ECE GATE PAPER 24 ANSWERS

7/25/2019 ECE GATE PAPER 24 ANSWERS

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1

Answer Keys

General Aptitude

1 A 2 D 3 B 4 14 5 D 6 D 7 C

8 B 9 D 10 7

Electronics and Communication Engineering

1 7 2 13.72 3 D 4 C 5 32 6 7.23 7 C

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

15 4 16 C 17 5 18 D 19 4.92 20 D 21 159.15

22 C 23 25 24 B 25 240 26 A 27 C 28 1

29 4.7 30 D 31 9.951 32 C 33 3 34 10 35 B

36 1.5 37 A 38 0.333 39 3.192 40 0.02 41 3.16 42 C

43 10 44 5.48 45 B 46 0.2 47 5 48 B 49 1.6779

50 1.67 51 B 52 A 53 0.6 54 B 55 D

Explanations:-

General Aptitude

3. The given analogy is a pair of antonyms. Cavort means to enjoy while sulk means to be sad.

Similarly, goad means to instigate while deter means to discourage or prevent.

4. The number can only be two digit number.

Let the number be xy.

10x + y = 8 x + y10x+y-8x-8y=0

2 x - 7 y = 0

2x = 7y

x = 7 & y = 2

Product = 72 = 14


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2

5. The pattern of number series

56 1.5 84

84 1.5 126

126 1.5 189

189 1.5 283.5

8. There are 4 people on all 3 boards, let's call them A, B, C and D.

So, to start we have:

Charity 1: ABCD

Charity 2: ABCD

Charity 3: ABCD

That accounts for 4 of the 5 people that the boards have in common with each other, but eachboard needs 1 more person in common with each other charity. So, we need 1 more person forcharities 1/2, 1 more person for charities 1/3 and 1 more person for charities 2/3; let's call them E,F and G.

Now we're up to:

Charity 1: ABCDEF

Charity 2: ABCDEG

Charity 3: ABCDFG

Finally, we need to round out each board with 2 more people to get up to 8, so our completedroster is:

Charity 1: ABCDEFHI

Charity 2: ABCDEGJK

Charity 3: ABCDFGLMA to M is 13 letters, so we have 13 board members in total.

9. We're choosing 2 items out of 4. So, the total # of possibilities is:4C2 = 4!/2!2! = 24/4 = 6.

Therefore, the basic denominator is 6. Eliminate 1/4 and 1/5 (the answer could be 1/2 or 1/3 aftercancelling).

Next, we need to calculate the # of products that can be written as the question demands.

If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier.The only way to form a difference of squares is to multiply:

(a + b) and (a - b).The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressionsthat give us what we want.

Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.

10. We're given the slope of a line and one point on the line (the origin: 0,0). From this, we candetermine every other point on the line. The most direct way to find x and y involves solving for

each individually.


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3

We know that slope (1/2, in this case), is equal to the change in y divided by the change in x.Since we know that the line passes through (0,0) and (x,1), we can solve as follows:

1/2 = (1 0)/(x 0)

1/2 = 1/x

x = 2

Use the same approach to solve for y:

1/2 = (y 0)/(10 0)

1/2 = y/10

y = 5

Thus, x + y = 2 + 5 = 7

Electronics and Communication Engineering

1. D D1 D2 D3eq 1 2 3

W W W WBeacuse I I I I

L L L L

2 2 3

7

Here MOSFETs is in saturation (or) linear region, the result doesnt effect.

2.10 19 3

i r D

n G

n 1.5 10 ;n N 6.3 10 cm

P _______; E 0.035

For heavily doped semiconductor

GE kT2

n n i

210

0.035 0.026

n 19

20

n 19

n p n e

1.5 10p e

6.3 10

8.646 10p 13.72

6.3 10

3. State transition matrix A is Ate at 0 I.

4.2(a,b ) for f(x) = lx + mn + n

a + bSatisfactory

By language's mean value theorem, the value o

mean value theorem is x =2

1 2x =

x

1 5

.

f

2


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4

5. No. of complex additions in Direct DFT = N N 1 8 7 56

No. of complex additions in FFT =2 2N log N 8 log 8 8 3 24

Difference = 56-24 = 32

6. If the waveform comprises of linear combination of sinusoid of different frequency such as

0 1 1 1 2 2 2 3 3i t I i sin t i sin t i sin t

then

2 2

2 1 2rms 0

i iI I

2 2

but it is valid if and only if frequency are different

If same frequency components are there, to apply above formula we should convert it intostandard form.

5sin 2t 30 4sin 2t 40

5 120 4 50 7.39 89.44

7.39cos 2t 89.44

2

2

rm

i t 5 7.39cos 2t 89.44

7.39I 5 7.23

2

7.

2 2

2

Differentating with respect to x

dy 2Ax + 2By = 0 (1)dx

Again on differenciating, we get

dy d y2A+2B y 0 (2)

dx dx

on solving equations (1) & (2), we get

22

2

d y dy y dyy 0

dx dx x dx

order 2 , degree1

8.

2

1

KG

s

For unit ramp input vs 0

K Lt SG s


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5

9. 1 12 2

1 1

2

1 1 sL L

s(s 1) s s 1

1 sL L 1 cos t

s s 1

10.

Now ceX t is inverse Fourier transform of cex t

22 j2 ft

j2 ft

ce

j2 2 t j2 t

3 3j2 t j2 t

2 2j2 t2

2.ex t 2.e df

j2 t

2.e e

j2. t

e e2.e

j2. t

j2 t

j2 t

j t

3sin 2 t

22.e

t

sin 3 t2.e

t6 e sin c 3 t

Cf Cf cf 2

2

f

X f

2

2

f

ce cX f X f f


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6

11.

From Given informationiV is

Always i x iT

V V , b at t , V2

becomes 10V,

0 satV V 15V

12. The given spectrum is periodic with 2 and continuous in frequency. Thus the correspondingtime domain signal is discrete and aperiodic in nature.

13.

F 0 P x 0 P x 0 0.6

F 1 P x 1 P x 0 P x 1

0.6 0.4 1

0V

15

t

xV

iV

15V

15V

10k5k

OV

x

i

15 10k V 10

15k

V peak to peak 20V

iV

10

10

T

2tT


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7

14.

Let mark the various nodes of the circuit as A, B, C, D.

If someone take C as reference then VB= 6V

If someone take D as reference then VB= 1V

If someone take A as reference then VB= 4V

If someone take B as reference then VB= 0V.

So we can conclude that as long as the reference node is not defined we cant have a uniqueanswer for absolute node potential. In this case reference node is not given so we dont have any

answer unique.

15.

A B C m 1,2,4,7

16. Of course since all poles are on left of s-plane for all bounded input output it always bounded.

But notice the presence of right half zero i.e. if we give a unbounded input such that the polesof input canceled by zero of transfer function. The output is still bounded.

So in the above case if we choose specifically input as e2tu(t) output is still bounded but if wego for 4te u(t),

6t ate ....e if a 2, then it is always bounded.

17. Twisted ring counter frequency CLKf 60k

10 kHzmod-value 2 3

0

10kf 5kHz

2

18. z = a is a pole of order 3 of f(z)

22

2z az a

22 z

2z a

z 2 a 2

z a

1 dResf z lim z a .f z

2! dz

1 dlim 2z .e

2 dz

lim e z 4z 2 e a 4a 2

1 1

11

ABC

2V

4V 1V

5V

6V

BA C

D


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8

19. o1

Z n b a 752

o0 r

1n b a 75

2

m, r 4.92

20. The most fundamental assumption in any 2 port network is they should not have any independentsource.

In this case we cant defined 11

V

Iratio

21. The oscillation of frequency is

3

3

R 2 10f 159.15 kHz

2 L 2 2 10

22. DRAM contains only one transistor along with one capacitor, so their packing density ismore.

23. If the load happens to be purely resistive then for maximum power transfer the condition is

L

L in

2

RL rms L

rm

2

rm

2

R rm L

R Z 3 j4 5

P I R

2010I 5 16.56

8 j4

I 5

P I R 5 5 25W

24. Characteristic equation of A is

A I 0

1 3 23 2 1 0

2 1 3

3 26 3 18 0 By Caley-Hamilton theorem

3 2A 6A 3A 18I 0

2090

99

j9

1

2010r.m.s 5

3 j4


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9

25. Duty cycle

Average power

26. Considering MAP criteria,

1

2

H

HP x 0 / y a P{x 1/ y a}

1H decision in favour of x=0

2H decision in favour of x=1.

1

2

H

HP{y a / x 0}.P{x 0} P{y a/ x 1}.P{x 1}

2 4 1 1. .

3 5 3 5

on receiving y=a, we should decide in favour of x=0.Similarly,

1

2

H

H

>P x 0 y b P x 1 y b

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ECE GATE PAPER 24 ANSWERS