ECE GATE PAPER 21 ANSWERS

7/25/2019 ECE GATE PAPER 21 ANSWERS

1/14

ECTest ID: 161271 TarGATE16 www.gateforum.com

ICPIntensive Classroom Program eGATE-Live Internet Based ClassesDLPTarGATE-All India Test SeriesLeaders in GATE Preparation65+ 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 A 2 C 3 D 4 28 5 B 6 C 7 A

8 120 9 12 10 0.71

Electronics and Communication Engineering

1 C 2 A 3 C 4 -1 5 D 6 D 7 B

8 10 9 1.96 10 2.66 11 3.021 12 A 13 6.6 14 C

15 0.2 16 A 17 C 18 2 19 55 20 A 21 A

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

29 D 30 0 31 4 32 B 33 D 34 B 35 A36 B 37 B 38 1 39 A 40 10 41 18 42 1.1

43 B 44 0.5 45 -1 46 C 47 C 48 2.009 49 D

50 A 51 38.45 52 0.25 53 148 54 10.67 55 4

Explanations:-

General Aptitude

1. The diagrams show that some suitors are blazers whereas there is no relationship between

tunics and blazers or bumpers and blazers.

4. Let the price of mixed variety be x per kg.Cost of 1 kg of type 1 sugar = Rs. 20 => its proportional cost is 3(x-20)

Cost of 1 kg of type 2 sugar = Rs. 40 => its proportional cost is 2 (40 x)

3(x-20) = 2(40-x) => 3x 60 = 80 2x => 5x = 140 => x = 28

5. Sample space = c264

Now there are 7 unique adjacent square sets in each row & each column. Favourable cases

will be 7 8rows 8columns 112

Hence required probability =c2

Favourable cases 112 1

Sample space 64 18

8. So you have unlimited jewels essentially, so there's 5 possibilities for the first slot, 5 for the

second, and 5 for the third slot.

5 x 5 x 5 = 125 = Number of possibilities.

However, at least two gems must be different. So we subtract out the possibilities where all

the gems are the same. There are five types of gems, so there are five possibilities where all


2/14


3/14




3

2. If f is irrotational

curl f 0

i j k

0x y z

x 2y az bx 3y 4z 2x cy 5z

i (c 4) j(2 a) k(b 2) 0

a 2, b 2, c 4

3. i (yz) j(xz) k(xy)

div( ) 0 0 0 0

curl( ) 0 (s tan dard result)

is both solenoidal & irrotational

4. Since Real part of an analytic function f z u iv, is harmonic i.e., satisfies Laplacesequations.

2 2

2 2

u u0

x y

Where .xu e cos y

2 x x

x 2

2

e cos y e cos y 0

e cos y. 1 0

1 0

1

5. A dice is tossed thrice n = 3

Probability of success P = probability of getting even number

1 1P q

2 2

Probability at least two success = P(x = 2) + P (x = 3)

2 3

2 1 3 0

C C

1 1 1 1 3 1 13 3

2 2 2 2 8 8 2

6. T parameters are given by


4/14




4

1 2

1

2

2

1

n 0T Where N : N n :1

10n

or

N0

NTN

0N

By comparing we get 1

2

N4

N , to find N1we need N2

Hence data insufficient

8. By the source transformation we obtain the thevenin equivalent to circuit

9. Small signal equivalent circuit is given by

Output impedance 0 d DZ r ll R

100k 2k 1.96 k

10. 0 8

I 1m A8k

28 0I 2 m A

4k

1 2I I I 1I 1mA

D1 is Reverse Bias and D2 is Forward bias

0 0V 8 V 8 04k 8

0

8V V

3

th

10V

V

th5 , R

100k 2 k5m50 m

G D


5/14




5

11. Above circuit is half wave rectifier circuit.

In half wave rectifier, average value of output voltage is

0(avg) D TV 0.318 V V 0.318(10 0.5) 3.021V

13.1

f 6.6 MHz3 50

14. Ripple counter 7 3 21ns

Synchronous counter = 7 ns.

15. d n E 3 82000 10 10 cm s 0.2 10 cm s

16. Dynamic resistance Td dV 1

r rI I

17. Linearity: Means both additive and homogeneity property

1 1ax t ay t [hom ogenity]

1 2 1 2x t x t x t x t [Linearity]

1 2 1 2ax t bx t ax t bx t

18.

19. Phase Margin (PM) 0180 0 0 0180 125 55

20. Initial slope = - 20dB/dec

2

1

K (1+T S)

T.F S (1+TS)

y = mx + c

0 20log(1) 20logk

k=1

1

1 2 3 2 5 2 t 2x(2t) 2 1 2t 1 / 2

x(2t) t

s 100T.F

10s(s 10)


6/14




6

22.m

f 755

f 15

23B 75

2 2

O ISNR FOM SNR

4

I

4

i

7510 SNR

2

2SNR 10 24dB

75

23. O C a1 V A 1 K m t

OV 0

24. Impedance for TM waves

2

cg

fZ 1

f

So impedance increases with increase in frequency.

25.

In rectangular cavity

TE (mnp)

m=0, 1, 2

n=0, 1, 2

p=1, 2

TM (mnp)

m=1, 2, 3

n=1, 2, 3

p=0, 1, 2


7/14




7

26.

2

i i i i ix y x x y

2 1 4 2

1 2 1 2

0 3 0 0

1 3 1 3

2 4 4 8

0 13 10 7

2

i i in 5, x 0 y 13 x 10 i ix y 7

The normal equations are i ina b x y 2

i i i ia x b x x y

13

5a b 0 13 a 2.65

a 0 b 10 7 b 0.7

27. Taking Laplace transform of given equation

2

2

2s F(s) sf(0) f (0) 4F(s)

s 4

Applying initial conditions,

1 1

2 2 2 2

1

2 2 2 3

2 2F(s) L [F(s)] L

(s 4) (s 4)

1 t 1 1f (t) sin(2 t) cos(2t) L sin at at cos at

8 4 (s a ) 2a

28. D.E can be written as

2

2

x y xy y 0 cauchy Euler 'sequation

d.( 1) 1 .y 0 where and z ln x

dz

1 .y 0

A.E:- 2m 1 0 m i


8/14




8

1 2

solution is y C.F

c cos z c sin z

1 2

12

y c cos(ln x) c sin(ln x) ...(1)

c sin(lnx) cos(ln x)&y c ...(2)

x x

1 2

Usin g y(1) 0 and y (1) 1, we get

0 c and 1 c

y sin(ln x )

y(e ) sin ln e 0

29. Taking Laplace transform of given equation

2s Y(s) sy(0) y (0) 5[sY(s) y(0)] 6Y(s) 0

Applying initial conditions

1 1

2 2

1

2t 3t

2s 13 2s 13Y(s) L [Y(s)] L

s 5s 6 s 5s 6

9 7L

s 2 s 3

y(t) 9 e 7e

30. 2z iz 2 0 z 2i, i Singularities lie outside the curve C2 2C : x 2y 1

By Cauchys integral theorem,

2C

coszdz 0

z iz 2

31. We can treat the network on the RHS as another network B.

Therefore the circuit reduces to

2Network

A

Network

B

6A

2A

+ _V1


9/14




9

Total current leaving Network A = 4A

By KCL, total current entering network B = 4A

32. given that element 2 and 4 are connected in parallel

Voltages across element 2 and 4 are equalx y ...(1)

According to Tellegens theoremn

K K

K 1

V i 0 ...(2)

Where n is no. of elements and VKand iK are corresponding voltages and currents

From (1) and (2)

1.1 x( 5) 2.2 x(4) 9 0 ( x y)

1 5x 4 4x 9 0 x 14 y

33. Express V2, I2in terms of V1and I1

From network V2 = V1 (1)

And also 1 1 2V I I 3

12 1

VI I ...(2)

3

From (1) and (2)

2 1

2 1

1 0V V

1 1I I3

34.

1 2 n

F0

L

1 2 n F

L

0 1 2 n

F

L

F L

R R RV V V

Rn-1 n-1 n-1V = + .... 1

R R R RR+ R+ R+

n-1 n-1 n-1

V +V .... V R 1

n R

By summing amplifier

V V +V .... V

R 1So 1+ 1

R n

R = (n - 1)R

1V

2V

2I

1I

3


10/14




10

35. First half is low filter with cut-off frequency

H

1 1

1f

2 R C

Second half work as a high pass filter with cut off frequency

L

2 2

1f2 R C

For circuit to work as band pass filter

H L

1 1 2 2

1 1 2 2

f f

1 1

2 R C 2 R C

R C R C

36. Above circuit is voltage series connection

Output impedance decreases by factor of 1 A

' 00

Z 2kZ 2

1 A (1 500 2)

Input impedance increases by factor of 1 A

'in inZ Z 1 A 200k(1 500 2) 200M

37. Clk 1100

1 0110

2 1011

38. Instruction ORA B resets carry flag. So, loop START will be executed only once.

40. Drift current Voltage applied

2

2

I 2.5

8 2

I 10mA

41. BE1 TV VE1 C1 SI I I e

BE1 T BE 2 T

BE1 BE 2 T

V V V V

E 2 E1 S S

V V V

BE 2 BE1

I 2I 2I e I e

2 e

V V 18mV

42. 19 19bieV 1.76 10 ; e 1.6 10 C


11/14




11

biV 1.1V

43. x t x f

2.5s1

x 2 t 2.5 X f 2 .exp2

1

X f 2 exp 2.5s2

44.

45.

46..

1 1 2x = - 4x + x

.

2 3x = x + 2u

.

3 3x = - 3x + u

.

11

.

2 2

.3

3

xx-4 1 0 0

x = 0 0 1 x + 2 u

0 0 -3 1xx

47. The circuit equations can be expressed as

o p

2

1

1

2

t 1

t

u t

r t

1

u t y t

1

dy

r tdt

2

1


12/14




12

cL C

dv (t)= i(t) - i (t) -3V (t)

dt

The state is defined as

L 1i (t) = x (t)

c 2v (t) = x (t) The circuit equation in terms of defined state can be written as

.

1 1 2x (t) = - x (t) + 2x (t) + i(t)

.

2 1 2x (t) = - x (t) -3x (t) + i(t)

..

1 1

.2

2

x (t) x (t)-1 2 1+ i(t)

-1 -3 x (t) 1x (t)

48. Slope = -40 =1

24 0

log log8

1

1

1

log log8 0.6

log 0.303

2.009rad / sec

49. Char eq 1 G 0

3 2

3 2

s bs 3s 1 ks k 0

s bs s 3 k k 1 0

at marginal stability

3

2

1

s : 1 3 k

s : b k 1

b 3 k k 1s : 0

6

b 3 k k 1 0

k 1b

k 3

1 1 2 1b

1 3 4 2

b 0.5

Frequency of oscillation

Ggiven


13/14




13

2

2

bs k 1 0

k 1s

b

k 12 j j

bk 1

4b

k 1 k 34

k 1

k 1

50. 2 2bE 0.15 0.15 0.0225

2 2

C Cb S 2 b

A AE T log 4 T2 2

2

Cb

A0.0225 2.T

2

2

C C

b

0.0225A A 9.48

T

51. 1

H f j2 f 1j2 f

2

22

O

i

1H f 1 2 f

2 f

PSD 1H f 1 2 38.45

PSD 2

52. No error when bit 0.is transmitted, but when bit 1 is transmitted

1 1 1 12

2 2 2 4

1 1 1P bit 1 p bit 2 bit 1 22 2 2

53. encQ

The charge enclosed from the plane is

6Q 4 40 10 160 c

0.15

0.15


14/14




And from the line

6Q 2 6 10 12 C

Then, encQ 160 12 148 C.

54. x y z2 2

i j k

Curl F

2y 3x 2x z

2 j 6x 4y k

Here N k

By stokes theorem, we have

c s sF.dr curl F.N ds 6x 4 y ds 2 x

0 0

32

6x 4y dydx 10.673

55. for approximately normal incidence of wave onto dielectric surface.

t t1 2 1

i 1 2 i 1 2

H H 2,

H H

So,

t 1

r 2 1

H 2

H

So, t

t

1 22

H 9 3 41 1H 1 1

2 34 9

y 2,2

x 0,0 2,00

Documents

ECE GATE PAPER 21 ANSWERS