ECE GATE PAPER 19 ANSWERS

7/25/2019 ECE GATE PAPER 19 ANSWERS

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1

Answer Key

General Aptitude

1 A 2 C 3 C 4 544 5 C 6 D 7 B

8 0.2 9 A 10 C

Electronics and Communication Engineering

1 A 2 B 3 B 4 0.06 5 0 6 C 7 D

8 B 9 1 10 3.766 11 2 12 5 13 64 14 31

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

22 0.029 23 B 24 D 25 A 26 -2.5 27 A 28 A

29 C 30 A 31 B 32 B 33 A 34 B 35 16.4336 3.5 37 D 38 B 39 C 40 5 41 1923 42 A

43 A 44 C 45 1 46 B 47 D 48 1.73 49 B

50 A 51 9 52 A 53 58.1 54 188.49 55 3.324

Explanations:-

General Aptitude

4. Required value 2 2 327 11 4 729 121 64 544

5. Anshu takes x days to do the work

4 8xAtul takes 2 x = days

5 5

1But given (Atul+ Anshu) =1daywork =

16

1 1 1 1 5 1+ = + =

8xx 16 x 8x 16

5

1 5 1 13x 1

+ = =x 8x 16 8 16

13x =16 = 26

8

8. We know that without any restriction she has to go 3 blocks south and 3 blocks east to get to

work. Hence, we need to find the number of arrangements of the anagram SSSEEE

Total arrangements : 6!/3!*3! = 20


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2

we know that she has to take 2 blocks south 1st. So, 1st 2 are fixed SS. The remaining 4 steps

we will have to find the number of arrangements of SEEE

= 4!/3!*1! = 4

Hence Probability = No of fav/total = 4/20 = 1/5

9. We know that 2x/y is a prime greater than 2. Therefore, x/y must be greater than 1, soeliminate (I).

We can quickly eliminate II and III by picking numbers. If we set 2x/y = 3 (a prime number

greater than 2), we get:

x/y = 3/2

So, any two numbers in a ratio of 3:2 will suffice. We can let x=6 and y=4 to see that neither

statement II nor III (nor I, for that matter) mustbe true.

As an aside, although it wasn't necessary for this question, we should remember to be careful

about the assumptions we make. Nowhere in the question stem does it say that x and y have

to be integers - we certainly could have picked non-integer values to eliminate all 3

statements as well.

The answer is A (none).

10. P(only 1 letter in the correct envelope) = P(1st correct and 2,3,4 in wrong) OR P(2nd correct

and 1,2,3 in wrong) OR P(3rd correct and 1,2,4 in wrong) OR P(4th correct and 1,2,3 in

wrong)

P(1st correct) = 1/4 (1 correct out of 4 envelopes)

P(2nd wrong) = 2/3

P(3rd wrong) = 1/2

P(4th wrong) = 1/1

Hence, P(1st correct and 2,3,4 in wrong) = 1/4*2/3*1/2 = 1/12

Same will be the probability for the other 3 casesHence required Probability = 4*1/12 = 1/3.

Electronics and Communication Engineering

1. Since both lines of regression pass through the point x, y , we have

8 x 10 y 66 0

40x 18y 214

Solving these we get x 13, y 17

2. Given 3f (x) x 2x 5

A=2; b=2;

By regula falsi method; we have

1stapproximation of the root is


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3

1

af (b) bf (a) 2f (3) 3f(2) 2(16) 3( 1)x

f (b) f (a) f (3) f(2) 16 ( 1)

32 32.0588

17

3. The directional derivative of f in the direction ofp

aa is f .

a

Where P is given point,

2 2

p(1, 1,1)

f f ff i j k i (y z) j (2xyz) k(xy )

x y z

f i 2 j k

Directional derivative ( i j 2k) 3

i 2 j k .1 1 4 6

4.2

1L[sin t]

s 1

222

3t

s 30

d 1 2sL[tsin t]

ds s 1 s 1

e (tsin t)dt L[tsint]

By definition of Laplace transform

30.06

50

5. t 0 sX(0) limX(t) lims.X(s) 2ss(808)

lim 0s(s 2s 101)

8. When the switch is at 1, the steady state becomes

i = 100/10+20 = 3.33 A

As soon as the switch is moved to position 2 at t = 0, the LR circuit starts decaying and the

decay current is given by

t

i Ie

Where I is the initial steady state current and is time constant L/Rt

40t0.025i 3.33e 3.33 e A

9. Diode D1will turn on first

0

5 1I 2mA

2k

V 1V


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4

10. Transistor Q2will be off as BEV 1.3V

E E2 0.7 3k(I ) 10 0 I 3.766mA

11. TH 1V (T ) 4V

TH 2

TH

V (T ) 6V

R 2.4k

B

5

B1

E1

E2 C1

B2 CE1 E1

E1 CE1

CE1

4 0.7 (2.4k 100 1k)I 0

3.3I 3.22265 10

102400

I 3.2226mA

I I 3.1903mA

6 2.4kI 0.7 V 1kI 0

6 0.0765 0.7 1k(I ) V

V 2V

CE2 CE210 3.3(1) V 2 3.3(1) 0 V 1.4V

12. For Half adder

Sum S A Bandcarry,C A.B

13. Conversion time N clockperiod

clockperiod 8 s

conversion time 8 8 s 64 s

14. n2 1 32 1 31

S A B

C AB

B

A AB


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5

15.

16.GE

2 kTi Cn N N e

17. y t h x t d

10

6

y 3 h x 3 d=4

5 10

4 5

y 1 2+ 1=7

18. M=2 M+N-1=5 N=4

19. An open loop system can be changed into stable/unstable using closed loop technique

21. s

3

term is not present. So, the given system is Unstable.

22. 2 2H 0.4log 1 0.4 0.6log 1 0.6

0.971bits symbol

C 1 H 0.029

0.6

0.4 0.4

0.6

drift

310

510 E

a a b b c c d d

a b c d

a b c d

a b c d

-1 1


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6

24. c

1 c

2

1

c c

c

120

2 2

120 120

25.

x y

o

x y

E 5cos t z a 5sin t z a

5cos t z a 5cos t z 90 a

We can see that x & y component of electric field have same magnitude and y componentleads x-component by 90

o. So it is left circular polarized.

26. 1 1 3

A B2.5 b 7.5

2 2 1R R 2.5 R ; then

1 1 3

A B ~0 b 2.5 0

Per the system to have infinitely many solutions;

b 2.5 0 b 2.5

27. Given P=

1 1 1 3

1 2 3 4

2 3 4 9

Reduce the matrix A to Echelon form.

Apply 2 2 1 3 3 1R R R ; R R 2R ;

P ~

1 1 1 3

0 1 2 3

0 1 2 3

Apply 3 3 2R R R ;

P ~

1 1 1 3

0 1 2 1

0 0 0 2

Number of non-zero rows = 3 = no. of given vectors

Rank of P 3

So, the given vectors are linearly independent


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7

28. 2 2 2f z x y 2ixy z is analytic Everywhere

1 i

0f z dz

is independent of the path joining

Z = 0 and z = I + i

29. 3 3 2 2x D 3x D xD 1 .y 0 cauchy Euler equation

3

3

1.z

z 21 2 3

1 /2

1 2

d d( 1)( 2) 3. ( 1) 1 .y 0where , D

dz dx

1 y 0 L.D.Ewith constan t coefficients

1 3A.E : m 1 0 m 1, i

2 2

solution is y

3 3c e e c cos z c sin .z

2 2

3y c x x c cos n x

2

l 3

3c sin n x

2

l

30. 2f x 1 2cosx 2sin x in 0,

f ' x 2sinx 4sin x cosx

f ' x 0 2sin x 1 2cos x 0

1sin x 0; cosx2

x 0 , ,3

f 0 1 2 1 ; f 1 2 1 3

and 3f 3 1 2 1 2 2 3 4 2

f x has absolute maximum at x

Now f" x 2cosx 4cos2x

1 1at x , f " x 2 43 2 2

f x has local minimum at x 0,3

31. Taking the laplace transform of the whole circuit


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8

Applying KVL,

15 I S S

S S

25

I S

S 1

i t 5sin t

32. 3O

5

1 12 20 10 L 6.33 H

LC L 10

3

5

1 1BW 10 10 R 10

RC R 10

33.

A B

A

B

[T] [T ][T ] (cascade connection)

J5 J51 J5 J51

T 11 1

1 1

J51.5 J51 J5

J2T J

1 11 2

J2

1.5 J51 J5 J5

[T] J 11 1

2

1.5 J7.5 2.5 J10 25

3 J

2 2 J5 1

4 J7.5 J10 25

1.5 0.5J J5 1

5s

I S

S

1S

J5

J2

J5

1

AT

BT


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9

35.

H-12

1 2 1 2

1 1f = = 16.43kHz

2 R R C C 2 30 25 0.5 0.25 10

36. 0 11L 1 2

0 1

1 1 1L

V - V7 - VI = I +I

2K 2K

V 2V

7 - V 2V - V 7I 3.5mA

2K 2K 2K

38. J X

k Y

n 1

n

n

J K Q

0 0 Q

0 1 0

1 0 1

1 1 Q

39. BC+D2

1011 11001101 0010

1 1000 1110

Z = 0, CY = 1, MSB = 1 so S = 1

Number of 1s = 4 so P = 1.

40.

15 3

d B

11

2142B

n 19 15

d

N 5 10 cm & V 100V

2 V 2(11.7)(8.85 10 )(100)x

eN 1.6 10 5 10

n(min)x 5.09 m.

41.2 1

R k k 12516 8

I

I

I

Loa

7

2k

2k

2k

2k

V


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10

2

0

15

0

12

RA

2 2125 10 ohmcm

2

2.5ohmcm

1

en

n 1.923 10

1923 10

43.

-1

-1

-1

-1

b[n]=x[n] b[n-1]

b[n] 1

x[n] 1 z

1y[n]=b[n] b[n-1]

2

y[n] 11 zb[n] 2

11 z

y[n] 2

x[n] 1 z

44. 2

2

S y(S) Sy(S)-2y(S) x(S)

y(S) 1

x(S) S -S-2

poles are 2,-1

It contains j axis, So ROC should be between two poles.

So assertion is true.

Reason: It is not always true that for stable system

ROC should lie between two poles for causal stable system. ROC should be right side of right

most poles

x n y n+ +

1/ 2

-1z

b(n)


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11

45. 1x 2t 1 x 2 t 2

46. 1 2 1 3 4 2 2 3 3 1 2 1 3 4 2dr 1 g g h g g h g g h g g h g g h

k k 1 2 3 4P g g g g

1 2 3 4g g g g

R dr

47. Char equation 1+ G = 0

2 2

2

s 2 s 4 k s 3 s 5 0

s 6s 8 k s 8s 15 0

s 1 k 3 8k 6 8 18k 0

for all coefficient ve

1 k 0

k 1

k 6 8

8 15k 0

k 18 5

k 6 8

8 15k 01 k 0 8k 6 0

k 1 k 6 8 0.75k 1

2

1

1 22

1.50.5 2.5

x 2t

x 2 t 1 2 1


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12

So k< - 1 and k>0.75

Char equation: all coefficient must have same sign either +ve orve

48. t

c t k 1 e Given ; k 1 2.5

dat t t ; c t 0.5

dt 1.732sec

49. O O 1 OGH 180 90 3tan 1802

1 O pc3tan 2 270

pcM| 0 GM

2which will not be satisfied for any frequency

4

From 0 to

gc does not exist PM

51. 2

ONNoise power1 H f 2B2

0j4 ftON

9e 2B2

O9N B

Noise power 2 = ON

2B2

ON B

Therefore Ratio of 0

0

Noise power 1

Noise

9

power

N

B

B

N2 = 9

52. BPSK Rx

c2

sin t6T

mB

mA

detection

c

2sin t

T 6


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When received signal is

1 c

2ES sin t

T c

2sin t

T 6

S

O

m C c C

T

2E 2 EA sin t 30 sin t dt cos30 cos 2 t 30 dt

T T T

S

O

M C c C

T T

2E 2 EB sin t 30 sin t dt cos 2 t 30 cos30 dt

T T T

When2S is received

m

m

A E cos50

B E cos30

1

e

0 0

d E cos30P G Q

N 2 N / 2

0 0

3 2 EQ E Q

2 N 2N

53. Wave gets attenuated to 37% of its original value when it travels distance equal to depth ofpenetration

2

Here

6 6

0

258.11 m

3 2 5 10 5 10

54. For 0

Wave impedance is given by

r

r

120 12016

60 188.49

55. Reflection coefficient

L 0

L 0

Z Z 50 75j 100s 0.53748 97.12

Z Z 50 75j 100

1 s 1 0.53748VSWR 3.324

1 s 1 0.53748

m

m

A E cos30

B E cos30

d '

x

x

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ECE GATE PAPER 19 ANSWERS