GENERAL APTITUDE - Gateforum · 2/14/2020 · Average number of students enrolled in school P Difference of the number of students enrolled in schools P and Q from 2014 to 2018 are

|EC-2020|

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1. The following figure shows the data of students enrolled in 5 years (2014 to 2018) for two schools P and

Q. During this period, the ratio of the average number of the students enrolled in school P to the average

of the difference of the number of students enrolled in schools P and Q is_________.

(A) 23: 8 (B) 31: 23 (C) 23:31 (D) 8:23

Key: (A)

Exp: From the figure,

Number of students enrolled in school P from 2014 to 2018 are 3000, 5000, 5000, 6000, 4000

Average number of students enrolled in school P

Difference of the number of students enrolled in schools P and Q from 2014 to 2018 are 1000, 2000,

3000, 1000, 1000

Average of difference of the number of students enrolled in schools

( ) ( )From 1 & 2 ,

The required ratio 4600 :1600 46 :16 23:8= = =

GENERAL APTITUDE

2014 2015 2016 2017 2018

Year

9

8

7

6

5

4

3

2

1

0

Nu

mb

er o

f st

ud

ents

(in

th

ou

san

ds)

School P

School Q

( )3000 5000 5000 6000 4000 23000

4600 ... 15 5

+ + + += = =

( )1000 2000 3000 1000 1000 8000

P and Q 1600 ... 25 5

+ + + += = =

|EC-2020|


2. It is quarter past three in your watch. The angle between the hour hand and the minute hand is _______.

(A) 0° (B) 15° (C) 22.5° (D) 7.5°

Key: (D)

Exp: The angle between the hands can be found using the formula:

( )1

Angle 60H 1 M ;2

= − where H is the hour, M is the minute

( ) ( )(( ) ( )1 1 15

60 3 11 15 180 165 7.52 2 2

= − = − = =

3. The global financial crisis in 2008 is considered to be the most serious world-wide financial crisis,

which started with the sub-prime lending crisis in USA in 2007. The subprime lending crisis led to the

banking crisis in 2008 with the collapse of Lehman Brothers in 2008. The sub-prime lending refers to

the provision of loans to those borrowers who may have difficulties in repaying loans, and it arises

because of excess liquidity following the East Asian crisis.

Which one of the following sequences shows the correct precedence as per the given passage?

(A) East Asian crisis→ subprime lending crisis →banking crisis→global financial crisis.

(B) Subprime lending crisis→global financial crisis→banking crisis→East Asian crisis.

(C) Global financial crisis→East Asian crisis→banking crisis→ subprime lending crisis.

(D) Banking crisis→ subprime lending crisis→global financial crisis →East Asian crisis.

Key: (A)

4. The Canadian constitution requires that equal importance be given to English and French.

Last year, Air Canada lost a lawsuit, and had to pay a six-figure fine to a French-speaking couple after

they filed complaints about formal in-flight announcements in English lasting 15 seconds, as opposed to

informal 5 second messages in French.

The French-speaking couple were upset at

(A) the English announcements being clearer than the French ones.

(B) the English announcements being longer than the French ones.

(C) equal importance being given to English and French.

(D) the in-flight announcements being made in English

Key: (A)

|EC-2020|


5. A superadditive function ( )f g satisfies the following property

( ) ( ) ( )1 2 1 2f x x f x f x+ +

Which of the following functions is a superadditive function for x>1?

(A) xe (B) xe− (C) x (D) 1 x

Key: (A)

6. Select the word that fits the analogy:

Explicit: Implicit:: Express:___________

(A) Compress (B) Suppress (C) Impress (D) Repress

Key: (D)

7. He was not only accused of theft_______of conspiracy.

(A) but also (B) rather (C) but even (D) rather than

Key: (A)

8. a, b, c are real numbers. The quadratic equation 2ax bx c 0− + = has equal roots, which is , then

(A) b

a = (B) 2b 4ac (C)

( )3

2

bc

2a = (D)

2 ac =

Key: (D)

Exp: Given that; the quadratic equation 2ax bx c 0− + = has equal roots, which is

Let , be the roots of the equation

( )( )

( )

b bThen 1

a a

c... 2

a

− − + = = →

=

( )

( )

bFrom 1 ;

a

b2 ... 3

a

+ = =

=

Q

|EC-2020|


( ) ( )

( ) ( )

2

3

2

2

2

cFrom 2 ; . c a ... 4

a

bc3 4 2

a

bc

2a

= =

=

=

9. A circle with centre O is shown in the figure. A rectangle PQRS of maximum possible area is inscribed

in the circle. If the radius of the circle is a, then the area of the shaded portion is_________.

(A) 2 2a 2a − (B) 2 2a 2a − (C) 2 2a 3a − (D) 2 2a a −

Key: (A)

Exp: Area of rectangle ( )( )2x 2y=

( ) ( )

( )

2 2 2

2 2 2

2 3

2 2

2 2

A 4xy

A 16x y

f x 16x a x

f x 0 32xa 64x 0

32x a 2x 0

2x a

ax

2

ax only possible

2

'

=

=

= −

= − =

− =

=

=

=

If a

x2

= then

O

P Q

S R

x

Y

S

Xa

2x

a

x

2y

y

y−

(0,0)

|EC-2020|


2 22 2 2 2 2a a

y a y a x2 2

ay

2

ay only possible

2

= − = − =

=

=

Q

Area of rectangle is maximum only when a a

x & y2 2

= =

Area of rectangle =4xy 2a a4 2a

2 2= =

Required shaded portion area = Area of circle - Area of rectangle

2 2a 2a .= −

10. The untimely loss of life is a cause of serious global concern as thousands of people get killed_____

accidents every year while many other die______ diseases like cardio vascular disease, cancer, etc.

(A) from, from (B) during, from (C) from, of (D) in, of

Key: (D)

|EC-2020|


Q. No. 1 to 25 Carry One Mark Each

1. The components in the circuit shown below are ideal. If the op-amp is in positive feedback and the input

voltage iV is a sine wave of amplitude 1V, the output voltage 0V is

(A) a non-inverted sine wave of 2V amplitude.

(B) a constant of either +5V or -5V.

(C) an inverted sine wave of 1 V amplitude.

(D) a square wave of 5 V amplitude

Key: (B)

Sol: Given,

( )( )1

0

V 1sin t Volts

V ?

=

=

For the above circuit to act as Schmitt trigger TLV (Lower threshold voltage) and THV (Higher threshold

voltage). Values should not go beyond positive peak and negative peak values of “V1”.

ELECTRONICS AND COMMUNICATION ENGINEERING

iV

2R 1k=

satV 5+ =

satV 5− = −

oV

+

−

1R 1k=

V+

5V+

+

−

5V−

0V

1k

1k

iV

1V

1V−

0

|EC-2020|


Threshold voltages for the above circuit is given by

1TL sat

2

RV V 5V

R= − = −

And 1TH sat

2

RV V 5V

R= = +

We got TL THV and V goes beyond 1V (positive and negative peak values of V1) the above circuit

can’t acts as a Schmitt trigger, so, it cannot convert the input sine wave to square wave.

In the above circuit there is no scope for the output to change from ( ) ( )sat satV 5V to V 5V+ − − (or) vice

versa.

Hence the output will remain either at +5V (or) –5V.

Hence the correct option is (B).

2. The partial derivative of the function

( )( )21 1 y1 x cos yf x,y,z e xze

− +−= +

With respect to x at the point (1, 0, e) is

(A) 1

e (B) 1 (C) –1 (D) 0

Key: (D)

Exp: ( )( )2

1

1 y1 xcos yf x,y,z e xze

−

+−= +

( )

( )

( )

2

1

1 x cos y 1 y

11 1.cos 0 1 0

1,0,e

0 1

1,0,e

fe cos y z e

x

fe cos (0) e.e

x

e [ 1] e.e 1 1 0

f0

x

−

− +

−− +

−

= − +

= − +

= − + = − + =

=

3. The general solution of 2

2

d y dy6 9y 0

dx dx− + = is

(A) 3x 3x

1 2y C e C e−= + (B) ( ) 3x

1 2y C C x e= +

(C) ( ) 3x

1 2y C C x e−= + (D) 3x

1y C e=

Key: (B)

|EC-2020|


Sol: Given D.E is

( )

2

2

2

d y dy6 9y 0

dx dx

D 6D 9 y 0

− + =

− + =

Auxiliary equation is 2m 6m 9 0− + =

( )2

m 3 0 m 3,3 real & repeated roots − = =

The required solution is

( )3x

1 2y e C C x= +

4. The loop transfer function of a negative feedback system is

( ) ( )( )

( )( )

K s 11G s H s

s s 2 s 8

+=

+ +

The value of K, for which the system is marginally stable, is ________.

Key: (160)

Sol: Given,

( ) ( )( )

( )( )

k s 11G s H s

s s 2 s 8

+=

+ + and we need to find the value of ‘k’ for marginally stable system.

( ) ( )1 G s H s 0+ =

( )( ) ( )

( )

( )

2

3 2

s s 2 s 8 k s 11 0

s s 10s 16 ks 11k 0

s 10s 16 k s 11k 0

+ + + + =

+ + + + =

+ + + + =

Since, it is 3rd order characteristics equation,

Internal product = External product (for marginally stable system)

( )

max

10 16 k 11k

160 10k 11k

160 k

k 160

+ =

+ =

=

=

Hence, the value of ‘k’ for marginally stable is 160.

|EC-2020|


5. For a vector field A , which one of the following is FALSE?

(A) A is another vector field.

(B) A is irrotational if 2A 0. =

(C) ( ) ( ) 2A .A A = −

(D) A is solenoidal if .A 0. =

Key: (B)

6. In the circuit shown below, all the components are ideal, and the input voltage is sinusoidal. The

magnitude of the steady-state output 0V (rounded off to two decimal places) is ______V.

Key: (650.538)

Sol: Given circuit

The above circuit acts as voltage doubler circuit

( )o PV 2 Peak value of the input voltage V =

Peak value of the input voltage PV 230 2=

( )OV 2 230 2 650.538 volts= =

1C

1D oV

2D

2C

+

−

0V

+

−

230V

rms

+

−

Positive clamper Positive peak detector

1C 0.1 F= 2D

1D2C 0.1 F=

0V( )230 V rms

+

−

|EC-2020|


7. In the circuit shown below, all the components are ideal. If iV is 2V,+ the current oI sourced by the op-

amp is _________mA.

Key: (6)

Sol: Given,

From virtual short concept N P iV V V= =

Applying KCL at node “N” we get

1 2I I=

N N 0N i

1 f

0 V V Vwhere V V

R R

− −= =

f

o i

1

o i

RV 1 V

R

V 2V

= +

=

Applying KCL at node “O” we get

1k

5V

5V−

−

+

1k

1k

iV

oI

fR 1k=

5V

5V−

oV

−

+

LR 1k=

1R 1k=

1I

iV0A

0A

LI

oI

2I

P

N O

|EC-2020|


2 0 L

N 0 00

f L

i i i0 0 i N i

i0 3

I I I

V V VI

R R

V 2V 2VI V 2V and V V

1k 1k

3V 3 2I 6mA

1k 10

+ =

−+ =

−+ = = =

= = =

8. A binary random variable X takes value +2 or -2. The probability ( )P X 2 .= + = The value of

(rounded off to one decimal place), for which the entropy of X is maximum, is ________.

Key: (0.5)

Sol: Given ( )

( )

P X 2

P X 2 1

= + =

= − = −

( ) ( ) ( )

( )( )

2 2

2 2

H X log 1 log 1

dH Xlog log 1

d

= − − − −

= − + −

The maximum value of H(X) requires that

( )

( )2

0

dH X0

d

1log 0

12 1

1

2 1

1

2

=

− =

− = =

− =

=

=

( )

( )

( ) 2 2

0 H X 0

1 H X 0

1 1 1H X log 2 log 2 1 bit symbol

2 2 2

= =

= =

= = + =

( )H X

0 1

2

1

|EC-2020|


9. The figure below shows a multiplexer where 1 0S and S are the select lines, 0 3I to I are the input data

lines, EN is the enable line, and F (P, Q, R) is the output. F is

(A) PQR PQ.+ (B) P QR+ (C) Q PR+ (D) PQ QR+

Key: (D)

Sol: In this case we need to obtain the output expression of the multiplexer circuit given below

( )

F PQR PQ.0 PQR PQ.1

PQR PQR PQ

QR P P PQ

PQ QR

= + + +

= + +

= + +

= +

Hence the correct answer is option (D).

10. Which one of the following pole-zero plots corresponds to the transfer function of an LTI system

characterized by the input-output difference equation given below?

( ) 3

k

k 0

y n 1 x n k=

= − −

0

R

0

1 1S 0S

EN

F

P Q

R

0I

1I

2I

3I

MUX

R

0

R

1 1S 0S

EN

F

P Q

|EC-2020|


(A) (B)

(C) (D)

Key: (D)

Sol: The given equation is

( ) ( ) ( )3

k

k 0

y n 1 x n k=

= − −

We need to obtain the pole zero plot of its corresponding transfer function

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )( )

( )

0

0 1 2 3

n1 2 3

0

1 2 3

y n 1 x n 0 1 x n 1 1 x n 2 1 x n 3

Y z X z z X z z X z z X z x n n z X z

Y z1 z z z

X z

−− − −

− − −

= − − + − − + − − + − −

= − + − −

= − + −

( )( )

( )( )

( )( )

2 3

23 2

3 3

Y z1 1 1H z 1 H z

z z z X z

z 1 z 1z z z 1

z z

= − + − =

− +− + −= =

1m

1−

1Re

1−

1rd3 order pole

1m

1−

1Re

1−

1th4 order pole

1m

1−

1Re

1−

1rd3 order pole

1m

1−

1Re

1−

1rd3 order pole

|EC-2020|


So, the transfer function having 3 zeros located at z = 1, z = + j, z = –j and it has 3 poles and all of them

lies on origin. So, the pole-zero plot is as shown below

Hence the correct answer is option (B).

11. A transmission line of length 3 4 and having a characteristic impedance of 50 is terminated with a

load of 400 . The impedance (rounded off to two decimal places) seen at the input end of the

transmission line is________ .

Key: (6.25)

Sol: Given, oZ 50=

Length of transmission line 3

m.4

=

Load impedance LZ 400=

Input impedance in

3Z ?

4

= =

( ) 2

oin

L

2m 1 ZZ

4 Z

+ = =

Where m = 0, 1, 2 …..

2

oin

L

Z3 50 50Z 6.25

4 Z 400

= = = =

input impedance in

3Z 6.25

4

= =

Img

Re×

rd3 order pole

Unit circle

oZ 50= LZ 400=

0=3

4

=

inZ ?=

|EC-2020|


12. A single crystal intrinsic semiconductor is at a temperature of 300K with effective density of states for

holes twice that of electrons. The thermal voltage is 26 mV.

The intrinsic Fermi level is shifted from mid-bandgap energy level by

(A) 9.01 meV (B) 18.02 meV (C) 13.45 meV (D) 26.90 meV

Key: (A)

Sol: VFi midgap

C

N1E E kT n

2 N

− =

l

Given, V CN 2N=

In question they asked, Fi midE E 9.01meV= +

13. In an 8085 microprocessor, the number of address lines required to access a 16 K byte memory bank

Is _______.

Key: (14)

Sol: In this case we need to obtain number of address lines needed to access a 16 kB memory using 8085 p.

4 10 14 P16kB 16k 1B 2 2 8 2 8 2 m= = = =

So, P = 14 i.e., we need minimum 14 address line to access the memory chip.

14. The random variable

( ) ( ) ( )1; 5 t 7

Y W t t dt, where t0; otherwise

−

= =

And W (t) is a real white Gaussian noise process with two-sided power spectral density

( )wS f 3W Hz,= for all f. The variance of Y is __________.

Key: (6)

Sol: Given ( ) ( )Y W t t dt

−

=

(i) y E y =

(ii) ( )2

2

yE Y = −

(iii) ( )t 1, 5 t 7 =

|EC-2020|


( ) ( )( ) ( ) ( )E Y E W t t dt E W t t dt 0

− −

= = =

( ) ( )WR 3 =

( )2 2E Y 0 E Y − =

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )

W

2

7

5

E W t t dt. W t ' ' t dt

E W t .W t ' . t ' t dt

R t ' t ' t t dt

3 t ' t ' t . t dt

3 t dt

3 1dt

3 2 6

− −

− −

− −

− −

−

=

=

= −

= −

=

=

= =

15. A digital communication system transmits a block of N bits. The probability of error in decoding a bit is

. The error event of each bit is independent of the error events of the other bits. The received block is

declared erroneous if at least one of its bits is decoded wrongly. The probability that the received block

is erroneous is

(A) ( )N 1− (B) N1− (C) ( )N

1 1− − (D) N

Key: (C)

Sol: N-bits transmitted independently,

Y = Probability of no-error ( )( )

( )N

1 1 ... N times

1

= − −

= −

Probability of received block is erroneous is

( )N

1 Y 1 1 .− = − −

|EC-2020|


16. A 10-bit D/A converter is calibrated over the full range from 0 to 10 V. If the input to the D/A converter

is 13A (in hex), the output (rounded off to three decimal places) is ________V.

Key: (3.066)

Sol: It is given that for a 10-bit DAC, the voltage range is (0-10) V, we need to obtain the output for input

(13A)16.

FSO

n 10

V 10Resolution

2 2= =

Analog output = Resolution × (Decimal equivalent of input)

( ) ( )10 16 10

10314 13A 314

2

3.066

= =

=

Hence the correct answer is 3.066V.

17. In the given circuit, the two-port network has the impedance matrix 40 60

Z .60 120

=

The value of LZ

for which maximum power is transferred to the load is _________ .

Key: (48)

Sol: The given circuit is as shown below

We need to obtain LZ so that it will receive maximum power from the circuit

Using Thevenin theorem the above circuit can be replaced by Thevenin equivalent as show below

Z

10

120V +−

+

−

+

−

1I 2I

LZ1V2V

40 60Z

60 120

=

10

120V +−

+

−

+

−

1I 2I

LZ1V2V

|EC-2020|


By maximum power transfer theorem, for maximum power transfer to LZ , the condition is

L thZ Z .=

To obtain thZ of the 2 port network given from output port, the independent voltage source (120V)

should be short circuited, and then ratio of 2V to

2I is thZ .

The Z matrix equation is

1 1 2V 40I 60I= + …(1)

2 1 2V 60I 120I= + …(2)

From the circuit 1 1V 10I ,= − the equation 1 becomes

1 1 2

1 2

1 2

10I 40I 60I

50I 60I

6I I

5

− = +

− =

− =

Now equation 2 becomes

2 2 2

2

2

th L

L

6V 60 I 120I

5

V72 120

I

Z 48 Z

So, Z 48

− = +

= − +

= =

=

Hence the correct answer is 48.

+−thV

thZ 2I

LZ

+

−

2V

40 60Z

60 120

=

10

+−

+

−

1I

2I

2V1V

|EC-2020|


18. The output y[n] of a discrete-time system for an input x[n] is

k n

y n max x k .−

=

The unit impulse response of the system is

(A) unit step signal u [n]. (B) unit impulse signal n .

(C) 1 for all n. (D) 0 for all n.

Key: (A)

Sol: It is given that

( ) ( )k n

y n max x k−

=

We need to obtain its unit impulse response.

If input is unit impulse i.e. ( ) ( )x k k= then output is

Impulse response ( ) ( )y n h n=

So replacing ( )x k by ( )k and y(n) by h(n) in above equation we have

( ) ( )

( ) ( )

( ) ( ) ( )

k n

k 2

h n max k

h 2 max k

max ,.... 3 , 2

max 0 , 0 , 0 .. 0 0

−

− −

=

− =

= − − −

= =

From this we can say if n 1 − then ( )h n 0=

If n 0 then ( )h n 1= for all n,

( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

k 5h 5 max k

max ... 11 , 0 , 1 , 2 , 3 , 4 , 5

max ... 0 , 1 , 0 , 0 , 0 , 0 , 0 max 0,1,0,0,0 2

− =

= −

= = =

Overall ( )

( )

( ) ( )

h n 0; n 0

h n 1; n 0

h n u n i.e., unit step signal

=

=

=

Hence the correct answer is option (A).

|EC-2020|


19. If 1 2 6v ,v ,....,v are six vectors in 4R , which one of the following statements is FALSE?

(A) 4

1 3 5 6If v , v ,v ,v spans R , then it forms a basis for 4R .

(B) Any four of these vectors form a basis for 4R .

(C) These vectors are not linearly independent.

(D) It is not necessary that these vectors span 4R .

Key: (B)

Exp: Given

1 2 6V , V , ....,V are six vectors in 4R

For a 4-dimensional vector space,

(i) any four linearly independent vectors form a basis (or)

(ii) Any set of four vectors in 4R spans 4R , then it forms a basis.

Clearly options (A), (C), (D) are true.

option (B) is FALSE.

20. In the circuit shown below, the Thevenin voltage THV is

(A) 4.5V (B) 2.4V (C) 3.6V (D) 2.8V

Key: (C)


−+

1A

2V2 4

22A1

+

−

THV

−+

1A

2V2 4

22A1

+

−

thV

|EC-2020|


We need to obtain the value of th.V

If we do source transformation to above practical current sources, then the circuit becomes

x

th x

th

1 2 4 1I A

1 2 2 5

2V 2I 4 4 3.6V

5

So V 3.6V

+ − −= =

+ +

−= + = + =

=

Hence the correct answer is option (C).

21. The current in the RL-circuit shown below is ( ) ( )i t 10 cos 5t 4 A.= −

The value of the inductor (rounded off to two decimal places) is ________H.

Key: (2.82)

Sol: The given circuit is shown below

~( )200cos 5t V

R

L

( )i t

+

−

−+

2 4

2

4VxI

+

−

thV

+−

+−1V

1

2V

~200cos5t

R

L

( )i t 10cos 5t4

= −

|EC-2020|


We need to obtain the value of inductor L

The source voltage and the loop current is known, so the impedance of the network is

( )

o

Lo

L

V 200 0 20 20Z 20 45 j R jX

I 10 45 2 2

20 20 20by comparision X L L 5

2 2 5 2

L 2.82H

So, the value of inductor is 2.82H

= = = = + = +

−

= = = =

=

22. The impedances Z jX,= for all X in the range ( ), ,− map to the Smith chart as

(A) a line passing through the centre of the chart.

(B) a point at the centre of the chart.

(C) a circle of radius 0.5 with centre at (0.5,0).

(D) a circle of radius 1 with centre at (0, 0).

Key: (D)

Sol: Given,

Load impedance Z jX= (purely reactive)

On the smith chart, as you move on the constant resistance “R” = 0 circle whose centre is (0, 0) and

radius is “1”, the impedance seen on this circle is purely reactive.

Hence the correct option is (D).

23. Consider the recombination process via bulk traps in a forward biased pn homojunciton diode. The

maximum recombination rate is max.U If the electron and the hole capture cross-sections are equal,

which one of the following if FALSE?

(A) With all other parameters unchanged, maxU decreases if the intrinsic carrier density is reduced.

(B) maxU depends exponentially on the applied bias.

(C) With all other parameters unchanged, maxU increases if the thermal velocity of the carriers

increases.

(D) maxU occurs at the edges of the depletion region in the device

Key: (D)

|EC-2020|


Sol: qV

2kTmax 0 Th t i

1u V N n e

2=

(i) max Thu V

(ii) max iu n

(iii) qV

2kTmaxu e

And also maxu occurs in the depletion region when (n + P) is minimum, not at the edges of depletion.

24. The two sides of a fair coin are labelled as 0 and 1. The coin is tossed two times independently. Let M

and N denote the labels corresponding to the outcomes of those tosses. For a random variable X, defined

as X=min (M,N) , the expected value E(X) (rounded off to two decimal places) is ______.

Key: (0.25)

Sol: Given that, the two sides of a fair coin are labelled as 0 and 1. Let M and N denote the labels

corresponding to the outcomes of those tosses.

Random variable

( )

( ) ( ) ( ) ( )

( )

X min M, N

X min 0,0 0; min 0,1 0;min 1,0 0;min 1,1 1

X 0,1. Discrete r.v

=

= = = = =

=

Probability distribution table is

( )

X 0 1

3 1P X

4 4

( ) ( )

( )

3 1 1E X x P x 0. 1. 0.25

4 4 4

Expected value E X 0.25

= = + = =

=

|EC-2020|


25. The pole-zero map of a rational function G(s) is shown below. When the closed contour is mapped

into the G(s)- plane, then the mapping encircles

(A) the point -1+j0 of the G(s)-plane once in the clockwise direction.

(B) the point -1+j0 of the G(s)-plane once in the counter-clockwise direction.

(C) the origin of the G(s)-plane once in the clockwise direction

(D) the origin of the G(s)-plane once in the counter-clockwise direction.

Key: (D)

Sol: Given, P = 4, Z = 3

We need to find the mapping of G(s), N = P – Z = 4 – 3 = 1

Hence, origin of G(s) are in CCW.

1m

1−

Re1−

S plane−

x

x

o

o

|EC-2020|


Q. No. 26 to 55 Carry Two Marks Each

26. ( )X is the Fourier transform of x(t) shown below. The value of ( )2

X d

− (rounded off to two

decimal places) is______________.

Key: (58.61)

Sol: The signal x(t) is as shown below

We need to obtain value of ( )2

x d

−

By Parseval’s theorem

( ) ( ) ( )( )2 2 2

X d 2 x t dt 2 x t dt

− − −

= =

[Since the given time domain signal x(t) is real, then the mod function does not have any role].

We know that shifting does not change the energy of signal so if we define ( ) ( )y t x t 1 ,= + it means

energy of x(t) and y(t) are same. Now y(t) becomes

3

( )x t

01−

1

1 2 3t

2

42−3−

3

( )x t

01−

1

1 2 3t

|EC-2020|


( ) ( ) ( )

( ) ( ) ( )

2 2 2

2 2 0

2 2 2

2 0 2

X d 2 x t dt 2 y t dt

2 y t dt 2 2 y t dt 4 y t dt

− − −

− −

= =

= = =

( ) ( )( )

( ) ( ) ( )

2

a a 0

e e e

a 0 a

y t is even then y t is also even, for even signal

y t dt 2 y t dt 2 y t dt− −

= =

( ) ( )

( ) ( )

( ) ( )

02 2

2

1 02 2

2 1

1 0

2 2

2 1

1 03 2 3 2

2 1

X d 4 y t dt

4 t 2 dt 2t 3 dt

4 t 4t 4 dt 4t 12t 9 dt

t t 4t t4 4 4t 12 9t

3 2 3 2

− −

−

− −

−

− −

−

− −

=

= + + +

= − + + + +

= + + + + +

( ) ( ) ( ) ( ) ( ) ( )1 4

4 1 8 2 1 4 4 1 2 0 1 6 0 1 9 0 13 3

2 4 114 6 4 6 9 4 1

3 3 3

14 564 58.61

3 3

= − + + − + − + + + + − + +

= − + + − + = +

= = =

Hence the correct answer 58.6%.

3( ) ( )y t x t 1= +

1− 0 1 2t

1−

2−

|EC-2020|


27. A one-sided abrupt pn junction diode has a depletion capacitance DC of 50 pF at a reverse bias of 0.2V.

The plot of 2

D1 C versus the applied voltage V for this diode is a straight line as shown in the figure

below. The slope of the plot is _____________20 2 110 F V .− −

(A) –3.8 (B) –1.2 (C) –5.7 (D) –0.4

Key: (A)

Sol: Given D Rc 50PF, V 0.2V= =

( )

( )

2

bi R

D s

2

s D bi R

1 2V V

C q N

2 1 1slope

q N C V V

= + +

= − = −

+

bi

20 2 1

If V 0.85V then, we will get

slope 3.8 10 F V ,Option A is correct− −

=

= −

28. In the voltage regulator shown below, 1V is the unregulated input at 15V. Assume BEV 0.7V= and the

base current is negligible for both the BJTs. If the regulated output oV is 9V, the value of 2R is

_________ .

1V 15V= oV 9V=

1R 1k=

2R

ZV 3.3V=

3R 1k=

2

D1 C

0V

|EC-2020|


Key: (800)

Sol: Given the following circuit

ONBEV 0.7V;= is very large implies

BI 0A

2R ?=

Apply KVL around loop (1) we get

L

L

V 0.7 3.3 0

V 4V

− − =

=

Applying KCL at node “L” we get

LI I=

Where 0 L

1

V V 9 4I 5mA

R 1k

− −= = =

LI I 5mA = =

L2 3

L

V 4R 800

I 5 10− = = =

29. Consider the following closed loop control system

( )R s Y(s)( )C s ( )G s

−

+

15VoV 9V=

1R 1k=

I

LI

2R

+

− 0.7BI 0A

+

−ZV 3.3V=

LV

loop (1)

3R

|EC-2020|


( )( )

( )1 s 1

where G s and C s Ks s 1 s 3

+= =

+ + If the steady state error for a unit ramp input is 0.1, then the

value of K is _________.

Key: (30)

Sol: Given,

( )( )

( )( )

( )ss

k s 11G s and E s and e 0.1

s s 1 s 3

+= = =

+ +for ramp

We need to find k value

( ) ( )

( )

( )

( )

ss vs 0

v

vs 0

v

1e and k limsG s E s

k

s 1s 1k lim k

s s 1 s 3

kk

3

30.1

k

3k 30

0.1

→

→

= =

+=

+ +

=

=

= =

Hence, the value of k is 30.

( )R s Output( )E s ( )G s

−+

|EC-2020|


30. The state diagram of a sequence detector is shown below. State 0S is the initial state of the sequence

detector. If the output is 1, then

(A) the sequence 01010 is detected (B) the sequence 01110 is detected.

(C) the sequence 01001 is detected (D) the sequence 01011 is detected

Key: (A)

Sol: The state diagram of a sequence detector is given where the initial state is S0. If output is 1 then we need

to obtain the input sequence.

In general

This transition means the state will transit from A BS to S when the input is 1 and at the end of transition it

will produce output as 0.

The given state diagram is

1 0 0 0

0 0

0 0 1 0 0 0

1 01 0

1 0

0S 1S 2S 3S

4S

0 1

AS BS1 0

input output

0S1S 1S

3S

4S

Initial state

1 0 0 0

1 0 0 0

1 0

Final target1 0

0 0

1 0

|EC-2020|


The above dotted line shows the desired sequence to get output 1. So, the state transition sequence is

0 0 1 0 0 0 1 0 0 1

0 1 2 3 4 3S S S S S S⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→ ⎯⎯→

The corresponding input sequence is 0, 1, 0, 1, 0

The corresponding output sequence is 0, 0, 0, 0, 1

So, the input sequence detected is 01010

Hence the correct answer is option (A)

31. The characteristic equation of a system is

( )3 2s 3s K 2 s 3K 0+ + + + =

In the root locus plot for the given system, as K varies from 0 to , the break-away or break-in point(s)

lie within

(A) ( ), 3− − (B) ( )3, 2− − (C) ( )1, 0− (D) ( )2, 1− −

Key: (C)

Sol: Given,

( )

( )

2

3 2

3 2

s 3s k 2 s 3k 0

We need to calculate break away / break in point

s 3s ks 2s 3k 0

s 3s 2s s 3 k 0

+ + + + =

+ + + + =

+ + + + =

( )

( )( )

2 2

3 2

s 3 k1

s 3s 2s

s 3s 2sk

s 3

++

+ +

− + +=

+

For breakaway|breakin we need to find dk

ds

( )( )

( )

( )( )

( )

2 3 2

2

3 2 2 3 2

2

3 2

2

3s 6s 2 s 3 s 3s 2sdk

ds s 3

3s 6s 2s 9s 18s 6 s 3s 2s

s 3

2s 2s 18s 6

s 3

+ + + − + + = −

+

+ + + + + − − − =

+

− + + +=

+

|EC-2020|


dk

0 s 0.46, 3.87, 1.65ds

= = − − −

Since –0.46 is a valid point. Hence correct option is (C).

32. For the given circuit, which one of the following is the correct state equation?

(A) 1

2

iv 4 4 v 4 4d

2 4 0 4i i idt

− − = +

− (B)

1

2

iv 4 4 v 4 0d

2 4 0 4i i idt

− − = +

− −

(C) 1

2

iv 4 4 v 0 4d

2 4 4 0i i idt

− = +

− − (D)

1

2

iv 4 4 v 0 4d

2 4 4 4i i idt

− = +

− −

Key: (C)

Sol: Given,

We need to select the correct state equation of given circuit diagram

KCL at P point

2

2 2

dvi 0.25 V i

dt

dv4V 4i 4i 4V 4i 4i ...(i)

dt

= + −

= − + + = − + +

KCL at point A

1i1v

2i

0.5H

+

−2

i

0.25F

1i1v

2i

0.5H+

−2

iA

P

0.25F

V

|EC-2020|


A1 A

1

1

1

1

V dii i V V 0.5

2 dt

diV 0.5

dti i2

di2i V 0.5 2i

dt

di0.5 2i V 2i

dt

di4i 2V 4i ...(ii)

dt

= + = +

+

= +

= + +

= − − +

= − − +

State equation

1

2

iv 4 4 V 0 4d

i2 4 i 4 0idt

− = +

− −

33. The band diagram of a p-type semiconductor with a band-gap of 1 eV is shown.

Using this semiconductor, a MOS capacitor having '

TH oxV of 0.16V, C− of 2100 nF cm and a metal

work function of 3.87 eV is fabricated. There is no charge within the oxide. If the voltage across the

capacitor is THV , the magnitude of depletion charge per unit area ( )2in C cm is

(A) 80.93 10− (B) 81.41 10−

(C) 80.52 10− (D) 81.70 10−

Key: (D)

Vacuum level

4 eV

0.5 eV

0.2 eV

CE

iE

FsE

VE

|EC-2020|


Sol: In MOS semiconductor

g

ms m s s

Eq q qX q ; q 1 0.7 0.3

2

3.87eV 4 0.5 0.3 eV

3.87 4.8 0.93eV

= − + + = − =

= − + +

= − = −

We know that, ( )( ) oxTN SD ss ms p

ox

't

V Q max Q' q 2q= − + +

Here ssQ' 0= (there is no charge in oxide )

SD

ox

SD

ox

9 8

SD

Q0.16 0.093 0.6

C

Q0.17

C

Q 0.17 100 10 1.7 10− −

− = − + +

=

= =

34. The base of an npn BJT T1 has a linear doping profile ( )BN x has shown below.

The base of another npn BJT T2 has a uniform doping 17 3

BN of 10 cm .− All other parameters are

identical for both the devices. Assuming that the hole density profile is the same as that of doping, the

common-emitter current gain of T2 is

Metalmq

iqX

Bq

qXSq

PqgE 2

doxide

VE

Vaccumlevel

CE

iE

FEFE

Semi conductor−

|EC-2020|


(A) approximately 2.0 times that of T1 (B) approximately 2.5 times that of T1

(C) approximately 0.3 times that of T1 (D) approximately 0.7 times that of T1

Key: (A)

35. A finite duration discrete-time signal x[n] is obtained by sampling the continuous-time signal

( ) ( )x t cos 200 t= at sampling instants t n 400,= n=0.1,….,7. The 8-point discrete Fourier transform

(DFT) of x[n] is defined as

kn7 j4

n 0

X k x n e , k 0,1,...,7.

−

=

= =

Which one of the following statements is TRUE?

(A) Only X[3] and X[5] are non-zero (B) Only X[4] is non-zero.

(C) Only X[2] and X[6] are non-zero (D) All X[k] are non-zero.

Key: (C)

Sol: It is given that a finite duration discrete signal x(n) is obtained by sampling ( )x t cos200 t= , with

s

1T

400= we need to comment about its 8 point DFT.

The discrete time signal is

( )

( )

200x n cos n cos n; n 0,1,2...7

400 2

2 3 4 5 6 7x n cos0,cos ,cos ,cos , cos ,cos ,cos ,cos

2 2 2 2 2 2 2

1,0, 1,0,1,0, 1,0

= = =

=

= − −

Let ( ) y n 1, 1,1, 1= − −

The its 4-point DFT will be

0 W

( )14 310 cm−

Collector

( )17 310 cm−

( )BN x

BaseEmitter

|EC-2020|


( )

1 1 1 1 1

1 j 1 j 1y k 0 0 4 0

1 1 1 1 1

1 j 1 j 1

− − − = = − −

− − −

By comparing x(n) and y(n) we can say

( )n

x n y2

=

By property

( ) ( )

( ) ( ) ( )

( ) ( )

( )

( )( ) ( )X 2 X 6

y n Y k

ny Y k ,Y k ,Y k ...m times

m

ny Y k ,Y k

2

Sin ce y n 0,0,4,0

nx n y 0,0, 4 ,0,0,0, 4 ,0

2

=

So only X (2) and X (6) are non-zero

Hence the correct answer is (C).

36. For the modulated signal x(t) = ( )cm(t) cos 2 f t , the message signal ( ) ( )m t 4cos 1000 t= and the

carrier frequency cf is 1MHz, The signal x(t) is passed through a demodulator, as shown in the figure

below. The output y(t) of the demodulator is

(A) ( )cos 920 t (B) ( )cos 460 t (C) ( )cos 1000 t (D) ( )cos 540 t

Key: (A)

Ideal LPF with cut off 510 Hz−

1

( )y t

510− 510( )f Hz

( )( )ccos 2 f 40 t +

|EC-2020|


Sol: ( ) ( ) ( )( )cs t x t cos 2 f 40 t= +

( ) ( )( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )( ) ( )( )

( ) ( )

c c

c

m t cos2 f t.cos 2 f 40 t

1m t cos 4 f t 2 40t cos 2 40t

2

1m t cos 2 40t

2

cos 1000 t cos 2 40t

1cos 2 540 t cos 2 460 t

2

cos 2 460 t cos 920 t

LPF

= +

= + +

=

=

= +

= =

37. Consider the following system of linear equations

1 2 1 1 2 2 1 2 3 1 2 4x 2x b ; 2x 4x b ; 3x 7x b ; 3x 9x b+ = + = + = + =

Which one of the following conditions ensures that a solution exists for the above system?

(A) 3 1 1 3 4b 2b and 3b 6b b 0= − + = (B) 3 1 1 3 4b 2b and 6b 3b b 0= − + =

(C) 2 1 1 3 4b 2b and 3b 6b b 0= − + = (D) 2 1 1 3 4b 2b and 6b 3b b 0= − + =

Key: (D)

Sol: Given, the system of linear equations:

1 2 1

1 2 2

1 2 3

1 2 4

x 2x b

2x 4x b

3x 7x b

3x 9x b

+ =

+ =

+ =

+ =

Consider, the augmented matrix

1

2

3

4

2 2 1 3 3 1 4 4 1

1 2 b

2 4 bA | B

3 7 b

3 9 b

R R 2R ; R R 3R ; R R 3R ;

=

→ − → − → −

1

2 1

3 1

4 1

4 4 3

1 2 b

0 0 b 2bA | B ~

0 1 b 3b

0 3 b 3b

R R 3R ;

− −

−

→ −

|EC-2020|


1

2 1

3 1

1 3 4

1 2 b

0 0 b 2bA | B ~

0 1 b 3b

0 0 6b 3b b

− −

− +

1

3 1

2 1

1 3 4

1 2 b

0 1 b 3bA | B ~ Echelon form

0 0 b 2b

0 0 6b 3b b

− → −

− +

( )

clearly,

A 2

=

Solution exists for the above system if

2 1 1 3 4

2 1

2 1 1 3 4

if b 2b 0 and 6b 3b b 0

b 2b

If b 2b and 6b 3b b 0 then

− = − + =

=

= − + =

( ) ( )A AB , = i.e., solution exists for the above system.

38. The magnetic field of a uniform plane wave in vacuum is given by

( ) ( ) ( )x y zH x, y,z, t a 2a ba cos t 3x y z .= + + + − −

The value of b is ________.

Key: (1)

Sol: Given

( ) ( )( )x y zˆ ˆ ˆH a 2a ba cos t 3x y z A m= + + + − −

We have to determine value of “b”.

By comparing the given field with the following general expression

( ) ( ) ( )x x y y z z x y zx,y,z

H

ˆ ˆ ˆH H a H a H a cos t x y z

= + + − + +

We get x y zH 1, H 2 and H b= = = and x y z3, 1, 1 = − = =

For a UPW H field H− and wave vector “ ” must be mutually perpendicular to each other

i.e., ( )( ) ( )x y z x y zˆ ˆ ˆ ˆ ˆ ˆH. 0 i.e., a 2a ba . 3a a a 0 = + + − + + =

3 2 b 0

b 1

− + + =

=

|EC-2020|


39. For a 2-port network consisting of an ideal lossless transformer, the parameter 21S (rounded off to two

decimal places) for a reference impedance of 10 , is _________.

Key: (0.8)

Sol: Given,

2-port network consisting of an ideal lossless transformer

1 1

2 2

V In 1and

V 1 I n = =

Reference input and output impedances 01 02R R 10= =

We have to determine forward transmission coefficient 21S ?=

01 02If R R ,then= the scattering matrix for an ideal transformer is given by

11 12

21 22

S S2

2 2

2

2 2

S S

21 2 2

n 1 2n

n 1 n 1S

2n 1 n

n 1 n 1

2n 2 2S n 2

n 1 2 1

0.8

−

+ +=

− + +

= = =

+ +

=

Forward transmission coefficient 21S 0.8=

1I 2I2 :1

+

−

1V 2V+

−

2 :1

2 Port Network

Port1 Port 2

|EC-2020|


40. P, Q and R are the decimal integers corresponding to the 4-bit binary number 1100 considered in signed

magnitude, 1’s complement, and 2’s complement representations, respectively. The 6-bit 2’s

complement representation of (P+Q+R) is

(A) 110101 (B) 111001 (C) 110010 (D) 111101

Key: (A)

Sol: It is given that

P represent decimal equivalent of signed magnitude number: 1100

Q represent decimal equivalent of 1’s complement number: 1100

R represent decimal equivalent of 2’s complement number: 1100

We need to obtain 6 digits, 2’s complement form of P + Q + R.

1100 in signed magnitude form = –[Decimal equivalent of (binary 100)] = –4

So, P = –4.

1100 in 1’s complement form

= – [Decimal equivalent of (1’s complement of 1100)]

= – [Decimal equivalent of 0011]

= –3

So, Q = –3

1100 in 2’s complement form

= – [Decimal equivalent of (2’s complement of 1100)]

= – [Decimal equivalent of 0100]

= –4

So, R = –4

P + Q + R = (–4) + (–3) + (–4) = (–11)10

+ 11 →01011 (in 2’s complement form)

– 11 →10101 (in 2’s complement form)

Now to increase the bit size to 6 from 5 bits.

We need to copy the sign bit once towards left.

So, 6 bit 2’s complement representation is 110101.

Hence the correct answer is 110101.

|EC-2020|


41. The components in the circuit given below are ideal. If R 2k and C 1 F,= = the -3 dB cut-off

frequency of the circuit in Hz is

(A) 79.58 (B) 34.46 (C) 14.92 (D) 59.68

Key: (A)

Sol: Given the following circuit

From virtual short concept N PV V 0V= =

The current flowing through parallel combination of 2R and 2C is

NN

VI 0A V 0V

12R ||

2sC

= = =

Applying KCL at node “N” we get

R

oV (j )

−

+

RC

2C

iV (j )

2R

R

oV

−

+

R

N

2I1I

0A

P

C

0A

2RI 0A= 2C

iV

|EC-2020|


1 2

i N N o

oN

i

o

i

I I

V V V V 1 RWhere Z R ||

R Z sC sCR 1

V ZV 0V

V R

RV 1sCR 1

V R sCR 1

Put s j

=

− − = = = +

− = =

−−+ = =+

=

( )

o

2i

V 1 1

V 1 j RC 1 RC

= =+ +

At 3-dB cut off frequency ( ) oc

i

V 1f ,

V 2=

(i.e.,)

( )2

C

1 1

21 RC

=

+

C

1rad sec

RC =

Cut off frequency c 3 6

1 1f 79.58Hz

2 RC 2 2 10 10−= = =

Hence the correct option is (A)

42. Using the incremental low frequency small-signal model of the MOS device, the Norton equivalent

resistance of the following circuit is

(A) ds

m ds

r R

1 g r

+

+ (B) ds m dsr R g r R+ + (C) ds

m

1r R

g+ + (D) dsr R+

DDV

R

m, dsg r

|EC-2020|


Key: (A)

Sol: The small signal equivalent model for the given circuit can be represented as follows

x gsV = − …(1)

xin

x

VR

I = …(2)

Applying KCL at node “S” we get x dI i 0+ = …(3)

Apply KVL around loop (1) we get

d 1 ds x0 i R i r V 0− − − =

Where ( ) ( )1 d m gs d m xi i g i g V= − = +

( )

( )

x d d m x ds

x m ds

d

ds

V i R i g V r

V 1 g ri

R r

− = + +

− + =

+

Substituting “id” in equation (3) we get

dsxin

x m ds

R rVR

I 1 g r

+= =

+

Hence the equivalent resistance looking into source terminal is dsin

m ds

R rR

1 g r

+=

+

Hence the correct option is (A).

+−

+

−

gs m gsg dsr

diD

R1i

diS

G

xV

xI loop (1)inR

|EC-2020|


43. For the BJT in the amplifier shown below, BEV 0.7V, kT q 26 mV.= = Assume that BJT output

resistance ( )or is very high and the base current is negligible. The capacitors are also assumed to be

short circuited at signal frequencies. The input iV is direct coupled. The low frequency voltage gain

0 iv v of the amplifier is

(A) –178.85 (B) –128.21 (C) -256.42 (D) -89.42

Key: (D)

Sol: Small signal equivalent model for the given circuit is shown in following figure

( ) ( )

be i

0 c C L m be C L

V

V i R || R g R || R

=

= − = −

( ) ( )om C L

i

Vg R || R 1

V = − →

~iV

+

−

be rm beg

( )C LR || R

+

−

ci

oV

CCV 10V=

LR

10 k

0V

CR

10 k

ER

20 k

EC

100 F

1C

1 F

fV

EEV 10V= −

|EC-2020|


DC equivalent model for the given circuit is

Applying KVL around loop (1) we get

( )E

E

0 0.7 I 20k 10 0

I 0.465mA

− − + =

=

= implies C EI I 0.465 mA= =

C

m T

T

3

Ig Where V 26mV

V

38.46 0.465 10 17.8m−

= =

= =

Substituting value of “mg ” in equation (1) we get

C3o

Li

R 10k andV17.88 10 10k ||10k

R 10kV

89.42

−=

= − =

= −

Hence the correct option is (D).

44. In a digital communication system, a symbol S randomly chosen from the set 1 2 3 4s ,s ,s ,s is

transmitted. It is given that 1 2 3s 3, s 1, s 1= − = − = + and 4s 2.= + The received symbol is T S W.= +

W is a zero-mean unit-variance Gaussian random variable and is independent of iS. P is the conditional

probability of symbol error for the maximum likelihood (ML) decoding when the transmitted symbol

iS s .= The index if for which the conditional symbol error probability iP is the highest is ______.

Key: (3)

Sol: ML Decision rule for AWGN channel

The ML decision rule can be simplified as, here oNˆi m, 1, 0

2= = =

CCV 10V=

10k

CB

+−0.7

20k EI

EEV 10V= −

Loop (1)

E

|EC-2020|


( )

( )( )

( )

m̂m̂

m̂m̂

N2

ˆo k mkm̂

k 1o

42

ˆk mkm̂

k 1

2

m̂m̂

m̂ arg max P r s

arg max n P r s

N 1arg max n N r s

2 N

arg min r s

arg min r s

=

=

=

=

−= − −

= −

= −

l

l

We select that vector ms which has minimum Euclidean distance

We can choose

3s , such that it has minimum Euclidean distance m = 3 = i

45. Which one of the following options contains two solutions of the differential equation ( )dy

y 1 x ?dx

= −

(A) 2ln y 1 2x C and y 1− = + = (B)

2ln y 1 2x C and y 1− = + = −

(C) 2ln y 1 0.5x C and y 1− = + = (D)

2ln y 1 0.5x C and y 1− = + = −

Key: (C)

Sol: Given D.E is ( )dy

y 1 xdx

= −

2

1dy xdx

y 1

Variable separable D.E

1dy xdx

y 1

xn y 1 c

2

=−

−

=−

− = +

2D.E is n y 1 0.5x c

Also y 1 will satisfy given D.E.

− = +

=

1s 2 2 12s 3s 4s

3− 1− 1 2

|EC-2020|


46. The current I in the given network is

(A) 2.38 23.63 A.− (B) 2.38 143.63 A

(C) 2.38 96.37 A− (D) 0A.

Key: (B)


We need to obtain Io

By KCL at node A

( )o 1 2

120 90 120 30I I I 2.38 36.3

80 j35 80 j35

− −= − + = − + = −

− −

So, magnitude of oI 2.38=

Hence the correct answer is 2.38 143.63 A

47. A system with transfer function ( )( )( )

1G s , a 0

s 1 s a=

+ + is subjected on an input 5cos 3t. The steady

state output of the system is ( )1

cos 3t 1.892 .10

− The value of a is ________.

Key: (4)

120 90 V−

120 30 V−

I

Z

Z

( )Z 80 j35= −

+−

−+

120 90−

120 30−

oI

Z

Z

Z 80 j35= −

~

~

+

−

+

−

A

1I

2I

−

+

−

+

|EC-2020|


Sol: Given,

Input = 5cos(3t) and output ( )1

cos 3t 1.89210

= −

We need to find a for ( )( )( )

1G s

s 1 s a=

+ +

2

2

1 5 1

10 10 9 a

9 a 25 a 4

=

+

+ = =

Hence, the correct value of ‘a’ is 4.

48. For the solid S shown below, the value of S

xdxdydz (rounded off to two decimal places) is_____.

Key: (2.25)

Sol: ( )3

0

32

0

x dxdy dz xdx A

x0.5

2

90.5

2

2.25

=

=

=

=

Hence the value of x dxdy dz is 2.25

3

x

z

1

1y0

3

x

z

1

1y0

1A 1 1 0.5

2= =

|EC-2020|


49. A pn junction solar cell of area 21.0 cm , illuminated uniformly with

2100 mW cm ,− has the following

parameters: Efficiency = 15%, open circuit voltage =0.7V, fill factor = 0.8, and thickness 200 m.= The

charge of an electron is 91.6 10 C.− The average optical generation rate ( )3 1in cm s− − is

(A) 191.04 10 (B) 190.84 10 (C) 195.57 10 (D) 1983.60 10

Key: (B)

Sol: We know that in solar cell,

Fill factor m m

sc oc

V IFF

I V= =

m m

in

V I100

P =

oc sc

in

sc

1

sc

FF V .ISo, 100

P

0.8 0.7I15 100

100m 1

15I 10 A

56

−

=

=

=

Optical generation rate can be calculated by

t

scsc L

0

L

IJ e G dt, where t thickners

A

eG .t

= = =

=

sc LI AeG t=

scL

1

19 4

1

5 2 19

19

19 3 1

19 3 1

IG

e t

15 10

56 1.6 10 200 10

15 10

56 16 2 10 10 10

15 10010

56 32

0.837 10 cm s

0.84 10 cm s

−

− −

−

− −

− −

− −

=

=

=

=

=

=

|EC-2020|


50. ( ) ( )PM FMS t and S t as defined below, are the phase modulated and the frequency modulated waveforms,

respectively, corresponding to the message signal m(t) shown in the figure.

( ) ( )( )

( ) ( )( )PM p

t

FM f

S t cos 1000 t K m t

and S t cos 1000 t K m d−

= +

= +

Where pK is the phase deviation constant in radians/volt and fK is the frequency deviation constant in

radians/second/volt. If the highest instantaneous frequencies of ( ) ( )PM FMS t and S t are same, then the

value of the ratio p

f

K

K is ____________ seconds.

Key: (2)

Sol: The PM and FM equations are shown below

( )

( )

PM c p

t

FM c f

S A cos 1000 t k m t

S A cos 1000 t k m t dt−

= +

= +

Given,

In ( ) ( )p

i max

max

k dPM f 500 m(t)

2 dt= +

10

1 3 6 7 ( )t sec

m(t)V

0

10

1 3 6 7 ( )t sec→

m(t)

|EC-2020|


In ( ) ( )f

i max

max

kFM f 500 m t

2= +

Both are equal when

( ) ( )p f

max

p

f

k kdm t m t

2 dt 2

k 102

k 5

=

= =

51. For an infinitesimally small dipole in free space, the electric field E in the far field is proportional to

( )jkre / r sin , where k 2 .− = A vertical infinitesimally small electric dipole ( ) l is placed at a

distance h (h>0) above an infinite ideal conducting plane, as shown in the figure. The minimum value of

h, for which one of the maxima in the far field radiation pattern occurs at 60 ,= is

(A) 0.5 (B) (C) 0.25 (D) 0.75

Key: (B)

Sol: Given,

An infinite & small dipole ( ) l

jkre 2

E sin where kr

− =

We have to determine minimum value of “h” for which one of the maxima in the far field radiation

pattern occurs at 60 =

h

Z

0

y

Il

|EC-2020|


Current phase shifts between Idl and its image is o0 =

When ever we place a dipole at height “h” above the conducting plane, there forms its image at a depth

“h” below the conducting plane and thus it forms a two-element array.

For a 2-element array, array factor is given by

A.F 2cos2

=

Array factor is maximum when m i.e. 2m2

= =

Where kdcos 2m= +=

Where 2

0 and k

= =

and we want maximum at o60=

o2dcos60 0 2m m 0,1,2....

dm

2

d 0, 2 , 4 ,..........

Minimum value of d 2

d 2

h

Minimum value of h

+ = =

=

=

=

=

=

=

z

60 =

Image

Infinite conducting plane

y

h

h

d 2h=

oI 0

oI 0

|EC-2020|


52. The transfer function of a stable discrete-time LTI system is ( )( )K z

H zz 0.5

−=

+ where K and are real

numbers. The value of (rounded off to one decimal place) with 1, for which the magnitude

response of the system is constant over all frequencies, is _________.

Key: (-2)

Sol: The transfer function of a stable discrete time system is ( )( )k z

H z ,z 0.5

−=

+where 1, if magnitude

response is constant for all frequency we need to obtain .

Magnitude response is constant for all frequency means H(z) represent transfer function of an all pass

filter.

In an all pass filter oP is pole then its zero must be

o

*1.

P

Here oP 0.5= −

So ( )* *

*

o

1 12 2

P 0.5

= = = − = −

−

So, value of 2. = −

Hence the correct answer is –2.

53. X is a random variable with uniform probability density function in the interval 2,10 .For Y 2X 6,− = −

the conditional probability ( )P Y 7 X 5 (rounded off to three decimal places) is ________.

Key: (0.3)

Sol: Pdf of X

Given, Y 2X 6= −

Y in range 10 14 ,−

2− 10 X→

1

12

10− 14 Y

1

24

|EC-2020|


( )( )

( )

P Y 7 X 5P Y 7 X 5

P X 5

=

Y 6

X 52

y 6 10

y 4

+=

+

( )

( )7

4

10

5

Y 4 Y 7P

P X 5

1 3dY24 24

51dX

1212

30.3

10

=

= =

= =

54. For the components in the sequential circuit shown below, pdt is the propagation delay, setupt is the setup

time, and holdt is the hold time. The maximum clock frequency (rounded off to the nearest integer), at

which the given circuit can operate reliably, is _______MHz.

Key: (76.92)

Flip Flop1

pd

setup

hol

t 3 ns

t 5 ns

t 1ns

=

=

=

pd

setup

hol

t 8 ns

t 4 ns

t 3ns

=

=

=

IN

Clk

pdt 2ns=pdt 2ns=

|EC-2020|


Sol: The given circuit diagram is as shown below

We need to obtain the CLK frequency for reliable operation.

Set up time: ( )st : Flip flop inputs ( )1 2I , I should be stable at least for st interval, before arrival of

clock.

Hold time: ( )ht : Flip flop inputs ( )1 2I , I should be stable at least for ht interval after arrival of clock.

Let as assume all states are stable and a clock trigger arrives.

(i) 1Y will be available at 3nsec ( )pd1t 3nsec=Q

(ii) 2Y will be available at 8nsec ( )

2pdt 8nsec=Q

(iii) 2I will be available after delay of XOR, NAND gates since Y, is available at 3nsec we will get

2I

after 3 + 2 + 2 = 7nsec (delay of both logic gates).

(iv) 1I will be available when 2Y will be available i.e., at 8nsec.

Finally, if all states are stable and a clock arrives, then to process the flip flops i.e. the new inputs to flip

flop are available at 1 2I 8nsec, I 7nsec= =

But before arrival of next clocks these new 1 2I , I should be stable before there corresponding flip flops

set up time.

So, for flip flop 1, the clock period should be

pd

setup

hold

t 3 ns

t 5 ns

t 1ns

=

=

=

pd

setup

hold

t 8 ns

t 4 ns

t 3ns

=

=

=

INCLK

pdt 2ns=pdt 2ns=

1Y

2I

1I

( )2I

2Y

FF1

FF2

|EC-2020|


s18nsec t 8nsec 5nsec 13nsec+ = + =

For flip flop 2 the clock period should be

2s7nsec t 7nsec 4nsec 11nsec+ = + =

Since both flip flops are simultaneously triggered from a single clock, for reliability on both flip flops.

( )CLKT max 13nsec,11nsec 13nsec =

(If we take 11nsec, FF1 will have errors)

CLK CLK CLK9

CLK

1 1f f f 76.92 MHz

T 13 10−

So maximum value of clock frequency can be 76.92 MHz

Hence the correct answer is option 76.92.

55. An enhancement MOSFET of threshold voltage 3 V is being used in the sample and hold circuit given

below. Assume that the substrate of the MOS device is connected to -10V. If the input voltage IV lies

between 10V, the minimum and the maximum values of GV required for proper sampling and holding

respectively, are

(A) 3V and -3V (B) 13V and -7V (C) 10V and -10V (D) 10 V and -13 V

Key: (B)

Sol: Given following circuit

( )Th sub 1V 3V, V 10V; V : 10V 10V= = − − +

1V 0V

GV

1V 0V

GV

|EC-2020|


During sampling GS THV V (MOSFET must be ON)

( )

( )

max max

max

s 1

G s Th

Th

G

G

V V 10Vi.e V V V

and V 3V

V 10 3

V 13V ... 1

= = −

=

−

During hold Gs ThV V (MOSFET is in OFF state)

( )

( )

( )

min min

min

s 1

G S Th

Th

G

G

V V 10Vi.e V V V

& V 3V

V 10 3

V 7V ... 2

= = − −

=

− −

−

From (1) & (2) we can say option (B) is correct

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