Solution For Gate Paper 2009

ECGATE-2009

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Answer Keys 1

B

2

A

3

B

4

C

5

C

6

C

7

A

8

D

9

B

10

A

11

C

12

B

13

D

14

A

15

A

16

D

17

C

18

C

19

C

20

D

21

C

22

D

23

B

24

B

25

A

26

D

27

A

28

B

29

C

30

A

31

D

32

D

33

A

34

D

35

D

36

D

37

A

38

C

39

A

40

A

41

B

42

D

43

B

44

A

45

B

46

B

47

C

48

B

49

C

50

D

51

B

52

B

53

C

54

C

55

C

56

D

57

A

58

59

B

60

D

Explanations:1.

The order of differential equation is two

3.

f (t) =

1 − cos 2t + cos 2t 1 , then it has o and Hz frequency component 2 π

n

7.

1 1  3  u (n ) ↔ 1   1 − z −1 3

z >1 , 3

n

1 1 −   u ( −n − 1) ↔ 1 2   1 − z −1 2 1 < z <1 2 3

z >1 , 2

8.

Since magnitude plot shows both increasing as well as decreasing plot, it is leadlag compensator

9.

Since Bandwidth is 10 kHz, thus output power is 10 × 10−11 × 10 × 103 = 1 × 10−6 W

11.

Use ncr (p ) ( q)

n −r

r

2

10

8

for any two losses which yield head 10

1 1 1 C 2     =10 C 2   2 2     2

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But in present case it is required only for first two tosses. Thus in this case

1 1 1 . K 10 times 2 2  2 

12.

13.

Since autocorrelation function and power spectral density bears a Fourier transform relation, then sin c required in frequency domain will five rectangular convolutions in time domain, thus it is a triangular function

−1 1 + c0 + z  f′ (z) =   z  

f′ (z) =

(1 + c0 ) + c1z−1

z d 2  (1 + c0 ) z + c1   = 2πi (1 + c0 ) ∫ f ′ ( z ) = d2 z  z2  

14.

Since 12A current is coming from one source and it is also known that 60V source is absorbing power i.e. current is flowing inside 60V source.

12 = x + I ⇒ I = 12 − x, thus possible option is ( A )

q µ D ⇒   = V −1 kT D 

15.

µ=

17.

1 SC zC = 1 RL + SC RL

zC =

RL SCR L + 1

RL SCR L + 1 RL RL = = = ⇒ RL = R RL Vi ( S ) R L + RSCR L + R SCRR L + R + R L +R SCR L

V0 ( S )

18.

Use the condition of controllability

20.

Apply right hand thumb rule

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22.

sin x = x −

x 3 x5 + L 3! 5!

sin ( x − π ) = ( x − π )

(x − π) −

3

3!

(x − π) +

5

L

5!

(x − π) + (x − π) − L sin x =1− 3! 5! (x − π) 2

−

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2

(x − π) sin x ⇒ = −1 + 3! ( x − π)

2

24.

Since it is one-sided Laplace transform

25.

( i)

dy dx = ⇒ ln y = ln x + C y x

y = eC ⇒ straight line x (ii) xy = constant

(iii) (iv )

x2 − y2 = constant x2 + y2 = constant

P-2 Q-3 R-3 S-1

36.

1 + z {y + z + y} 0 + z  = 1, z = 1, z = 0

40. 1− c

1

2

3

−1 −

( )

H e jω = e− j2ω − e− j3ω , It is FIR high pass filter

−ω2 + 1 = 0 . Thus output is zero for all sampling frequencies −ω2 + 2jω + 1 ω=1

44.

ω = 1, H ( j) =

46.

The mean is 3

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47.

µ=

48.

C = B log2 1 + SNR 

2

, efficiency =

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1 2 = 20% 2+ 1 2

C = B log2 SNR  C′ = B log2 2SNR  = B log2 SNR + B log2 2 C′ = C1 + B

(100 )

2

50.

z1 =

50

z′ = z1 zin =

= 200,

z2 = 200

z2 = 100

50 × 50 = 25 Ω 100

P1 P2 0 0 59.

g 0

c e d 0 0

From the figure 0

1

1 ⇒ g = P1 + P2 , 0 1 1

1 1

0 1

1 1

d=c+e

1 0 1 1 1 1

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4