gate ece 2013 paper

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No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical,photocopying, or otherwise without the prior permission of the author.

GATE SOLVED PAPER

Electronics & Communication

2013

Copyright By NODIA & COMPANY

Information contained in this book has been obtained by authors, from sources believes to be reliable. However,neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor itsauthors shall be responsible for any error, omissions, or damages arising out of use of this information. This bookis published with the understanding that Nodia and its authors are supplying information but are not attemptingto render engineering or other professional services.

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Q.1 to Q.25 carry one mark each.

Q. 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

Sol. 1 Let A denotes the position of switch at ground floor and Bdenotes 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 Bare 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

Y A B5=

i.e., the XOR gate

Hence correct option is (C)

Q. 2 Consider a vector field A rv v^ h. The closed loop line integral A dl:v v# can beexpressed as

(A) A ds# :d v v^ h## over the closed surface bounded by the loop(B) A dv:d v

^ h###over the closed volume bounded by the loop

(C) A dv:d v^ h### over the open volume bounded by the loop(D) A ds# :d v v^ h## over the open surface bounded by the loop

Sol. 2 Stokes theorem states that the circulation a vector field Avaround a closed path

lis equal to the surface integral of the curl of Avover the open surface Sbounded

by l.

i.e., A dl:v v# A ds# :d= v v^ h##Here, line integral is taken across a closed path which is denoted by a small

circle on the integral notation where as, the surface integral of A#d v^ his takenover open surface bounded by the loop.Hence correct option is (D)


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GATE SOLVED PAPER - EC 2013

Q. 3 Two systems with impulse responses h t1^ hand h t2^ hare connected in cascade.Then the overall impulse response of the cascaded system is given by

(A) product of h t1^ hand h t2^ h(B) sum of h t

1^ hand h t

2^ h(C) convolution of h t1^ hand h t2^ h(D) subtraction of h t2^ hfrom h t1^ h

Sol. 3 If the two systems with impulse response h t1^ h and h t2^ h are connected incascaded configuration as shown in figure, then the overall response of the system

is the convolution of the individual impulse responses.


Q. 4 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

Sol. 4 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.

Hence correct option is (A)

Q. 5 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

Sol. 5 In IC technology, dry oxidation as compared to wet oxidation produces superior

quality oxide with a lower growth rate

Hence correct option is (D)

Q. 6 The maximum value of quntil which the approximation sin .q qholds to

within 10% error is

(A) 10c (B) 18c

(C) 50c (D) 90c

Sol. 6 Here, as we know

sinLim0

q"q

0.

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


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q 18 .18180

0314#c cc

p= = =

sinq .sin18 0 309c= =

%error.

. . % . %0309

0314 0309 100 049#= - =

Now, we check it for 50cq =

q .50 50180

0873#c cc

p= = =

sinq .sin50 077c= =

%error.

0.77 0.873 . %0873

1225= - =-

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

the approximation

sinq. q

Hence correct option is (B)

Q. 7 The divergence of the vector field A xa ya za x y z= + +v

t t t is(A) 0 (B) 1/ 3

(C) 1 (D) 3

Sol. 7 Given, the vector field

Av xa ya za x y z= + +v v v

so,

A$d v(Divergence of Av)x

A

y

A

z

Ax y z2

2

2

2

2

2= + +

1 1 1 3= + + =


Q. 8 The impulse response of a system is h t tu t =^ ^h h. For an input u t 1-^ h, theoutput is

(A) t u t2

2 ^ h (B) t t u t21

1-

-^ ^h h

(C)t

u t2

11

2--

^ ^h h (D) t u t2 1 12- -^ h

Sol. 8 Given, the input

x t^ h u t 1= -^ hIts laplace transform is

X s^ h se

s

=

-

The impulse response of system is given

h t^ h t u t= ^ hIts Laplace transform is

H s^ hs

12=

Hence, the overall response at the output is

Y s^ h X s H s = ^ ^h h

s

e s

3=-

its inverse laplace transform is


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y t^ h t u t21

12

= -

-^ ^h h


Q. 9 The Bode plot of a transfer function G s^ his shown in the figure below.

The gain log G s20 ^_ hiis 32dBand 8dB- at 1 /rad sand 10 /rad srespectively.The phase is negative for all w. Then G s^ his(A) .

s398 (B) .

s

3982

(C)s32 (D)

s

322

Sol. 9

From the given plot, we obtain the slope as Slope

log loglog log

w w

G G20 20

2 1

2 1=--

From the figure

logG20 2 8dB=-

logG20 1 32dB=

and 1w 1 /rad s=

2w 10 /rad s=

So, the slope is

Slopelog log

8 3210 1

=-

- -

40 /dB decade=-Therefore, the transfer function can be given as

G s^ hS

k2=

at 1w =

G jw^ h w

kk2= =

In decibel,

log G j20 w^ h logk20 32= =or, k .10 398

3220

= =

Hence, the Transfer function is

G s^ h .s

k

s

3982 2= =



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Q. 10 In the circuit shown below what is the output voltage Vout^ hif a silicontransistor Qand an ideal op-amp are used?

(A) 15V- (B) 0.7V-

(C) 0.7V+ (D) 15V+

Sol. 10 For the given ideal op-amp, negative terminal will be also ground (at zero voltage)

and so, the collector terminal of the B JT will be at zero voltage.

i.e., VC 0volt=

The current in 1kWresistor is given by

I 1k5 0

W= - 5mA=

This current will flow completely through the B JT since, no current will flow

into the ideal op-amp ( /I Presistance of ideal op-amp is infinity). So, for B JT

we have

VC 0=

VB 0= IC 5mA=

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 0.7volts=

& V VB E- .07=

& V0 out- .07=

or, Vout 0.7volt=-


Q. 11 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

Sol. 11 In the equivalent star connection, the resistance can be given as


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RC R R RR R

a b c

b a=+ +

RB R R RR R

a b c

a c=+ +

RA R R RR R

a b c

b c=+ +

So, if the delta connection components Ra, Rband Rcare scaled by a factor k

then

RAl kR kR kR k R k R

a b c

b c=

+ +^ ^h h

kk

R R RR R

a b c

b c2

=+ +

k RA=

hence, it is also scaled by a factor k


Q. 12 For 8085 microprocessor, the following program is executed.

MVI A, 05H;

MVI B, 05H;

PTR: ADD B;

DCR B;

J NZ PTR;

ADI 03H;

HLT;

At the end of program, accumulator contains(A) 17H (B) 20H

(C) 23H (D) 05H

Sol. 12 The program is being executed as follows

MVI A, 0.5H; A H05=

MVI B, 0.5H; B H05=

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 03 14 03 17H HA= + = + =



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Q. 13 The bit rate of a digital communication system is /kbits sR . T he modulation

used is 32-QAM. The minimum bandwidth required for ISI free transmission is

(A) / 10HzR (B) / 10kHzR(C) /5HzR (D) / 5kHzR

Sol. 13 In ideal Nyquist Channel, bandwidth required for ISI (Inter Symbol reference)

free transmission is

W 2

Rb=

Here, the used modulation is QA M32- (Quantum Amplitude modulation

i.e., q 32=

or 2v 32=

v 5bits=

So, the signaling rate (sampling rate) is

RbR5

= (R"given bit rate)

Hence, for ISI free transmission, minimum bandwidth is

W kHzR R2 10

b= =


Q. 14 For a periodic signal /sin cos sinv t t t t 30 100 10 300 6 500 4p= + + +^ ^h h, thefundamental frequency in /rad s

(A) 100 (B) 300

(C) 500 (D) 1500

Sol. 14 Given, the signal

v t^ h sin cos sint t t30 100 10 300 6 500 4= + + + p^ hSo we have

1w 100 /rad s=

2w 00 /rad s3=

3w 00 /rad s5=

Therefore, the respective time periods are

T1 sec2

100

2

1w

p p= =

T2 sec2

3002

2wp p= =

T3 sec5002p=

So, the fundamental time period of the signal is

L.C.M. ,T T T1 2 3^ h, ,

2 ,2 ,2

H CF

L CM

100 300 500

p p p= ^

h

h

or, T0 1002p=

Hence, the fundamental frequency in rad/ sec is

0w 100 /rad s10

2p

= =Hence correct option is (A)


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Q. 15 In a voltage-voltage feedback as shown below, which one of the following

statements is TRUE if the gain kis increased?

(A) The input impedance increases and output impedance decreases

(B) The input impedance increases and output impedance also increases

(C) The input impedance decreases and output impedance also decreases

(D) The input impedance decreases and output impedance increases

Sol. 15 The /i pvoltage of the system is given as

Vin V Vf1= +

V k Vout1= +

V k A V 1 0 1= + V A Vout 0 1=^ h V k A11 0= +^ hTherefore, if kis increased then input voltage is also increased so, the input

impedance increases. Now, we have

Vout A V0 1=

Ak A

V

1in

00

=+^ h

k A

A V

1in

0

0=+^ h

Since, Vi nis independent of kwhen 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.


Q. 16 A band-limited signal with a maximum frequency of 5kHzis to be sampled.According to the sampling theorem, the sampling frequency which is not valid

is

(A) 5kHz (B) 12kHz

(C) 15kHz (D) 20kHz

Sol. 16 Given, the maximum frequency of the band-limited signal

fm 5kHz=

According to the Nyquist sampling theorem, the sampling frequency must be

greater than the Nyquist frequency which is given as

fN 2 2 5 10kHzfm #= = =

So, the sampling frequency fsmust satisfy


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fs fN$

fs 10kHz$

only the option (A) doesnt satisfy the condition therefore, 5kHzis not a valid

sampling frequency.Hence correct option is (A)

Q. 17 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

Sol. 17 In a MOSFET operating in the saturation region, the channel length modulation

effect causes a decrease in output resistance.Hence correct option is (D)

Q. 18 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 jwaxis

(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 jwaxis.

Sol. 18 For a system to be casual, the R.O.C of system transfer function H s

^ hwhich is

rational should be in the right half plane and to the right of the right most pole.

For the stability of L T I 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 L T I 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.


Q. 19 The minimum eigen value of the following matrix is3

5

2

5

12

7

2

7

5

R

T

SS

SS

V

X

WW

WW(A) 0 (B) 1

(C) 2 (D) 3

Sol. 19 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 among

the given options is

l 0=

We check the characteristic equation of matrix for this eigen value

A Il- A= (for 0l = )

3

5

2

5

12

7

2

7

5

=


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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 value

is

l 0=


Q. 20 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

Sol. 20 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 sign

change from a1^ hto a0^ hin the Routh matrix 1stcolumn. Also, there will be acorresponding 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.PHence correct option is (D)

Q. 21 Assuming zero initial condition, the response y t^ hof the system given below toa unit step input u t^ his

(A) u t^ h (B) tu t^ h(C)

tu t2

2

^ h (D) e u tt-

^ hSol. 21 The Laplace transform of unit step funnis

U s^ hs1=

So, the O/ P of the system is given as

Y s^ hs s1 1= b bl l

s

12=

For zero initial condition, we check

u t

^ h

dt

dy t=

^ h& U s^ h SY s y 0= -^ ^h h


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& U s^ h ss

y1 02= -c ^m h

or, U s^ hs1= y 0 0=^^ h h

Hence, the O/ P is correct which is Y s^ h

s

12=

its inverse Laplace transform is given by

y t^ h t u t= ^ hHence correct option is (B)

Q. 22 The transfer functionV s

V s

1

2

^^

hhof the circuit shown below is

(A) .s

s1

05 1+

+ (B)ss

23 6

++

(C)ss

12

++ (D)

ss

21

++

Sol. 22 For the given capacitance, C F100m= in the circuit, we have the reactance.

XC sc1=

s 100 101 6# #=-

s

104=

So,

V s

V s

1

2

^^

hh

10 10

s s

s10 10 10

44

44

4=+ +

+

ss

21=

++


Q. 23 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 ofthe source, its value in Wshould be

(A) 3 (B) 4

(C) 5 (D) 7

Sol. 23 For the purely resistive load, maximum average power is transferred when

RL R XT h T h 2 2= +

where R j XT h T h + is the equivalent thevinin (input) impedance of the circuit.

Hence, we obtain

RL 4 32 2= +

5WHence correct option is (C)


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Q. 24 The return loss of a device is found to be20dB. 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

Sol. 24 Given, the return loss of device as 20dB

i.e.,in dB

G ^ h 20dB=- (loss)

or, 20logG 20=-

& G .10 011= =-

Therefore, the standing wave ration is given by

VSWR1

1

G

G=

-

+

.

1 ... .

1 0101

0911 122=

-+ = =

Hence correct option is (A)Q. 25 Let g t e t

2

= p-^ h , and h t^ his a filter matched to g t^ h. If g t^ his applied as inputto h t^ h, then the Fourier transform of the output is(A) e f

2p- (B) e /f 22p-

(C) e fp- (D) e f22p-

Sol. 25 The matched filter is characterized by a frequency response that is given as

H f^ h * expG f j fT 2p= -^ ^h hwhere g t^ h G ff ^ hNow, consider a filter matched to a known signal g t

^ h. The fourier transform of

the resulting matched filter output g t0^ hwill be G f0^ h H f G f = ^ ^h h * expG f G f j fT 2p= -^ ^ ^h h h expG f j fT 22 p= -^ ^h hT is duration of g t^ hAssume exp j fT2 1p- =^ hSo, G f0^ h G f 2= _ iSince, the given Gaussian function is

g t

^ h e t

2

= p-

Fourier transform of this signal will be

g t^ h e et f f2 2= p p- - G f= ^ hTherefore, output of the matched filter is

G f0^ h e f 22= p-Hence No Option is correct

Q.26 to Q.55 carry two marks each.

Q. 26 Let U andV be two independent zero mean Gaussain random variables of

variances

4

1and

9

1respectively. The probability P V U3 2F

^ his

(A) 4/ 9 (B) 1/ 2

(C) 2/ 3 (D) 5/ 9


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Sol. 26 Given, random variables U andV with mean zero and variances41and

91

i.e., U V 0= =

u2s

41=

and v2s

91=

so, P U 0$^ h 21=and P V 0$^ h 21=The distribution is shown in the figure below

f uu h2

e1

u

u

2 2 u2

p s= s

-

f vv h e2

1

v

v

2 2 v2

p s= s

-

We can express the distribution in standard form by assuming

X 2u

Y

uU

0

u 2s

= - = =

and Y vYv

V0 3

v 3s= - = =

for which we have

X U2 0= =

Y V2 0= =

and X2 4U 12= =

also, Y2 9V 12= =

Therefore, X Y- is also a normal random variable with

X Y- 0=

Hence,

P X Y 0$-^ h P X Y 0 21#= - =^ hor, we can say

P U V2 3 0#-^ h 21=Thus, P V U3 2$^ h 21=Hence correct option is (B)

Q. 27 Let A be an m n# matrix and Ban n m# matrix. It is given that

determinant I A Bm + =^ h determinant I B An +^ h, where Ikis the k k# identitymatrix. Using the above property, the determinant of the matrix given below is

21

1

1

12

1

1

11

2

1

11

1

2

R

T

SSSSSS

V

X

WWWWWW


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(A) 2 (B) 5

(C) 8 (D) 16

Sol. 27 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

SSSSSS

V

X

WWWWWW

1 1 1 16 @

Hence, we get

A

11

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

SSSSSS

V

X

WWWWWW

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.


Q. 28 In the circuit shown below, if the source voltage 100 53.13 VVS c+= then the

Thevenins equivalent voltage in Volts as seen by the load resistance RL is


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

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

Sol. 28 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

I2 0=

or, 40j I2 0=

i.e., the dependent source in loop 1is short circuited. Therefore,

VL 1 jj V

4 34 s

=+

^ h

VT h V10 L 1= 100 53.13jj

4 340

c=+

.

100 53.135 5313

40 90

c

cc=

800 90c=


Q. 29 The open-loop transfer function of a dc motor is given asV s

s

s1 1010

a

w=

+^^

hh

.

When connected in feedback as shown below, the approximate value of Kathat

will reduce the time constant of the closed loop system by one hundred times as

compared to that of the open-loop system is

(A) 1 (B) 5

(C) 10 (D) 100

Sol. 29 Given, open loop transfer function


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G s^ hs

K

s

K1 1010 a a

101= +

=+

By taking inverse Laplace transform, we have

g t

^ h e t10

1

= -

Comparing with standard form of transfer function, A e /t t- , we get the open

loop time constant,

olt 10=

Now, we obtain the closed loop transfer function for the given system as

H s^ h1 10 10G s

G s

s KK

110

a

a=+

=+ +^

^h

h

s K

K

a

a

101= + +^ h

By taking inverse laplace transform, we get

h t^ h .k ea k ta 101= - +^ hSo, the time constant of closed loop system is obtained as

clt k

1a 10

1= +

or, clt k1

a

= (approximately)

Now, given that kareduces open loop time constant by a factor of 100. i.e.,

clt 100olt=

or,k1

a

10010=

Hence, ka 10=


Q. 30 In the circuit shown below, the knee current of the ideal Zener dioide is 10mA

. To maintain 5Vacross RL , the minimum value of RL in Wand the minimum

power rating of the Zener diode in mW, respectively, are

(A) 125 and 125 (B) 125 and 250

(C) 250 and 125 (D) 250 and 250

Sol. 30


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From the circuit, we have

Is I IZ L= +

or, IZ I Is L= - (1)

Since, voltage across zener diode is 5V so, current through 100Wresistor is

obtained as

Is 0.05A10010 5= - =

Therefore, the load current is given by

IL R5

L

=

Since, for proper operation, we must have

IZ Iknes$

So, from Eq. (1), we write

0.05AR5

L

- 10mA$

50mAR5

L

- 10mA$

40mAR5

L

$

40 10 3# -

R5

L

$

140 10 3#

- R5

L#

40 10

53

# - RL#

or, 125W RL#

Therefore, minimum value of 125RL W=

Now, we know that power rating of Zener diode is given by

PR V I

maxZ Z= ^ hI maxZ his maximum current through zener diode in reverse bias. Maximumcurrrent through zener diode flows when load current is zero. i.e.,

I maxZ h .I 10010 5 005s= = - =

Therefore, PR 5 0.05W#=

250mW=


Q. 31 The following arrangement consists of an ideal transformer and an attenuator

which attenuates by a factor of 0.8. An ac voltage 100VVW X1 = is applied

across WX to get an open circuit voltageVY Z1across YZ. Next, an ac voltage

100VVY Z2 = is applied across YZ to get an open circuit voltageVW X2acrossWX. Then, /V VY Z W X 1 1, /V VW X Y Z 2 2are respectively,


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

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

Sol. 31 For the given transformer, we have

V

V

W X

.1

125=

Since,V

VY Z .08= (attenuation factor)

So,VV

W X

Y Z . .08 125 1= =^ ^h hor, VY Z VW X=

at VW X1 100V= ; VV

100100

W X

Y Z

1

1 =

at VW Z2 100V= ;

V

V

100

100

Y Z

W X

2

2 =


Q. 32 Two magnetically uncoupled inductive coils haveQfactors q1and q2at the

chosen operating frequency. Their respective resistances are R1and R2. When

connected in 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 h

Sol. 32 The quality factor of the inductances are given by

q1 RL

1

1w=

and q2 RL2

2w=

So, in series circuit, the effective quality factor is given by

QR

X

R RL L

eq

L eq

1 2

1 2w w= =++

R R

R RL

R RL

1 12 1

1 2

1

1 2

2w w

=+

+

R R

R

q

R

q

1 12

1

2

2

2 1

=

+

+

R R

q R q R

1 2

1 1 2 2=++


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Q. 33 The impulse response of a continuous time system is given by

h t t t 1 3d d= - + -

^ ^ ^h h h. The value of the step response at t 2= is

(A) 0 (B) 1

(C) 2 (D) 3

Sol. 33 Given, the impulse response of continuous time system

h t^ h t t1 3d d= - + -^ ^h hFrom the convolution property, we know

x t t t 0d -*^ ^h h x t t0= -^ hSo, for the input

x t^ h u t= ^ h(Unit step funn)The output of the system is obtained as

y t

^ h u t h t =

*^ ^h h u t t t 1 3d d= - + -*^ ^ ^h h h6 @ u t u t 1 3= - + -^ ^h hat t 2=

y 2^ h u u2 1 2 3= - + -^ ^h h 1=


Q. 34 The small-signal resistance (i.e., /dV dI B D) in kWoffered by the n-channel

MOSFET M shown in the figure below, at a bias point of 2VVB= is (device

data for M: device transconductance parameter / 40 /A Vk C W L'

N n x02

mm= =^ h, threshold voltage 1VVT N= , and neglect body effect and channel lengthmodulation effects)

(A) 12.5 (B) 25

(C) 50 (D) 100

Sol. 34 Given,

VB V2=

VT N V1=

So, we have

Drain voltage VD 2volt=

VG 2volt=

VS 0= (Ground)

Therefore, VGS V2 > T N=

and VDS 2 V V> GS T N = -So, the MOSFET is in the saturation region. Therefore, drain current is

ID k V VN GS T N 2= -^ h


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or, ID k V 1N B2

= -^ hDifferentiating both side with respect to ID

1 k VdI

dV2 1N BD

B= -

^ hSince, VBQ 2volt= (at D.C. Voltage)Hence, we obtain

dIdV

D

B k V2 1

1

N B

=-^ h

2 40 10 2 1

16

# # #=

-- ^ h 12.5 103 W#=

12.5kW=


Q. 35 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 then-channel MOSFET M, the transconductance 1 /mA Vgm= , and body effect

and channel length modulation effect are to be neglected. T he lower cutoff

frequency in HZ of the circuit is approximately at

(A) 8 (B) 32

(C) 50 (D) 200

Sol. 35 For the given circuit, we obtain the small signal model as shown in figure below :

We obtain the node voltage at V1as

RV

RsC

Vg V

1DL

m i1 1+

++ 0=

& V1

RR

sC

g V

11

1D

L

m i=+

+

-

Therefore, the output voltageV0is obtained as

V0R

sC

V R1

L

L1=+


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R

sC

R

RR

sC

g V

1 11

1L

L

DL

m i=+ +

+

-J

L

KKK

N

P

OOO

so, the transfer function is

VV

i

0sC R R

R R sC g

1 D LD L m=

+ +

-

^ hThen, we have the pole at

C R R

1

D L

w =+^ h

It gives the lower cutoff frequency of transfer function.

i.e., 0w C R R

1

D L

=+^ h

or, f0C R R2

1

D Lp=

+^ h

2 10 20 101

6 3# # #p

= -

.797=

8Hz.


Q. 36 A system described by the differential equationdt

d y

dt

dyy t x t 5 62

2

+ + =^ ^h h. Letx t^ hbe a rectangular pulse given byx t

t

otherwise

1

0

0 2< , we need to(A) change the initial condition to y 0-^ hand the forcing function to x t2^ h(B) change the initial condition to y2 0^ hand the forcing function to x t-^ h(C) change the initial condition to j y2 0^ hand the forcing function to

j x t2^ h(D) change the initial condition to y2 0- ^ hand the forcing function to x t2- ^ h

Sol. 37 The solution of a system described by a linear, constant coefficient, ordinary, first

order differential equation with forcing function x t^ his y t^ hso, we can define afunction relating x t^ hand y t^ has below P

dt

dyQy K+ + x t= ^ h

where P, Q, K are constant. Taking the Laplace transform both the sides, we

get

P sY s Py Q Y s 0- +^ ^ ^h h h X s= ^ h ....(1)Now, the solutions becomes

y t1^ h 2y t=- ^ hor, Y s1^ h Y s2=- ^ hSo, Eq. (1) changes to

P sY s P y Q Y s 01 1 1- +^ ^ ^h h h X s1= ^ hor, 2 2P SY s P y QY s 01 1- - -^ ^ ^h h h X s1= ^ h ....(2)Comparing Eq. (1) and (2), we conclude that

X s1^ h X s2=- ^ h y 01^ h y2 0=- ^ hWhich makes the two equations to be same. Hence, we require to change the

initial condition to y2 0- ^ hand the forcing equation to x t2- ^ hHence correct option is (D)

Q. 38 Consider two identically distributed zero-mean random variables U andV

. Let the cumulative distribution functions ofU and V2 be F x^ hand G x^ hrespectively. Then, for all values of

x

(A) F x G x 0#-^ ^h h (B) F x G x 0$-^ ^h h(C) ( ) ( ) .F x G x x 0#-^ h (D) ( ) ( ) .F x G x x 0$-^ h

Sol. 38 The mean of random variablesU andV are both zero

i.e., U V 0= =

Also, the random variables are identical

i.e., f uU h f vV= ^ hor, F uU h F vV= ^ hi.e., their cdf are also same. So,

F uU h F v2V2= ^ hi.e., the cdf of random variable V2 will be also same but for any instant

V2 U$

Therefore,


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G x^ h F x= ^ hbut, x G x^ h xF x$ ^ hor, F x G x x -^ ^h h6 @ 0#Hence correct option is (C)

Q. 39 The DFT of a vector a b c d 8 Bis the vector a b g d 8 B. Consider the product

p q r s 8 B a b c d a

d

c

b

b

a

d

c

c

b

a

d

d

c

b

a

=

R

T

SSSSSS

8V

X

WWWWWW

B

The DFT of the vector p q r s 8 Bis a scaled version of(A) 2 2 2 2a b g d 9 C (B) a b g d 9 C(C) a b b d d g g a+ + + +8 B (D) a b g d 8 B

Sol. 39 Given, the DFT of vector a b c d 8 Bas . . .D F T a b c d 8 B% / a b g d = 8 BAlso, we have

p q r s 8 B a b c d a

d

c

b

b

a

d

c

c

b

a

d

d

c

b

a

=

R

T

SSSSSS

8V

X

WWWWWW

B ...(1)

For matrix circular convolution, we know

x n h n *6 6@ @h

hh

h

hh

h

hh

x

xx

0

1

2

2

0

1

1

2

0

0

1

1=

R

T

S

SSS

R

T

S

SSS

V

X

W

WWW

V

X

W

WWW

where , ,x x x0 1 2" ,are three point signals for x n6 @and similarly for h n6 @, h0, h1and h2are three point signals. Comparing this transformation to Eq(1), we get

p q r s 6 @a

b

c

d

d

a

b

c

c

d

a

b

a b c d

T

=

R

T

SSSSSS

8V

X

WWWWWW

B

a b c d a b c d T T= *6 6@ @ *

a

b

c

d

a

b

c

d

=

R

T

SSSSSS

R

T

SSSSSS

V

X

WWWWWW

V

X

WWWWWW

Now, we know that

x n x n 1 2*6 6@ @ X k X k ,D F T D F T 1 2= 6 6@ @So,

*

a

b

c

d

a

b

c

d

R

T

SSSS

SS

R

T

SSSS

SS

V

X

WWWW

WW

V

X

WWWW

WW

*

a

b

g

d

a

b

g

d

=

R

T

SSSS

SS

R

T

SSSS

SS

V

X

WWWW

WW

V

X

WWWW

WW 2 2 2 2a b g d = 9 C


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Q. 40 The signal flow graph for a system is given below. The transfer function

U s

Y s

^^

hh

for this system is

(A)s s

s

5 6 21

2+ ++ (B)

s s

s

6 21

2+ ++

(C)s s

s

4 2

1

2+ +

+ (D)s s5 6 2

1

2+ +Sol. 40 For the given SFG, we have two forward paths

Pk1 s s s1 11 1 2= =- - -^ ^ ^ ^h h h h

Pk2 s s1 1 11 1= =- -^ ^ ^ ^h h h h

since, all the loops are touching to the paths Pk1and Pk2so,

k1D 1k2D= =

Now, we have

D 1= -(sum of individual loops)

+(sum of product of nontouching loops)

Here, the loops are L 1 4 1 4= - =-^ ^h h L 2 s s4 4

1 1= - =- -^ ^h h L 3 s s s2 2

1 1 2= - =-- - -^ ^ ^h h h L 4 s s2 1 2

1 1= - =-- -^ ^ ^h h hAs all the loop , ,L L L1 2 3and L 4are touching to each other so,

D L L L L1 1 2 3 4= - + + +^ h s s s1 4 4 2 21 2 1= - - - - -- - -^ h s s5 6 21 2= + +

From Masons gain formulae

U s

Y s

^ hh Pk kD

S D=

s s

s s

5 6 21 22 1

=+ +

+- -

- -

s s

s

5 6 21

2= + ++


Q. 41 In the circuit shown below the op-amps are ideal. Then, Voutin Volts is


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(A) 4 (B) 6

(C) 8 (D) 10

Sol. 41

For the given ideal op-Amps we can assume

V2- V V2 2= =

+ (ideal)

V1+ V V1 1= =

- (ideal)

So, by voltage division

V1V

21out #=

Vout V2 1=

and, as the I/ P current in Op-amp is always zero therefore, there will be novoltage drop across 1KWin I I op-amp

i.e., V2 1V=

Therefore,

V V1

1 2- V

122

= - -^ h

& V 11- 1 2= +

or, V1 4=

Hence,

Vout 2 8voltV1= =


Q. 42 In the circuit shown below, Q1has negligible collector-to-emitter saturation


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voltage and the diode drops negligible voltage across it under forward bias. I f

Vccis 5V+ , X andY are digital signals with 0Vas logic 0 and Vccas logic 1,

then the Boolean expression for Zis

(A) X Y (B) X Y

(C) X Y (D) X Y

Sol. 42 For the given circuit, we can make the truth table as below

X Y Z

0

0

1

1

0

1

0

1

0

1

0

0

Logic 0means voltage is 0voltv = and logic 1means voltage is 5volt

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, zdirectly 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 X Y=


Q. 43 A voltage sin t1000 w Volts is applied across Y Z. Assuming ideal diodes, the

voltage measured acrossW X in Volts, is


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(A) sin tw (B) /sin sint t 2w w+_ i(C) /sin sint t 2w w-^ h (D) 0 for all t

Sol. 43 Given, the input voltage

VY Z sin t100 w=

For ve+ half cycle

VY Z 0>

i.e., VY is a higher voltage than VZSo, the diode will be in cutoff region. Therefore, there will no voltage difference

between X andW node.

i.e., VW X 0=

Now, for ve- half cycle all the four diodes will active and so, X andW

terminal is short circuited

i.e., VW X 0=

Hence, VW X 0= for all t

Hence correct option is (D)Q. 44 Three capacitorsC1, C2and C3whose values are 10 Fm , 5 Fm , and 2 Fm

respectively, have breakdown voltages of 10V, 5Vand 2Vrespectively. For the

interconnection shown below, the maximum safe voltage in Volts that can be

applied across the combination, 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

Sol. 44


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Consider that the voltage across the three capacitors C1, C2and C3areV1, V2and

V3respectively. So, we can write

VV

3

2CC

2

3= ....(1)

Since, Voltage is inversely proportional to capacitanceNow, given that

C1 F10m= ; 10VV max1 =^ h C2 F5m= ; 5VV max2 =^ h C3 F2m= ; 2VV max3 =^ hSo, from Eq (1) we have

VV

3

252=

for Vmax3^ h 2=

We obtain,

V2 0.8volt52 2

5 >6 H H@ (B) u

y u

X X

X

1

1

0

1

1

1

1 1

=

-

- - +

-

= - - +

o

> >6 H H@(C)

u

y u

X X

X

1

1

0

1

1

1

1 1

=-

- - +

-

= - - -

o > >6

H H@

(D)u

y u

X X

X

1

0

1

1

1

1

1 1

=- -

- +

-

= - -

o > >6

H H@

Sol. 54 For the shown state diagram we can denote the states x1, x2as below


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So, from the state diagram, we obtain

x1o x u1=- -

x2o x u x1 1 1 1 1 1 12 1=- + - - + - -^ ^ ^ ^ ^ ^h h h h h h h x2o x x u2 1=- + +and y x x u1 1 1 1 1 1 1 1 1 12 1= - + - - + - -^ ^ ^ ^ ^ ^ ^h h h h h h h h h h x x u1 2= - +

Hence, in matrix form we can write the state variable equations

x

x

1

2

o

o> H x

x u

1

1

0

1

1

11

2=

-

- +

-> > >H H Hand y

x

x u1 1

1

2= - +8 >B H

which can be written in more general form as

Xo

X

1

1

0

1

1

1=

-

- +

-

> >H H y X u1 1= - +8 BHence correct option is (A)

Q. 55 The state transition matrix eA t of the system shown in the figure above is

(A)e

te e

0t

t t

-

- -> H (B) ete e

0t

t t-

-

- -> H

(C)e

e e

0t

t t

-

- -> H (D) e tee0

t t

t

-- -

-> HSol. 55 From the obtained state-variable equations

We have

A 1

1

0

1=

-

-> H

So, SI A- S

S

1

1

0

1=

+

- +> H

and SI A 1- -^ h S

S

S11 1

1

0

12=

+

+

+^ h> H

S

S S

11

11

0

11

2

= +

+ +^ h

R

T

SSSS

V

X

WWWW

Hence, the state transition matrix is obtained as

eA t L SI A1 1= -- -^ h

LS

S S

11

11

0

11

1

2

= +

+ +

-

^ h

R

T

SSSS

V

X

WWWW

Z

[

\

]]

]]

_

`

a

bb

bb

e

te e

0t t

1

=-

- -> HHence correct option is (A)


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General Aptitude (GA) Questions

Q.56 to Q.60 carry one mark each.

Q. 56 Choose the grammatically CORRECT sentence:

(A) Two and two add four

(B) Two and two become four

(C) Two and two are four

(D) T wo and two make four

Sol. 56 Two and two make four


Q. 57 Statement: You can always give me a ring whenever you need.

Which one of the following is the best inference from the above statement?

(A) Because I have a nice caller tune.

(B) Because I have a better telephone facility

(C) Because a friend in need is a friend indeed

(D) Because you need not pay towards the telephone bills when you give me a

ring

Sol. 57 You can always given me a ring whenever you need. Because a friend is need is a

friend indeed


Q. 58 In the summer of 2012, in New Delhi, the mean temperature of Monday to

Wednesday was 41C and of Tuesday to Thursday was 43 Cc . I f the temperatureon Thursday was 15% higher than that of Monday, then the temperature in Cc

on Thursday was

(A) 40 (B) 43

(C) 46 (D) 49

Sol. 58 Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively

as , , ,T T T T M T U W T H

So, from the given data we have

T T T

3H T U W + + 41= ....(1)

and T T T3

T U W T H + + 43= ....(2)

also, as the temperature on Thursday was %15 higher than that of Monday

i.e. TT H . T115 M= ....(3)

solving eq (1), (2) and (3), we obtain

TT H 46 Cc=


Q. 59 Complete the sentence: Dare .................. mistakes.

(A) commit (B) to commit

(C) committed (D) committing

Sol. 59 Dare to commit mistakes


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Q. 60 They were requested not to quarrel with others.

Which one of the following options is the closest in meaning to the word quarrel?

(A) make out (B) call out(C) dig out (D) fall out

Sol. 60 They were requested not to quarrel with others.

Quarrel has a similar meaning to fall out


Q. No. 61-65 Carry Two Marks Each

Q. 61 A car travels 8kmin the first quarter of an hour, 6kmin the second quarter

and 16kmin the third quarter. The average speed of the car in km per hour

over the entire journey is(A) 30 (B) 36

(C) 40 (D) 24

Sol. 61 Given, the distance travelled by the car in each quarter intervals as

Distance Time Duration

8 km hr41

6 km hr41

16 km hr41

Therefore, the total time taken hr41

41

41

43

= + + +

Total distance travelled 8 6 16 30km= + + =

Hence, average speedTotal timetaken

Total distancetravelled=

/3 430=

/km hr40=


Q. 62 Find the sum to nterms of the series ...10 84 734+ + +

(A) 10

9 9 1

1

n +

+

^ h (B) 8

9 9 1

1

n -

+

^ h

(C) n8

9 9 1n -+

^ h (D) n

89 9 1n 2- +^ h

Sol. 62 It will be easy to check the options for given series. From the given series.

10 84 73 ......4+ + +

We get

Sum of 1term S 101= =

Sum of 2terms S 10 84 942= = + =

and sum of 3terms S 10 84 734 8283= = + + =

Checking all the options one by one, we observe that only (D) option satisfies as

Sn n8

9 9 1n 2=

-

+^ h


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so, S1 289 9 1

102

2- + =^ h

S2 289 9 1

942= -

+ =^ h

S3 389 9 1

8283

2= -

+ =^ hHence correct option is (D)

Q. 63 Statement: There were different streams of freedom movements in colonial India

carried out by the moderates, liberals, radicals, socialists, and so on.

Which one of the following is the best inference from the above statement?

(A) The emergence of nationalism in colonial India led to our Independence

(B) Nationalism in India emerged in the context of colonialism

(C) Nationalism in India is homogeneous

(D) Nationalism in India is heterogeneous

Sol. 63 Nationalism in India is heterogeneous


Q. 64 The set of values of pfor which the roots of the equation

x x p p 3 2 1 02+ + - =^ h are of opposite sign is(A) ,03-^ h (B) ,0 1^ h(C) ,1 3^ h (D) ,0 3^ h

Sol. 64 Given, the quadratic equation

x x P P 3 2 12+ + -^ h 0=It will have the roots with opposite sign if P P 1-^ h 0-

or P 0> and 0P 1


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Thursday, Friday

Friday, Saturday

Out of these seven possibilities, only two consist a saturday. Therefore, the

probability of saturday is given as

P72=



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Answer Sheet

1. (C) 13. (B) 25. (*) 37. (D) 49. (B) 61. (C)

2. (D) 14. (A) 26. (B) 38. (C) 50. (D) 62. (D)

3. (C) 15. (A) 27. (B) 39. (A) 51. (C) 63. (D)

4. (A) 16. (A) 28. (C) 40. (A) 52. (C) 64. (B)

5. (D) 17. (D) 29. (C) 41. (C) 53. (A) 65. (A)

6. (B) 18. (C) 30. (B) 42. (B) 54. (A)

7. (D) 19. (A) 31. (C) 43. (D) 55. (A)

8. (C) 20. (D) 32. (C) 44. (C) 56. (D)

9. (B) 21. (B) 33. (B) 45. (D) 57. (C)

10. (B) 22. (D) 34. (A) 46. (C) 58. (C)

11. (B) 23. (C) 35. (A) 47. (C) 59. (B)

12. (A) 24. (A) 36. (B) 48. (D) 60. (D)

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gate ece 2013 paper

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