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GATE FORUM 110, Shopper's Point 1st Floor , S. V. Road Opp. Andheri Railway Station. Andheri (W) Mumbai. Tel: (022) 2623 7471 / 72

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• Hubli / Dharwad

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written permission. Discuss this questions paper at www.gateforum.com. Page 1 of 22

Q. 1 to 20 Carry One Mark Each

1. All the four entries of the 2×2 matrix 11 12

21 22

p pP

p p

=

are nonzero, and one of its

eigenvalues is zero. Which of the following statements is true?

(A) 11 22 12 21

p p p p 1− = (B) 11 22 12 21

p p p p 1− = −

(C) 11 22 12 21

p p p p 0− = (D) 11 22 12 21

p p p p 0+ =

2. The system of linear equations

4x 2y 7

2x y 6

+ =+ =

has

(A) a unique solution (B) no solution

(C) an infinite number of solutions (D) exactly two distinct solutions

3. The equation ( )sin z 10has=

(A) no real or complex solution

(B) exactly two distinct complex solutions

(C) a unique solution

(D) an infinite number of complex solutions

4. For real values of x, the minimum value of the function ( ) ( ) ( )f x exp x exp x is= + −

(A) 2 (B) 1 (C) 0.5 (D) 0

5. Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x=0?

(A) ( )3sin x (B) ( )2sin x (C) ( )3cos x (D) ( )2cos x

6. Which of the following is a solution to the differential equation ( ) ( )dx t

3x t 0?dt

+ =

(A) ( ) tx t 3e−= (B) ( ) 3tx t 2e−= (C) ( ) 23x t t

2= − (D) ( ) 2x t 3t=

7. In the following graph, the number of trees (P) and the number of cut-sets (Q)

are

(A) P=2, Q=2

(B) P=2, Q=6

(C) P=4, Q=6

(D) P=4, Q=10

( )4

( )1

( )2 ( )3




8. In the following circuit, the switch S is closed at t=0. The rate of change of

current ( )di0

dt+ is given by

(A) 0 (B) S SR I

L (C)

( )S SR R I

L

+ (D) ∞

9. The input and output of a continuous time system are respectively denoted by

x(t) and y(t). Which of the following descriptions corresponds to a causal system?

(A) ( ) ( ) ( )y t x t 2 x t 4= − + + (B) ( ) ( ) ( )y t t 4 x t 1= − +

(C) ( ) ( ) ( )y t t 4 x t 1= + − (D) ( ) ( ) ( )y t t 5 x t 5= + +

10. The impulse response h(t) of a linear time-invariant continuous time system is

described by ( ) ( ) ( ) ( ) ( )h t exp t u t exp t u t ,= α + β − where u(t) denotes the unit step

function, and α and β are real constants. This system is stable if

(A) α is positive and β is positive (B) α is negative and β is negative

(C) α is positive and β is negative (D) α is negative and β is positive

11. The pole-zero plot given below corresponds to a

(A) Low pass filter (B) High pass filter (C) Band pass filter (D) Notch filter

12. Step responses of a set of three second-order underdamped systems all have the

same percentage overshoot. Which of the following diagrams represents the poles of the three systems?

S R

( )i tLSRSI

σ

jω

O

O

XX




(A) (B)

(C) (D)

13. Which of the following is NOT associated with a p-n junction?

(A) Junction capacitance (B) Charge Storage Capacitance

(C) Depletion Capacitance (D) Channel Length Modulation

14. Which of the following is true?

(A) A silicon wafer heavily doped with boron is a p+ substrate

(B) A silicon wafer lightly doped with boron is a p+ substrate

(C) A silicon wafer heavily doped with arsenic is a p+ substrate

(D) A silicon wafer lightly doped with arsenic is a p+ substrate

15. For a Hertz dipole antenna, the half power beam width (HPBW) in the E-plane is

(A) 360º (B) 180º (C) 90º (D) 45º

16. For static electric and magnetic fields in an inhomogeneous source-free medium,

which of the following represents the correct form of two of Maxwell’s equations?

(A) .E 0

B 0

∇ =∇ × =

(B) .E 0

.B 0

∇ =∇ =

(C) E 0

B 0

∇ × =∇ × =

(D) E 0

.B 0

∇ × =∇ =

σ

jω

.

.

.

.

.

.

.

.

X

X

X

X

X

X

σ

jω

⋅ ⋅ ⋅X X X ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅X X X ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

σ

jω

⋅⋅

⋅ ⋅ ⋅ ⋅

XX

X

..

..

XX

X

..

σ

jω

X

.X

X

X

X

X

.

..

.

.

.. .

.. .




17. In the following limiter circuit, an input voltage i

V 10sin100 t= π applied. Assume

that the diode drop is 0.7V when it is forward biased. The Zener breakdown

voltage is 6.8V.

The maximum and minimum values of the output voltage respectively are

(A) 6.1V, 0.7V− (B) 0.7V, 7.5V− (C) 7.5V, 0.7V− (D) 7.5V, 7.5V−

18. A silicon wafer has 100nm of oxide on it and is inserted in a furnace at a

temperature above 1000ºC for further oxidation in dry oxygen. The oxidation rate

(A) is independent of current oxide thickness and temperature

(B) is independent of current oxide thickness but depends on temperature

(C) slows down as the oxide grows

(D) is zero as the existing oxide prevents further oxidation

19. The drain current of a MOSFET in saturation is given by ( )2p GS rK V V= −l where K

is a constant. The magnitude of the transconductance gm is

(A) ( )2GS T

DS

K V V

V

− (B) ( )GS T2K V V− (C) d

GS DS

I

V V− (D)

( )2GS T

GS

K V V

V

−

20. Consider the amplitude modulated (AM) signal C c m c

A cos t 2cos t cos t.ω + ω ω For

demodulating the signal using envelope detector, the minimum value of Ac should

be

(A) 2 (B) 1 (C) 0.5 (D) 0

Q. 21 to 75 carry two Marks Each

21. The Thevenin equivalent impedance Zth between the nodes P and Q in the

following circuit is

(A) 1 (B) 1

1 ss

+ + (C) 1

2 ss

+ + (D) 2

2

s s 1

s 2s 1

+ ++ +

1K D1

0D2 V

6.8VZ

iV

+−

1Ω

10V

1H1 F

1Ω

1AQ

P




22. The driving point impedance of the following network

is given by ( ) 2

0.2sZ s .

s 0.1s 2=

+ + The component values are

(A) L 5H, R 0.5 , C 0.1F= = Ω = (B) L 0.1H, R 0.5 , C 5F= = Ω =

(C) L 5H, R 2 , C 0.1F= = Ω = (D) L 0.1H, R 2 , C 5F= = Ω =

23. The circuit shown in the figure is used to charge the capacitor C alternately from

two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows

( ) ( )

( ) ( ) ( )For 2nT t 2n 1 T, n 0,1,2,... S1 to P1 and S2 to P2

For 2n 1 T t 2n 2 T, n 0,1,2,... S1 to Q1 and S2 to Q2

≤ < + =

+ ≤ < + =

Assume that the capacitor has zero initial charge. Given that u(t) is a unit step function, the voltage Vc(t) across the capacitor is given be

(A) ( ) ( )n

n 0

1 tu t nT∞

=

− −∑ (B) ( ) ( ) ( )n

n 0

u t 2 1 u t nT∞

=

+ − −∑

(C) ( ) ( ) ( ) ( )n

n 0

tu t 2 1 t nT u t nT∞

=

+ − − −∑ (D) ( ) ( )t 2nT t 2nT T

n 0

0.5 e 0.5e∞

− − − − −

=

− + ∑

24. The probability density function (PDF) of a random variable X is as shown below

R( )Z sL O

( )CV t

+

−1A 1A

1F

C 1Ω

S2S1

0.5Ω1Ω

Q1 P1 Q2 P2

1Ω

1

1− 0 1 x

PDF




The corresponding cumulative distribution function (CDF) has the form

(A) (B)

(C) (D)

25. The recursion relation to solve x=e-x using Newton Raphson method is

(A) nx

n 1x e−+ = (B) nx

n 1 nx x e−+ = −

(C) ( )n

n

x

n 1 n x

ex 1 x

1 e

−

+ −= +

+ (D)

( )n

n

x2

n n

n 1 x

n

x e 1 x 1x

x e

−

+ −

− + −=

−

26. The residue of the function ( )( ) ( )2 2

1f z at z 2 is

z 2 z 2= =

+ −

(A) 1

32− (B)

1

16− (C)

1

16 (D)

1

32

27. Consider the matrix 0 1

P .2 3

= − −

The value of eP is

(A) 2 1 1 2

2 1 2 1

2e 3e e e

2e 2e 5e e

− − − −

− − − −

− − − −

(B) 1 2 2 1

1 2 1 2

e e 2e e

2e 4e 3e 2e

− − − −

− − − −

+ − − +

(C) 2 1 1 2

2 1 2 1

5e e 3e e

2e 6e 4e e

− − − −

− − − −

− − − +

(D) 1 2 1 2

1 2 1 2

2e e e e

2e 2e e 2e

− − − −

− − − −

− − − + − +

28. In the Taylor series expansion of exp(x)+sin(x) about the point x=π, the

coefficient of ( )2x − π is

(A) ( )exp π (B) ( )0.5exp π (C) ( )exp 1π + (D) ( )exp 1π −

1

1− 0 1 x

CDF

1

1− 0 1 x

CDF

1

1− 0 1 x

CDF1

1− 0 1x

CDF

1−




29. ( ) ( ) ( )xP x M exp 2 x Nexp 3 x= − + − is the probability density function for the real

random variable X, over the entire x axis. M and N are both positive real

numbers. The equation relating M and N is

(A) 2

M N 13

+ = (B) 1

2M N 13

+ = (C) M N 1+ = (D) M N 3+ =

30. The value of the integral of the function ( ) 3 4g x,y 4x 10y= + along the straight

line segment from the point (0, 0) to the point (1, 2) in the x-y plane is

(A) 33 (B) 35 (C) 40 (D) 56

31. A linear, time-invariant, causal continuous time system has a rational transfer

function with simple poles at s=-2 and s=-4, and one simple zero at s=-1. A unit

step u(t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is

(A) ( ) ( ) ( )exp 2t exp 4t u t − + −

(B) ( ) ( ) ( ) ( )4exp 2t 12exp 4t exp t u t − − + − − −

(C) ( ) ( ) ( )4exp 2t 12exp 4t u t − − + −

(D) ( ) ( ) ( )0.5exp 2t 1.5exp 4t u t − − + −

32. The signal x(t) is described by

( ) 1 for 1 t 1x t

0 otherwise

− ≤ ≤ +=

Two of the angular frequencies at which its Fourier transform becomes zero are

(A) ,2π π (B) 0.5 ,1.5π π (C) 0, π (D) 2 ,2.5π π

33. A discrete time linear shift-invariant system has an impulse response h[n] with

h[0]=1, h[1]=-1. h[2]-2, and zero otherwise. The system is given an input

sequence x[n] with x[0] – x[2] -1, and zero otherwise. The number of nonzero samples in the output sequence y[n], and the value of y[2] are, respectively

(A) 5, 2 (B) 6, 2 (C) 6, 1 (D) 5, 3

34. Consider points P and Q in the x-y plane, with P=(1,0) and Q=(0,1). The line

integral ( )Q

P

2 xdx ydy+∫ along the semicircle with the line segment PQ as its

diameter

(A) is -1

(B) is 0

(C) is 1

(D) depends on the direction (clockwise or anti-clockwise) of the semicircle




35. Let x(t) be the input and y(t) be the output of a continuous time system. Match

the system properties P1, P2 and P3 with system relations R1, R2, R3, R4.

Properties Relations

P1: Linear but NOT time-invariant R1: ( ) ( )2y t t x t=

P2: Time-invariant but NOT linear ( ) ( )R2 : y t t x t=

P3: Linear and time-invariant ( ) ( )R3: y t x t=

( ) ( )R4 : y t x t 5= −

(A) ( ) ( ) ( )P1,R1 , P2,R3 , P3,R4 (B) ( ) ( ) ( )P1,R2 , P2,R3 , P3,R4

(C) ( ) ( ) ( )P1,R3 , P2,R1 , P3,R2 (D) ( ) ( ) ( )P1,R1 , P2,R2 , P3,R3

36. A memoryless source emits n symbols each with a probability p. The entropy of

the source as a function of n

(A) increases as log n (B) decreases as log (1/n)

(C) increases as n (D) increases as n log n

37. x(n) is a real-valued periodic sequence with a period N. x(n) and X(k) form N-

point. Discrete Fourier Transform (DFT) pairs. The DFT Y(k) of the sequence

( ) ( ) ( )N 1

r 0

1y n x r x n r is

N

−

=

= +∑

(A) ( ) 2

X k (B) ( ) ( )N 1

r 0

1X r X * k r

N

−

=

+∑

(C) ( ) ( )N 1

r 0

1X r X k r

N

−

=

+∑ (D) 0

38. Group I lists a set of four transfer functions. Group II gives a list of possible step

responses y(t). Match the step responses with the corresponding transfer

functions

Group I

2 2 2 2

25 36 36 49P Q R S

s 25 s 20s 36 s 12s 36 s 7s 49= = = =

+ + + + + + +

Group II

(1) (2)

( )y t

1

t

( )y t

1

t




(3) (4)

(A) P-3, Q-1, R-4, S-2 (B) P-3, Q-2, R-4, S-1

(C) P-2, Q-1, R-4, S-3 (D) P-3, Q-4, R-1, S-2

39. A certain system has transfer function ( ) 2

s 8G s ,

s s 4

+=

+ α − α is a parameter.

Consider the standard negative unity feedback configuration as shown below

Which of the following statements is true?

(A) The closed loop system in never stable for any value of α

(B) For some positive values of α, the closed loop system is stable, but not for all

positive values

(C) For all positive values of α, the closed loop system is stable

(D) The closed loop system is stable for all values of α, both positive and

negative

40. A single flow graph of a system is given below

( )y t

1

t

( )y t

1

t

−

( )G s+

α

1 / s

γ

− γ

1 / s

1 / s

− β β

1

1

− α




The set of equations that correspond to this signal flow graph is

(A)

1 1

1

2 2

2

3 3

x x0 1 0ud

x 0 x 0 0dt u

0 0 1x x

β −γ = γ α + −α −β

(B)

1 1

1

2 2

2

3 3

x x0 0 0ud

x 0 x 0 1dt u

0 1 0x x

α γ = −α −γ + β −β

(C)

1 1

1

2 2

2

3 3

x x0 1 0ud

x 0 x 0 1dt u

0 0 0x x

−α −β = −β −γ + α γ

(D)

1 1

1

2 2

2

3 3

x x0 0 1ud

x 0 x 0 0dt u

0 1 0x x

−γ β = γ α + −β −α

41. The number of open right half plane poles of ( ) 5 4 3 2

10G s is

s 2s 3s 6s 5s 3=

+ + + + +

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

42. The magnitude of frequency response of an underdamped second order system is

5 at 0rad/sec and peaks to 10

at 5 2rad / sec.3

The transfer function of the

system is

(A) 2

500

s 10s 100+ + (B)

2

375

s 5s 75+ +

(C) 2

720

s 12s 144+ + (D)

2

1125

s 25s 225+ +

43. Group 1 gives two possible choices for the impedance Z in the diagram. The

circuit elements in Z satisfy the condition R2C2 >R1C1. The transfer function 0

i

V

V

represents a kind of controller. Match the impedances in Group I with the types of controllers in Group II.

1V 1R

−

+

Z1

C

0V




Group I Group II

P.

1. PID controller

2. Lead compensator

3. Lag compensator

Q.

(A) Q 1,R 2− − (B) Q 1,R 3− − (C) Q 2,R 3− − (D) Q 3,R 2− −

44. For the circuit shown in the following figure, transistors M1 and M2 are identical

NMOS transistors. Assume that M2 is in saturation and the output is unloaded

The current Ix is related to Ibias as

(A) x bias sI I I= + (B)

x biasI I=

(C) x bias sI I I= − (D) out

x bias DD

E

VI I V

R

= − −

45. The measured transconductance gm of an NMOS transistor operating in the linear

region is plotted against the gate voltage VG at constant drain voltage VD. Which of the following figures represents the expected dependence of gm on VG?

(A) (B)

2R

2C

ER

outV

xI

DDV

biasI

aV

M1 M2

SI

mg

GV

mg

GV

2R

2C




(C) (D)

46. Consider the following circuit using an ideal OPAMP. The I-V characteristics of the

diode is described by the relation T

V

V

0 T 0I I e 1 where V 25mV, I 1 A

= − = = µ

and V

is the voltage across the diode (taken as positive for forward bias).

For an input voltage 1

V 1V,= − the output voltage V0 is

(A) 0 V (B) 0.1V (C) 0.7V (D) 1.1V

47.

The OPAMP circuit shown above represents a

(A) high pass filter (B) low pass filter

(C) band pass filter (D) band reject filter

mg

GV

mg

GV

4 KD

iV 1V= −

100K

−

+0V

−

+

C

2R

1R LiV

0V




48. Two identical NMOS transistors M1 and M2 are connected as shown below. Vbias is

chosen so that both transistors are in saturation. The equivalent gm of the pair is

defined to be out

i

I

V

∂∂

at constant Vout.

The equivalent gm of the pair is

(A) The sum of individual gm’s of the transistors

(B) The product of individual gm’s of the transistors

(C) Nearly equal to the gm of M1

(D) Nearly equal to gm/g0 of M2

49. An 8085 executes the following instructions

2710 LXI H, 30A0H

2713 DAD H

2714 PCHL

All addresses and constants are in Hex. Let PC be the contents of the program

counter and HL be the contents of the HL register pair just after executing PCHL.

Which of the following statements is correct

(A) PC 2715H= (B) PC 30A0H= (C) PC 6140H= (D) PC 6140H=

HL 30A0H= HL 2715H= HL 6140H= HL 2715H=

50. An astable multivibrator circuit using IC 555 timer is shown below. Assume that

the circuit is oscillating steadily

outI

outV

2M

1MiV

biasV

9V

( )8

Supply( )4

Re set

( )6 Threshold

( )2 Trigger

( )Output 3

( )Disch arge7

( )Gnd1

12 K

.01 FµCV

10K

30K




The voltage VC across the capacitor varies between

(A) 3V to 5V (B) 3V to 6V (C) 3.6V to 6V (D) 3.6V to 5V

51. Silicon is doped with boron to a concentration of 4×1017 atoms/cm3. Assuming

the intrinsic carrier concentration of silicon to be 1.5×1010/cm3 and the value of

kT

q to be 25mV at 300K

Compared to undoped silicon, the Fermi level of doped silicon

(A) Goes down by 0.13eV (B) Goes up by 0.13eV

(C) Goes down by 0.427eV (D) Goes up by 0.427eV

52. The cross section of a JFET is shown in the following figure. Let Va be -2V and let

Vp be the initial pinch-off voltage. If the width W is doubled (with other

geometrical parameters and doping levels remaining the same), then the ratio

between the mutual transconductances of the initial and the modified JFET is

(A) 4

(B)

( )p

p

1 2 / V1

2 1 1 / 2V

− −

(C)

( )p

p

1 2 / V

1 1 / 2V

−

−

(D) ( )

( )( )p

p

1 2 / V

1 1 / 2 V

−

−

53. Consider the Schmidt trigger circuit shown below.

A triangular wave which goes from -12V to 12V is applied to the inverting input of

the OPAMP. Assume that the output of the OPAMP sings from +15V to -15V. The voltage at the non-inverting input switches between

(A) -12V and +12V (B) -7.5V and +7.5V (C) -5V and +5V (D) 0V and 5V

W Drain

p+

n

p+

GateGV

Source

GateGV

−+

10k

10k

iV

10k

0V

15V+

15V−




54. The logic function implemented by the following circuit at the terminal OUT is

(A) P NOR Q

(B) P NAND Q

(C) P OR Q

(D) P AND Q

55. Consider the following assertions

S1: For Zener effect to occur, a very abrupt junction is required

S2: For quantum tunneling to occur, a very narrow energy barrier is required

Which of the following is correct?

(A) Only S2 is true

(B) S1 and S2 are both true but S2 is not a reason for S1

(C) S1 and S2 are both true and S2 is a reason for S1

(D) Both S1 and S2 are false

56. The two numbers represented in signed 2’s complement form are P 11101101 and Q 11100110= = . If Q is subtracted from P, the value obtained in

signed 2’s complement form is

(A) 100000111 (B) 00000111 (C) 11111001 (D) 111111001

57. Which of the following Boolean Expression correctly represents the relation between P, Q, R and M1?

(A) ( )1M P OR Q XOR R= (B) ( )1M P AND Q XOR R=

(C) ( )1M P NOR Q XOR R= (D) ( )1M P XOR Q XOR R=

58. For the circuit shown in the following figure I0-I3 are inputs to the 4:1 multiplexer R(MSB) and S are control bits

ddV

OUTP− Q−

P

Q

R

X

Y

Z

1M




The output Z can be represented by

(A) PQ PQS QRS+ + (B) PQ PQR PQS+ +

(C) PQR PQR PQRS QRS+ + + (D) PQR PQRS PQRS QRS+ + +

59. For each of the positive edge-triggered J-K flip flop used in the following figure, the propagation delay is T∆

Which of the following waveforms correctly represents the output at Q1?

(A)

(B)

(C)

(D)

Z

R S

P

P

P

Q

P

Q

3I

2I

1I

0I

4 :1 Mux

CLK

1

1

1

1

1Q0Q0J

0K

1J

1K

T

1t

t0

1

CLK

2T

1t T+ ∆

0

1

4T

1t 2 T+ ∆

0

1

2T

1t 2 T+ ∆

0

1

4T

1t T+ ∆

0

1




60. For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible

Which of the following statements is true?

(A) Q goes to 1 at the CLK transition and stays at 1

(B) Q goes to 0 at the CLK transition and stays at 0

(C) Q goes to 1 at the CLK transition and goes to 0 when D goes to 1

(D) Q goes to 0 at the CLK transition and goes to 1 when D goes to 1

61. A rectangular waveguide of internal dimensions ( )a 4cm and b 3= = is to be

operated in TE11 mode. The minimum operating frequency is

(A) 6.25GHz (B) 6.0GHz (C) 5.0GHz (D) 3.75GHz

62. One of a loss-less transmission line having the characteristic impedance of 75Ω

and length of 1cm is short-circuited. At 3GHz, the input impedance at the other end of the transmission line is

(A) 0 (B) Resistive (C) Capacitive (D) Inductive

63. A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant r 9ε = ). The magnitude of the reflection

coefficient is

(A) 0 (B) 0.3 (C) 0.5 (D) 0.8

64. In the design of a single mode step index optical fiber close to upper cut-off, the single mode operations is NOT preserved if

(A) Radius as well as operating wavelength are halved

(B) Radius as well as operating wavelength are doubled

(C) Radius is halved and operating wavelength is doubled

(D) Radius is doubled and operating wavelength is halved

65. At 20GHz, the gain of a parabolic dish antenna of 1 meter diameter and 70% efficiency is

(A) 15dB (B) 25dB (C) 35dB (D) 45dB

1

0

1

0

CLK

D

Q

Q




66. Noise with double-sided power spectral density of K over all frequencies is passed

through a RC low pass filter with 3dB cut-off frequency of fc. The noise power at the filter output is

(A) K (B) cKf (C) cK fπ (D) ∞

67. Consider a Binary Symmetric Channel (BSC) with probability of error being p. To

transit a bit, say 1, we transmit a sequence of three 1s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is

(A) ( )3 2p 3p 1 p+ − (B) 3p (C) ( )31 p− (D) ( )3 2p p 1 p+ −

68. Four messages band limited to W, W, 2W and 3W respectively are to be

multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is

(A) W (B) 3W (C) 6W (D) 7W

69. Consider the frequency modulated signal

( ) ( )510cos 2 10 t 5sin 2 1500t 7.5sin 2 1000t π × + π × + π ×

with carrier frequency of

510 Hz . The modulation index is

(A) 12.5 (B) 10 (C) 7.5 (D) 5

70. The signal c m ccos t 0.5cos t sin tω + ω ω is

(A) FM only (B) AM only

(C) both AM & FM (D) neither AM nor FM

Common Data Questions 71, 72 & 73

A speech signal, band limited to 4kHz and peak voltage varying between +5V and

-5V is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.

71. If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is

(A) 64kHz (B) 32kHz (C) 8kHz (D) 4kHz

72. Assuming the signal to be uniformly distributed between its peak values, the signal to noise ratio at the quantizer output is

(A) 16dB (B) 32dB (C) 48dB (D) 64dB

73. The number of quantitization levels required to reduce the quantization noise by a factor of 4 would be

(A) 1024 (B) 512 (C) 256 (D) 64




Common Data Questions 74 & 75

The following series RLC circuit with zero initial conditions is excited by a unit

impulse function ( )tδ

74. For t > 0, the output voltage ( )cV t is

(A)

1 3t t

2 22

e e3

− −

(B)

1t

22

te3

−

(C)

1t

22 3

e cos t23

−

(D)

1t

22 3

e sin t23

−

75. For t > 0, the voltage across the resistor is

(A)

3 1t t

2 21

e e3

− − −

(B)

1t

23t 1 3t

e cos sin2 23

− −

(C)

1t

22 3t

e sin23

−

(D)

1t

22 3t

e cos23

−

Linked Answer Questions: Q.76 to 85 Carry Two Marks Each

Statement for Linked Answer Questions: 76 & 77

A two-port network shown below is excited by external dc sources. The voltages

and the currents are measured with voltmeters V1, V2 and ammeter A1, A2 (all assumed to be ideal), as indicated. Under following switch conditions, the readings obtained are:

i) 1 2 1 1 2 2S Open, S Closed A 0A, V 4.5V, V 1.5V, A 1A− − = = = =

ii) 1 2 1 1 2 2S Closed, S Open A 4A, V 6V, V 6V, A 0A− − = = = =

+−( )tδ

1H 1Ω

1F ( )cV t

1A

6V

+−

+ − 1S

1V

+

−2V

+

−

2S

2A− +

1 2

2′1′

1.5V

+−Two Port

Network




76. The z-parameter matrix for this network is

(A) 1.5 1.5

4.5 1.5

(B) 1.5 4.5

1.5 4.5

(C) 1.5 4.5

1.5 1.5

(D) 4.5 1.5

1.5 4.5

77. The h-parameter matrix for this network is

(A) 3 3

1 0.67

− −

(B) 3 1

3 0.67

− −

(C) 3 3

1 0.67

(D) 3 1

3 0.67

− −


In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10Hz.

78. The samples ( ) ( )x n n 0, 1, 2, ...= are given by

(A) ( )0.05n5 1 e−− (B) 0.05n5e− (C) ( )5n5 1 e−− (D) 5n5e−

79. The expression and the region of convergence of the z-transform of the sampled signal are

(A) 5

5

5z, z e

z e

−−

<−

(B) 0.05

0.05

5z, z e

z e

−−

<−

(C) 0.05

0.05

5z, z e

z e

−−

>−

(D) 5

5

5z, z e

z e

−−

>−

+− ( )s

Sampler

f 10Hz=z transform−

( )X z

5V

s10 Fµ

200k

( )x n





In the following transistor circuit BE c EV 0.7V, r 25mV / I , and= = β and all the

capacitances are very large

80. The value of DC current IE is

(A) 1mA (B) 2mA (C) 5mA (D) 10mA

81. The mid-band voltage gain of the amplifier is approximately

(A) -180 (B) -120 (C) -90 (D) -60


In the following circuit, the comparator output is logic “I” if V1 > V2 and is logic “0” otherwise. The D/A conversion is done as per the relations

( ) ( )3

n 2DAC 3 2 1 0

n 0

V 2 Volts, where b MSB ,b , b and b LSB are the counter outputs−

== ∑

The counter starts from the clear state

82. The stable reading of the LED display is

(A) 06 (B) 07 (C) 12 (D) 13

20k

10k

C1C

ccV 9V=

3k

c2C

3kEC2.3k

EI

−+

inV 6.2V

Clock

=

4 bit

Up Counter

5V+

Clk

Clr

4 bit

Up Counter

Binaryto

BCD

2 Digit

LED

Display




83. The magnitude of the error between DAC inV and V at steady state in volts is

(A) 0.2 (B) 0.3 (C) 0.5 (D) 1.0


The impulse response h(t) of a linear time invariant continuous time system is

given by ( ) ( ) ( )h t exp 2t u t= − , where u(t) denotes the unit step function

84. The frequency response ( )H ω of this system in terms of angular frequency ω is

given by ( )H ω

(A) 1

1 f2+ ω (B)

( )sin ω

ω (C)

1

2 j+ ω (D)

j

2 j

ω+ ω

85. The output of this system to the sinusoidal input ( ) ( )x t 2cos 2t= for all time t, is

(A) 0 (B) ( )0.252 cos 2t 0.125− − π

(C) ( )0.52 cos 2t 0.125− − π (D) ( )0.52 cos 2t 0.25− − π

GATE PAPERECGATE-2009 www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission. Discuss this question paper at www.gatementor.com.

EC GATE 2010 www.gateforum.com

© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this document may be reproduced or utilized in any form without the written permission. Discuss GATE 2010 question paper at www.gatementor.com. 1

Q. No. 1 – 25 Carry One Mark Each

1. The eigen values of a skew-symmetric matrix are

(A) always zero (B) always pure imaginary

(C) either zero or pure imaginary (D) always real

2. The trigonometric Fourier series for the waveform f(t) shown below contains

(A) only cosine terms and zero value for the dc component

(B) only cosine terms and a positive value for the dc component

(C) only cosine terms and a negative value for the dc component

(D) only sine terms and a negative for the dc component

3. A function n(x) satisfied the differential equation ( ) ( )2

2 2

d n x n x0

dx L− = where L is a

constant. The boundary conditions are: n(0)=K and n ( ∞ ) = 0. The solution to this equation is

(A) n(x) = K exp(x/L) (B) n(x) = K exp(-x/ L )

(C) n(x) = K2 exp(-x/L) (D) n(x) = K exp(-x/L)

4. For the two-port network shown below, the short-circuit admittance parameter matrix is

4 2

(A) S2 4

−⎡ ⎤⎢ ⎥−⎣ ⎦

1 0.5

(B) S0.5 1

−⎡ ⎤⎢ ⎥−⎣ ⎦

1 0.5

(C) S0.5 1

⎡ ⎤⎢ ⎥⎣ ⎦

4 2

(D) S2 4

⎡ ⎤⎢ ⎥⎣ ⎦

f(t)

A

0.5 Ω

0.5 Ω 0.5 Ω

2’ 1’

1 2



5. For parallel RLC circuit, which one of the following statements is NOT correct? (A) The bandwidth of the circuit deceases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value. 6. At room temperature, a possible value for the mobility of electrons in the inversion

layer of a silicon n-channel MOSFET is (A) 450 cm2/V-s (B) 1350 cm2/V-s (C) 1800 cm2/V-s (D) 3600 cm2/V-s

7. Thin gate oxide in a CMOS process in preferably grown using (A) wet oxidation (B) dry oxidation (C) epitaxial deposition (D) ion implantation 8. In the silicon BJT circuit shown below, assume that the emitter area of transistor

Q1 is half that of transistor Q2.

The value of current I0 is approximately (A) 0.5 mA (B) 2mA (C) 9.3 mA (D) 15mA 9. The amplifier circuit shown below uses a silicon transistor. The capacitors CC and

CE can be assumed to be short at signal frequency and the effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is TRUE?

~

VCC=9V

RC=2.7K Ω

CC

CC

RB=800K Ω

VS

RE=0.3 K Ω CE

R0

V0

β=100

V|

R=9.3K Ω I0

Q2

(β2=715) (β1=700)

Q1

-10 V



(A) The input resistance Ri increases and the magnitude of voltage gain AV decreases

(B) The input resistance Ri decreases and the magnitude of voltage gain AV decreases

(C) Both input resistance Ri and the magnitude of voltage gain AV decrease (D) Both input resistance Ri and the magnitude of voltage gain AV increase

10. Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

2

1

R(A)

R− 3

1

R(B)

R−

2 3

1

R R(C)

R− 2 3

1

R R(D)

R

⎛ ⎞+− ⎜ ⎟

⎝ ⎠

11. Match the logic ga5tes in Column A with their equivalents in Column B.

(A) P–2, Q-4, R-1, S-3 (B) P-4, Q-2, R-1, S-3

(C) P–2, Q-4, R-3, S-1 (D) P-4, Q-2, R-3, S-1

V0

+

R1

R2

R3

+

- Vi

Column B Column A

P

Q

R

S

1

2

3

4



12. For the output F to be 1 in the logic circuit shown, the input combination should be

(A) A = 1, B= 1. C = 0 (B) A = 1, B= 0,C = 0 (C) A = 0, B= 1. C = 0 (D) A = 0, B= 0, C = 1 13. In the circuit shown, the device connected to Y5 can have address in the range

(A) 2000 - 20FF (B) 2D00 – 2DFF (C) 2E00 – 2EFF (D) FD00 - FDFF

14. Consider the z-transform X(z) = 5z2 + 4z-1 + 3; 0<|z| < ∞ . The inverse z-transform x[n] is

(A) 5δ[n + 2] + 3δ[n] + 4δ[n – 1] (B) 5δ[n - 2] + 3δ[n] + 4δ[n + 1]

(C) 5 u[n + 2] + 3 u[n] + 4 u[n – 1] (D) 5 u[n - 2] + 3 u[n] + 4 u[n + 1]

15. Two discrete time systems with impulse responses h1[n] = δ[n -1] and h2[n] = δ[n – 2] are connected in cascade. The overall impulse response of the cascaded system is

(A) δ[n - 1] + δ[n - 2] (B) δ[n - 4]

(C) δ[n - 3] (D) δ[n - 1] δ[n - 2]

16. For an N-point FFT algorithm with N = 2m which one of the following statements is TRUE?

(A) It is not possible to construct a signal flow graph with both input and output in normal order

(B) The number of butterflies in the mth stage is N/m

F

A

B

C

To device Chip select

VCC GND

A11 A12 A13 A14 A15

A

B

A8 A9 A10

Y5

74LS138 3-to-8

decoder

G2

G2

G1 10 /M



(C) In-place computation requires storage of only 2N node data (D) Computation of a butterfly requires only one complex multiplication

17. The transfer function Y(s)/R(s) of the system shown is

(A) 0 1

(B)s 1+

2

(C)s 1+

2

(D)s 3+

18. A system with transfer function ( )( )

Y s

X s=

ss p+

has an output y(t) = cos 2t3π⎛ ⎞−⎜ ⎟

⎝ ⎠for

the input signal x(t) = p cos 2t2π⎛ ⎞−⎜ ⎟

⎝ ⎠. Then, the system parameter ‘p’ is

(A) 3 (B) 2

3 (C) 1 (D)

32

19. For the asymptotic Bode magnitude plot shown below, the system transfer function can be

10s 1

(A)0.1s 1

++

100s 1

(B)0.1s 1

++

100s

(C)10s 1+

0.1s 1

(D)10s 1

++

20. Suppose that the modulating signal is m(t) = 2cos (2π fmt) and the carrier signal is xC(t) = AC cos(2πfct), which one of the following is a conventional AM signal without over-modulation? (A) x(t) = Acm(t) cos(2πfct)

(B) x(t) = Ac[1 + m(t)]cos(2πfct)

(C) x(t) = Ac cos(2πfct) + CA4

m(t) cos(2πfct)

(D) x(t) = Ac cos(2πfmt) cos(2πfct) + Ac sin(2πfmt) sin(2πfct)

40.

0

0.001 0.1 10 1000

Mag

nitude

(db)

R(s) +

-

Σ

Σ

Y(s) 1

s 1+

1s 1+

- +



21. Consider an angle modulated signal x(t) = 6cos[2πx106t+2sin(8000πt) + 4cos(8000pt)] V. The average power of x(t) is.

(A) 10W (B) 18W (C) 20W (D) 28W

22. If the scattering matrix [S] of a two port network is 0 0

0 0

0.2 0 0.9 90S

0.9 90 0.1 90

⎡ ⎤∠ ∠=⎡ ⎤ ⎢ ⎥⎣ ⎦ ∠ ∠⎣ ⎦

then

the network is

(A) lossless and reciprocal (B) lossless but not reciprocal

(C) not lossless but reciprocal (D) neither lossless nor reciprocal

23. A transmission line has a characteristic impedance of 50 Ω and a resistance of 0.1 Ω /m. if the line is distortion less, the attenuation constant (in Np/m) is

(A) 500 (B) 5 (C) 0.014 (D) 0.002

24. Consider the pulse shape s(t) as shown. The impulse response h(t) of the filter matched to this pulse is

s(t)

1

0

t T

h(t)

(B)

t T 0

1

h(t)

(A)

1

t

0 -T

(C) (D)

h(t) h(t)

1

T 2T 0 T

t

1

0



25. The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2) is

(A) 1

30π (B)

160π

(C) 1

120π (D)

1240π

Q. No. 26 – 51 Carry Two Marks Each

26. If ey = 1xX , then y has a

(A) maximum at x= e (B) minimum at x= e

(C) maximum at x= e-1 (D) minimum at x= e-1

27. A fair coin is tossed independently four times. The probability of the event “the number of time heads shown up is more than the number of times tails shown up” is

(A) 116

(B) 18 (C)

14 (D)

516

28. If 2x y

c

A xy a x a then A.dl→

= + ∫ over the path shown in the figure is

(A) 0 (B)

2

3 (C) 1 (D) 2 3

29. The residues of a complex function X(z) = ( ) ( )

1 12zz z 1 z 2

−− −

at its poles are

(A) 1 1

, and 12 2

− (B) 1 1

, and 12 2

− −

(C) 1 3

, 1 and2 2

− − (D) 1 3

, 1 and2 2

−

y

3 -

1 -

0 1 3 2 3

C

x



30. Consider differential equation ( )dy x

dx-y (x) = x with the initial condition y(0) = 0.

Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is

(A) 0.01 (B) 0.031 (C) 0.0631 (D) 0.1

31. Given 1

x

3s 1f(t) L . If lim f(t) 1, then the value of K is

s3 4s2 (K 3)s−

→∞

⎡ ⎤+= =⎢ ⎥+ + −⎣ ⎦

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

32. In the circuit shown, the switch S is open for a long time and is closed at t=0. The current i(t) for t≥0+ is

(A) i(t)=0.5-0.125e-1000tA (B) i(t)=1.5-0.125e-1000tA

(C) i(t)=0.5-0.5e-1000tA (D) i(t)=0.375e-1000tA

33. The current I in the circuit shown is

(A) -j1A (B) J1A (C) 0A (D) 20A

1.5A 10 Ω

10 Ω

10 Ω

S t=0

i(t)

15mH

20mH

o

3

20 0 V

10 rad/s

∠

ω =

I 50μF 1 Ω



34. In the circuit shown, the power supplied by the voltage source is

(A) 0W (B) 5W (C) 10W (D) 100W

35. In a uniformly doped BJT, assume that NE, NB and NC are the emitter, base and collector dopings in atoms/cm3, respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following conditions is TRUE?

(A) NE=NB=NC (B) NE NB and NB>NC

(C) NE=NB and NB<NC (D) NE<NB<NC

36. Compared to a p-n junction with NA=ND=1014/cm3, which one of the following statements is TRUE for a p-n junction with NA=ND=1020/cm3?

(A) Reverse breakdown voltage is lower and depletion capacitance is lower

(B) Reverse breakdown voltage is higher and depletion capacitance is lower

(C) Reverse breakdown voltage is lower and depletion capacitance is higher

(D) Reverse breakdown voltage is higher and depletion capacitance is higher

37. Assuming that flip-flops are in reset condition initially, the count sequence observed at QA in the circuit shown is

(A) 0010111… (B) 0001011… (C) 0101111… (D) 0110100…

1 Ω

1 Ω

1 Ω

1 Ω

1 Ω

1A

2A

10V + -

DA QA QB Qc DB Dc

CQBQAQ

Clock

Output



38. The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)

(A) (B)

(C) (D)

39. The Boolean function realized by the logic circuit shown is

(A) F=∑m(0,1,3,5,9,10,14) (B) F=∑m(2,3,5,7,8,12,13)

(C) F=∑m(1,2,4,5,11,14,15) (D) F=∑m(2,3,5,7,8,9,12)

+20V

R

4R

R

D1

D2

ν0 ν1 +

−

ν1

ν0

10

−10 −5 0 ν1

ν0

5

0 −5 −10

ν1

ν0

0

5

+5

ν1

ν0

0

10

+5

C

D

A B

F(A,B,C,D) Y

I0

I1

I2

I3

4x1 MUX



40. For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

(A) 00H (B) 45H (C) 67H (D) E7H

41. A continuous time LTI system is described by

2

2d y(t) dy(t) dx(t)

4 3y(t) 2 4x(t)dt dtdt

+ = +

Assuming zero initial conditions, the response y(t) of the above system for the input x(t)=e-2tu(t) is given by

(A) (et-e3t)u(t) (B) (e-t-3-3t)u(t)

(C) (e-t+e-3t)u(t) (D) (et+e3t)u(t)

42. The transfer function of a discrete time LTI system is given by

1

1 2

32 z

4H(z)3 1

1 z z4 8

−

− −

−=

− +

Consider the following statements:

S1: The system is stable and causal for ROC:|z|>½

S2: The system is stable but not causal for ROC:|z|<¼

S3: The system is neither stable nor causal for ROC: ¼<|z|<½

Which one of the following statements is valid?

(A) Both S1 and S2 are true (B) Both S2 and S3 are true

(C) Both S1 and S3 are true (D) S1, S2 and S3 are all true

43. The Nyquist sampling rate for the signal sin(500 t) sin(700) t

s(t) xt t

π π=

π π is given by

(A) 400 Hz (B) 600 Hz (C) 1200Hz (D) 1400 Hz 44. A unity negative feedback closed loop system has a plant with the transfer function

21

G(s)s 2s 2

=+ +

and a controller Gc(S) in the feed forward path. For a unit set

input, the transfer function of the controller that gives minimum steady state error is

3000 MVI A, 45H 3002 MOV B, A 3003 STC 3004 CMC 3005 RAR 3006 XRA B



(A) cs 1

G (s)s 2

+=

+ (B) c

s 2G (s)

s 1+

=+

(C) c(s 1)(S 4)

G (s)(s 2)(s 3)

+ +=

+ + (D) c

2G (s) 1 3s

s= + +

45. X(t) is a stationary process with the power spectral density Sx(f)>0 for all f. The

process is passed through a system shown below.

Let Sy(f) be the power spectral density of Y(t). Which one of the following statements is correct?

(A) Sy(f)>0 for all f (B) Sy(f)=0 for |f|>1kHz (C) Sy(f)=0 for f=nf0, f0=2kHz, n any integer (D) Sy(f)=0 for f=(2n+1)f0=1kHz, n any integer 46. A plane wave having the electric field component z

8iE 24cos(3x10 t y)a= − β V/m

and traveling in free space is incident normally on a lossless medium with m= m0 and e=9e0 which occupies the region y≥0. The reflected magnetic field component is given by

(A) 8x

1cos(3x10 t y)a A/m

10+

π (B) 8

x1

cos(3x10 t y)a A/m20

+π

(C) 8x

1cos(3x10 t y)a A/m

20− +

π (D) 8

x1

cos(3x10 t y)a A/m10

− +π

47. In the circuit shown, all the transmission line sections are lossless. The Voltage

Standing Wave Ration (VSWR) on the 60W line is

(A) 1.00 (B) 1.64 (C) 2.50 (D) 3.00

Delay=0.5ms

ddt

Y(t) X(t) +

+

+

Z0=30 Ω

Short

λ/8

λ/4

Z0=60 ZL=300Z 30 2= Ω



Common Data Questions: 48 & 49

Consider the common emitter amplifier shown below with the following circuit parameters:

b=100, gm=0.3861 A/V, r0= ∞ , rp=259 W, RS=1k W, RB=93K W, RC=250 W, RL=1k W, C1= ∞ and C2=4.7mF.

48. The resistance seen by the source Vs is

(A) 258 Ω (B) 1258 Ω (C) 93 KΩ (D) ∞

49. The lower cut-off frequency due to C2 is

(A) 33.9 Hz (B) 27.1 Hz (C) 13.6 Hz (D) 16.9 Hz

Common Data Questions: 50 & 51

The signal flow graph of a system is shown below.

U(s)

2 1/s 1/s

1

0.5

1

-1

-1

Y(s)

+10V

RC

RB

RS

RL

C2

C1

+ νs

-

+ ν0

-



50. The state variable representation of the system can be

(A) 1 1 0

x x u1 0 2

y [0 0.5]x

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦=

(B) 1 1 0

x x u1 0 2

y [0 0.5]x

−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦=

(C) 1 1 0

x x u1 0 2

y [0.5 0.5]x

⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦=

(D) 1 1 0

x x u1 0 2

y [0.5 0.5]x

−⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦=

51. The transfer function of the system is

(A) 2s 1

s 1

++

(B) 2s 1

s 1

−+

(C) 2s 1

s s 1

++ +

(D) 2s 1

s s 1

−+ +

Linked Answer Questions: Q.52 to Q.55 Carry Two Marks Each


The silicon sample with unit cross-sectional area shown below is in thermal

equilibrium. The following information is given: T=300K, electronic charge=1.6x10-

19C, thermal voltage=26mV and electron mobility = 1350cm2/V-s

52. The magnitude of the electric field at x=0.5 μm is

(A) 1kV/cm (B) 5kV/cm (C) 10 kV/cm (D) 26kV/cm

53. The magnitude of the electron drift current density at x=0.5 μm is

(A) 2.16x104 A/cm2 (B) 1.08x104 A/cm2

(C) 4.32x103 A/cm2 (D) 6.48x102 A/cm2

ND=1016/cm3

x=0 x=1μm

1V




Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk).

Yk=Nk if transmitted bit bk=0

Yk=a+Nk if transmitted bit bk=1

Where Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5αe-α|n| (This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6V.

54. The value of the parameter α (in V-1) is (A) 1010 (B) 107 (C) 1.414x10-10 (D) 2x10-20

55. The probability of bit error is (A) 0.5xe-3.5 (B) 0.5xe-5 (C) 0.5xe-7 (D) 0.5xe-10

Q. No. 56 – 60 Carry One Mark Each

56. Which of the following options is the closest in meaning to the world below: Circuitous (A) Cyclic (B) indirect (C) confusing (D) crooked 57. The question below consists of a pair of related of related words followed by four

pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed: Worker (A) fallow : land (B) unaware : sleeper (C) wit : jester (D) renovated : house 58. Choose the most appropriate word from the options given below to complete the

following sentence: If we manage to ________ our natural resources, we would leave a better planet

for our children. (A) uphold (B) restrain (C) Cherish (D) conserve

Receiver

LPF S/H +

Sampling time Threshold z

x(t)

n(t)

r(t) y(t) y(tk) k

kk

1 if y(t ) zb

0 if y(t )< Z≥⎧

= ⎨⎩



59. Choose the most appropriate word from the options given below to complete the following sentence:

His rather casual remarks on politics _______ his lack of seriousness about the subject.

(A) masked (B) belied (C) cherish (D) conserve 60. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10

of them play both hockey and football. Then the number of persons playing neither hockey nor football is:

(A) 2 (B) 17 (C) 13 (D) 3

Q. No. 61 – 65 Carry Two Marks Each

61. Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare; and regretfully, there exist people in military establishments who think that chemical agents are useful tools for their cause.

which of the following statements best sums up the meaning of the above passage:

(A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in warfare would be undesirable. (D) People in military establishments like to use chemical agents in war. 62. If 137+276=435 how much is 731+672? (A) 534 (B) 1403 (C) 1623 (D) 1513 63. 5 skilled workers can build a wall in 20 days; 8 semi-skilled worker can build a wall

in 25days; 10 unskilled workers can build a wall in 30 days. If a team has 2 killed, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?

(A) 20 days (B) 18 days (C) 16 days (D) 15 days 64. Given digits 2,2,3,3,3,4,4,4,4 how many distinct 4 digit numbers greater than

3000 can be formed? (A) 50 (B) 51 (C) 52 (D) 54 65. Hari (H), Gita (G), Irfan (I) and Saira (S) are sibiligs (i.e. brothers and sisters). All

were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than3 years. Given the following facts:

i. Hair’s age + Gita’s age > Irfan’s age + Saira’s age. ii. The age difference between Gita and Saira is 1 year. However, Gita is not the

oldest and Saira is not the youngest. iii. There are not twins. In what order were they born (0ldest first)? (A) HSIG (B) SGHI (C) IGSH (D) IHSG


Answer keys

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

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

15 16 D 17 C 18 19 B 20 A 21 A

22 A 23 24 A 25 C 26 A 27 D 28 B

29 A 30 A 31 D 32 A 33 D 34 B 35 A

36 C 37 A 38 D 39 C 40 C 41 C 42 A

43 B 44 B 45 C 46 B 47 B 48 C 49 C

50 B 51 C 52 53 B 54 D 55 A 56 C

57 D 58 59 B 60 C 61 A 62 D 63 C

64 65 66 67 A 68 C 69 B 70

71 D 72 C 73 B 74 D 75 76 C 77 A

78 B 79 C 80 A 81 D 82 D 83 C 84 C

85 D

Explanation:

1. Use I P 0λ − =⎡ ⎤⎣ ⎦

( ) ( )

11 12

21 22

11 12

21 22

11 22 21 12

11 22 21 12

P P00

P P0

P P0

P P

P P P P 0

Putting =0, we get P P P P 0

λ ⎛ ⎞⎡ ⎤− =⎜ ⎟⎢ ⎥λ⎣ ⎦ ⎝ ⎠

λ − −⎡ ⎤=⎢ ⎥− λ −⎣ ⎦

⇒ λ − λ − − =

λ − =

2. ( )( )

4x 2y 7 1

2x y 6 2

+ = →

+ = →

( ) ( )1 can be written as 2x+y=3.5 3→

Now LHS of equation (2) and (3) are same but RHS is different which is not possible. Hence no solution

3. Sine value lies between -1 and +1. Therefore no real or complex solution exists

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4. ( ) x xf x e e−= +

( )( )( )

( )

x x

x x

x x

0 0

f x e e

f x 0 e e 0

f x e e ve for x 0

Thus minimum.

Minimum f x e e 2

−

−

−

′ = −

′ = ⇒ − =

′′ = + ⇒ + =

= + =

5. 3 5x x

sinx x ..........3! 5!

= − + +

( )2 4x x

cos x 1 .............2! 4!

= − + +

Thus sin(x3) will have odd powers

sin(x2) will have even powers

cos(x3) will have even powers

cos(x2) will have even powers

6. ( ) ( )

dx t3x t 0

dt+ =

( )( )( )

( ) 3t

dx t3dt

x t

logx t 3t

x t k.e−

=

=

=

∫ ∫

Thus option (B) satisfies the solution

8. At t = 0, inductor acts like a open circuit

We know di

L Vdt

=

( )+

S S

S S

at t=o , inductor path is open circuit

V=I .R at t 0

I .RdiThus

dt L

+=

⎛ ⎞= ⎜ ⎟⎝ ⎠

9. Only option (C) satisfies the condition of causality

10. ( ) ( ) ( )t th t e u t e u t+α β= + −

For h(t) to be stable ( )h t dt < ∞∫

It will happen when α is negative and β is positive.



16. Curl of electric field is zero.

Divergence of magnetic field is zero.

17. During +ve part of Vi

D1 will be forward biased

Zener diode is reverse biased

Thus net voltage = 6. 8 0.7 7.5V+ =

During –ve part of Vi

D2 will be forward biased

D1 will be reverse biased

Thus net voltage = -0.7V

19. , ( )2D GS TI k V V= − ( )DGS T

GS

I2k V V gm

Vδ

= − =δ

20. C C mA cos t 2 cos t cos tω + ω ωC

C C m

C

CC

2A cos t 1 cos t

A

2for envelop detection <1 1 A should be at least 2

A

⎡ ⎤ω + ω⎢ ⎥

⎣ ⎦

μ ⇒ < ⇒

21. For finding Thevenin equivalent

Short voltage source

Open circuit source

Now Thevenin equivalent = ( ) ( )1

1 S 1S11 1 SS 1

1 1 SS

+ + +1+ + = =

+ + +

22. 1

Y R SLCS

= + +

2

2 2

CSZ

S LC RCS 1Comparing it with given equation

0.25 0.1S1S 0.1S 2 S 0.05S 12

C 0.1RC 0.05R 0.5

1LC

21

L 50.2

=+ +

=+ + + +

==

=

=

= =



24. Cumulative distribution function = ( )x

Pdf dx−∞∫

Thus when we integrate the line from (-1, 0) to (0, 1) we get a parabolic curve.

The maximum value of Pdf can be 1

Thus option (A) satisfies the solution

25. ( ) nxn nf x x e−=

( )

( )

n

nn n n

n n

xn

xx x xnn n n n

n 1 n x x

f x 1 e

x 1 ex e x x e x ex x

1 e 1 e 1 e

−

−− − −

+ − −

′ = +

+− + − += − = =

+ + + nx−

26. Residue at z = 2

( )( ) ( )

22z 2

1 d 1 1lim z 2 Residue

1! dz 32z 2 z 2→

−− ⇒

+ −=

27. 0 1

P2 3

⎡ ⎤= ⎢ ⎥− −⎣ ⎦

( ) ( )

11

1

1

1

find L SI A

S 0 0 10 S 2 3

S 12 S 3

S 3 11L find inverse laplace transform of the expression

S 1 S 2 2 S

−−

−

−

−

−⎡ ⎤⎣ ⎦

⎡ ⎤⎡ ⎤ ⎡ ⎤−⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦⎣ ⎦

−⎡ ⎤⎢ ⎥+⎣ ⎦

⎡ ⎤+⎡ ⎤⎢ ⎥⎢ ⎥+ + −⎢ ⎥⎣ ⎦⎣ ⎦

28. ( ) xf x e sinx= +

( ) ( )

( )( )( )

( )

2

x

x

x

2

1Coefficient of x- f x

2!

f x e cos x

f x e sinx

f x e

Thus coefficient of x- 0.5

π=π

′′π =

′ = +

′′ = −

′′ =

π =

29. Using property of probability density of xP dx 1∞

−∞

=∫



( ) ( ) 2Mexp 2 x N exp 3 x dx 1 M N 1

3

∞

−∞

⎡ ⎤− + − = ⇒ + =⎣ ⎦∫

30. ( )3 4

C

4x 10y+∫

( )0,0 1

2

( )1

43

0

y 2x

4x 10 2x dx 33

=

⎡ ⎤⇒ + =⎢ ⎥⎣ ⎦∫

1

1

1+1−

32.

( )

( )( )( )

1j t

1

2 sinX j 1.e

At =0 X j0 2

At X j 0

2 X j2 0

− ω

−

ωω = =

ω

ω =

ω = π π =

ω = π π =

∫

33. Use convolution to get the result

35. ( ) ( )21R y t t x t=

( ) ( )( )

( ) ( )( )

( ) ( )

2

3

4

Linear time variant

R y t t x t

Non linear non time invariant

R y t x t

Non linear time invariant

R y t x t 5

Linear time invariant

=

=

= −

36. 1

H nP logP

=

Since they all have same probability

Thus it increases with n

37. Direct property

Convolution in time domain



Multiplication in frequency domain

38. 2 2

25 36P , 0, R

S 25 S 12S 36= 1ζ = = ζ =

+ + +

2 2

36 20 49Q , 1 S

12S 20S 36 S 7S 49= 1ζ = > = ζ <

+ + + +

39. Closed loop system transfer function

( )( )

( )

2

2

0

S 8C S

S S 1 4

Use Routh criteria

1 4SS 02 1

4S

+=

+ α + +

+

Thus for all positive value of ‘α’ this will be stable

40. 11 2

dxx x

dt= −α + β + u

21 2

31 2

dxx x u

dtdx

x x 0.dt

= −β + γ +

= α + γ + u

Thus option (C) satisfies the solution

41. Use Routh criteria.

42. Satisfy the given condition with different option given

Option (A) satisfy the solution

43. 0

i 1

V ZV Z

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

11 1

1 1 1

R1Z R

SC R C S 1

⎛ ⎞= = ⎜ ⎟

+⎝ ⎠

Now first case

( ) ( )

2 22

2 2

2 2 1 1o

i 1 2

R C S 11Z R

SC SC

R C S 1 R C S 1VV R C S

⎛ ⎞+= + = ⎜ ⎟

⎝ ⎠

+ +=

Thus it is PID controller



Second case

22

2 2 2

o 1 1 2

i 2 2 1

-1 11 1 2 2

1 1 2 2

R1Z R

SC R C S 1

V R C S 1 RThus

V R C S 1 R

Taking phase = tan R C tan R C

Given that R C R C

−

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

⎛ ⎞ ⎛ ⎞+= ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

ω − ω

<

Thus phase is –ve

Hence this is phase lag compensator

44. Given circuit is current mirror

G

x bia

I 0

I I

=

= s

47. Given circuit is low pass filter

50. Charging and discharging level of capacitor will be the voltage across it

This is equal to CC CC1 2

V and V3 3

Thus 3V to 6V is the voltage VC across the capacitor

51. Use VF V n

A

NkTE E

q N

⎛ ⎞− = ⎜ ⎟

⎝ ⎠

Since it is doped with acceptor impurity, Fermi level will shift down

53. V0 is switching between +15 to -150

Thus voltage at non inverting input switches between 15 10 15 10

to , 7.5 to 7.510 10 10 10

× − ×−

+ +

54. AND gate

56. P 11101101 25= = −

( )Q 11100110 18

Thus P Q 25 18 7

Thus ' 7 's two complement representation is 11111001

= = −

− = − − − = −

−

57. 1M Z XOR R=



( )

( ) ( ) ( )1

1

M X.Y XOR R

M P.Q. P Q XOR R PQ PQ XOR R P XOR Q XOR R

=

⎡ ⎤= + = + =⎣ ⎦

58. Use this to find solution

R S Z0 0 P Q0 1 P1 0 P.Q1 1 P

+

59. Two JK flip flop. Thus propagation delay will be 2 TΔ

Frequency is getting reduced by a factor e

61. 2 28 8

2

3 10 1 1 3 10 5f 6

2 4 3 2 4 3 10−

× × ×⎛ ⎞ ⎛ ⎞= × + = =⎜ ⎟ ⎜ ⎟ × × ×⎝ ⎠ ⎝ ⎠.25 GHz

62. L 0in 0

0 L

Z jZ tanZ Z

Z jZ tan

⎛ ⎞+ β= ⎜ ⎟

+ β⎝ ⎠

L

0in 0

0

2X 10cm, Z 0, 1

10

0 jZ tan 5Thus Z is Z which is inductiveZ

π= = β = ×

π⎛ ⎞+⎜ ⎟⎜ ⎟⎝ ⎠

63. 0 02 1

0 0

1n , n

3μ μ

= =ε ε

Thus reflection coefficient =

11

3 0.51

13

−= −

+

Thus magnitude is 0.5

67. Probability of error = P

Thus probability of no error = (1-P)

Now probability that transmitted bit, received in error

= all bits are with error + one bit is with error

( ) ( )3 2 3 21P 3C P 1 P P 3P 1 P= + − = + −

68. In TDM minimum bandwidth is twice the maximum frequency present



Thus 6W is the answer

69. Total frequency deviation = 5 1500 7.5 1000 15000× + × =

Thus modulation index = m

f 1500010

f /max 1500Δ

= =

71. According to Nyquist Criteria, minimum sampling frequency = 2 4 k 8kHz× =

But each cycle can accommodate two bits

Thus minimum frequency = 4 kHz

72. ( )SNR 6n 1.7 6 8 1.7 48 1.7. Thus correct answer is C= + = × + = +

73. To reduce quantization noise by 4

No. of levels should increase by 2

Thus 512 levels are required.

74. ( )C

1SCV t

1 SL RSC=

+ +

( )1 t2

2 22

31 1 2 2S 2. e1 2S S 1 3 3S 1 31S S 2 2

−= = = =

+ ++ + ⎛ ⎞+ + ⎜ ⎟⎝ ⎠

3sin t

2

2

75. Similar way this question can be solved

76. 1 11 1 12V Z I Z I= +

2 21 1 22V Z I Z I= +

Using the given information: S1-open, S2-closed

12 12

22 22

4.5 Z 1 Z 4.5

1.5 Z 1 Z 1.5

= ⇒ =

= ⇒ =

S1-closed, S2-open

11 11

21 21

6 Z 1 Z 1.5

6 Z 4 Z 1.5

1.5 4.5Thus Z matrix

1.5 1.5

= ⇒ =

= ⇒ =

⎡ ⎤= ⎢ ⎥⎣ ⎦



77. ( )1 11 1 12 2V Z I Z I 1= + →

( )

( )

21 1 22

22

212 1 2

22 22

1 1

Z I VI 2

Z

Z 1I I V 2

Z Z

Substitute (2) in (1) to get V in terms of I and V

3 3Thus H matrix

1 0.67

− += →

−= + →

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

2

78. Initial voltage across R = 5Vn

Final voltage across R = 0V

Thus 0.5tRV 5.e−=

When we sample this at f 102∂ =

We get ( )0.5

n 0.05n10x n 5.e 5.e− −= =

79. ( ) 0.05n n

n 0

X Z 5.e .Z∞

− −

== ∑

0.050.05 1

0.050.05

5Z e

1 e Z5Z

Z eZ e

−− −

−−

= >−

= >−

80. Design is independent of IB as β is very high B

I 0

9 10V 3

20 10

β

β

≈

×= =

+V

Thus applying Kirchhoff’s Law in base emitter, we get

3E E3 0.7 I 2.3 10 I 1mA= + × × ⇒ =

81. Mid band voltage gain = L

e

Rr

′−

eE

L C L

V

25r 25

I

R 3k 3k R R 1.5

1500Thus A 60

25

= = Ω

′ = = =

= − = −

k



82. 3 2 1 0 DACb b b b V

DAC c

0 0 0 0 0

0 0 0 1 0.5

0 0 1 1 15

1 1 0 1 6.5

1 1 1 0 7

1 1 1 1 7.5

Using the given V relation we can get t

Thus stable reading of LED is 6.5×2=13

83. Magnitude error = 6. 5 6.2 0.3− =

84. ( ) 2t j t

0

1H e e dt

j

∞− − ω ⎛ ⎞

ω = = +⎜ ⎟ω⎝ ⎠∫ 2

85. When input is sinusoidal then output is also sinusoidal with same frequency but amplitude and phase changes.

Thus amplitude is 0.5

at 2

1 22 2

j 2 2 2−

ω=

= =ω +

Phase is 1

2

tan 0.252

−

ω=

ω⎛ ⎞ = π⎜ ⎟⎝ ⎠

Thus ( ) ( )0.5y t 2 cos 2t 0.25−= − π


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Answer Keys

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

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

15 A 16 D 17 C 18 C 19 C 20 D 21 C

22 D 23 B 24 B 25 A 26 D 27 A 28 B

29 C 30 A 31 D 32 D 33 A 34 D 35 D

36 D 37 A 38 C 39 A 40 A 41 B 42 D

43 B 44 A 45 B 46 B 47 C 48 B 49 C

50 D 51 B 52 B 53 C 54 C 55 C 56 D

57 A 58 59 B 60 D

Explanations:-

1. The order of differential equation is two

3. ( ) 1 cos2t cos2t 1f t , then it has o and Hz frequency component

2

− +=π

7. ( )n

1

1 1 1u n z ,313

1 z3

−

↔ > −

( )n

1

1 1 1u n 1 z ,212

1 z2

1 1z23

−

− − − ↔ > −

< <

8. Since magnitude plot shows both increasing as well as decreasing plot, it is lead-

lag compensator

9. Since Bandwidth is 10 kHz, thus output power is 11 3 610 10 10 10 1 10 W− −× × × = ×

11. Use ( ) ( )r

r n r

cn p q for any two losses which yield head

−

2 8 10

10 10

2 2

1 1 1C C

2 2 2

=




But in present case it is required only for first two tosses. Thus in this case

1 1 1. 10 times

2 2 2

K

12. Since autocorrelation function and power spectral density bears a Fourier

transform relation, then sin c required in frequency domain will five rectangular

convolutions in time domain, thus it is a triangular function

13. ( )1

01 c zf z

z

− + + ′ =

( ) ( )

( ) ( ) ( )

1

0 1

0 12

02 2

1 c c zf z

z

1 c z cdf z z 2 i 1 c

d z

−+ +′ =

+ + ′ = = π +

∫

14. Since 12A current is coming from one source and it is also known that 60V source

is absorbing power i.e. current is flowing inside 60V source.

( )12 x I I 12 x, thus possible option is A= + ⇒ = −

15. 1qD V

kT D

−µ µ = ⇒ =

17. L

C

L

1R

SCz1

RSC

=+

( )( )

LC

L

L

0 L L LL

Li L L L L

L

Rz

SCR 1

R

V S SCR 1 R RR R

RV S R RSCR R SCRR R RR

SCR

=+

+= = = ⇒ =

+ + + ++

18. Use the condition of controllability

20. Apply right hand thumb rule




22. 3 5x x

sinx x3! 5!

= − + L

( ) ( ) ( ) ( )

( )( ) ( )

( )( )

3 5

2 2

2

x xsin x x

3! 5!

x xsinx1

x 3! 5!

xsinx1

x 3!

− π − π− π = − π − +

− π − π− = − + −

− π

− π⇒ = − +

− π

L

L

24. Since it is one-sided Laplace transform

25. ( ) dy dxi lny lnx C

y x= ⇒ = +

( )( )( )

C

2 2

2 2

ye straight line

x

ii xy constant

iii x y constant

iv x y constant

P-2 Q-3 R-3 S-1

= ⇒

=

− =

+ =

36. 1 z y z y 0 z 1, z 1, z 0 + + + + = = =

40.

( )j j2 j3H e e e , It is FIR high pass filterω − ω − ω= −

44. ( )2

2

1

11, H j 0

2j 1 ω=

−ω +ω = = =−ω + ω +

. Thus output is zero for all sampling frequencies

46. The mean is 3

12 3c

1 −

1− −




47. 1

1 2, efficiency = 20%12 2+2

µ = =

48. 2

C Blog 1 SNR= +

2

2 2 2

1

C Blog SNR

C Blog 2SNR Blog SNR Blog 2

C C B

=

′ = = +

′ = +

50. ( )2

1 2

100z 200, z 200

50= = =

1 2

in

z z z 100

50 50z 25

100

′ = =×= = Ω

59. From the figure

1 2

1 2

P P g

0 0 0

g P P0 1 1

1 0 1

1 1 1

⇒ = + ,

c e d

0 0

d c e0 1 1

1 0 1

1 1 1

= +

EC GATE 2010 ww.gateforum.com


EC GATE 2010 Answer Keys

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

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

15 C 16 D 17 B 18 B 19 A 20 C 21 B

22 C 23 D 24 C 25 C 26 A 27 D 28 C

29 C 30 B 31 D 32 A 33 A 34 A 35 B

36 C 37 D 38 B 39 D 40 C 41 B 42 C

43 C 44 D 45 D 46 A 47 B 48 B 49 A

50 D 51 C 52 C 53 A 54 B 55 D 56 B

57 A 58 D 59 C 60 D 61 C 62 C 63 D

64 B 65 B

Explanations:-

1. For a real skew symmetric matrix the non-zero eigen values are all pure imaginary and thus occurs in complex conjugate pair.

2. Since given waveform is even, thus it will have only cosine terms. Function f(t) has more negative value then positive, so it will have negative DC Component.

3. 2

2 2

d x n(x)0

dx L− =

x x2 2 2

2

1S n(x) 0, n(x) Ae Be

Ln( ) 0 B 0, n(0) k B k

−⎛ ⎞− = = +⎜ ⎟⎝ ⎠∞ = ∴ = = ∴ =

4.

1 1 10.5 0.5 0.5

1 1 10.5 0.5 0.5

−⎡ ⎤+⎢ ⎥⎢ ⎥−⎢ ⎥+⎢ ⎥⎣ ⎦

=4 22 4

−⎡ ⎤⎢ ⎥−⎣ ⎦

5. Bandwidth1

RC= ,

i L C

1 1 1 1Z R X X

= + + Thus it is maximum at resonance.

9. Given circuit after removing CE will behave as current-series feedback. Overall voltage gain will decrease as feedback signal comes into picture and since it is current-series feedback, input impedance increases.



10. By rearranging

0 2

i 1

V RV R

= −

11. A B A . B+ =

A B A B

A B AB AB

A B AB AB

A B AB AB

A B A B AB

⋅ = +

⊕ = +

= +

= +

⊕ = +

12. B A=

F A B C A A C C so, F 1 when C 1= = = = =

14. ( ) 2 1 5,0,3,4X z 5z 4z 3− ⎧ ⎫

= + + = ⎨ ⎬↑⎩ ⎭

15. ( )1h n delay by 1→ , ( ) ( ) ( )2 1 2h n delay by 2, h n *h n delay by 3→ →

17. Construct a signal flow graph

15 14 13 12 11 10 9 87 6 5 4 3 2 1 0

o o o o o o o oo o2 D

A A A A A A A A13. A A A A A A A A

0 0 1 0 1 1 0 1

1111 1111F F

+

−v

−

+

3R

0V

2R

1R

1−

1s 1

−+

1

1s 1+( )R s ( )Y s



( )( )

[ ]11 0y s 1s 1

R s 1 1 s 11

s 1 s 1

−+= =

+⎡ ⎤− −⎢ ⎥+ +⎣ ⎦

18. ( ) ( ) ( )y t M x t= ω

2 2

2 2 2 2 2 2

2

wp1

p

p p , 4 p 4p2

3p 4, p3

=ω +

ω + = ω + =

= =

19. 2

20R 20j

2 20jR Cs LCRs= = −

++ +

20. ( )( )

m m

c

A DSB V Vm

V 0D SSB

⎫→ ⎪ = = = ∞⎬→ ⎪⎭

( )

( )

2B m 2

1C is without over modulation

→ = =

21. 2 2Ac 6 36

Power 182 2 2

= = = =

22. 11 22S S 0 lossy≠ ≠ [ ] [ ]1S S Reciprocal=

23. R G

Distribution lessL C

=

0

0.1R G and Z R G G

2500

0.1 0.10.002NP /m

2500

α = = → =

×α = =

24. ( ) ( )h t s T t= −

25. 2

AV

r

r

E 1 1 1P

2 120u 12 1202 120

4

= = = =η π

× π ×× π ×ε

26. ( )2 2

1 dy 1 lnxy lnx, 0, x e max

x dx x x= = − = =

27. 4 heads & 0 tail or 3 head & 1 tail 4 0 3 1

4 40 3

1 1 1 1 5P C C

2 2 2 2 16⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠



28. ( ) ( )2x y x

C C

ˆ ˆ ˆ ˆA.d xy a x a a dx ay dy= + +∫ ∫

2 13 13 3

2

1 1C 1 33 3

4 1(xy dx x dy) x dx 3x dx dy dy 1

3 3= + = + + + =∫ ∫ ∫ ∫ ∫

29. ( ) ( ) ( ) ( ) ( )z 0 z 1 z 2

1 3z z , z 1 z 1, z 2 z

2 2= = =

−× = − × = − × =

31. ( )

( )3 2

s 0

s 3s 11

s 4s k 3 s→

+=

+ + −

( )2

3s 1 1 k 4

k 3s 4s k 3+

= = =−+ + −

32. ( ) ( )i f 0.5, i i 0.75 = =

( ) ( ) 1 1000tf i ji t V i i e 0.5 0.125e

− −π= + − = −

33.

s

20LsV

1 1C

Ls R

=⎡ ⎤+ +⎢ ⎥⎣ ⎦

2

20R 20j

2 20jR Cs LCRs= = −

++ +

34.

( ) ( )KVL 10 2 i 3 2 i 2

10 10 4i i 0 Power 0

→ = + + +

= + ⇒ =∴ =

37.

Ls

~20C0310ω =

R

v

s

1C

1 2+

1

1

2

+

−1 0 v

3

1

i 3+1 i

1

i

0 0 0

A B CQ Q Q0 0 01 1 00 1 11 0 10 1 00 0 10 0 1



38. iWhen V 10= − 0V 5 Only B option matches=

39.

40. A 0100 0101→

B 0100 0101Carry flag 0RAR will A 0010 0010XRA B 0110 0111

6 7

→→→

41. ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )t 3t2 s 2 1 2

y s , y t e e u ts 1 s 3 s 2 s 1 s 3

− −+= × = = −

+ + + + +

42. S1 true , ( )S2 False meter stable nor causal S3 True

43. fs 2fm=

500 700

2 1200Mz2 2

⎡ ⎤= + =⎢ ⎥⎣ ⎦

44. ( ) ( )( ) ( ) ( )c

sR sE s

1 G s G s H s=

⎡ ⎤+ +⎣ ⎦

( ) ( )( ) ( )

C s 0

C Cs 0 s 0

11 G s G s

so, G s should be maximum for low error G s→

→ →

=⎡ ⎤+ +⎣ ⎦

= ∞

46. y 0 y 0< >

[ ][ ]

1 2120 40

40 120 140 120 2

η = π η = π

− π −ρ = =

+ π

Ve E will change direction & M will not change its direction∴ ρ = − ,

Direction of M for incident field is in x, so direction of M in the reflected field is also is in x.

iC

124E 12M120 10

×ρ= = =

η π π

0 0 1 1

0 1 1 0

1 0 0 0

1 1 0 0

CDAB

( )F m 2,3,5,7,8,9,12= Σ



47. 60 30j 60 30j 160 30j 60 120 300 17

+ −ρ = = =

+ + +

1

USWR 1.641

+ ρ= =

− ρ

48. ( )s s B eZ R R ||Br= +

cr 2.475

1.258kV

=

=

50. Let us take two state variable as shown x1 and x2

( )

( )

( ) [ ]

2 2 1 1 2

2 2

2 1

2

1 2

1

x x x , x x 2v s

x x 01 1U s

1 0x x 2

xy s 0.5x 0.5x 0.5 0.5

x

= − + = − +

⎡ ⎤ ⎡ ⎤ ⎡ ⎤−⎡ ⎤⎢ ⎥ ⎢ ⎥⇒ = + ⎢ ⎥⎢ ⎥−⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

⎡ ⎤⎢ ⎥= + =⎢ ⎥⎣ ⎦

54. ( )

( ) ( )

0n

0

NS f

2 exist in fourier transform pairN

R2

⎤= ⎥

⎥ →⎥τ = δ τ ⎥⎦

( ) 16 140 0

72

N NR 0 , Power 2 1 10 2 10

2 22

Variance which is equal to power power since mean is zero 10d

− −= = × × = ×

= ⇒ α =

55. Probability of error

7 6n a / 2 10 10 10

a / 2

0.5 e 0.5 e 0.5 e 0.5 e−

∞−α −α × − × −= α = = =∫

56. Circuitous means round about or not direct. Therefore the closest in meaning will be indirect

57. A worker who is inactive or not working is termed as unemployed, similarly land which is inactive or not in use is called Fallow.

1−

2 0.5

1−

( )U s1x 11s 2x 11s 2x

( )y s

1

j30

60

20 60Ω

60 30j+

sR

BR reB CR LR



58. The clue in this sentence is ‘If we manage to ______our natural resources’ and ‘better planet’. This implies that the blank should be filled by a word which means ‘preserve’ or ‘keep for long time’. Therefore the word ‘conserve’ is the right answer.

59. The key words in the statement are ‘casual remarks’ and ‘lack of seriousness’. The blank should be filled with a word meaning ‘showed’ or ‘revealed’. Hence, ‘betrayed’ is the correct answer.

60. Representing the given information in the Venn diagram, we have

Let the number of people who play only hockey = a

The number of people who play only football = b

Now, a = n(H) – 10 = 15 – 10 = 5

b = n(F) – 10 = 17– 10 = 7

Clearly, a + b + 10 + n = 25

⇒ n = 25 – 7 - 5 -10 ⇒ n = 3

∴ The number of people who play neither Hockey nor Football is 3

61. Among the answer choices, the three options B, C and D can be inferred from the passage. But the main essence of the passage is that chemical agents are being used my military establishments in warfare which is not desirable. Therefore option C is the statement which best sums up the meaning of the passage.

62. Given,

137 + 276 = 435

Adding units digits i.e. 7 + 6 = 13, but given as 5, which is 13 – 8 and also 1 is carry forwarded to the tens place.

i.e., +1

7 6

3 7

-------------

05

Here, 7 + 3 + 1 = 11 i.e., 11 – 8 = 3 and 1 is carry forwarded to hundreds place

+1 +1

1 7 6

2 3 7

------------------------

3 5

Now, the sum of digits in hundred’s place is 1 + 1 + 2 = 4

n(H) = 15

μ = 25

n(F) = 17

b 10 a

n = ?



i.e.,

1 7 6

2 3 7

------------------------

4 3 5

-------------------------

Using the same logic, we have

+1

+1 7 3 1

6 7 2

---------------------------------

1 6 2 3

---------------------------------

Sum of units digits 1+2 = 3, sum of tens digit = 3 + 7 = 10 i.e. 10 -2 and + 1 carry forward and

Sum of hundreds digits= 1 + 7 + 6 = 14 i.e., 14 – 8 = 6 and one carry forward.

63. Given, 5 skilled workers can build a wall in 20 days i.e., 1 skilled worker can build the same wall in 100 days

∴ The capacity of each skilled worker is 1

100

8 semi-skilled workers can build a wall in 25 days i.e., 1 semi-skilled worker can build the same wall in 200 days

∴ the capacity of each semi-skilled worker is 1

200

Similarly, the capacity of 1 unskilled worker is1

300.

Now, the capacity of 2 skilled+6 semi-skilled+5 unskilled workers is 21

100⎛ ⎞⎜ ⎟⎝ ⎠

+ 6 1

200⎛ ⎞⎜ ⎟⎝ ⎠

+ 5

300

= 2 3 5 20 1

100 100 100 300 15+ + = =

∴ The required numbers of days is 15

64. The given digits are 2,2,3,3,3,4,4,4,4 we have to find the numbers that are greater than 3000 ∴ The first digit can be 3 or 4 but not 2. Now, let us fix the first, second and third digits as 3, 2, 2 and then the fourth place can be filled in 3

ways. i.e., 3 ways ∴ the number of ways is 3. Similarly, we fix first, third and fourth places as 3, 2 and 2 respectively, so the second place can be

filled in 3 ways again i.e., The number of ways is 3. Now, we fix first, second and fourth places just as previous cases and we obtain the same result. ∴ The number of ways is 3 so; the total number of ways is 9. Similarly this can do by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the 9 ways

in each case. ∴ The number of numbers greater than 3000 starting with 3 is 27 Similarly by taking 4 as the first digit and applying the same process, we get 27 numbers

3 2 or 3 or 4 2 2

2 2 or 3 or 43 2



∴ The total number of numbers that are greater than 3000 is 27 + 27 = 54 But, 3222 and 4222 is not possible as there are only two 2’s (given), 3333 is also not possible as there

are only three 3’s (given) ∴ The total number of numbers that are greater than 3000 is 54 – 3 = 51.

65.

i) H + G > I + S

ii) |G - S| = 1

Meaning G & S will be next to each other in the order. So the option A is ruled out.

G not oldest

S not youngest

iii) No twins.

Going by the options, we will try to solve the equation,

Taking an example with youngest aged 1, we can try to solve the equation,

and correct the age (started with ages 4,3,2,1) to suit condition (i) and (ii) which gives 5,4,3,1

S I +4 Generalizing, we can take their ages in

terms of I's age,

G I + 3 In this case, H + G > I + S

H I + 2 Since 2I + 5 > 2I + 4

I I

I In this order, G is always less than I and H is always less than S.

G So G<I and H<S

S Implies G+H < I + S, all values are positive

H Defies condition i) Hence incorrect.

I In this order H< I, G<S

H Hence H +G < I + S

S Defies Condition i)

G Hence incorrect.

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