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Electronics & Telecommunication Set - 1

General Aptitude :

Q.1 to Q.5 carry one mark each

Q.1 Which of the following is CORRECT with respect to grammar and usage?

Mount Everest is ____________.

(A) the highest peak in the world (B) highest peak in the world

(C) one of highest peak in the world (D) one of the highest peak in the world

Ans. (A)

Sol. Before superlative article ‘the’ has to be used. ‘one of ’ the expression should take plural noun and so

option ‘C’ and ‘D’ can’t be the answer.

Hence, the correct option is (A).

Q.2 The policeman asked the victim of a theft, “What did you ____________ ?”

(A) loose (B) lose (C) loss (D) louse

Ans. (B)

Sol. “lose” is verb.

Hence, the correct option is (B).

Q.3 Despite the new medicine’s ______________ in treating diabetes, it is not ______________widely.

(A) effectiveness --- prescribed (B) availability --- used

(C) prescription --- available (D) acceptance --- proscribed

Ans. (A)

Sol. ‘effectiveness’ is noun and ‘pre scribed’ is verb. These words are apt and befitting with the word

‘medicine.’


Q.4 In a huge pile of apples and oranges, both ripe and unripe mixed together, 15% are unripe fruits. Of the

unripe fruits, 45% are apples. Of the ripe ones, 66% are oranges. If the pile contains a total of 5692000

fruits, how many of them are apples?

(A) 2029198 (B) 2467482 (C) 2789080 (D) 3577422

Ans. (A)

Sol. According to the question, following tree diagram can be made.

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GATE 2016 [EC] Set - 1

Total number of apples = Ripe apples + Unripe apples

= (0.85 × 0.45 + 0.15 × 0.34) 569200

= 2029198


Q.5 Michael lives 10 km away from where I live. Ahmed lives 5 km away and Susan lives 7 km away from

where I live. Arun is farther away than Ahmed but closer than Susan from where I live. From the

information provided here, what is one possible distance (in km) at which I live from Arun's place?

(A) 3.00 (B) 4.99 (C) 6.02 (D) 7.01

Ans. (C)

Sol. Following line with respective distances can be drawn

Arun can reside anywhere between Ahmed and Susan i.e. between 5 km and 7 km from I.5 < 6.02 < 7

Hence, the correct option is (C).

Q.6 to Q.10 carry two marks each

Q.6 A person moving through a tuberculosis prone zone has a 50% probability of becoming infected. However

only 30% of infected people develop the disease. What percentage of people moving through a

tuberculosis prone zone remains infected but does not show symptoms of disease?

(A) 15 (B) 33 (C) 35 (D) 37

Ans. (C)

Sol. According to the question, following information can is obtained.

Total fruits(5692000)

Unripe (15%)Ripe (85%)

(45%)apples

(34%)apples

(55%)oranges

(66%)oranges

Total = 10 kms

Arun MichaelSusan5 km I

7 km

Tuberculosic

Infected50%

Afected50%

Develop the disease30% of 50%

Does notDevelop the disease

70% of 50%

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Hence, percentage of people moving through a tuberculosis prone zone remains infected but does not

show symptoms of disease is70 50 35

35%100 100 100


Q.7 In a world filled with uncertainty, he was glad to have many good friends. He had always assisted them

in times of need and was confident that they would reciprocate. However, the events of the last week proved him wrong.

Which of the following inference(s) is/are logically valid and can be inferred from the above passage?

(i) His friends were always asking him to help them.

(ii) He felt that when in need of help, his friends would let him down.

(iii) He was sure that his friends would help him when in need.

(iv) His friends did not help him last week.

(A) (i) and (ii) (B) (iii) and (iv) (C) (iii) only (D) (iv) only

Ans. (B)

Sol. The words ‘was confident that they would reciprocate’ and ‘las t week proved him wrong’ lead to

statements iii and iv as logically valid inferences.


Q.8 Leela is older than her cousin Pavithra. Pavithra's brother Shiva is older than Leela. When Pavithra and

Shiva are visiting Leela, all three like to play chess. Pavithra wins more often than Leela does.

Which one of the following statements must be TRUE based on the above?

(A) When Shiva plays chess with Leela and Pavithra, he often loses.

(B) Leela is the oldest of the three.

(C) Shiva is a better chess player than Pavithra.(D) Pavithra is the youngest of the three

Ans. (D)

Sol. From the given data, following relations are possible.

L > P (Leela is older thean Pavithra)

S > L (Shiv is older than Leela)

So Pavithra is youngest

Hence, the correct option is (D).

Q.9 If 1aqr

and 1br s and 1cs q

, the value of abc is __________.

(A) 1( )rqs (B) 0 (C) 1 (D) r q s

Ans. (C)

Sol. Given : , , a b cq r r s s q

Taking log on both sides

log log , log log , log log a q r b r s c s q

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Hence,log log log

, ,log log log

r s q

a b cq r s

So,log log log

1log log log

r s q

a b cq r s


Q.10 P, Q, R and S are working on a project. Q can finish the task in 25 days, working alone for 12 hours a

day. R can finish the task in 50 days, working alone for 12 hours per day. Q worked 12 hours a day but

took sick leave in the beginning for two days. R worked 18 hours a day on all days. What is the ratio of

work done by Q and R after 7 days from the start of the project?

(A) 10 : 11 (B) 11 : 10 (C) 20 : 21 (D) 21 : 20

Ans. (C)

Sol. Given : Q can do work in 25 × 12 = 300 hrs

R can do work in 50 × 12 = 600 hrs

So we can say Q is twice efficient as R

Now Q worked only for 5 days at a rate of 12 hrs/day. So for 60 units of his work (Total work for Q i.e

300 hrs) he will do only1

5 of work

60 1

300 5

.

While R worked for all 7 days at a rate of 18 hrs/day

So he will do 18 × 7 = 126 of his work (Total work for 600 hrs)

He will do126

0.21300

of his work

So required ratio

1 126

: 120 :1265 600

20 : 21


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Technical Part

Q.1 to Q.25 carry one mark each

Q.1 Let 4 M I , (where I denotes the identity matrix) and 2, M I M I and 3 M I . Then, for any natural

number 1,k M equals :

(A) 4 1k M (B) 4 2k M (C) 4 3k M (D) 4k M

Ans. (C)

Sol. Given : 4 M I

Hence 4 k M I

Or 4( 1) k M I

Multiply 1 M on both side

1 4( 1) 1 k M M M I 4 3 1 k M M


Q.2 The second moment of a Poisson-distributed random variable is 2. The mean of the random variable

is_____________.

Ans. 1

Sol. Given : Second moment = 2

In Poisson distribution,

Mean = first moment

Second moment 2

Hence 2 2 2 2 0

( 2) ( 1) 0

1, 2

Since cannot be negative.

Hence, the correct answer is 1.

Q.3 Given the following statements about a function f : R R , select the right option :

P : If ( ) f x is continuous at0

x x , then it is also differentiable at0

x x .

Q : If ( ) f x is continuous at 0 x x , then it may not be differentiable at 0 x x .

R : If ( ) f x is differentiable at 0 x x , then it is also continuous at 0 x x .

(A) P is true, Q is false, R is false (B) P is false, Q is true, R is true

(C) P is false, Q is true, R is false (D) P is true, Q is false, R is true

Ans. (B)

Sol. If ( ) f x is continuous at 0 x x then it is not necessary that it will be differential also at 0 x x .

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Example 1 :

For y x , it is continuous at 0 x but not differential at 0 x .

Example 2 :

For y x , this function is continuous as well as differentiable at 0 x .

Hence, if ( ) f x is continuous at 0 x x then it may not be differential at 0 x and if ( ) f x is differentiable

at 0 x x then it is also continuous at 0 x x .


Q.4 Which one of the following is a property of the solutions to the Laplace equation : 2 0 f ?

(A) The solutions have neither maxima nor minima anywhere except at the boundaries.

(B) The solutions are not separable in the coordinates.

(C) The solutions are not continuous.

(D) The solutions are not dependent on the boundary conditions.

Ans. (A)

Q.5 Consider the plot of ( ) f x versus x as shown below.

Suppose5

( ) ( )

x

F x f y dy

. Which one of the following is a graph of ( )F x ?

(A) (B)

x x = 0

y x

x0

y

( ) f x

2

5 x

5

2

0

( ) F x

5 x

5

0

( ) F x

5 x

5

0

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(C) (D)

Ans. (C)

Sol. Given : 5

( ) ( ) x

F x f y dy

Then '( ) ( )F x f x which is a density function.From the given figure,

'( ) ( ) 0F x f x when 0 x Hence ( )F x is decreasing function for 0 x .

'( ) ( ) 0F x f x when 0 x Hence ( )F x is increasing function for 0 x .


Note : You can easily get the correct answer by elimination method. Option (A), (B) and (D) can be easily

eliminated because line passing through the origin is a straight line. [Integration of straight line results in

a parabola].

Q.6 Which one of the following is an Eigen function of the class of all continuous-time, linear, time-invariant

systems (u(t ) denotes the unit-step function)?

(A) 0 ( ) j t

e u t

(B) 0cos( )t (C)0 j t e

(D) 0sin( )t

Ans. (C)Sol. If the input to the system is Eigen signal then output is also the Eigen signal.


Q.7 A continuous-time function x(t ) is periodic with period T . The function is sampled uniformly with a

sampling periods

T . In which one of the following cases is the sampled signal periodic?

(A) 2s

T T (B) 1.2s

T T (C) Always (D) Never

Ans. (B)

Sol. A signal is said to be periodic ifs

T

T is a rational number.

[Rational number : Any number that can be represented in the form of pq

, then it is called a rational

number where p and q are any integer]

For option (B),

1.2s

T T

61.2

5s

T

T which is a rational number.


( ) F x

5 x

5

0

( ) F x

5 x

5

0

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Q.8 Consider the sequence n n x n a u n b u n , where u n denotes the unit-step sequence and

0 1a b . The region of convergence (ROC) of the z-transform of x n is

(A) z a (B) z b (C) z a (D) a z b

Ans. (B)

Sol. Given : n n x n a u n b u n

Given sequence is a right sided sequence.

Hence ROC will be right side of right most pole.

0 1a b

Hence right most pole is b .

ROC z b


Q.9 Consider a two-port network with the transmission matrix : A B

T C D

. If the network is reciprocal

then

(A) 1T T (B) 2T T (C) Determinant (T ) = 0 (D) Determinant (T ) = 1

Ans. (D)

Sol. Given : A B

T C D

For reciprocal network

1 A B

C D

1 AD BC


Q.10 A continuous-time sinusoid of frequency 33 Hz is multiplied with a periodic Dirac impulse train of

frequency 46 Hz. The resulting signal is passed through an ideal analog low-pass filter with a cutoff

frequency of 23 Hz. The fundamental frequency (in Hz) of the output is _________.

Ans. 13

Sol. Given : 33 Hz, 46 Hz m s

f f

Let ( ) x t message signal

and ( ) y t sampled signal

Then ( ) ( ) ( )sn

y t x t t nT

( ) ( )

s sn

Y f f X f nf

Spectrum of ( ) X f and ( )Y f are as shown

a

z pl n z- plane

b

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Cutoff frequency of LPF = 23 Hz.Hence, frequency at the output = 13 Hz.


Q.11 A small percentage of impurity is added to an intrinsic semiconductor at 300 K. Which one of the

following statements is true for the energy band diagram shown in the following figure?

(A) Intrinsic semiconductor doped with pentavalent atoms to form n-type semiconductor.

(B) Intrinsic semiconductor doped with trivalent atoms to form n-type semiconductor.

(C) Intrinsic semiconductor doped with pentavalent atoms to form p-type semiconductor.

(D) Intrinsic semiconductor doped with trivalent atoms to form p-type semiconductor.

Ans. (A)

Sol. Given diagram :

Energy level created just below the conduction energy level is called donor energy level, which is created

when a pentavalent impurity is introduced in an intrinsic semiconductor.


( ) X f

f 0 33 Hz – 33 Hz

( )Y f

f 0 33 Hz – 33 Hz 79 Hz – 79 Hz – 13 Hz 13 Hz

– 23 Hz 23 Hz

0.01 eV New Energy Level

Conduction Band

Valence Band

E C

E S

0.01 eV New Energy Level

Conduction Band

Valence Band

E C

E S

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Q.12 Consider the following statements for a metal oxide semiconductor field effect transistor (MOSFET) :

P : As channel length reduces, OFF-state current increases.

Q : As channel length reduces, output resistance increases.

R : As channel length reduces, threshold voltage remains constant.

S : As channel length reduces, ON current increases.

Which of the above statements are INCORRECT?(A) P and Q (B) P and S (C) Q and R (D) R and S

Ans. (C)

Sol. When channel length reduces then output resistance also reduces and threshold voltage will change.


Q.13 Consider the constant current source shown in the figure below. Let represent the current gain of the

transistor.

The load current 0 I through L R is

(A) 01 ref V

I R

(B) 01

ref V

I R

(C) 01

2

ref V

I R

(D) 01 2

ref V

I R

Ans. (B)

Sol.

R

2 R

1 R 1 R

CC V

R

2 R

1 R 1 R

CC V

AV

BV

ref V BV

C I

0 E I I

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From the figure,

A CC ref V V V

and A BV V (Due to virtual ground)

( )CC CC ref ref CC BC

V V V V V V I

R R R

01

ref C E

V I I I

R


Q.14 The following signal iV of peak voltage 8 V is applied to the non-inverting terminal of an ideal op-amp.

The transistor has 0.7V, 100, 1.5V, 10V BE LED CC

V V V and 10VCC

V .

The number of times the LED glows is ________.

Ans. 3

Sol.

2 k

CC V

CC V

iV 15 k

100

8 k

10V 10V

2 V

4 V

6 V

V i

– 2 V

– 4 V

– 6 V

t

2 k

CC V

CC V

iV 15 k

100

8 k

10V 10V

BV

BJT

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From the above figure

210 2 Volt

2 8

B

V

Comparator output is CC V , when i BV V

When the output will beCC

V then BJT will be in saturation and hence shorted and LED will glow.

iV exceeds 2 V, 3 times. Therefore LED will glow three times.


Q.15 Consider the oscillator circuit shown in the figure. The function of the network (shown in dotted lines)

consisting of the 100 k Ω resistor in series with the two diodes connected back-to-back is to

(A) introduce amplitude stabilization by preventing the op amp from saturating and thus producing

sinusoidal oscillations of fixed amplitude.

(B) introduce amplitude stabilization by forcing the op-amp to swing between positive and negative

saturation and thus producing square wave oscillations of fixed amplitude.

(C) introduce frequency stabilization by forcing the circuit to oscillate at a single frequency.

(D) enable the loop gain to take on a value that produces square wave oscillations.

Ans. (A)

Sol. Given circuit is a Wein-bridge oscillator, which produce sinusoidal signal.

The amplitude of output wave is decided by feedback through inverting input terminal of Op-Amp.


2 V

V i

t

CC V

CC V 22.1k

100k

1 D

2 D

158k 1nF

158k 1nF

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Q.16 The block diagram of a frequency synthesizer consisting of a Phase Locked Loop (PLL) and a divide-by-

N counter (comprising ÷ 2 ,÷ 4, ÷ 8, ÷ 16 outputs) is sketched below. The synthesizer is excited with a 5

kHz signal (Input 1). The free-running frequency of the PLL is set to 20 kHz. Assume that the commutator

switch makes contacts repeatedly in the order 1-2-3-4.

The corresponding frequencies synthesized are :

(A) 10 kHz, 20 kHz, 40 kHz, 80 kHz (B) 20 kHz, 40 kHz, 80 kHz, 160 kHz

(C) 80 kHz, 40 kHz, 20 kHz, 10 kHz (D) 160 kHz, 80 kHz, 40 kHz, 20 kHz

Ans. (A)

Sol. Given circuit diagram can be redrawn as

in f VCO output ( in N f ) Divide by N counter

5 kHz 10 kHz 2

5 kHz 20 kHz 4

5 kHz 40 kHz 8

5 kHz 80 kHz 16


Amplifier

Low PassFilter

PhaseDetector

VCO

Input 1

CC V

CC V

1 »2

2

3

4

»4

»8

»10

Synthesizer output

R

Phasedetector Input

f in Amplifier & filter

Divide by N counter

VCO

Output( ) N f in

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Q.17 The output of the combinational circuit given below is

(A) A + B + C (B) A (B + C) (C) B (C + A) (D) C (A + B)

Ans. (C)

Sol. Given logic circuit

Output Y is given by,

Y ABC AB BC

Y ABC AB ABC AB BC

Y (A B C) AB ABC (A B) BC ABC BC

Y ABC BC ABC BC

Y (A B C) BC ABC (B C) ABC BC ABC

Y BC(A 1) ABC BC ABC

Y B(C AC) B(C A)


Q.18 What is the voltage out V in the following circuit?

(A) 0 V (B) ( of PMOS of NMOS ) / 2T T V V

(C) Switching threshold of inverter (D) DDV

A

B CY

A

B CY

out V

10 k

DDV

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Ans. (C)

Sol.

Apply KVL at PMOS

0out GS DDV V V

GS out DDV V V

(NMOS) 0 V (NMOS) (NMOS) DG DS GS out

V V V V

(PMOS) 0 V (PMOS) (PMOS) DG DS GS DD out

V V V V V

i.e. both NMOS and PMOS are in saturation region

2

DDout

V V Switching threshold of inverter.

Q.19 Match the inferences X, Y, and Z, about a system, to the corresponding properties of the elements of first

column in Routh’s Table of the system characteristic equation.

X : The system is stable … P : … when all elements are positive

Y : The system is unstable … Q : … when any one element is zero

Z : The test breaks down … R : … when there is a change in sign of coefficients

(A) X → P, Y → Q, Z → R (B) X → Q, Y → P, Z → R

(C) X → R, Y → Q, Z → P (D) X → P, Y → R, Z → Q

DS V

10 k

DDV

out V

0 A

0 A

0 ADrop across

10 k 0 Volt

out V

DDV

out V

DD out V V

D I out V

out V

D I

S

S

G

G

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Ans. (D)

Sol. When all elements are positive then number of sign change in 1st row is zero. Hence number of poles

which lies on right side is zero and system is stable (X → P).

When there is change in sign of coefficient then system is unstable (Y → R).

When any one element is zero then test breaks down (Z → Q).


Q.20 A closed-loop control system is stable if the Nyquist plot of the corresponding open-loop transfer function

(A) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number

of right-half s-plane poles.

(B) encircles the s-plane point (0 − j1) in the clockwise direction as many times as the number of right-

half s-plane poles.

(C) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number

of left-half s-plane poles.

(D) encircles the s-plane point (−1 + j0) in the counterclockwise direction as many times as the number

of right-half s-plane zeros.

Ans. (A)

Sol. We know that,

N P Z

Where N = Number of encirclement of ( 1 0) j and it is positive when Nyquist plot encircle the

point ( 1 0) j in counter clock wise direction.

P = Number of open loop poles lying in the right half of s-plane.

Z = Number of closed loop poles lying in the right half of s-plane.

For stability, 0 Z N P


Q.21 Consider binary data transmission at a rate of 56 kbps using baseband binary pulse amplitude modulation

(PAM) that is designed to have a raised-cosine spectrum. The transmission bandwidth (in kHz) required

for a roll-off factor of 0.25 is ________.

Ans. 35

Sol. Given : Bit rate 56 Kbpsb R

Roll-off factor 0.25

Transmission56

BW ( 1) (1.25) 35 kHz2 2

b R


Q.22 A superheterodyne receiver operates in the frequency range of 58 MHz − 68 MHz. The intermediate

frequency IF f and local oscillator frequency LO f are chosen such that IF LO f f . It is required that the

image frequencies fall outside the 58 MHz − 68 MHz band. The minimum required IF f (in MHz) is ____

Ans. 5

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Sol. Given : 58 MHz 68 MHzs

f

and si f should fall outside the range 58 MHz – 68 MHz

Hencemin

58 MHzs f

2 68 MHz is s

f f IF

58 MHz 2 68 MHz IF

min( ) 5 MHz IF


Q.23 The amplitude of a sinusoidal carrier is modulated by a single sinusoid to obtain the amplitude modulated

signal ( ) 5cos1600 20 cos1800 5cos 2000s t t t t . The value of the modulation index is _________

Ans. 0.5

Sol. Given : ( ) 5cos1600 20cos1800 5cos 2000s t t t t

( ) 20cos1800 5cos1600 5cos 2000s t t t t Compare with

( ) cos 2 cos 2 ( ) cos 2 ( )2 2

c cc c c m c m

A As t A f t f f t f f t

Hence 20 V,c A

5 V2

c A

205 0.5

2

Hence, the correct answer is 0.5.

Q.24 Concentric spherical shells of radii 2 m, 4 m, and 8 m carry uniform surface charge densities of 20 nC/m2

− 4 nC/m2 ands

respectively. The value ofs

(nC/m2 ) required to ensure that the electric flux density

0 D

at radius 10 m is _________.

Ans. – 0.25

Sol. Given : 1 2 32 m, 4 m, 8 m r r r

2 2

1 220 nC/m , 4 nC/m s s

From Gauss Law,

D ds Q [Charge enclosed]

If 0 D

, then 0Q

1 2 3 0Q Q Q

2 2 2

1 1 2 2 3 34 4 4 0 s s sr r r

2 2 2

1 2 34 2 4 4 4 8 0 s s s

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2 2

380 4 4 4 8 0s

2

380 64 8 0s

2

3

160.25 nC/m

64s

Hence, the correct answer is – 0.25.

Q.25 The propagation constant of a lossy transmission line is (2 + j5) m−1 and its characteristic impedance is

(50 + 0) Ω at = 106 rad s−1. The values of the line constants L, C, R, G are respectively

(A) L = 200 µH/m, C = 0.1 µF/m, R = 50 Ω/m, G = 0.02 S/m

(B) L = 250 µH/m, C = 0.1 µF/m, R = 100 Ω/m, G = 0.04 S/m

(C) L = 200 µH/m, C = 0.2 µF/m, R = 100 Ω/m, G = 0.02 S/m

(D) L = 250 µH/m, C = 0.2 µF/m, R = 50 Ω/m, G = 0.04 S/m

Ans. (B)

Sol. Given : Propagation constant = (2 + j5) m−1

Characteristic impedance 0 (50 0) Z j and = 106 rad s−1

The Propagation constant of a Transmission line is given by,

( ) ( ) R j L G j C

and characteristics equation of transmission line is given by,

0

R j L Z

G j C

0 (2 5)(50 0) 100 250 Z R j L j j j

Hence 100 /m R

and 250 L

6

250 250250 H/m

10

L

and0

(2 5)0.04 0.1

50

jG j C j

Z

Hence, 0.04 S/mG and 0.1, C

6

0.1 0.10.1 F/m

10C

Hence the correct option is (B).

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Q.26 to Q.55 carry two marks each

Q.26 The integral1

( 10)2 D

x y dx dy , where D denotes the disc :

2 2 4 x y , evaluate to _________.

Ans. 20

Sol. Let

1

( 10)2 D

I x y dxdy Put cos , sin , x r y r dxdy rdrd

2 2

0 0

1(cos sin ) 10

2 I r r dr d

2 2

2

0 0

1(cos sin ) 10

2 I r r dr d

2 22 23 2

0 00 0

1 10(cos sin )

2 3 2 2

r r I d d

2 2

0 0

1 8 1(cos sin ) 5 (4)

2 3 2 I d d

2

0

1 8 1(sin cos ) 20(2 )

2 3 2

I

1 8 1

(0 0) (1 1) 20 (2 )2 3 2

I

20 I


Q.27 A sequence x[n] is specified as

1 1 1, for 2

1 1 0 0

n x n

n x n

The initial conditions are x[0] = 1, x[1] = 1, and x[n] = 0 for n < 0. The value of x[12] is ______

Ans. 233

Sol. Given :1 1

1 0 A

Characteristics equation is given by,

0 A I

1 10

1

2 1 0 2 1 0

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By Caley Hamilton Theorem, we have

2 0 A A I 2 ( ) A A I

Squaring both sides

4 2

( ) ( ) 2 A A I A I A A I [Note :2

I I ]4 ( ) 2 3 2 A A I A I A I [ 2 A A I ]

Again squaring both sides,

8 2 2(3 2 ) 9 4 12 A A I A I A 8 9( ) 4 12 21 13 A A I I A A I

Now, 12 4 8 (3 2 ) (21 13 ) A A A A I A I

121 1 1 0

144 89 144 891 0 0 1

A A I

Put n = 12 in given question

1212 1 1 1

11 1 0 0

x

x

12 233 144 1

11 144 89 0

x

x

12 233 x


Q.28 In the following integral, the contour C encloses the points 2π j and −2π j

3

1 sin

2 ( 2 )C

zdz

z j

The value of the integral is ________.

Ans. (– 133.87)

Sol. Given : 3

1 sin

2 ( 2 )C

z I dz

z j

We know that,

1

( ) 2( )

( ) !

n

n

z aC

f z jdz f z

z a n

1 2"(2 )

2 2!

j I f j

( ) sin f z z

'( ) cos f z z

"( ) sin f z z

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1 2

sin(2 )2 2!

j I j

1sinh 2 133.87

2 I

Hence, the correct answer is – 1.33.87.

Q.29 The region specified by ( , , ) :3 5, , 3 4.58 4

z z

in cylindrical coordinates has

volume of _______.

Ans. 4.712

Sol. Given :

For cylindrical coordinates the volume is given by,

V

V d d dz

5 /4 4.5

3 /8 3

z

V d d dz

54.5 /4 2

3 /8 32

V d dz

4.5 /4

/4 4.5

/8 3

3 /8

8 8V d dz z

8 (4.5 3) 8 1.54 8 8

V

4.712V


Q.30 The Laplace transform of the causal periodic square wave of period T shown in the figure below is

(A)/2

1( )1 sT

F se

(B)/2

1( )(1 )sT

F ss e

(C) 1( )(1 )sT

F ss e

(D) 1( )1 sT

F se

Ans. (B)

Sol. Laplace transform of periodic signal is given by,

/2

0

( )( )

( )1 1

T

sT

sT sT

f t e dt F s

L f t e e

0 T /2 T 3 /2T 2T

1

t

( ) f t

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/2

/2

0

0

11

( )1 (1 )

T

sT T

sT

sT sT

e dt e

L f t e e s

/2

/2

/2 /2

1 1 1( ) (1 )

(1 ) (1 ) (1 )

sT sT

sT sT sT

e L f t e

s e s e e

/21 1

( )1 sT

L f t s e


Q.31 A network consisting of a finite number of linear resistor (R), inductor (L), and capacitor (C) elements

connected all in series or all in parallel, is excited with a source of the form

3

0

1

cos( ),k

k

a k t

where 00, 0k a

The source has nonzero impedance. Which one of the following is a possible form of the output measured

across a resistor in the network?

(A)3

0

1

cos( ),k k k

b k t

where ,k k b a k (B)4

0

1

cos( ),k k k

b k t

where 0,k b k

(C)3

0

1

cos( )k k k

a k t

(D)2

0

1

cos( )k k k

a k t

Ans. (A)

Sol. Let a network contains single resistor, inductor and capacitor all in series.

Total impedance, j

Z R j LC

1 Z R j L

C

Current,

3

01 cos( )

1

k s k

a k t V I

Z R j L

C

3

0

1

2

2 1

cos( )

1

1tan

k k

a k t

I

LC

R LC R

R

3

0

1

cos( )

s k k

V a k t

L C

I

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Let 1

1

tan s

L

R

,

3

0

1

2

2

cos( )

1

k k

a k t

I R L

C

3

0

1 2

cos( )1

k

k

a I k t

R LC

3

0

1

cos( )

k k k

I b k t where ,k k

b a k

where 22 1

k

k

a

b R L

C

and k


Note :

When a sinusoidal input is given to LTI system the output is also sinusoidal with change in magnitude

and phase shift offered by LTI system.

Q.32 A first-order low-pass filter of time constant T is excited with different input signals (with zero initial

conditions up to t = 0). Match the excitation signals X, Y, Z with the corresponding time responses for

t ≥ 0

X : Impulse P : /1 t T e

Y : Unit step Q : /(1 )t T t T e

Z : Ramp R : /t T e

(A) X → R, Y → Q, Z → P (B) X → Q, Y → P, Z → R

(C) X → R, Y → P, Z → Q (D) X → P, Y → R, Z → Q

Ans. (C)

Sol. For first order system

1( ) ; ( ) 1G s H s

sT

Close loop transfer function is given by,

( ) ( )

( ) 1 ( ) ( )

Y s G s

R s G s H s

( )( ) ( )

1 ( ) ( )

G sY s R s

G s H s

LTI System0sin A t 0sin( ) B t

1( ) G s

sT

( ) 1 H s

( )Y s( ) R s

Input Output

Fig. First order system

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For impulse response ( ) 1 R s

( ) 1/

( )11 ( ) ( )

1

G s sT Y s

G s H s

sT

1

( ) 1Y s sT

Taking inverse Laplace transform, we get

/1( ) t T y t eT

for 0t ( ) X R

For step response1

( ) R ss

1 (1 ) ( )( )

(1 ) (1 )

sT sT Y s

s sT s sT

1 1( )1(1 )

T T Y ss sT s

T sT


/( ) (1 )t T y t e for 0t ( )Y P

For ramp response,2

1( ) R s

s

2 2

1 1( )

1(1 ) 1

T T Y s

s sT s sT


/( ) (1 )t T y t t T e for 0t ( ) Z Q


Trick :

You can directly check by

d

dt

(ramp response) = step response

d

dt (step response) = impulse response

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Q.33 An AC voltage source V = 10 sin(t ) volts is applied to the following network. Assume that 1 R 3 k Ω

2 R 6 k Ω and 3 R 9 k Ω, and that the diode is ideal.

RMS current rms I (in mA) through the diode is ________.

Ans. 1

Sol.

Apply KVL in the loop shown, we get

1 2 33 6 3

I I I V R R R

3 6 9 53 6 3

I I I V I

R1

R2

R1 R2

d c

ba

R2

R3

R2 R3

e h

g f

R3

R2

R1

R2

rms I

10sin( )V t

R1

R2

R1 R2

R2

R3

R2 R3

R3 R2

R1

R2

rms I

10sin( )V t

/ 3 I

/ 3 I

/ 6 I

/ 3 I

I

/ 3 I

/ 3 I

/ 3 I

/ 6 I

/ 6 I

/ 6 I / 6 I

/ 6 I

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10sin2sin mA

5 5

V t I t

For half wave rectifier,

21 mA

2 2

mrms

I I

Hence, the correct answer is 1.Q.34 In the circuit shown in the figure.

The maximum power (in watt) delivered to the resistor R is __________.

Ans. 0.8 W

Sol. Calculation of thV : open circuit the resistance R.

0

25 2 V

2 3

v

0

40 40

100 100 2 160 V40 10 50 thV v

Calculation of th

R : Short the independent voltage source of 5 volt. Therefore 0 0v and as a result

dependent voltage source 0100 0v .

Hence both dependent and independent voltage source will acts as a short circuit.

40 1010 k || 40 k k

40 10th

R

4008 k

50th

R

Maximum power, maxP2

16 16 1000.8 W

4 4 8 1000

th

th

V

R


3 k

2 k 5 V

10 k

40 k 0v 0100v R

3 k

2 k 5 V

10 k

40 k 0v 0100v thV

3 k

2 k

10 k

40 k th R

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Q.35 Consider the signal 6 2 3 1 8 7 1 4 2 x n n n n n n .

If ( ) j X e is the discrete-time Fourier transform of x[n], then 21

( )sin (2 ) j X e d

is equal to ______

Ans. 8

Sol. From the definition of DTFT,

( ) j j nn

X e x n e

And 1

( )2

j j n x n X e e d

1

0 ( )2

j x X e d

Hence 1

( ) ( ) 0 * 0

2

j j X e Y e d x y

2( ) sin (2 ) jY e 4 41 cos 4 1 ( )

( )2 2 2 2

j j j e eY e

4 41 1 1( )2 4 4

j j jY e e e

Taking inverse Fourier transform,

1 1 1

4 42 4 4

y n n n n

1 1 1

, 0, 0, 0, , 0, 0, 0,4 2 4

y n

1

02

y

6, 3, 8, 7, 4 x n

0 8 x

1

( ) ( ) 2 0 * 0

j j X e Y e d x y

12 8 8

2


Q.36 Consider a silicon p-n junction with a uniform acceptor doping concentration of 1017 cm−3 on the p-side

and a uniform donor doping concentration of 1016 cm−3 on the n-side. No external voltage is applied to the

diode. Given : kT /q = 26 mV, in = 1.5 × 1010 cm−3 ,

14

0 012 , 8.85 10Si F/m and q = 1.6 × 10−19

C. The charge per unit junction area (nC cm−2) in the depletion region on the p-side is ___________.

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Ans. 4.83

Sol. Given diagram,

Build in potential of p-n junction diode is given as16 17

3

0 2 10 2

10 10ln 26 10 ln

(1.5 10 )

A DT

i

N N V V

n

0 0.757 VV

Depletion width of p-n junction diode is given by

0

2 1 1

A D

W V q N N

16

19 19 17

2 8.854 10 12 1 10.757 3.3255 cm

1.6 10 10 10

W

Now6 16

16 17

3.3255 10 10

10 10

D

P

A D

WN W

N N

0.3023 μcmPW

Charge per unit junction area in the depletion layer on the p-side is A P

qN W 19 17 6 21.6 10 10 0.3023 10 4.8368 nC/cm


Q.37 Consider an n-channel metal oxide semiconductor field effect transistor (MOSFET) with a gate-to-source

voltage of 1.8 V. Assume that 6 24, 70 10 AV N oxW

C L

, the threshold voltage is 0.3 V, and the

channel length modulation parameter is 0.09 V−1 . In the saturation region, the drain conductance (in micro

seimens) is __________.

Ans. 28.35Sol. Given :

In the saturation region, drain conductance is given by,

d DS g I

2 6 21 1( ) 0.09 70 10 4(1.8 0.3)2 2

d N ox GS T

W g C V V

L

28.35 S d g


P N

P W N W

W

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Q.38 The figure below shows the doping distribution in a p-type semiconductor in log scale.

The magnitude of the electric field (in kV/cm) in the semiconductor due to non-uniform doping is

_________.

Ans. 0.0133

Sol. Applying the current density equation

drif t Dif fusion J J J

There is no net flow of current

Thus, 0 J Hence, for hole we can write

0 P PdP

qD q pE dx

P P

dPqD q pE

dx

P P

dP D pE

dx

P

P

D dP E

p dx

We know PT

P

DV

Hence, T V dP

E p dx

Here A p N

T A A

V d N E

N dx

ln ( )T Ad

E V N xdx

Units are in log scale, hence we can write as

10 1log 1 m x 1

1 10 m 0.001 cm x

10 2log 2 m x 2

2 10 m 0.01 cm x 14ln(10 ) 32.23

1014

1016

1 2

Position ( m)

N A(cm

)–3

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And 16ln(10 ) 36.84

36.84 32.230.026

0.01 0.001 E

0.0133 kV/cm E


Q.39 Consider a silicon sample at T = 300 K, with a uniform donor density d N = 5 × 1016 cm−3, illuminated

uniformly such that the optical generation rate isopt

G = 1.5 × 1020 cm−3−1 throughout the sample. The

incident radiation is turned off at t = 0. Assume low-level injection to be valid and ignore surface effects

The carrier lifetimes are 0 p = 0.1 µs and 0n = 0.5 µs.

The hole concentration at t = 0 and the hole concentration at t = 0.3 µs, respectively, are

(A) 13 3 11 31.5 10 cm and 7.47 10 cm (B) 13 3 11 31.5 10 cm and 8.23 10 cm

(C) 13 3 11 37.5 10 cm and 3.73 10 cm (D) 13 3 11 37.5 10 cm and 4.12 10 cm

Ans. (A)

Sol. Given : 20 3 11.5 10 cm sopt G

20

61.5 10

0.1 10

A Aopt

p

N N G R

13 31.5 10 /cm A N

0.3/ 13 0.1( ) 1.5 10 p

t

no p t p e e

11 3( ) 7.46 10 /cm p t


Q.40 An ideal op-amp has voltage sources 1 3 5 1, , ,......, N V V V V connected to the non-inverting input and

2 4 6, , ,......, N V V V V connected to the inverting input as shown in the figure below ( CC V = 15 volt, CCV

= −15 volt). The voltages 1 2 3 4 5 6, , , , , ,......V V V V V V are 1, − 1/2, 1/3, −1/4, 1/5, −1/6, .… volt, respectively

As N approaches infinity, the output voltage (in volt) is ___________.

1014

1016

1 2m (log scale)

N A(cm ) –3

32.24

36.84

0.001 0.01

n-type Si

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Ans. 15

Sol.

Apply nodal analysis at node A,

3 11 ..... 01 k 1 k 1 k 1 k

A A N A AV V V V V V V

1 3 11 .....2

A N

N V V V V

B A

V V [ Virtual ground concept]

Apply nodal analysis at node B,

02 4 ..... 0

10 k 10 k 10 k 10 k

A N A A A V V V V V V V V

[We used B A

V V ]

0 2 4 61 ( ..... )2

A N

N V V V V V V

1 3 10 2 4 6

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

21

2

N N

V V V N V V V V V

N

0 1 2 3 4

1 1 1..... 1 .....

2 3 4V V V V V

CC V

CC V

10 k

10k

10k

10k

1k

1k

1k

1k

2V

4V

N V

1V

3V

1 N V

0V

CC V

CC V

10k

10k

10k

10k

1k

1k

1k 1k

2V

4V

N V

1V

3V

1 N V

0V BV

AV

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0

1V

N

Output of op-amp goes to saturation

0 15 Voltsat CC V V V


Q.41 A p-i-n photodiode of responsivity 0.8 A/W is connected to the inverting input of an ideal op-amp as

shown in the figure,CC

V = 15 V,CC

V = −15 V, Load resistor L

R = 10 k Ω. If 10 µW of power is

incident on the photodiode, then the value of the photocurrent (in µA) through the load is ________.

Ans. 800

Sol.

Responsivity 0Generated photo current

Incident light power i

I

P

0

60.8 A/W

10 10

I

0 8 A I

Hence, 6 60 0 1 M 8 10 1 10 8 VV I

Photo current through load 10 k L

R is given by,

0

3

8800 A

10 10 L

L

V I

R

[In upward direction]

CC V

CC V

1M

0V

10k

CC V

1M

CC V

CC V

1M

0V

10k L R

CC V

1M

0 V

0 A

0 A

0 I

L I

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Q.42 Identify the circuit below.

(A) Binary to Gray code converter (B) Binary to XS3 converter

(C) Gray to Binary converter (D) XS3 to Binary converter

Ans. (A)

Sol. Truth table of the circuit is shown below.

2X 1X 0X 2Y 1Y 0Y

0 0 0 0 0 0

0 0 1 0 0 1

0 1 0 0 1 1

0 1 1 0 1 0

1 0 0 1 1 0

1 0 1 1 1 1

1 1 0 1 0 0

1 1 1 1 0 1

None of the option are correct but answer is close to option (A).Hence, the correct option is (A).

Q.43 The functionality implemented by the circuit below is

OP0

OP1

OP2

OP3

OP4

OP5

OP6

OP7

IP0

IP1

IP2

IP3

IP4

IP5

IP6

IP7

3 : 8Decoder

8 : 3Encoder

X2

X1

X0

Y2

Y1

Y0

2 : 4Decoder

O0

O1

O2

O3

C1

C0

P

Q

R

S

Enable = 1

is a tristate buffer

Y

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(A) 2-to-1 multiplexer (B) 4-to-1 multiplexer

(C) 7-to-1 multiplexer (D) 6-to-1 multiplexer

Ans. (B)

Sol. When the outputs 0 1 2 3(O , O , O , O ) of the decoder are at logic 1, corresponding tristate buffer is activated

In that case, whatever data is applied at the input of a buffer, becomes its output.

Hence, when

1 0C C 00 Then 0O 1

Y = P

1 0C C 01 Then 1O 1

Y = Q

1 0C C 10 Then 2O 1

Y = R

1 0C C 11 Then 3O 1

Y = SThis behavior is property of 4 to 1 multiplexer.


Q.44 In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the

following options is the correct reason for this?

(A) For POP, the data transceivers remain in the same direction as for instruction fetch (memory to

processor), whereas for PUSH their direction has to be reversed.

(B) Memory write operations are slower than memory read operations in an 8085 based system.

(C) The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP

operation uses the address already in the stack pointer.(D) Order of registers has to be interchanged for a PUSH operation, whereas POP uses their natural order

Ans. (C)

Sol. For PUSHP

R instruction in 8085 machine cycles are Fetch (F), Write (W) and Write (W).

i.e. 6 + 3 + 3 = 12 T-state/clock cycle.

Stack pointer holds the address of previously stored temporary data, so to store new data SP is decreased

by ‘1’ after decoding on code, hence fetch has 6 T-state unlike 4 T-states for most of the instruction

But for POPP

R Fetch (F), Read (R) and Read (R)

i.e. 4 3 3 10 T-state/clock cycle


Q.45 The open-loop transfer function of a unity-feedback control system is

2( )

5 5

K G s

s s

The value of K at the breakaway point of the feedback control system’s root-locus plot is ________.

Ans. 1.25

Sol. Given :2

( )5 5

K G s

s s

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1 ( ) ( ) 0G s H s

21 0

5 5

K

s s

2 5 5 K s s

For break away point 0dK

ds

2 5 0dK

sds

2.5s

To find system gain, apply magnitude condition,

2.5( ) ( ) 1

sG s H s

2.5 2( ) ( ) 1

( 2.5) 5 ( 2.5) 5s

K G s H s

(6.25 5 12.5) 1.25K


Q.46 The open-loop transfer function of a unity-feedback control system is given by

( )( 2)

K G s

s s

For the peak overshoot of the closed-loop system to a unit step input to be 10%, the value of K is ________

Ans. 2.8

Sol. Given : ( ) ; ( ) 1( 2)

K G s H s

s s


1 ( ) ( ) 0G s H s

1 0( 2)

K

s s

2 2 0s s K

Compare with 2 22 0n n

s s

2 2n and n K

1 1

n K

Peak overshoot is given by,

210.1

P M e

2ln(0.1)

1

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22.3

1

2 2 2 2(2.3) (1 ) 2 215.16 (2.3)

0.59

Also,2 2

1 12.8

(0.59)K


Q.47 The transfer function of a linear time invariant system is given by,4 3( ) 2 5 5 2 H s s s s

The number of zeros in the right half of the s-plane is ________.

Ans. 3

Sol. Given :4 3( ) 2 5 5 2 H s s s s

To find the number of zeroes in the right half of s-plane, apply Routh Hurwitz criteria4 32 5 5 2 0s s s

By Routh array4s 2 0 – 2

3s – 5 5

2s 2 – 2

1s 0(2)

0s – 2

Number of sign change in first column

= Number of Roots (zeros) in right half of s-plane

= 3


Q.48 Consider a discrete memoryless source with alphabet 0 1 2 3 4, , , , ,....S s s s s s and respective

probabilities of occurrence1 1 1 1 1

, , , , ,....2 4 8 16 32

P

. The entropy of the source (in bits) is _______.

Ans. 2

Sol. Given :

Entropy of source is given by,

2

0

1log

N

i

i i

H PP

2 2 2 2

1 1 1 1log 2 log 4 log 8 log 16 .....

2 4 8 16 H

2 3 41 1 1 1

2 3 4 .....2 2 2 2

H

……. (i)

Auxiliary equation is given by,

2AE 2 2 s

(AE) 2 0 d

sds

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Divide both side by 2,

Hence

2 3 41 1 1

2 3 .....2 2 2 2

H

……. (ii)

Subtracting (ii) from (i),

2 3

1 1 1 .....2 2 2 2

H

1

21

121

2

H

2 bits/symbol H


Q.49 A digital communication system uses a repetition code for channel encoding/decoding. During

transmission, each bit is repeated three times (0 is transmitted as 000, and 1 is transmitted as 111). It is

assumed that the source puts out symbols independently and with equal probability. The decoder operates

as follows: In a block of three received bits, if the number of zeros exceeds the number of ones, the decoder

decides in favor of a 0, and if the number of ones exceeds the number of zeros, the decoder decides in

favor of a 1. Assuming a binary symmetric channel with crossover probability p = 0.1, the average

probability of error is ________.

Ans. 0.028

Sol. Given :

Crossover probability p = 0.1

Average probability of error2 33 3 p p

2 33(0.1) 3(0.1) 0.028


Q.50 An analog pulse s(t ) is transmitted over an additive white Gaussian noise (AWGN) channel. The received

signal is r (t ) = s(t ) + n(t ), where n(t ) is additive white Gaussian noise with power spectral density 0

2

N

The received signal is passed through a filter with impulse response h(t ). Let S E and h E denote the

energies of the pulse s(t ) and the filter h(t ), respectively. When the signal-to-noise ratio (SNR) ismaximized at the output of the filter ( maxSNR ), which of the following holds?

(A) max0

2; S S h

E E E SNR

N (B) max

0

;2

S S h

E E E SNR

N

(C) max0

2; S

S h

E E E SNR

N (D) max

0

2; h

S h

E E E SNR

N

Ans. (A)

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Sol. The signal to noise ratio is maximum when

( ) ( )h t s T t

But shifting does not change the energy

Hence, h s E E

and max0 0

2

( ) / 2

s s E E

SNR N N


Q.51 The current density in a medium is given by

2

2

400sinˆ Am

2 ( 4)

r J a

r

The total current and the average current density flowing through the portion of a spherical surface

r = 0.8 m, , 0 212 4

are given, respectively, by

(A) 15.09 A, 12.86 Am 2 (B) 18.73 A, 13.65 Am 2

(C) 12.86 A, 9.23 Am 2 (D) 10.28 A, 7.56 Am 2

Ans. (D)

Sol. Given :

Total current is given by,/4 2

2

2

0

12

400sinsin

2 ( 4)

I J ds r d d r

/4

22 2

2 0

/12

400sin

2 ( 4) I r d

r

/42

2

/12

400 2 1 cos 2

2 ( 4) 2

r

I d r

/42 2

2 2

/12

11

400 sin 2 4004 12 2

( 4) 2 4 ( 4) 12 4

r r I

r r

Given r = 0.8 m,2400 (0.8)

0.13 7.56 Amp4 64 I

Total area 2 sin ds r d d

/4 2

2 2

0

12

sin

r r d d

/4

2

12

sin 2r d

2 0.82 0.259 r r

2 2(0.8) 2 0.259 1.041 m

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39


Average current 2 27.56

7.26 A/m 7.56 A/m1.041


Q.52 An antenna pointing in a certain direction has a noise temperature of 50 K. The ambient temperature is

290 K. The antenna is connected to a pre-amplifier that has a noise figure of 2 dB and an available gain

of 40 dB over an effective bandwidth of 12 MHz. The effective input noise temperature eT for the amplifier

and the noise power aoP at the output of the preamplifier, respectively, are

(A) 10169.36K and 3.73 10 We ao

T P (B) 10170.8K and 4.56 10 We aoT P

(C) 10182.5K and 3.85 10 We aoT P (D) 10160.62K and 4.6 10 We aoT P

Ans. (A)

Sol. Given : Noise temperature of antenna 50 K ant

T

Ambient temperature 0 290 K T

Noise figure, F = 2 dB or 2/1010F

Gain = 40 dB = 410 ,

Effective bandwidth = 612 MHz 12 10 Hz

(i) The effective input temperaturee

T for the amplifier is given by,

2/10

0( 1) (10 1)290eT F T

169.6 K eT

(ii) Noise power at the input is given by,

(T )i ant e N K T B

23 61.38 10 (50 169.6) 12 10i N

143.63 10 Wi

N

0 Gaini N N

10Gain(dB) 40 10 log (Gain)

4Gain 10

14 4 10

0 3.63 10 10 3.63 10 W N


Q.53 Two lossless X-band horn antennas are separated by a distance of 200λ . The amplitude reflection

coefficients at the terminals of the transmitting and receiving antennas are 0.15 and 0.18, respectively

The maximum directivities of the transmitting and receiving antennas (over the isotropic antenna) are 18

dB and 22 dB, respectively. Assuming that the input power in the lossless transmission line connected to

the antenna is 2 W, and that the antennas are perfectly aligned and polarization matched, the power ( in

mW) delivered to the load at the receiver is ________.

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4


Ans. 3

Sol. Given :

Power delivered to the load at the receiver is given by,

2 2

2

(1 ) (1 )

4

t r t r

r t

G GP P

d

2 2 1.8 2.2

2

(1 0.15 ) (1 0.18 )10 10 2

4 200

r P

2.995 mW 3 mWr P


Q.54 The electric field of a uniform plane wave travelling along the negative z direction is given by the

following equation :

0ˆ ˆ( )i jkz

w x y E a ja E e

This wave is incident upon a receiving antenna placed at the origin and whose radiated electric field

towards the incident wave is given by the following equation :

1ˆ ˆ( 2 ) jkr

a x y l E a a E e

r

The polarization of the incident wave, the polarization of the antenna and losses due to the polarization

mismatch are, respectively,

(A) Linear, Circular (clockwise), − 5 dB (B) Circular (clockwise), Linear, − 5 dB

(C) Circular (clockwise), Linear, − 3 dB (D) Circular (anti clockwise), Linear, − 3 dB

Ans. (C)

Sol. Given : 0ˆ ˆ( )i jkz

w x y E a ja E e

Wave contains two orthogonal components with equal amplitude and y-component leads X-component

by 090 and also wave is travelling in negative z-direction.

Hence, circular (clock wise) polarization

1ˆ ˆ( 2 )

jkr

a x y l E a a E er

Wave contains orthogonal components with unequal amplitude and both are in-phase.

Hence, linear polarization2

ˆ ˆPLFinc ant

P P

Transmitter

0.15t

18 dB

P t = 2 W

Gt = 101.8

200d

Receiver

0.18r

22 dB

P r = ?

Gr = 102.2

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41


Whereˆ ˆ

ˆ2

x y

inc

a jaP

= Polarizing vector of incident wave.

ˆ ˆ2ˆ

5

x y

ant

a aP = Polarizing vector of radiated field.

2ˆ ˆ ˆ ˆ2 5 1PLF

2 10 25

x y x ya ja a a

221 5 1

PLF10 22 5

j

1PLF(dB) 10log 3 dB

2


Q.55 The far-zone power density radiated by a helical antenna is approximated as :

4

0 2

1ˆ cos

rad average r W W a C

r

The radiated power density is symmetrical with respect to and exists only in the upper hemisphere

00 ; 0 2 ;2

C

is a constant. The power radiated by the antenna (in watts) and the maximum

directivity of the antenna, respectively, are

(A) 1.5 0C , 10 dB (B) 1.256 0C , 10 dB (C) 1.256 0C , 12 dB (D) 1.5 0C , 12 dB

Ans. (B)

Sol. Given :4

0 2

1ˆ cosrad average r W W a C

r

Power radiated by the antenna is given by,

rad P

rad W ds /2 2

4 2

0 2

0 0

1cos sinC r d d

r

/2

2 40

2

0

2 cos ( cos )C

r d r

( cos ) sin d d

/25

0 0 0

0

cos 22 1.256

5 5C C C

Alternate Method for

2

4

0

cos sin d

:

Put cos t

sin dt d 2 0

4 4

0 1

cos sin

t

d t dt

05

15

1

5

t

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42


Directivity4

rad

U

W d

4

0

/2 2

4

0

0 0

4 cos

cos sin

C

C d d

442cos 10cos

1/ 5

Max value = 10

Max value (in dB) 1010log 10 10 dB


gate 2016 ec [set-1]

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