ece gate paper 9answers

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ECTest ID: 161259 TarGATE16 www.gateforum.com

ICPIntensive Classroom Program eGATE-Live Internet Based ClassesDLPTarGATE-All India Test Series

All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any formwithout the written permission. 1

Answer Keys

General Aptitude

1 A 2 B 3 A 4 A 5 240 6 B 7 B

8 A 9 40 10 A

Electronics and Communication Engineering

1 D 2 A 3 D 4 A 5 3 6 10.35 7 14

8 B 9 D 10 200 11 A 12 C 13 D 14 D

15 D 16 B 17 1 18 D 19 A 20 A 21 A

22 D 23 0.25 24 1.13 25 0.515 26 A 27 A 28 C

29 B 30 A 31 D 32 1.1167 33 82.1 34 5 35 A

36 7.65 37 82 38 C 39 C 40 0.022 41 4.7 42 A

43 C 44 1 45 D 46 A 47 C 48 18.42 49 A

50 B 51 C 52 B 53 4.77 54 0.74 55 2.9

Explanations:-

General Aptitude

1. Accordion, banjo and flute are musical instruments while skillet is used in the kitchen.

2. Blue blood refers to aristocratic ancestry.

3. Calumny means defamation or lie while acclaim means to praise.

4. Third no400

805

Smallest no=76

And largest no=84

Required answer=76+84/2=80

5. Speed of carDistancecovered

Time taken

450

75kmph

6

2speed of tractor 75 30kmph

5

Distancecovered by tractor in 8 hours 30 8 240km

6. * D- Despotic, T- Tyrant B- Bureaucrat


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It can be inferred from the diagram that tyrant and bureaucrat are not related, henceconclusion I is correct. Conclusion II is wrong because some despotic are tyrants but not allof them.

7. In option A with his prayer or thank-offering is wrongly placed, this modifier refers to theindividual who approaches god. Option C is wrong because community is a collective nounand so their should be replaced with it. Option D is incorrect because after the wordapproach it is appropriate that the place (altar of god) is mentioned.

8. Required population

1 2R RP 1 1

100 100

12 15245000 1 1

100 100

3 3245000 1 1

25 20

28 23245000

25 20

315560

9. Since 70% of the employees received bonus of at least 10,000, 30% of the employeesreceived bonus of less than 10,000. We know that 60 employees received bonus of less than10,000. If E is the number of employees, we can set up the following equation:

.30E = 60

E = 200

40% of the employees received bonus of at least 50,000. Thus, (40% * 200) or 80 employeesreceived bonus of at least 50,000.

20% of the employees received bonus of at least 1,00,000. Thus, (20% * 200) or 40employees received bonus of at least 1,00,000.

If 80 employees received at least 50,000, and 40 employees received at least 1,00,000, then

(80 - 40) or 40 employees received bonus of at least 50,000 but less than 1,00,000.

10. Arc Length = Circumference * ((Arc Measure)/360)

Solving for Arc Measure,

24 cm = 144 cm * ((Arc Measure)/360)

24 cm/144 cm = ((Arc Measure)/360)

24 cm * 360 = (Arc Measure) * 144 cm

8640 cm = (Arc Measure) * 144 cm

DT

B


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8640 cm/144 cm = Arc Measure

Arc Measure = 60

The answer is A.

Electronics and Communication Engineering

2. P = probability of defection item = 2% = 0.02

n = 100

=np = 100 0.02 = 2

Let x be random variable of defective items

Required Probability = 0 2 0

2

2

e e 2 1P x 0 e

0! 0! e

3. D.E can be written as

xy

2 2

xy 1

xdy ydxe (ydx xdy) (bydx axdy) 0x y

d e d tan y / x bydx axdy 0 ...(1)

equation(1) is exact if (by) (ax) b ay x

4.(6)

5

6 6

5!L t F(s)

s s

(say)

5 3t at

6

120

L t e L e f (t) F(s a)(s 3)

5. We know that 0

f x dx 1

3x

0

3x

0

0

k e dx 1

ek 1

3

ke e 1

3

k 3


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6.V = V - V

a 1 3

V = 18V2

V V V

1 1 2 1.56 3

V = 15V1

V V - V3 3 2

+ + 1.5 = 04.5 3

V = 8 V3

V = V V 15 8.1 6.9Va 1 3

P = V 1.5 = 6.91.5 =10.35W1.5A a

7. V - V = 12V (super node)1 3

VV - V31 2 + + 4 = 0

20 2

V = 10 V2

VV 1 31 - + + 4 = 020 2 2

V +10 V 1 7 71 3 = - 4 =20 2 2 20

V 10V 701 3V V 12V

1 3

11V = -70 - 12 = -823

V 7.45V3

V = -19.45V1

V V V4 4 2

+ - 4 = 04 8

3V V 324 2

V 14V4

12V 8

20

4A

V1

V4V3

V2

2 4

10V


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10. T

T

f f

200f f 201 200MHz.

1 201

11. 10 24.8Number of steps 12 11000.4

12. Clk J K Q Q

1 1 0

1 0 1 0 1

2 1 1 1 0

3 0 1 0 1

13. RVI 0.125mA8 12k

17.

u t y t

dyu t

dt

19. O 2 111 1 2 1 1

V z RT F ; z

V z z 1 R C s

2 1 1

1 2 2 2 1 1

R 1 R C sT.F

R 1 R C s R 1 R C s

21.

2

n2

Y s 6060

R s s 16s 60

16 8

2 60 60

Damping ratio = 1.033

Damping constant = n2 16 rad sec

22

2 2

Rz

1 R C s

1

1 t

t1


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22. For DSBSC, Bandwidth is 20 KHz

DSBLC, Bandwidth is 20 KHz

SSB Bandwidth is 10 KHz

For FM

f 12 f w , f k m t max2 100 10 220 kHz

23. For a continuous Random variable probabilities can be found using area under pdf for

the region greater than 2 and less than -2.

24.a a

22Ob b

Qdr Qdr V

104 r4 r

r

OQ Qbln . C 1.13nFa40 V

25.

2

0

0 0r r 0 0

0

2 2

1x 1

3

16x 1 x 1

2 c 2

x 9 / 8

91 tan Loss tangent

8 =0.5154

26.x 40 4.2 4.4 4.6 4.8 5.0 5.2

log x 1.3863 1.4351 1.4816 1.5261 1.5686 1.6094 1.6487

By simpsons1

rule;3

5.2

4

0.2log xdx (1.3863 1.6487) 4(1.4351 1.5261 1.602) 2(1.4816 1.5686)

3

0.2

3.035 18.2824 6.10043

0.2[27.4178] 1.82785 1.83

3

27. By Greens theorem


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c

ydx cosxdy ( sin x 1)dxdy

12

yy 0 x

2

(sinx 1)dx dy

1 2

yy 02

1

y 0

1

2

0

2

cos x x dy

y ycos dy

2 2 2

y

y2y sin2 2 2

2

2 21

4 8

28. We know that v v

f z iy x

if f z u iv is analytic.

2 2

22 2

x y 1 x y 2xv

x x y

And

2 2

22 2

x y 1 x y 2yv

y x y

Replacing x by z and y by 0 (Milne Thomson method)

in (1), we get

2 2

4 4

2

z zf z i

z z

1 i

z

,1

2

x2

y 2x /


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Integrating, we get

1 i

f z cz

29. f(x), g(x) satisfy the conditions of cauchys mean value theorem

f (c) f(b) f(a)

g (c) g(b) g(a)

cos c sin b sin a

sinc cosb cosa

a bc

2

30. f(x)= 3x 2x 5

Given that 0x 2

By newton-raphson method, we have

3nn 1 n

n

2

f (x )x x ; where f (x) x 2x 5

f (x )

f (x) 3x 2

01 0

0

12 1

1

23 2

2

f(x ) f(2)x x 2 2.1

f (x ) f (2)

f (x ) f (2.1)x x 2.1 2.0946

f (x ) f (2.1)

f(x ) f (2.0946)

x x 2.0946 2.0946f (x ) f (2.0946)

32. 1122 1 = 0

Vh = |

IV

I2/4

I2V2/5

+

_

+

_

I1

V1

5 I2

20KV2

I2/4

+


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2 2 2

2 22 1

2 2 1

22 1

22 1

2 1

1

2

3V = I .20 15I

4

I VV = 5 V

4 5

6 5 V I V5 4

V6 5 V . +V5 4 15

V6 V +V

5 12

67V V

60

V 671.1167

V 60

33.L

L

L 2 2

2 2

24 j20 24 j5 j120 j120(6 + j5)Z = 8 + 8 + = 8 + 8 + j11.8+17.83

24 j20 6 j5 6 + j5 6

Z = 17.83 j11.8

17.83 j11.8y

17.83 11.8

j11.8j (314)C = C=82.16F

17.83 11.8

34. Maximum Ac output power is

CQ R

CE Q CC

CEQ CQ

max.

10VI I 1A

10

V V 10V

V I 10 1P 5W

2 2

36.C 3 3 9 9

3 4 1 2

1 1f

2 R R C C 2 8 10 9 10 3 10 2 10

Cf 7.65kHZ

37. Instruction PCHL will load program counter with HL register pair. So here PCHL

will load PC with 1009 H and program execution will shift to 1009 H skipping ADD

B.

So accumulator = 40H+42H = 82H.


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38.

39. If control input is 0, then

1

2

F x yHalf adder

F xy

if control input is 1, then

1

2

F x yHalf substractor

F xy

40.i F

10.3

E E1 exp

KT

i FE E 0.022eV

41. The zener diode initially is in the break down region. So, further increase in current will notchange the voltage across the zener diode.

0

A B

1

A B

C

y c.0 c.1 c

1F 0

0

1

2F 1

4 :1

MUX

4:1

MUX

2:1

MUX


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43. i p

So h(t) 0 t < 0 sonot casual

44.

45. 4 4 40

2 12 u t dt 1

4 2

t

tt t t

e e e

46. 2

k s 3T . Fs 1 s 4s 5

s 0

d . c gain T S

3k 408 ; k 13.33

5 3

213.33 s 3

T Fs 1 s 4s 5

48. T 0.13 ; T 0.25 0.13

0.520.25

Maximum phase shift

1 1

m

1 1 0.52Sin sin

1 1 0.52

o p

0 2

0 p

0 11

t

t 1

h t

0

o p

2

1

1

1

t 2 t

t

2 (t)

u(t)


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1 10.48

sin sin 0.3161.52

18.42

50. For finding the probability of error, let us find the probability of correct reception as per the

decision rule.

Total No. of decision rules will be 8.

Rule I

Decide in favour of X = 0

Rule II

Rule III

Rule IV

Rule V

Rule VI

Rule VII

Rule VII

Now

Y a b c

X 0 0 0

Y a b c

X 0 0 1

Y a b c

X 0 1 0

Y a b c

X 0 1 1

Y a b c

X 0 1 1

Y a b c

X 1 0 1

Y a b c

X 1 1 0

Y a b c

X 1 1 1

P Y a X 0 p Y b X 0 P Y c X 0

P Y a X 1 p Y b X 1 p Y c X 1

a b c

0

1

0.12 0.16 0.12

0.12 0.18 0.3


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Then as per the decision rule given in equation

c

e

P 0.58

P 0.42

51. If a constellation point is located at a distanced from decision boundary and noise in

AWGMwith variance z , then probability of error is given byd

Q .

Using this principle, we can find the probability of error in above case.

Probability of correction reception

c s

c s

P n t d and n t d

P n t d .P n t d

Since cn t and sn t are independent of each other.

c s

2

2

e

2

2

1 P n t d 1 P n t d

d d1 Q 1 Q

d d1 2Q Q

d dP 2Q Q

1 12Q Q22

2Q 0.5 Q 0.5

52. From the transformation it appears that continuous Random variable X is becoming

discrete Random variable Y

Y takes value 1 and 2

P{X=1} = P{0X


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2

2 2Oavg

I d1U , P r sin W rad

2 4

2 2

2

rad O

0 0

P U , sin d d I d124

2rar

avg O2

PU I d1

4 96

2

d

U ,G , 3sin

Uavg

d maxD G , 3 at 2

10D 1010g 3 4.77dB

54. Directivity (D) =

4L 4 17

174

D 12.3dB

Beam solid Angle

4 40.74 r

D 17

55. Maximum average power density is

O

avg

1S EH along 0 elevation

2

Power density along 30Oelevation

avg

1S EHcos30

2

6 6 2

avg

2S 8 10 9.24 10 w m

3

2

2 3

rad avg avgP S 4 r S 4 5 10

2.9kW

ece gate paper 9answers

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