Kanodia EC Solved Paper

7/27/2019 Kanodia EC Solved Paper

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GATEELECTRONICS & COMMUNICATION

Solved Paper ( 2013-1996 )

RK Kanodia

Ashish Murolia

For GATE 2014 All session paper visit

www.nodia.co.in

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GATE Electronics & CommunicationSolved 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 ofuse of this information. This book is published with the understanding that NODIA & COMPANY and its author

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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 anddata 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 andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series.Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gaussand Greens theorems.

Differential equations : First order equation (linear and nonlinear), Higher order lineardifferential equations with constant coefficients, Method of variation of parameters, Cauchysand Eulers equations, Initial and boundary value problems, Partial Differential Equations andvariable separable method.

Complex variables : Analytic functions, Cauchys integral theorem and integral formula,Taylors and Laurent series, Residue theorem, solution integrals.

Probability and Statistics :Sampling theorems, Conditional probability, Mean, median, modeand 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 methodsfor 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 andfundamental circuit matrices. Solution methods: nodal and mesh analysis. Network theorems:superposition, Thevenin and Nortons 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 Laplacetransform: 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 insilicon: diffusion current, drift current, mobility, and resistivity. Generation and recombinationof 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,


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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 andFET 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 andwave-shaping circuits, 555 Timers. Power supplies.

Digital circuits :Boolean algebra, minimization of Boolean functions; logic gates; digital ICfamilies (DTL, TTL, ECL, MOS, CMOS). Combinatorial circuits: arithmetic circuits, codeconverters, multiplexers, decoders, PROMs and PLAs. Sequential circuits: latches and flip-flops, counters and shift-registers. Sample and hold circuits, ADCs, DACs. Semiconductormemories. 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 andFFT, z-transform. Sampling theorem. Linear Time-Invariant (LTI) Systems: definitions andproperties; causality, stability, impulse response, convolution, poles and zeros, parallel andcascade structure, frequency response, group delay, phase delay. Signal transmission throughLTI systems.

Control Systems :Basic control system components; block diagrammatic description, reductionof block diagrams. Open loop and closed loop (feedback) systems and stability analysis ofthese 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. Toolsand techniques for LTI control system analysis: root loci, Routh-Hurwitz criterion, Bode and

Nyquist plots. Control system compensators: elements of lead and lag compensation, elementsof Proportional-Integral-Derivative (PID) control. State variable representation and solutionof state equation of LTI control systems.

Communications :Random signals and noise: probability, random variables, probability densityfunction, autocorrelation, power spectral density. Analog communication systems: amplitudeand angle modulation and demodulation systems, spectral analysis of these operations,superheterodyne receivers; elements of hardware, realizations of analog communicationsystems; signal-to-noise ratio (SNR) calculations for amplitude modulation (AM) and frequencymodulation (FM) for low noise conditions. Fundamentals of information theory and channelcapacity theorem. Digital communication systems: pulse code modulation (PCM), differentialpulse code modulation (DPCM), digital modulation schemes: amplitude, phase and frequencyshift keying schemes (ASK, PSK, FSK), matched filter receivers, bandwidth consideration andprobability of error calculations for these schemes. Basics of TDMA, FDMA and CDMA andGSM.

Electromagnetics : Elements of vector calculus: divergence and curl; Gauss and Stokestheorems, Maxwells equations: differential and integral forms. Wave equation, Poynting vector.Plane waves: propagation through various media; reflection and refraction; phase and groupvelocity; 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. Basicsof propagation in dielectric waveguide and optical fibers. Basics of Antennas: Dipole antennas;radiation pattern; antenna gain.

***********


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PREFACE

This book doesnt make promise but provides complete satisfaction to the readers. Themarket scenario is confusing and readers dont find the optimum quality books. This book

provides complete set of problems appeared in competition exams as well as fresh set ofproblems.

The book is categorized into units which are then sub-divided into chapters and theconcepts of the problems are addressed in the relevant chapters. The aim of the book isto avoid the unnecessary elaboration and highlights only those concepts and techniqueswhich are absolutely necessary. Again time is a critical factor both from the point of viewof preparation duration and time taken for solving each problem in the examination. Sothe problems solving methods is the books are those which take the least distance to thesolution.

But however to make a comment that this book is absolute for GATE preparation will bean inappropriate one. The theory for the preparation of the examination should be followedfrom the standard books. But for a wide collection of problems, for a variety of problemsand the efficient way of solving them, what one needs to go needs to go through is therein there in the book. Each unit (e.g. Networks) is subdivided into average seven number ofchapters on an average each of which contains 40 problems which are selected so as to avoidunnecessary redundancy and highly needed completeness.

I shall appreciate and greatly acknowledge the comments and suggestion from the users ofthis book.

R. K. KanodiaAshish Murolia


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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

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CHAPTER 1ENGINEERING MATHEMATICS

nodia

2013 ONE MARK

MCQ 1.1 The maximum value of quntil which the approximation sin .q qholds to within10% error is(A) 10c (B) 18c

(C) 50c (D) 90c

MCQ 1.2 The minimum eigen value of the following matrix is3

5

2

5

12

7

2

7

5

R

T

SSSS

V

X

WWWW

(A) 0 (B) 1

(C) 2 (D) 3

MCQ 1.3 A polynomial ( )f x a x a x a x a x a 44

33

22

1 0= + + + - with all coefficients positive has

(A) no real roots

(B) no negative real root

(C) odd number of real roots

(D) at least one positive and one negative real root

2013 TWO MARKS

MCQ 1.4 Let A be an m n# matrix and Ban n m# matrix. It is given that determinantI A Bm+ =^ h determinant I B An+^ h, where Ikis the k k# identity matrix. Using

the above property, the determinant of the matrix given below is2

1

11

1

2

11

1

1

21

1

1

12

R

T

SS

SSSS

V

X

WW

WWWW

(A) 2 (B) 5

(C) 8 (D) 16

2012 ONE MARK

MCQ 1.5 With initial condition ( ) .x 1 0 5= , the solution of the differential equation

tdtdx

x t+ = , is

(A) x t21

= - (B) x t212= -

(C) x t2

2

= (D) x t

2=

MCQ 1.6 Given ( )f zz z1

13

2=+

-+

.


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Page 2 Engineering Mathematics Chapter 1

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For online test series visit www.gatehelp.com

If Cis a counter clockwise path in the z-plane such that z 1 1+ = , the value

of ( )j

f z dz 21

Cp # is

(A) 2- (B) 1-(C) 1 (D) 2

MCQ 1.7 If ,x 1= - then the value of xxis

(A) e /2p- (B) e /2p

(C) x (D) 1

2012 TWO MARKS

MCQ 1.8 Consider the differential equation

( ) ( ) ( )dt

d y tdt

dy t y t22

2

+ + ( )td= with ( ) 2 0andy tdtdy

tt

00

=- ==

=-

-

The numerical value ofdt

dy

t 0= +is

(A) 2- (B) 1-

(C) 0 (D) 1

MCQ 1.9 The direction of vector A is radially outward from the origin, with krA n= .

where r x y z 2 2 2 2= + + and kis a constant. The value of nfor which A 0:d = 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 thenumber 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 x x x 9 24 53 2= - + + in the interval [ , ]1 6 is(A) 21 (B) 25

(C) 41 (D) 46

MCQ 1.12 Given that

andA I52

30

10

01

= - - => >H H, the value of A 3is(A) 15 12A I+ (B) 19 30A I+

(C) 17 15A I+ (D) 17 21A I+

2011 ONE MARK

MCQ 1.13 Consider a closed surface Ssurrounding volume V . If rvis the position vector of a

point inside S, with ntthe unit normal on S, the value of the integral r n dS 5S

$v t## is

(A) V3 (B) V5

(C) V10 (D) V15

MCQ 1.14 The solution of the differential equation , (0)dx

dyky y c = = is

(A) x ce ky= - (B) x kecy=


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Chapter 1 Engineering Mathematics Page 3

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(C) y cekx= (D) y ce kx= -

MCQ 1.15 The value of the integral

( )z z

zdz

4 5

3 4

c

2

+ +

- +# where cis the circle z 1= is givenby(A) 0 (B) 1/10

(C) 4/5 (D) 1

2011 TWO MARKS

MCQ 1.16 A numerical solution of the equation ( )f x x 3 0+ - = can be obtained usingNewton- Raphson method. If the starting value is x 2= for the iteration, thevalue of xthat 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

4 6 20

4

x y z

x y y

x y z

6

ml

+ + =

+ + =

+ + =

has NO solution for values of land given by(A) 6, 20ml= = (B) 6, 20ml= =Y

(C) 6, 20ml= =Y (D) 6, 20ml= =Y

MCQ 1.18 A fair dice is tossed two times. The probability that the second toss results in avalue 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


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MCQ 1.21 A function ( )n x satisfied the differential equation( ) ( )

dx

d n x

L

n x02

2

2- =

where L is a constant. The boundary conditions are : (0)n K= and ( )n 03 = .The solution to this equation is

(A) ( ) ( / )expn x K x L= (B) ( ) ( / )expn x K x L= -

(C) ( ) ( / )expn x K x L2= - (D) ( ) ( / )expn x K x L= -

2010 TWO MARKS

MCQ 1.22 If e x /y x1= , then yhas a(A) maximum at x e= (B) minimum at x e=

(C) maximum at x e 1= - (D) minimum at x e 1= -

MCQ 1.23 A fair coin is tossed independently four times. The probability of the event thenumber of time heads shown up is more than the number of times tail shown up

(A) 1/16 (B) 1/3

(C) 1/4 (D) 5/16

MCQ 1.24 If A xya x a x y2= +v t t , then A dl

C

$v v# over the path shown in the figure is

(A) 0 (B)3

2

(C) 1 (D) 2 3

MCQ 1.25 The residues of a complex function

( )( )( )

x zz z z

z

1 21 2=

- --

at its poles are

(A) ,21

21- and 1 (B) ,

21

21- and 1-

(C) , 121 and

23- (D) ,

21 1- and

23

MCQ 1.26 Consider differential equation( )

( )dx

dy xy x x- = , with the initial

condition ( )y 0 0= . Using Eulers 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

MCQ 1.27 Given ( )( )

f t Ls s k s

s

4 33 11

3 2= + + -+- ; E. If ( ) 1lim f tt =" 3 , then the value

of kis


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(A) 1 (B) 2

(C) 3 (D) 4

2009 ONE MARK

MCQ 1.28 The order of the differential equationdt

d y

dt

dyy e

t

2

2 34+ + = -c m is

(A) 1 (B) 2

(C) 3 (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?

(A)

2

1 2

c m (B) C

2

1102

2

b l(C)21 10c m (D) C 2110 2 10b l

MCQ 1.30 If ( )f z c c z 0 11= + - , then

( )z

f zdz

1

unitcircle

+

# is given by(A) c2 1p (B) ( )c2 1 0p +

(C) j c2 1p (D) ( )c2 1 0p +

2009 TWO MARKS

MCQ 1.31 The Taylor series expansion of sinx

xp-

at x p= is given by

(A)!

( )...

x1

3

2p+

-+ (B)

!( )

...x

13

2p- -

-+

(C)!

( )...

x1

3

2p-

-+ (D)

!( )

...x

13

2p- +

-+

MCQ 1.32 Match each differential equation in Group I to its family of solution curves from

Group IIGroup I Group II

A. dx

dy

x

y

= 1. Circles

B.dx

dy

x

y=- 2. Straight lines

C.dx

dy

yx= 3. Hyperbolas

D.dx

dy

yx=-

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

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

(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

1

3

0

3

1

0

5

6

3

-

- -

R

T

SSSS

V

X

WWWW


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

(C) 3 , 3 , 5j j j+ - + (D) 3, 1 3 , 1 3j j- + - -

2008 ONE MARKS

MCQ 1.34 All the four entries of the 2 2# matrix Pp

p

p

p

11

21

12

22= = Gare nonzero,

and one of its eigenvalue is zero. Which of the following statements is true?(A) p p p p 111 12 12 21- = (B) p p p p 111 22 12 21- =-

(C) p p p p 011 22 12 21- = (D) p p p p 011 22 12 21+ =

MCQ 1.35 The system of linear equations

x y4 2+ 7=

x y2 + 6= has

(A) a unique solution

(B) no solution

(C) an infinite number of solutions

(D) exactly two distinct solutions

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( ) ( ) ( )exp expf x x 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 xin its Taylorseries expansion about the point x 0= ?

(A) ( )sin x3 (B) ( )sin x2

(C) ( )cos x

3

(D) ( )cos x

2

MCQ 1.39 Which of the following is a solution to the differential equation( )

( )dt

dx tx t3 0+ = ?

(A) ( )x t e3 t= - (B) ( )x t e2 t3= -

(C) ( )x t t23 2=- (D) ( )x t t3 2=

2008 TWO MARKS

MCQ 1.40 The recursion relation to solve x e x= - using Newton - Raphson method is

(A) x en x1n

=+ - (B) x x en n x1n

= -+ -

(C) (1 )x xe

e

1n n x

x

1n

n

= ++

+ -

-

(D)( )

xx e

x e x1 1n

nx

nx

n1

2

n

n

=-

- - -+ -

-

MCQ 1.41 The residue of the function ( )f z ( ) ( )z z2 2

12 2

=+ -

at z 2= is


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(A)321- (B)

161-

(C)

16

1 (D)

32

1

MCQ 1.42 Consider the matrix P0

2

1

3=

- -= G. The value of epis

(A)e e

e e

e e

e e

2 3

2 2 5

2 1

2 1

1 2

2 1

-

-

-

-

- -

- -

- -

- -> H (B) e ee e

e e

e e2 4

2

3 2

1 1

1 2

2 1

1 2

+

-

-

+

- -

-

- -

- -> H(C)

e e

e e

e e

e

5

2 6

3

4 6

2 1

2 1

1 2

2 1

-

-

-

+

- -

- -

- -

- -> H (D) e ee e

e e

e e

2

2 2 2

1 2

1 2

1 2

1 2

-

- +

-

- +

- -

- -

- -

- -> HMCQ 1.43 In the Taylor series expansion of ( ) ( )exp sinx x+ about the point x p= , the

coefficient of ( )x 2p- is

(A) ( )exp p (B) . ( )exp0 5 p

(C) ( )exp 1p + (D) ( )exp 1p -

MCQ 1.44 The value of the integral of the function ( , )g x y x y 4 103 4= + along the straightline 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 Pand Qin the x y- plane, with ( , )P 1 0= and ( , )Q 0 1= . The

line integral 2 ( )xdx ydy P

Q

+# along the semicircle with the line segment PQasits diameter(A) is 1-

(B) is 0

(C) is 1

(D) depends on the direction (clockwise or anit-clockwise) of the semicircle

2007 ONE MARK

MCQ 1.46 The following plot shows a function which varies linearly with x. The value of the

integral I ydx 1

2

=# is

(A) 1.0 (B) 2.5

(C) 4.0 (D) 5.0

MCQ 1.47 For x 1


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MCQ 1.48 limsin

20 q

q

"q

b lis

(A) 0.5 (B) 1(C) 2 (D) not defined

MCQ 1.49 Which one of following functions is strictly bounded?

(A) /x1 2 (B) ex

(C) x2 (D) e x2-

MCQ 1.50 For the function e x- , the linear approximation around x 2= is

(A) ( )x e3 2- - (B) x1 -

(C) ( )x e3 3 2 1 2 2+ - - -6 @ (D) e 2-

2007 TWO MARKS

MCQ 1.51 The solution of the differential equation kdx

d yy y

22

2

2= - under the boundaryconditions(i) y y1= at x 0= and

(ii) y y2= at x 3= , where ,k y1and y2are constants, is

(A) ( )expy y yk

xy1 2 2 2= - - +a k (B) ( )expy y y kx y2 1 1= - - +a k

(C) sinhy y y kx y1 2 1= - +^ ah k (D) expy y y kx y1 2 2= - - +^ ah kMCQ 1.52 The equation x x x4 4 03 2- + - = is to be solved using the Newton - Raphson

method. If x 2= is taken as the initial approximation of the solution, then nextapproximation using this method will be(A) 2/3 (B) 4/3

(C) 1 (D) 3/2

MCQ 1.53 Three functions ( ), ( )f t f t 1 2 and ( )f t3 which are zero outside the interval [ , ]T0 areshown in the figure. Which of the following statements is correct?


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(A) ( )f t1 and ( )f t2 are orthogonal (B) ( )f t1 and ( )f t3 are orthogonal

(C) ( )f t2 and ( )f t3 are orthogonal D) ( )f t1 and ( )f t2 are orthonormal

MCQ 1.54 If the semi-circular control Dof radius 2 is as shown in the figure, then the valueof the integral

( )sds

11

D

2 -# is

(A) jp (B) jp-

(C) p- (D) p

MCQ 1.55 It is given that , ...X X XM1 2 at M non-zero, orthogonal vectors. The dimension ofthe vector space spanned by the M2 vectors , ,... , , ,...X X X X X X M M1 2 1 2- - - is

(A) M2 (B) M 1+

(C) M

(D) dependent on the choice of , ,...X X XM1 2

MCQ 1.56 Consider the function ( )f x x x 22= - - . The maximum value of ( )f x in the closedinterval [ , ]4 4- is

(A) 18 (B) 10(C) 225- (D) indeterminate

MCQ 1.57 An examination consists of two papers, Paper 1 and Paper 2. The probabilityof 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 astudent failing in both the papers is(A) 0.5 (B) 0.18

(C) 0.12 (D) 0.06

2006 ONE MARK

MCQ 1.58 The rank of the matrix

1

1

1

1

1

1

1

0

1

-

R

T

SSSS

V

X

WWWWis

(A) 0 (B) 1

(C) 2 (D) 3

MCQ 1.59 P4#4# , where Pis a vector, is equal to

(A) P P P2#4# 4- (B) ( )P P24 4 4 #+

(C) P P2

4 4#+ (D) ( )P P

24 4 $ 4

-

MCQ 1.60 ( )P ds4 # $# # , where Pis a vector, is equal to(A) P dl$# (B) P dl4 4 $# ##(C) P dl4 $## (D) Pdv4$# # #


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MCQ 1.61 A probability density function is of the form ( ) , ( , )p x K e x x 3 3!= -a- . Thevalue of K is(A) 0.5 (B) 1

(C) .0 5a (D) a

MCQ 1.62 A solution for the differential equation ( ) ( ) ( )x t x t t 2 d+ =o with initial condition

(0 ) 0x =- is

(A) ( )e u tt2- (B) ( )e u tt2

(C) ( )e u tt- (D) ( )e u tt

2006 TWO MARKS

MCQ 1.63 The eigenvalue and the corresponding eigenvector of 2 2# matrix are given by

Eigenvalue Eigenvector

81l = v1

11 = = G

42l = v1

12=

-= G

The matrix is

(A)6

2

2

6= G (B)4

6

6

4= G(C)

2

4

4

2= G (D)4

8

8

4= GMCQ 1.64 For the function of a complex variable lnW Z= (where,W u j v = + and Z x j y = +

, 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 integralz 4

1

z j

2

2+

- =

# dz is positive sense is(A)

j

2p

(B)2p-

(C)j

2

p- (D)

2

p

MCQ 1.66 The integral sin d30

q qp# is given by

(A)21 (B)

32

(C)34 (D)

38

MCQ 1.67 Three companies ,X Y and Zsupply computers to a university. The percentage

of computers supplied by them and the probability of those being defective aretabulated below

Company % of Computer Supplied Probability of being supplied defective

X %60 .0 01

Y %30 .0 02

Z %10 .0 03


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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 For the matrix4

2

2

4= Gthe eigenvalue corresponding to the eigenvector101

101= Gis(A) 2 (B) 4

(C) 6 (D) 8

MCQ 1.69 For the differential equationdx

d yk y 0

2

22+ = the boundary conditions are

(i) y 0= for x 0= and (ii) y 0= for x a=The form of non-zero solutions of y(where mvaries over all integers) are

(A) siny A am xmm

p= / (B) cosy A am xmm p= /(C) y A xm a

m

m

= p/ (D) y A em am x

m

= p-/

MCQ 1.70 As xincreased from 3- to 3, the function ( )f xe

e

1 xx

=+

(A) monotonically increases

(B) monotonically decreases

(C) increases to a maximum value and then decreases

(D) decreases to a minimum value and then increases

2005 ONE MARK

MCQ 1.71 The following differential equation has

dt

d y

dt

dyy3 4 2

2

3 2+ + +c cm m x=(A) degree 2= , order 1= (B) degree 1= , order 2=

(C) degree 4= , order 3= (D) degree 2= , order 3=

MCQ 1.72 A fair dice is rolled twice. The probability that an odd number will follow an evennumber is

(A) 1/2 (B) 1/6

(C) 1/3 (D) 1/4

MCQ 1.73 A solution of the following differential equation is given bydx

d y

dx

dyy5 6 0

2

2

- + =

(A) y e ex x2 3= + - (B) y e ex x2 3= +

(C) 3y e x x2 3= +- (D) y e ex x2 3= +- -

2005 TWO MARKS

MCQ 1.74 In what range should ( )Re s remain so that the Laplace transform of the function

e( )a t2 5+ + exits.

(A) ( ) 2Re s a> + (B) ( )Re s a 7> +

(C) ( )Re s 2< (D) ( )Re s a 5> +


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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 2E. Newton-Raphson method 1. Solving nonlinear equationsF. Runge-kutta method 2. Solving linear simultaneous

equationsG. Simpsons 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- - - -

MCQ 1.77 Given the matrix4

4

2

3

-= G, the eigenvector is(A)

3

2= G (B)4

3= G(C)

2

1-= G (D)1

2

-= G

MCQ 1.78 Let,.

A2

0

0 1

3=

-

= Gand A

a

b0

1 21

=-

= G. Then ( )a b+ =

(A) 7/20 (B) 3/20

(C) 19/60 (D) 11/20

MCQ 1.79 The value of the integral expI x dx21

8

2

0p= -

3 c m# is(A) 1 (B) p

(C) 2 (D) 2p

MCQ 1.80 Given an orthogonal matrix

A

1

11

0

1

11

0

1

10

1

1

10

1

=-

- -

R

T

SSSSS

V

X

WWWWW

A AT 1-6 @ is


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(A)0

0

0

0

0

0

0

0

0

0

0

0

41

41

21

21

R

T

SSS

SSS

V

X

WWW

WWW

(B)0

0

0

0

0

0

0

0

0

0

0

0

21

21

21

21

R

T

SSS

SSS

V

X

WWW

WWW

(C)

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

R

T

SSSSS

V

X

WWWWW

(D)0

0

0

0

0

0

0

0

0

0

0

0

41

41

41

41

R

T

SSSSSS

V

X

WWWWWW

***********


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SOLUTIONS

SOL 1.1 Option (B) is correct.

Here, as we know

sinLim0

q"q

0.

but for %10 error, we can check option (B) first,

q 18 .18180

0 314#c cc

p= = =

sin q .sin 18 0 309c= =

%error.

. . % . %0 309

0 314 0 309 100 0 49#= - =

Now, we check it for 50cq=

q .50 50180

0 873#c cc

p= = =

sin q .sin 50 0 77c= =

%error.

0.77 0.873 . %0 873

12 25= - =-

so, the error is more than %10 . Hence, for error less than 10%, 18cq= can have

the approximation sin q. q

SOL 1.2 Option (A) is correct.

For, a given matrix A6 @the eigen value is calculated as A Il- 0=where lgives the eigen values of matrix. Here, the minimum eigen value amongthe given options is

l 0=We check the characteristic equation of matrix for this eigen value A Il- A= (for 0l= )

3

52

5

127

2

75

=

3 5 260 49 25 14 35 24= - - - + -^ ^ ^h h h 33 55 22= - + 0=Hence, it satisfied the characteristic equation and so, the minimum eigen valueis

l 0=

SOL 1.3 Option (D) is correct.

Given, the polynomial

f x^ h a x a x a x a x a 4 4 3 3 2 2 1 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 signchange from a1^ hto a0^ hin the Routh matrix 1stcolumn. Also, there will be a


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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


Consider the given matrix be

I A Bm+

2

1

1

1

1

2

1

1

1

1

2

1

1

1

1

2

=

R

T

SSSSSS

V

X

WWWWWW

where m 4= so, we obtain

A B

2

1

1

1

1

2

1

1

1

1

2

1

1

1

1

2

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

= -

R

T

SSSSSS

R

T

SSSSSS

V

X

WWWWWW

V

X

WWWWWW

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

=

R

T

SSSSSS

V

X

WWWWWW

1

1

1

1

=

R

T

SSS

SSS

V

X

WWW

WWW

1 1 1 1

6 @Hence, we get

A

1

1

1

1

=

R

T

SSSSSS

V

X

WWWWWW

, B 1 1 1 1= 8 B

Therefore, B A 1 1 1 1= 8 B 11

1

1

R

T

SSS

SSS

V

X

WWW

WWW 4=From the given property Det I A Bm+^ h Det I B Am= +^ h

& Det

2

1

1

1

1

2

1

1

1

1

2

1

1

1

1

2

R

T

SSSSSS

V

X

WWWWWW

Det

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

4= +

R

T

SSSSSS

V

X

WWWWWW

Z

[

\

]]

]]

_

`

a

bb

bb

1 4= +

5=Note : Determinant of identity matrix is always 1.


tdtdx

x+ t=


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dtdx

tx+ 1=

dt

dxPx+ Q= (General form)

Integrating factor, I F e e e t ln tPdt tdt1

= = = =# #

Solution has the form,

x I F# Q I F dt C #= +^ h# x t# ( )( )t dt C 1= +# xt t C

2

2

= +

Taking the initial condition,

( )x 1 .0 5=

0.5 C21

= + & C 0=

So, xt t2

2

= x t2

& =

SOL 1.6 Option (C) is correct.

( )f z z z1

13

2=

+ -

+

( )j

f z dz 21

Cp

# =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.

( )f z ( )( )z z

z1 3

1=+ +

- + ( )( )

( )( )lim

z z

z z

1 31 1

22 1

z 1=

+ +

+ - += =

"-

So ( )j

f z dz 21

Cp # 1=


x i1= - = cos sini2 2p p= +

So, x ei2= p

xx ei x

2= p^ h & ei i2p^ h e 2= p-


( ) ( )

( )dt

d y t

dt

dy ty t

22

2

+ + ( )td=

By taking Laplace transform with initial conditions

( ) ( ) [ ( ) ( )] ( )s Y s sy dt

dysy s y Y s 0 2 0

t

2

0

- - + - +=

; E 1=& ( ) 2 ( ) ( )s Y s s sY s Y s 2 0 22 + - + + +

6 6@ @ 1=

( ) [ ]Y s s s 2 12 + + s1 2 4= - -

( )Y s s s

s

2 12 3

2= + +- -

We know that, If, ( )y t ( )Y sL


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then,( )

dt

dy t ( ) ( )sY s y 0

L-

So, ( ) ( )sY s y

0- ( )

( )

s s

s s

2 1

2 3

22= + +

- -

+

( )s s

s s s s

2 12 3 2 4 2

2

2 2

=+ +

- - + + +

( ) ( )sY s y 0- ( ) ( ) ( )s

s

s

s

s12

11

11

2 2 2= ++

=++

++

( )s s1

11

12= +

++

Taking inverse Laplace transform

( )

dt

dy t ( ) ( )e u t te u t t t= +- -

At t 0= +,dtdy

t 0= + e 0 10= + =


Divergence of A in spherical coordinates is given as

A:d ( )r r

r A1

r22

2

2= ( )r r

kr1 n2

2

2

2= +

( )r

kn r2 n2

1= + +

( )k n r2 0n 1= + =- (given)

n 2+ 0= & n 2=-


Probability of appearing a head is /1 2. If the number of required tosses is odd,

we have following sequence of events.

,H ,T T H ,...........T T T T H

Probability P .....21

21

213 5= + + +b bl l

P1 3

2

41

21

=-

=


( )f x x x x9 24 53 2= - + +

( )

dx

df x x x3 18 24 02= - + =

&( )

dx

df x 6 8 0x x2= - + = x 4= , x 2=

( )

dx

d f x2

2

x6 18= -

For ,x 2=

( )

dx

d f x

12 18 6 0


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Characteristic equation.

A Il- 0=

5

2

3l

l

- - -

- 0=

5 62l l+ + 0=

5 62l l+ + 0=

Since characteristic equation satisfies its own matrix, so

5 6A A2 + + 0= 5 6A A I2& =- -

Multiplying with A

5 6A A A3 2+ + 0=

5( 5 6 ) 6A A I A3 + - - + 0=

A 3 19 30A I= +


From Divergence theorem, we have

A dv4$v v### A n ds s

$= v t#The position vector

rv u x u y u z x y z= + +t t t^ hHere, 5A r=v v, thus

A4$ v ux

uy

uz

u x u y u z x y z x y z :2

2

2

2

2

2= + + + +t t t t t tc ^m h 5

dxdx

dy

dy

dzdz

= + +c m 3 5#= 15=So, 5r n ds

s$v t## 15dv V15= =###


We havedx

dy ky=

Integratingy

dy# k dx A= +#

or ln y kx A= +

Since ( )y 0 c= thus ln c A=So, we get, ln y lnkx c= +

or ln y ln lne ckx= +

or y cekx=


C R Integrals isz z

zdz

4 53 4

C

2 + +- +

# where C is circle z 1=

( )f z d z C# 0= if poles are outside C.

Now z z4 52 + + 0=

( )z 2 12+ + 0=

Thus z ,1 2 j z2 1>,1 2&!=-

So poles are outside the unit circle.



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We have ( )f x x x 3 0= + - =

( )f xl x

12

1= +

Substituting x0 2= we get

( )f x0l .1 35355= and ( ) .f x 2 2 3 0 4140 = + - =

Newton Raphson Method

x1 ( )( )

xf x

f x0

0

0= -

l

Substituting all values we have

x1 ..2

1 35350 414= - .1 694=


Writing :A Bwe have

:

:

:

1

1

1

1

4

4

1

6

6

20

l m

R

T

SSSS

V

X

WWWW

Apply R R R3 3 2" -

:

:

: 20

1

1

0

1

4

0

1

6

6

6

20

l m- -

R

T

SSSS

V

X

WWWW

For equation to have solution, rank of A and :A Bmust be same. Thus for nosolution; 6, 20!ml=


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 total outcomes

No of favourable outcomes3615

125

= = =


Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in

conjugate pairs.


For a function ( )x t trigonometric fourier series is

( )x t [ ]cos sinA A n t B n t o n nn 1

w w= + +3

=

/

Where, A o ( )T x t dt

1

T0

0

= # T0 "fundamental period

A n ( )cosT x t n t dt

2

T0

0

w= #

Bn ( )sinT x t n t dt

2

T0

0

w= #For an even function ( ),x t B 0n=

Since given function is even function so coefficient B 0n= , only cosine and constant

terms are present in its fourier series representation.

Constant term :


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A 0 ( )T x t dt

1/

/

T

T

4

3 4=

-#

T

A dt A dt 1 2

/

/

/

/

T

T

T

T

4

4

4

3 4= + -

-: D# #

T

T AA

T12

22

= -: D A2=-Constant term is negative.


Given differential equation

( ) ( )

dx

d n x

L

n x2

2

2- 0=

Let ( )n x A e x= l

So, A e LA ex

x

2 2l -l

l

0=

L

122l - L

0 1& !l= =

Boundary condition, ( )n 03 = so takeL1l= -

( )n x A e Lx

= -

( )n 0 A e K A K 0 &= = =

So, ( )n x K e ( / )x L= -


Given that ey x x1

=

or ln ey ln x x1

=

or y lnx

x1=

Nowdx

dy ln

x x x x

1 1x

12= + - -^ h ln

x x

12 2= -

For maxima and minima :

dx

dy (1 ) 0ln

xx

12= - =

lnx 1=

"

x

e1=

Nowdx

d y2

2

lnx

xx x x

2 2 1 13 3 2=- - - -b bl l

lnx x

x

x

2 2 12 3 3=- + -

dy

d x

at x e

2

2

1=

e e e

2 2 1 0


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Av xya x a x y2= +t t

d lv dxa dya x y= +t t

A d lC :v v

# ( ) ( )xya x a dxa dya x yC x y2

:= + +t t t t

# ( )xydx x dy

C

2= +#

xdx xdx dy dy 334

31

/

/

/

/

2 3

1 3

1

3

3

1

1 3

2 3= + + +# # ##

[ ] [ ]21

34

31

23

31

34

34 3 1

31 1 3= - + - + - + -: :D D

1=


Given function

( )X z ( )( )z z z

z1 2

1 2=- -

-

Poles are located at 0, 1, 2andz z z= = =

At Z 0= residues is

R0 ( )z X zZ 0

:==

( )( )0 1 0 2

1 2 0#=- -

- 21

=

at z 1= , R1 ( 1) ( )Z X ZZ 1

:= -=

( )1 1 2

1 2 1 1#=-

-=

At z 2= , R2 ( ) ( )z X z2 z 2:

= - =

( )2 2 1

1 2 223#=

--

=-


Taking step size h 0.1= , ( )y 0 0=

x ydx

dyx y= + y y h

dx

dyi i1 = ++

0 0 0 . ( )y 0 0 1 0 01 = + =

0.1 0 0.1 . ( . ) .y 0 0 1 0 1 0 012= + =

0.2 0.01 0.21 . . . .y 0 01 0 21 0 1 0 0313 #= + =

0.3 0.031

From table, at . , ( . ) .x y x0 3 0 3 0 031= = =


Given that

( )f t ( )s s K s

s

4 33 1

L1

3 2= + + -+- ; E

( )lim f tt " 3

1=

By final value theorem

( )lim f tt " 3

( )lim sF s 1s 0

= ="

or( )

( )lim

s s K s

s s

4 3

3 1s 0 3 2

:

+ + -

+"

1=


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or[ ( )]

( )lim

s s s K

s s

4 3

3 1s 0 2 + + -

+"

1=

K 3

1

- 1=

or K 4=


The highest derivative terms present in DE is of 2nd order.


Number of elements in sample space is 210. Only one element

, , , , , , , , ,H H T T T T T T T T " ,is event. Thus probability is2110


We have

( )f z c c z0 11= + -

( )f z1 ( )

z

f z

z

c c z1 1 0 11

= +

= + + -

( )

z

z c c120 1

= + +

Since ( )f z1 has double pole at z 0= , the residue at z 0= is

Res ( )f z z1 0= . ( )lim z f zz 0

21=

" .lim z

z

z c c

z 0

220 1

="

c m c1=Hence

( )f z dz 1unitcircle

# [ ( )]z

f zdz

1

unitcircle

= +# j2p= [Residue at z 0= ]

j c2 1p=


We have ( )f x sinx

xp

=-

Substituting x p- y= ,we get

( )f y p+ ( )sin sin

y

y

y

yp=

+=- ( )sin

y y1=-

! !

...y

y y y1

3 5=- - + -c m

or ( )f y p+ ! ! ...

y y

1 3 5

2 4

=- + - +

Substituting x yp- = we get

( )f x !

( )!

( )...

x x1

3 5

2 4p p=- +

--

-+


(A)dx

dy

x

y=

ory

dy# x

dx=#

or log y log logx c= +or y cx= Straight Line

Thus option (A) and (C) may be correct.

(B)dx

dy

x

y=-


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ory

dy# x

dx=-#or log y log logx c=- +

or log y log logx

c1= +

or yxc= Hyperbola


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.


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 p p p p 11 22 12 21- 0=


The given system is

x

y

4

2

2

1= =G G7

6= = G

We have A 4

2

2

1= = G

and A 42

21

0= = Rank of matrix ( )A 2 , e 1>x and 0 1e<


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We have x e x= -

or ( )f x x e x= - -

'( )f x e1 x= + -

The Newton-Raphson iterative formula is

xn 1+ '( )( )

xf x

f xn

n

n= -

Now ( )f xn x enxn= - -

'( )f xn e1 xn= + -

Thus xn 1+ xe

x e

1n x

nx

n

n

= -+

--

-

( )

e

x e

1

1x

nx

n

n

=+

+-

-

SOL 1.41

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

( ) ( )n dz

d z a f z 1

1n

nn

z a1

1

=-

--

-

=6 @Here we have n 2= and a 2=

Thus Res ( )f z z 2= ( )!( )

( ) ( )dzd z

z z2 11 2

2 21

z a

22 2

=-

-- + =

; E

( )dz

d

z 21

z a

2=

+ =; E ( )z 2

z a

3=

+

-

=; E

64

2=- 32

1=-


eP ( )L sI A1 1= -- -6 @ L

s

s0

0 0

2

1

31

1

= -- -

--e o= =G G

Ls

s2

1

31

1

=-

+-

-e o= G L

( )( )

( )( )

( )( )

( )( )

s s

s

s s

s s

s s

s1 1 2

3

1 22

1 21

1 2

= -+ +

+

+ +-

+ +

+ +

f p> H

e e

e e

e e

e e

2

2 2 2

1 2

1 2

1 2

1 2= -

- +

-

- +

- -

- -

- -

- -= GSOL 1.43 Option (B) is correct.

Taylor series is given as

( )f x ( )!

'( )!

( )"( ) ...f a x a f a

x af a

1 2

2

= + - + -

+

For x p= we have

Thus ( )f x ( )!

'( )!

( )"( )...f x f

xf x

1 2

2

p p p p

= + - + -

Now ( )f x sine xx= +

'( )f x cose xx= +

"( )f x sine xx= -

"( )f p sine ep= - =p p


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Thus the coefficient of ( )x 2p- is!

"( )f2

p


The equation of straight line from ( , )0 0 to ( , )1 2 is y x2= .

Now ( , )g x y x y4 103 4= +

or, ( , )g x x2 x x4 1603 4= +

Now ( , )g x x20

1# ( )x x dx 4 1603 40

1= +#

[ ]x x32 334 5 01= + =


I ( )xdx ydy 2P

Q

= +# xdx ydy 2 2

P

Q

P

Q

= + ## xdx ydy 2 2 0

0

1

1

0= + =##


The given plot is straight line whose equation is

x y

1 1- + 1=

or y x 1= +

Now I ydx1

2

=# ( )x dx112

= +#

( )x21 2

= +; E .

29

24 2 5= - =


coth xsinhcosh

xx=

as 1, 1coshx x


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( )x e1 2 2= - - -6 @ Neglecting higher powers ( )x e3 2= - -


We have kdx

d y22

2

y y2= -

ordx

d y

k

y2

2

2-

k

y22=-

A.E. Dk

122

- 0=

or Dk1

!=

C.F. C e C e 1 2k

x

k

x

= +-

P.I.D k

yy1

k

2 1 222

22

=-

-=c m

Thus solution is

y C e C e y 1 2 2kx

k

x

= + +-

From ( )y y0 1= we get

C C1 2+ y y1 2= -

From ( )y y23 = we get that C1must be zero.

Thus C2 y y1 2= -

y ( )y y e y 1 2 2kx

= - +-


We have

( )f x x x x4 43 2= - + -

'( )f x x x3 2 42= - +

Taking x 20= in Newton-Raphosn method

x1 '( )( )

xf x

f x0

0

0= -

( ) ( )

( )2

3 2 2 2 4

2 2 4 2 42

3 2

= -- +

- + -

34=

SOL 1.53 Option (C) is correct.For two orthogonal signal ( )f x and ( )g x

( ) ( )f x g x dx 3

3

-

+# 0=i.e. common area between ( )f x and ( )g x is zero.


We know that

s

ds1

1

D

2 -

# j2p= [sum of residues]Singular points are at s 1!= but only s 1=+ lies inside the given contour,

Thus Residue at s 1=+ is

( ) ( )lim s f s1s 1

-"

( )lim ss

11

121

s 1 2= -

-=

"

s

ds1

1

D

2 -# j j2

2p p= =` j


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For two orthogonal vectors, we require two dimensions to define them and

similarly for three orthogonal vector we require three dimensions to define them.

M2 vectors are basically M orthogonal vector and we require M dimensions todefine them.


We have

( )f x x x 22= - +

'( )f x x2 1 0= - = x21

" =

"( )f x 2=

Since "( )f x 2 0>= , thus x21= is minimum point. The maximum value in

closed interval ,4 4-6 @will be at x 4=- or x 4=Now maximum value

[ ( 4), (4)]max f f= -

( , )max 18 10=

18=


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 0.6P BA =^ hWe know that

PBAb l ( )( )P BP B+=

or ( )P A B+ ( )P B PBA

= b l . . .0 6 0 2 0 12#= =SOL 1.58 Option (C) is correct.

We have

A

1

1

1

1

1

1

1

0

1

1

1

0

1

1

0

1

0

0

+= - -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW R R3 1-

Since one full row is zero, ( )A 3


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We know ( )p x d x 3

3

-# 1=

Thus K e dx x

3

3a-

-

# 1=

or K e dx K e dx x x0

0

+3

3

a a-

-## 1=

or( )

Ke

ke

x x00a a

+-3

3a a

-

-6 6@ @ 1=or K K

a a+ 1=

or K 2a=


We have ( ) ( )x t x t 2+o ( )s t=

Taking Laplace transform both sides

( ) ( ) ( )sX s x X s 0 2- + 1=

or ( ) ( )sX s X s 2+ 1= Since ( )x 0 0=-

( )X s s 2

1=+

Now taking inverse Laplace transform we have

( )x t ( )e u tt2= -


Sum of the Eigen values must be equal to the sum of element of principal diagonal

of matrix.

Only matrix6

2

2

6= Gsatisfy this condition.SOL 1.64 Option (B) is correct.

We have W ln z=

u j v+ ( )ln x j y= +

or eu j v+ x j y= +

or e eu j v x j y= +

( )cos sine v j v u + x j y= +

Now cosx e vu= and siny e vu=

Thus x y2 2+ e u2= Equation of circle


We have

z

dz4

1

z j

2

2 +

- =

# ( )( )z i z i

dz2 2

1

z j 2

=+ -

- =

#

( , )P 0 2 lies inside the circle z j 2- = and ( , )P 0 2- does not lie.

Thus By cauchys integral formula

I ( ) ( )( )limi z i z i z i 2 2 2 21z i2p= - + -" i ii2 22 2C

p p= + =#SOL 1.66 Option (C) is correct.

I sin d30

q q= p#

sin sin d4

3 30

q q q= -p ` j# sin sin sin3 3 4 3q q q= -


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cos s43

123

0 0

q w q= - =p p: :D D 43 43 121 121 34= + - + =8 8B B

SOL 1.67 Option (D) is correct.Let d "defective and y "supply by Y

pd

ya k ( )( )P dP y 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# # #= + + =

Pd

ya k .. .0 0150 006 0 4= =SOL 1.68 Option (C) is correct.

We have A 42 24= = GNow [ ]A I Xl-6 @ 0=

or4

2

2

4

101

101

l

l

-

-= =G G 00= = G

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

or l 6=


We have dx

d y

k y2

22

+ 0=

or D y k y 2 2+ 0=

The AE is m k2 2+ 0=

The solution of A Eis m i k!=

Thus sin cosy A kx B kx = +

From x 0= , y 0= we get B 0= and ,x a y 0= = we get

sinA ka 0=

or sin ka 0=

k am xp

=

Thus y sinAa

m xm

m

p= ` j/SOL 1.70 Option (A) is correct.

We have ( )f x e

e

1 xx

=+

For x " 3, the value of ( )f x monotonically increases.


Order is the highest derivative term present in the equation and degree is the

power of highest derivative term.Order 2= , degree 1=


Probability of coming odd number is 21 and the probability of coming even number

is 21 . Both the events are independent to each other, thus probability of coming

odd number after an even number is 21

21

41

# = .


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We havedx

d y

dx

dyy5 6

2

2

- + 0=

The . .A E is m m5 62 - + 0=

m ,3 2=

The CF is yc C e C e x x

13

22= +

Since Q 0= , thus y C e C e x x13

22= +

Thus only (B) may be correct.


We have ( )f t e( )a t2 5= + + .e e( )a t5 2= +

Taking Laplace transform we get

( )F s ( )e s a 215= - +; 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.


Newton - Raphson "Method-Solving nonlinear eq.

Runge - kutta Method "Solving ordinary differential eq.

Simpsons Rule "Numerical Integration

Gauss elimination "Solving linear simultaneous eq.


We have A 4

4

2

3=

-= GCharacteristic equation is

A Il- 0=

or4

4

2

3

l

l

-

- 0=

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

or 12 82l l- + + - 0=

or 202l l+ - 0=

or l ,5 4=- Eigen values

Eigen vector for 5l= -

( )A I Xil- 0=

( ) x

x

1 5

4

2

8 41

2

- -

-= =G G 00= = G

x

x

1

0

2

01

2= =G G 00= = G R R42 1-

x x21 2+ 0=Let x x2 11 2&- = =- ,

Thus X 2

1=

-= G Eigen vector



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We have

A .2

0

0 1

3=

-= Gand A ab0

1 21

=- = GNow A A 1- I=

or. a

b

2

0

0 1

3 021-= =G G 10

0

1= = G

or.a b

b

1

0

2 0 1

3

-= G 100

1= = G

or .a2 0 1- 0= and b3 1=

Thus solving above we have b31= and a

601=

Therefore a b+ 31 601 207= + =


Gaussian PDF is

( )f x e dx21 ( )x

2 2

2

p s=

3

3-

-

s

m-# for x3 3# #-

and ( )f x d x 3

3

-# 1=

Substituting 0m= and 2s= in above we get

e dx

2 2

1 x8

2

p 3

3-

-# 1=or 2 e dx

2 21

0

x

8

2

p

3-# 1=

or e dx21

0

x

8

2

p

3-# 1=


From orthogonal matrix

[ ]A A T I=

Since the inverse of I is I , thus

[ ]A A T 1- I I1= =-

*********


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CHAPTER 2NETWORKS

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2013 ONE MARK

MCQ 2.1 Consider a delta connection of resistors and its equivalent star connection asshown 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) /k1 (D) k

MCQ 2.2 The transfer functionV s

V s

1

2

^^

hhof the circuit shown below is

(A) .s

s1

0 5 1+

+ (B)ss

23 6

++

(C)ss

12

++ (D)

ss

21

++

MCQ 2.3 A source cosv t V t 100s p=^ h has an internal impedance of j4 3 W+^ h . If a purelyresistive load connected to this source has to extract the maximum power out of

the source, its value in Wshould be(A) 3 (B) 4

(C) 5 (D) 7

2013 TWO MARKS

MCQ 2.4 In the circuit shown below, if the source voltage 100 53.13 VVS c+= then theThevenins equivalent voltage in Volts as seen by the load resistance RL is

(A) 100 90c+ (B) 800 0c+


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Chapter 2 Networks Page 33

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(C) 800 90c+ (D) 100 60c+

MCQ 2.5 The following arrangement consists of an ideal transformer and an attenuator

which attenuates by a factor of 0.8. An ac voltage 100 VVW X1 = is appliedacross WX to get an open circuit voltage VY Z1across YZ. Next, an ac voltage

100 VVY Z2= is applied across YZ to get an open circuit voltage VW X2across WX.Then, /V VY Z W X 1 1, /V VW X Y Z 2 2are respectively,

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

(C) 100/100 and 100/100 (D) 80/100 and 80/100

MCQ 2.6 Two magnetically uncoupled inductive coils have Qfactors q1and q2at the chosenoperating frequency. Their respective resistances are R1and R2. When connectedin series, their effective Qfactor at the same operating frequency is(A) q q1 2+

(B) / /q q1 11 2+^ ^h h(C) /q R q R R R 1 1 2 2 1 2+ +^ ^h h(D) /q R q R R R 1 2 2 1 1 2+ +

^ ^h hMCQ 2.7 Three capacitors C1, C2and C3whose values are 10 Fm , 5 Fm , and 2 Fm respectively,have breakdown voltages of 10 V, 5 Vand 2 Vrespectively. For the interconnectionshown below, the maximum safe voltage in Volts that can be applied across thecombination, and the corresponding total charge in Cm stored in the effective

capacitance across the terminals are respectively,

(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


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MCQ 2.8 The current ISin Amps in the voltage source, and voltage VSin Volts across the

current source respectively, are(A) ,13 20- (B) ,8 10-

(C) ,8 20- (D) ,13 20-

MCQ 2.9 The current in the 1Wresistor in Amps is(A) 2 (B) 3.33

(C) 10 (D) 12

2012 ONE MARK

MCQ 2.10 In the following figure, C1and C2are ideal capacitors. C1has been charged to12 Vbefore the ideal switch Sis closed at .t 0= The current ( )i t for all tis

(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 3)j W- by a current

5 (100 100)cos Atp + is(A) 44.2 W (B) 50 W

(C) 62.5W (D) 125 W

MCQ 2.12 In the circuit shown below, the current through the inductor is

(A) Aj1

2+

(B) Aj11

+-

(C) Aj1

1+

(D) 0 A

2012 TWO MARKS

MCQ 2.13 Assuming both the voltage sources are in phase, the value of Rfor which maximumpower is transferred from circuit A to circuit Bis


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(A) 0.8 W (B) 1.4 W

(C) 2 W (D) 2.8 W

MCQ 2.14 If 6 VV VA B- = then V VC D- is

(A) 5 V- (B) V2

(C) V3 (D) V6

Common Data For Q. 15 and 16 :

With 10 Vdc connected at port A in the linear nonreciprocal two-port network

shown below, the following were observed :(i) 1 Wconnected at port Bdraws a current of 3 A

(ii) 2.5 Wconnected at port Bdraws a current of 2 A

MCQ 2.15 With 10 Vdc connected at port A , the current drawn by 7 Wconnected at portBis

(A) 3/7 A (B) 5/7 A

(C) 1 A (D) 9/7 A

MCQ 2.16 For the same network, with 6 Vdc connected at port A , 1 Wconnected at portBdraws 7/3 .A If 8 Vdc is connected to port A , the open circuit voltage at portBis(A) 6 V (B) 7 V

(C) 8 V (D) 9 V

2011 ONE MARK

MCQ 2.17 In the circuit shown below, the Norton equivalent current in amperes with respectto the terminals P and Q is


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(A) 6.4 .j 4 8- (B) 6.56 .j 7 87-

(C) 10 0j+ (D) 16 0j+

MCQ 2.18 In the circuit shown below, the value of RL such that the power transferred toRL is maximum is

(A) 5 W (B) 10 W

(C) 15 W (D) 20 W

MCQ 2.19 The circuit shown below is driven by a sinusoidal input ( / )cosv V t RC i p= . Thesteady state output vois

(A) ( /3) ( / )cosV t RC p (B) ( /3) ( / )sinV t RC p

(C) ( /2) ( / )cosV t RC p (D) ( /2) ( / )sinV t RC p

2011 TWO MARKS

MCQ 2.20 In the circuit shown below, the current I is equal to

(A) 1.4 A0c+ (B) 2.0 A0c+

(C) 2.8 0 Ac+ (D) 3.2 0 Ac+

MCQ 2.21 In the circuit shown below, the network N is described by the following Y matrix:.

.

.

.

S

S

S

SY

0 1

0 01

0 01

0 1=

-> H. the voltage gainVV

1

2 is


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(A) 1/90 (B) 1/90

(C) 1/99 (D) 1/11

MCQ 2.22 In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with thevoltage polarity as indicated. The switch is closed at time 0t= . The current ( )i t

at a time tafter the switch is closed is

(A) ( ) 15 ( 2 10 )exp Ai t t3#= -

(B) ( ) 5 ( 2 10 )exp Ai t t3#= -

(C) ( ) 10 ( 2 10 )exp Ai t t3#= -

(D) ( ) 5 ( 2 10 )exp Ai t t3#=- -

2010 ONE MARK

MCQ 2.23 For the two-port network shown below, the short-circuit admittance parametermatrix is

(A) S4

2

2

4-

-> H (B) . . S10 5 0 51- -> H(C)

.

.S

1

0 5

0 5

1> H (D) S4

2

2

4> HMCQ 2.24 For parallel RL Ccircuit, which one of the following statements is NOT correct ?

(A) The bandwidth of the circuit decreases if Ris 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 minimumvalue.


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2010 TWO MARKS

MCQ 2.25 In the circuit shown, the switch Sis open for a long time and is closed at t 0= .The current ( )i t for t 0$ +is

(A) ( ) 0.5 0.125 Ai t e t1000= - - (B) ( ) 1.5 0.125 Ai t e t1000= - -

(C) ( ) 0.5 0.5 Ai t e t1000= - - (D) ( ) 0.375 Ai t e t1000= -

MCQ 2.26 The current I in the circuit shown is

(A) 1 Aj- (B) 1 Aj

(C) 0 A (D) 20 A

MCQ 2.27

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 ONE MARK

MCQ 2.28 In the interconnection of ideal sources shown in the figure, it is known that the60 V source is absorbing power.

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


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(A) 10 A (B) 13 A

(C) 15 A (D) 18 A

MCQ 2.29 If the transfer function of the following network is

( )( )

V s

V s

i

o sCR21=

+

The value of the load resistance RL is

(A)R

4 (B)R

2

(C) R (D) R2

MCQ 2.30 A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time the battery delivers a constant current of2 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?

(A) 220 J (B) 12 kJ

(C) 13.2 kJ (D) 14.4 J

GATE 2009 TWO MARK

MCQ 2.31 An AC source of RMS voltage 20 V with internal impedance ( )Z j1 2s W= + feeds a load of impedance ( )Z j7 4L W= + in the figure below. The reactive powerconsumed by the load is

(A) 8 VAR (B) 16 VAR

(C) 28 VAR (D) 32 VAR

MCQ 2.32 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


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(A) 0.2 ( )e u tt125- mA (B) 20 ( )e u tt1250- mA

(C) 0.2 ( )e u tt1250- mA (D) 20 ( )e u tt1000- mA

MCQ 2.33 In the circuit shown, what value of RL maximizes the power delivered to RL ?

(A) .2 4 W (B)38 W

(C) 4 W (D) 6 W

MCQ 2.34 The time domain behavior of an RL circuit is represented by

L dtdi

Ri+ (1 ) ( )sinV B e t u t /Rt L

0= + -

.

For an initial current of ( )iR

V0 0= , the steady state value of the current is given

by

(A) ( )i tR

V0" (B) ( )i tR

V2 0"

(C) ( ) ( )i tR

VB10" + (D) ( ) ( )i t

R

VB

2 10" +

GATE 2008 ONE MARK

MCQ 2.35 In the following graph, the number of trees ( )P and the number of cut-set ( )Q are

(A) ,P Q2 2= = (B) ,P Q2 6= =

(C) ,P Q4 6= = (D) ,P Q4 10= =

MCQ 2.36 In the following circuit, the switch Sis closed at t 0= . The rate of change ofcurrent ( )

dtdi 0+ is given by


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(A) 0 (B)L

R Is s

(C)( )

L

R R Is s+ (D) 3

GATE 2008 TWO MARKS

MCQ 2.37 The Thevenin equivalent impedance Zth between the nodes P and Q in thefollowing circuit is

(A) 1 (B) ss

1 1+ +

(C) ss

2 1+ + (D)s s

s s

2 11

2

2

+ +

+ +

MCQ 2.38 The driving point impedance of the following network is given by

( )Z s ..

s s

s

0 1 20 2

2= + +

The component values are

(A) 5 , 0.5 , 0.1L R CH FW= = =

(B) . , 0.5 ,L R C0 1 5H FW= = =

(C) 5 , , 0.1L R C2H FW= = =

(D) . , ,L R C0 1 2 5H FW= = =

MCQ 2.39 The circuit shown in the figure is used to charge the capacitor C alternatelyfrom two current sources as indicated. The switches S1and S2are mechanicallycoupled and connected as follows:

For ( )nT t n T 2 2 1# # + , ( , , ,..)n 0 1 2= S1to P1and S2to P2For ( ) ( ) ,n T t n T 2 1 2 2# #+ + ( , , ,...)n 0 1 2= S1to Q1and S2to Q2


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Assume that the capacitor has zero initial charge. Given that ( )u t is a unit step

function , the voltage ( )v tc across the capacitor is given by

(A) ( ) ( )tu t nT 1 nn 1

- -3

=

/

(B) ( ) ( ) ( )u t u t nT 2 1 nn 1

+ - -3

=

/

(C) ( ) ( ) ( )( )tu t u t nT t nT 2 1 nn 1

+ - - -3

=/

(D) . .e e T0 5 0 5( ) ( )t nT t nT n

2 2

1

- + -3

- - - -

=6 @/

Common Data For Q.40 and 41 :

The following series RL Ccircuit with zero conditions is excited by a unitimpulse functions ( )td .

MCQ 2.40 For t 0> , the output voltage v tC his(A) e e

32 t t21 23-

-^ h (B) te t3

221-

(C) cose t3

223t

21- c m (D) sine t

32

23t

21- c m

MCQ 2.41 For t 0> , the voltage across the resistor is

(A) e e3

1 t t23

21

- -_ i(B) cos sine t t

23

31

23t

21

-- c cm m= G

(C) sine t3

223t

21- c m

(D) cose t3

223t

21- c m

Statement for linked Answers Questions 42 and 43:

A two-port network shown below is excited by external DC source. The voltageand the current are measured with voltmeters ,V V1 2and ammeters. ,A A1 2(all

assumed to be ideal), as indicated


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

MCQ 2.42 The z-parameter matrix for this network is

(A).

.

.

.

1 5

4 5

1 5

1 5= G (B).

.

.

.

1 5

1 5

4 5

4 5= G(C)

.

.

.

.

1 5

1 5

4 5

1 5= G (D)

.

.

.

.

4 5

1 5

1 5

4 5= GMCQ 2.43 The h-parameter matrix for this network is(A)

.

3

1

3

0 67

-

-= G (B) .

3

3

1

0 67

- -= G(C)

.

3

1

3

0 67= G (D) .3

3

1

0 67- -= G

GATE 2007 ONE MARK

MCQ 2.44 An independent voltage source in series with an impedance Z R j X s s s= + delivers

a maximum average power to a load impedance ZL when(A) Z R j X L s s= + (B) Z RL s=

(C) Z j XL s= (D) Z R j X L s s= -

MCQ 2.45 The RCcircuit shown in the figure is

(A) a low-pass filter (B) a high-pass filter

(C) a band-pass filter (D) a band-reject filter

GATE 2007 TWO MARKS

MCQ 2.46 Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of

Filter 1 be B1and that of Filter 2 be .B2 the valueB

B

2

1 is


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(A) 4 (B) 1

(C) 1/2 (D) 1/4

MCQ 2.47 For the circuit shown in the figure, the Thevenin voltage and resistance lookinginto X Y- are

(A) , 2V34 W (B) 4 ,V 3

2 W

(C) V,34

32 W (D) 4 , 2V W

MCQ 2.48 In the circuit shown, vCis 0 volts at t 0= sec. For t 0> , the capacitor current( )i tC , where tis in seconds is given by

(A) . ( )exp t0 50 25- mA

(B) . ( )exp t0 25 25- mA(C) 0.5 ( 12.5 )exp t0 - mA

(D) . ( . )exp t0 25 6 25- mA

MCQ 2.49 In the ac network shown in the figure, the phasor voltage VAB(in Volts) is

(A) 0 (B) 5 30c+

(C) .12 5 30c+ (D) 17 30c+

GATE 2006 TWO MARKS

MCQ 2.50 A two-port network is represented by A B C D parameters given by

V

I

1

1= G A

C

B

D

V

I

2

2=

-= =G G

If port-2 is terminated by RL , the input impedance seen at port-1 is given by(A)

C D R

A B R

L

L

++ (B)

B R D

A R C

L

L

++

(C)B R C

D R A

L

L

++ (D)

D CR

B A R

L

L

++


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MCQ 2.51 In the two port network shown in the figure below, Z12and Z21and respectively

(A) reand r0b (B) 0 and r0b-

(C) 0 and rob (D) reand r0b-

MCQ 2.52 The first and the last critical frequencies (singularities) of a driving pointimpedance function of a passive network having two kinds of elements, are a poleand a zero respectively. The above property will be satisfied by

(A) RL network only (B) RCnetwork only(C) L Cnetwork only (D) RCas well as RL networks

MCQ 2.53 A 2 mH inductor with some initial current can be represented as shown below,where sis the Laplace Transform variable. The value of initial current is

(A) 0.5 A (B) 2.0 A

(C) 1.0 A (D) 0.0 A

MCQ 2.54 In the figure shown below, assume that all the capacitors are initially uncharged.

If ( ) ( )v t u t 10i = Volts, ( )v to is given by

(A) 8e / .t 0 004- Volts (B) 8(1 )e / .t 0 004- - Volts

(C) ( )u t8 Volts (D) 8 Volts

MCQ 2.55 A negative resistance Rnegis connected to a passive network N having driving

point impedance as shown below. For ( )Z s2 to be positive real,


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(A) ( ),ReR Z jneg 1 6# w w (B) ( ) ,R Z jneg 1 6# w w

(C) ( ),ImR Z jneg 1 6# w w (D) ( ),R Z jneg 1 6+# w w

GATE 2005 ONE MARK

MCQ 2.56 The condition on ,R L and Csuch that the step response ( )y t in the figure hasno oscillations, is

(A) RCL

21

$ (B) RCL

$

(C) RCL2$ (D) R

L C

1=

MCQ 2.57 The A B C D parameters of an ideal :n 1transformer shown in the figure are

n

x0

0> H

The value of xwill be

(A) n (B)n1

(C) n2 (D)n

12

MCQ 2.58 In a series RL C circuit, 2R kW= , L 1= H, and C4001 Fm= The resonant

frequency is

(A) 2 104# Hz (B) 101 4p

# Hz

(C) 104Hz (D) 2 104p # Hz

MCQ 2.59 The maximum power that can be transferred to the load resistor RL from thevoltage source in the figure is

(A) 1 W (B) 10 W

(C) 0.25 W (D) 0.5 W


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MCQ 2.60 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 TWO MARKS

MCQ 2.61 For the circuit shown in the figure, the instantaneous current ( )i t1 is

(A) 902

10 3c A (B)

210 3 90c- A

(C) 5 60c A (D) 5 60c- A

MCQ 2.62 Impedance Zas shown in the given figure is

(A) j29 W (B) j9 W

(C) j19 W (D) j39 W

MCQ 2.63 For the circuit shown in the figure, Thevenins voltage and Thevenins equivalent

resistance at terminals a b- is

(A) 5 V and 2 W (B) 7.5 V and .2 5 W

(C) 4 V and 2 W (D) 3 V and .2 5 W

MCQ 2.64 If R R R R 1 2 4= = = and .R R1 13= in the bridge circuit shown in the figure, thenthe reading in the ideal voltmeter connected between a and b is


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(A) .0 238V (B) 0.138 V

(C) .0 238- V (D) 1 V

MCQ 2.65 The hparameters of the circuit shown in the figure are

(A).

.

.

.

0 1

0 1

0 1

0 3-= G (B) .10

1

1

0 05

-= G

(C)

30

20

20

20= G (D) .10

1

1

0 05-= GMCQ 2.66 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 V2at time t 2= sec is

(A) 3 V (B) 3- V

(C) 4 V (D) 4- V

GATE 2004 ONE MARK

MCQ 2.67 Consider the network graph shown in the figure. Which one of the following is

NOT a tree of this graph ?

(A) a (B

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