ECE GATE PAPER 22 ANSWERS

7/21/2019 ECE GATE PAPER 22 ANSWERS

http://slidepdf.com/reader/full/ece-gate-paper-22-answers 1/13

ECTest ID: 161272 TarGATE’16 www.gateforum.com

ICP–Intensive Classroom Program eGATE-Live Internet Based Classes DLP TarGATE-All India Test SeriesLeaders in GATE Preparation 65+ Centers across India

© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

1

Answer Keys

General Aptitude

1 C 2 B 3 B 4 24 5 15876 6 D 7 D

8 4 9 54 10 C

Electronics and Communication Engineering

1 B 2 -1 3 0.5 4 B 5 B 6 -0.1538 7 0.333

8 D 9 20 10 0.4 11 D 12 C 13 A 14 C

15 2 16 D 17 C 18 B 19 B 20 2000 21 A

22 B 23 D 24 0 25 A 26 6 27 B 28 D

29 D 30 C 31 C 32 15.62 33 1.231 34 D 35 A36 D 37 1.53 38 C 39 C 40 B 41 A 42 46

43 C 44 B 45 22.89 46 7 47 1.95 48 1.15 49 1393

50 -165.3 51 50 52 B 53 0.001 54 A 55 53.31

Explanations:-

General Aptitude

4.

4

5

6

5

(?) 12

48(?)

4 6

5 5

6

25 5

(?) (?) 48 12

(?) 48 12 (?) 48 12

? 48 12 12 2 24

5. Let the side of the square be x cms

4x 504

504

x 126cm4

2 2Area of the Square x 126 126 15876cm

8. While we could use formulas, a lot of combination questions are easier to answer through a bit of common sense and brute force. On this question, it's much quicker to just list the possible committees than it is to use a formula approach.






2

No Paul means no Jane, which leaves us with only Joan/Stuart/Jessica. Sadly, Stuart cannotappear in any other committees.

Paul then needs 2 co-members and has 3 possible buddies for those 2 spots. We could use3C2=3 to calculate or we could just brute force:

Paul/Joan/Jessica

Paul/Joan/JanePaul/Jane/Jessica

So, as others have noted, there are 4 possible committees

9. Let's start by breaking 210 down into primes:

210 = 10 * 21 = 2*5*3*7

So, to start, we know that all permutation of 2,3,5,7 will fulfill the requirements: 4! = 4*3*2 =

24

however, we have to recognize that we may be able to use 1 and another number in place of a

pair of our numbers.

Since each digit has to be less than or equal to 9, the only substitution we can do is to use 1*6

instead of 2*3 (2*5, the next smallest pair of primes, would give us 10).

So, we can also use the digits 1,5,6,7. Same as last time, this gives us 4! = 24 possible

Arrangements

So, 24+24=48 nowhere does it say we need 4 digits. So, we can also make some 3 digit

numbers using 5, 6 and 7.

That's 3! more possibilities.. Add another 6 to our 48=54

10. You have 20 shoes, and are looking for a matching pair. The first shoe you take could be

anyone, so probability of getting one shoe is 1. Now, you took 1 shoe out of the 20 original,

and you have 19 left, but there is only 1 shoe in the 19 that matches the first one. So the

probability to get the matching shoe is 1/19.

So multiply your probabilities: 1 * 1/19 = 1/19

Electronics and Communication Engineering

1. We have

2x y

2 2

x y

1 r

tan r

For r 1

tan 0

0






3

2. Let y xf x, y x y C Implict function

Then y 1 x

y x 1

f yx y log ydy x

f dx x log x x.yy

Atdy

x 1, y 1 ; 1dx

3. For final value, finding,s 0limsF(s)

s 0

1 1lims. 0.5

s(s 2) 2

4. Taking Laplace transform of given equation

2 10s y(s) sy(0) y (0) 4 sy(s) y(0) 5y(s)

s 1

Substituting initial conditions

2 2

s 6 10y(s)

s 4s 5 (s 1)(s 4s 5)

Using partial fractions

2

1 1 2t t

2

1 1y(s)

(s 2) 1 s 1

1 1L [y(s)] L e sin t e

(s 2) 1 s 1

5. y 3 yD.E e dx y xe dy … (1)

(1) Can be written as

y 3dxx e .y linear equation in x

dy

Also, (1) can be written as

y y 3e dx xe y dy 0 isexact

M N

M NSince

y x

Equation (1) is not a homogeneous D.E.






4

6. Using source transformations:

x x x

x

x

10 5i 50i 12.5i 5

5 32.5i

i 0.1538A

8. The circuit is ambiguous and hence cannot exist. Because two voltage sources of different

values are connected in paralleled across the same points

9. 10 5 2 K 550 2K 10

K 20

10.

11.

C G G

R 1 ( 1) G G 2 1 G

E

F 20E

10 5

F 50xE 20E

x 20 50 0.4

x5 i

x5 i

5 5 2A

10

10V

5 x50 i

5V

10

10V

2.5

xi






5

12.

t2t 2t

2 0

0

e 1 eu t+ e u d

2 2

18.

For SR flip-flop S=R=1, will never occur, hence it can be used as slave

19.

0111 0011

RAR 0011 1001

0100 1010

20. From Einstein’s Relation

n

n

2

n

D KT

q

502000cm V s

0.025

21.D A

N N

So, the sample is n-type

23. Synchronous Detector uses LPF.

24. for r 2,

2

120 0

r r

25.

O OO

broad

114.6 114.6

FNBW 6.37L 18

O O O

end

2 2FNBW 114 114 38.2

L 18

J nQ J nQ

nQnQK K

Clock

J nQ S nQ

nQnQK R

Clock

JK JK

(1)(2)






6

26.

6 2 2 2

AX x 2 a 1 1

2 1 b 1

6 2 2 2 2

2 a 1 1 12 1 b 1 1

16 2

4 a 1 1 2 16 8

5 b 1

5 a 5 a 8

a 3 a 3

5 b b 3

a b 6

27: Given

2

0 0

dy0.025y 1; x 0; y 1;h 1

dx

Here f(x, y) =0.025y2+1.

By implicit Euler’s method; we have

n 1 n n 1 n 1

2

1 0 1 1 1 1

2

1 12

1 1

y y hf x , y

y y hf (x ,y ) y 1 1[0.025 y 1]

y 0.025y 2

0.025y y 2 0

1y 20 8 5 37.88,2.12

28. Given that 0 0

dyx y & x 0; y 1

dx

1 0 1 2 3

11 0 0 2 0 0

Let h 0.2; f(x, y) x y

By3 rd order runge kutta method; we have

1y y k 4k k ;

6k h

where k hf (x , y ), k hf x , y2 2

12 0 0

3 0 0 2 1

k hk hf x ,y

2 2

k hf x h, y 2k k






7

1 0 0

2

3

k 0.2[x y ] 0.2[0 1] 0.2

0.2 0.2k 0.2 0 1 0.2[1.2] 0.24

2 2

k 0.2 0 0.2 1 2(0.24) 0.2 0.296

11y 1 0.2 4(0.24) 0.296 1.2436

29. z = 0,1 (singularities) of f(z), lies inside

C : z 1 3

2z 2z2z

z z 1 e ef z e

z z 1 z 1 z

Also, e2z is analytic in and on ‘C’

2z 2z

C C C

2z 2z

z 1 z 0

2

e e

f z dz dz dzz 1 z

2 j e 2 j e using cachy's integral formula

2 j e 1 , where j 1

30.c

(3x 4 y)dx (2x 3y)dy

2 2

(2 4)dxdy (by Green 's theorem)

2 dxdy 2( r ) 2( 2 ) 8

31.O

O

L C

4 0I I I 4 90

1 j10

2

4 1 5 1 16 j j4 j 1 4 j 4 I j A

5 5 5 5

32. Applying KCL at Node A

OA A

V V5 30 4 j3 4 j3

O

AV 15.625 30

33.

t 1

L 1 1 L

t t

L 1 L L L

8I t Ke 10sin 100t tan , I 3A

6

I t 0 I I t I e






8

If transients go out

L LI t I

So, 1 o

11000t tan 8 6 17.45

1000t 70.587

t 1.2319 m sec

34. O

pc : GH 180

1 O

pc90 tan 1803

2

2 2

3M 0

9

pc1GMM

35.2

3 1M 1 1

4

4 2 24 9 0 1.60, 5.60

gc 1.26

O 1

pc

180180 3 90 tan 2 0.449 rads sec

36.

1

1

dMslope

d log

6 3220 2

log log 0.1

2

2

0 620 5

log 10 log

37. 3 1 1 1Y s = . =s s+3 s s+3

3tY t = 1 e u t

Steady state value y 1

3t1 e 0.99 1






9

t 1.53sec

38. 2

1 1H s = =

s +5s+6 s+3 s+2

1/2 1 3 / 2Y s = +s s+3 s 2

3t 2t1 3Y t = e e u t

2 2

39.

41.

' inin

'

out out

R R

1 A

R R (1 A )

42. In –ve cycle capacitor will charged to 26V.

In +ve cycle, after capacitor got charged diode will never turn ON.

and

0 iV V 26V

0 peak inV V 26 20 26 46V .

43. Let P be even parity bit and d be difference output.

a b c d

1 -1

a b c d -a -b -c -d

a b-a c-b d-c -d

1 0 -1 0 0

0

1

1

d

a

c b

b






10

2 1 0

Input p d

d d d

0 0 0 0 0

0 0 1 1 1

0 1 0 1 1

0 1 1 0 0

1 0 0 1 1

1 0 1 0 0

1 1 0 0 0

1 1 1 1 1

So p = d

' ' ' ' '

2 1 0 1 0 2 1 0 1 0

'

2 1 0 2 1 0

2 2 2

p d d d d d d d d d d d

d d d d d d

d d d

F

44.

3 2 1 0

Input

x x x x

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

3 2 1 0

Output

y y y y

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

So y3 y2 y1 y0 is excess-3 equivalent of x3 x2 x1 x0

45. a R nn 16

N 65535V V 3 V

2 2

Step size R 16

1V

2

00 01 11 10

0

1

2d1 0d d

1

1 1

1






11

3

Step size V 45.77 V65535

step sizeQuantisation error 22.89 V

2

46.

A B C D

B C

Serial inQ Q Q Q

Q Q

X 1 0 0 0 1

1 1 1 0 0 0

2 0 1 1 0 1

3 1 0 1 1 0

4 0 1 0 1 0

5 0 0 1 0 0

6 0 0 0 1 1

7 1 0 0 0

47. At punch through,

1

2 bi PE c

0

B c B

26 V V N 1x

e N N N

PE bi

14 164

19 16B B

16 3

B

V 25V& neglect V

2 11.7 8.85 10 25 10 1

0.75 10 1.602 10 N N 10

N 1.95 10 cm

48. Since,ec e b c d

b e

4

dd 7

s

100ps 25ps

x 1.2 1012ps

V 10

c c c

ec

T 12ec

T

r C 10 0.1pF 1ps

so, 100 25 12 1

138ps

1 1f

2nT 2n 138 10

f 1.15GHz






12

49. PT

P T

23

P

DV

p

D p V

cmD 500 25.9 10 V

V sec

2

P

cmD 12.95

sec.

So, diffusion length of holes

P P

9

P

3

P

P

L D

L 12.95 150 10 cm

L 1.393 10

L 1393 microns

50. Thermal Noise Power = KTB

23

23

1.38 10 306 7000

2955960 10

Converting into dB scale

2310log 2955960 10

165.3dBw

51. si sf f 2IF

SS

f f 2

2

si sf 2f

si s s s

s si s s

f f 2f f

f f f 2f

2 0.5

1.5

2 2

RR I 1 Q

2 2 21 Q

2 2275 1 Q 1.5

5625 1Q 50

2.25

52. The AGC allows listening to station without constantly monitoring the volume control.

53. OO

I dl cos t Br A

4 r






13

6

8

2 100.02 rad m

c 3 10

67 10 0.1 cos 2 10 t 2A 10A

4 100

6cos 2 10 t 2 wb m 0.001 wb m

55. 2avg 0

1P H

2

f x,z x z 1 0

x zn n

f a aa ; ds dsa

f 2

x z2avg xt 0

1 a aP P .ds H a .

2 2

53.31 mW.

Documents

ECE GATE PAPER 22 ANSWERS