Paper 2

PAPER\ue000-2

1 Paper-2

Questions\ue000Q1.\ue000to\ue000Q20.\ue000carry\ue000one\ue000mark\ue000each. Q1.

Q2.

a is \u00b4 such\ue000th = 5B and Aa= t Aa 4\Bu,\ue000then If A and B are\ue000square\ue000matrices\ue000of\ue000order 40 0 d 7 (A)\ue0005

(B)\ue00025

(C)\ue000625

(D)\ue000None\ue000of\ue000these

d 3 -x x The\ue000particular\ue000integral\ue000of\ue000the\ue000differential\ue000equation is ( D - D ) y =e + e , D \u00ba dx 1 x -x (A) ( e +e ) 2 1 2 (C) x ( ex +e x ) 2

Q3.

(C)

2 1

x( e

x

+e

2

(D) x ( ex 2

1

s( s +1) e

-

(B)

2

s

(D)

s +1

-

-x

)

-

x

e )

s

( s + )1 e

-

2

s

( s +1) 2

Consider\ue000the ( ) as\ue000shown\ue000in\ue000fig.\ue000Q4.\ue000The\ue000FT\ue000of x t ( ) is x t x (t )

Fig\ue000Q4

1 1

-1

(A)

2 sin w- 2

t

(B)

w (C) 2 jwcos w Q5.

1

d -t If x( t) = { te u( t )},\ue000then\ue000laplace\ue000transform\ue000of ( ) is x t dt (A)

Q4.

(B)

2cos w- 2

j w (D) 2 jwsin w

If\ue000a\ue000resistor\ue000of\ue00010 W is\ue000placed\ue000in\ue000parallel\ue000with\ue000vo i

vs

Fig\ue000Q5 Linear Resistive Network

(A)\ue000increased

(B)\ue000decreased

(C)\ue000unchanged

(D)\ue000It\ue000is\ue000not\ue000possible\ue000to\ue000s

All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.

2 Paper-2

Q6.

Q7.

The\ue000current\ue000carrying\ue000capacity\ue000of\ue000a\ue0001\ue000W,\ue0004\ue000MW (A)\ue0000.5 kA

(B)\ue0002 kA

(C)\ue0002\ue000mA

(D)\ue0000.5\ue000mA

The\ue000equation\ue000governing\ue000the\ue000diffusion\ue000of\ue000neutral\ue000atom\ue000is

\u00b6\u00b6 (A) N =D N 2 \u00b6 t \u00b6 x

\u00b6\u00b6 (B) N =D N 2 \u00b6 x \u00b6 t

(C)

(D)

2

Q8.

2

2 \u00b6 \u00b6 N =D N 2 \u00b6 \u00b6 x t

2 \u00b6 \u00b6 N =D N 2 \u00b6 \u00b6 t x

The\ue000p-type\ue000substrate\ue000in\ue000a\ue000monolithic\ue000circuit\ue000should\ue000be\u (A)\ue000any\ue000dc\ue000ground\ue000point (B)\ue000the\ue000most\ue000negative\ue000voltage\ue000available\ue000in\ue000the\ue000circuit (C)\ue000the\ue000most\ue000positive\ue000voltage\ue000available\ue000in\ue000the\ue000circuit (D)\ue000no\ue000where, i.e.\ue000be\ue000floating

Q9.

Consider\ue000the\ue000List\ue000I\ue000and\ue000List\ue000II List\ue000I\ue000(\ue000Oscillator)

List\ue000II\ue000(\ue000Characteristic)

P.\ue000Colpitts\ue000Oscillator

1.\ue000RC\ue000Oscillator

Q.\ue000Phase\ue000shift\ue000Oscillator2.\ue000LC\ue000Oscillator R.\ue000Tunnel\ue000diode\ue000Oscillator 3.\ue000Negative\ue000resistance\ue000Oscillator S.\ue000Relaxation\ue000Oscillator

4.\ue000Sweep\ue000Circuits

The\ue000correct\ue000match\ue000is P

Q

R

S

(A)

1

2

3

4

(B)

2

1

3

4

(C)

1

2

4

3

(D)

2

1

4

3

Multiple Choice Questions

G AT E Electronics & Communications By

R. K. Kanodia Price 425.00

Pages 602

Published by NODIA & COMPANY

For\ue000previous\ue000years\ue000papers\ue000&\ue000sample\ue000chapter\ue000of\ue0 Feel\ue000free\ue000to\ue000call\ue000at\ue00009350292376\ue000for\ue000any\u

3 Paper-2

Q10.

For\ue000the\ue000circuit\ue000shown\ue000in\ue000fig.\ue000Q10, VCB = 05 . V\ue000and b = 100.\ue000The\ue000value\ue000of I Q is +5\ue000V

5\ue000kW

Vo

Fig\ue000Q10

IQ

-5\ue000V

Q11.

(A)\ue0001.68\ue000mA

(B)\ue0000.909\ue000mA

(C)\ue0000.134\ue000mA

(D)\ue000None\ue000of\ue000the\ue000above

A\ue000four-variable\ue000switching\ue000function\ue000has minterms m6 and m9 .\ue000If\ue000the literals complemented,\ue000the\ue000corresponding minterm numbers\ue000are (A) m3 and m0 (B) m9 and m6 (C) m2 and m0 (D) m6 and m9

Q12.

The\ue000diode\ue000logic\ue000circuit\ue000of\ue000fig.\ue000Q12\ue000is\ue000a D2

V1 V2

Q13.

Fig\ue000Q12

D1

Vo

(A)\ue000AND

(B)\ue000OR

(C) NAND

(D)\ue000NOR

The\ue000even\ue000part\ue000of\ue000a\ue000function xn [ ] = [ ]- [ - 4 ]is

un

un

1 (A) {1 +d[ n] - u[ n- 4]- u[- n- 4]} 2 1 (B) {u[n +3] -u[ n -4] 2

+ [dn ]}

1 (C) {u [n]+ u[ -n] - u[ n-4] -u[ -n -4]} 2 (D)\ue000Above\ue000all All questions are drawn from the book GATE ECE by R. K. Kanodia published by NODIA & COMPANY. For full solution refer the same.

4 Paper-2

Q14.

The trigonometricFourierseriesforthewaveformshowninfigQ14willbe

x( t)

Fig Q14

A

-p

(A)

(B)

(C)

(D) Q15.

+2A (sint 2 p

1

A 2A

(cost -

1

(cost -

1

A

2

+

p

A 2A 2

+

p

A 2A 2

+

p

-p 2

p

2

p

t

1 sin3t + sin5t .... ) 3 5 1 cos2t + cos3t .... ) 2 3

1 cos3t + cos5t .... ) 5 3

1 1 (sint +cost + sin3t + cos3t .... ) 3 3

Thepoll–zeroconfigurationofaphase–leadcompensatorisgivenby jw

jw

s

s

(A)

(B) jw

jw

s

s

(C) Q16.

(D)

While designing controller, the advantage of pole– zero cancellation is (A) The system order is increased (B) The system order is reduced (C) The cost of controller becomes low (D) System’s error reduced to optimum levels GATE ECE By R. K. Kanodia

MCQs : The book contains only solved Multiple choice questions (MCQ) which is the main requirement of the GATE exam. Each and every problem has its complete solution. We understand the fact that theoretical studies should be done from the standard book, that one has studied for the semester exams and thus one should use the same material to understand the concepts of the same. We have deliberately excluded theoretical matter in the book so as not to mislead the students. However, wherever needed, satisfactory explanation of the formula has been included in the solution.

5 Paper-2

Q17.

Assertion (A): PSKisinferiortoFSK. Reason (R): PSKrequirelessbandwidththanFSK.

Choose correct option: (A) Both A and R individually true and R is the correct explanation of A. (B) Both A and R individually true and but R is not the correct explanation of A. (C) A is true but R is false (D) A is false Q18.

Q19.

Q20.

In a certain telemetry system, the measured values are converted to digital form. The digital values can then be transmitted via FSK (binary or quaternary) , or BPSK or QPSK systems. Out of these the best noi immunity can be obtained with (A) binary FSK

(B) quaternary FSK

(C) BPSK

(D) QPSK

An antenna, when radiating, has a highly directional radiation pattern. When the antenna is receiving, its radiation pattern (A) is more directive

(B) is less directive

(C) is the same

(D) exhibits no directivity at all

The beamwidth between first null of uniform linear array of N equally spaced ( element spacing equally excited antenna is determined by (A) N alone and not by d

(B) d alone and not by N

(C) the ratio N d

(D) the product Nd

= d)

GATE ECE By R. K. Kanodia

Adherence to Pattern: All Multiple choice questions are strictly according to the GATE pattern. Every problem selected and included in the book is a model problem for the preparation of the exam which would thus prepare and equip the students better. Kindly not the standard of Multiple choice questions and their solution in every unit is much better than the ones available in a famous serie problems & solutions as far as GATE is concerned.

6 Paper-2

Questions Q21. to Q75. carry two marks each.

Q21.

Q22.

Q23.

é 0 ê If A =ê êtana êë 2

tan 0

aù 2 úthen ú ú ûú

(A) I

+A

(B) I

-A

(C) I

+ 2A

(D) I

-2 A

æ è

Forwhatvalueof x0ç £ x

p £ ö÷, the function=y x 2 ø (1 +tanx

(A) tan x

(B) 0

(C) cot x

(D) cos x 2a

Thevalueof

f ( x)

òf ( x) +f (2 a0

Q24.

-A

(I

écos a - sin a ù )×ê 2 úisequalto ê ú cosa û ësina

x)

dx is

(A) 0

(B) 1

(C) a

(D) 2a

Theintegratingfactorforthedifferentialequation (xy ) dx+2y dy=0 is given by x + 3

(A) e

-x

4

(B) e x

(C) e x Q25.

has a maxima ? )

(D) e

cospz

òz - 1 dz =? where

c is the circle z

-

3

2

2

x

=3

c

Q26.

(A) i2 p

(B) - i2 p

(C) i6 p2

(D) - i6 p2

If3isthemeanand

3 2

isthestandarddeviationofabinomialdistribution,thenthedistributionis

12

æ1 3 ö (B) ç + ÷ 2è 2 ø

60

æ1 4 ö (D) ç + ÷ 5è 5 ø

æ3 1 ö (A) ç + ÷ 4è 4 ø æ4 1 ö (C) ç + ÷ 5 è 5ø

12

5


Levels of MCQs: The Multiple choice questions included in the guide are in a conceptually evolving method, allowing the student to progress from one level of complexity to another but always aiding in understanding the basic foundation of the s Thus, the MCQs gradually and scientifically advance from the basic level to a more complex level, helping in the systematic understanding of the problem rather than an abrupt one.

7 Paper-2

Q27.

Forthedifferentialequation dy = dx

- x2

giventhat y

x:

0

0.2

0.4

0.6

y:

0

0.02

0.0795

0.1762

Using Milne predictor–correction method, the y at next value of x is

Q28.

(A) 0.2498

(B) 0.3046

(C) 0.4648

(D) 0.5114

d æ4sin 4wsin2wö w) = TheinverseFouriertransformof( X j ç ÷is w dwè ø (A) t

2

- 2e- trect {2(t -

(B) t

2

+2e- trect (2( t +4))

4)}

(C) - t{rect (2t +8) + rect (2t - 8)} (D) t{rect (2t +4) -rect (2t - 8)} Q29.

2

d Thetimesignal (x )t correspondingto X ( s) = s 2 ds

æ1 ö 1 ç ÷ çs2 +9 ÷ +s +3 is è ø

2 æ- 3t 2t ö t + sin3t + cos3t ÷u( t) ç ÷ 3 9 è ø

(A) eç (B) ( e

- 3t

+ 2t sin3t +t2 cos3t )u (t

)

æ- 3t + 2t sin3t +t2 cos3t öu( t) ÷ 3 è ø

(C) eç (D) ( e Q30.

- 3t

+ t 2 sin3t +2t cos3t )u (t

)

Theincidencematrixofagraphisasgivenbelow

é1 ê0 ê A = ê0 ê0 ë

0

0

0

1

0

0

1

0

0

-1

1

0

0

1

0

0

-1

0

0

1

0

0

1

-1

1ù

ú ú - 1ú ú 0û 0

The number of possible tree are (A) 40

(B) 70

(C) 50

(D) 240


8 Paper-2

Q31.

InthefigQ31thevalueof v 1 is 2W

1W

3W

1W

Fig Q31

+ 2W

8 V

6W

v1

6W

18 V

–

Q32.

(A) 6 V

(B) 7 V

(C) 8 V

(D) 10 V

Inthecircuitoffig.Q32the30Vsourcehasbeenappliedforalongtime.Theswitchisopenedat t = 1 ( ) is ms. At t = 4 ms the 4 v Cms t=1 ms

1 kW

30u (-t ) V

Q33.

0.6 mF

6.25 kW

Fig Q32

+

25 kW

vC

–

(A) 8.39 mV

(B) 2.59 V

(C) 1.13 mV

(D) 2.77 V

Fora RLC seriescircuit= R20W ,L =06 . H,thevalueof C willbe [CD =critically damped, OD =over damped, UD =under damped]. CD

Q34.

OD

>6

UD

(A)

C

= 6 mF

C

C < 6 mF

(B)

C

= 6 mF

C < 6 mF

(C)

C

>6

mF

C

= 6 mF

C < 6 mF

(D)

C < 6 mF

C

=6

C

mF

mF -

InthecircuitshowninFig.Q340v(

C

> 6 mF

> 6 mF

= )8V

( ) tfor t ³ 0 is and iin ( t) = 4(dt) . TheC v Fig Q34

iin

(A) 164e

-t

V

(C) 208(1 - e

- 3t

50 W

20 mF

+ vC –

-

(B) 208e )V

t

V

-3 t

(D) 164e

V


9 Paper-2

Q35.

Inthecircuitshowninfig.Q35,whenthevoltage V1 is10V,thecurrent I is1A.Iftheappliedvoltage at port-2 is 100 V, the short circuit current flowing through at port 1 will be

Resistive

V1

Q36.

Fig Q35

I

Linear Network

(A) 0.1 A

(B) 1 A

(C) 10 A

(D) 100 A

V Thenetworkfunctionofcircuitshowninfig.Q36is H (w )= o V1 2 kW

vi

~

= 1

15 kW

C

4

+j 001 . w Fig Q36

+

+ vC –

vo

AvC

–

The value of the C and A is

Q37.

(A) 10 mF, 6

(B) 5 mF, 10

(C) 5 mF, 6

(D) 10 mF, 10

-3 In germanium(n i 2= .4 1´013 cm ) semiconductor at T

cm -3 anddonorconcentrationis N d

= 300 K, the acceptor concentrations is N a = 1013

= 0.Thethermalequilibriumconcentration

p 0 is

. ´10 9 cm -3 (A) 297 . ´ 1012 cm -3 (B) 268 . ´ 1013 cm -3 (C) 295 (D) 2.4 cm Q38.

-3

-3 Asilicon(n i = 15 . ´ 1010 cm ) pn junctionat T

= 300 Khas N d = 1014 cm -3 and N a = 1017 cm -3 .The

built-in voltage is

Q39.

(A) 0.63 V

(B) 0.93 V

(C) 0.026 V

(D) 0.038 V

The maximum electric field in reverse-biased silicon pn junction is E max

concentration are N d4 =10´16 cm -3 and N a

|=3 ´|10

5

V cm. The doping

=4 ´1017 cm -3 . The magnitude of the reverse bias voltage is

(A) 3.6 V

(B) 9.8 V

(C) 7.2 V

(D) 12.3 V


10 Paper-2

Q40.

MOSFET is biased in the saturation required ratio W L is

Q41.

2 m n = 525 cm - ,V V TN s = .0 75V, t ox =400 A°. When = 6 mA. region at VGS = 5 V, the required rated ( current is The I D sat )

A ideal n-channel MOSFET has parameters

(A) 14.7

(B) 11.2

(C) 9.61

(D) 7.2

¢ =8 Fora n - channelenhancement-modeMOSFETtheparametersare V. TNV,=k08 n

mA V 2 and

05 required v GS is W L = 5. If the transistor is biased in saturation region w.ithmA, ID = then (A) 1.68 V (B) 2.38 V (C) 4.56 V (D) 3.14 V

Q42.

06 The cutin voltageforeachdiodeinfig.Q42is V g. = V.Eachdiodecurrentis0.5mA.Thevalueof R1 , R2 and R 3 will be respectively +10 V

R1

Fig Q42

+5 V R2 0 V R3

-5 V

Q43.

(A) 10 kW, 5 kW, 2.93 kW

(B) 6 kW, 3 kW, 3.43 kW

(C) 5 kW, 6 kW, 4.933 kW

(D) 6 kW, 8 kW, 6.43 kW

Inthecircuitoffig.Q43 Zener voltageis V Z

=5

Vand b = 100.Thevalueof I CQ and VCEQ are

+12 V

500 W

Fig Q43

(A) 12.47 mA , 43 . V

(B) 12.47 mA , 5.7 V

(C) 1 0. 4 3A , 5.7 V

(D) 10.43 A , 43 . V


11 Paper-2

Q44.

¢ =k08 . V, =4 0 mA / V 2 and l = 0.The Thetransistorsinthecircuitoffig.Q44haveparameter V TN n

width-to-length ratio of M 2W is L

(=1). If V

2

o

W . 010 Vwhen V i = 5 V,then ( = L

+5 V

) for M

1

1

is

Fig Q44

M1 Vo M2

Vi

Q45.

(A) 47.5

(B) 28.4

(C) 40.5

(D) 20.3

Inthecircuitoffig.Q45the CMRR oftheop-ampis60dB.Themagnitudeofthe v o is 2 V

Fig Q45 100 kW

R

R

1 kW

R

R

vo

1 kW

100 kW

Q46.

Q47.

(A) 1 mV

(B) 100 mV

(C) 200 mV

(D) 2 mV

Ifthe X and Y logicinputsareavailableandtheircomplements X and Y arenotavailable,theminimum Å is X Y number of two-input NAND required to implement (A) 4

(B) 5

(C) 6

(D) 7

TherearefourBooleanvariables ,x 1

2

x ,x 3 and x 4 . The following function are defined on sets of them

) =Sm (3, 4, 5) f ( x3 , x2 , x 1

g( x4 , x3 , x2 ) =Sm (1, 6, 7) h( x4 , x3 , x 2 , x1 ) = fg Then h( x4 , x3 , x2 , x1 ) is (A) Sm(3, 12, 13)

(B) Sm(3, 6)

(C) Sm(3, 12)

(D) 0 GATE ECE By R. K. Kanodia

Unit Division: Each unit has been further sub-divided into separate chapters and not clustered together. Thus the non-combination of all the problems in a single unit makes the reader, to remain focused and able to manage his time d preparation.

12 Paper-2

Q48.

The ideal inverter in fig. Q48 has a reference voltage of 2.5 V. The forward voltage of the diode is 0.75 V. The maximum number of diode logic circuit, that may be cascaded ahead of the inverter without produc logic error, is +5 V

+5 V

+5 V

Fig Q48 +5 V

A

Z

B C

n

Q49.

D

Stages of Diode Logic

(A) 3

(B) 4

(C) 5

(D) 9

Considerthefollowingsetof8085 mPinstruction MVI RLC MOV RLC RLC ADD

A, BYTE1 B, A B

If BYTE1 = 07H, then content of accumulator, after the execution of program will be

Q50.

(A) 46H

(B) 70H

(C) 38H

(D) 68H

Considertheexecutionofthefollowinginstructionby8085 mp MVI SHLD

H, 01FFH 2050H

After execution the contents of memory location 2050H, 2051H and registers H, L will be respe (A) 01H, FFH, FFH, (B) FFH,

01H, FFH,

(C) FFH,

01H,

(D) 01H, FFH,

01H 01H

01H, FFH 01H, FFH


Includes Previous Exam Questions: This books contain questions on earlier IES, IAS & GATE exams that might be relevant to learn some concepts but we have purposely not mentioned them in our book. We believe and strongly advocate th year GATE contains new and unique problems.

13 Paper-2

Q51.

Thesystemshowninfig.Q51is

x[ n]

1 4

y[ n]

+

D

y[ n-2]

D

+

+

Fig Q51

1 4

+

-1 2

Q52.

(A) Stable and causal

(B) Stable but not causal

(C) Causal but unstable

(D) unstable and not causal

Thetransferfunctionofasystemisgivenas

æ 1ö 2 zç + ÷ è 2ø . H( z ) = æz - 1 öæz -1 ö ç ÷ç ÷ è 2øè 3ø Consider the two statements Statement(i) :Systemiscausalandstable. Statement(ii) :Inversesystemiscausalandstable.

The correct option is

Q53.

(A) (i) is true

(B) (ii) is true

(C) Both (i) and (ii) are true

(D) Both are false

= x- t , Acausal LTI filterhasthefrequencyresponse H) sho j wn in fig. Q53. For the input si(gn)al ( w

e

jt

output will be H( j w) j2

-1

Fig Q53

1

w

-j2

(A) - 2 je(C) 4pje-

jt

jt

(B) 2 je

-

jt

(D) - 4pje-

jt


Less Erroneous: The book has very few errors [less than 5%] compared to the other books available in the market which have upto 40% errors. This puts the students in a better and more com fortable situation as all the errors are traceable due to availability of the complete solution and moreover, the errors are never conceptual but data or typo mistakes. Kindly note t all the errata will be soon available at our website www.nodia.co.in

14 Paper-2

Q54.

Q55.

æ1 ö = a å n ç ÷will 4 è ø n=0 ¥

Thenumericvalueof

(A)

16

(C)

4

n

be

9

9

(B)

9

(D)

9

Eachoftwosequence[x ]nand y[ n] hasaperiod N

4

16

= 4.TheFScoefficientare

1 1 ]1 X [0] =X [3] = X [ 1] = X [2] =1 And Y[0 ],Y [1 ],Y [2 ],Y 3[ = 2 2 The FS coefficient [ ]Zforthesignal k z[ n] =x[ n] y [n ]will be (A) 6 (C) 6 Q56.

Q57.

(B) 6| k|

|k |

(D) e

j

p

k 2

A Routh tableisshownbelow.ThelocationofpoleonRHP, LHP andimaginaryaxisare s7

1

2

-1

-2

s5

1

2

-1

-2

s5

3

4

-1

s4

1

-1

-8

s3

7

8

s2

-18

-21

s1

-9

s0

-21

(A) 1,

2,

4

(B) 1,

6,

0

(C) 1,

0,

6

(D) None of the above

Theopen-looptransferfunctionofa ufb controlsystemis G( s) =

K (1 +2s)( 1+ 4s) 2

2

s (s

+2s + 8)

The position, velocity and acceleration error constants are respectively (A) 0,

0, 4K

(C) ¥ , K , 0 8

(B) 0, 4K ,

¥

(D) ¥ ,

K

¥,

8


15 Paper-2

Q58.

Theforward-pathtransferfunctionofa ufb systemis G( s) =

K ( s +1)( s + 2) ( s +5)( s +6)

The break points are

Q59.

Break-in

Breakaway

(A)

-1.563

-5.437

(B)

-5.437

-1.563

(C)

-1.216

-5.743

(D)

-5.743

-1.216

The Nyquist plotofasystemisshowninfig.Q59.Theopen-looptransferfunctionis 4s +1 G( s) H ( s)= 2 s ( s +1)( 2s +1) Im

Fig Q59

10.64 w

w=

=

¥

Re

0

The no. of poles of closed loop system in RHP are (A) 0

(B) 1

(C) 2

D) 4


Time Management: Time is a very important factor in any competitive exam and the same applies for GATE too. It has been observed and concluded that if students can manage time, they can get a better score in GATE. The solutions provided a extremely logical yet tricky so that they save time when the student solve them in the examination, as they have already b used to solving difficult and tricky problems.

16 Paper-2

Q60.

( i). The state space representation is Forthenetworkshowninfig.Q60.Theoutputis t R

i1

1W

1H

v1

i3

iR

v2

Fig Q60

i2 vi

Q61.

4v1

1 F

1W

(A)

év&1 ù é 1 - 1ù év1 ù 1é ù = ê&i ú =ê- 3 1 ú êi ú + 0ê v iú R , i [4 û ë1 û ë û ë3 û ë

év1 1] ê ëi3

ù ú û

(B)

év&1 ù é 1- 1-ù év1 ù 1é ù = ê&i ú =ê- 3 - 1 ú êi ú + 0ê v iú R , i [4 û ë3 û ë û ë3 û ë

év1 1] ê ëi3

ù ú û

(C)

év&1 ù é1 êv& ú =1 ê ë2 û ë

év 4] ê 1 ëv 2

ù ú û

(D)

év&1 ù é 1 3ù év1 ù é1 ù êv& ú =ê 1- 6 ú êv ú + ê -1 vúi ,R i =[1 ûë 2 û ë û ë2 û ë

3-ù év1

ù é1 ù ú ê ú + ê vúi ,R i =[1 6 û ëv 2 û ë -1 û

év ù 4] ê 1 ú ëv 2 û

2 ( ) is Thepowerspectraldensityofabandpasswhitenoise nN t 2 asshowninfig.Q61.Thevalueof n

is SX( w )

4pB

Fig Q61

N

2

-wc

Q62.

Q63.

w c

w

(A) NB

(B) 2NB

(C) 2pNB

(D)

NB p

Inareceivertheinputsignalis100 mV,whiletheinternalnoiseattheinputis10 mV.Withamplification the output signal is 2 V, while the output noise is 0.4 V. The noise figure of receiver is (A) 2

(B) 0.5

(C) 0.2


12signalseachband-limitedto5kHzaretobetransmittedoverasinglechannelbyfrequencydivision multiplexing . If AM-SSB modulation guard band of 1 kHz is used, then the band width of the multiplexed signal will be (A) 131 kHz

(B) 81 kHz

(C) 121 kHz

(D) 71 kHz


17 Paper-2

Q64.

A carrier wave of 1 GHz and amplitude 3 V is frequency modulated by a sinusoidal modulating signal frequency of 500 Hz and of peak amplitude of 1 V. The frequency deviation is 1 kHz. The peak level of the modulating wave form is changed to 5 V and the modulating frequency is changed to 2 kHz. The expression for the new modulated wave form is (A) cos [2p´ 10 t +25 . cos (4p´ 10 t)] 6

3

(B) cos [2p´ 10 t +5 cos (4p´ 10 t)] 6

3

(C) 3cos 2 [ p´ 10 t +25 . cos(4p´ 10 t)] 6

3

(D) 3cos 2 [ p´ 10 t +5 cos (4p´ 10 t)] 6

Q65.

3

Let message signal m ( )t=cos(4p10 3t) and carrier signal ( )c=t 5cos (2p10 6t) are used to generate a FM

signal. It the peak frequency deviation of the generated FM signal is three times the transmissi p 1008 ´ (10)3( t) in the FM signal would bandwidth of the AM signal, then the coefficient of the 2term cos be

Q66.

Q67.

Q68.

Q69.

5

(A) 5J 4( 3)

(B)

5 (C) J 8( 4) 2

(D) 5J 4( 6)

Thecurlofvectorfield A=rz sinf u r

+3rz 2 cosf

2

J 8( 3)

ur

at point (5, 90°, 1) is

(A) 0

(B) 12u q

(C) 6u r

(D) 5u f

A 150 MHz uniform plane wave is normally incident from air onto a material. Measurements yield a SWR of 3 and the appearance of an electric field minimum at 0.3l in front of the interface. The impedance material is (A) 502 - j 641 W

(B) 641 - j 502

j (C) 641 + 502

j W (D) 502 + 641

W

W

The quarter-wave lossless 100 W line is terminated by load Z L 60 V, the voltage at the sending end is (A) 126 V

(B) 28.6 V

(C) 21.3 V

(D) 169 V

= 210 W. If the voltage at the receiving end is

An antenna can be modeled as an electric dipole of length 4 m at 3 MHz. If current is uniform over its length, then radiation resistance of the antenna is (A) 1.974 W

(B) 1.263 W

(C) 2.186 W

(D) 2.693 W


18 Paper-2

Q70.

Anarraycomprisestwodipolesthatareseparatedbyhalfwavelength.Ifthedipolesarefedbycurrents, that are 180° out of phase with each other, then array factor is (A) sin

æpcos q + pö ç ÷ 4 4 ø è

(B) cos


(C) cos


(D) sin


Common Data Questions Common Data for Questions Q.71-73:

= 2/ V For the circuit shown in fig. Q71-73 transistor parameters are V. TNmA V, 2Kand = 0. n =l05 The transistor is in saturation. +10 V

Fig Q71-73

10 kW

vo

vi

~

VGG

Q71.

Q72.

Q73.

If I DQ istobe0.4mA,thevalueof VGSQ is (A) 5.14 V

(B) 4.36 V

(C) 2.89 V

(D) 1.83 V

Thevaluesof g m and ro are (A) 0.89 mS, ¥

(B) 0.89 mS, 0

(C) 1.48 mS, 0

(D) 1.48 mS, ¥

Thesmallsignalvoltagegain Av is (A) 14.3

(B) -14.3

(C) -8.9

(D) 8.9 GATE ECE By R. K. Kanodia

Variety: The book carries in it a large variety of problems. The words of one of the senior educators of a reputed coaching institute bear testimony to the fact wherein he comments that “We can’t expect so much variety of problems in a sin available in the market.”

19 Paper-2

Common Data for Questions Q74-75:

Consider three continuous-time periodic signals whose Fourier series representation are as follow x1 ( t) = Q74.

k æ 1ö å ç ÷e k =0 è3 ø 100

jk

2p 50

t

,

x 2 ( t) =

k

å

100

-

coskp e

jk

2p 50

t

,

=100 -

ækp öex 3 ( t) = å j sin ç ÷ è2 ø k = -100 100

jk

2p 50

t

Theevensignalsare (A) x 2 ( t ) only (B) x2 ( t) and x 3 ( t) (C) x1 ( t) and x3 ( t) (D) x1 ( t) only

Q75.

Therealvaluedsignalsare (A) x 1 ( t ) and x2 ( t) (B) x2 ( t) and x 3 ( t) (C) x3 ( t) and x1 ( t) (D) x1 ( t) and x3 ( t)

Linked Answer Questions: Q76. to Q85. carry two marks each.

Statement for Linked

Answer Questions: Q76. and Q77:

08m, L = 4 mm, The parameters of an n - channel enhancement-mode MOSFET. are V, WV= 6= 4m TN 2 t ox =450 A°, m n = 650 cm -V . s Q76.

Theconductionparameter K n is (A) 0.8 mA V2 (B) 0.8 mA V 2 (C) 0.4 m A V2 (D) 0.4 mA V 2

Q77.

If VGS

=V DS = 3 V, then current I D is

(A) 1.94 mA (B) 2.87 mA (C) 5.68 mA (D) 3.84 mA


20 Paper-2


Answer Questions: Q78 and

Q79:

The 8-bit left shift register and D -flip-flop shown in fig. Q78-79 is synchronized with same clock. D flip-flop is initially cleared. b7

b6

D

b5

b4

Fig Q78-79

b3 b2 b1 b0

Q

CLK Q

Q78.

Thecircuitactas (A) Binary to 2’s complement converter (B) Binary to Gray code converter (C) Binary to 1’s complement converter (D) Binary to Excess–3 code converter

Q79.

IfinitiallyregistercontainsbyteB7,thenafter4clockpulsecontentsofregisterwillbe (A) 73

(B) 72

(C) 7E

(D) 74



Q81:

A block diagram is shown in fig. Q80-81. 5 R 1(s )

+

1 s

Fig Q80-81 2s

+ +

C 2(s )

2

Q80.

Thetransferfunctionforthissystemis (A)

(C)

2s(2 s +1) 2s

2

+3s + 5

2s( 2s +1) 4s

2

+13s +5

(B)

(D)

2s( 2s +1) 2

2s

+13s +5

2s( 2 s +1) 2

4s

+3s + 5


Attractive Format: We understand student psychology and the fact that if the book is in an attractive format, the student would feel good in reading the book. This fact also heightens the interest to study in a student. Thus the sty le of the book is so that it appeals to its readers, yet is expressive and detailed.

21 Paper-2

Q81.

Thepoleofthissystemare (A) - 075 . ± j 139 . (C) - 05 . ,

(B) - 041 . , - 609 . (D) - 025 . ± j 088 .

- 167 .



Q83:

Ten telemetry signals, each of bandwidth 2 kHz, are to be transmitted simultaneously by binar The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The must be sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. Q82.

Q83.

Theminimumpossibledataratemustbe (A) 272.64 kbits/sec

(B) 436.32 kbits/sec

(C) 936.64 kbits/sec

(D)None of the above

Theminimumtransmissionbandwidthis (A) 218.16 kHz

(B) 468.32 kHz

(C) 136.32 kHz




Q85:

In an air-filled waveguide, a TE mode operating at 6 GHz has Ey Q84.

Q85.

py =15sinæ2çpx ö÷cos æ ç bè èa ø

ö ø

sin÷ (wt - 12z) V m

Thecutofffrequencyis (A) 4.189 GHz

(B) 5.973 GHz

(C) 8.438 GHz

(D) 7.946 GHz

Theintrinsicimpedanceis (A) 35.72 W

(B) 3978 W

(C) 1989 W

. W (D) 7144


Aim : The aim of the book is to provide quality material, a fact which can easily be seen in books available for the preparation of IIT-JEE, AIEEE, CPMT & CAT, but till date never observed in the material available GATE preparation. In other wor we want to provide ELITE material but which is also economical. E : Expressive L : Less Erroneous I : Individualistic T : Targeted approach E : Exhaustive content

22 Paper-2

Answers Paper-2 1.

(C)

2.

(B)

3.

(B)

4.

(B)

5.

(C)

6.

(A)

7.

(A)

8.

(B)

9.

(B)

10.

(B)

11.

(B)

12.

(B)

13.

(D)

14.

(C)

15.

(A)

16.

(B)

17.

(C)

18.

(B)

19.

(C)

20.

(D)

21.

(A)

22.

(D)

23.

(C)

24.

(B)

25.

(B)

26.

(A)

27.

(B)

28.

(C)

29.

(C)

30.

(A)

31.

(A)

32.

(D)

33.

(A)

34.

(B)

35.

(C)

36.

(C)

37.

(C)

38.

(A)

39.

(C)

40.

(A)

41.

(B)

42.

(A)

43.

(B)

44.

(D)

45.

(B)

46.

(A)

47.

(A)

48.

(A)

49.

(A)

50.

(C)

51.

(A)

52.

(C)

53.

(B)

54.

(C)

55.

(A)

56.

(A)

57.

(D)

58.

(A)

59.

(C)

60.

(B)

61.

(B)

62.

(A)

63.

(D)

64.

(C)

65.

(D)

66.

(D)

67.

(C)

68.

(A)

69.

(B)

70.

(B)

71.

(C)

72.

(A)

73.

(C)

74.

(A)

75.

(B)

76.

(C)

77.

(A)

78.

(B)

79.

(C)

80.

(C)

81.

(C)

82.

(B)

83.

(A)

84.

(B)

85.

(B)

A Concern Publishers, who claim to be the partners in the progress of students, use cheap tactics to sell their material to them. A publis famous for providing books on GATE, which are incidentally also erroneous & copied, has also imitated our book this year. They have gone to the extent this year and introduced a revised edition of their book on GATE on Elect. & Comm., which also includes a large nu of Multiple Choice Questions [MCQ]. All the MCQ’s & their solutions have been copied from our book [GATE by RK Kanodia]. This is condemnable act of plagiarism. They are trying to restrict our book from reaching the students, by giving high incentives to retailers distributors. They are using all possible unfair means of the publishing field, to make money and monetary exploit students, yet the b fails to reach the students. An Appeal

We would like to make this Appeal to you against plagiarism and exploitation, keeping in mind the interest of stu publishing field, to support us in our fight. Our company has been started with very few resources and but have large vision. The greatest challenge for us is to reach e part of India. We realize that educators are busy people in times of change and thus they are not able to rate the books given to them able to suggest the same to the students. Thus students, who are dependent on such educators for the knowledge of books, are not a the resources available to the maximum. The restriction of erroneous and plagiarized work, in reaching the hands of students thus b a huge task. Therefore, we request and ask for your support against this condemnable act of plagiarism.

23 Paper-2

MCQ GATE-ECE by RK Kanodia Kindly note that our publication GATE-ECE by RK Kanodia, has the following features that make it an excellent study material in comparison to other books available on the GATE exam:

1. MCQs: The book contains only solved Multiple choice questions (MCQ) which is the main requirement of the GATE exam. Each and every problem has its complete solution. We understand that theoretical studies should be done from the standard book, that one has studied for the semester exams and thus one should use the same material to understand the concepts of the same. We have deliberately excluded theoretical matter in the guide book so as not to mislead the students. However, wherever needed, satisfactory explanation of the formula has been included in the solution. 2. Adherence to Pattern: All Multiple choice questions are strictly according to the GATE pattern. Every problem selected and included in the book is a model problem for the preparation of the exam which would thus prepare and equip the students better. Kindly note, that the standard of Multiple choice questions and their solution in every unit is much better than the ones available in a famous series of problems & solutions as far as GATE is concerned. 3. Levels of MCQs: The Multiple choice questions included in this book are in a conceptually evolving method, allowing the student to progress from one level of complexity to another but always aiding in understanding the basic foundation of the subject. Thus, the MCQs gradually and scientifically advance from the basic level to a more complex level, helping in the systematic understanding of the problem rather than an abrupt one. 4. Unit Division: Each unit has been further sub-divided into separate chapters and not clustered together. Thus the non-combination of all the problems in a single unit makes the reader, to remain focused and able to manage his time during his preparation. 5. Time Management: Time is a very important factor in any competitive exam and the same applies for GATE too. It has been observed and concluded that if students can manage time, they can get a better score in GATE. The solutions provided are extremely logical yet tricky so that they save time when the student solve them in the examination, as they have already been used to solving difficult and tricky problems.

6. Variety: The book carries in it a large variety of problems. The words of one of the senior educators of a reputed coaching institute bear testimony to the fact wherein he comments that “We can’t expect so much variety of problems in a single book available in the market.” 7. Includes Previous Exam Questions: This book contains questions on earlier IES, IAS & GATE exams that might be relevant to learn some concepts but we have purposely not mentioned them in our book. We believe and strongly advocate that every year GATE contains new and unique problems. 8. Less Erroneous: The book has very few errors [less than 5%] compared to the other books available in the market which have upto 40% errors. This puts the students in a better and more com fortable situation as all the errors are traceable due to availability of the complete solution and moreover, the errors are never conceptual but data or ty po mistakes. Kindly note that, all the errata will be soon available at our website www.nodia.co.in 9. Attractive Format: We understand student psychology and the fact that if the book is in an attractive format, the student would feel good in reading the book. This fact also heightens the interest to study in a student. Thus the style of the book is so designed that it appeals to its readers, yetisexpressiveanddetailed. 10. Aim : The aim of the book is to provide quality material, a fact which can easily be seen in books available for the preparation of IIT-JEE, AIEEE, CPMT & CAT, but till date never observed in the material available GATE preparation. In other words, we want to provide ELITE material but which is also economical.

E L I T E

: : : : :

Expressive Less Erroneous Individualistic Targeted approach Exhaustive content

We have received feedback, which state that the book fulfills more than what is stated above and thus it has been a great success last year, on all aspects. Everyone who got through, due to this book, has given excellent feedback. Reviews can be read at our website www.nodia.co.in. However, nothing in the world can be achieved without the help of constructive criticism and thus we would be obliged if you can send across your feedback to make our book, a GUIDE in true sense of the word.

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Paper 2

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