gate mock test ece solution

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EC : ELECTRONICS AND COMMUNICATION ENGINEERING

GATE 2013 Mock Test SolutionDuration: Three Hours Maximum Marks: 100

Test Booklet Series : C Test Type : 5 Paper Code : EC101

Solution 1: b

Time required for n bit ABC

2n

f

Time required for 8 bit8

6

2 2560.064 sec.

4 10 ABC m

f

Solution 2: c

3 30

1 1 1 110 10 1 10

2 4 8 16v

1510 9.31

16volt

Solution 3: c IFT

PSD ACF For white Noise PSD is

So ACF will be impulse

Solution 4: a

Solution 5: b

0

sin 2 sin . [ sin ]]c t dt c t dt c is even function

Central ordinate theorem

(0) ( ).

sin

FT

x x t dt

t sinct

t

sinct

. 1sinct dt

So0

. 1/ 2sinct dt

Solution 6: b

24 36 z z




GATE 2013 Mock Test Solution2

1 11 1

1

1

9( )

1 1 2(1 2 ) 1 1

4 4

35 1 16( ) .

1112 1 21 4

z z A B H z

z z z z

H z

z z

For system to be stable

1| | 2

4 z

35 .1[ ] [ ] 16(2) [ 1]

112 4

nnh n u n u n

Solution 7: a

Lorentz force equation: F (E B)q v

When q experiences no force

Then, E ( B)q q v

E ( B)v ^ ^ ^ ^[( ) ( 2 )] x y x z ^ ^ ^

2 2 V/m x y z

Solution 8: d

E 0, H 0 TE Mode z z

E 0, H 0 TM Mode z z

E H 0 TEM Mode z z

E 0, H 0 Hybrid Mode z z

Solution 9: d

Since the input is a step i.e. 10 U(t )

P

10 10

1 K 6sse

and 12 t U(t ) is a ramp input

Hence sse

Since sse = constant value for step input this suggests that the system is a type-0 system and

1

K ss

v

e for ramp input.




GATE 2013 Mock Test Solution

Where,1

ss

v

eK

1ss

v

eK

Solution 10: b

1 + G(s) H(s) = 0 2 8 64 0s s On comparing it with a 2nd order standard C-R equation is

2 22 0n ns s

8n

2 8n

0.5

and the frequency at Resonance peak is given by21 2

r n

2 2 8

1 2(0.5) 8 1 2 4 2 rad/sec4 2n

Solution 11: c

Area 2 total widthS

4

wa

then applying this technique, it becomes

2 22T T2AT S 2 AT S4 2

ww a

Solution 12: d

By KCL at node “C”

3.5 = 22 I

2I 1.5AKCL At node “D”

A

– T T

FT

1

2

3

4

R

4A

2A

I2

1A

3.5

A

D

C

BE





3I 1.5 1 0.5 A KCL At node “E”

2+0.5 = 1I 4

1I 1.5A

Solution 13: b

Let the equivalent resistance across 5A source Be R then circuit becomes.

The switch is connected from a long time thus inductor acts as a short circuit.

So current through inductor at 0t is

L L(0 ) 5A (0 )i i (As inductor does not allows a sudden change in current)

At 0t

L (0 )i will flows through the resistance of 20

R v = 5 × 20 = 100 volts

Solution 14: c

The open loop transfer function of a closed loop system is =K

( 1)s s

Thus T.F. =2

K

K ( 1)

K K 1( 1)

s s

s s

s s

On comparing the T.F with a standard transfer function of 2 nd order system

i.e.,2

21 2n

n ns

K n

and 2 1n 1 12 .

2 K

Thus,1

Gain

1 2 11

2 1 1

2

1

K K 4K 2

1 K K K

12

2

Hence is reduced by 2 times.

VR

i1

10H

20

+ –

5A

t = 0



EC : ELECTRONICS AN

GA

Power consumed by resistive

Power consumed =

and PP 2

P P

400I

Z R (

v

2P

4003840R (2

2P P0.024 R R 0.

P

80R ,1

3

Solution 16: c

Result is 1,0A. Hence answer isSolution 17: b

Max. Slope of the signal is

max

3

4min

( )

3

10 10

3 10

d m t SFS

dt

S

S

Solution 18 : a Solution 1

Solution 22 : c

Since 3-branches teSo P – Z = 3 [P = 3]

0 z

Thus the transfer function of

T.F =

K

( 2) ( 4)s s s

As K is marked at imaginary

K = marginalK

From C – R equation of T.F.2( 2 ) ( 4) K s s s

3 26 8 K 0s s s Routh Array

D COMMUNICATION ENGINEERI

E 2013 Mock Test Solution

part only

2P PI R

2 2 2P P

400

) R (20)

2

P2R

)

2024 (20) 0

5

3 to avoid slope over load error

: a Solution 20 : c Solution 21: a

rminating at infinityP = Open loop pole, Z = Open loop zero’s

he graph is

axis i.e.

0

G





2

1

0

1 8

6 K

48 K 0

6

K

s

s

s

s

For K = marginalK

48 K 0

6m

K 48m

Hence ‘c’ is correct

Solution 23: c Solution 24: d

Solution 25 : a

For the impedance to independent on supply frequency the current in the two branches must bein quadrature.

1 1L 1

tan tan 90R RC

Thus,

1 1L 1tan tan tan

R RCL 1

1 .R RC

i.e.

L 11 . 0

R RC

LR

C

Power loss =

2 22 C

R LL

C

v vv

Solution 26: a

+ –

R

j

wc jwL

v

R

ZL ZC





1 1 2 21 2

( ).( 1).( ); 1 ;

( ) ( 1)

d G s s sP PC sP P

U s s s

2 ( ).( 1)d P G s s

1 21; 1

2( 1) 1

1 . ( 1)1

s s

xs s

2

1 ( )( 1)( )0

( ) 1 ( 1)( 1)

d G s sC s

U s s s

1( )

1d G s

s

( ) ( )t d G t e u t

Solution 27: c

2 16( )

( 1)( 3)( 2) 1 3 2

s A B C X s

s s s s s s

15 1 7 12( ) .6 1 10 15( 2)

X ss s

For Fourier transform to define if ROC is 1 Re( ) 2s

Then,3 215 7 12

( )6 10 15

t t t x t e e e

Solution 28: c

1( )

2

be

E P PSK erfc

1( )

2 2

be

E P FSK erfc

Given that

( ) ( )e eP PSK P FSK

( )( )

2

bb

E FSK E PSK

For sinusoidal

2

2

A2

( )2

b

AT PSK

1( )

2bT FSK

( ) 1

( ) 2

( ) 1

( ) 2

b

b

b

b

T PSK

T FSK

R FSK

R PSK

Solution 29: a Solution 30: d

Solution 31 : a



EC : ELECTRONICS AN

GA

11 1

22 2

23 3

44 4

; 1

; 1

; 1

; 1

P z

P z

P z

P z

1 2 2( )

( )

Y z z z z z

X z

1 2 4( ) 2 H z z z z

[ ] [ 1] 2 [ 2]h n n n

After 3 clock pulse

0 .1 .1 ( )

( )( )

Y AB AB AB B A B

B A B B A B

Solution 32: d



4

[ 4]n

AB B AB

G





Solution 33: c General expression is^

E cos( ]o x t z

4 rad/m.C 3

r r

^ 4 8 4 1E 10 cos 2 (10 ) V/m

3 8 x t z

^ 4 8 410 cos 2 (10 )

3 6 x t z

a phase delay of 1

m8

along z is given such that E is maximum as z =1

m8

as given.

Solution 34: c A charge q moving with a velocity v experiences a force F in the magnetic field.

F V Bq

If we know two non-zero forces 1 2F and F for two velocities, then we have

B = 2 1

1 2

F F

(F . V )q

Substituting the values , we get B =^ ^ ^ 2( ) Wb/m x y z

Solution 35 : a Solution 36 : d Solution 37 : a

Solution 38 : cSolution: Total no. of sample space = 6 × 6 = 36The sum of the numbers on the two dice can be ‘8’ in 5 ways i.e. (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)and sum of two numbers can be 9 in 4 ways i.e. (3, 6), (4, 5), (6, 3)Required probability of coming sum as 8 or 9= 9/36 = ¼

So, Probability of not coming sum of 8 or 9 =1-1/4=3/4

Solution 39 : b Solution 40 : a Solution 41 : a

Solution 42 : a Solution 43 : b Solution 44 : a

Solution 45 : dP. State N. State QB QA

QB QA QB+ QA+ JB JA

1 1 1 0 X 11 0 0 1 X X0 1 0 0 0 10 0 1 1 1 X



EC : ELECTRONICS AN

GA

1 A J

Solution 46 : d

1 2 3I I I 0

5 12

12 2

A A

A

v v

v

2 24 5 A A Av v v

52 Av

Solution 47 : c

Q

VC

Threshold newV

gateoxideCd

(per unit ar

1

8

3.9 8.854 10

400 10

Now putting values is above

191.6 101 1

0.086

19

6

1.6 102

0.080 10

N 1.08

Due to –ve threshold vol

Solution 48 : c

Solution 49 : aWrite the given function in time co

9 1

3[1 ]

s

s s

Put s j



B A B J Q Q

1602

Av

60 16 0 Av

[general formula]

doping concentration

Threshold old

gate oxideC

qN

ea)

460.086 10

ormula

doping

60

12 210 /cm

ltage it is a p -type MOSFET.

stant

G





9 13G( )

[1 ]

j

j j j

2 2

2 2

9 1 9 13 3

| G( ) |

[ 1 ] [ 1 ]

j

1 1G( ) 180 tan 90 tan3

j

0

| G( ) | 0

G( ) 90 270

j

j

The plot will cut the real axis when imaginary part of G( ) 0 j

9 13G( ) H( )

[1 ]

j

j j

j j

2 2

2

9 13 [ ]

j

j j j

23 . 9 0 j j 23 9 0

3 r/s

The Nyquist plot is

Hence system is “stable”

and 3 r/s pc

Solution 50 : a Solution 51 : d

Solution 52 : d

At the output of balance modulator

cos ( ). . cosC c f C c A t K m t dt A t 2

cos(2 ( ). ) cos( ( ) )2C

c f f

At K m t dt K m t dt

–1 + 0 j

j

= 3

/ rs





( ) cos 2 ( ).2

C c f

A y t t K m t dt

Solution 53 : c

Solution 54: b

The given is a transformer so

2 1 2

1 2 1

I N

I N

v

v

1 25v v …(i)

21

II

5

…(ii)

and from the circuit

2 2 22

V VI

16 16 8

v

…(iii)

By KVL to input side

1 2 1

50 2500[I ( 0.02 )]

2v v

22 2

I50 2 500 0.02 552

v v

22 250 500 0.02 5

40

vv v

50 × 40 = 2 2 2500[ 0.8 ] 200v v v

2 1.81voltsv

Solution 55: c2 2

2L

(1.81)P 0.4095 watts

8 8

v

Solution 56: a Solution 57: b Solution 58: b Solution 59: a

Solution 60 : b

The correct word to be used instead of illusion is delusion.Illusion means false picture. Delusion means false idea.

The sentence means the friend should not have any false idea about his scholarship.

Solution 61: b

Solution 62: b

Each compound is compared with all the other, giving 20 comparisons or results, but since each comparison has

been counted twice, we have =20

102

comparisons.

Solution 63: b

Solution 64: d

Therefore, required sugar = 300*60%-300*40% = 60 gram.

Solution 65: a

The enemy aircraft will be brought down even if one of the four shots hits the aircraft.

The opposite of this situation is that none of the four shots hit the aircraft.

The probability that none of the four shots hit the aircraft is given by (1-0.7)(1-0.6)(1-0.5)(1-0.4) =

0.3*0.4*0.5*0.6 = 0.036

So, the probability that at least one of the four hits the aircraft = 1 – 0.036 = 0.964.

gate mock test ece solution

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