 2E sin (wc t) over the symbol interval (0, T). If the local oscillator in a T coherent receiver is ahead in phase by 45c with respect to the received signal, the probability of error in the resulting system is (A) Q c 2E m (B) Q c E m N0 N0




(C) Q c

E 2N 0 m

(D) Q c

Chapter 9

E 4N 0 m

MCQ 9.12

A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary symbol X is such that P (X = 0) = 9/10, then the probability of error for an optimum receiver will be (A) 7/80 (B) 63/80 (C) 9/10 (D) 1/10

MCQ 9.13

The signal m (t) as shown is applied to both a phase modulator (with k p as the phase constant) and a frequency modulator (with k f as the frequency constant) having the same carrier frequency.

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The ratio k p /k f (in rad/Hz) for the same maximum phase deviation is (B) 4p (A) 8p (C) 2p (D) p

Statement for Linked Answer Question 14 and 15 :

MCQ 9.14

MCQ 9.15

The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =- 3, b =- 1 (D) a = 3, b = 1 The phase of the above lead compensator is maximum at (B) 3 rad/s (A) 2 rad/s (C)

2011

6 rad/s

(D) 1/ 3 rad/s

ONE MARK

MCQ 9.16

An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and the samples are quantized into 4 levels. The quantized levels are assumed to be independent and equally probable. If we transmit two quantized samples per second, the information rate is (A) 1 bit/sec (B) 2 bits/sec (C) 3 bits/sec (D) 4 bits/sec

MCQ 9.17

The Column -1 lists the attributes and the Column -2 lists the modulation systems. Match the attribute to the modulation system that best meets it.




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Column -1

Column -2

P.

Power efficient transmission of signals

Q.

Most bandwidth efficient transmission of 2. voice signals

FM

R.

Simplest receiver structure

VSB

S.

Bandwidth efficient transmission of signals 4. with significant dc component

1.

3.

Conventional AM

SSB-SC

(A) P-4, Q-2, R-1, S-3 (B) P-2, Q-4, R-1, S-3 (C) P-3, Q-2, R-1, S-4 (D) P-2, Q-4, R-3, S-1 2011 MCQ 9.18

TWO MARKS

X (t) is a stationary random process with auto-correlation function RX (t) = exp (- pt 2). This process is passed through the system shown below. The power spectral density of the output process Y (t) is

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(A) (4p 2 f 2 + 1) exp (- pf 2) (C) (4p 2 f 2 + 1) exp (- pf ) MCQ 9.19

(B) (4p 2 f 2 - 1) exp (- pf 2) (D) (4p 2 f 2 - 1) exp (- pf )

A message signal m (t) = cos 2000pt + 4 cos 4000pt modulates the carrier c (t) = cos 2pfc t where fc = 1 MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy (A) 0.5 ms < RC < 1 ms (B) 1 μs << RC < 0.5 ms (C) RC << 1 μs (D) RC >> 0.5 ms

Statement for Linked Answer Questions: 20 and 21 A four-phase and an eight-phase signal constellation are shown in the figure below.




Chapter 9

MCQ 9.20

For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r 1 , and r 2 of the circles are (B) r 1 = 0.707d, r 2 = 1.932d (A) r 1 = 0.707d, r2 = 2.782d (C) r 1 = 0.707d, r 2 = 1.545d (D) r 1 = 0.707d, r 2 = 1.307d

MCQ 9.21

Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is (A) 11.90 dB (B) 8.73 dB (C) 6.79 dB

(D) 5.33 dB

2010 MCQ 9.22

ONE MARK

Suppose that the modulating signal is m (t) = 2 cos (2pfm t) and the carrier signal is xC (t) = AC cos (2pfC t), which one of the following is a conventional AM signal without over-modulation (A) x (t) = AC m (t) cos (2pfC t) (B) x (t) = AC [1 + m (t)] cos (2pfC t) (C) x (t) = AC cos (2pfC t) + AC m (t) cos (2pfC t) 4

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(D) x (t) = AC cos (2pfm t) cos (2pfC t) + AC sin (2pfm t) sin (2pfC t) MCQ 9.23

Consider an angle modulated signal

x (t) = 6 cos [2p # 106 t + 2 sin (800pt)] + 4 cos (800pt) The average power of x (t) is (A) 10 W (B) 18 W (C) 20 W (D) 28 W MCQ 9.24

Consider the pulse shape s (t) as shown below. The impulse response h (t) of the filter matched to this pulse is




2010

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TWO MARKS

Statement for linked Answer Question : 25 and 26 : Consider a baseband binary PAM receiver shown below. The additive channel noise n (t) is with power spectral density Sn (f ) = N 0 /2 = 10-20 W/Hz . The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y (tk ). Yk = Nk , if transmitted bit bk = 0 Yk = a + Nk if transmitted bit bk = 1 Where Nk represents the noise sample value. The noise sample has a probability density function, PNk (n) = 0.5ae- a n (This has mean zero and variance 2/a 2 ). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V .

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MCQ 9.25

The value of the parameter a (in V - 1 ) is (A) 1010 (B) 107 (C) 1.414 # 10-10 (D) 2 # 10-20

MCQ 9.26

The probability of bit error is (A) 0.5 # e-3.5 (C) 0.5 # e-7

(B) 0.5 # e-5 (D) 0.5 # e-10

MCQ 9.27

The Nyquist sampling rate for the signal sin (500pt) sin (700) pt is given by s (t) = # pt pt (A) 400 Hz (B) 600 Hz (C) 1200 Hz (D) 1400 Hz

MCQ 9.28

X (t) is a stationary process with the power spectral density Sx (f ) > 0 , for all f . The process is passed through a system shown below

Let Sy (f ) be the power spectral density of Y (t). Which one of the following statements is correct (A) Sy (f ) > 0 for all f (B) Sy (f ) = 0 for f > 1 kHz




Chapter 9

(C) Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer (D) Sy (f ) = 0 for f = (2n + 1) f0 = 1 kHz , n any integer 2009 MCQ 9.29

ONE MARK

For a message siganl m (t) = cos (2pfm t) and carrier of frequency fc , which of the following represents a single side-band (SSB) signal ? (A) cos (2pfm t) cos (2pfc t) (B) cos (2pfc t) (C) cos [2p (fc + fm) t] (D) [1 + cos (2pfm t) cos (2pfc t) 2009

MCQ 9.30

TWO MARKS

Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities 12 , 14 and 1 4 respectively. What is the conditional probability P (X + Y = 2 X - Y = 0) ? (A) 0

(B) 1/16

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(C) 1/6 MCQ 9.31

(D) 1

A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true ? k

1

2

3

4

5

P (X = k)

0.1

0.2

0.3

0.4

0.5

(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong MCQ 9.32

A message signal given by m (t) = ( 12 ) cos w1 t - ( 12 ) sin w2 t amplitude - modulated with a carrier of frequency wC to generator s (t)[ 1 + m (t)] cos wc t . What is the power efficiency achieved by this modulation scheme ? (A) 8.33% (B) 11.11% (C) 20% (D) 25%

MCQ 9.33

A communication channel with AWGN operating at a signal to noise ration SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping constant, the resulting capacity C2 is given by (B) C2 . C1 + B (A) C2 . 2C1 (C) C2 . C1 + 2B (D) C2 . C1 + 0.3B

Common Data For Q. 34 and 35 : The amplitude of a random signal is uniformly distributed between -5 V and 5 V. MCQ 9.34

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step of the quantization is approximately (A) 0.033 V (B) 0.05 V




(C) 0.0667 V MCQ 9.35

(D) 0.10 V

If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0.1 V, the resulting signal to quantization noise ration is approximately (A) 46 dB (B) 43.8 dB (C) 42 dB

(D) 40 dB

2008 MCQ 9.36

ONE MARK

Consider the amplitude modulated (AM) signalAc cos wc t + 2 cos wm t cos wc t . For demodulating the signal using envelope detector, the minimum value of Ac should be (A) 2 (B) 1 (C) 0.5 2008

MCQ 9.37

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(D) 0

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TWO MARKS

The probability density function (pdf) of random variable is as shown below

The corresponding commutative distribution function CDF has the form

MCQ 9.38

A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n (A) increases as log n (B) decreases as log ( n1 ) (C) increases as n (D) increases as n log n

MCQ 9.39

Noise with double-sided power spectral density on K over all frequencies is passed through a RC low pass filter with 3 dB cut-off frequency of fc . The noise power at the filter output is (A) K (B) Kfc (C) kpfc (D) 3

MCQ 9.40

Consider a Binary Symmetric Channel (BSC) with probability of error being p . To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will




Chapter 9

interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is (B) p3 (A) p3 + 3p2 (1 - p) (C) (1 - p3) (D) p3 + p2 (1 - p) MCQ 9.41

Four messages band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is (A) W (B) 3W (C) 6W (D) 7W

MCQ 9.42

Consider the frequency modulated signal 10 cos [2p # 105 t + 5 sin (2p # 1500t) + 7.5 sin (2p # 1000t)] with carrier frequency of 105 Hz. The modulation index is (A) 12.5 (B) 10 (C) 7.5 (D) 5

MCQ 9.43

The signal cos wc t + 0.5 cos wm t sin wc t is (A) FM only (B) AM only (C) both AM and FM (D) neither AM nor FM

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Common Data For Q. 40 to 46 :

A speed signal, band limited to 4 kHz and peak voltage varying between +5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. MCQ 9.44

If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (A) 64 kHz (B) 32 kHz (C) 8 kHz (D) 4 kHz

MCQ 9.45

Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is (A) 16 dB (B) 32 dB (C) 48 dB (D) 4 kHz

MCQ 9.46

Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is (A) 1024 (B) 512 (C) 256 (D) 64 2007

MCQ 9.47

ONE MARK

If R (t) is the auto correlation function of a real, wide-sense stationary random process, then which of the following is NOT true (A) R (t) = R (- t) (B) R (t) # R (0) (C) R (t) =- R (- t) (D) The mean square value of the process is R (0)



MCQ 9.48


If S (f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (B) S (f) $ 0 (A) S (0) # S (f) (C) S (- f) =- S (f)

MCQ 9.49

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(D)

#- 3 S (f) df = 0 3

If E denotes expectation, the variance of a random variable X is given by (A) E [X2] - E2 [X] (B) E [X2] + E2 [X] (C) E [X2]

(D) E2 [X]

2007 MCQ 9.50

MCQ 9.51

TWO MARKS

A Hilbert transformer is a (A) non-linear system (C) time-varying system

In delta modulation, the slope overload distortion can be reduced by (A) decreasing the step size (B) decreasing the granular noise

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(C) decreasing the sampling rate MCQ 9.52

(B) non-causal system (D) low-pass system

(D) increasing the step size

The raised cosine pulse p (t) is used for zero ISI in digital communications. The expression for p (t) with unity roll-off factor is given by sin 4pWt p (t) = 4pWt (1 - 16W2 t2) The value of p (t) at t = 1 is 4W (A) - 0.5 (C) 0.5

(B) 0 (D) 3

MCQ 9.53

In the following scheme, if the spectrum M (f) of m (t) is as shown, then the spectrum Y (f) of y (t) will be

MCQ 9.54

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in




error in a block of n bits is given by (A) pn

(B) 1 - pn

(C) np (1 - p) n - 1 + (1 + p) n

(D) 1 - (1 - p) n

Chapter 9

MCQ 9.55

In a GSM system, 8 channels can co-exist in 200 kHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of 1 , i.e. a five-cell repeat pattern, the maximum number 5 of simultaneous channels that can exist in one cell is (A) 200 (B) 40 (C) 25 (D) 5

MCQ 9.56

In a Direct Sequence CDMA system the chip rate is 1.2288 # 106 chips per second. If the processing gain is desired to be AT LEAST 100, the data rate (A) must be less than or equal to 12.288 # 103 bits per sec (B) must be greater than 12.288 # 103 bits per sec (C) must be exactly equal to 12.288 # 103 bits per sec (D) can take any value less than 122.88 # 103 bits per sec

nodia


Two 4-array signal constellations are shown. It is given that f1 and f2 constitute an orthonormal basis for the two constellation. Assume that the four symbols in both the constellations are equiprobable. Let N0 denote the power 2 spectral density of white Gaussian noise.

MCQ 9.57

The if ratio or the average energy of Constellation 1 to the average energy of Constellation 2 is (A) 4a2 (B) 4 (C) 2 (D) 8

MCQ 9.58

If these constellations are used for digital communications over an AWGN channel, then which of the following statements is true ? (A) Probability of symbol error for Constellation 1 is lower (B) Probability of symbol error for Constellation 1 is higher (C) Probability of symbol error is equal for both the constellations (D) The value of N0 will determine which of the constellations has a lower probability of symbol error

Statement for Linked Answer Question 59 and 60 : An input to a 6-level quantizer has the probability density function f (x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to




Page 447

maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are’ - 1'.'0' and '1' .

MCQ 9.59

MCQ 9.60

The values of a and b are (A) a = 1 and b = 1 (B) a = 1 and b = 3 6 12 5 40 1 1 1 (C) a = and b = (D) a = and b = 1 4 16 3 24 Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is (A) 152 (B) 64 9 3 (C) 76 (D) 28 3 2006

MCQ 9.61

nodia

A low-pass filter having a frequency response H (jw) = A (w) e jf (w) does not produce any phase distortions if (A) A (w) = Cw3, f (w) = kw3 (B) A (w) = Cw2, f (w) = kw (C) A (w) = Cw, f (w) = kw2

(D) A (w) = C, f (w) = kw- 1

2006 MCQ 9.62

ONE MARK

TWO MARKS

A signal with bandwidth 500 Hz is first multiplied by a signal g (t) where g (t) =

3

/ (- 1) k d (t - 0.5 # 10- 4 k) R =- 3

The resulting signal is then passed through an ideal lowpass filter with bandwidth 1 kHz. The output of the lowpass filter would be (B) m (t) (A) d (t) (C) 0 (D) m (t) d (t) MCQ 9.63

The minimum sampling frequency (in samples/sec) required to reconstruct the following signal from its samples without distortion 3 2 x (t) = 5` sin 2p100t j + 7` sin 2p100t j would be pt pt (B) 4 # 103 (A) 2 # 103 (C) 6 # 103 (D) 8 # 103

MCQ 9.64

The minimum step-size required for a Delta-Modulator operating at 32k samples/ sec to track the signal (here u (t) is the unit-step function) x (t) = 125[ u (t) - u (t - 1) + (250t)[ u (t - 1) - u (t - 2)] so that slope-overload is avoided, would be (A) 2 - 10 (B) 2 - 8




(C) 2 - 6

Chapter 9

(D) 2 - 4

MCQ 9.65

A zero-mean white Gaussian noise is passes through an ideal lowpass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period ts = 0.03 msec. The samples so obtained would be (A) correlated (B) statistically independent (C) uncorrelated (D) orthogonal

MCQ 9.66

A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate is (A) 6000 bits/sec (B) 4500 bits/sec (C) 3000 bits/sec

MCQ 9.67

MCQ 9.68

(D) 1500 bits/sec

The diagonal clipping in Amplitude Demodulation (using envelop detector) can be avoided it RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and w is carrier frequency both in rad/ sec) (A) RC < 1 (B) RC > 1 W W 1 (C) RC < (D) RC > 1 w w A uniformly distributed random variable X with probability density function fx (x) = 1 pu (x + 5) - u (x - 5)] 10

nodia

where u (.) is the unit step function is passed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be

(A) fy (y) = 1 [u (y + 2.5) - u (y - 2.25)] 5 (B) fy (y) = 0.5d (y) + 0.5d (y - 1) (C) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 5d (y) (D) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 1 [u (y + 2.5) - u (y - 2.5)] 10

MCQ 9.69

In the following figure the minimum value of the constant "C" , which is to be added to y1 (t) such that y1 (t) and y2 (t) are different , is

(A) 3

(B) 3 2



MCQ 9.70


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2 (C) 3 (D) 3 12 L A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then passed through a Narrow-Band Frequency Modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be (A) 4 # 10 4 Hz (B) 2 # 106 Hz (C) 2 # 109 Hz (D) 2 # 1010 Hz

Common Data For Q. 71 and 72 : Let g (t) = p (t)*( pt), where * denotes convolution & p (t) = u (t) - u (t - 1) lim z"3 with u (t) being the unit step function MCQ 9.71

The impulse response of filter matched to the signal s (t) = g (t) - d (1 - 2)* g (t) is given as : (A) s (1 - t) (B) - s (1 - t) (C) - s (t) (D) s (t)

MCQ 9.72

An Amplitude Modulated signal is given as

nodia

xAM (t) = 100 [p (t) + 0.5g (t)] cos wc t in the interval 0 # t # 1. One set of possible values of modulating signal and modulation index would be (B) t, 1.0 (A) t, 0.5 (C) t, 2.0 (D) t2, 0.5

Common Data For Q. 73 and 74 : The following two question refer to wide sense stationary stochastic process MCQ 9.73

It is desired to generate a stochastic process (as voltage process) with power spectral density S (w) = 16/ (16 + w2) by driving a Linear-Time-Invariant system by zero mean white noise (As voltage process) with power spectral density being constant equal to 1. The system which can perform the desired task could be (A) first order lowpass R-L filter (B) first order highpass R-C filter (C) tuned L-C filter (D) series R-L-C filter

MCQ 9.74

The parameters of the system obtained in previous Q would be (A) first order R-L lowpass filter would have R = 4W L = 1H (B) first order R-C highpass filter would have R = 4W C = 0.25F (C) tuned L-C filter would have L = 4H C = 4F (D) series R-L-C lowpass filter would have R = 1W , L = 4H , C = 4F

Common Data For Q. 75 an 76 : Consider the following Amplitude Modulated (AM) signal, where fm < B XAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t




Chapter 9

MCQ 9.75

The average side-band power for the AM signal given above is (A) 25 (B) 12.5 (C) 6.25 (D) 3.125

MCQ 9.76

The AM signal gets added to a noise with Power Spectral Density Sn (f) given in the figure below. The ratio of average sideband power to mean noise power would be :

25 8N0 B (C) 25 2N0 B

25 4N0 B (D) 25 N0 B

(A)

2005 MCQ 9.77

(B)

nodia

ONE MARK

Find the correct match between group 1 and group 2. Group 1 Group 2 P. {1 + km (t) A sin (wc t)} W. Phase modulation X. Frequency modulation Q. km (t) A sin (wc t) R. A sin {wc t + km (t)} Y. Amplitude modulation t S. A sin ;wc t + k # m (t) dt E Z. DSB-SC modulation -3

(A) P - Z, Q - Y, R - X, S - W (B) P - W, Q - X, R - Y, S - Z (C) P - X, Q - W, R - Z, S - Y (D) P - Y, Q - Z, R - W, S - X MCQ 9.78

Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth ? (A) VSB (B) DSB-SC (C) SSB (D) AM 2005

TWO MARKS

MCQ 9.79

A device with input X (t) and output y (t) is characterized by: Y (t) = x2 (t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is (A) 370 kHz (B) 190 kHz (C) 380 kHz (D) 95 kHz

MCQ 9.80

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this matched filter ?




Page 451

nodia

MCQ 9.81

Noise with uniform power spectral density of N0 W/Hz is passed though a filter H (w) = 2 exp (- jwtd ) followed by an ideal pass filter of bandwidth B Hz. The output noise power in Watts is (A) 2N0 B (B) 4N0 B (C) 8N0 B (D) 16N0 B

MCQ 9.82

An output of a communication channel is a random variable v with the probability density function as shown in the figure. The mean square value of v is

(A) 4 (C) 8 MCQ 9.83

(B) 6 (D) 9

A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is (A) 21 kHz (B) 22 kHz (C) 42 kHz (D) 44 kHz

Common Data For Q. 84 and 85 : Asymmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.



MCQ 9.84

MCQ 9.85


If the probability density function is divide into three regions as shown in the figure, the value of a in the figure is (B) 2 (A) 1 3 3 (C) 1 (D) 1 2 4 The quantization noise power for the quantization region between - a and + a in the figure is (B) 1 (A) 4 81 9 (C) 5 (D) 2 81 81 2004

MCQ 9.86

Chapter 9

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ONE MARK

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor (A) 8 (B) 12 6 (C) 16

(D) 8

MCQ 9.87

An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelop detector is (B) 20m sec (A) 500m sec (C) 0.2m sec (D) 1m sec

MCQ 9.88

An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by (A) broadband FM (B) SSB with carrier (C) DSB-SC (D) SSB without carrier

MCQ 9.89

In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t1 # t # t2 . This indicates that during this interval (A) the input to the modulator is essentially constant (B) the modulator is going through slope overload (C) the accumulator is in saturation (D) the speech signal is being sampled at the Nyquist rate

MCQ 9.90

The distribution function Fx (x) of a random variable x is shown in the figure. The probability that X = 1 is




(A) zero (C) 0.55

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(B) 0.25 (D) 0.30

2004

TWO MARKS

MCQ 9.91

A 1 mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through cable that has 40 dB loss. If the effective one-side noise spectral density at the receiver is 10 - 20 Watt/Hz, then the signal-to-noise ratio at the receiver is (A) 50 dB (B) 30 dB (C) 40 dB (D) 60 dB

MCQ 9.92

Consider the signal x (t) shown in Fig. Let h (t) denote the impulse response of the filter matched to x (t), with h (t) being non-zero only in the interval 0 to 4 sec. The slope of h (t) in the interval 3 < t < 4 sec is

MCQ 9.93

nodia

(A) 1 sec - 1 (B) - 1 sec - 1 2 (C) - 1 sec - 1 (D) 1 sec - 1 2 A source produces binary data at the rate of 10 kbps. The binary symbols are represented as shown in the figure. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let B1 and B2 be the bandwidth requirements of the above rectangular pulses is 10 kHz, B1 and B2 are

(A) B1 = 20 kHz, B2 = 20 kHz (C) B1 = 20 khz, B2 = 10 kHz MCQ 9.94

(B) B1 = 10 kHz, B2 = 20 kHz (D) B1 = 10 kHz, B2 = 10 kHz

A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The ourput of the modulator is passed through an ideal high-pass filter with cut-off frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 90c phase




Chapter 9

shift as shown in the figure. The envelope of the resultant signal is

(A) constant

(B)

1 + sin (2p # 106 t)

5 - sin (2p - 106 t) 5 + cos (2p # 106 t) (D) 4 4 Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is (A) 0.1 kHz sinusoid (B) 20.1 kHz sinusoid (C) a linear function of time (D) a constant (C)

MCQ 9.95

MCQ 9.96

Consider a binary digital communication system with equally likely 0’s and 1’s. When binary 0 is transmitted the detector input can lie between the levels - 0.25 V and + 0.25 V with equl probability : when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V, the bit is taken as 1), the average bit error probability is (A) 0.15 (B) 0.2

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(C) 0.05

(D) 0.5

MCQ 9.97

A random variable X with uniform density in the interval 0 to 1 is quantized as follows : If 0 # X # 0.3 , xq = 0 If 0.3 < X # 1, xq = 0.7 where xq is the quantized value of X. The root-mean square value of the quantization noise is (A) 0.573 (B) 0.198 (D) 0.266 (C) 2.205

MCQ 9.98

Choose the current one from among the alternative A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2. Group 1 Group 2 1. FM P. Slope overload 2. DM Q. m-law 3. PSK R. Envelope detector 4. PCM S. Hilbert transform T. Hilbert transform U. Matched filter (A) 1 - T, 2 - P, 3 - U, 4 - S (B) 1 - S, 2 - U, 3 - P, 4 - T (C) 1 - S, 2 - P, 3 - U, 4 - Q (D) 1 - U, 2 - R, 3 - S, 4 - Q

MCQ 9.99

Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed. The bit rate for the multiplexed signal is (A) 115.2 kbps (B) 28.8 kbps




(C) 57.6 kbps MCQ 9.100

Page 455

(D) 38.4 kbps

Consider a system shown in the figure. Let X (f) and Y (f) and denote the Fourier transforms of x (t) and y (t) respectively. The ideal HPF has the cutoff frequency 10 kHz.

The positive frequencies where Y (f) has spectral peaks are (A) 1 kHz and 24 kHz (B) 2 kHz and 24 kHz (C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHz 2003

nodia

ONE MARK

MCQ 9.101

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is (A) the in-phase component (B) the quadrature - component (C) zero (D) the envelope

MCQ 9.102

The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is (A) raised - cosine (B) flat (C) parabolic (D) Gaussian

MCQ 9.103

At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by. (A) 6 dB (B) 3 dB (C) 2 dB (D) 0 dB 2003

MCQ 9.104

TWO MARKS

Let X and Y be two statistically independent random variables uniformly distributed in the ranges (- 1, 1) and (- 2, 1) respectively. Let Z = X + Y . Then the probability that (z #- 1) is (A) zero (B) 1 6 1 (C) (D) 1 3 12





Chapter 9

X (t) is a random process with a constant mean value of 2 and the auto correlation function Rxx (t) = 4 (e - 0.2 t + 1). MCQ 9.105

Let X be the Gaussian random variable obtained by sampling the process at t = ti and let 3 Q (a) = # - 1 e dy a 2p The probability that 6x # 1@ is (A) 1 - Q (0.5) (B) Q (0.5) 1 (C) Q c (D) 1 - Q c 1 m 2 2m 2 2 Let Y and Z be the random variable obtained by sampling X (t) at t = 2 and t = 4 respectively. Let W = Y - Z . The variance of W is (A) 13.36 (B) 9.36 (C) 2.64 (D) 8.00 x2 2

MCQ 9.106

MCQ 9.107

A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into 128 levels using a mid-rise uniform quantizer. The quantization-noise power is (A) 0.768 V (B) 48 # 10 - 6 V2

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(B) 12 # 10 - 6 V2 MCQ 9.108

Let x (t) = 2 cos (800p) + cos (1400pt). x (t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

(A) 2.7, 3.4 (C) 2.6, 2.7, 3.3, 3.4, 3.6 MCQ 9.109

(D) 3.072 V

(B) 3.3, 3.6 (D) 2.7, 3.3

A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input-output characteristic V0 = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band-pass filter. Let Vi = Aci cos (2pfi ct) + m (t) is the message signal. Then the value of fci (in MHz) is (A) 1.0 (B) 0.333 (B) 0.5 (D) 3.0

Common Data For Q. 110 and 111 : Let m (t) = cos [(4p # 103) t] be the message signal & c (t) = 5 cos [(2p # 106 t)] be the carrier. MCQ 9.110

c (t) and m (t) are used to generate an AM signal. The modulation index of the Total sideband power generated AM signal is 0.5. Then the quantity is Carrier power (A) 1 (B) 1 2 4




(C) 1 3 MCQ 9.111

MCQ 9.112

Page 457

(D) 1 8

c (t) and m (t) are used to generated an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p (1008 # 103 t)] in the FM signal (in terms of the Bessel coefficients) is (A) 5J4 (3) (B) 5 J8 (3) 2 5 (C) J8 (4) (D) 5J4 (6) 2 Choose the correct one from among the alternative A, B, C, D after matching an item in Group 1 with most appropriate item in Group 2. Group 1 Group 2 P. Ring modulator 1. Clock recovery Q. VCO 2. Demodulation of FM R. Foster-Seely discriminator 3. Frequency conversion S. Mixer 4. Summing the two inputs 5. Generation of FM 6. Generation of DSB-Sc (B) P - 6; Q = 5; R - 2; S - 3 (A) P - 1; Q - 3; R - 2; S - 4 (C) P - 6; Q - 1; R - 3; S - 2 (D) P - 5; Q - 6; R - 1; S - 3

nodia

MCQ 9.113

A superheterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = Cmax /Cmin denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then (B) R = 2.10, I - 1150 (A) R = 4.41, I = 1600 (C) R = 3.0, I = 600 (D) R = 9.0, I = 1150

MCQ 9.114

If Eb , the energy per bit of a binary digital signal, is 10 - 5 watt-sec and the onesided power spectral density of the white noise, N0 = 10 - 6 W/Hz, then the output SNR of the matched filter is (A) 26 dB (B) 10 dB (C) 20 dB (D) 13 dB

MCQ 9.115

The input to a linear delta modulator having a step-size 3= 0.628 is a sine wave with frequency fm and peak amplitude Em . If the sampling frequency fx = 40 kHz, the combination of the sine-wave frequency and the peak amplitude, where slope overload will take place is Em fm (A) 0.3 V 8 kHz (B) 1.5 V 4 kHz (C) 1.5 V 2 kHz (D) 3.0 V 1 kHz

MCQ 9.116

If S represents the carrier synchronization at the receiver and r represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is (B) r = 1.0, S is required (A) r = 0.5, S is required (C) r = 0.5, S is not required (D) r = 1.0, S is not required



MCQ 9.117


Chapter 9

A signal is sampled at 8 kHz and is quantized using 8 - bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is (A) R = 32 kbps, SNRq = 25.8 dB (B) R = 64 kbps, SNRq = 49.8 dB (C) R = 64 kbps, SNRq = 55.8 dB (D) R = 32 kbps, SNRq = 49.8 dB 2002

ONE MARK

MCQ 9.118

A 2 MHz sinusoidal carrier amplitude modulated by symmetrical square wave of period 100 m sec . Which of the following frequencies will NOT be present in the modulated signal ? (A) 990 kHz (B) 1010 kHz (C) 1020 kHz (D) 1030 kHz

MCQ 9.119

Consider a sample signal y (t) = 5 # 10 - 6 # (t)

+3

/ d (t - nTs)

nodia n =- 3

where x (t) = 10 cos (8p # 103) t and Ts = 100m sec. When y (t) is passed through an ideal lowpass filter with a cutoff frequency of 5 KHz, the output of the filter is (A) 5 # 10 - 6 cos (8p # 103) t (b) 5 # 10 - 5 cos (8p # 103) t (C) 5 # 10 - 1 cos (8p # 103) t

(D) 10 cos (8p # 103) t

MCQ 9.120

For a bit-rate of 8 Kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are (A) 16 kHz and 20 kHz (C) 20 kHz and 32 kHz (C) 20 kHz and 40 kHz (D) 32 kHz and 40 kHz

MCQ 9.121

The line-of-sight communication requires the transmit and receive antennas to face each other. If the transmit antenna is vertically polarized, for best reception the receiver antenna should be (A) horizontally polarized (B) vertically polarized (C) at 45c with respect to horizontal polarization (D) at 45c with respect to vertical polarization 2002

MCQ 9.122

TWO MARKS

An angle-modulated signal is given by s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t). The maximum frequency and phase deviations of s (t) are (B) 6 kHz, 80p rad (A) 10.5 kHz, 140p rad (C) 10.5 kHz, 100p rad

MCQ 9.123

(D) 7.5 kHz, 100p rad

In the figure m (t) = 2 sin 2pt , s (t) = cos 200pt and n (t) = sin 199pt . t t The output y (t) will be



MCQ 9.124

MCQ 9.125


(A) sin 2pt (B) sin 2pt + sin pt cos 3pt t t t sin sin 2 p t pt cos 0.75pt sin 2 p t sin 0 . 5 p t (C) (D) + + cos 1.5pt t t t t A signal x (t) = 100 cos (24p # 103) t is ideally sampled with a sampling period of 50m sec ana then passed through an ideal lowpass filter with cutoff frequency of 15 kHz. Which of the following frequencies is/are present at the filter output ? (A) 12 kHz only (B) 8 kHz only (C) 12 kHz and 9 kHz (D) 12 kHz and 8 kHz If the variance ax2 of d (n) = x (n) - x (n - 1) is one-tenth the variance ax2 of stationary zero-mean discrete-time signal x (n), then the normalized autocorrelation function Rxx (k) at k = 1 is 2 a x (A) 0.95 (B) 0.90

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(C) 0.10 2001 MCQ 9.126

Page 459

(D) 0.05

ONE MARK

A bandlimited signal is sampled at the Nyquist rate. The signal can be recovered by passing the samples through (A) an RC filter (B) an envelope detector (C) a PLL (D) an ideal low-pass filter with the appropriate bandwidth

MCQ 9.127

The PDF of a Gaussian random variable X is given by (x - 4) px (x) = 1 e - 18 . The probability of the event {X = 4} is 3 2p 1 1 (B) (A) 2 3 2p (C) 0 (D) 1 4 2

2001

TWO MARKS

MCQ 9.128

A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 # 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is (A) 16 Mbps (B) 100 Mbps (C) 600 Mbps (D) 6.4 Gbps

MCQ 9.129

The Nyquist sampling interval, for the signal sin c (700t) + sin c (500t) is (A) 1 sec (B) p sec 350 350 1 (C) (D) p sec sec 700 175 During transmission over a communication channel, bit errors occur independently

MCQ 9.130




Chapter 9

with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to (B) p + (n - 1)( 1 - p) (A) 1 - (1 - p) n (C) np (1 - p) n - 1 MCQ 9.131

(D) (1 - p) n + np (1 - p) n - 1

The PSD and the power of a signal g (t) are, respectively, Sg (w) and Pg . The PSD and the power of the signal ag (t) are, respectively, (A) a2 Sg (w) and a2 Pg (B) a2 Sg (w) and aPg (C) aSg (w) and a2 Pg (D) aSg (w) and aPs 2000

ONE MARK

MCQ 9.132

The amplitude modulated waveform s (t) = Ac [1 + Ka m (t)] cos wc t is fed to an ideal envelope detector. The maximum magnitude of K0 m (t) is greater than 1. Which of the following could be the detector output ? (A) Ac m (t) (B) Ac2 [1 + Ka m (t)] 2 (C) [Ac (1 + Ka m (t)] (D) Ac [1 + Ka m (t)] 2

MCQ 9.133

The frequency range for satellite communication is (A) 1 KHz to 100 KHz (B) 100 KHz to 10 KHz (C) 10 MHz to 30 MHz (D) 1 GHz to 30 GHz 2000

nodia

TWO MARKS

MCQ 9.134

In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 KHz and 25 KHz respectively. These waveforms will be orthogonal for a bit interval of (A) 45m sec (B) 200m sec (C) 50m sec (D) 250m sec

MCQ 9.135

A message m (t) bandlimited to the frequency fm has a power of Pm . The power of the output signal in the figure is

(A) Pm cos q 2

MCQ 9.136

MCQ 9.137

(B) Pm 4

2 2 (C) Pm sin q (D) Pm cos q 4 4 The Hilbert transform of cos w1 t + sin w2 t is (B) sin w1 t + cos w2 t (A) sin w1 t - cos w2 t (C) cos w1 t - sin w2 t (D) sin w1 t + sin w2 t

In a FM system, a carrier of 100 MHz modulated by a sinusoidal signal of 5 KHz. The bandwidth by Carson’s approximation is 1 MHz. If y (t) = (modulated waveform) 3 , than by using Carson’s approximation, the bandwidth of y (t) around 300 MHz and the and the spacing of spectral components are, respectively. (A) 3 MHz, 5 KHz (B) 1 MHz, 15 KHz




(C) 3 MHz, 15 KHz

(D) 1 MHz, 5 KHz

1999 MCQ 9.138

Page 461

ONE MARK

The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by y (t) = (1/100) cos (100t - 10-6) cos (106 t - 1.56) The group delay (tg) and the phase delay (t p) in seconds, of the channel are (A) tg = 10-6, t p = 1.56 (B) tg = 1.56, t p = 10-6 (C) tg = 108, t p = 1.56 # 10-6 (D) tg = 108, t p = 1.56

MCQ 9.139

A modulated signal is given by s (t) = m1 (t) cos (2pfc t) + m2 (t) sin (2pfc t) where the baseband signal m1 (t) and m2 (t) have bandwidths of 10 kHz, and 15 kHz, respectively. The bandwidth of the modulated signal, in kHz, is (A) 10 (B) 15 (C) 25 (D) 30

MCQ 9.140

nodia

A modulated signal is given by s (t) = e-at cos [(wc + Dw) t] u (t), where a wc and Dw are positive constants, and wc >> Dw . The complex envelope of s (t) is given by (A) exp (- at) exp [j (wc + Dw) t] u (t) (B) exp (- at) exp (jDwt) u (t) (C) exp (jDwt) u (t) (D) exp [jwc + Dw) t] 1999

TWO MARKS

MCQ 9.141

The Nyquist sampling frequency (in Hz) of a signal given by 6 # 10 4 sin c2 (400t) * 106 sin c3 (100t) is (A) 200 (B) 300 (C) 500 (D) 1000

MCQ 9.142

The peak-to-peak input to an 8-bit PCM coder is 2 volts. The signal power-toquantization noise power ratio (in dB) for an input of 0.5 cos (wm t) is (A) 47.8 (B) 49.8 (C) 95.6 (D) 99.6

MCQ 9.143

The input to a matched filter is given by 6 -4 "10 sin (2p # 10 t) 0 < 1 < 10 sec s (t) = 0 otherwise The peak amplitude of the filter output is (B) 5 volts (A) 10 volts (C) 10 millivolts (D) 5 millivolts

MCQ 9.144

Four independent messages have bandwidths of 100 Hz, 200 Hz and 400 Hz , respectively. Each is sampled at the Nyquist rate, and the samples are time division multiplexed (TDM) and transmitted. The transmitted sample rate (in Hz) is




(A) 1600 (C) 400

Chapter 9

(B) 800 (D) 200

1998

ONE MARK

MCQ 9.145

The amplitude spectrum of a Gaussian pulse is (A) uniform (B) a sine function (C) Gaussian (D) an impulse function

MCQ 9.146

The ACF of a rectangular pulse of duration T is (A) a rectangular pulse of duration T (B) a rectangular pulse of duration 2T (C) a triangular pulse of duration T (D) a triangular pulse of duration 2T

MCQ 9.147

The image channel selectivity of superheterodyne receiver depends upon (A) IF amplifiers only (B) RF and IF amplifiers only (C) Preselector, RF and IF amplifiers (D) Preselector, and RF amplifiers only

MCQ 9.148

In a PCM system with uniform quantisation, increasing the number of bits from 8 to 9 will reduce the quantisation noise power by a factor of (A) 9 (B) 8 (C) 4 (D) 2

MCQ 9.149

Flat top sampling of low pass signals (A) gives rise to aperture effect (C) leads to aliasing

nodia

(B) implies oversampling (D) introduces delay distortion

MCQ 9.150

A DSB-SC signal is generated using the carrier cos (we t + q) and modulating signal x (t). The envelope of the DSB-SC signal is (B) x (t) (A) x (t) (C) only positive portion of x (t) (D) x (t) cos q

MCQ 9.151

Quadrature multiplexing is (A) the same as FDM (B) the same as TDM (C) a combination of FDM and TDM (D) quite different from FDM and TDM

MCQ 9.152

The Fourier transform of a voltage signal x (t) is X (f). The unit of X (f) is (A) volt (B) volt-sec (C) volt/sec (D) volt 2

MCQ 9.153

Compression in PCM refers to relative compression of (A) higher signal amplitudes (B) lower signal amplitudes (C) lower signal frequencies (D) higher signal frequencies

MCQ 9.154

For a give data rate, the bandwidth B p of a BPSK signal and the bandwidth B 0




Page 463

of the OOK signal are related as (A) B p = B 0 4

(B) B p = B 0 2

(C) B p = B 0

(D) B p = 2B 0

MCQ 9.155

The spectral density of a real valued random process has (A) an even symmetry (B) an odd symmetry (C) a conjugate symmetry (D) no symmetry

MCQ 9.156

The probability density function of the envelope of narrow band Gaussian noise is (A) Poisson (B) Gaussian (C) Rayleigh (D) Rician 1997

MCQ 9.157

ONE MARK

The line code that has zero dc component for pulse transmission of random binary data is (A) Non-return to zero (NRZ)

nodia

(B) Return to zero (RZ) (C) Alternate Mark Inversion (AM) (D) None of the above

2

MCQ 9.158

A probability density function is given by p (x) = Ke-x /2 - 3 < x < 3. The value of K should be 2 (B) (A) 1 p 2p (C) 1 (D) 1 2 p p 2

MCQ 9.159

A deterministic signal has the power spectrum given in the figure is, The minimum sampling rate needed to completely represent this signal is

(A) 1 kHz (C) 3 kHz MCQ 9.160

(B) 2 kHz (D) None of these

A communication channel has first order low pass transfer function. The channel is used to transmit pulses at a symbol rate greater than the half-power frequency of the low pass function. Which of the network shown in the figure is can be used to equalise the received pulses?



MCQ 9.161


The power spectral density of a deterministic signal is given by [sin (f) /f 2] where f is frequency. The auto correlation function of this signal in the time domain is (A) a rectangular pulse (B) a delta function (C) a sine pulse (D) a triangular pulse 1996

MCQ 9.162

Chapter 9

nodia

ONE MARK

A rectangular pulse of duration T is applied to a filter matched to this input. The out put of the filter is a (A) rectangular pulse of duration T (B) rectangular pulse of duration 2T (C) triangular pulse (D) sine function

MCQ 9.163

The image channel rejection in a superheterodyne receiver comes from (B) RF stages only (A) IF stages only (D) detector RF and IF stages (C) detector and RF stages only 1996

MCQ 9.164

TWO MARKS

The number of bits in a binary PCM system is increased from n to n + 1. As a result, the signal to quantization noise ratio will improve by a factor (A) n + 1 (B) 2(n + 1)/n n (C) 22 (n + 1)/n

MCQ 9.165

(D) which is independent of n

The auto correlation function of an energy signal has (A) no symmetry (B) conjugate symmetry (C) odd symmetry (D) even symmetry ***********




Page 465

SOLUTIONS SOL 9.1

SOL 9.2

Option (B) is correct. In ideal Nyquist Channel, bandwidth required for ISI (Inter Symbol reference) free transmission is W = Rb 2 Here, the used modulation is 32 - QAM (Quantum Amplitude modulation i.e., q = 32 or 2v = 32 v = 5 bits So, the signaling rate (sampling rate) is (R " given bit rate) Rb = R 5

nodia

Hence, for ISI free transmission, minimum bandwidth is W = Rb = R kHz 2 10 Option (B) is correct.

Given, random variables U and V with mean zero and variances 1 and 1 9 4 i.e., U =V=0 su2 = 1 4 1 2 and sv = 9 1 so, P Û $ 0h = 2 1 and P ^V $ 0h = 2 The distribution is shown in the figure below

fu û h = fv ^v h =

1 e -u 2s 2p su2

2 u

1 e -v 2s 2p sv2 We can express the distribution in standard form by assuming X = u - 0 = u = 2U su Y2 v 0 and Y = = v = 3V sv Y3 2 v




Chapter 9

for which we have X = 2U = 0 Y = 2V = 0 and X2 = 4U2 = 1 also, Y2 = 9V2 = 1 Therefore, X - Y is also a normal random variable with X-Y = 0 Hence, P ^X - Y $ 0h = P ^X - Y # 0h = 1 2 or, we can say P ^2U - 3V # 0h = 1 2 1 Thus, P ^3V $ 2U h = 2 SOL 9.3

Option (C) is correct. The mean of random variables U and V are both zero i.e., U =V=0 Also, the random variables are identical i.e., fU û h = fV ^v h or, FU û h = FV ^v h i.e., their cdf are also same. So, FU û h = F2V ^2v h i.e., the cdf of random variable 2V will be also same but for any instant 2V $ U Therefore, G ^x h = F ^x h but, x G ^x h $ xF ^x h or, 6F ^x h - G ^x h@ x # 0

nodia

SOL 9.4


P Û =+ 1h = P Û =- 1h = 1 2 where U is a random variable which is identical to V i.e., P ^V =+ 1h = P ^V =- 1h = 1 2 So, random variable U and V can have following values U =+ 1, - 1; V =+ 1, - 1 Therefore the random variable U + V can have the following values, - 2 When U = V =- 1 U + V = *0 When U = 1,V = 1 or u =- 1, v = 1 2 When U = V = 1 Hence, we obtain the probabilities for U + V as follows Given,

P Û + V h

U+V -2 0

1 1=1 2#2 4 1 1 1 1 1 b2 # 2l+b2 # 2l = 2

2 Therefore, the entropy of the Û + V h is obtained as

1 1=1 2#2 4




Page 467

H Û + V h = / P Û + V h log 2 '

1 P Û + V h 1 = 1 log 2 4 + 1 log 2 2 + 1 log 2 4 2 4 4 2 1 2 = + + 4 2 4 3 = 2

SOL 9.5

Option (D) is correct. For the shown received signal, we conclude that if 0 is the transmitted signal then the received signal will be also zero as the threshold is 1 and the pdf of bit 0 is not crossing 1. Again, we can observe that there is an error when bit 1 is received as it crosses the threshold. The probability of error is given by the area enclosed by the 1 bit pdf (shown by shaded region)

nodia

P (error when bit 1 received) = 1 # 1 # 0.25 = 1 2 8 1 received 1 or Pb = transmitted 0 l 8 Since, the 1 and 0 transmission is equiprobable: i.e., P ^ 0 h = P ^1 h = 1 2 Hence bit error rate (BER) is BER = P b received 0 l P ^0 h + P b received 1 l P ^1 h transmitted 1 transmitted 0 1 1 = 0+ # 8 2 1 = 16 SOL 9.6

Option (B) is correct. The optimum threshold is the threshold value for transmission as obtained at the intersection of two pdf. From the shown pdf. We obtain at the intersection (transmitted, received) = b 4 , 1 l 5 5 we can obtain the intersection by solving the two linear eqs pdf of received bit 0 x+y = 1 0 . 5 y = x pdf of received bit 1 2 Hence for threshold = 4 , we have 5 BER = P b received 1 l P ^0 h + P b received 0 l P ^1 h transmitted 0 transmitted 1 1 1 1 1 1 4 = b # # l# +b # # 1l# 1 2 5 2 2 2 5 5 2 = 1 <(BER for threshold = 1) 20



SOL 9.7


Chapter 9

Hence, optimum threshold is 4 5 Option (A) is correct. The mean square value of a stationary process equals the total area under the graph of power spectral density, that is E [X 2 (t)] = or,

#S 3

-3

E [X 2 (t)] = 1 2p

X

(f ) df

#S 3

-3

X

(w) dw

3 (Since the PSD is even) E [X 2 (t)] = 2 # 1 # SX (w) dw 2p 0 = 1 [area under the triangle + integration of delta function] p = 1 ;2 b 1 # 1 # 103 # 6 l + 400E 2 p = 1 66000 + 400@ = 6400 p p E [X (t)] is the absolute value of mean of signal X (t) which is also equal to value of X (w) at (w = 0). From given PSD

or,

nodia SX (w) w = 0 = 0

SX (w) = X (w) 2 = 0

X (w) 2w = 0 = 0 X (w) w = 0 = 0

SOL 9.8

Option (C) is correct. For raised cosine spectrum transmission bandwidth is given as BT = W (1 + a) BT = Rb (1 + a) 2 3500 = Rb (1 + 0.75) 2 3500 # 2 = 4000 Rb = 1.75

SOL 9.9

a " Roll of factor Rb " Maximum signaling rate

Option (D) is correct. Entropy function of a discrete memory less system is given as H =

N-1

/ P log b P1 l k

k

k=0

where Pk is probability of symbol Sk . For first two symbols probability is same, so N-1

H = P1 log b 1 l + P2 log b 1 l + / Pk log b 1 l P1 P2 Pk k=3 =-e P1 log P1 + P2 log P2 +

N-1

/ P log P o k

k

k=3

=-e 2P log P +

N-1

/ P log P o k

k

(P1 = P2 = P)

k=3




P1 = P + e, P2 = P - e

Now, So,

Page 469

Hl =-=(P + e) log (P + e) + (P - e) log (P - e) +

N-1

/ P log P G k

k

k=3

By comparing, SOL 9.10

Hl < H , Entropy of source decreases.

Option (B) is correct. Probability density function of uniformly distributed variables X and Y is shown as

P &[max (x, y)] < 1 0 2 Since X and Y are independent. P &[max (x, y)] < 1 0 2 PbX < 1 l 2 Similarly for Y : P bY < 1 l 2 1 So P &[max (x, y)] < 0 2

nodia = P b X < 1 l P bY < 1 l 2 2 = shaded area = 3 4 =3 4 =3#3= 9 4 4 16

Alternate Method: From the given data since random variables X and Y lies in the interval [- 1, 1] as from the figure X , Y lies in the region of the square ABCD .

Probability for max 6X, Y @ < 1/2 : The points for max 6X, Y @ < 1/2 will be inside the region of square AEFG . So, P &max 6X, Y @ < 1 0 = Area of 4AEFG 2 Area of square ABCD 3 3 #2 2 = = 9 2#2 16 SOL 9.11





Chapter 9

In a coherent binary PSK system, the pair of signals s1 (t) and s2 (t) used to represent binary system 1 and 0 respectively. s1 (t) = 2E sin wc t T 2E sin w t c T where 0 # t # T , E is the transmitted energy per bit. General function of local oscillator 2 sin (w t), 0 # t < T f1 (t) = c T But here local oscillator is ahead with 45c. so, 2 sin (w t + 45c) f1 (t) = c T The coordinates of message points are s2 (t) =-

s11 = =

T

# s (t) f (t) dt 1

0

#

1

2E sin w t c T

T

2 sin (w t + 45c) dt c T

nodia 0

=

2E T

=

2E T

= 1 T

E

Similarly, Signal space diagram

T

# sin (w t) sin (w t + 45c) dt c

0

2 T

#

0

T

#

0

T

c

1 [sin 45c + sin (2w t + 45c)] dt c 2

1 dt + 1 E Tsin (2w t + 45c) dt #0 c T 2 1444444 42 4 44444 3 0 = E 2 s21 =-

E 2

Now here the two message points are s11 and s21 . The error at the receiver will be considered. When : (i) s11 is transmitted and s21 received (ii) s21 is transmitted and s11 received So, probability for the 1st case will be as : P b s21 received l = P (X < 0) (as shown in diagram) s11 transmitted = P _ E/2 + N < 0i = P _N < - E/2 i Taking the Gaussian distribution as shown below :




Page 471

Mean of the Gaussian distribution =

E/2 Variance = N 0 2 Putting it in the probability function : P bN < -

Taking,

E = 2l

#

0

=

#

0

1

`x + E/2 j

2

e-

2N 0 /2

2p N 0 2 2 1 e- `x + NE/2 j dx pN 0

-3

dx

0

-3

x + E/2 =t N 0 /2

nodia N 0 dt 2

dx =

So, P _N < - E/2 i =

#

E N0 m

1 e- t2 dt Q c 2p 2

3

E/N 0

where Q is error function. Since symbols are equiprobable in the 2 nd case So, P b s11 received l = Q c E m N0 s21 transmitted So the average probability of error = 1 ;P b s21 received l + P b s11 received lE 2 s11 transmitted s21 transmitted = 1 =Q c 2

E +Q c N0 m

E =Q c N 0 mG

SOL 9.12

Option ( ) is correct.

SOL 9.13

Option (B) is correct. General equation of FM and PM waves are given by fFM (t) = Ac cos ;wc t + 2pk f

E N0 m

t

# m (t) dtE 0

fPM (t) = Ac cos [wc t + k p m (t)] For same maximum phase deviation. t

k p [m (t)] max = 2pk f ; # m (t) dtE 0

max

k p # 2 = 2pk f [x (t)] max where,

x (t) =

t

# m (t) dt 0




[x (t)] max = 4 k p # 2 = 2p k f # 4 kp = 4p kf

So,

SOL 9.14

Chapter 9

Option (A) is correct. jw + a GC (s) = s + a = s+b jw + b Phase lead angle f = tan-1 a w k - tan-1 a w k a b Jw - wN b O = tan-1 w (b - a) f = tan-1 K a c ab + w 2 m 2 KK OO w 1+ ab P L For phase-lead compensation f > 0

nodia b-a > 0

b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true. SOL 9.15


f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 w dw 1 +a k 1 +awk a b 1 + w2 = 1 + 1 w2 a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l w = ab = 1 # 2 =

2 rad/ sec

SOL 9.16

Option (D) is correct. Quantized 4 level require 2 bit representation i.e. for one sample 2 bit are required. Since 2 sample per second are transmitted we require 4 bit to be transmitted per second.

SOL 9.17

Option (B) is correct. In FM the amplitude is constant and power is efficient transmitted. No variation in power. There is most bandwidth efficient transmission in SSB- SC. because we transmit only one side band. Simple Diode in Non linear region ( Square law ) is used in conventional AM that is simplest receiver structure.




Page 473

In VSB dc. component exists. SOL 9.18

Option (A) is correct. We have Sx (f) = F {Rx (t)} = F {exp (- pt2)} 2

= e- pf The given circuit can be simplified as

Power spectral density of output is Sy (f) = G (f) 2 Sx (f) = j2pf - 1 2 e- pf

2

= ( (2pf) 2 + 1) 2 e- pf Sy (f) = (4p2 f 2 + 1) e- pf

or SOL 9.19

2

2

Option (B) is correct. Highest frequency component in m (t) is fm = 4000p/2p = 2000 Hz Carrier frequency fC = 1 MHz For Envelope detector condition

nodia 1/fC << RC << 1/fm 1 μs << RC << 0.5 ms

SOL 9.20


Four phase signal constellation is shown below

Now

d2 = r 12 + r 12 d2 = 2r 12 r1 = d/ 2 = 0.707d

q = 2p = 2p = p 8 4 M Applying Cooine law we have




Chapter 9

d2 = r 22 + r 22 - 2r 22 cos p 4 = 2r 22 - 2r 22 1/ 2 = (2 d r2 = = 1.3065d 2- 2

or SOL 9.21

2 ) r 22

Option (D) is correct. Here Pe for 4 PSK and 8 PSK is same because Pe depends on d . Since Pe is same, d is same for 4 PSK and 8 PSK.

Additional Power SNR

nodia

= (SNR) 2 - (SNR) 1 = 10 log b ES2 l - 10 log b ES1 l No No E = 10 log b S2 l ES1 2 = 10 log a r2 k & 20 log a r2 k = 20 log 1.3065d r1 r1 0.707d

Additional SNR = 5.33 dB

SOL 9.22

SOL 9.23

SOL 9.24

Option (C) is correct. Conventional AM signal is given by x (t) = AC [1 + mm (t)] cos (2pfC t) Where m < 1, for no over modulation. In option (C) x (t) = AC :1 + 1 m (t)D cos (2pfC t) 4 Thus m = 1 < 1 and this is a conventional AM-signal without over-modulation 4 Option (B) is correct. (6) 2 Power P = = 18 W 2 Option (C) is correct. Impulse response of the matched filter is given by h (t) = S (T - t)



SOL 9.25


Page 475

Option (B) is correct. Let response of LPF filters 1, H (f ) = * 0,

f < 1 MHz elsewhere

Noise variance (power) is given as P = s2 =

#0

or SOL 9.26

#0

fo

2 H (f ) No df = 22 (given) a

2 # 10-20 df = 22 a -20 6 2 # 10 # 10 = 22 a 2 a = 1014 a = 107

1 # 106

nodia

Option (D) is correct. Probability of error is given by Pe = 1 [P (0/1) + P (1/0)] 2 P (0/1) =

a/2

#- 3 0.5e- a n - a dn = 0.5e-10

where a = 2 # 10-6 V and a = 107 V - 1 P (1/0) =

#a/32 0.5e- a n dn

= 0.5e-10

Pe = 0.5e-10 SOL 9.27

Option (C) is correct. S (t) = sin c (500t) sin c (700t) S (f ) is convolution of two signals whose spectrum covers f 1 = 250 Hz and f 2 = 350 Hz . So convolution extends f = 25 + 350 = 600 Hz Nyquist sampling rate N = 2f = 2 # 600 = 1200 Hz

SOL 9.28

Option (D) is correct. For the given system, output is written as y (t) = d [x (t) + x (t - 0.5)] dt dx (t) dx (t - 0.5) y (t) = + dt dt Taking Laplace on both sides of above equation Y (s) = sX (s) + se-0.5s X (s) Y (s) H (s) = = s (1 + e-0.5s) X (s)




Chapter 9

H (f ) = jf (1 + e-0.5 # 2pf ) = jf (1 + e- pf ) Power spectral density of output SY (f ) = H (f ) 2 SX (f ) = f 2 (1 + e- pf ) 2 SX (f ) For SY (f ) = 0 , 1 + e- pf = 0 f = (2n + 1) f0 or f0 = 1 KHz SOL 9.29

Option (C) is correct. cos (2pfm t) cos (2pfc t) $ DSB suppressed carrier cos (2pfc t) $ Carrier Only cos [2p (fc + fm) t] $ USB Only [1 + cos (2pfm t) cos (2pfc t)] $ USB with carrier

SOL 9.30

Option (C) is correct. We have p (X = 0) = p (Y = 0) = 1 2 p (X = 1) = p (Y = 1) = 1 4 1 p (X = 2) = p (Y = 2) = 4

Let and Now

nodia X+Y = 2 $ A X-Y = 0 $ B

P (X + Y = 2 X - Y = 0) =

P (A + B) P (B)

Event P (A + B) happen when X + Y = 2 and X - Y = 0 . It is only the case when X = 1 and Y = 1. Thus P (A + B) = 1 # 1 = 1 4 4 16 Now event P (B) happen when X - Y = 0 It occurs when X = Y , i.e. X = 0 and Y = 0 or X = 1 and Y = 1 or X = 2 and Y = 2 Thus P (B) = 1 # 1 + 1 # 1 + 1 # 1 = 6 2 2 4 4 4 4 16 1/16 P (A + B) Now = =1 6 P (B) 6/16 SOL 9.31

Option (B) is correct. The mean is X = Sxi pi (x) = 1 # 0.1 + 2 # 0.2 + 3 # 0.4 + 4 # 0.2 + 5 # 0.1 = 0.1 + 0.4 + 1.2 + 0.8 + 0.5 = 3.0 2

X = Sxi2 pi (x) = 1 # 0.1 + 4 # 0.2 + 9 # 0.4 + 16 # 0.2 + 25 # 0.1




Page 477

= 0.1 + 0.8 + 3.6 + 3.2 + 2.5 = 10.2 sx2

Variance

= X2 - ^X h2 = 10.2 - (3) 2 = 1.2

SOL 9.32

Option (C) is correct. m (t) = 1 cos w1 t - 1 sin w2 t 2 2 Modulation index

sAM (t) = [1 + m (t)] cos wc t m (t) max = Vc 2 2 m = `1j +`1j = 1 2 2 2

2 ` 2j h = m # 100% = 1 2 # 100% = 20% 2 m +2 ` 2j +2 1

SOL 9.33

2


C1 = B log2 `1 + S j N . B log2 ` S j N

nodia

As S >> 1 N

If we double the S ratio then N

C2 . B log2 ` 2S j N

. B log2 2 + B log2 S N . B + C1

SOL 9.34

Option (C) is correct. We have SNR = 1.76 + 6n or 43.5 = 1.76 + 6n 6n = 43.5 + 1.76 6n = 41.74 $ n . 7 No. of quantization level is 27 = 128 Step size required is = VH - VL = 128

5 - (- 5) = 10 128 128

= .078125 . .0667 SOL 9.35

Option (B) is correct. For positive values step size s+ = 0.05 V For negative value step size s- = 0.1 V No. of quantization in + ive is = 5 = 5 = 100 s+ 0.05




Chapter 9

2n + = 100 $ n+ = 7 No. of quantization in - ve Q1 = 5 = 5 = 50 s0.1 Thus

2n = 50 $ n - = 6 -

Thus

+ S ` N j+ = 1.76 + 6n = 1.76 + 42 = 43.76 dB S ` N j- = 1.76 + 6n = 1.76 + 36 = 37.76 dB

S ` N j0 = 43.76 dB

Best SOL 9.36

Option (A) is correct. We have xAM (t) = Ac cos wc + 2 cos wm t cos wc t = AC c1 + 2 cos wm t m cos wc t Ac For demodulation by envelope demodulator modulation index must be less than or equal to 1. 2 #1 Thus Ac

nodia Ac $ 2

Hence minimum value of Ac = 2 SOL 9.37

Option (A) is correct. CDF is the integration of PDF. Plot in option (A) is the integration of plot given in question.

SOL 9.38

Option (A) is correct. The entropy is

m

Since

H = / pi log2 1 bits pi i=1 p1 = p2 = ... = pn = 1 n n

H = / 1 log n = log n n i=1 SOL 9.39

Option (C) is correct. PSD of noise is

N0 = K 2

...(1)

The 3-dB cut off frequency is fc =

1 2pRC

...(2)

Output noise power is = N0 = c N0 m 1 = Kpfc 2 2RC 4RC SOL 9.40

Option (D) is correct. At receiving end if we get two zero or three zero then its error. Let p be the probability of 1 bit error, the probability that transmitted bit error is = Three zero + two zero and single one




Page 479

= 3 C3 p3 + 3C2 p2 (1 - p) = p3 + p2 (1 - p) SOL 9.41

Option (D) is correct. Bandwidth of TDM is = 1 (sum of Nyquist Rate) 2 1 = [2W + 2W + 4W + 6W] = 7W 2

SOL 9.42

Option (B) is correct. We have qi = 2p105 t + 5 sin (2p1500t) + 7.5 sin (2p1000t) wi = dqi = 2p105 + 10p1500 cos (2p1500t) + 15p1000 cos (2p1000t) dt Maximum frequency deviation is 3wmax = 2p (5 # 1500 + 7.5 # 1000) 3 fmax = 15000 3f Modulation index is = max = 15000 = 10 fm 1500

nodia

SOL 9.43


SOL 9.44


fm = 4 KHz

fs = 2fm = 8 kHz Bit Rate Rb = nfs = 8 # 8 = 64 kbps The minimum transmission bandwidth is BW = Rb = 32 kHz 2 SOL 9.45

Option (C) is correct. S0 c N m = 1.76 + 6n dB 0 = 1.76 + 6 # 8 = 49.76 dB

SOL 9.46

We have n = 8

Option (B) is correct. Noise \ 12 L To reduce quantization noise by factor 4, quantization level must be two times i.e. 2L . Now L = 2n = 28 = 256 As

Thus

2L = 512

SOL 9.47

Option (C) is correct. Autocorrelation is even function.

SOL 9.48

Option (B) is correct. Power spectral density is non negative. Thus it is always zero or greater than zero.

SOL 9.49

Option (A) is correct. The variance of a random variable x is given by E [X2] - E2 [X]




SOL 9.50

Option (A) is correct. A Hilbert transformer is a non-linear system.

SOL 9.51

Option (D) is correct. Slope overload distortion can be reduced by increasing the step size 3 $ slope of x (t) Ts

SOL 9.52


p (t) =

Chapter 9

sin (4pWt) 4pWt (1 - 16W2 t2)

at t = 1 it is 0 form. Thus applying L' Hospital rule 0 4W 4pW cos (4pWt) p( ) = 4pW [1 - 48W2 t2] 1 4W

= SOL 9.53

cos (4pWt) = cos p = 0.5 1-3 1 - 48W2 t2

Option (B) is correct. The block diagram is as shown below

nodia

Here

M1 (f) = Mt (f) Y1 (f) = M (f) c e

j 2p B

Y2 (f) = M1 (f) c e

j 2p B

+ e -j2pB m 2 - e -j2pB m 2

Y (f) = Y1 (f) + Y2 (f) All waveform is shown below



SOL 9.54


Page 481

Option (C) is correct. By Binomial distribution the probability of error is pe = n Cr pr (1 - p) n - r Probability of at most one error = Probability of no error + Probability of one error = n C0 p0 (1 - p) n - 0 + n C1 p1 (1 - p) n - 1 = n (1 - p) n + np (1 - p) n - 1

SOL 9.55

Option (B) is correct. Bandwidth allocated for 1 Channel = 5 M Hz Average bandwidth for 1 Channel 5 = 1 MHz 5 Total Number of Simultaneously Channel = 1M # 8 = 40 Channel 200k

SOL 9.56

Option (A) is correct. Chip Rate RC = 1.2288 # 106 chips/sec Data Rate Rb = RC G

nodia

Since the processing gain G must be at least 100, thus for Gmin we get 6 Rb max = RC = 1.2288 # 10 = 12.288 # 103 bps 100 Gmin SOL 9.57

Option (B) is correct. Energy of constellation 1 is Eg1 2

2 a) 2 + ( 2 a) 2 + (- 2 2 a) 2 = 2a2 + 2a2 + 2a2 + 8a2 = 16a2 Energy of constellation 2 is = (0) + (-

2

2 a) + (-

Eg2 = a2 + a2 + a2 + a2 = 4a2 2 E Ratio = g1 = 16a2 = 4 Eg2 4a SOL 9.58

Option (A) is correct. Noise Power is same for both which is N0 . 2 Thus probability of error will be lower for the constellation 1 as it has higher signal energy.

SOL 9.59

Option (A) is correct. Area under the pdf curve must be unity 2a + 4a + 4b = 1 2a + 8b = 1 For maximum entropy three region must by equivaprobable thus

Thus

2a = 4b = 4b

...(1) ...(2)

From (1) and (2) we get




Chapter 9

b = 1 and a = 1 12 6 SOL 9.60

Option (*) is correct.

SOL 9.61

Option (B) is correct. A LPF will not produce phase distortion if phase varies linearly with frequency. f (w) \ w f (w) = kw

i.e. SOL 9.62

Option (B) is correct. Let m (t) is a low pass signal, whose frequency spectra is shown below

nodia

Fourier transform of g (t)

3

1 / d (f - 20 # 103 k) 0.5 # 10-4 k =- 3 Spectrum of G (f ) is shown below G (t) =

Now when m (t) is sampled with above signal the spectrum of sampled signal will look like.

When sampled signal is passed through a LP filter of BW 1 kHz, only m (t) will remain. SOL 9.63

Option (C) is correct. The highest frequency signal in x (t) is 1000 # 3 = 3 kHz if expression is expanded. Thus minimum frequency requirement is f = 2 # 3 # 103 = 6 # 103 Hz

SOL 9.64

Option (B) is correct. We have x (t) = 125t [u (t) - u (t - 1)] + (250 - 125t) [u (t - 1) - u (t - 2)]




Page 483

The slope of expression x (t) is 125 and sampling frequency fs is 32 # 1000 samples/sec. Let 3 be the step size, then to avoid slope overload 3 $ slope x (t) Ts 3 fc $ slope x (t) 3# 32000 $ 125 3 $ 125 32000 3 = 2- 8 SOL 9.65

Option (A) is correct. The sampling frequency is fs =

1 = 33 kHz 0.03m

Since fs $ 2fm , the signal can be recovered and are correlated. SOL 9.66

Option (B) is correct. We have p1 = 0.25 , p2 = 0.25 and p3 = 0.5

nodia 3

H = / p1 log2 1 bits/symbol p1 i=1 = p1 log2 1 + p2 log2 1 + p3 log2 1 p1 p2 p3 = 0.25 log2 1 + 0.25 log2 1 + 0.5 log2 1 0.25 0.25 0.5 = 0.25 log2 4 + 0.25 log2 4 + 0.5 log2 2 = 0.5 + 0.5 + 1 = 3 bits/symbol 2 2

Average bit rate

SOL 9.67

Rb = 3000 symbol/sec = Rb H = 3 # 3000 = 4500 bits/sec 2

Option (A) is correct. The diagonal clipping in AM using envelop detector can be avoided if 1 << RC < 1 wc W 1 $ Wm sin Wt But from 1 + m cos Wt RC We can say that RC depends on W , thus RC < 1 W

SOL 9.68


SOL 9.69

Option (B) is correct. When 3 /2 is added to y (t) then signal will move to next quantization level. Otherwise if they have step size less than 3 then they will be on the same 2 quantization level.

SOL 9.70

Option (C) is correct. After the SSB modulation the frequency of signal will be fc - fm i.e.




Chapter 9

1000 - 10 kHz . 1000 kHz The bandwidth of FM is BW = 2 (b + 1) 3 f For NBFMb << 1, thus BWNBFM . 2 3 f = 2 (109 - 106) . 2 # 109 SOL 9.71

Option (A) is correct. We have p (t) = u (t) - u (t - 1) g (t) = p (t)* p (t) s (t) = g (t) - d (t - 2)* g (t) = g (t) - g (t - 2) All signal are shown in figure below :

nodia

The impulse response of matched filter is

h (t) = s (T - t) = s (1 - t) Here T is the time where output SNR is maximum. SOL 9.72

Option (A) is correct. We have xAM (t) = 10 [P (t) + 0.5g (t)] cos wc t where p (t) = u (t) - u (t - 1) and g (t) = r (t) - 2r (t - 1) + r (t - 2) For desired interval 0 # t # 1, p (t) = 1 and g (t) = t , Thus we have, xAM (t) = 100 (1 - 0.5t) cos wc t Hence modulation index is 0.5

SOL 9.73

Option (A) is correct. We know that SYY (w) = H (w) 2 .SXX (w) Now SYY (w) = 16 2 and SXX (w) = 1 white noise 16 + w Thus or or

16 = H (w) 2 16 + w2 4 H (w) = 16 + w2 H (s) = 4 4+s

which is a first order low pass RL filter. SOL 9.74


R = 4 R + sL 4+s R L

= 4 4+s +s Comparing we get L = 1 H and R = 4W or

SOL 9.75

R L





Page 485

We have xAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t The modulation index is 0.5

SOL 9.76

Carrier power

Pc =

(10) 2 = 50 2

Side band power

Ps =

(10) 2 = 50 2

Side band power

2 (0.5) 2 (50) Ps = m Pc = = 6.25 2 2

Option (B) is correct. Mean noise power = Area under the PSD curve = 4 ; 1 # B # No E = BNo 2 2 The ratio of average sideband power to mean noise power is Side Band Power = 6.25 = 25 Noise Power N0 B 4No B

SOL 9.77


nodia {1 + km (t)} A sin (wc t) $ Amplitude modulation dm (t) Asin (wc t) $ DSB-SC modulation A sin {cos t + km (t)} $ Phase Modulation A sin [wct + k] t- 3 m (t) dt $ Frequency Modulation

SOL 9.78


VSB $ fm + fc DSB - SC $ 2fm SSB $ fm

AM $ 2fm Thus SSB has minimum bandwidth and it require minimum power. SOL 9.79

Option (A) is correct. Let x (t) be the input signal where x (t) = cos (cos t + b1 cos wm t) cos (2wc t + 2b1 cos wm t) y (t) = x2 (t) = 1 + 2 2 3f Here b = 2b1 and b1 = = 90 = 18 fm 5 BW = 2 (b + 1) fm = 2 (2 # 18 + 1) # 5 = 370 kHz

SOL 9.80

Option (C) is correct. The transfer function of matched filter is h (t) = x (t - t) = x (2 - t) The output of matched filter is the convolution of x (t) and h (t) as shown below



SOL 9.81


Option (B) is correct. H (f) = 2e - jwt H (f) = 2 G0 (f) = H (f) 2 Gi (f) = 4No W/Hz

We have

d

= 4No # B

The noise power is SOL 9.82

Chapter 9

Option (C) is correct. As the area under pdf curve must be unity 1 (4 # k) = 1 $ k = 1 2 2

nodia

Now mean square value is

sv2 =

+3

#- 3

v2 p (v) dv

4

v2 ` v j dv 8 3 4 = # c v m dv = 8 8 0 =

SOL 9.83

#0

as p (v) = 1 v 8

Option (D) is correct. The phase deviation is b =

3f = 10 = 10 fm 1

If phase deviation remain same and modulating frequency is changed BW = 2 (b + 1) fm' = 2 (10 + 1) 2 = 44 kHz SOL 9.84

Option (B) is correct. As the area under pdf curve must be unity and all three region are equivaprobable. Thus are under each region must be 13 . 2a # 1 = 1 $ a = 2 4 3 3

SOL 9.85

Option (A) is correct. Nq =

+a

#- a

x2 p (x) dx = 2

#0

3 a 3 x $ 1 dx = 1 ; x E = a 4 2 3 0 6

a 2

Substituting a = 2 we have 3 Nq = 4 81 SOL 9.86

Option (C) is correct. When word length is 6 2 #6 12 S ` N jN = 6 = 2 = 2




Page 487

When word length is 8 2 #8 16 S ` N jN = 8 = 2 = 2

16 ^ N hN = 8 = 212 = 2 4 = 16 S ^ N hN = 6 2 S

Now

Thus it improves by a factor of 16. SOL 9.87

Option (B) is correct. Carrier frequency Modulating frequency

fc = 1 # 106 Hz fm = 2 # 103 Hz

For an envelope detector 2pfc > 1 > 2pfm Rc 1 < RC < 1 2pfc 2pfm 1 < RC < 1 2pfc 2pfm 1 < RC < 1 6 2p10 2 # 103 1.59 # 10 - 7 < RC < 7.96 # 10 - 5 so, 20 msec sec best lies in this interval. SOL 9.88

nodia


SAM (t) = Ac [1 + 0.1 cos wm t] cos wm t sNBFM (t) = Ac cos [wc t + 0.1 sin wm t]

s (t) = SAM (t) + SNB fm (t) = Ac [1 + 0.1 cos wm t] cos wc t + Ac cos (wc t + 0.1 sin wm t) = Ac cos wc t + Ac 0.1 cos wm t cos wc t + Ac cos wc t cos (0.1 sin wm t) - Ac sin wc t. sin (0.1 sin wm t) As 0.1 sin wm t ,+ 0.1 to - 0.1 so, cos (0.1 sin wm t) . 1 As when q is small cos q . 1 and sin q , q, thus sin (0.1 sin wm t) = 0.1 sincos wc t cos wm t + Ac cos wc t - Ac 0.1 sin wm t sin wc t = 2Ac cos wc t + 0.1Ac cos (wc + wm) t 1 44 2 44 3 1 4 4 4 4 424444 43 cosec USB

Thus it is SSB with carrier. SOL 9.89

Option (A) is correct. Consecutive pulses are of same polarity when modulator is in slope overload. Consecutive pulses are of opposite polarity when the input is constant.

SOL 9.90

Option (D) is correct. F (x1 # X < x2) = p (X = x2) - P (X = x1) or

P (X = 1) = P (X = 1+) - P (X = 1 -)




Chapter 9

= 0.55 - 0.25 = 0.30 SOL 9.91

Option (A) is correct. The SNR at transmitter is SNRtr = Ptr NB 10 - 3 = 109 10 - 20 # 100 # 106 In dB SNRtr = 10 log 109 = 90 dB Cable Loss = 40 db At receiver after cable loss we have SNRRc = 90 - 40 = 50 dB

SOL 9.92

Option (B) is correct. The impulse response of matched filter is h (t) = x (T - t) Since here T = 4 , thus

nodia

h (t) = x (4 - t) The graph of h (t) is as shown below.

From graph it may be easily seen that slope between 3 < t < 4 is - 1. SOL 9.93

Option (C) is correct. The required bandwidth of M array PSK is BW = 2Rb n where 2n = M and Rb is bit rate For BPSK, M = 2 = 2n $ n = 1 Thus B1 = 2Rb = 2 # 10 = 20 kHz 1 For QPSK, Thus

SOL 9.94

M = 4 = 2n $ n = 2 B2 = 2Rb = 10 kHz 2


fc = 100 MHz = 100 # 106 and fm = 1 MHz = 1 # 106 The output of balanced modulator is VBM (t) = [cos wc t][ cos wc t] = 1 [cos (wc + wm) t + cos (wc - wm) t] 2 If VBM (t) is passed through HPF of cut off frequency fH = 100 # 106 , then only (wc + wm) passes and output of HPF is




Page 489

VHP (t) = 1 cos (wc + wm) t 2 V0 (t) = VHP (t) + sin (2p # 100 # 106) t = 1 cos [2p100 # 106 + 2p # 1 # 106 t] + sin (2p # 100 # 106) t 2 = 1 cos [2p108 + 2p106 t] + sin (2p108) t 2 = 1 [cos (2p108 t) t cos (2p106 t)] - sin [2p108 t sin (2p106 t) + sin 2p108 t] 2 = 1 cos (2p106 t) cos 2p108 t + `1 - 1 sin 2p106 t j sin 2p108 t 2 2 This signal is in form Now

= A cos 2p108 t + B sin 2p108 t The envelope of this signal is = =

nodia =

=

SOL 9.95

A2 + B2 2 6 6 2 1 1 ` 2 cos (2p10 t)j + `1 - 2 sin (2p10 t j 1 cos2 (2p106 t) + 1 + 1 sin2 (2p106 t) - sin (2p106 t) 4 4 1 + 1 - sin (2p106 t) = 5 - sin (2p106 t) 4 4


s (t) = A cos [2p10 # 103 t] + A cos [2p10.1 # 103 t] 1 Here T1 = = 100m sec 10 # 103 1 and = 99m sec T2 = 10.1 # 103 Period of added signal will be LCM [T1, T2] Thus T = LCM [100, 99] = 9900m sec Thus frequency f = 1 = 0.1 kHz 9900m SOL 9.96

Option (A) is correct. The pdf of transmission of 0 and 1 will be as shown below :

Probability of error of 1 P (0 # X # 0.2) = 0.2 Probability of error of 0 : P (0.2 # X # 0.25) = 0.05 # 2 = 0.1 P (0 # X # 0.2) + P (0.2 # X # 0.25) 2 = 0.2 + 0.1 = 0.15 0

Average error =



SOL 9.97


Chapter 9

Option (B) is correct. The square mean value is s2 =

#- 3 (x - xq) 2 f (x) dx

=

#0 (x - xq) 2 f (x) dx

=

#0

3

1

0. 3

(x - 0) 2 f (x) dx +

0. 1

#0.3 (x - 0.7) 2 f (x) dx

3 0. 3 3 2 1 = ; x E + ; x + 0.49x - 14 x E 3 0 3 2 0. 3

s2 = 0.039

or

RMS = SOL 9.98

s2 =

0.039 = 0.198

Option (C) is correct. FM $ Capture effect DM $ Slope over load PSK $ Matched filter PCM $ m - law

SOL 9.99

nodia

Option (C) is correct. Since fs = 2fm , the signal frequency and sampling frequency are as follows fm1 = 1200 Hz $ 2400 samples per sec fm2 = 600 Hz $ 1200 samples per sec fm3 = 600 Hz $ 1200 samples per sec Thus by time division multiplexing total 4800 samples per second will be sent. Since each sample require 12 bit, total 4800 # 12 bits per second will be sent Thus bit rate Rb = 4800 # 12 = 57.6 kbps

SOL 9.100

Option (B) is correct. The input signal X (f) has the peak at 1 kHz and - 1 kHz. After balanced modulator the output will have peak at fc ! 1 kHz i.e. : 10 ! 1 $ 11 and 9 kHz 10 ! (- 1) $ 9 and 11 kHz 9 kHz will be filtered out by HPF of 10 kHz. Thus 11 kHz will remain. After passing through 13 kHz balanced modulator signal will have 13 ! 11 kHz signal i.e. 2 and 24 kHz. Thus peak of Y (f) are at 2 kHz and 24 kHz.

SOL 9.101

Option (A) is correct. The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component nq (t) of the noise n (t) is completely rejected by the detector.

SOL 9.102

Option (C) is correct. The noise at the input to an ideal frequency detector is white. The PSD of noise at the output is parabolic

SOL 9.103


Pe = 1 erfc c 2

Ed 2h m




Page 491

Since Pe of Binary FSK is 3 dB inferior to binary PSK SOL 9.104

Option (D) is correct. The pdf of Z will be convolution of pdf of X and pdf of Y as shown below. p [Z # z] =

Now

p [Z #- 2] =

z

#- 3 fZ (z) dz -2

#- 3fZ (z) dz

= Area [z #- 2] = 1 # 1 #1 = 1 2 6 12

nodia SOL 9.105


RXX (t) = 4 (e - 0.2 t + 1) RXX (0) = 4 (e - 0.2 0 + 1) = 8 = s2 s =2 2 m =0

or mean Now

P (x # 1) = Fx (1) X-m s m = 1 - Qc 1 - 0 m = 1 - Qc 1 m 2 2 2 2

= 1 - Qc

SOL 9.106

Given

at x = 1

Option (C) is correct. W = Y-Z E [W2] = E [Y - Z] 2 = E [Y2] + E [Z2] - 2E [YZ] = sw2 We have

E [X2 (t)] = Rx (10) = 4 [e - 0.2 0 + 1] = 4 [1 + 1] = 8 E [Y2] = E [X2 (2)] = 8




Chapter 9

E [Z2] = E [X2 (4)] = 8 E [YZ] = RXX (2) = 4 [e-0.2 (4 - 2) + 1] = 6.68 E [W2] = sw2 = 8 + 8 - 2 # 6.68 = 2.64 SOL 9.107

Option (C) is correct. Step size d =

2mp = 1.536 = 0.012 V L 128

2 (0.012) 2 Quantization Noise power = d = 12 12

= 12 # 10-6 V2 SOL 9.108

Option (D) is correct. The frequency of pulse train is f 1- 3 = 1 k Hz 10 The Fourier Series coefficient of given pulse train is -T /2 Ae-jnw t dt Cn = 1 # To -T /2 o

o

nodia o

= 1 To

#

-To /6

-To /6

Ae-jhw t dt o

A [e-jw t] --TT //66 To (- jhwo) A (e-jw t - e jhw T /6) = (- j2pn) = A (e jhp/3 - e-jhp/3) j 2p n or Cn = A sin ` np j pn 3 From Cn it may be easily seen that 1, 2, 4, 5, 7, harmonics are present and 0, 3, 6, 9,.. are absent. Thus p (t) has 1 kHz, 2 kHz, 4 kHz, 5 kHz, 7 kHz,... frequency component and 3 kHz, 6 kHz.. are absent. The signal x (t) has the frequency components 0.4 kHz and 0.7 kHz. The sampled signal of x (t) i.e. x (t)* p (t) will have =

o

o o

o

o

o

1 ! 0.4 and 1 ! 0.7 kHz 2 ! 0.4 and 2 ! 0.7 kHz 4 ! 0.4 and 4 ! 0.7 kHz Thus in range of 2.5 kHz to 3.5 kHz the frequency present is 2 + 0.7 = 2.7 kHz 4 - 0.7 = 3.3 kHz SOL 9.109

Option (C) is correct. vi = Ac1 cos (2pfc t) + m (t) v0 = ao vi + avi3 v0 = a0 [Ac' cos (2pfc' t) + m (t)] + a1 [Ac' cos (2pfc' t) + m (t)] 3 = a0 Ac' cos (2pfc' t) + a0 m (t) + a1 [(Ac' cos 2pfc' t) 3 + (Ac' cos (2pfc') t) 2 m (t) + 3Ac' cos (2pfc' t) m2 (t) + m3 (t)] = a0 Ac' cos (2pfc' t) + a0 m (t) + a1 (Ac' cos 2fc' t) 3 1 + cos (4pfc' t) + 3a1 Ac'2 ; Em (t) 2




Page 493

= 3a1 Ac' cos (2pfc' t) m2 (t) + m3 (t) The term 3a1 Ac' ( cos 42pf t ) m (t) is a DSB-SC signal having carrier frequency 1. MHz. Thus 2fc' = 1 MHz or fc' = 0.5 MHz ' c

SOL 9.110

Option (D) is correct. 2 PT = Pc c1 + a m 2

or SOL 9.111

Option (D) is correct. AM Band width = 2fm

2 P (0.5) 2 Psb = Pc a = c 2 2 Psb = 1 Pc 8

Peak frequency deviation = 3 (2fm) = 6fm 6f Modulation index b = m = 6 fm The FM signal is represented in terms of Bessel function as

nodia xFM (t) = Ac

3

/ Jn (b) cos (wc - nwn) t

n =- 3

wc + nwm = 2p (1008 # 103) 2p106 + n4p # 103 = 2p (1008 # 103), n = 4 Thus coefficient = 5J4 (6) SOL 9.112

Option (B) is correct. Ring modulation $ Generation of DSB - SC VCO $ Generation of FM Foster seely discriminator $ Demodulation of fm mixer $ frequency conversion

SOL 9.113

Option (A) is correct. fmax = 1650 + 450 = 2100 kHz fmin = 550 + 450 = 1000 kHz 1 or f = 2p LC frequency is minimum, capacitance will be maximum

or

f2 R = Cmax = max = (2.1) 2 2 Cmin fmin R = 4.41 fi = fc + 2fIF = 700 + 2 (455) = 1600 kHz

SOL 9.114

Option (D) is correct. Eb = 10 - 6 watt-sec No = 10 - 5 W/Hz o (SNR) matched filler = E = N

106 = .05 2 # 10 - 5 2 (SNR)dB = 10 log 10 (0.05) = 13 dB o

SOL 9.115





Chapter 9

3 fs 2pfm This is satisfied with Em = 1.5 V and fm = 4 kHz For slopeoverload to take place Em $

SOL 9.116

Option (A) is correct. If s " carrier synchronization at receiver r " represents bandwidth efficiency then for coherent binary PSK r = 0.5 and s is required.

SOL 9.117

Option (B) is correct. Bit Rate = 8k # 8 = 64 kbps (SNR)q = 1.76 + 6.02n dB = 1.76 + 6.02 # 8 = 49.8 dB

SOL 9.118

Option (C) is correct. The frequency of message signal is fc = 1000 kHz 1 The frequency of message signal is 1 fm = = 10 kHz 100 # 10 - 6 Here message signal is symmetrical square wave whose FS has only odd harmonics i.e. 10 kHz, 30 kHz 50 kHz. Modulated signal contain fc ! fm

nodia

frequency component. Thus modulated signal has

fc ! fm = (1000 ! 10) kH = 1010 kHz, 990 kHz fc ! 3fm = (1000 ! 10) kH = 1030 kHz, 970 kHz Thus, there is no 1020 kHz component in modulated signal. SOL 9.119


y (t) = 5 # 10 - 6 x (t)

+3

/ d (t - nTs) n =- 3

x (t) = 10 cos (8p # 103) t Ts = 100m sec The cut off fc of LPF is 5 kHz We know that for the output of filter x (t) y (t) = Ts 10 cos (8p # 103) t # 5 # 10 - 6 100 # 10 - 6 = 5 # 10 - 1 cos (8p # 103) t =

SOL 9.120

Option (C) is correct. Transmitted frequencies in coherent BFSK should be integral of bit rate 8 kHz.

SOL 9.121

Option (B) is correct. For best reception, if transmitting waves are vertically polarized, then receiver should also be vertically polarized i.e. transmitter and receiver must be in same polarization.

SOL 9.122

Option (D) is correct. s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t)




Page 495

= cos {4p106 t + 100p sin (150t + q)} Angle modulated signal is s (t) = A cos {wc t + b sin (wm t + q)} Comparing with angle modulated signal we get Phase deviations b = 100p Frequency deviations 3 f = bfm = 100p # 150 = 7.5 kHz 2p SOL 9.123

Option (*) is correct. We have m (t) s (t) = y1 (t) 2 sin (2pt) cos (200pt) = t sin (202pt) - sin (198pt) = t y1 (t) + n (t) = y2 (t) = sin 202pt - sin 198pt + sin 199pt t t

nodia y2 (t) s (t) = u (t) =

[sin 202pt - sin 198pt + sin 199pt] cos 200pt t

= 1 [sin (402pt) + sin (2pt) - {sin (398pt) - sin (2pt)} + sin (399pt) - sin (pt)] 2 sin (2pt) + sin (2pt) - sin (pt) 2t sin (2pt) + 2 sin (0.5t) cos (1.5pt) = 2t = sin 2pt + sin 0.5pt cos 1.5pt 2t t

y (t) =

After filtering

SOL 9.124

Option (B) is correct. The signal frequency is 3 fm = 24p10 = 12 kHz 2p Ts = 50m sec " fs = 1 = 1 # 106 = 20 kHz Ts 50

After sampling signal will have fs ! fm frequency component i.e. 32 and 12 kHz At filter output only 8 kHz will be present as cutoff frequency is 15 kHz. SOL 9.125

Option (A) is correct. d (n) = x (n) - x (n - 1) E [d (n)] 2 = E [x (n) - x (n - 1)] 2 E [d (n)] 2 = E [x (n)] 2 + E [x (n - 1)] 2 - 2E [x (n) x (n - 1)] or as k = 1 sd2 = sx2 + sx2 - 2Rxx (1) 2 s 2 x As we have been given sd = , therefore 10 or

or

sx2 = s2 + s2 - 2R (1) x x xx 10 2Rxx (1) = 19 sx2 10




Chapter 9

Rxx = 19 = 0.95 20 sx2

or SOL 9.126

Option (A) is correct. An ideal low - pass filter with appropriate bandwidth fm is used to recover the signal which is sampled at nyquist rate 2fm .

SOL 9.127

Option (A) is correct. For any PDF the probability at mean is 1 . Here given PDF is Gaussian random 2 variable and X = 4 is mean.

SOL 9.128

Option (C) is correct. We require 6 bit for 64 intensity levels because 64 = 26 Data Rate = Frames per second # pixels per frame # bits per pixel = 625 # 400 # 400 # 6 = 600 Mbps sec

SOL 9.129

Option (C) is correct. We have sin c (700t) + sin c (500t) =

sin (700pt) sin (500pt) + 700pt 500pt

nodia

Here the maximum frequency component is 2pfm = 700p i.e. fm = 350 Hz Thus Nyquist rate fs = 2fm = 2 (350) = 700 Hz Thus sampling interval = 1 sec 700 SOL 9.130

Option (D) is correct. Probability of error = p Probability of no error = q = (1 - p) Probability for at most one bit error

= Probability of no bit error + probability of 1 bit error = (1 - p) n + np (1 - p) n - 1 SOL 9.131

Option (A) is correct. FT

If g (t) then PSD of g (t) is

G (w) Sg (w) = G (w) 2

and power is Pg = 1 2p Now PSD of ag (t) is

ag (t)

#- 3Sg (w) dw

FT

3

aG (w)

Sag (w) = a (G (w)) 2 or Similarly SOL 9.132

= a2 G (w) 2 Sag (w) = a2 Sg (w) Pag = a2 Pg

Option (C) is correct. The envelope of the input signal is [1 + ka m (t)] that will be output of envelope




Page 497

detector. SOL 9.133

Option (D) is correct. Frequency Range for satellite communication is 1 GHz to 30 GHz,

SOL 9.134

Option (B) is correct. Waveform will be orthogonal when each bit contains integer number of cycles of carrier. Bit rate Rb = HCF (f1, f2) = HCF (10k, 25k) = 5 kHz Tb = 1 = 1 = 0.2 msec = 200 msec Rb 5k

Thus bit interval is SOL 9.135

Option (D) is correct. We have The input to LPF is

Pm = m2 (t) x (t) = m (t) cos wo t cos (wo t + q)

nodia m (t) [cos (2wo t + q) + cos q] 2 m (t) cos (2wo t + q) m (t) cos q = + 2 2

=

The output of filter will be

y (t) =

m (t) cos q 2

Power of output signal is

2 Py = y2 (t) = 1 m2 (t) cos2 q = Pm cos q 4 4

SOL 9.136

Option (A) is correct. Hilbert transformer always adds - 90c to the positive frequency component and 90c to the negative frequency component. Hilbert Trans form cos wt " sin wt sin wt " cos wt Thus

SOL 9.137

cos w1 t + sin w2 t " sin w1 t - cos w2 t


x (t) = Ac cos {wc t + b sin wm t} y (t) = {x (t)} 3

= Ac2 cos (3wc t + 3b sin wm t) + 3 cos (wc t + b sin wm t) Thus the fundamental frequency doesn’t change but BW is three times. BW = 2 (3 f') = 2 (3 f # 3) = 3 MHz SOL 9.138


SOL 9.139

Option (C) is correct. This is Quadrature modulated signal. In QAM, two signals having bandwidth. B 1 & B 2 can be transmitted simultaneous over a bandwidth of (B 1 + B 2) Hz




B.W. = (15 + 10) = 25 kHz

so SOL 9.140

Chapter 9

Option (B) is correct. A modulated signal can be expressed in terms of its in-phase and quadrature component as S (t) = S1 (t) cos (2pfc t) - SQ (t) sin (2pfc t) S (t) = [e-at cpsDwt cos wc t - eat sin Dwt sin wc t] m (t) = [e-at cos Dwt] cos 2pfc t - [e-at sin Dwt] sin 2pfc t = S1 (t) cos 2pfc t - SQ (t) sin 2pfc t

Here

Complex envelope of s (t) is S (t) = S1 (t) + jSQ (t) = e-at cos Dwt + je-at sin Dwt = e-at [cos Dwt + j sin Dwt] = exp (- at) exp (jDwt) m (t) SOL 9.141

Option (B) is correct. Given function Let

nodia

g (t) = 6 # 10 4 sin c2 (400t) ) 106 sin c3 (100t) g1 (t) = 6 # 10 4 sin c2 (400t)

g2 (t) = (106) sin c3 (100t) We know that g1 (t) ) g2 (t) ? G1 (w) G2 (w) occupies minimum of Bandwidth of G1 (w) or G2 (w) Band width of G1 (w) = 2 # 400 = 800 rad/ sec or = 400 Hz Band width of G2 (w) = 3 # 100 = 300 rad/ sec or 150 Hz Sampling frequency = 2 # 150 = 300 Hz SOL 9.142

Option (B) is correct. For a sinusoidal input SNR (dB) is PCM is obtained by following formulae. SNR (dB) = 1.8 + 6n n is no. of bits Here n =8 So, SNR (dB) = 1.8 + 6 # 8 = 49.8

SOL 9.143

Option (D) is correct. We know that matched filter output is given by

# g (l) g (T - t + l) dl at t = T = # g (l) g (l) dl = # g (t) dt

g 0 (t) =

6g 0 (t)@max

=

3

-3

0

0

3

3

-3

-3

#

1 # 10-4

0

2

[10 sin (2p # 106) 2] dt

[g 0 (t)] max = 1 # 100 # 10-4 = 5 mV 2 SOL 9.144

Option (B) is correct. Sampling rate must be equal to twice of maximum frequency. f s = 2 # 400 = 800 Hz

SOL 9.145

Option (C) is correct. The amplitude spectrum of a gaussian pulse is also gaussian as shown in the fig. -y 2 fY (y) = 1 exp c 2 m 2p



SOL 9.146


Page 499

Option (C) is correct. Let the rectangular pulse is given as

nodia

Auto correlation function is given by T/2 Rxx (t) = 1 # x (t) x (t - t) dt T -T/2 When x (t) is shifted to right (t > 0), x (t - t) will be shown as dotted line.

Rxx (t) = 1 T

#

T +t 2

A2 dt

T - +t 2

2 2 = A :T + T - tD = A :T - tD 2 T 2 T 2 (t) can be negative or positive, so generalizing above equations 2 Rxx (t) = A :T - t D T 2

Rxx (t) is a regular pulse of duration T .




Chapter 9

SOL 9.147

Option (B) is correct. Selectivity refers to select a desired frequency while rejecting all others. In super heterodyne receiver selective is obtained partially by RF amplifier and mainly by IF amplifier.

SOL 9.148

Option (C) is correct. In PCM, SNR a 22n so if bit increased from 8 to 9 2#8 (SNR) 1 = 22 # 9 = 22 = 1 4 (SNR) 2 2 so SNR will increased by a factor of 4

SOL 9.149

Option (A) is correct. In flat top sampling an amplitude distortion is produced while reconstructing original signal x (t) from sampled signal s (t). High frequency of x (t) are mostly attenuated. This effect is known as aperture effect.

SOL 9.150

Option (A) is correct. Carrier C (t) = cos (we t + q) Modulating signal = x (t) DSB - SC modulated signal = x (t) c (t) = x (t) cos (we t + q) envelope = x (t)

SOL 9.151

Option (D) is correct. In Quadrature multiplexing two baseband signals can transmitted or modulated using I 4 phase & Quadrature carriers and its quite different form FDM & TDM.

SOL 9.152

Option (A) is correct. Fourier transform perform a conversion from time domain to frequency domain for analysis purposes. Units remain same.

SOL 9.153

Option (A) is correct. In PCM, SNR is depends an step size (i.e. signal amplitude) SNR can be improved by using smaller steps for smaller amplitude. This is obtained by compressing the signal.

SOL 9.154

Option (C) is correct. Band width is same for BPSK and APSK(OOK) which is equal to twice of signal Bandwidth.

SOL 9.155

Option (A) is correct. The spectral density of a real value random process symmetric about vertical axis so it has an even symmetry.

SOL 9.156


SOL 9.157

Option (C) is correct. It is one of the advantage of bipolar signalling (AMI) that its spectrum has a dc null for binary data transmission PSD of bipolar signalling is

nodia



SOL 9.158


Page 501

Option (A) is correct. Probability Density function (PDF) of a random variable x defined as Px (x) = 1 e-x /2 2p 1 so here K = 2p 2

SOL 9.159

Option (C) is correct. Here the highest frequency component in the spectrum is 1.5 kHz [at 2 kHz is not included in the spectrum] Minimum sampling freq. = 1.5 # 2 = 3 kHz

SOL 9.160

Option (B) is correct. We need a high pass filter for receiving the pulses.

SOL 9.161

Option (D) is correct. Power spectral density function of a signal g (t) is fourier transform of its auto correlation function

nodia

Sg (w) Rg (t) here Sg (w) = sin c2 (f) so Rg (t) is a triangular pulse. f [triang.] = sin c2 (f) F

SOL 9.162

Option (C) is correct. For a signal g (t), its matched filter response given as h (t) = g (T - t) so here g (t) is a rectangular pulse of duration T .

output of matched filter y (t) = g (t) ) h (t)




Chapter 9

if we shift g (- t) for convolution y (t) increases first linearly then decreases to zero.

SOL 9.163

Option (C) is correct. The difference between incoming signal frequency (fc) and its image frequency (fc) is 2I f (which is large enough). The RF filter may provide poor selectivity against adjacent channels separated by a small frequency differences but it can provide reasonable selectivity against a station separated by 2I f . So it provides adequate suppression of image channel.

SOL 9.164

Option (C) is correct. In PCM SNR is given by SNR = 3 22n 2

nodia

if no. of bits is increased from n to (n + 1) SNR will increase by a factor of 22 (n + 1)/n SOL 9.165

Option (D) is correct. The auto correlation of energy signal is an even function. auto correlation function is gives as R (t) = put Let

R (- t) =

3

# x (t) x (t + t) dt -3 3

# x (t) x (t - t) dt

-3

t-t = a dt = da R (- t) =

3

# x (a + t) x (a) da -3

Changing variable a " t R (- t) =

3

# x (t) x (t + t) dt = R (t) -3

R (- t) = R (t) even function

***********


Kanodia EC Solved Paper

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