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EC-GATE-2014 PAPER-01| www.gateforum.com

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1

Q. No. 1 – 5 Carry One Mark Each

1. Choose the most appropriate phrase from the options given below to complete the following

sentence.

The aircraft_______ take off as soon as its flight plan was filed.

(A) is allowed to (B) will be allowed to

(C) was allowed to (D) has been allowed to

Answer: (C)

2. Read the statements:

All women are entrepreneurs.

Some women are doctors

Which of the following conclusions can be logically inferred from the above statements?

(A) All women are doctors (B) All doctors are entrepreneurs

(C) All entrepreneurs are women (D) Some entrepreneurs are doctors

Answer: (D)

3. Choose the most appropriate word from the options given below to complete the following

sentence.

Many ancient cultures attributed disease to supernatural causes. However, modern science

has largely helped _________ such notions.

(A) impel (B) dispel (C) propel (D) repel

Answer: (B)

4. The statistics of runs scored in a series by four batsmen are provided in the following table,

Who is the most consistent batsman of these four?

Batsman Average Standard deviation

K 31.2 5.21

L 46.0 6.35

M 54.4 6.22

N 17.9 5.90

(A) K (B) L (C) M (D) N

Answer: (A)

Exp: If the standard deviation is less, there will be less deviation or batsman is more consistent

5. What is the next number in the series?

12 35 81 173 357 ____

Answer: 725



2

Exp:

⇒357368725


6. Find the odd one from the following group:

W,E,K,O I,Q,W,A F,N,T,X N,V,B,D

(A) W,E,K,O (B) I,Q,W,A (B) F,N,T,X (D) N,V,B,D

Answer: (D)

Exp:

Difference of position: D

7. For submitting tax returns, all resident males with annual income below Rs 10 lakh should fill

up Form P and all resident females with income below Rs 8 lakh should fill up Form All

people with incomes above Rs 10 lakh should fill up Form R, except non residents with

income above Rs 15 lakhs, who should fill up Form S. All others should fill Form T. An

example of a person who should fill Form T is

(A) a resident male with annual income Rs 9 lakh

(B) a resident female with annual income Rs 9 lakh

(C) a non-resident male with annual income Rs 16 lakh

(D) a non-resident female with annual income Rs 16 lakh

Answer: (B)

Exp: Resident female in between 8 to 10 lakhs haven’t been mentioned.

8. A train that is 280 metres long, travelling at a uniform speed, crosses a platform in 60 seconds

and passes a man standing on the platform in 20 seconds. What is the length of the platform

in metres?

Answer: 560

Exp: For a train to cross a person, it takes 20 seconds for its 280m.

So, for second 60 seconds. Total distance travelled should be 840. Including 280 train length

so length of plates =840-280=560

12 35 81 173 357 ________

23 46 92 184 368

difference

W E K O

8 6 4

1 Q W A

8 6 4

F N T X

8 6 4

N V B D

8 6 2



3

9. The exports and imports (in crores of Rs.) of a country from 2000 to 2007 are given in the

following bar chart. If the trade deficit is defined as excess of imports over exports, in which

year is the trade deficit 1/5th of the exports?

(A) 2005 (B) 2004 (C) 2007 (D) 2006

Answer: (D)

Exp: imports exports 10 1

2004,exports 70 7

−= =

26 22005,

76 7

20 12006,

100 5

10 12007,

100 11

=

=

=

10. You are given three coins: one has heads on both faces, the second has tails on both faces,

and the third has a head on one face and a tail on the other. You choose a coin at random and

toss it, and it comes up heads. The probability that the other face is tails is

(A) 1/4 (B) 1/3 (C) 1/2 (D) 2/3

Answer: (B)

120

110

100

90

80

70

60

50

40

30

20

10

02000 2001 2002 2003 2004 2005 2006 2007

Exports Im ports



4


1. For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT

ALWAYS hold?

(A) (MT)

T = M (B) (cM

T)

T = c(M)

T

(C) (M + N)T = M

T + N

T (D) MN = NM

Answer: (D)

Exp: Matrix multiplication is not commutative in general.

2. In a housing society, half of the families have a single child per family, while the remaining

half have two children per family. The probability that a child picked at random, has a sibling

is _____

Answer: 0.667

Exp: Let 1E = one children family

2E = two children family and

A = picking a child then by Baye’s theorem, required probability is

( )2

1.x

2E 2P 0.667A 1 x 1 3

. .x2 2 2

= = =+

(Here ‘x’ is number of families)

3. C is a closed path in the z-plane given by z 3.= The value of the integral 2

C

z z 4j

z 2 j

− +→ + ∫

dz is

(A) ( )4 1 j2− π + (B) ( )4 3 j2π − (C) ( )4 3 j2− π + (D) ( )4 1 j2π −

Answer: (C)

Exp:

Z 2j= − is a singularity lies inside C : Z 3=

∴ By Cauchy’s integral formula,

[ ] [ ]

22

CZ 2 j

Z Z 4jdz 2 j. Z Z 4j

Z 2j

2 j 4 2 j 4 j 4 3 j2

=−

− + = π − + +

= π − + + = − π +

∫

4. A real (4 × 4) matrix A satisfies the equation A2 = I, where I is the (4 × 4) identity matrix.

The positive eigen value of A is __________.

Answer: 1

Exp:

2 1A I A A if−= ⇒ = ⇒ λ is on eigen value of A then 1

λ is also its eigen value. Since, we

require positive eigen value. 1∴λ = is the only possibility as no other positive number is self

inversed



5

5. Let X1, X2, and X3 be independent and identically distributed random variables with the

uniform distribution on [0, 1]. The probability PX1 is the largest is ________

Answer: 0.32-0.34

6. For maximum power transfer between two cascaded sections of an electrical network, the

relationship between the output impedance Z1 of the first section to the input impedance Z2 of

the second section is

(A) 2 lZ Z= (B) 2 lZ Z= − (C) 2 1Z Z∗= (D) 2 1Z Z∗= −

Answer: (C)

Exp: Two cascaded sections

Z1 = Output impedance of first section

Z2 = Input impedance of second section

For maximum power transfer, upto 1st section is

*

L 1

*

L 2 1

Z Z

Z Z Z

=

= ⇒

7. Consider the configuration shown in the figure which is a portion of a larger electrical

network

For R 1= Ω and currents i1 = 2A, i4 = -1A, i5 = -4A, which one of the following is TRUE?

(A) i6 = 5 A

(B) 3i 4A= −

(C) Data is sufficient to conclude that the supposed currents are impossible

(D) Data is insufficient to identify the current 2 3 6i , i , and i

Answer: (A)

1Z LZ 2ZSection

1

Section

2

5i

2i

4i

1i

6i

3i

R R

R



6

Exp: Given 1

4

5

i 2A

i 1A

i 4A

== −= −

KCL at node A, 1 4 2

2

i i i

i 2 1 1A

+ =⇒ = − =

1. KCL at node B, 2 5 3

3

i i i

i 1 4 3A

+ =⇒ = − =−

KCL at node C,

( )3 6 1

6

i i i

i 2 3 5A

+ =

⇒ = − − =

8. When the optical power incident on a photodiode is 10 Wµ and the responsivity is 0.8A / W,

the photocurrent generated ( )in Aµ is ________.

Answer: 8

Exp: ( ) p

0

IResponsivity R

P=

p

6

8

I0.8

10 10

I 8 A

−=×

⇒ = µ

9. In the figure, assume that the forward voltage drops of the PN diode D1 and Schottky diode

D2 are 0.7 V and 0.3 V, respectively. If ON denotes conducting state of the diode and OFF

denotes non-conducting state of the diode, then in the circuit,

(A) both D1 and D2 are ON (B) D1 is ON and D2 is OFF

(C) both D1 and D2 are OFF (D) D1 is OFF and D2 is ON

Answer: (D)

Exp: Assume both the diode ON.

Then circuit will be as per figure (2)

( )

2

1 2

D

D D

1 2

10 0.7I 9.3mA

1k

0.7 0.3I 20mA

20

Now, I I I

10.7 mA Not possible

D is OFF and hense D ON

−∴ = =

−= =

= −

= −

∴ −

1kΩ 20Ω

10Ω1

D 2D

1K

10V

Ω20

1DI

2DI

I

0.7V 0.3V

1K

10V

Ω20

1D

2D

( )Figure 1

5i

1 Ω

1 Ω

1 Ω

4i

3i

2i

6ii1A

B

C



7

10. If fixed positive charges are present in the gate oxide of an n-channel enhancement type

MOSFET, it will lead to

(A) a decrease in the threshold voltage (B) channel length modulation

(C) an increase in substrate leakage current (D) an increase in accumulation capacitance

Answer: (A)

11. A good current buffer has

(A) low input impedance and low output impedance

(B) low input impedance and high output impedance

(C) high input impedance and low output impedance

(D) high input impedance and high output impedance

Answer: (B)

Exp: i

Ideal current Buffer has Z 0=

0

Z = ∞

12. In the ac equivalent circuit shown in the figure, if in

i is the input current and RF is very large,

the type of feedback is

(A) voltage-voltage feedback (B) voltage-current feedback

(C) current-voltage feedback (D) current-current feedback

Answer: (B)

Exp: Output sample is voltage and is added at the input or current

∴ It is voltage – shunt negative feedback i.e, voltage-current negative feedback

13. In the low-pass filter shown in the figure, for a cut-off frequency of 5kHz, the value of R2

( )in kΩ is ____________.

2R

C

10nF1kΩ

1R o

Vi

V −+

DR

DR

2M

1M

FR

small signal

input i in

outυ



8

Answer: 3.18

Exp: f 5KHz=

( )2

2 3 9

1Cut off frequency LPF 5KHz

R C

1R 3.18k

2 5 10 10 10−

= =2π

⇒ = = Ωπ× × × ×

14. In the following circuit employing pass transistor logic, all NMOS transistors are identical

with a threshold voltage of 1 V. Ignoring the body-effect, the output voltages at P, Q and R

are,

(A) 4 V, 3 V, 2 V (B) 5 V, 5 V, 5 V

(C) 4 V, 4 V, 4 V (D) 5 V, 4 V, 3 V

Answer: (C)

Exp: Assume al NMOS are in saturation

( )

( ) ( )( ) ( )

( )

( ) ( )1

1

DS GS T

1

p p

p p

2

D GS T

2

D p

V V V

For m

5 V 5 V 1

5 V 4 V Sat

I k V V

I K 4 V ........ 1

∴ ≥ −

− ≥ − −

− > − ⇒

∴ = −

= −

( )( ) ( )

( ) ( )

1

2

1 2

2

2

D Q

2

D Q

D D

2 2

p Q

p Q p Q

p Q

For m ,

I K 5 V 1

I K 4 V ...... 2

I I

4 V 4 V

V V & V V 8

V V 4V

= − −

= −

∴ =

− = −

⇒ = + =

⇒ = =

( )

( ) ( )

3

2 3

3

2

D R

D D

2 2

Q R

R Q

p Q R

For m ,

I K 5 V 1

I I

4 V 4 V

V V 4V

V V V 4V

= − −

∴ =

− = −

⇒ = =

∴ = = =

5V 5V 5V

RQP

5V

5V

1M

2M

3M

P

Q

R

5V

5V

5V



9

15. The Boolean expression ( ) ( ) ( )X Y X Y X Y X+ + + + + simplifies to

(A) X (B) Y (C) XY (D) X+Y

Answer: (A)

Exp: Given Boolean Expression is ( )( )X Y X Y XY X+ + + +

As per the transposition theorem

( ) ( )( )( ) ( )

( )( ) ( )( )

( )

A BC A B A C

so, X Y X Y X YY X 0

X Y X Y XY X X XY .X

X X Y .X X XX. Y.X X 0 Y.X

Applyabsorption theorem X 1 Y X.1 X

+ = + +

+ + = + = +

+ + + + = +

= + + = + + = + +

= + = =

16. Five JK flip-flops are cascaded to form the circuit shown in Figure. Clock pulses at a

frequency of 1 MHz are applied as shown. The frequency (in kHz) of the waveform at Q3 is

__________ .

Answer: 62.5

Exp: Given circuit is a Ripple (Asynchrnous) counter. In Ripple counter, o/p frequency of each

flip-flop is half of the input frequency if their all the states are used otherwise o/p frequency

of the counter is input frequency

modulus of the counter=

So, the frequency at 3

6

z

input frequencyQ

16

1 10H 62.5kHz

16

=

×= =

17. A discrete-time signal [ ] ( )2x n sin n ,n beingan integer,is= π

(A) periodic with period π . (B) periodic with period 2π .

(C) periodic with period / 2π . (D) not periodic

Answer: (D)

1

1

J4 Q4clk>

K4

1

1

J3 Q3

K2

J2 Q2clk>

K2

1

1

1

1

J1 Q1clk>

K1clk>

J0

K0

1

1

clk>

clock



10

Exp: Assume [ ]x n to be periodic, (with period N)

[ ] [ ]

( ) ( )( )2 2

x n x n N

sin n sin n N

⇒ = +

⇒ π = π +

Every frigonometric function repeate after 2π interval.

( ) ( )2 2 2

2

sin n 2 k sin h N

2k2 k N N

⇒ π + π = π + π

⇒ π = π ⇒ = π

Since ‘k’ is any integer, there is no possible value of ‘k’ for which ‘N’ can be an integer, thus

non-periodic.

18. Consider two real valued signals, x(t) band-limited to [ ]500 Hz, 500Hz− and ( )y t band-

limited to [ ]1kHz, 1kHz− . For ( ) ( ) ( )z t x t . y t ,= the Nyquist sampling frequency (in kHz) is

__________

Answer: 3

Exp: ( )x t is band limited to [ ]500Hz, 500Hz−

( ) [ ]y t is band limited to 1000Hz, 1000Hz−

( ) ( ) ( )z t x t .y t=

Multiplication in time domain results convolution in frequency domain.

The range of convolution in frequency domain is [ ]1500Hz, 1500Hz−

So maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz

19. A continuous, linear time-invariant filter has an impulse response h(t) described by

( ) 3 for 0 t 3

0 otherwiseh t ≤ ≤=

When a constant input of value 5 is applied to this filter, the steady state output is _______.

Answer: 45

Exp:

( ) ( ) ( )y t x t * h t=

( )x t =

( )h t =

( ) ( )3

0

y t 3.5.d 45 steady state output= τ =∫

( )x t( )y t( )h t

5

t

3

3t



11

20. The forward path transfer function of a unity negative feedback system is given by

( ) ( ) ( )K

G ss 2 s 1

=+ −

The value of K which will place both the poles of the closed-loop system at the same

location, is ______.

Answer: 2.25

Exp: Given ( ) ( )( )( )

KG s

s 2 s 1

H s 1

=+ −

=

Characteristic equation: ( ) ( )

( )( )

1 G s H s 0

K1 0

s 2 s 1

+ =

+ =+ −

The poles are 1,2

9s 1 4K

4= − ± −

If 9

K 0,4

− = then both poles of the closed loop system at the same location.

So, 9

K 2.254

= ⇒

21. Consider the feedback system shown in the figure. The Nyquist plot of G(s) is also shown.

Which one of the following conclusions is correct?

(A) G(s) is an all-pass filter

(B) G(s) is a strictly proper transfer function

(C) G(s) is a stable and minimum-phase transfer function

(D) The closed-loop system is unstable for sufficiently large and positive k

Answer: ( D)

Exp: For larger values of K, it will encircle the critical point (-1+j0), which makes closed-loop

system unstable.

( )ImG jω

1+ ( )ReG jω1−

+

−k ( )G s



12

22. In a code-division multiple access (CDMA) system with N = 8 chips, the maximum number

of users who can be assigned mutually orthogonal signature sequences is ________

Answer: 7.99 to 8.01

Exp: Spreading factor(SF)=chip rate

symbol rate

This if a single symbol is represented by a code of 8 chips

Chip rate =80×symbol rate

S.F (Spreading Factor)8 symbol rate

8symbol rate

×= =

Spread factor (or) process gain and determine to a certain extent the upper limit of the total

number of uses supported simultaneously by a station.

23. The capacity of a Binary Symmetric Channel (BSC) with cross-over probability 0.5 is

________

Answer: 0

Exp: Capacity of channel is 1-H(p)

H(p) is entropy function

With cross over probability of 0.5

( ) 2 2

1 1 1 1H p log log 1

2 0.5 2 0.5

Capacity 1 1 0

= + =

⇒ = − =

24. A two-port network has sattering parameters given by [ ] 11 12

21 22

S SS

S S

=

. If the port-2 of the

two-port is short circuited, the 11

S parameter for the resultant one-port network is

( ) 11 11 22 12 21

22

s s s s sA

1 s

− ++

( ) 11 11 22 12 21

22

s s s s sB

1 s

− −+

( ) 11 11 22 12 21

22

s s s s sC

1 s

− +−

( ) 11 11 22 12 21

22

s s s s sD

1 s

− +−

Answer:(B)

Exp:

1 11 1 12 2

2 21 1 22 2

b s a s a

b s a s a

= += +

1 11 12 1

2 21 22 2

b s s a

b s s a

=

;

2

11

1 a 0

bs

a=

=

By verification Answer B satisfies.

Two port

Network

1a2a

1b2b



13

25. The force on a point charge +q kept at a distance d from the surface of an infinite grounded

metal plate in a medium of permittivity ∈ is

(A) 0 (B) 2

2

qaway from the plate

16 dπ ∈

(C) 2

2

qtowards the plate

16 dπ ∈ (D)

2

2

qtowards the plate

4 dπ ∈

Answer:(C)

Exp:

( )

1 2

2

2 2

2 2

Q Q1F

4 R

1 9 9F

4 16 d2d

=π∈

= =π∈ π∈

Since the charges are opposite polarity

the force between them is attractive.

Q.No. 26 – 55 Carry Two Marks Each

26. The Taylor series expansion of 3 sin x + 2 cos x is

( )3

2 xA 2 3x x .......

2+ − − + ( )

32 x

B 2 3x x .......2

− + − +

( )3

2 xC 2 3x x .......

2+ + + + ( )

32 x

D 2 3x x .......2

− − + +

Answer: (A)

Exp:

3 2x x3sin x 2cos x 3 x ... 2 1 ...

3! 2!

+ = − + + − +

3

2 x2 3x x ...

2= + − − +

27. For a Function g(t), it is given that ( ) 2j t 2g t e dt e+∞ − ω − ω

−∞= ω∫ for any real value ω . If

( ) ( ) ( )t

y t g d , then y t dt+∞

−∞ −∞= τ τ∫ ∫ is

(A)0 (B)-j (C) -j

2 (D)

j

2

Answer: (B)

Exp: Given

( ) ( )( )

( )

2jwt 2wg t .e dt .e let G j

g t dt 0

∞− −

−∞

∞

−∞

= ω ω

⇒ =

∫

∫

d

d

q−

q+

metal plate



14

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

( ) ( ) ( )

( ) ( )

( ) ( ) ( )2

t

j t

2w

y t g z .dz y t g t * u t u t in unit step function

Y j G j .U j

Y j y t .e dt

1Y j0 y t dt .e 0

j

1j

j

−∞

∞− ω

−∞

∞−

−∞

= ⇒ =

⇒ ω = ω ω

ω =

⇒ = = ω + πδ ω ω = ω

= = −

∫

∫

∫

28. The volume under the surface z(x, y) = x + y and above the triangle in the x-y plane defined

by 0 y x and 0 x 12≤ ≤ ≤ ≤ is___________.

Answer: 864

Exp:

( ) ( )12 x

R x 0 y 0

Volume Z x, y dydx x y dydx= =

= = +∫∫ ∫ ∫

x 1212 122 32

x 0 00 0

y 3 3 xxy .dx x dx 864

2 2 2 3=

= + = = =

∫ ∫

29. Consider the matrix:

6

0 0 0 0 0 1

0 0 0 0 1 0

0 0 0 1 0 0J

0 0 1 0 0 0

0 1 0 0 0 0

1 0 0 0 0 0

=

Which is obtained by reversing the order of the columns of the identity matrix 6

I .

Let 6 6

P I J ,= + α where α is a non-negative real number. The value of α for which det(P) =

0 is ___________.

Answer: 1

Exp: ( ) 2 2Consider, i Let P I J= + α

1 0 0 1 1

0 1 1 0 1

α = + α = α

( )

2

4 4

P 1

1 0 0

ii Let P I J

⇒ = − α

α 0 1 α 0 = + α = 0 α 1 0 α 0 0 1



15

( ) ( )

( ) ( ) ( ) ( )

( )

22 2 2

6 6

32

1 0 0 1

P 1 1 0 0 1

0 0 1 0 0

1 1 1

S im ilarly , if P I J then w e get

P 1

P 0 1, 1

is non negative

1

α α= α − α α

α

= − α − α α − α = − α

= + α

= − α

∴ = ⇒ α = −

α

∴ α =

∵

30. A Y-network has resistances of 10Ω each in two of its arms, while the third arm has a

resistance of 11Ω in the equivalent network,∆ − the lowest value ( )inΩ among the three

resistances is ______________.

Answer: 29.09Ω

Exp:

X 29.09

y 32

z 32

= Ω= Ω= Ω

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

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

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

10 10 10 11 10 11X

11

10 10 10 11 10 11y

10

10 10 10 11 10 11z

10

+ += Ω

+ += Ω

+ += Ω

i.e, lowest value among three resistances is 29.09Ω

31. A 230 V rms source supplies power to two loads connected in parallel. The first load draws

10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power

factor. The complex power delivered by the source is

(A) (18 + j 1.5) kVA (B) (18 - j 1.5) kVA

(C) (20 + j 1.5) kVA (D) (20 - j 1.5) kVA

Star Connection

10Ω

10Ω11Ω

X

Y

Z10Ω

10Ω 11Ω

Delta Connection



16

Answer: (B)

Exp:

Load 1:

I

P 10kw

cos 0.8 S P jQ 10 j7.5 KVA

Q P tan 7.5KVAR

= φ = = − = −= φ =

Load 2: S 10 KVA=

Qcos 0.8 sin

S

Pcos

S

P0.8 P 8kw Q 6KVAR

10

φ = φ =

φ =

= → = =

IS P jQ 8 j6= + = +

Complex power delivered by the source is I IIS S 18 j1.5 KVA+ = −

32. A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms)

value of x is_______.

Answer: 0.408

x

1

0

T / 2 T / 2

t

LoadI

LoadII

+

−

230V



17

Exp:

( )( )T

2

rms

0

1x x t dt

T= ∫

( )

( )2T 2 T

2

T02

T3 2

2

0

3

rms 3

2 Tt 0 t2Tx t

T0 t T2

21.t .dt 0 .dt

T T

1 4 t.

T T 3

4 T 1x . 0.408

3T 8 6

≤ ≤= ≤ ≤

= +

=

= ⇒ ⇒

∫ ∫

33. In the circuit shown in the figure, the value of capacitor C(in mF) needed to have critically

damped response i(t) is____________.

Answer: 10mF

Exp: By KVL,

( ) ( ) ( ) ( )di t 1v t Ri t L. i t dt

dt C= + + ∫

Differentiate with respect to time,

( ) ( ) ( )

( ) ( ) ( )2

2

2

2

1,2

2

1,2

R.di t di ti i tR0 . 0

dt L dt LC

d i t di t i tR. 0

dt L dt LC

R R 4

L L LCD

2

R R 1D

2L 2L LC

= + + =

+ + =

− ± − =

− = ± −

For critically damped response,

2

2

R 1 4LC F

2L LC R

= ⇒ =

Given, L=4H; R= 40Ω

( )2

4 4C 10mF

40

×= ⇒

40Ω 4 H C

( )i t

+ −O

V

X

1

( )0,0 T2

Tt



18

34. A BJT is biased in forward active mode, Assume BE

V 0.7V,kT / q 25mV= = and reverse

saturation current 13

SI 10−= A. The transconductance of the BJT (in mA/V) is ________.

Answer: 5.785

Exp: 13

BE s

KTV 0.7V, 25mV, I 10

q

−= = =

BE T

Cm

T

V /V

C S

13 0.7/25mV

Cm

T

ITransconductance, g

V

I I e 1

10 e 1 144.625mA

I 144.625mAg 5.785A / V

V 25mV

−

=

= −

= − =

∴ = = =

35. The doping concentrations on the p-side and n-side of a silicon diode are 16 31 10 cm−× and 17 31 10 cm−× , respectively. A forward bias of 0.3 V is applied to the diode. At T = 300K, the

intrinsic carrier concentration of silicon 10 3

in 1.5 10 cm−= × and kT

26mV.q

= The electron

concentration at the edge of the depletion region on the p-side is

(A) 9 32.3 10 cm−× (B) 16 31 10 cm−× (C) 17 31 10 cm−× (D) 6 32.25 10 cm−×

Answer:(A)

Exp: bi T

2V /Vi

A

nElectron concentration, n e

N

( )210

0.3/26mV

16

9 3

1.5 10e

1 10

2.3 10 / cm

×=

×= ×

36. A depletion type N-channel MOSFET is biased in its linear region for use as a voltage

controlled resistor. Assume threshold voltage

TH GS DSV 0.5V,V 2.0V, V 5V,W / L 100,= = = = 8 2

OXC 10 F / cm−= and 2

n 800cm / V sµ = − .

The value of the resistance of the voltage controlled resistor ( )in isΩ ________.

Answer:500

Exp: x

8

T GS DSWGiven V 0.5V; V 2V; V 5V; 100; C 10 f / cm

L−

θ= − = = = =

( )

( )

2

n

2

D n 0x GS T DS DS

1 1

2Dds n 0x GS T DS DS

DS DS

800cm / v s

1 WI C 2 V V V V

2 L

I 1 Wr C 2 V V V V

V V 2 L

− −

µ = −

= µ − −

∂ ∂ = µ − − ∂ ∂



19

( )1

n 0x GS T n 0x DS

W WC V V C V

L L

− = µ − − µ

( )

( )

ds

n 0x GS T Ds

8

1r

WC V V V

L

1500

800 10 100 2 0.5 5−

⇒ =µ − −

= = Ω× × + −

37. In the voltage regulator circuit shown in the figure, the op-amp is ideal. The BJT has

BEV 0.7V= and 100,β = and the zener voltage is 4.7V. For a regulated output of 9 V, the

value of ( )R inΩ is ______ .

Answer:1093

Exp: BE Z 0

Given V 0.7V, 100, V 4.7V, V 9V= β = = =

( )

R

R z

RV 9

R 1k

R4.7 9 V V

R 1k

R 1093

= ×+

= × =+

= Ω

∵

38. In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero

input offset voltage. The output voltage out

V is

(A) ( )2 1 2I R R− +

(B) 2 2

I R

(C) 1 2

I R

(D) ( )1 1 2I R R− +

2R

1R

1l

2l

−

+ outV

IV 12V=

0V 9 V=

1kΩ

zV 4.7 V=

+

−

1kΩ

R

=i

V 12V 9V

1K

RV

R

+−

zV



20

Answer: (C)

Exp: i

Given, Z = ∞

( )

( )

L

0

0

i

2 1 2 1

1 21

1 2

A

V 0

V R / /R I

R RI ...... 1

R R

= ∞

=

=

=+

KCL at inverting node

( )2 02i

1 2

02

2 1 2

0 1 2 2 11

2 1 2 1 2

0 1 2

V VV0 Z

R R

V 1 1V

R R R

V R R R RI

R R R R R

V I R

−+ = ∴ = ∞

= +

+= +

⇒ =

39. For the amplifier shown in the figure, the BJT parameters are BE

V 0.7V, 200,= β = and

thermal voltage T

V 25mV.= The voltage gain ( )0 iv / v of the amplifier is _______.

Answer: -237.76

Exp: BE T

V 0.7V, 200, V 25mV= β = =

DC Analysis:

B

E

E

11kV 12 3V

11k 33k

V 3 0.7 2.3V

2.3I 2.277mA

10 1k

= × =+

= − =

= =+

1R

2R

0V2V

1V

+

−1I

CCV 12V= +

CR

5kΩ1R

33kΩ1 Fµ

iv

2R

11kΩ

1 Fµo

v

SR

10Ω

1ER

1k Ω

EC

1mF



21

( )( ) ( )

B

C

e

0 CV

i e s

V

I 11.34 A

I 2.26mA

25mVr 10.98

2.277mA

V R 200 5kA

V r 1 R 200 10.98 201 10

A 237.76

= µ=

= = Ω

−β − ×= = =

β + + β × +

= −

40. The output F in the digital logic circuit shown in the figure is

( )A F XYZ XYZ= + ( )B F XYZ XYZ= +

( )C F XYZ XYZ= + ( )D F XYZ XYZ= +

Answer: (A)

Exp:

Assume dummy variable K as a output of XOR gate K X Y XY XY= ⊕ = +

( )( )

( )

F K. K Z

KZ K.Z

K. KZ K.K.Z

0 K.Z K. K 0 and K.K K

=

= +

= +

= + = =

∵

Put the value of K in above expression

( )F XY XY Z

XYZ XYZ

= +

= +

XOR

X

Y AND

F

Z

XNOR

X

Y

Z

XOR

XNOR

K

F



22

41. Consider the Boolean function, ( )F w,x,y,z wy xy wxyz wxy xz xyz.= + + + + + which one

of the following is the complete set of essential prime implicants?

(A) w,y, xz, x z (B) w,y,xz (C) y,x yz (D) y,xz,xz

Answer: (D)

Exp: Given Boolean Function is

( )F w,x, y,z wy xy wxyz wxy xz xyz= + + + + +

By using K-map

So, the essential prime implicants (EPI ) are y, xz, xz

42. The digital logic shown in the figure satisfies the given state diagram when Q1 is connected

to input A of the XOR gate.

Suppose the XOR gate is replaced by an XNOR gate. Which one of the following options

preserves the state diagram?

(A) Input A is connected to Q2

(B) Input A is connected to Q2

(C) Input A is connected to Q1 and S is complemented

(D) Input A is connected to Q1

Answer: (D)

D1 Q1

Q1CLK

>

A

S

D2 Q2

Q2>

S 0=

S 1=00 01

10 11

S 1=S 0=

S 0=

S 1=

S 0=

S 1=

00 01 11 10

00 1 1 1

01 1 1 1

11 1 1 1

10 1 1 1

x z

xz

y

wxyz



23

Exp: The input of D2 flip-flop is

( )2 1 1 1D Q s Q s A Q= + =∵

The alternate expression for EX-NOR gate is A B A B A B= ⊕ = ⊕ = ⊕

So, if the Ex-OR gate is substituted by Ex-NOR gate then input A should be connected to 1Q

( )2 1 1 1 1 1

i 1

D Q S Q S Q S Q .S A Q

Q S Q .S

= + = + =

= +

∵

43. Lex [ ] ( ) ( )n n

1 1x n u n u n 1

9 3

= − − − − − . The Region of Convergence (ROC) of the z-

transform of x[n]

(A) 1

is z9

> (B) 1

is z3

< (C) 1 1

is z3 9

> > (D) does not exist.

Answer: (C)

Exp: [ ] [ ] [ ]n n

1 1Given x n u n u n 1

9 3

− − = − − −

[ ]h

oc

1 1for u n R in z

9 9

− >

(Right sided sequence, oc

R in exterior of circle of radius 19

)

Thus overall oc

1 1R in z

9 3< <

44. Consider a discrete time periodic signal x[n] = n

sins

π

. Let k

a be the complex Fourier

series coefficients of x[n]. The coefficients ka are non-zero when k = Bm 1,± where m is

any integer. The value of B is_________.

Answer: 10

Exp: [ ] nGiven x n sin ; N 10

5

π = =

⇒ Fourier series co-efficients are also periodic with period N 10=

[ ]2 2

j n i n10 10

1 1x n e e

2j 2 j

π π−−

=

1 1 1 1 10 9

1 1 1 1

1 11 1

1 1 1a ; a a a a

2j 2 j 2 j

a a 10 a a 20or

a a 20a a 10

k 10m 1 or k 10.m 1 B 10

− − − +

− −− −

− −= = ⇒ = = =

= + = + = += +

⇒ = + = − ⇒ =



24

45. A system is described by the following differential equation, where u(t) is the input to the

system and y(t) is the output of the system.

( ) ( ) ( ).

y t 5y t u t+ =

When y(0) = 1 and u(t) is a unit step function, y(t) is

(A) 5t0.2 0.8e−+ (B) 5t0.2 0.2e−− (C) 5t0.8 0.2e−+ (D) 5t0.8 0.8e−−

Answer: (A)

Exp: Given ( ) ( ) ( ) ( ) ( )y t 5y t u t and y 0 1; u t isa unit stepfunction.+ = =

Apply Laplace transform to the given differential equation.

( ) ( ) ( )

( )[ ] ( ) ( ) ( ) ( )

( ) ( )

( ) ( )( )

( ) ( )

1S y s y 0 5y s

s

1 dy 1y s s 5 y 0 L s y s y 0 L u tss dt

11

sy ss 5

s 1 A By s

s s 5 s s 5

1 4A ; B5 5

1 4y s

5s 5 s 5

− + =

+ = + = − =

+=

+

+= ⇒ +

+ +

= =

= ++

Apply inverse Laplace transform,

( )

( )

5t

5t

1 4y t e

5 5

y t 0.2 0.8e

−

−

= +

= +

46. Consider the state space model of a system, as given below

[ ]

.

1

1 1.

2 2 2

.3 3

3

x1 1 0 x 0 x

x 0 1 0 x 4 u; y 1 1 1 x

0 0 2 x 0 xx

− = − + = −

The system is

(A) controllable and observable

(B) uncontrollable and observable

(C) uncontrollable and unobservable

(D) controllable and unobservable

Answer: (B)



25

Exp: From the given state model,

[ ]

2

c

c

c c

0

2

0

0

01 1 0

A 0 1 0 B 4 c 1 1 1

0 0 2 0

Controllable: Q c B AB A B

if Q 0 controllable

0 4 8

Q 4 4 4 Q 0

0 0 0

uncontrollable

C

Observable : Q CA

CA

If Q 0 observable

1 1 1

Q 1 0 2

1 1 4

− = − = = −

= =

≠ →

− = − ⇒ =

∴

=

≠ →

= − −

−

0Q 1

Observable

⇒ =

∴

The system is uncontrollable and observable

47. The phase margin in degrees of ( ) ( ) ( ) ( )10

G ss 0.1 s 1 s 10

=+ + + +

calculated using the

asymptotic Bode plot is_______.

Answer: 48

Exp:

( ) ( )( )( )

( )[ ]

( ) [ ][ ][ ]

10G s

s 0.1 s 1 s 10

10G s

s s0.1 1 1 s 1 .10

0.1 10

10G s

1 10s 1 s 1 0.1s

=+ + +

= + + +

=+ + +

By Approximation, ( ) [ ]10

G s10s 1

=+



26

Phase Margin = gc

1

180 GH

10 0.99180 tan

1

Phase Margin =95 .73

ω=ω

−

θ= +

× = −

°

gc

2

2

2

gc

101

100 1

99100

1

990.9949r / sc

1

ω = =ω +

= ω =ω

⇒ ω ⇒ ω =ω

Asymptotic approximation, Phase margin = 45 48φ− °

48. For the following feedback system ( ) ( ) ( )1

G ss 1 s 2

=+ + +

. The 2% settling time of the step

response is required to be less than 2 seconds.

Which one of the following compensators C(s) achieves this?

( ) 1A 3

s 5

+

( ) 0.03B 5 1

s

+ ( ) ( )C 2 s 4+ ( ) s 8

D 4s 3

+ +

Answer: (C)

Exp: By observing the options, if we place other options, characteristic equation will have 3rd

order

one, where we cannot describe the settling time.

( ) ( )If C s 2 s 4= + is considered

The characteristic equation, is

2

2

s 3s 2 2s 8 0

s 5s 10 0

+ + + + =

⇒ + + =

Standard character equation 2 2

n n

2

n n

s 2 s 0

10; 2.5

+ ξω +ω =

ω = ξω =

Given, 2% settling time, n

n

42 w 2

w< ⇒ ξ >

ξ

49. Let x be a real-valued random variable with E[X] and E[X2] denoting the mean values of X

and X2, respectively. The relation which always holds true is

( ) [ ]( )2 2A E X E X > ( ) [ ]( )22B E X E X ≥

( ) [ ]( )22C E X E X = ( ) [ ]( )22D E X E X >

Answer: (B)

Exp: ( ) ( ) ( ) 22V x E x E x 0 i.e., var iance cannot be negative= − ≥

( ) ( ) 22E x E x∴ ≥

r + ( )C s ( )G sy

−



27

50. Consider a random process ( ) ( )X t 2 sin 2 t ,= π + ϕ where the random phase ϕ is uniformly

distributed in the interval [ ]0,2π . The auto-correlation ( ) ( )1 2E X t X t

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

1 2 1 2

1 2 1 2

A cos 2 t t B sin 2 t t

C sin 2 t t D cos 2 t t

π + π −

π + π −

Answer: (D)

Exp: Given ( )X(t) 2 sin 2 t= π + φ

φ in uniformly distributed in the interval [ ]0,2π

[ ] ( )2

1 2 1 20

E x(t )x(t ) 2 sin(2 t ) 2 sin 2 t f ( )dπ

φ= π + θ π + θ θ θ∫

( ) ( )

2

1 20

2 2

1 2 1 20 0

12 sin 2 t sin 2 t . .d

21 1

sin(2 (t t ) 2 )d cos(2 (t t )d2 2

π

π π

= π + θ π + θ θπ

= π + + θ θ + π − θπ π

∫

∫ ∫

First integral will result into zero as we are integrating from 0 to 2 .π

Second integral result into 1 2cos 2 (t t )π −

[ ] ( )1 2 1 2E X(t )X(t ) cos 2 (t t⇒ = π −

51. Let ( )Q γ be the BER of a BPSK system over an AWGN channel with two-sided noise

power spectral density N0/2. The parameter γ is a function of bit energy and noise power

spectral density.

A system with tow independent and identical AWGN channels with noise power spectral

density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of

both the channels.

If the BER of this system is ( )Q b ,γ then the value of b is _____________.

Answer: 1.414

Exp: O

2E EBit error rate for BPSK Q . Q

NNO2

=

O

2EY

N⇒ =

0 /1 BPSKModulator

AWGNChannel1

AWGNChannel 2

BPSKDemodulator

0 /1+

1

2π

f ( )φ θ

0 2π θ



28

Function of bit energy and noise OSD

NP

2

Counterllation diagram of BPSK

Channel is WGN

A which implies noise sample as independent

1 1

1 2

1

1

1 2

1

1

O

1 1

1 1

O

Let 2x n n x n

where x 2x

n n n

2ENow Bit error rate Q

N

E is energy in x

N is PSD of h

+ + = +

=

= +

=

1E 4E= [as amplitudes are getting doubled]

1

O ON N= [independent and identical channel]

O O

4E 2EBit error rate Q Q 2 b 2 or 1.414

N N

⇒ = = ⇒ =

52. A fair coin is tossed repeatedly until a ‘Head’ appears for the first time. Let L be the number

of tosses to get this first ‘Head’. The entropy H(L) in bits is _________.

Answer: 2

Exp: In this problem random variable is L

2 2 2

L can be 1,2,..............

1P L 1

2

1P L 2

4

1P L 3

8

1 1 1 1 1 1 1 1 1H L log lgo log ......... 0 1. 2. 3. .........

1 1 12 4 8 2 4 82 4 8

= =

= =

= =

= + + + = + + + +

[ Arithmatic gemometric series summation]

2

1 .12 2 211 1

2 12

= + =− −

( )φ2 t

−a a ( )φ1 t

+

+

+x

+1

x n

+2

x n

noise in channel 1

+ +1 2

2x n n

noise in channel 2



29

53. In spherical coordinates, let ˆ ˆa .aθ φ denote until vectors along the ,θ φ directions.

( )100ˆE sin cos t r a V / m

rθ= θ ω − β and

( )0.265ˆH sin cos t r a A / m

rφ= θ ω − β

represent the electric and magnetic field components of the EM wave of large distances r

from a dipole antenna, in free space. The average power (W) crossing the hemispherical shell

located at r 1km,0 / 2is _______= ≤ θ ≤ π

Answer: 55.5

Exp:

J r100E sin e

r

− βθ = θ

( )

( )

( )( )

J r

Q

*

avg Qs

2 2

2s

2

avts

223

0 Q 0

0.265H sin e

r

1P E H .ds

2

100 0.2651sin r sin d d

2 r

1P 26.5 sin d d

2

213.25 sin d d 13.25. 23

P 55.5w

− β

θ

ππ

θ= =

= θ

=

= θ θ θ φ

= θ φ

= θ θ φ = π

=

∫

∫

∫

∫ ∫

54. For a parallel plate transmission line, let v be the speed of propagation and Z be the

characteristic impedance. Neglecting fringe effects, a reduction of the spacing between the

plates by a factor of two results in

(A) halving of v and no change in Z (B) no change in v and halving of Z

(C) no change in both v and Z (D) halving of both v and Z

Answer: (B)

Exp:

o

r

276 dZ log

r

= ∈

d → distance between the two plates

so, zo – changes, if the spacing between the plates changes.

1

VLC

= → independent of spacing between the plates



30

55. The input impedance of a 8

λ section of a lossless transmission line of characteristic

impedance 50Ω is found to be real when the other end is terminated by a load

( )LZ R jX .= + Ω if X is 30 ,Ω the value of R ( )inΩ is _________

Answer: 40

Exp:

( )o

L oin o

o L

L L Lin

L L L

Given,s

Z 50

Z JZZ Z

8 Z KZ

Z J50 Z J50 50 JZZ 50 50

50 JZ 50 JZ 50 JZ

λ=

= Ω

+λ= = +

+ + −= = × + + −

( )

( )

2 2

L L L

in 2 2

L

in

mg in

2 2

L

2 2

L

2 2 2

2 2 2 2 2

50Z 50Z J 50 ZZ 50

50 Z

Given , Z Real

So, I Z 0

50 Z 0

Z 50

R X 50

R 50 X 50 30

R 40

+ + − =

+

→

=

− =

=

+ =

= − = −= Ω



1


1. Choose the most appropriate word from the options given below to complete the following

sentence.

Communication and interpersonal skills are_____ important in their own ways.

(A) each (B) both (C) all (D) either

Answer: (B)

2. Which of the options given below best completes the following sentence?

She will feel much better if she ________________.

(A) will get some rest (B) gets some rest

(C) will be getting some rest (D) is getting some rest

Answer: (B)

3. Choose the most appropriate pair of words from the options given below to complete the

following sentence.

She could not _____ the thought of _________ the election to her bitter rival.

(A) bear, loosing (B) bare, loosing (C) bear, losing (D) bare, losing

Answer: (C)

4. A regular die has six sides with numbers 1 to 6 marked on its sides. If a very large number of

throws show the following frequencies of occurrence: 1 → 0.167; 2→ 0.167; 3→ 0.152; 4 →

0.166; 5→ 0.168; 6 → 0.180. We call this die

(A) irregular (B) biased (C) Gaussian (D) insufficient

Answer: (B)

Exp: For a very large number of throws, the frequency should be same for unbiased throw. As it

not same, then the die is baised.

5. Fill in the missing number in the series.

2 3 6 15 ___ 157.5 630

Answer: 45

Exp:

2nd number

is in increa sin g order as shown above1st number

2 3 6 15 45 157.5 630

1.5 2 2.5 3 3.5 4



2


6. Find the odd one in the following group

Q,W,Z,B B,H,K,M W,C,G,J M,S,V,X

(A) Q,W,Z,B (B) B,H,K,M (C) W,C,G,J (D) M,S,V,X

Answer: (C)

Exp:

7. Lights of four colors (red, blue, green, yellow) are hung on a ladder. On every step of the

ladder there are two lights. If one of the lights is red, the other light on that step will always

be blue. If one of the lights on a step is green, the other light on that step will always be

yellow. Which of the following statements is not necessarily correct?

(A) The number of red lights is equal to the number of blue lights

(B) The number of green lights is equal to the number of yellow lights

(C) The sum of the red and green lights is equal to the sum of the yellow and blue lights

(D) The sum of the red and blue lights is equal to the sum of the green and yellow lights

Answer: (D)

8. The sum of eight consecutive odd numbers is 656. The average of four consecutive even

numbers is 87. What is the sum of the smallest odd number and second largest even number?

Answer: 163

Exp: Eight consecutive odd number =656

a-6, a-1, a-2, a ,a+2 ,a+4, a+6

a+8=656

a=81

Smallest m=75 … (1)

Average consecutive even numbers

a 2 a a 2 a 487

4

a 86

− + + + + +⇒ =

⇒ =

Second largest number =88

1+2=163

9. The total exports and revenues from the exports of a country are given in the two charts

shown below. The pie chart for exports shows the quantity of each item exported as a

percentage of the total quantity of exports. The pie chart for the revenues shows the

percentage of the total revenue generated through export of each item. The total quantity of

exports of all the items is 500 thousand tonnes and the total revenues are 250 crore rupees.

Which item among the following has generated the maximum revenue per kg?

a W Z B

6 3 2

17 23 26 2B H K N

6 3 2

M S V X

6 3 2

W C G J

6 4 3



3

(A) Item 2 (B) Item 3 (C) Item 6 (D) Item 5

Answer: (D)

Exp: Item:2 Item:3

7

3

4 3

20250 10

10020

500 10100

0.5 10 5 10 1 Item 2

× ×

× ×

× = × =

7

3

23 250 10

19 500 10

1.2 Item3

× ×× ×

=

Item: 6 Item:5

19

1.18 Item 616

= = 20 5

1.6 1.6 Item 512 3

= = ⇒ =

10. It takes 30 minutes to empty a half-full tank by draining it at a constant rate. It is decided to

simultaneously pump water into the half-full tank while draining it. What is the rate at which

water has to be pumped in so that it gets fully filled in 10 minutes?

(A) 4 times the draining rate (B) 3 times the draining rate

(C) 2.5 times the draining rate (D) 2 times the draining rate

Answer: (A)

Exp: halfV 30(s) drawing rate s= =

Total volume =60 S tank

1

1

1

(s )(10) (s)10 30s

s (s) s 3s

s1 4s

s 4drawing rate

− =− =

==

Item 6

16%

Item 1

11%

Item5

12%

Item4

22%

Item3

19%

Item2

20%

Exports

Item 6

19%

Item 1

12%

Item 5

20%

Item3

23%

Item 2

20%

Revenues

Item 4

6%



4


1. The determinant of matrix A is 5 and the determinant of matrix B is 40. The determinant of

matrix AB is ________.

Answer: 200

Exp:

( ) ( )AB A . B 5 . 40 200= = =

2. Let X be a random variable which is uniformly chosen from the set of positive odd numbers

less than 100. The expectation E[X]is __________.

Answer:50

Exp:

( )X 1,3,5,....,99 n 50 number of observations= ⇒ =

( ) [ ] ( )n

2

i

i 1

1 1 1E x x 1 3 5 .... 99 50 50

n 50 50=

∴ = = + + + + = =∑

3. For 0 t ,≤ < ∞ the maximum value of the function ( ) t 2 tf t e 2e− −= − occurs at

(A) t = loge4 (B) t = loge2 (C) t = 0 (D) t = loge8

Answer: (A)

Exp: ( )' t 2tf t e 4e 0− −= − + =

( )

t t t 4

e

'' 4

e

1e 4e 1 e t log

4

and f t 0 at t log

− − − ⇒ − ⇒ = ⇒ =

< =

4. The value of

x

x

1lim 1

x→∞

+ is

(A) ln2 (B) 1.0 (C) e (D) ∞

Answer: (C)

Exp:

( )

x

x

1lim 1 e standard limit

x→∞

+ =

5. If the characteristic equation of the differential equation

2

2

d y dy2 y 0

dx dx+ α + =

has two equal roots, then the values of α are

( )A 1± ( )B 0,0 ( )C j± ( )D 1/ 2±

Answer: (A)

Exp:

2For equal roots, Discriminant B 4AC 0− =

2

4 4 0

1

⇒ α − =⇒ α = ±



5

6. Norton’s theorem states that a complex network connected to a load can be replaced with an

equivalent impedance

(A) in series with a current source (B) in parallel with a voltage source

(C) in series with a voltage source (D) in parallel with a current source

Answer: (D)

Exp: Norton’s theorem

7. In the figure shown, the ideal switch has been open for a long time. If it is closed at t=0, then

the magnitude of the current (in mA) through the 4kΩ resistor at t = 0+ is _______.

Answer: 1.2 mA

Exp: For t = o+

( )

( )

10i o 1.11mA

9K

i o 1.2mA

+ = ⇒

+

8. A silicon bar is doped with donor impurities ND = 2.25 x 1015

atoms / cm3. Given the intrinsic

carrier concentration of silicon at T = 300 K is ni = 1.5 x 1010

cm-3

. Assuming complete

impurity ionization, the equilibrium electron and hole concentrations are

(A) n0 = 1.5 x 1016

cm-3

, p0 = 1.5 x 105 cm

-3

(B) n0 = 1.5 x 1010

cm-3

, p0= 1.5 x 1015

cm-3

(C) n0 = 2.25 x 1015

cm-3

, p0 = 1.5 x 1010

cm-3

(D) n0 = 2.25 x 1015

cm-3

, p0 = 1 x 105 cm

-3

Answer: (D)

Exp:

15 3

D

10 3

i

N 2.25 10 Atom / cm

h 1.5 10 / cm

= ×

= ×

Since complete ionization taken place,

( )

15 3

0 D

2102

5 3i0 15

0

h N 2.25 10 / cm

1.5 10nP 1 10 / cm

n 2.25 10

= = ×

×= = = ×

×

5kΩ 4 kΩ 1kΩ

1mHt 0=

10 Fµ10V +−

i

equZ LoadNI

5 kΩ 4 kΩ

( )i 0 +

10V +− •

•



6

9. An increase in the base recombination of a BJT will increase

(A) the common emitter dc current gain β

(B) the breakdown voltage BVCEO

(C) the unity-gain cut-off frequency fT

(D) the transconductance gm

Answer: (B)

10. In CMOS technology, shallow P-well or N-well regions can be formed using

(A) low pressure chemical vapour deposition

(B) low energy sputtering

(C) low temperature dry oxidation

(D) low energy ion-implantation

Answer: (D)

11. The feedback topology in the amplifier circuit (the base bias circuit is not shown for

simplicity) in the figure is

(A) Voltage shunt feedback

(B) Current series feedback

(C) Current shunt feedback

(D) Voltage series feedback

Answer: (B)

Exp: By opening the output feed back signed becomes zero. Hence it is current sampling.

As the feedback signal fv is subtracted from the signal same sv it is series mixing.

12. In the differential amplifier shown in the figure, the magnitudes of the common-mode and

differential-mode gains are Acm and Ad, respectively. If the resistance RE is increased, then

(A) Acm increases

(B) common-mode rejection ratio increases

(C) Ad increases

(D) common-mode rejection ratio decreases

Answer: (B)

Exp: dA does not depend on ER

cmA decreases as ER is increased

d

cm

ACMRR Increases

A∴ = =

CCV

CRCR

+

−0V+

iV

−

ER

EEV−

0I

CCV

oI

oV

ER

SR

SV ~

cR



7

13. A cascade connection of two voltage amplifiers A1 and A2 is shown in the figure. The open-

loop gain Av0, input resistance Rin, and output resistance RO for A1 and A2 are as follows:

v0 in 0

v0 in 0

A1:A 10,R 10k ,R 1k

A2 : A 5,R 5k , R 200

= = Ω = Ω= = Ω = Ω

The approximate overall voltage gain out inV / V is __________.

Answer: 34.722

Exp: 2

1 2

2 1 2

i0 Lv V V

i i 0 L 0

ZV ROverall voltage gain,A A A

V Z Z R Z

= =

+ +

V

5k 1k10 5

5k 1k 1k 200

A 34.722

= × + +

=

14. For an n-variable Boolean function, the maximum number of prime implicants is

(A) 2(n-1) (B) n/2 (C) 2n (D) 2

(n-1)

Answer: (D)

Exp: For an n-variable Boolean function, the maximum number of prime implicants ( )n 1

2−=

15. The number of bytes required to represent the decimal number 1856357 in packed BCD

(Binary Coded Decimal) form is __________ .

Answer: 4

Exp: In packed BCD (Binary Coded Decimal) typically encoded two decimal digits within a single

byte by taking advantage of the fact that four bits are enough to represent the range 0 to 9.

So, 1856357 is required 4-bytes to stored these BCD digits

16. In a half-subtractor circuit with X and Y as inputs, the Borrow (M) and Difference (N = X - Y)

are given by

(A) M = X, ⊕ Y, N = XY (B) M = XY, N = X⊕ Y

(C) M = X Y , ⊕ N = X⊕ Y (D) M = XY N = X Y⊕

Answer: (C)

+

−

inV A1 A2 LR

1kΩ

+

−

outV



8

Exp: Function Table for Half-subtractor is

Hence, N X Y and m XY= ⊕ =

Hence, N X Y and m XY= ⊕ =

17. An FIR system is described by the system function

( ) 1 27 3H z 1 z z

2 2

− −= + + The system is

(A) maximum phase (B) minimum phase (C) mixed phase (D) zero phase

Answer: (C)

Exp: Minimum phase system has all zeros inside unit circle maximum phase system has all zeros

outside unit circle mixed phase system has some zero outside unit circle and some zeros

inside unit circle.

( ) 1 27 3for H s 1 z z

2 2

− −= + +

One zero is inside and one zero outside unit circle hence mixed phase system

18. Let x[n] = x[-n]. Let X(z) be the z-transform of x[n]. If 0.5 + j 0.25 is a zero o X(z), which

one of the following must also be a zero of X(z).

(A) 0.5 - j0.25 (B) 1/(0.5 + j0.25)

(C) 1/(0.5 - j0.25) (D) 2 + j4

Answer: (B)

Exp: [ ] [ ]Given x n x n= −

( ) ( ) [ ]1x z x z Time reversal property in z transform

if one zero is 0.5 j0.25

1then other zero will be

0.5 j0.25

−⇒ = −

⇒ +

+

19. Consider the periodic square wave in the figure shown.

X Y Difference (N) Borrow (M)

0 0 0 0

0 1 1 1

1 0 1 0

1 1 0 0

x

1

0 1 2 3 4 t

1−



9

The ratio of the power in the 7th harmonic to the power in the 5

th harmonic for this waveform

is closest in value to _______.

Answer: 0.5

Exp: For a periodic sequence wave, nth harmonic component is 1

nα

⇒ power in nth harmonic component is 2

1

nα

⇒ Ratio of the power in 7th harmonic to power in 5

th harmonic for given waveform is

2

2

1257 0.5

1 495

= ≈

20. The natural frequency of an undamped second-order system is 40 rad/s. If the system is

damped with a damping ratio 0.3, the damped natural frequency in rad/s is ________.

Answer: 38.15 r / sec

Exp: nGiven 40 r / secω =

( )

2

d n

2

d

d

0.3

1

40 1 0.3

38.15 r / sec

ξ =

ω = ω − ξ

ω = −

ω =

21. For the following sytem,

When ( )1X s 0= , the transfer function ( )( )2

y s

x s is

( ) 2

s 1A

s

+ ( ) 1

Bs 1+

( ) ( )s 2

Cs s 1

++

( ) ( )s 1

Ds s 2

++

Answer: (D)

Exp: ( )1If X s 0=

( )( )2

Y s; The block diagram becomes

X s

( )( )

( )( )

( )( )2

1 1Y s s 1s s

1 sX s s 2 / s 1 s s 21 .

s s 1

+= = ⇒

+ + +++

( )1x s +

−

s

s 1+

( )2x s

++ 1

s

( )Y s

( )2X s+

−

1

s( )Y s

( )++S

S 1



10

22. The capacity of a band-limited additive white Gaussian noise (AWGN) channel is given by

2 2

PC Wlog 1

w

= + σ bits per second (bps), where W is the channel bandwidth, P is the

average power received and 2σ is the one-sided power spectral density of the AWGN. For a

fixed 2

P1000,=

σ, the channel capacity (in kbps) with infinite bandwidth ( )W → ∞ is

approximately

(A) 1.44 (B) 1.08 (C) 0.72 (D) 0.36

Answer: (A)

Exp: 2

2 2w

Pln 1

PC lim log 1 lim

ln 2→∞ ω→∞

ω + σ ω = ω + = σ ω

[ ]

2 2

2 2

22 2

This limit is equivalent to

22 2

2

P Pln 1 ln 1

1 P Plim . lim

P Pln 2 ln

ln 1 x P Plim 1 ln e 1.44KGpa

x .ln

ω→∞ ω→∞

↓

ω→∞

+ + σ ω σ ω = =σ σ

σ ω σ ω

+= = = =

σ σ

23. Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the

modulation index ( )µ when the maximum and minimum values of the envelope,

respectively, are 3 V and 1 V, is ________.

Answer: 0.5

Exp: ( ) ( )( ) ( )

max

max

A t A t min

A t A t min

−µ =

+

3 1 1

0.53 1 2

−µ = = =

+

24. To maximize power transfer, a lossless transmission line is to be matched to a resistive load

impedance via a / 4λ transformer as shown.

The characteristic impedance ( )inΩ of the / 4λ transformer is _________.

Answer: 70.7Ω

lossless transmission line

/ 4 transformerλ

LZ 100= Ω

LZ 50= Ω



11

Exp: Here impedance is matched by using QWT ( )4λ

L in

'

0

'0Z Z Z

100 50 50 2

Z 70.7

∴ =

= × =

= = Ω

25. Which one of the following field patterns represents a TEM wave travelling in the positive x

direction?

( ) ˆ ˆA E 8y, H 4z= + = − ( ) ˆ ˆB E 2y, H 3z= − = −

( ) ˆ ˆC E 2z, H 2y+ = + ( ) ˆ ˆD E 3y, H 4z= − = +

Answer: (B)

Exp: For TEM wave

Electric field (E), Magnetic field (H) and

Direction of propagation (P) are orthogonal to each other.

Here xP a= +

By verification

y z

y z x

E 2a , H 3a

E H a a a P

= − =−

× = − ×− = + →

Q. No. 26 – 55 Carry Two Marks Each

26. The system of linear equations

2 1 3 a 5

3 0 1 b 4

1 2 5 c 14

= −

has

(A) a unique solution (B) infinitely many solutions

(C) no solution (D) exactly two solutions

Answer: (B)

Exp: [ ]2 1 3 5

A / B 3 0 1 4

141 2 5

= −

( ) ( )

3 22 2 1 R R

3 3 1

2 1 3 2 1 35 5R 2R 3R

0 3 7 23 0 3 7 23R 2R R

23 00 3 7 0 0 0

Since, rank A rank A / B number of unknowns

+

→ − − − − → − − − → −

= <

∴ Equations have infinitely many solutions.



12

27. The real part of an analytic function f(z) where z =.x + jy is given by e-y

cos(x). The

imaginary part of f(z) is

(A) eycos(x) (B) e

-ysin(x) (C) -e

ysin(x) (D) –e

-ysin(x)

Answer: (B)

Exp:

yreal part u e cos x and V ?−= =

( ) y y y

y

v vdv dx dy

x y

u udx dy Usin g C R equations e cos xdx e sin xdy d e sin x

y x

Integrating, we get V e sin x

− − −

−

∂ ∂= +

∂ ∂∂ ∂

= − + − = − = ∂ ∂

=

28. The maximum value of the determinant among all 2×2 real symmetric matrices with trace 14

is ________.

Answer: 49

Exp:

y xGeneral 2 2 real symmetric matrix is

x z

×

( )( )

( ) ( )

2

2 2 2

det yz x and trace is y z 14 given

z 14 y .............. *

Let f yz x det x y 14y u sin g *

⇒ = − + =

⇒ = −

= − = − − +

Using maxima and minima of a function of two variables, we have f is maximum at

x 0, y 7= = and therefore, maximum value of the determinant is 49

29. ( )( )2

x y zˆ ˆ ˆIf r xa ya za and r r, then div r ln r= + + = ∇

= _______.

Answer: 3

Exp:

( ) ( )( ) ( )2

2

rln r div r ln r div r 3

r∇ = ⇒ ∇ = =

( ) ( )x x x2 2

1 x 1 rˆ ˆ ˆln r a ln r a a x

x r r r r

∂ ∇ = = = = ∂ ∑ ∑ ∑

30. A series LCR circuit is operated at a frequency different from its resonant frequency. The

operating frequency is such that the current leads the supply voltage. The magnitude of

current is half the value at resonance. If the values of L, C and R are 1 H, 1 F and 1Ω ,

respectively, the operating angular frequency (in rad/s) is ________.

Answer: 0.45 r/sec

Exp: The operating frequency (wx), at which current leads the supply.

i.e., x rω <ω

again magnitude of current is half the value at resonance



13

x x

x resonance

resonancex

Vi,e.,. at I

z

Vat I

R

II

2

V Vi.e., Z 2R

z 2R

ω = ω ⇒ =

ω=ω ⇒ =

=

= = =

Given R = 1Ω; L=1H; C=1F

2

2

c

2

2

c

1Z R L 2

1R L 4

= + − ω = ω

= + − ω = ω

By substituting R, L & C values,

2

2

2

2

2

1,2

x r

1 11 4 5

1Assume x, then, x 5

x

x 5x 1 0

x 4.791, 0.208

if x 4.791 2.18r sec

if x 0.208 0.45r sec

But

⇒ + −ω = ⇒ ω = = ω ω

ω = + =

⇒ − + ==

= ⇒ ω== ⇒ ω=

ω <ω

So, operating frequency x 0.45 r secω =

31. In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in

S) is _____ .

Answer: 1.24

Exp: If two, n wsπ− are connected in parallel,

The y-parameter are added

3Ω

3Ω 3Ω1 2

2 '1'

2Ω 2Ω

2Ω



14

equ 1 2

1 2

equ

12

11 11

21

11 11

11 22 12 21

i.e., y y y

2 1 113 3 2

y y1 2 1 1

3 3 2

5 53 6

y5 5

6 3

y1y y

hy y

y y

where y y y y y

= +

− − = = −−

− = −

−

= ∆

∆ = − −

The value of 22

11 22

5 5 5 5h y

3 3 6 6

y 2.0833

5y h 1.243

− − = ∆ = −

∆ =

= ∴ =

32. In the figure shown, the capacitor is initially uncharged. Which one of the following

expressions describes the current I(t) (in mA) for t > 0?

( ) ( ) ( )t /5 2A I t 1 e , msec

3 3

− τ= − τ = ( ) ( ) ( )t /5 2B I t 1 e , msec

2 3

− τ= − τ =

( ) ( ) ( )t /5C I t 1 e , 3msec

2

− τ= − τ = ( ) ( ) ( )t /5D I t 1 e , 3 msec

2

− τ= − τ =

Answer: (A)

Exp:

( ) ( ) [ ]t

c R 2 final initial final

3 6

equ equ

t V t V V V e

2R .C 10 10

3

−τ

−

ν = = + −

τ = ⇒ × ×

equ

equ

2R 2K 1K K

3

C 1 F

= ⇒ Ω

= µ

1R

2R

1kΩ

+−5V

2 kΩ

IC

1 Fµ



15

( )

( ) ( )

initial

final s.s

t

R 2

t tR 2

R 2 R 2

2msec

3

V 0volts

2 10V V 5. volts

3 3

10 10t e

3 3

t10 5t 1 e volts i (t) 1 e mA

3 2K 3

−τ

− −τ τ

τ =

=

= = =

ν = −

ν ν = − ⇒ = = −

33. In the magnetically coupled circuit shown in the figure, 56 % of the total flux emanating from

one coil links the other coil. The value of the mutual inductance (in H) is ______ .

Answer: 2.49 Henry

Exp: Given 56% of the total flux emanating from one coil links to other coil.

i.e, K 56% 0.56= ⇒

We have,

( )

1 2

1 2

MK

L L

L 4H; L 5H

M 0.56 20 m 2.50H

=

= =

= ⇒ =

34. Assume electronic charge q = 1.6×10-19

C, kT/q = 25 mV and electron mobility n 1000µ =

cm2/V-s. If the concentration gradient of electrons injected into a P-type silicon sample is

1×1021

/cm4, the magnitude of electron diffusion current density (in A/cm

2) is _________.

Answer: 4000

Exp: 19 2

n

kJGiven q 1.6 10 ; 2.5mV, 1000cm / v s

q

−= × = µ = −

n

n

2

n

2

n

19 21

2

D kJFrom Einstein relation,

q

D 25mV 1000cm / v S

25cm / s

dnDiffuion current Density J q D

dx

1.6 10 25 1 10

4000A / cm

−

=µ

⇒ = × −⇒

=

= × × × ×=

10 ΩM

( )060cos 4t 30 V+

4 H 5H ( )1/16 F~



16

35. Consider an abrupt PN junction (at T = 300 K) shown in the figure. The depletion region

width Xn on the N-side of the junction is 0.2 µm and the permittivity of silicon ( )siε is

1.044×10-12

F/cm. At the junction, the approximate value of the peak electric field (in kV/cm)

is _________.

Answer: 30.66

Exp: 12

n Si nGiven x 0.2 m, 1.044 10 F /−= µ ∈ = × µ

16 3

DN 10 / cm=

D nq N xPeak Electric field, E =

∈

19 16

12

1.6 10 10 0.0000230.66KV / cm

1.044 10

−

−

× × ×= =

×

36. When a silicon diode having a doping concentration of NA = 9 × 1016

cm-3

on p-side and ND =

1 × 1016

cm-3

on n-side is reverse biased, the total depletion width is found to be 3 mµ . Given

that the permittivity of silicon is 1.04 × 10–12

F/cm, the depletion width on the p-side and the

maximum electric field in the depletion region, respectively, are

(A) 2.7 mµ and 2.3 × 105 V/cm (B) 0.3 mµ and 4.15 × 10

5 V/cm

(C) 0.3 mµ and 0.42 × 105 V/cm (D) 2.1 mµ and 0.42 × 10

5 V/cm

Answer: (B)

Exp: 16 3 16 3

A DGiven N 9 10 / cm ; N 1 10 / cm= × = ×

( )

n p

12

16

n A

16

p D

n p

Total depletion width x x x 3 m

1.04 10 F / cm

x N 9 10

x N 1 10

x 9x ......... 1

−

= + = µ

∈= ×

×= =

×

=

n p

p p

p

19 16A p

12

5

Total Depletion width, x x 3 m

9x x 3 m

x 0.3 m

qN x 1.6 10 9 10 0.3 mMax. Electric field, E

1.04 10

4.15 10 V / cm

−

−

+ = µ

+ = µ

= µ

× × × × µ= =

∈ ×= ×

A D

P region

N N

+ −>> 16 3

D

N region

N 10 / cm

−

=11

X



17

37. The diode in the circuit shown has Von = 0.7 Volts but is ideal otherwise.

If Vi = 5sin ( )tω Volts, the minimum and maximum values of VO (in Volts) are, respectively,

(A) -5 and 2.7 (B) 2.7 and 5 (C) -5 and 3.85 (D) 1.3 and 5

Answer: (C)

Exp: i 0 iWhen V makes Diode 'D' OFF, V V=

( )

( )

( ) ( )

0

i

i

0 on

1 2

0

V min 5V

When V makes diode 'D ' ON,

V 0.7 2V V 2V

R R

5 0.7 2 1kV max 0.7 2V

1k 1k

3.85V

∴ = −

− −= + +

+

− −∴ = + +

+

=

38. For the n-channel MOS transistor shown in the figure, the threshold voltage VTh is 0.8 V.

Neglect channel length modulation effects. When the drain voltage VD = 1.6 V, the drain

current ID was found to be 0.5 mA. If VD is adjusted to be 2 V by changing the values of R

and VDD, the new value of ID (in mA) is

(A) 0.625 (B) 0.75 (C) 1.125 (D) 1.5

Answer: (C)

iV

1kΩ

1R

oV

2R

1kΩ

+− 2 V

DDV

R

D

SG



18

Exp: ThGiven V 0.8V=

( )

[ ]

2

D D n DS Th

1 wWhen V 1.6V, I 0.5mA cos V V

2 L

Device is in sat

= = = µ −

∵

( )

( )

3 2

n

D

2

D n DS Th

3

1cos 0.78125 10 A / V

2 L

When V 2V

1I cos V V

2 L

078125 10 2 0.8 1.125mA

−

−

ω⇒ µ = ×

=ω

= µ −

= × −

39. For the MOSFETs shown in the figure, the threshold voltage tV 2V= and

21 WK C 0.1 mA / V

2 L∞ = µ =

. The value of ID (in mA) is ________.

Answer: 0.9

Exp: 2

t

1 WGiven V 2V, K cos 0.1A / V

2 L= = µ =

( )( )

1 2 1

2

D D n Gs t

22

1 WI I cos V V

2 L

0.1mA / V 5 2

0.9mA

= = µ −

= −

=

DDV 12 V= +

1R

10 kΩ

10 kΩ2

R

DDV 5V= −

DI

12V

DI

10K

10K

−5V



19

40. In the circuit shown, choose the correct timing diagram of the output (y) from the given

waveforms W1, W2, W3 and W4.

(A) W1 (B) W2 (C) W3 (D) W4

Answer: (C)

Exp: This circuit has used negative edge triggered, so output of the D-flip flop will changed only

when CLK signal is going from HIGH to LOW (1 to 0)

1X D Q

FF1

QClk

( )output y

D Q

Q>

>

2X

ClK

1X

2X

W1

W2

W3

W4

C LK 0 0 0 0

1 1 1 1

1X

2X

( )3wY



20

This is a synchronous circuit, so both the flip flops will trigger at the same time and will

respond on falling edge of the Clock. So, the correct output (Y) waveform is associated to w3

waveform.

41. The outputs of the two flip-flops Q1, Q2 in the figure shown are initialized to 0,0. The

sequence generated at Q1 upon application of clock signal is

(A) 01110… (B) 01010… (C) 00110… (D) 01100…

Answer: (D)

Exp:

So, the output sequence generated at Q1 is 01100….

42. For the 8085 microprocessor, the interfacing circuit to input 8-bit digital data (DI0 – DI7)

from an external device is shown in the figure. The instruction for correct data transfer is

(A) MVI A, F8H

(B) IN F8H

(C) OUT F8H

(D) LDA F8F8H

Clock

Initial →

1st CP →

2nd

CP →

3rd

CP →

4th CP→

( )1 2J Q ( )1 2K Q ( )2 1J Q ( )2 1K Q 1Q Q2

- - - - 0 0

1 0 0 1 1 0

1 0 1 0 1 1

0 1 1 0 0 1

0 1 0 1 0 0

>

J1 Q1

K1 Q1

Q1J2 Q2

K2Q2

CLK

>

3 to 8− −Decoder

2A

1A

0A

C

B

A

7

6

5

4

3

2

1

02A

G2B

G1

G

10 / M

RD

3A

4A5

A

6A

7A

8A

9A

10A

11A

12A

13A

14A

15A

Digital

Inputs

I / O Device

0 7DI DI− 0 7

DO D−

1DS

2DS

( )0 7

Data Bus

D D−



21

Answer: (D)

Exp: This circuit diagram indicating that it is memory mapped I/O because to enable the 3-to-8

decoder 2AG is required active low signal through ( )oI m and 2BG is required active

low through ( )DR it means I/o device read the status of device LDA instruction is

appropriate with device address.

Again to enable the decoder o/p of AND gate must be 1 and 2Ds signal required is 1 which is

the o/p of multi-i/p AND gate to enable I/O device.

So,

11 10 9 8 3 2 1 015 14 13 12 7 6 5 4

FF 8 8

A A A A A A A AA A A A A A A A

1 0 0 0 1 0 0 01 1 1 1 1 1 1 1

Device address = F8F8H

The correct instruction used → LDA F8F8H

43. Consider a discrete-time signal

[ ] n for 0 n 10x n

0 otherwise

≤ ≤=

If y[n] is the convolution of x[n] with itself, the value of y[4] is _________.

Answer: 10

Exp: [ ] n for 0 n 10Given x n

0 elsewhere

≤ ≤ =

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ]

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

n

k 0

4

k 0

y n x n * x n

y n x k .x n k

y 4 x k .x G k

x 0 .x 4 x 1 x 3 x 2 x 2 x 3 x 1 x 4 .x 0

0 3 4 3 0 10

=

=

=

= −

⇒ = −

= + + + +

= + + + + =

∑

∑

44. The input-output relationship of a causal stable LTI system is given as

[ ] [ ] [ ]y n y n 1 x n= α − + β

If the impulse response h[n] of this system satisfies the condition [ ]n 0h n 2,∞=Σ = the

relationship between andα β is

( )A 1 / 2α = − β ( )B 1 / 2α = + β ( )C 2α = β ( )D 2α = − β



22

Answer: (A)

Exp: Given system equation as

[ ] [ ] [ ]

( )( )

( )

[ ] ( ) [ ] [ ]

[ ]

1

1

h

h 0

y n y n 1 x n

y z

x z 1 z

H z1 z

h n u n causal system

Also given that h n 2

−

−

∞

=

= α − + β

β⇒ =

− α

β⇒ =

− α

= β α

=∑

12

1

12

12

β = − α

β− α =

βα = −

45. The value of the integral ( )2sinc 5t∞

−∞∫ dt is ____________.

Answer: 0.2

Exp: We can use pasrevalis theorem

( ) ( ) sin 5 t

Let x t sin 5t5 t

in frequency domain

π=

π⇒

( ) ( )22.5

2 2

2.5

1Now, x t dt x t df

5

1 15 0.2

25 5

∞ ∞

−∞ −∞ −

= =

= × = =

∫ ∫ ∫

( )1 X f5

−2.5 −2.5 f



23

46. An unforced liner time invariant (LTI) system is represented by

11

22

x1 0x

x0 2x

− = −

If the initial conditions are x1(0) = 1 and x2(0) = -1, the solution of the state equation is

( ) ( ) ( )1 2A x t 1, x t 2= − = ( ) ( ) ( )t t

1 2B x t e , x t 2e− −= − =

( ) ( ) ( )t 2t

1 2C x t e , x t e− −= = − ( ) ( ) ( )t t

1 2D x t e , x t 2e− −= − = −

Answer: (C)

Exp: Solution of state equation of ( ) ( )1 1X t L SI A .X 0− − = −

( )

[ ]

( )( )

1

1

1 1 0X 0 A

1 0 2

S 1 0SI A

0 S 2

S 2 01

0 S 1S 1 S 2

−−

− = = − −

+ − = +

+ = ++ +

[ ]

( )

( )

( )( )

( )( )

( )( )

1

1

11

1

t11

2t

t1

2t2

t t1 1

2t22

10

S 1SI A

10

S 2

1L 0

S 1L SI A

10 L

S 2

e 0L SI A

0 e

X t e 0 1

1X t 0 e

X t e X t e

X t eX t e

−

−

−−

−

−−−

−

−

−

−

−

+− = +

+ − =

+

− =

= −

− == ∴

= −− 2t−



24

47. The Bode asymptotic magnitude plot of a minimum phase system is shown in the figure.

If the system is connected in a unity negative feedback configuration, the steady state error of

the closed loop system, to a unit ramp input, is_________.

Answer: 0.50

Exp:

→ Due to initial slope , it is a type-1 system, and it has non zero velocity error coefficient

( )VK

→ The magnitude plot is giving 0dB at 2r/sec.

Which gives vk

v

ss

v

ss

ss

k 2

AThe steady state error e

k

given unit ramp input; A 1

1e

2

e 0.50

∴ =

=

=

=

=

26.02

6.02( )G jω

( )dB0

6.02−0.1 1 2 10 20

( )rad / s in log scaleω

26.02

6.02

( )G jw

dB

( )−20dB / dec

−6.02

( )w r / sec0.1 1 2 10

20

( )( )−20db dec



25

48. Consider the state space system expressed by the signal flow diagram shown in the figure.

The corresponding system is

(A) always controllable (B) always observable

(C) always stable (D) always unstable

Answer: (A)

Exp: From the given signal flow graph, the state model is

[ ]

[ ]

1 1

2 2

3 2 1 33

1

1 2 3 2

3

1 2 3

3 2 1

X X0 1 0 0

X 0 0 1 X 0 u

a a a 1XX

X

Y C C C X

X

0 1 0 0

A 0 0 1 ;B 0 ;C C C C

a a a 1

= +

=

= = =

Controllability:

2

c

C 1

2

1 2 1

C

Q B AB A B

0 0 1

Q 0 1 a

1 a a a

Q 1 0

=

= +

= ≠

Observability

( ) ( ) ( )

1 2 3

O 3 3 1 2 3 2 1 3

2 2

2 3 3 1 3 2 2 3 1 2 3 1 1 2 3 1 2

C C CC

Q CA a c c a c c a c

CA c a c a a a c c a a a c a c c a a

= ⇒ + + +

+ + +

+ +

0 1 2 3 1 2 3

Q depends on a ,a ,a & c & c & c⇒.

It is always controllable

3C

1S−1

u 3x

1S− 1S−

1x 1

c2

x

2a

1a

y

3a



26

49. The input to a 1-bit quantizer is a random variable X with pdf ( ) 2x

xf x 2e for x 0−= ≥ and

( )xf x 0 for x 0= < , for x 0< For outputs to be of equal probability, the quantizer threshold

should be _____.

Answer: 0.35

Exp:

One bit quantizer will give two levels.

Both levels have probability of 1

2

Pd of input X is

Let T

x be the thsuhold

( ) 1 T

2 T

x x xQ x

x x x

≥ = <

Where 1 2

x and x are two levels

( )

T

T

T

T

1

2x

x

2x

x

2x2

2x

T

T

T

1P Q r x

2

12.e dx

2

e 12.

2 2

1e e

2

1e

2

12x ln

2

2x 0.693

x 0.35

∞−

∞−

−− ∞

−

= =

⇒ =

=−

− + =

=

− =

− = −=

∫

( )πxf

Tx

X ( )Q xone bit

Quantizer



27

50. Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol

waveforms ( )1 1s t cos2 f t= α π and ( )2 2s t cos2 f t,= π where 4mV.α = Assume an AWGN

channel with two-sided noise power spectral density 120N0.5 10 W / Hz.

2

−= × Using an

optimal receiver and the relation ( ) 2u / 2

v

1Q v e du

2

∞ −=π ∫

the bit error probability for a data

rate of 500 kbps is

( ) ( )A Q 2 ( ) ( )B Q 2 2 ( ) ( )C Q 4 ( ) ( )D Q 4 2

Answer: (C)

Exp: For Binary SK

F

O

EBit error probability Q

N

=

E → Energy per bit [No. of symbols = No. of bits]

[ ]

( )

23

3

6 612

12

0

12

e 12

A T 1E ,A 4 10 ,T inverse of data rate

2 500 10

16 10 2 10E 16 10

2

N 1 10

16 10P Q Q 4

1 10

−

− −−

−

−

−

= = × =×

× × ×⇒ = = ×

= ×

×⇒ = =

×

51. The power spectral density of a real stationary random process .X(t) is given by

( )1

, f wwx 0, f w

S f≤>

=

The value of the expectation ( ) 1E X t t

4w

π −

is ___________.

Answer: 4

Exp: ( )x

1, f w

wGiven S f

0 , f w

≤ = ≥

( )

( )

( )

w

j2 ft

x

w

j2 wt j2 wt

x

1R .e df

w

sin 2 wt1 e e 1

w j2 t w t

1sin 2 w.

1 1 1 44wNow, E t .x t R . .

14w 4w w 1.4w

π

−

π − π

τ =

π −= = π π

π π× − = π ⇒ π = π

∫



28

52. In the figure, M(f) is the Fourier transform of the message signal .m(t) where A = 100 Hz and

B = 40 Hz. Given v(t) = cos ( )c2 f tπ and ( ) ( )( )cw t cos 2 f A t= π + , where c

f A> The cutoff

frequencies of both the filters are C

f

The bandwidth of the signal at the output of the modulator (in Hz) is _____.

Answer: 60

Exp: ( ) ( )m t M f↔

After multiplication with ( ) ( )cV t cos 2 f t= π

( ) ( ) ( )

( ) ( )( )1

1 1

Let w t m t .V t

W f specturm of w t is

=

⇒

After high pass filter

After multiplication with ( )( )ccos 2 f A tπ + and low pass filter of cut off c

f

( )M f

1−

A− B− B A f

( )m t

( )v t

High Pass

Filter

( )w t

Low Pass

Filter

( )s t

1

AB−A −B

( )M f

f

−cf

− − − − − + +c c c cf A f B f B f A − − − − − + +

c c c cf A f B f B f A

cf

− − − −C Cf A f B + +

C C Cf f B f A



29

Bandwidth A B

100 40 60

= −= − =

53. If the electric field of a plane wave is

( ) ( ) ( ) ( )O OE Z, t x3cos t kz 30 y4sin t kz 45 mV / m ,= ω − + − ω − +

the polarization state of the plane wave is

(A) left elliptical (B) left circular

(C) right elliptical (D) right circular

Answer: (A)

Exp:

( ) ( ) ( )( )

( )( )

( )

1 x y

x

y

o

x

o

y

x y

E z t 3cos cot kz 3o a 4 sin t kz 45 a

E 3cos t kz 30

E 4cos t kz 45

At z 0 E 3cos t 30

E 4sin t 45

E E so Elliptical polarization

Q 30 135 105

= − + ° − − ω − + °

= ω − + °

= − ω − + °

= = ω +

= − ω +

≠ →

= ° − °=− °

∴ left hand elliptical (LEP)

54. In the transmission line shown, the impedance Zin (in ohms) between node A and the ground

is _________.

Answer: 33.33Ω

Exp: Here

( )( )

in L

in

2

Z Z 502

100Z 100 50 33.33

3

λ=

λ= = = Ω

∴ = = = Ω

0Z 50 , L 0.5= Ω = λ

100Ω 50Ωin

Z ?=

A

( )− −A B ( )−A Bf

O



30

55. For a rectangular waveguide of internal dimensions ( )a b a b× > , the cut-off frequency for

the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE20

mode. If a 5 cm,= the value of b (in cm) is _____.

Answer: 2

Exp:

2

c10

c10 c20

c11 2 2

C 1t

2 a

1 2t K ; t K

a a

1 1t K

a b

=

= =

= +

( )

c10 c20c11

2 2

2 2

2 2

2

2 2

f fGiven t

2

1 1 K 1 2K

a b 2 a a

1 1 3

a b 2a

1 1 9 1 9 1

5 b 4 5 5 20 b

10.2 0.45

b

1 1b 2cm

b 2

+=

+ = +

+ =

+ = ⇒ − + =

− + =

∴ = ⇒ =



1


1. “India is a country of rich heritage and cultural diversity.” Which one of the following facts

best supports the claim made in the above sentence?

(A) India is a union of 28 states and 7 union territories.

(B) India has a population of over 1.1 billion.

(C) India is home to 22 official languages and thousands of dialects.

(D) The Indian cricket team draws players from over ten states.

Answer: C

Exp: Diversity is shown in terms of difference language

2. The value of one U.S. dollar is 65 Indian Rupees today, compared to 60 last year. The Indian

Rupee has ____________.

(A) Depressed (B) Depreciated (C) Appreciated (D) Stabilized

Answer: B

3. 'Advice' is ________________.

(A) a verb (B) a noun

(C) an adjective (D) both a verb and a noun

Answer: B

4. The next term in the series 81, 54, 36, 24 … is ________

Answer: 16

Exp: 2

81 54 27;27 183

− = × =

254 36 18;18 12

3

236 24 12;12 8

3

24 8 16

− = × =

− = × =

∴ − =

5. In which of the following options will the expression P < M be definitely true?

(A) M < R > P > S (B) M > S < P < F

(C) Q < M < F = P (D) P = A < R < M

Answer: D


6. Find the next term in the sequence: 7G, 11K, 13M, ___

(A) 15Q (B) 17Q (C) 15P (D) 17P

Answer: B



2

7. The multi-level hierarchical pie chart shows the population of animals in a reserve forest. The

correct conclusions from this information are:

(i) Butterflies are birds

(ii) There are more tigers in this forest than red ants

(iii) All reptiles in this forest are either snakes or crocodiles

(iv) Elephants are the largest mammals in this forest

(A) (i) and (ii) only (B) (i), (ii), (iii) and (iv)

(C) (i), (iii) and (iv) only (D) (i), (ii) and (iii) only

Answer: D

Exp: It is not mentioned that elephant is the largest animal

8. A man can row at 8 km per hour in still water. If it takes him thrice as long to row upstream,

as to row downstream, then find the stream velocity in km per hour.

Answer: 4

Exp: 4 km/hr.

Speed of man=8

Left distance =d

Time taken=d

8

Upstream:

Speed of stream=s

speed upstream S' (8 s)

dt'

8 s

⇒ = = −

= −

Downstream:

Givend

speed downstream t ''8 s

= =+

Red ant−Beetle Tiger

Elephant

Mammal

In sec t

Honey

bee−

Moth

Hawk

Bird

Re ptile

Leopard

Snake

Crocadile

Butterfly

DrongoBulbul



3

3t ' t ''

3d d

8 s 8 s

3d d

8 s 8 s

s 4km / hr

⇒ =

⇒ =− +

⇒ =− +

⇒ =

9. A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the

share of raw material, labour, energy, plant & machinery, and transportation costs in the total

manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In

2013, the raw material expenses increased by 30% and all other expenses increased by 20%.

If the company registered a profit of Rs. 10 lakhs in 2012, at what price (in Rs.) was each air

purifier sold?

Answer: 20,000

Exp: Total expenditure=15

x 4,50,000100

= =

x=3×106

Profit=10 lakhs

So, total selling price =40,00,000 … (1)

Total purifies=200 … (2)

S.P of each purifier=(1)/(2)=20,000

10. A batch of one hundred bulbs is inspected by testing four randomly chosen bulbs. The batch

is rejected if even one of the bulbs is defective. A batch typically has five defective bulbs.

The probability that the current batch is accepted is _________

Answer: 0.8145

Exp: Probability for one bulb to be non defective is 95

100

∴ Probabilities that none of the bulbs is defectives

495

100

= 0.8145

Trans

portat

ion

10%

Labour

15%

Plant and

machinery

30%

Energy

25%

Raw Material

20%



4

Q.No. 1 – 25 Carry One Mark Each

1. The maximum value of the function f(x) = ln(1 + x)- x (where .x > - 1) occurs at x=______.

Answer: 0

Exp: ( )1 1f x 0 1 0

1 x= ⇒ − =

+

( )( )

11

2

x0 x 0

1 x

1and f x 0 at x 0

1 x

−⇒ = ⇒ =

+−

= < =+

2. Which ONE of the following is a linear non-homogeneous differential equation, where x and

y are the independent and dependent variables respectively?

( ) xdyA xy e

dx

−+ = ( ) dyB xy 0

dx+ =

( ) ydyC xy e

dx

−+ = ( ) y ydyD e e 0

dx

− −+ = =

Answer: A

Exp: (A) xdyxy e

dx

−+ = is a first order linear equation (non-homogeneous)

(B) dy

xy 0dx

+ = is a first order linear equation (homogeneous

(C), (D) are non linear equations

3. Match the application to appropriate numerical method.

Application Numerical |Method

P1: Numerical integration M1: Newton-Raphson Method

P2: Solution to a transcendental equation M2: Runge-Kutta Method

P3: Solution to a system of linear equations M3: Simpson’s 1/3-rule

P4: Solution to a differential equation M4: Gauss Elimination Method

(A) P1—M3, P2—M2, P3—M4, P4—M1 (B) P1—M3, P2—M1, P3—M4, P4—M2

(C) P1—M4, P2—M1, P3—M3, P4—M2 (D) P1—M2, P2—M1, P3—M3, P4—M4

Answer: B

Exp: P1 M3,P2 M1,P3 M4,P4 M2− − − −

4. An unbiased coin is tossed an infinite number of times. The probability that the fourth head

appears at the tenth toss is

(A) 0.067 (B) 0.073 (C) 0.082 (D) 0.091



5

Answer: C

Exp: P[fourth head appears at the tenth toss] = P [getting 3 heads in the first 9 tosses and one

head at tenth toss]

3

9

C

1 1 219 . 0.082

2 2 256

= × = =

5. If z = xyln(xy), then

( ) z zA x y 0

x y

∂ ∂+ =

∂ ∂ ( ) z z

B y xx y

∂ ∂=

∂ ∂

( ) z zC x y

x y

∂ ∂=

∂ ∂ ( ) z z

D y x 0x y

∂ ∂+ =

∂ ∂

Answer: C

Exp: ( )z 1y x y ln xy y 1 ln xy

x xy

∂= × × + = + ∂

( )z z zand x 1 ln xy x y

y x y

∂ ∂ ∂= + ⇒ =

∂ ∂ ∂

6. A series RC circuit is connected to a DC voltage source at time t = 0. The relation between

the

source voltage VS, the resistance R, the capacitance C, and the current i(t) is given below:

( ) ( )t

e0

1V Ri t i u du

c= + ∫

Which one of the following represents the current f(t)?

( )i t

t0

( )A

( )i t

0 t

( )B

( )C

( )i t

0 t

( )D

0 t

( )i t



6

Answer: A

Exp: In a series RC circuit,

→ Initially at t = 0, capacitor charges with a current of SV

R and in steady state at t = ∞ ,

capacitor behaves like open circuit and no current flows through the circuit

→ So the current i(t) represents an exponential decay function

7. In the figure shown, the value of the current I (in Amperes) is __________.

Answer: 0.5

Exp:

V 5 VApply KCL at node V, 1 0

5 15

30V volts

4

V 2current I 0.50 Amperes

15 4

−− + =

⇒ =

⇒ = ⇒ ⇒

8. In MOSFET fabrication, the channel length is defined during the process of

(A) Isolation oxide growth

(B) Channel stop implantation

(C) Poly-silicon gate patterning

(D) Lithography step leading to the contact pads

Answer: C

5Ω 5ΩI

10 Ω1A↑±5V

0

( )i t

→ t

5V+−

5Ω

1A

5ΩI

10Ω

V



7

9. A thin P-type silicon sample is uniformly illuminated with light which generates excess

carriers. The recombination rate is directly proportional to

(A) The minority carrier mobility

(B) The minority carrier recombination lifetime

(C) The majority carrier concentration

(D) The excess minority carrier concentration

Answer: D

Exp: ( )( )o o

' '

n n n nRecombination rate, R B n n P P= + +

0 0n nn & P = Electron and hole concentrations respectively under thermal equilibrium

' '

n nn & p = Excess elements and hole concentrations respectively

10. At T = 300 K, the hole mobility of a semiconductor 2

P

kT500cm / V s and 26mV.

qµ = − =

The hole diffusion constant PD in cm2/s is ________

Answer: 13

Exp: From Einstein relation,

P

p

2 2

P

D kJ

q

D 26mV 500cm / v s 13cm / s

=µ

⇒ = × − =

11. The desirable characteristics of a transconductance amplifier are

(A) High input resistance and high output resistance

(B) High input resistance and low output resistance

(C) Low input resistance and high output resistance

(D) Low input resistance and low output resistance

Answer: A

Exp: Transconductance amplifier must have i 0z and z= ∞ = ∞ ideally

12. In the circuit shown, the PNP transistor has BEV 0.7 and 50.= β = Assume that BR 100k= Ω

For V0 to be 5 V, the value of ( )CR in kΩ _________________

CR

BR

EEV 10V=

0V



8

Answer: 1.075

Exp: KVL in base loop gives,

B

C B

0 C C

0C

C

10 0.7I 93 A

100K

I I 50 93 A 4.65mA

from figure, V I R

V 5VR 1.075

I 4.65mA

−= = µ

⇒ = β = × µ =

=

⇒ = = = Ω

13. The figure shows a half-wave rectifier. The diode D is ideal. The average steady-state current

(in Amperes) through the diode is approximately ____________.

Answer: 0.09

Exp: dcdc m

IV V

4fc= −

dcdc L m

dc L m

dc

3

II R V

4fc

1I R V

4fc

10I 0.09A

1100

4 50 4 10−

= −

+ =

⇒ = =+

× × ×

14. An analog voltage in the range 0 to 8 V is divided in 16 equal intervals for conversion to 4-bit

digital output. The maximum quantization error (in V) is _________________

Answer: 0.25

Exp: Maximum quantization error is step size

2

−

8 0 1

step size 0.5V16 2

−− = = =

Quantization error = 0.25 V

~10sin t

f 50 Hz

ω=

R

100Ω

C

4 mF

D



9

15. The circuit shown in the figure is a

(A) Toggle Flip Flop (B) JK Flip Flop

(C) SR Latch (D) Master-Slave D Flip Flop

Answer: D

Exp: Latches are used to construct Flip-Flop. Latches are level triggered, so if you use two latches

in cascaded with inverted clock, then one latch will behave as master and another latch which

is having inverted clock will be used as a slave and combined it will behave as a flip-flop. So

given circuit is implementing Master-Slave D flip-flop

16. Consider the multiplexer based logic circuit shown in the figure.

Which one of the following Boolean functions is realized by the circuit?

( ) 1 2A F WS S= ( ) 1 2 1 2B F WS WS S S= + +

( ) 1 2C F W S S= + + ( ) 1 2D F W S S= ⊕ ⊕

Answer: D

Exp:

Output of first MUX = 1 1 1

1

ws ws w s

Let Y w s

+ = ⊕= ⊕

Output of second MUX = 2 2

2

1 2

Ys Ys

Y s

w s s

+= ⊕= ⊕ +

D

Clk

En

D LatchQ

Q En

D LatchQ

Q

W 0

MUX

1

1S

0

MUX

1

F

2S

W

1S

2S

F

Y0

MUX

1 0

MUX

1



10

17. Let x(t)= ( )cos 10 tπ + ( )cos 30 tπ be sampled at 20 Hz and reconstructed using an ideal low-

pass filter with cut-off frequency of 20 Hz. The frequency/frequencies present in the

reconstructed signal is/are

(A) 5 Hz and 15 Hz only (B) 10 Hz and 15 Hz only

(C) 5 Hz, 10 Hz and 15 Hz only (D) 5 Hz only

Answer: (A)

Explanation: ( ) ( ) ( ) sx t cos 10 t cos 30 t , F 20Hz= π + π =

Spectrum of x(t)

Spectrum of sampled version of x(t)

After LPF, signal will contain 5 and 15Hz component only

18. For an all-pass system ( ) ( )( ) ( )

1

j

1

z bH z , where H e 1,

1 az

−− ω

−

−= =

− for all ω .If

( ) ( )Re a 0, Im a 0,≠ ≠ then b equals

(A) a (B) a* (C) 1/a* (D) 1/a

Answer: (B)

Exp: For an all pass system, 1

polezero *

=*

1or zero

pole=

pole a

1zero

b

1 1or b a *

b a *

=

=

⇒ = =

19. A modulated signal is y(t) = m.(t)cos(40000 tπ ), where the baseband signal m(t) has

frequency components less than 5 kHz only. The minimum required rate (in kHz) at which

y(t) should be sampled to recover m(t) is __________________.

− − −35 25 15 5 5 15 25 35

t− −15 5 5 15



11

Answer: 10 KHz.

Exp: Since m(t) is a base band signal with maximum frequency 5 KHz, assumed spreads as

follows:

[ ]7 *1y(t) m(t)cos(40000 t) m(f ) (f 20k) (f 20k)

2= π → δ − + δ +∵

[ ]1y(f ) M(f 20k) M(f 20k)

2∴ = − + +

Thus the spectrum of the modulated signal is as follows:

If y(t) is sampled with a sampling frequency ‘fs’ then the resultant signal is a periodic

extension of successive replica of y(f) with a period ‘fs’.

It is observed that 10 KHz and 20 KHz are the two sampling frequencies which causes a

replica of M(f) which can be filtered out by a LPF.

Thus the minimum sampling frequency (fs) which extracts m(t) from g(f) is 10 KHz.

20. Consider the following block diagram in the figure.

The transfer function ( )( )

C s

R sis

( ) 1 2

1 2

G GA

1 G G+ ( ) 1 2 1B G G G 1+ + ( ) 1 2 2C G G G 1+ + ( ) 1

1 2

GD

1 G G+

( )R s1

G

+

+2

G+

+

( )C s

M(f )

5k− 5k+ f (Hz)

m(t)f

M(f )

y(f )

25k−f (Hz)

20k− 15k− 15k 20k 25k



12

Answer: C

Exp: By drawing the signal flow graph for the given block diagram

Number of parallel paths are three

1 1 2 2 2 3

Gains P G G , P G ,P 1= = =

By mason’s gain formula,

( )( ) 1 2 3

1 2 2

C sP P P

R s

G G G 1

= + +

⇒ + +

21. The input ( )2t3e u t− , where u(t) is the unit step function, is applied to a system with transfer

function . s 2

s 3

−+

. If the initial value of the output is -2, then the value of the output at steady

state is__________________.

Answer: 0

Exp: 1

( )( )

Y s S 2

X s S 3

−=

+

( ) ( ) ( ) ( )SY s 3Y s S s 2X s⇒ + = × −

Due to initial condition, we can write above equation as

( ) ( ) ( ) ( ) ( ) ( )Sy s y 0 3y s sx s x 0 2x s−− + = − −

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )( )( )

2t

3t

y 0 2, x 0 0 x t 3e u t

3Sy s 2 3y s s 2

s 2

5s 3 y s 3 2 y s

5 3

y t 5e u t

y steady sate 0

− −

−

= − = =

− ⇒ + + = − −

−+ = − − ⇒ =

+⇒ = −

∞ =

Exp: 2

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

( )

2t

at t s 0 s 0

s 2H s ;X t 3e .u t

s 3

3 3X s Y s

s 2 s 3

3sy t y lim S.y s lim

s 3

y 0

=∞ → →

−= = −

+− −

∴ = ⇒ =− +

−⇒ ∞ = =

+∞ =

1

11G

2G

11

( )R s( )C s



13

22. The phase response of a passband waveform at the receiver is given by

( ) ( )c cf 2 f f 2 fφ = − πα − − πβ

Where fc is the centre frequency, and andα β are positive constants. The actual signal

propagation delay from the transmitter to receiver is

( )Aα − βα + β

( )Bαβ

α + β ( )C α ( )D β

Answer: C

Exp: Phase response of pass band waveform

( ) ( )( )

c c

y

f 2 f f 2 f

d fGroup delay t

2 df

φ = − πα − − πβ

− φ= = α

π

Thus ' 'α is actual signal propagation delay from transmitter to receiver

23. Consider an FM signal ( ) [ ]c 1 1 2 2f t cos 2 f t sin 2 f t sin 2 f t.= π + β π + β π . The maximum

deviation of the instantaneous frequency from the carrier frequency fc is

( ) 1 1 2 2A f fβ + β ( ) 1 2 2 1B f fβ + β ( ) 1 2C β + β ( ) 1 2D f f+

Answer: A

Exp: Instantaneous phase ( )i c 1 1 2 2t 2 f t sin 2 f sin 2 f tφ = π + β π + β π

( ) ( )i i

c 1 1 1 2 2 2

d 1Instantaneous frequency f t t

dt 2

f f cos 2 f t f cos 2 f t

= φ ×π

= + β π + β π

Instantaneous frequency deviation 1 1 1 2 2 2f cos 2 f t f cos 2 f t= β π + β π

1 1 2 2Maximum f f f∆ = β + β

24. Consider an air filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this

waveguide, the cut-off frequency (in MHz) of TE21 mode is _________.

Answer: 7810MHz.

Exp: ( )2

c 21

2 210

10

10

C 2 1f TE

2 9 b

3 10 2 1

2 5 3

1.5 10 0.16 0.111

0.52 1.5 10

7.81GHz

7810 MHz.

2 = +

× = +

= × +

= × ×==



14

25. In the following figure, the transmitter Tx sends a wideband modulated RF signal via a

coaxial cable to the receiver Rx. The output impedance ZT of Tx, the characteristic

impedance Z0 of the cable and the input impedance ZR of Rx are all real.

Which one of the following statements is TRUE about the distortion of the received signal

due to impedance mismatch?

(A) The signal gets distorted if ZR ≠ Z0, irrespective of the value of ZT

(B) The signal gets distorted if ZT ≠ Z0, irrespective of the value of ZR

(C) Signal distortion implies impedance mismatch at both ends: ZT ≠ Z0 and ZR ≠ Z0

(D) Impedance mismatches do NOT result in signal distortion but reduce power transfer

efficiency

Answer: C

Exp: Signal distortion implies impedance mismatch at both ends. i.e.,

T 0

R 0

Z Z

Z Z

≠≠


26. The maximum value of f(x)=2x3 – 9x

2 +12x - 3 in the interval 0 x 3≤ ≤ is _______.

Answer: 6

Exp: ( ) [ ]1 2f x 6x 18x 12 0 x 1,2 0,3= − + = ⇒ = ∈

( ) ( ) ( ) ( )Now f 0 3 ; f 3 6 and f 1 2 ; f 2 1= − = = =

Hence, f(x) is maximum at x 3= and the maximum value is 6

27. Which one of the following statements is NOT true for a square matrix?

(A) If A is upper triangular, the eigenvalues of A are the diagonal elements of it

(B) If A is real symmetric, the eigenvalues of A are always real and positive

(C) If A is real, the eigenvalues of A and AT are always the same

(D) If all the principal minors of A are positive, all the eigenvalues of A are also positive

Answer: B

Transmitter

TZ

TX

Receiver

RZ

RX

0Characteristic Im pedance Z=



15

Exp: 1 1

Consider,A1 1

− −

which is real symmetric matrix

Characteristic equation is ( )2A I 0 1 1 0− λ = ⇒ + λ − =

( )1 1

0, 2 not positive

⇒ λ + = ±

∴λ = −

(B) is not true

(A), (C), (D) are true using properties of eigen values

28. A fair coin is tossed repeatedly till both head and tail appear at least once. The average

number of tosses required is __________________.

Exp: Let the first toss be Head.

Let x denotes the number of tosses( after getting first head) to get first tail.

We can summarize the even as:

Event x Probability(p(x))

(After getting first H)

T 1 1/2

HT 2 1/2*1/2=1/4

HHT 3 1/8

and so on………..

( ) ( )

( )

( )

( )

( )

x 1

1 1 1E x xp x 1x 2x 3x

2 4 8

1 1 1Let, S 1x 2x 3x I

2 4 8

1 1 1 1S 2x 3x II

2 4 8 16

I II gives

1 1 1 1 11 S

2 2 4 8 16

1

1 2S 112

12

S 2

E x 2

∞

=

= = + +

= + +

⇒ = + +

−

− = + + + +

⇒ = =−

⇒ =

⇒ =

∑

i.e. The expected number of tosses (after first head) to get first tail is 2 and same can be

applicable if first toss results in tail.

Hence the average number of tosses is 1+2 = 3.



16

29. Let X1, X2, and X3 be independent and identically distributed random variables with the

uniform distribution on [0, 1]. The probability PX1+ X2 ≤ X3 is ________.

Answer: 0.16

Exp: Given 1 2 3x x and x be independent and identically distributed with uniform distribution on

[ ]0,1

1 2 3

1 2 3 1 2 3

Let z x x x

P x x x P x x x 0

P z 0

= + −

⇒ + ≤ = + − ≤

= ≤

Let us find probability density function of random variable z.

Since Z is summation of three random variable 1 2 3x , x and x−

Overall pdf of z is convolution of the pdf of 1 2 3x x and x−

1 2pdf of x x+ is

3pdf of x is−

( ) ( )0

2 30

11

z 1 z 1 1P z 0 dz 0.16

2 6 6− −

+ +≤ = = = =∫

30. Consider the building block called ‘Network N’ shown in the figure.

Let C 100 F and R 10k= µ = Ω

Two such blocks are connected in cascade, as shown in the figure.

Network N

C+ +

R( )1V s

−

( )2V s

−

+

−( )1

V s Network N Network N

+

− −

+

( )3V s

1

O 1 2

1

−1



17

The transfer function ( )( )

3

1

v s

v s of the cascaded network is

( ) sA

1 s+ ( )

2

2

sB

1 3s s+ + ( )

2s

C1 s

+

( ) sD

2 s+

Answer: B

Exp: Two blocks are connected in cascade, Represent in s-domain,

( )( )

( ) [ ]

( )

( )( )

3

1

2 2

2 2 2

2 6 6 3 3

2 6 4 12 6 4

23

2

1

V s R . R

1 1 1V sR R R R

sc SC SC

R . R

1 1 R. 2R SC 1 1 RSC

SC SC SC

S C .R.R

1 2R SC RSC R S C

S .100 100 10 10 10 10 10 10

S 100 10 10 10 3S 100 10 10 1

V s S

V s 1 3S S

− −

− −

= + + + +

= + + +

= + + +

× × × × × × ×=

× × × × + + × × +

=+ +

31. In the circuit shown in the figure, the value of node voltage V2 is

(A) 22 + j 2 V (B) 2 + j 22 V (C) 22 – j 2 V (D) 2 – j 22 V

O10 0 V∠

+ −

1V

2V

4Ω

j3− Ω 6Ω j6ΩA

O4 0∠

+

−

R ( )3V sR

+

−

( )1V s

1

SC

1

SC



18

Answer: D

Exp:

1 2

KVL for V & V :

( )o

1 2

o

1 2

V V 10 0 .... 1

V V 10 0

− =

= +

KCL at super node:

( )

( ) ( )

( )

o 1 2 2

o1 2 2

6

oo2 2 2

o

2

2

V V V4 0 0 ... 2

j3 6 j6

V V V4 0

j3 V j6

V 10 0 V Vfrom 1 & 2 , 4 0

j3 6 j6

1 1 1 10V 4 0

j3 6 j6 j3

V 2 j22 Volts

− + + + =−

+ + =−

++ + =

−

+ + = + −

∴ = −

32. In the circuit shown in the figure, the angular frequency ω (in rad/s), at which the Norton

equivalent impedance as seen from terminals b-b' is purely resistive, is

__________________.

Answer: 2 r/sec

1Ω 1F

+~

−10cos tω( )Volts

0.5H

b

b '

6Ω Ωj6

2V

+ −

4Ω

V

o4 0 A− Ωj3

V010 0

Super node

1V2V

+ −+

−

+

−o10 0



19

Exp: Norton’s equivalent impedance

( ) ( )

( )

N

2 22

N N2 4 2

3

2

11* j .

12Z1 j .1

1 j .2

j 1

2 j j

2 j . 2 j2 jZ Z

2j 4

Equating imaginary term to zero i.e., 4 0

4 0 2r / sec

ω= +

ω+ ω

ω= +

+ ω ω

ω − − ω ω + ω− ω + ω = ⇒ = ω − ω ω + ω

ω − ω =

⇒ω ω − = ⇒ ω =

33. For the Y-network shown in the figure, the value of ( )1R inΩ in the equivalent ∆ -network is

_____________________.

Answer: 10Ω

Exp:

( )( ) ( )( ) ( )( )1

1

7.5 5 3 5 7.5 3R

7.5

R 10

+ += Ω

= Ω

34. The donor and accepter impurities in an abrupt junction silicon diode are 1 x 1016

cm-3

and 5

x 1018

cm-3

, respectively. Assume that the intrinsic carrier concentration in silicon ni = 1.5 x

1010

cm-3

at 300 K, kT

26mVq

= and the permittivity of silicon 12

si 1.04 10 F/ cm.−ε = × The

built-in potential and the depletion width of the diode under thermal equilibrium conditions,

respectively, are

(A) 0.7 V and 1 x 10-4

cm (B) 0.86 V and 1 x 10-4

cm

(C) 0.7 V and 3.3 x 10-5

cm (D) 0.86 V and 3.3 x 10-5

cm

1R

5Ω 3Ω

7.5Ω

1Ω 1F

0.5H

b

1b

uZ

1R

5Ω 3Ω

Ω7.5



20

Answer: D

Exp:

( )18 16

A Dbi T 22

10i

N N 5 10 1 10V V ln 26mv ln

n 1.5 10

× × × = = ×

5S bi A D

A D

0.859V

2 V N NW 3.34 10 cm

q N N

−

=

ε += = ×

35. The slope of the ID vs VGS curve of an n-channel MOSFET in linear regime is 3 110− −Ω at

DSV 0.1V.= . For the same device, neglecting channel length modulation, the slope of the

DI vs VGS curve ( )in A / V under saturation regime is approximately ___________.

Answer: 0.07

Exp: In linear region, ( )2

DSD GS T DS

VI k V V V

2

= − −

23 DSD

DS DS

GS

3

VI10 kV V is small, is neglected

V 2

10K 0.01

0.1

−

−

∂= =

∂

⇒ = =

∵

In saturation region, ( )2

D GS T

1I k V V

2= −

( )D GS T

D

GS

kI V V

2

I k 0.010.07

V 2 2

= −

∂= = =

∂

36. An ideal MOS capacitor has boron doping-concentration of 1015

cm-3

in the substrate. When a

gate voltage is applied, a depletion region of width 0.5 µm is formed with a surface (channel)

potential of 0.2 V. Given that -14

o8.854 × 10 F/cm ε = and the relative permittivities of silicon

and silicon dioxide are 12 and 4, respectively, the peak electric field (in V/µm) in the oxide

region is __________________.

Answer: 2.4

Exp: s

2 0.2E 0.8v / m

0.5

×= = µ

sox s

ox

EE E 2.4v / m

E= = µ



21

37. In the circuit shown, the silicon BJT has 50β = . Assume VBE =0.7 V and VCE(sat) = 0.2 V.

Which one of the following statements is correct?

(A) For RC = 1 k ,Ω the BJT operates in the saturation region

(B) For RC = 3 k ,Ω , the BJT operates in the saturation region

(C) For RC =20 k ,Ω , the BJT operates in the cut-off region

(D) For RC =20 k ,Ω , the BJT operates in the linear region

Answer: B

Exp: KVL in base loop,

( )

( )

B

B

C B

CE

C

C

C

5 I 50k 0.7 0

5 0.7I 80 A

50k

I I 50 86 A 4.3mA

10 V sat 10 0.2R

I 4.3mA

R 2279 and the BJT is in saturation

− − =

−= = µ

⇒ = β = × µ =

− −∴ = =

= Ω

38. Assuming that the Op-amp in the circuit shown is ideal, VO is given by

( ) 1 2

5A V 3V

2− ( ) 1 2

5B ZV V

2− ( ) 1 2

3 7C V V

2 2− + ( ) 1 2

11D 3V V

2− +

Answer: D

Exp: Virtual ground and KCL at inverting terminal gives

2 02 1 2

0 2 2 2 1

0 1 2

V VV V V0

R 2R 3R

V V V V V

3R R 3R 2R R

11V 3V V

2

−−+ + =

= + + −

= − +

10V

CR

50kΩ

BR

5V

3R

OV

R2

V2R

R1V −

+

+

−

•2V

2R

R

3R

R

•••

1V

2V

oV



22

39. For the MOSFET M1 shown in the figure, assume W/L = 2, VDD = 2.0 V, 2

n oxC 100 A / Vµ = µ

and VTH = 0.5 V. The transistor M1 switches from saturation region to linear region when Vin

(in Volts) is_________________.

Answer: 1.5

Exp: Transistor 1

m switch from saturation to linear

( )

( )

( ) ( )

DS GS T DS 0 GS i

DS 0 i T

2

D n GS T

26DD oGS

2i 6

i

i

V V V ;where V V and V V

V V V V

1 wDrain current I cos V V

2 L

V V 1100 10 2 V 0.5

10K 2

2 V 0.5100 10 V 0.5

10K

V 1.5V

−

−

⇒ = − = =

∴ = = −

= µ −

−= × × × −

− −= × −

⇒ =

40. If WL is the Word Line and BL the Bit Line, an SRAM cell is shown in

DDV

R 10k= Ω

outV

1Min

V

WL

DDVBLBL

( )A

WL

DDVBLBL

( )B



23

Answer: B

Exp: For an SRAM construction four MOSFETs are required (2-PMOS and 2-NMOS) with

interchanged outputs connected to each CMOS inverter. So option (B) is correct.

41. In the circuit shown, W and Y are MSBs of the control inputs. The output F is given by

( )A F W X WX Y Z= + + ( )B F W X WX YZ= + +

( )C F W XY WXY= + ( ) ( )D F W X YZ= +

Answer: C

Exp:

The output of the first MUX =

( )cc cc

cc

W V WX.V

WX WX V logic1

W X

× +

+ =

= ⊕

∵

Let Q W X= ⊕

4 :1 MUX 4 :1 MUX

0I

1I

2I

3I

CCV

Q

0I

1I

2I

3I

QF

W X Y Z

WL

DDVBL

BL

( )D

WL

DDV

BLBL

( )C

0I

1I

2I

3I

4 : 1

MUX Q

W X

ccV

0I

1I

2I

3I

4 : 1

MUX Q

Y Z

F



24

The output of the second MUX = Q.Y Z Q. Y Z+

( )Q.Y Z Z

Q.Y.1 Q.Y

= +

= =

Put the value of Q in above expression

( )WX WX .Y

W X.Y WX.Y

= +

= +

42. If X and Y are inputs and the Difference (D = X – Y) and the Borrow (B) are the outputs,

which

one of the following diagrams implements a half-subtractor?

Answer: A

Exp:

X Y D B

0

0

1

1

0

1

0

1

0

1

1

0

0

1

0

0

So, D X Y XY XY and B X.Y= ⊕ = + =

Y0

I

1I

2 :1MUX D

S

SX

Y 0I

1I

2 :1MUX B

( )A

X0

I

1I

2 :1MUX D

S

S

X

Y

0I

1I

2 :1MUX B

( )B

Y0

I

1I

2 :1MUX B

S

SX

Y 0I

1I

2 :1MUX D

( )C

X0

I

1I

2 :1MUX B

S

S

X

Y

0I

1I

2 :1MUX D

( )B



25

43. Let ( ) ( ) ( ) ( ) ( ) ( ) ( )1 11 1

1 2 1 2H z 1 pz ,H z 1 qz ,H z H z r H z− −− −= − = − = + . The quantities p, q, r

are real numbers. Consider 1 1

p ,q , r 1.2 4

= = < If the zero of H(z) lies on the unit circle, then

r = ________

Answer: -0.5

Exp: ( ) ( ) 11

1H z 1 Pz−−= −

( ) ( )

( ) ( )( )

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

( )

11

2

1 1 1

1 1 1 1 1 1

H z 1 qz

1 qz r 1 Pz 1 r q rp z1 1H z r

1 Pz 1 qz 1 Pz 1 Pz 1 Pz 1 Pz

q rpzero of H z

1 r

−−

− − −

− − − − − −

= −

− + − + − += + = =

− − − − − −

+=

+

Since zero is existing on unit circle

q rp q rp

1 or 11 r 1 r

+ +⇒ = = −

+ +

1 r 1 r

4 2 4 21 or 11 r 1 r

− + − += = −

+ +

1 r 1 r1 r or 1 r

4 2 4 2

5 r 5 3 3r 1r or r r 0.522 2 4 4 2

5r is not possible

2

− + = + − + = − −

−⇒ = − ⇒ = − = = − ⇒ = −

= −

X

X. Y XY

X Y

= += ⊕

0I

1I

2 :1

MUX

Y

B

0I

1I

2 :1

MUX

Y

D

X. Y X.0

X.Y 0

X .Y

= +

= +

=



26

44. Let h(t) denote the impulse response of a causal system with transfer function 1

s 1+. Consider

the following three statements.

S1: The system is stable.

S2: ( )

( )h t 1

h t

+ is independent of t for t 0.

S3: A non-causal system with the same transfer function is stable.

For the above system,

(A) Only S1 and S2 are true (B) only S2 and S3 are true

(C) Only S1 and S3 are true (D) S1, S2 and S3 are true

Answer: A

Exp: ( ) ( ) ( ) ( )t1h t H s h t e u t

s 1

−↔ = ⇒ =+

1S : System is stable (TRUE)

Because h(t) absolutely integrable

2S :

( )( )

h t 1

h t

+ is independent of time (TRUE)

( )t 1

1

t

ee

e

− +−

− ⇒ (independent of time)

3S : A non-causal system with same transfer function is stable

( )t1e u t

s 1

−↔ − −+

(a non-causal system) but this is not absolutely integrable thus

unstable.

Only 1 2S and S are TRUE

45. The z-transform of the sequence x[n] is given by ( )( )21

1X z ,

1 2z−=

−, with the region of

convergence [ ]z 2. Then, x 2> is ______.

Answer: 12

Exp(1):

( )( ) ( ) ( )2 1 11

1 1 1X z

1 2z 1 2z1 2z− −−

= =− −−

[ ] [ ] [ ]

[ ] ( )

[ ] ( )

n n

nn kK

k 0

22 kk 0 2 1 1 2 0

k 0

x n 2 u n * 2 u n

x n 2 .2

x 2 2 .2 2 .2 2 .2 2 .2 4 4 4 12

−

=

−

=

=

=

⇒ = = + + = + + =

∑

∑



27

Exp(2):

( )( ) ( )

2

2 21

1 ZX z

Z 21 2Z−= =

−−

( )

( )

( )

( )

( )

( )

1

u z v z

nn n

m n m n n

m 0

nm n m n

m 0

Z ZX n Z .

Z 2 Z 2

u .V usin g conduction theorem and u 2 ; v 2

2 .2 2 n 1

x 2 12

−

↓ ↓

−=

−

=

= − −

= = =

= = +

∴ =

∑

∑

46. The steady state error of the system shown in the figure for a unit step input is _______.

Answer: 0.5

Exp: ( ) ( )4 2Given G s ;H s

s 2 s 4= =

+ +

( ) ( )ps 0

ps 0

p

ss

p

ss

ss

For unit step input,

k limG s H s

4 2k lim

s 2 s 4

k 1

ASteady state error e

1 k

1e

1 1

1e 0.50

2

→

→

=

= + +

=

=+

=+

= ⇒

( )R s

( )r t

+

−

( )E s

( )e tK 4=

1

s 2+( )C s

( )C t

2

s 4+



28

47. The state equation of a second-order linear system is given by

( ) ( ) ( ) 0x t Ax t , x 0 x= =

( ) ( )t t 2 t

0 0t t 2 t

0e e e1For x , x t and for x , x t

11 e e 2e

− − −

− − −

− = = = = − − − +

when ( )0

3x , x t

5

=

is

( )t 2 t

t 2 t

8e 11eA

8e 22e

− −

− −

− + −

( )t 2t

t 2t

11e 8eB

11e 16e

− −

− −

− − +

( )t 2t

t 2t

3e 5eC

3e 10e

− −

− −

− − +

( )t 2 t

t 2 t

5e 3eD

5e 6e

− −

− −

− − +

Answer: B

Exp: Apply linearity principle,

( )

( )

t t 2 t

t t 2t

t 2t

t 2 t

3 1 0a b s

5 1 1

a 3 ; b 8

e e ex t 3

e e 2e

11e 8ex t

11e 16e

− − −

− − −

− −

− −

= + −

= =

−⇒ = +

− − +

−⇒ =

− +

48. In the root locus plot shown in the figure, the pole/zero marks and the arrows have been

removed. Which one of the following transfer functions has this root locus?

( ) ( ) ( ) ( )s 1

As 2 s 4 s 7

++ + +

( ) ( ) ( ) ( )s 4

Bs 1 s 2 s 7

++ + +

( ) ( ) ( ) ( )s 7

Cs 1 s 2 s 4

++ + +

( ) ( ) ( )( ) ( )s 1 s 2

Ds 7 s 4

+ ++ +

jω

2

1 σ



29

Answer: B

Exp:: ( )

( )( )( )s 4

For transfer functions 1 s 2 s 3

++ + +

From pole zero plot

49. Let X(t) be a wide sense stationary (WSS) random process with power spectral density SX(f).

If Y(t) is the process defined as ( ) ( )Y t X 2t 1= − , the power spectral density SY(f) is

( ) ( ) j f

Y X

1 fA S f S e

2 2

− π =

( ) ( ) j f / 2

Y X

1 fB S f S e

2 2

− π =

( ) ( )Y X

1 fC S f S

2 2

=

( ) ( ) j2 f

Y X

1 fD S f S e

2 2

− π =

Answer: C

Exp: Shifting in time domain does not change PSD. Since PSD is Fourier transform of

autocorrelation function of WSS process, autocorrelation function depends on time

difference.

( ) ( ) ( )

( ) ( ) ( )

x x

y x

X t R z S f

1 fY t X 2t 1 R 2 S

2 2

↔ ↔

= − ↔ ζ ↔

[time scaling property of Fourier transform]

50. A real band-limited random process X(t) has two-sided power spectral density

( ) ( ) 610 3000 f W atts / Hz for f 3 kHz

x 0 otherwiseS f− − ≤=

Where f is the frequency expressed in Hz. The signal X(t)modulates a carrier cos16000 tπ

and the resultant signal is passed through an ideal band-pass filter of unity gain with centre

frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is _______.

Answer: 2.5

Exp:

( )xS f

( )xS f

−3 +3 ( )f in KHz

−× 33 10 watts

σ

jω

1−2−4−7−



30

After modulation with ( )cos 16000 tπ

( ) ( ) ( )y x c x c

1S f S f f S f f

4= − + +

This is obtain the power spectral density Random process y(t), we shift the given power

spectral density random process x(t) to the right by fc shift it to be the left by fc and the two

shifted power spectral and divide by 4.

After band pass filter of center frequency 8 KHz and BW of 2 kHz

Total output power is area of shaded region

[ ][ ]

[ ]

3 3 3 3

2 Area of one side portion

2 Area of triangle Area of rec tan gle

12 2 10 0.5 10 2 10 1 10

2

2

0.5 2 2.5watts

− −

=

= +

− × × × × + × × × =

= + =

51. In a PCM system, the signal ( ) ( ) ( ) m t sin 100 t cos 100 t= π + π V is sampled at the Nyquist

rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum

data rate of the PCM system in bits per second is _____.

Answer: 200

Exp: Nyquist rate 2 50Hz= ×

( ) ( ) ( )

max min

100 samples / sec

2 2m t m tL

L 0.75

2 2L 3.77 4

0.75

=

− −−∆ = ⇒ =

= = =

No. of bits required to encode ‘4’ levels = 2 bits/level

Thus data rate 2 100 200 bits / sec= × =

− − −11 8 5 5 8 11

−× 31.5 10 /2

( )f in KHz

− − −9 8 7

−× 31.5 10 /2

7 8 9



31

52. A binary random variable X takes the value of 1 with probability 1/3. X is input to a cascade

of 2 independent identical binary symmetric channels (BSCs) each with crossover probability

1/2. The outputs of BSCs are the random variables Y1 and Y2 as shown in the figure.

The value of H(Y1) + H(Y2) in bits is___________.

Answer: 2

Exp: 1 2Let P x 2 , P x 0

3 3= = = =

( )

1 1 2

1 1 1 1 1 1 1

1

to find H Y we need to know P y 0 and P y 1

P Y 0 P Y 0 / x 0 P x 0 P y 0 / x 1 P x 1

1 1 1 2 1.

2 3 2 3 2

1P y 1

2

= =

= = = = = + = = =

= + × =

= =

( ) 2 2

1 2 2

1 1H y log log 1

2 2⇒ = + =

Similarly

2 2

2

1 2

1 1P y 0 and P y 1

2 2

H y 1

H y H y 2 bits

= = = =

⇒ =

⇒ + =

53. Given the vector ( ) ( ) ( ) ( )x yˆ ˆA cosx sin y a sin x cos y a= + , where

xa , ya denote unit vectors

along x,y directions, respectively. The magnitude of curl of A is ________

Answer: 0

Exp (1):

x y zˆ ˆ ˆa a a

Curl Ax y z

cos x sin y sin x cos y 0

∂ ∂ ∂=

∂ ∂ ∂

0

Curl A 0

=

∴ =

XBSC 1

YBSC

2Y



32

Exp(2):

( ) ( ) ( )

x y

x y z

x y z

ˆ ˆGiven A cos x sin ya sin x cos y a

a a a

A / x / y / z

cos x sin y sin x cos y 0

a 0 a 0 a cos x cos y cos x cos y 0

A 0

= +

∇ × = ∂ ∂ ∂ ∂ ∂ ∂

= − + − =

∴ ∇ × =

54. A region shown below contains a perfect conducting half-space and air. The surface current

sK

on the surface of the perfect conductor is sˆK x2=

amperes per meter. The tangential H

field in the air just above the perfect conductor is

(A) ( )ˆ ˆx z 2+ amperes per meter (B) x2 amperes per meter

(C) z2− amperes per meter (D) z2 amperes per meter

Answer: D

Exp: Given medium (1) is perfect conductor

Medium (2) is air

1

H 0∴ =

From boundary conditions

( )

( )

1 2 n S

1

S x

n y

2 y x

x x y y z z y x

x z z x x

z

z

H H a K

H 0ˆK 2a

a a

ˆH a 2a

H a H a H a a 2a

H a H a 2a

H 2

H 2a

− × =

==

=

− × =

− + + × =

− + =∴ =

=

y

SK

Airx

Perfect conductor



33

55. Assume that a plane wave in air with an electric field ( ) yˆE 10cos t 3x 3z a= ω − −

V/m is

incident on a non-magnetic dielectric slab of relative permittivity 3 which covers the region.

Z > 0 The angle of transmission in the dielectric slab is _________________ degrees.

Answer: 30

Exp: ( ) yGiven E 10cos t 3x 3z a= ω − −

( )x y zJ x cos ycos zcos

0

x x

y y

z z

2 2 2 2

x y z

2

z z z i

i 2t

t 1 t

o

t t

E E e

So, cos 3

cos 0

cos 3

9 3 13

3cos 3 cos 61.28

13

sin E sin 61.28 3 0.8769sin

sin E sin 1 3

30.4 30

− β θ + θ + θ=β = β θ =β = β θ =

β = β θ =

β + β + β = β

+ = β ⇒ β =

β θ = ⇒ θ = ⇒ θ = = θ

θ= ⇒ = ⇒ = θ

θ θ

θ = ⇒ θ



1


1. Which of the following options is the closest in meaning to the word underlined in the

sentence

below? In a democracy, everybody has the freedom to disagree with the government.

(A) Dissent (B) Descent (C) Decent (D) Decadent

Answer: A

Exp: Dissent is to disagree

2. After the discussion, Tom said to me, 'Please revert!’ He expects me to _________.

(A) Retract (B) Get back to him

(C) Move in reverse (D) Retreat

Answer: B

Exp: Revert means set back

3. While receiving the award, the scientist said, "I feel vindicated". Which of the following is

closest in meaning to the word ‘vindicated’?

(A) Punished (B) Substantiated (C) Appreciated (D) Chastened

Answer: B

Exp: Vindicate has 2 meanings

1. Clear of blain

2. Substantiate, justify

4. Let ( ) n mf x,y x y P. If x= = is doubled and y is halved, the new value of f is

( ) n mA 2 P− ( ) m nB 2 P− ( ) ( )C 2 n m P− ( ) ( )D 2 m n P−

Answer: A

Exp:

m

n n m n my(2x) 2 x y

2

− × = ×

5. In a sequence of 12 consecutive odd numbers, the sum of the first 5 numbers is 425. What is

the sum of the last 5 numbers in the sequence?

Answer: 495

Exp: Let consecutive odd numbers be a-10, a-8, a-6, a-4, a-2, a, ……a+12

Sum of 1st 5 number = 5a-30=425 a=91⇒

Last 5 numbers=(a+4)+(a+6)+…….+(a+12)

=(95+97+99+101+103)= 495



2


6. Find the next term in the sequence: 13M, 17Q, 19S, ___

(A) 21W (B) 21V (C) 23W (D) 23V

Answer: C

Exp:

13 M

17(13 4) Q(M 4)

19(17 2) S(Q 2)

23(19 4) W (s 4)

23W

+ +

+ +

+ = +

⇒

7. If ‘KCLFTSB’ stands for ‘best of luck’ and ‘SHSWDG’ stands for ‘good wishes’, which of

the following indicates ‘ace the exam’?

(A) MCHTX (B) MXHTC (C) XMHCT (D) XMHTC

Answer: B

Exp: KCLFTSB SHSWDG

Reverse order: Reverse order:

BCS TOF LUCK GO OD W I S HES

Ace the exam

Reverse order should be

MAXE EHT ECA

Looking at the options we have M X H T C

8 . Industrial consumption of power doubled from 2000-2001 to 2010-2011. Find the annual rate

of increase in percent assuming it to be uniform over the years.

(A) 5.6 (B) 7.2 (C) 10.0 (D) 12.2

Answer: B

Exp:

n

10

rA P 1

100

A 2P

r2 1

100

r 7.2

= +

=

= +

∴ =



3

9. A firm producing air purifiers sold 200 units in 2012. The following pie chart presents the

share of raw material, labour, energy, plant & machinery, and transportation costs in the total

manufacturing cost of the firm in 2012. The expenditure on labour in 2012 is Rs. 4,50,000. In

2013, the raw material expenses increased by 30% and all other expenses increased by 20%.

What is the percentage increase in total cost for the company in 2013?

Answer: 22%

Exp: Let total cost in 2012 is 100

Raw material increases in 2013 to 1.3 x 20=26

Other Expenses increased in 2013 to 1.2 x 80=96

Total Cost in 2013 =96+26=122

Total Cost increased by 22%

Hint:Labour cost (i.e, 4,50,000) in 2012 is redundant data.

10. A five digit number is formed using the digits 1,3,5,7 and 9 without repeating any of them.

What is the sum of all such possible five digit numbers?

(A) 6666660 (B) 6666600 (C) 6666666 (D) 6666606

Answer: B

Exp: 1 appears in units place in 4! Ways

Similarly all other positions in 4! Ways

Same for other digits.

Sum of all the numbers = (11111) X 4! (1+3+5+7+9) = 6666600

Labour

15%

Plant and

machinery

30%

Energy

25%

Raw Material

20%

Transportation30%

−



4

Q.No. 1 – 25 Carry One Mark Each

1. The series n 0

1

n!

∞

=∑ converges to

(A) 2 ln2 (B) 2 (C) 2 (D) e

Answer: D

Exp: n 0

1 1 11 .........

n! 1! 2!

∞

=

= + + +∑

2

x x xe as e 1 ......., x in R

1! 2!= = + + + ∀

2. The magnitude of the gradient for the function ( ) 2 2 3f x,y,z x 3y z= + + at the point (1,1,1) is

_________.

Answer: 7

Exp: ( ) ( ) ( ) ( ) ( )( ) ( )2

P 1,1,1 P 1,1,1f i 2x j 6y k 3z∇ = + +

( )

P

2i 6 j 3k

f 4 36 9 7

= + +

∇ = + + =

3. Let X be a zero mean unit variance Gaussian random variable. E X is equal to _____

Answer: 0.8

Exp: ( ) ( )2x

21

X ~ N 0,1 f x e , x2

−⇒ = −∞ < < ∞

π

( )2x

2 2

0

u

0

E x x .f x dx

1x x e dx

2

2 2e du 0.797 0.8

2

∞

−∞

∞ −

∞ −

∴ =

=π

= = =ππ

∫

∫

∫

4. If a and b are constants, the most general solution of the differential equation

2

2

d x dx2 x 0

dt dt+ + = is

( ) tA ae− ( ) t tB ae bte− −+ ( ) t tC ae bte−+ ( ) 2tD ae−

Answer: B

Exp: ( )

2

t

A.E : m 2m 1 0 m 1, 1

general solution is x a bt e−

− + + = ⇒ = − −

∴ = +



5

5. The directional derivative of ( ) ( ) ( )xyf x,y x y at 1,1

2= + in the direction of the unit vector at

an angle of 4

π with y-axis, is given by _____ .

Answer: 3

Exp: ( )2 2

2 21 2xy y x 2xyf x y xy f i j

2 2 2

+ += + ⇒∇ = +

( ) 3 3at 1,1 , f i j

2 2∇ = +

e = unit vector in the direction i.e., making an angle of 4

π with y-axis

sin i cos j4 4

3 1ˆdirectional derivative e. f 2 322

π π = +

∴ = ∇ = =

6. The circuit shown in the figure represents a

(A) Voltage controlled voltage source (B) Voltage controlled current source

(C) Current controlled current source (D) Current controlled voltage source

Answer: C

Exp:

The dependent source represents a current controlled current source

7. The magnitude of current (in mA) through the resistor R2 in the figure shown is_______.

iI

l iA I R

2R

1kΩ

1R

2 kΩ

4R 3kΩ

10mA3

R

4 kΩ 2 mA

1 1A I



6

Answer: 2.8

Exp: By source transformation

By KVL,

20 10k.I 8 0

28I

10k

I 2.8mA

− + =

⇒ =

⇒ =

8. At T = 300 K, the band gap and the intrinsic carrier concentration of GaAs are 1.42 eV and

106 cm

-3, respectively. In order to generate electron hole pairs in GaAs, which one of the

wavelength ( )Cλ ranges of incident radiation, is most suitable? (Given that: Plank’s constant

is 6.62 × 10-34

J-s, velocity of light is 3 × 1010

cm/s and charge of electron is 1.6 × 10-19

C)

(A) 0.42 mµ < Cλ < 0.87 µm (B) 0.87 mµ < Cλ < 1.42 µm

(C) 1.42 mµ < Cλ < 1.62 µm (D) 1.62 mµ < Cλ < 6.62 µm

Answer: A

Exp: 34 8

19

hC 6.62 10 3 10E 0.87 m

1.42 1.6 10

−

−

× × ×= ⇒ λ = = µ

λ × ×

9. In the figure ( )iln ρ is plotted as a function of 1/T, where iρ the intrinsic resistivity of silicon,

T is is the temperature, and the plot is almost linear.

The slope of the line can be used to estimate

(A) Band gap energy of silicon (Eg)

(B) Sum of electron and hole mobility in silicon ( )n pµ + µ

(C) Reciprocal of the sum of electron and hole mobility in silicon ( ) 1

n p

−µ + µ

(D) Intrinsic carrier concentration of silicon ( )in

( )iln ρ

1/ T

+− +

−8V20V

Ω2k

= Ω2

R 1k Ω4k

Ω3k

I



7

Answer: A

Exp:

3Eg/kT2

i

i

n T e and

1

−

ι

α

ρ αη

∴ From the graph, Energy graph of iS can be estimated

10. The cut-off wavelength (in µm) of light that can be used for intrinsic excitation of a

semiconductor material of bandgap Eg= 1.1 eV is ________

Answer: 1.125

Exp: hC

E =λ

34 8

19

6.6 10 3 101.125 m

1.1 1.6 10

−

−

× × ×⇒ λ = = µ

× ×

11. If the emitter resistance in a common-emitter voltage amplifier is not bypassed, it will

(A) Reduce both the voltage gain and the input impedance

(B) Reduce the voltage gain and increase the input impedance

(C) Increase the voltage gain and reduce the input impedance

(D) Increase both the voltage gain and the input impedance

Answer: B

Exp: When a CE amplifier’s emitter resistance is not by passed, due to the negative feedback the

voltage gain decreases and input impedance increases

12. Two silicon diodes, with a forward voltage drop of 0.7 V, are used in the circuit shown in the

figure. The range of input voltage Vi for which the output voltage 0 iV V= , is

( ) iA 0.3V V 1.3V− < < ( ) iB 0.3V V 2V− < <

( ) iC 1.0V V 2.0V− < < ( ) iD 1.7V V 2.7V− < <

Answer: D

Exp: i 1 2When V 1.7V ; D ON and D OFF< − − −

0

i 1 2

0

i 1 2

0 i

V 1.7V

When V 2.7V;D OFF & D ON

V 2.7V

When 1.7 V 2.7V, Both D & D OFF

V V

∴ = −> − −

∴ =− < <

∴ =

R

+

iV

−

1D

1V− ± ± 2 V

D2O

V

+

−



8

13. The circuit shown represents:

(A) A bandpass filter (B) A voltage controlled oscillator

(C) An amplitude modulator (D) A monostable multivibrator

Answer: D

14. For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then

(A) Droop rate decreases and acquisition time decreases

(B) Droop rate decreases and acquisition time increases

(C) Droop rate increases and acquisition time decreases

(D) Droop rate increases and acquisition time increases

Answer: B

Exp: Capacitor drop rate dv

dt=

For a capacitor, dv 1

dt c∝

∴ Drop rate decreases as capacitor value is increased

For a capacitor, Q cv i t t c= = × ⇒ ∝

∴ Acquisition time increases as capacitor value increased

15. In the circuit shown in the figure, if C=0, the expression for Y is

( )A Y A B A B= + ( )B Y A B= +

( )C Y A B= + ( )D Y A B=

iv

2C 12 V+

12 V−

1C

0v

1R

2R

2 V−

−

+

CAB

BA

Y



9

Answer: A

Exp:

Y 1.A B

A B

A B A B AB AB

=

=

= ⊕ = + +

16. The output (Y) of the circuit shown in the figure is

( )A A B C+ +

( )B A B . C A .C+ +

( )C A B C+ +

( )D A . B . C

Answer: A

Exp:

D0V

B C

A

B

C

( )Output Y

A

DDV

A B C

A

B

C

( )output Y

C 0=

A

B

A

B A B⋅

A B⋅

A B⋅

A B A B A B⋅ + ⋅ =

Y

1



10

This circuit is CMOS implementation

If the NMOS is connected in series, then the output expression is product of each input with

complement to the final product.

So, Y A.B .C

A B C

=

= + +

17. A Fourier transform pair is given by

[ ]n j6 fFT

j2 f

2 Aeu n 3

231 e

3

− π

− π

+ ⇔ −

where u[n] denotes the unit step sequence. The value of A is _________.

Answer: 3.375

Exp: [ ] [ ]n

2x n u n 3

3

= +

( )

3

j3n

j j n

jn 3

3

2.e

2 3X e .e

231 e

3

3 27A 3.375

2 8

−Ω

∞Ω − Ω

− Ω=−

= = −

⇒ = = =

∑

18. A real-valued signal x(t) limited to the frequency band w

f2

≤ is passed through a linear time

invariant system whose frequency response is

( )j4 f w

e , f2

H fw

0, f2

− π ≤= >

The output of the system is

( ) ( )A x t 4+ ( ) ( )B x t 4− ( ) ( )C x t 2+ ( ) ( )D x t 2−

Answer: D

Exp: Let ( )x t Fourier transform be ( )x t

( ) ( ) ( )[ ]( ) ( ) ( )( ) ( )( ) ( )

j4 f

y t x t * h t convolution

Y f X f .H f

Y f e .X f

y t x t 2

− π

=

⇒ =

⇒ =

⇒ = −

( )x t

( )y t( )h t



11

19. The sequence x[n] = 0.5n u[n], where u[n] is the unit step sequence, is convolved with itself to

obtain y[n]. Then ( )n

y n∞

=−∞∑ _________________.

Answer: 4

Exp: [ ] [ ] [ ]y n x n * x n=

( ) [ ]( ) ( ) ( )( )

( ) [ ]

[ ] ( )

j

j j j

j

j j

j j n

h

j0

n

Let Y e is F.T. pair with y n

Y e X e .X e

1 1Y e .

1 0.5e 1 0.5e

also Y e y n .e

1 1y n Y e . 4

0.5 0.5

Ω

Ω Ω Ω

Ω− Ω − Ω

∞Ω − Ω

=−∞

∞

=−∞

⇒ =

=− −

=

⇒ = = =

∑

∑

20. In a Bode magnitude plot, which one of the following slopes would be exhibited at high

frequencies by a 4th order all-pole system?

(A) – 80 dB/decade (B) – 40 dB/decade

(C) +40 dB/decade (D) +80 dB/decade

Answer: A

Exp: → In a BODE diagram, in plotting the magnitude with respect to frequency, a pole introduce a

line 4 slope 20dB / dc−

→ If 4th order all-pole system means gives a slope of ( )20 * 4 dB / dec i.e. 80dB / dec− −

21. For the second order closed-loop system shown in the figure, the natural frequency (in rad/s)

is

(A) 16 (B) 4 (C) 2 (D) 1

Answer: C

Exp: ( )( ) 2

Y s 4Transfer function

U s S 4s 4=

+ +

If we compare with standard 2nd

order system transfer function

2

n

2 2

n n

2

n n

wi.e.,

s 2 w s w

w 4 w 2 rad / sec

+ ξ +

= ⇒ =

( )U s +

− ( )4

S S 4+( )Y s



12

22. If calls arrive at a telephone exchange such that the time of arrival of any call is independent

of the time of arrival of earlier or future calls, the probability distribution function of the total

number of calls in a fixed time interval will be

(A) Poisson (B) Gaussian (C) Exponential (D) Gamma

Answer: A

Exp: Poisson distribution: It is the property of Poisson distribution.

23. In a double side-band (DSB) full carrier AM transmission system, if the modulation index is

doubled, then the ratio of total sideband power to the carrier power increases by a factor of

_________________.

Answer: 4

Exp: 2Ratio of total side band power

Carrier powerα µ

If it in doubled, this ratio will be come 4 times

24. For an antenna radiating in free space, the electric field at a distance of 1 km is found to be 12

mV/m. Given that intrinsic impedance of the free space is 120 ,π Ω the magnitude of average

power density due to this antenna at a distance of 2 km from the antenna (in nW/m2 )

is________________.

Answer: 47.7

Exp: Electric field of an antenna is

0

2 3

Radiation inductive Electrostaticfield field field

1 22

2 1

2 82

I dl J 1 JE sin

4 r r r

1E

r

E rE 6mv / m

E r

1 E 1 36 10P 47.7 nw / m

2 2 120

θ

↓ ↓ ↓

−

η β = θ + − π β

∴ α

= ⇒ =

×= = =

η π

25. Match column A with column B.

Column A Column B

(1) Point electromagnetic source (P) Highly directional

(2) Dish antenna (Q) End free

(3) Yagi-Uda antenna (R) Isotropic

(A)

1 P

2 Q

3 R

→→→

(B)

1 R

2 P

3 Q

→→→

(C)

1 Q

2 P

3 R

→→→

(D)

1 R

2 Q

3 P

→→→



13

Answer: B

Exp: 1. Point electromagnetic source, can radiate fields in all directions equally, so isotropic.

2. Dish antenna → highly directional

3. Yagi – uda antenna End fire→

Figure: Yagi-uda antenna


26. With initial values y(0) = y’(0)=1 the solution of the differential equation 2

2

d y dy4 4y 0

dx dx+ + =

at x = 1 is ________

Answer: 0.54

Exp: 2A.E : m 4m 4 0 m 2, 2+ + = ⇒ = − −

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

( ) ( ) ( ) ( )

( )

2x

' 2x 2x

'

2x

2

solutions is y a bx e ....... 1

y a bx 2e e b ........ 2

u sin g y 0 1; y 0 1, 1 and 2 gives

a 1 and b 3

y 1 3x e

at x 1, y 4e 0.541 0.54

−

− −

−

−

∴ = +

= + − +

= == =

∴ = += = =

27. Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-

office has a probability 1

5 of losing an incoming parcel, independently of all other parcels.

Given that a parcel is lost, the probability that it was lost by the second post-office

is_________.

Answer: 0.44

Exp: Parcel will be lost if

a. it is lost by the first post office

b. it is passed by first post office but lost by the second post office

Prob(parcel is lost) = 1 4 1 9

x5 5 5 25

+ =

P (Parcel lost by second post if it passes first post office)= P (Parcel passed by first post

office) x P(Parcel lost by second post office)

R F



14

=4 1 4

5 5 25× =

Prob(parcel lost by 2nd

post office | parcel lost)=4 / 25 4

0.449 / 25 9

= =

28. The unilateral Laplace transform of ( ) 2

1f t is

s s 1+ +. Which one of the following is the

unilateral Laplace transform of g(t) = t. f(t)?

( )( )22

sA

s s 1

−

+ + ( ) ( )

( )22

2s 1B

s s 1

− +

+ + ( )

( )22

SC

s s 1+ + ( )

( )22

2S 1D

s s 1

+

+ +

Answer: D

Exp: (1)

( ) ( )If f t F s↔

( ) ( )

( )( ) ( )

2

2 22 2

dThen tf t F s

ds

d 1

ds s s 1

2s 1 2s 1

s s 1 s s 1

−↔

− = + + − + +

= − =+ + + +

Exp: (2)

( ) 2

1F s

s s 1=

+ +

( ) ( ) ( ) ( )

( )22

dL g t t.f t F s usin g multiplication by t

ds

2s 1

s s 1

= = −

+=

+ +

29. For a right angled triangle, if the sum of the lengths of the hypotenuse and a side is kept

constant, in order to have maximum area of the triangle, the angle between the hypotenuse

and the side is

(A) 12O (B) 36

O (C) 60

O (D) 45

O

Answer: (C) ( As per IIT Website)

Exp: Let x (opposite side), y (adjacent side) and z (hypotenuse) of a right angled triangle.

( ) ( )Given Z y K cons tan t ...... 1+ = and angle between them say ' 'θ then Area,

( )( )

( )

21 1 zA xy zsin zcos sin 2

2 2 4

kNow 1 z zsin k z

1 sin

= = θ θ = θ

⇒ + θ = ⇒ =+ θ



15

( )

( ) ( ) ( ) ( )( )

2

2

22

4

o

k sin 2A

4 1 sin

dAIn order to have max imum area, 0

d

1 sin 2cos2 sin 2 cos .2 1 sink0

4 1 sin

30 , Answer obtained is different than officialkey6

θ∴ =

+ θ

=θ

+ θ θ − θ θ + θ⇒ =

+ θ

π⇒ θ = =

30. The steady state output of the circuit shown in the figure is given by

( ) ( ) ( )( )y t A sin t= ω ω + φ ω . If the amplitude ( )A 0.25,ω = then the frequency ω is

( ) 1A

3 R C ( ) 2

B3 R C

( ) 1C

R C ( ) 2

DR C

Answer: B

Exp:

By nodal method,

( ) ( )oV 1 0 V V

0R 1 2

j c j c

−+ + =

ω ω

( )

o

2 2 2

1 j c 1 0V j c

R 2 R

2V

2 3j RC

V 1Y

2 2 j 3RC

1 1given A

4 4 9R c .

2

3 RC

ω + ω + =

=+ ω

= ⇒+ ω

ω = ⇒+ ω

⇒ ω =

R

( )sin tω±

( )C y t

C

C

+−

V

C

CC

R

ωsin t

( )y t



16

31. In the circuit shown in the figure, the value of V0(t) (in Volts) for t → ∞ is ______.

Answer: 31.25

Exp:

For t → ∞ , i.e., at steady state, inductor will behave as a shot circuit and hence B X

V 5.i=

( ) ( ) ( )

B x x x

0 x 0

50By KCL at node B, 10 V 2i i 0 i

8

250V t 5i t V t 31.25 volts

8

− + − + = ⇒ =

= ⇒ = =

32. The equivalent resistance in the infinite ladder network shown in the figure is Re.

The value of Re/R is ________

Answer: 2.618

Exp:

→ For an infinite ladder network, if all resistance are having same value of R

Then equivalent resistance is 1 5

.R2

+

→ For the given network, we can split in to R is in series with Requivalent

xi

+− x

2i( )10u t A

2 H

5Ω+

−( )o

V t

5Ω

2 R R R R

eR R R R R

+−

5Ω+

−( )0V t

2H

xiB

x2i

5ΩA

( )10x t

R

R

R

R

R

R

R

R

e quR



17

equ

equ

RR R 1.618R 2.618

R⇒ = + ⇒ =

33. For the two-port network shown in the figure, the impedance (Z) matrix (in Ω ) is

( ) 6 24A

42 9

( ) 9 8B

8 24

( ) 9 6C

6 24

( ) 42 6D

6 60

Answer: C

Exp: For the two-part network

[ ] 1

matrix

1

1 1 1

30 10 30Y matrix

1 1 1

30 60 30

Z Y

0.1333 0.0333Z

0.0333 0.05

9 6Z

6 24

−

−

+ − =

− +

=

− = −

=

34. Consider a silicon sample doped with ND = 1×1015

/cm3 donor atoms. Assume that the

intrinsic carrier concentration ni = 1.5×1010

/cm3. If the sample is additionally doped with NA

= 1×1018

/cm3 acceptor atoms, the approximate number of electrons/cm

3 in the sample, at

T=300 K, will be _________________.

Answer: 225.2

Exp: 18 15 17

A DP N N 1 10 1 10 9.99 10= − = × − × = ×

( )2

1023i

17

1.5 10225.2 / cm

P 9.99 10

×ηη = = =

×

30Ω

10Ω 60Ω+

−

1

1'

2

2 '

+

−

equR

R

( )+=eq

1 5R R

2



18

35. Consider two BJTs biased at the same collector current with area 1

A 0.2 m 0.2 m= µ × µ and

2A 300 m 300 m= µ × µ . Assuming that all other device parameters are identical, kT/q = 26

mV, the intrinsic carrier concentration is 1 × 1010

cm-3

, and q = 1.6 × 10-19

C, the difference

between the base-emitter voltages (in mV) of the two BJTs (i.e., VBE1 – VBE2) is

_________________.

Answer: 381

Exp: ( )1 2C CI I Given=

( )

( )

BE1

BE TT 2

1 2

BE BE1 2

2T

1

2

1 2

1

1 2

V

V /VV

S S

V V

SV

S

S 3

BE BE T S

S

BE BE

I e I e

Ie

I

I 300 300V V V ln 26 10 ln I A

I 0.2 0.2

V V 381mV

−

−

=

=

× − = = × α ×

− =

∵

36. An N-type semiconductor having uniform doping is biased as shown in the figure.

If EC is the lowest energy level of the conduction band, EV is the highest energy level of the

valance band and EF is the Fermi level, which one of the following represents the energy

band diagram for the biased N-type semiconductor?

Answer: D

37. Consider the common-collector amplifier in the figure (bias circuitry ensures that the

transistor operates in forward active region, but has been omitted for simplicity). Let IC be the

collector current, VBE be the base-emitter voltage and VT be the thermal voltage. Also, m

g

and 0

r are the small-signal transconductance and output resistance of the transistor,

respectively. Which one of the following conditions ensures a nearly constant small signal

voltage gain for a wide range of values of RE?

V

N type semiconductor−

( )A

CE

FE

VE

( )B

CE

FE

VE

( )C

CE

FE

VE

( )D

CEF

E

VE



19

( ) m EA g R 1<< ( ) C E TB I R V>> ( ) m 0C g r 1>> ( ) BE rD V V>>

Answer: B

Exp: E E E EV

Te E T E EE

E

R R I RA

Vr R V I RR

I

= = =+ ++

( )C E

V C E

T C E

C E T V

I RA I I

V I R

I R U A in almost cons tan t

∴+

∴ >> ⇒

∵

38. A BJT in a common-base configuration is used to amplify a signal received by a 50Ω

antenna. Assume kT/q = 25 mV. The value of the collector bias current (in mA) required to

match the input impedance of the amplifier to the impedance of the antenna is________.

Answer: 0.5

Exp: Input impedance of CB amplifier, Ii e

E

Vz r

I= =

( )T

E

E

25mV50 signal is received from 50 antenna and V 25mV

I

25mVI 0.5mA

50

⇒ = Ω =

⇒ = =Ω

∵

39 . For the common collector amplifier shown in the figure, the BJT has high β , negligible

VCE(sat), and VBE = 0.7 V. The maximum undistorted peak-to-peak output voltage vo (in Volts)

is______.

CCV

outV

ER

inV

CCV 12V= +

1R

5kΩ1 Fµ

iv

2R

10kΩ

1 Fµ

ER1kΩ

ov



20

Answer: 9.4

Exp: B

high, I is neglectedβ =∵

( )

B

E B

CE

0

10kV 12 8V

10k 5k

V V 0.7 7.3V

V 12 7.3 4.7V

Maximum undistorted V p p 2 4.7V 9.4V

∴ = × =+

= − =∴ = − =

∴ − = × =

40. An 8-to-1 multiplexer is used to implement a logical function Y as shown in the figure. The

output Y is given by

( )A Y A B C A C D= + ( )B Y A B C A B D= +

( )C Y A B C A C D= + ( )D Y A B D A B C= +

Answer: C

Exp: Y ABCD ABCD ABC= + +

Remaining combinations of the select

lines will produce output 0.

So, ( )Y ACD B B ABC

ACD ABC

ABC ACD

= + +

= +

= +

41. A 16-bit ripple carry adder is realized using 16 identical full adders (FA) as shown in the

figure. The carry-propagation delay of each FA is 12 ns and the sum-propagation delay of

each FA is 15 ns. The worst case delay (in ns) of this 16-bit adder will be __________.

00

I

D 1I

02

I

D 3I

0 4I

05

I

16

I

07

I2

S1

S0

S

A B C

Y

0A

0B

0FA

0S

0C

1A

1B

1FA

1S

1C

14A 14

B

14FA

14S

14C

15A

15B

15FA 15

C

15S

oI

1I

2I

3I

4I

5I

6I

7I

0

D

0

D

0

0

1

0

A B C

2S 1S 0S

Y8 : 1

M U X



21

Answer: 195

Exp:

This is 16-bit ripple carry adder circuit, in their operation carry signal is propagating from 1st

stage FA0 to last state FA15, so their propagation delay is added together but sum result is

not propagating. We can say that next stage sum result depends upon previous carry.

So, last stage carry (C15) will be produced after 16 12ns 192ns× =

Second last stage carry (C14) will be produced after 180 ns.

For last stage sum result (S15) total delay = 180ns + 15ns = 195ns

So, worst case delay = 195 ns

42. An 8085 microprocessor executes “STA 1234H” with starting address location 1FFEH (STA

copies the contents of the Accumulator to the 16-bit address location). While the instruction

is fetched and executed, the sequence of values written at the address pins A15-A8 is

(A) 1FH, 1FH, 20H, 12H (B) 1FH, FEH, 1FH, FFH, 12H

(C) 1FH, 1FH, 12H, 12H (D) 1FH, 1FH, 12H, 20H, 12H

Answer: A

Exp: Let the opcode of STA is XXH and content of accumulator is YYH.

Instruction: STA 1234 H

Starting address given = 1FFEH

So, the sequence of data and addresses is given below:

43. A stable linear time invariant (LTI) system has a transfer function ( ) 2

1H s

s s 6=

+ −. To make

this system causal it needs to be cascaded with another LTI system having a transfer function

H1(s). A correct choice for H1(s) among the following options is

(A) s + 3 (B) s - 2 (C) s - 6 (D) s + 1

15 8 7 0A A A A

Address (in hex) : Data (in hex)

1F FE H XXH

1F FF H 34H

20 00 H 12 H

12 34 H YYH

− −

→→→→

0FA

0A0B

0S

1FA

1A1B

1S

1FA

1A1B

1S

0C ........ .1C

1FA

1A1B

1S

14C

14FA

14A14B

14S

15C

15FA

15A15B

15S



22

Answer: B

Exp: ( )2

1Given, H s

s s 6=

+ − ( )( )1

s 3 s 2=

+ −

It is given that system is stable thus its ROC includes j axisω . This implies it cannot be

causal, because for causal system ROC is right side of the rightmost pole.

Poles at s 2⇒ = must be removes so that it can be become causal and stable

simultaneously.

( )( ) ( )

( )1

1 1s 2

s 3 s 2 s 3

Thus H s s 2

⇒ − =+ − +

= −

44. A causal LTI system has zero initial conditions and impulses response h(t). Its input x(t) and

output y(t) are related through the linear constant-coefficient differential equation

( ) ( ) ( ) ( )

2

2

2

d y t dy ta a y t x t

dt dt+ + =

Let another signal g(t) be defined as

( ) ( ) ( ) ( )t

2

0

dh tg t a h d ah t

dt= τ τ + +∫

If G(s) is the Laplace transform of g(t), then the number of poles of G(s) is _______.

Answer: 1

Exp: Given differential equation

( ) ( ) ( ) ( )

( )( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( )

2 2

2 2

t

2

0

2

2

2 22 2 2 2

2 2

2 2

s y s sy s y s x s

y s 1H s

x s s s

dg t h z dz h t h t

dt

H sSH s H s

s

1 1s

s ss s s s 2s

s s 1

ss s s

No. of poles 1

+ α + α =

⇒ = =+ α + α

= α + + α

= α + + α

α= α + +

+ α + α+ α + α + + α

α + α += =

+ α + α

=

∫



23

45. The N-point DFT X of a sequence [ ]x n ,0 n N 1≤ ≤ − is given by

[ ] [ ]2N 1

j nk,N

n 0

1X k x n e 0 k N L

N

π− −

=

= ≤ ≤ −∑

Denote this relation as X = DFT(x). For N = 4, which one of the following sequences

satisfies DFT(DFT (x))=x ?

( ) [ ]A x 1 2 3 4= ( ) [ ]B x 1 2 3 2=

( ) [ ]C x 1 3 2 2= ( ) [ ]D x 1 2 2 3=

Answer: B

Exp: This can be solve by directly using option and satisfying the condition given in question

( )

( )( ) ( ) [ ]

[ ]

( ) [ ]

[ ]

2N 1 j nkN

FT FT

n 0

X DFT x

1D D x DFT X X n e

N

DFT y 1 2 3 4

1 1 1 1 1 10

1 j 1 j 2 2 2j1 1X

1 1 1 1 3 24 4

1 j 1 j 4 2 j2

DFT of x will not result in 1 2 3 4

Try with DFT of y 1 2 3 2

1 1 1 1

1 j 1 j1X

1 1 1 14

1 j 1 j

π− −

=

=

= =

− − + = = − −

+ − − − −

− −

=− −+ − −

∑

1 8 4

2 2 11

3 0 04

2 2 1

4 1 1 1 1 4 2 1

1 j 1 j 1 4 21 1DFT of

0 1 1 1 1 0 6 324

1 1 j 1 j 1 4 2

− − = =

− −

− − − − = = = − − − + − − −

Same as x

Then ‘B’ is right option

46. The state transition matrix ( )tφ of a system

.

1 1

.2

2

xx 0 1

x0 0x

=

( ) t 1A

1 0

( ) 1 0B

t 1

( ) 0 1C

1 t

( ) 1 tD

0 1



24

Answer: D

Exp: Given state model,

( )( )

( )( )

( )( ) ( )

[ ]

( )

( )

1 1

2 2

11

1

2

21

x t x t0 1

0 0x t x t

0 1A

0 0

t state transistion matrix

t L SI A

s 1 s 11SI A

0 s 0 ss

1 1s s

t L10

s

1 tt

0 1

−−

−1−

−

=

=

φ ⇒

φ = −

− − = ⇒

φ =

φ =

47. Consider a transfer function ( ) ( ) ( )2

p 2

ps 3ps 2G s

s 3 p s 2 p

+ −=

+ + + − with p a positive real parameter.

The maximum value of p until which GP remains stable is ________.

Answer: 2

Exp: ( ) ( ) ( )2

p 2

ps 3ps 2Given G s

s 3 p s 2 p

+ −=

+ + + −

( ) ( )( ) ( )

2

2

By R H criteria

The characteristic equation is s 3 p s 2 p 0

i.e. s 3 p s 2 p 0

−

+ + + − =

+ + + − =

By forming R-H array,

( )( )( )

2

1

0

1 2 ps

s 3 0

s 2 p

−

+ φ

−

For stability, first column elements must be positive and non-zero

( )( )( )( )

i.e. 1 3 p 0 p 3

and 2 2 p 0 p 2

i.e. 3 p 2

+ > ⇒ > −

− > ⇒ <

− < <

The maximum value of p unit which pG remains stable is 2



25

48. The characteristic equation of a unity negative feedback system 1 + KG(s) = 0. The open loop

transfer function G(s) has one pole at 0 and two poles at -1. The root locus of the system for

varying K is shown in the figure.

The constant damping ratio line, for 0.5ζ = , intersects the root locus at point A. The distance

from the origin to point A is given as 0.5. The value of K at point A is ________ .

Answer: 0.375

Exp: We know that the co-ordinate of point A of the given root locus i.e., magnitude condition

( ) ( )G s H s 1=

Here, the damping factor 0.5 and the length of 0A 5ξ = =

0.5ξ =

Then in the right angle triangle

OX OX 1cos cos60 OX

OA 0.5 4

AX AX 3sin sin 60 AX

OA 0.5 4

θ = ⇒ = ⇒ =

⇒ θ = ⇒ = ⇒ =

So, the co-ordinate of point A is j 31

4 4− +

Substituting the above value of A in the transfer function and equating to 1

i.e. by magnitude condition,

( )2

j 31s4 4

2

k1

s s 1

1 3 9 3k .

16 16 16 16

k 0.375

−= +

=+

= + +

=

jω

0.5ζ =

1−

A

1

3−

O

( )0,0σx x

***

*A

OX−1 −23 −1

3

θ



26

49. Consider a communication scheme where the binary valued signal X satisfies PX = +

1=0.75 and PX = -1= 0.25. The received signal Y = X + Z, where Z is a Gaussian random

variable with zero mean and variance 2σ . The received signal Y is fed to the threshold

detector. The output of the threshold detector X is:

1. Y

1. Y .X + > τ− ≤ τ=

To achieve a minimum probability of error ˆP X X≠ , the threshold τ should be

(A) Strictly positive

(B) Zero

(C) Strictly negative

(D) Strictly positive, zero, or strictly negative depending on the nonzero value of 2σ

Answer: C

Exp: C

( ) ( )

( ) ( )

( )( )

( )( )

2 2

2

2

2

2

1 0

1 0

Z 2

Z

11

21

11

20

H : x 1; H : x 1

P H 0.75;P H 0.25

Received signal =X+Z

1Where Z N 0, 2 ; f z e

2

1 Z if X 1Received signal

1 Z if X 1

1f y H e

2

1f y H e

2

− σ

− γ−σ

γ

− γ+σ

γ

= + = −

= =

γ

− =σ π

+ =γ = − + = −

=σ π

=σ π

∼

At optimum threshold yopt: for minimum probability of error

( )( )

( )( )

( ) ( ) ( )( )

( )( )

( )( )

opt

2 2

2

opt

2opt

1 0

0 1y y

11 1

02

1y

2y 0

1

2 20 2

opt n

1

opt

opt

f y H P H

f y H P H

P He

P H

P He

P H

P H 1.1y l 0.55

2 P H 2

y Optimum threshold

y 0 Threshold is negative.

γ

γ =

− γ− − γ+ σ

+ σ

=

=

=

σ − σ= = = − σ

=

< ∴



27

50. Consider the Z-channel given in the figure. The input is 0 or 1 with equal probability.

If the output is 0, the probability that the input is also 0 equals ______________

Answer: 0.8

Exp: Given channel

We have to det er min e, P x 0 / y 0

11.P y 0 / x 0 P x 0 42P x 0 / y 0 0.81P y 0 511. 0.25

2 2

= =

= = == = = = = =

= + ×

51. An M-level PSK modulation scheme is used to transmit independent binary digits over a

band-pass channel with bandwidth 100 kHz. The bit rate is 200 kbps and the system

characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value

of M is ________.

Answer: 16

Exp: Bandwidth requirement for m-level PSK ( )11

T= + α

[Where T is symbol duration. α is roll of factor]

( )

[ ]

( )

3

3

3 5 6

3

6

2 6

2

11 100 10

T

1 100% excess bandwidth

12 100 10

TBit duration

2T 1

100 10 0.5 10 5 10 sec200 10

20 sec

Symbol duration 20 10 secBit duration log m 4 M 16

log m 5 10

− −

−

−

⇒ + α = ×

α =

⇒ = ×

⇒ =× = = × = ×

×= µ

×= ⇒ = = ⇒ =

×

1.00 0

0.25

0.75

INPUT OUTPUT

1 1

1.0

0.25

0.75

=X 0

input

=X 1

=Y 0

=Y 1

output



28

52. Consider a discrete-time channel Y =- X + Z, where the additive noise z is signal-dependent.

In particular, given the transmitted symbol X a, a∈ − + . at any instant, the noise sample Z is

chosen independently from a Gaussian distribution with mean Xβ and unit variance. Assume

a threshold detector with zero threshold at the receiver.

When 0,β = the BER was found to be Q(a) = 1 × 10-8

( ) ( )2 2u / 2 v / 2

v

1Q v e du, and for v 1,use Q v e

2

∞ − − = > ≈ π ∫

When 0.3,β = − the BER is closest to

(A) 10-7

(B) 10-6

(C) 10-4

(D) 10-2

Answer: C

Exp: X∈[-a,a] and P(x = -a) = P(x = a) = ½

=X+Zγ → Received signal

( )

( ) ( )( )( )

( ) ( )( )

( ) ( )( )

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

22

2

2

2 2

8

1Z X

22Z

1

0

1y a 1

21

1y a 1

20

e 1 1 0 0

0 1 1y a 1 y a 1

2 2

0

Z N X,1Q a 1 10

1f z e Q a e

2

a z if x a

a z if x a

H : x a

H : x a

and Threshold 0

1f y H e

2

1f y H e

2

BER :

P P H P e H P H P e H

1 1 1 1e dy e dy

2 22 2

−

− −β −ϑ

− − +β

γ

− + +β

γ

∞− − +β − + +β

−∞ −

β = ×

= ≈π

− + = −γ = + = +

= += −

=

=π

=π

= +

= +π π∫ ∫

∼

( )( )

( )

( )( ) ( )( )

2

2

8 a 2

e

e

4.249 2 4

e

4

e

Q a 1

0

P Q a 1 10 e a 6.07

0.3

P Q 6.07 1 0.3 Q 4.249

P e 1.2 10

P 10 .

− −

− −

−

= + β

β =

= = × = ⇒ =

β = −

= − =

= = ×



29

53. The electric field (assumed to be one-dimensional) between two points A and B is shown. Let

Aψ and

Bψ be the electrostatic potentials at A and B, respectively. The value of

B Aψ − ψ in

Volts is ________.

Answer: -15

Exp: A B

( ) ( )40kV / cm, 20kV / cm 5 10 kV / cm, 40kV / cm−×

( )

( )

( )

( )

4/ cm

4

4

4

B 5 104

ABA 0

5 102

4 4 8 4

0

4 4 4

AB

40 20E 20 x 0 E 4 10 x 20

5 10

V E.dl 4 10 x 20 dx

x4 10 20x 2 10 25 10 20 5 10

2

50 10 100 10 150 10 kV

V 15V

−

−

−

×

×

− −

− − −

−− = − ⇒ = × +

×

= − = − × +

= − × + = − × × × + × ×

= − × + × = − ×

⇒ = −

∫ ∫

54. Given x y zˆ ˆ ˆF za xa ya= + +

. If S represents the portion of the sphere 2 2 2x y z 1+ + = for

z 0≥ , then S

F . ds∇ ×∫

is ___________.

Answer: 3.14

Exp: S C

F.ds F.dr(u sin g stoke 's theorem andC is closed curve i.e.,∇ × =∫ ∫

2 2x y 1, z 0

x cos , y sin and : 0 to 2

+ = =⇒ = θ = θ θ π

( )

C

2

C 0

2

0

zdx xdy ydz

xdy cos cos d

1 sin 23.14

2 2

π

π

= + +

= = θ θ θ

θ = θ + = π

∫

∫ ∫

20kV / cm

0kV / cmA

5 mµ

40kV / cm

B



30

55. If ( ) ( ) ( )3 2 2 2ˆ ˆ ˆE 2y 3yz x 6xy 3xz y 6xyz z= − − − − + is the electric field in a source free

region, a valid expression for the electrostatic potential is

( ) 3 2A xy yz− ( ) 3 2B 2xy xyz− ( ) 3 2C y xyz+ ( ) 3 2D 2xy 3xyz−

Answer: D

Exp: Given ( ) ( )3 2 2 2

x y zE 2y 3yz a 6xy 3xz a 6xyz.a= − − − − +

By verification option (D) satisfy

E V= −∇

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