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1 11/11

1/14 7EC (2)

General Aptitude

Q. No. 1 – 5 Carry One Mark Each

1. An apple costs Rs. 10. An onion costs Rs. 8.

Select the most suitable sentence with respect to grammar and usage.

(A) The price of an apple is greater than an onion.

(B) The price of an apple is more than onion.

(C) The price of an apple is greater than that of an onion.

(D) Apples are more costlier than onions.

Key: (C)

2. The Buddha said, “Holding on to anger is like grasping a hot coal with the intent of throwing it at

someone else; you are the one who gets burnt.”

Select the word below which is closest in meaning to the word underlined above.

(A) burning (B) igniting (C) clutching (D) flinging

Key: (C)

3. M has a son Q and a daughter R. He has no other children. E is the mother of P and daughter-in-law of M.

How is P related to M?

(A) P is the son-in-law of M. (B) P is the grandchild of M.

(C) P is the daughter-in law of M. (D) P is the grandfather of M.

Key: (B)

4. The number that least fits this set: (324, 441, 97 and 64) is ________.

(A) 324 (B) 441 (C) 97 (D) 64

Key: (C)

Exp: 2 2 2324 18 ;441 21 ;64 8 but

297 x for any positive integer

i.e. 97 is odd man out

5. It takes 10 s and 15 s, respectively, for two trains travelling at different constant speeds to completely pass

a telegraph post. The length of the first train is 120 m and that of the second train is 150 m. The

magnitude of the difference in the speeds of the two trains (in m/s) is ____________.

(A) 2.0 (B) 10.0 (C) 12.0 (D) 22.0

Key: (A)

Exp: Speed length

length speed timetime

1 1

2 2

1 2

120 10 s s 12

150 15 s s 10

s s 2




2 22/12

2/14 7EC (2)

Q. No. 6 – 10 Carry Two Marks Each

6. The velocity V of a vehicle along a straight line is measured in m/s and plotted as shown with

respect to time in seconds. At the end of the 7 seconds, how much will the odometer reading

increase by (in m)?

(A) 0 (B) 3 (C) 4 (D) 5

Key: (D)

7. The overwhelming number of people infected with rabies in India has been flagged by the World

Health Organization as a source of concern. It is estimated that inoculating 70% of pets and stray dogs

against rabies can lead to a significant reduction in the number of people infected with rabies.

Which of the following can be logically inferred from the above sentences?

(A) The number of people in India infected with rabies is high.

(B) The number of people in other parts of the world who are infected with rabies is low.

(C) Rabies can be eradicated in India by vaccinating 70% of stray dogs

(D) Stray dogs are the main sources of rabies worldwide.

Key: (A)

8. A flat is shared by four first year undergraduate students. They agreed to allow the oldest of them to enjoy

some extra space in the flat. Manu is two months older than Sravan, who is three months younger than

Trideep. Pavan is one month older than Sravan. Who should occupy the extra space in the flat?

(A) Manu (B) Sravan (C) Trideep (D) Pavan

Key: (C)

9. Find the area bounded by the lines 3x+2y=14, 2x-3y=5 in the first quadrant.

(A) 14.95 (B) 15.25 (C) 15.70 (D) 20.35

Key: (B)




3 33/13

3/14 7EC (2)

Exp

14A ,0

3

B 0, 7

5C , 0

2

5D 0,

3

E 4,1

Required area is area of

OAB – area of CEA

1 14 1 13

7 1 15.25 sq.units2 3 2 6

10. A straight line is fit to a data set (ln x, y). This line intercepts the abscissa at ln x = 0.1 and has a slope of

−0.02. What is the value of y at x = 5 from the fit?

(A) −0.030 (B) −0.014 (C) 0.014 (D) 0.030

Key: (A)

Exp:

n

ay a bx, where x l x and 0.1, b 0.02

b

a 0.02 x a 0.002

0.002 0.02 x

at x 5, y 0.002 0.02 1.609 0.03018

0.030

Electronics and Communication Engineering

Q. No. 1 – 25 Carry One Mark Each

1. Consider a 2 × 2 square matrix x

A ,

Where x is unknown. If the eigenvalues of the matrix A are j and j , the x is equal to

(A) j (B) j (C) (D)

Key: (D)

Exp: Product of eigen values = det

2

2 2 2

j j x

x

x

2. For 2

sin zf z ,

z the residue of the pole at z = 0 is __________.

Key: 1

O

y

x

2x 3y 5

C A

B

D

E

3x 2y 14

R




4 44/14

4/14 7EC (2)

Exp: 3 5 2

2 2

sin z 1 z z 1 z zf z z ..... ......

z z 3! 5! z 3! 5!

1Resideu coefficient of 1

z

3. The probability of getting a “head” in a single toss of a biased coin is 0.3. The coin is tossed repeatedly

till a “head” is obtained. If the tosses are independent, then the probability of getting “head” for the

first time in the fifth toss is .

Key: 0.07

Exp: Required probability = TTTTH

0.7 0.7 0.7 0.7 0.3

0.07203

4. The integral

1

0

dx

1 x is equal to _________.

Key: 2

Exp:

1 11 111 1221 12 2

0

0

0 0

1 x 1 x1 x dx 2 1 x 2

1 11

2 2

5. Consider the first order initial value problem

3y' y 2x x , y 0 1, 0 x

with exact solution 2 xy x x e . For x = 0.1, the percentage difference between the exact solution

and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-

size h = 0.1is

Key: 0.06 to 0.063

6. Consider the signal x t cos 6 t sin 8 t , where t is in seconds. The Nyquist sampling rate (in

samples/second) for the signal y(t) = x(2t+5) is

(A) 8 (B) 12 (C) 16 (D) 32

Key: (C)

Exp: Shifting doesn‟t effects the sampling rate due to scaling by a factor „2‟ spectral components are doubled.

Thus maximum frequency of Y(f ) 8.

Nyquist rate 8 2 16Hz

3

4

3 4

X(f )




5 55/15

5/14 7EC (2)

7. If the signal sin t sin t

x t *t t

with * denoting the convolution operation, then x(t) is equal to

(A) sin t

t (B)

sin 2t

2 t (C)

2sin t

t (D)

2

sin t

t

Key: (A)

Exp:

F

F

sin t 1sa t

t

sa t rect2

1sa t rect

2

Convolution in time domain leads to multiplication in frequency domain

1 sin t

rect rect X x t sa t2 2 t

8. A discrete-time signal x n n 3 2 n 5 has z-transform X(z). If Y(z) = X(-z) is the z-

transform of another signal y[n], then

(A) y[n] = x[n] (B) y[n] = x[-n] (C) y[n] = -x[n] (D) y[n] = -x[-n]

Key: (C)

Exp:

3 5

x n n 3 2 n 5

x z z 2z

3 5

3 5

x z z 2 2

y z z 2z

y n n 3 2 n 5

y n x n

9. In the RLC circuit shown in the figure, the input voltage is given by

i t 2cos 200t 4sin 500t

The output voltage 0 t is

(A) cos (200t) + 2sin (500t)

(B) 2 cos (200t) + 4sin (500t)

(C) sin (200t) + 2cos (500t)

(D) 2 sin (200t) + 4cos (500t)

Key: (B)




6 66/16

6/14 7EC (2)

Exp: Vi(t) = 2cos 200t + 4 sin 500t, since there are 2 frequency term output will also have 2 frequency term.

If we take 4sin500t first i.e. W = 500 then on the output section, this parallel LC combination have

LCZ , so it is open circuit and V0 = Vi

So w.r.t. 4sin500t output must be 4sin500t without any change in amplitude and phase, this is satisfied by

only option B.

10. The I-V characteristics of three types of diodes at the room temperature, made of semiconductors X, Y

and Z, are shown in the figure. Assume that the diodes are uniformly doped and identical in all respects

except their materials. If EgX, EgY and EgZ are the band gaps of X, Y and Z, respectively, then

(A) EgX > EgY > EgZ (B) EgX = EgY = EgZ

(C) EgX < EgY < EgZ (D) no relationship among these band gaps exists

Key: (C)

11. The figure shows the band diagram of a Metal Oxide Semiconductor (MOS). The surface region of this

MOS is in

(A) inversion (B) accumulation (C) depletion (D) flat band

Key: (A)

Exp: The semiconductor used in the MOSFET is n-type. At the surface the intrinsic level is above EF as it is

found at the distance of below EF, So, the surface is in inversion region.

X Y Z

I

V

j200 j200




7 77/17

7/14 7EC (2)

12. The figure shows the I-V characteristics of a solar cell illuminated uniformly with solar light of power 100 mW/cm

2. The solar cell has an area of 3 cm

2 and a fill factor of 0.7. The maximum efficiency

(in %) of the device is ________.

Key: 21

Exp: oc sc

in

FF.V I 0.7 0.5 180Efficiency 100%

P (100 3)

= 21%.

13. The diodes D1 and D2 in the figure are ideal and the capacitors are identical. The product RC is very

large compared to the time period of the ac voltage. Assuming that the diodes do not breakdown in the

reverse bias, the output voltage V0 (in volt) at the steady state is

Key: 0

Exp: oV 0volts

10V

10V

oV 0V




8 88/18

8/14 7EC (2)

14. Consider the circuit shown in the figure. Assuming VBE1 = VEB2 = 0.7 volt, the value of the dc

voltage VC2 (in volt) is

Key: 0.5

Exp: 2

2 2

2

2

2

E 2.5 BE1

B E

5

B

5

C

4 3

C

V V 1.8V

V V 0.7 1.1V

0.1I 10 A

10k

I 50 10 A

V 5 10 10 0.5V

15. In the astable multivibrator circuit shown in the figure, the frequency of oscillation (in kHz) at the

output pin 3 is .




9 99/19

9/14 7EC (2)

Key: 5.64

Exp: 6

A B

1.44 1.44f 5.64kHz

R 2R C [2200 2 4700] 0.022 10

16. In an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0,

respectively. If the instruction RLC is executed, then the contents of the accumulator (in hex) and the

carry flag, respectively, will be

(A) 4E and 0 (B) 4E and 1 (C) 4F and 0 (D) 4F and 1

Key: (D)

Exp: Accumulator

RLCRotate left accumulator content without carry

17. The logic functionality realized by the circuit shown below is

(A) OR (B) XOR (C) NAND (D) AND

Key: (D)

Exp: All the transistor are n-mos we know when input to gate is 0 n mos behave as open circuit. Input to

Gate is 1 n mos is short circuit.

we can redraw the circuit as

0 01 01 0 1 1 1

CY

0 1 0 10 1 1 1

CY

1 4FH

Accumulator

A

Y

1T

2T

B

B




10 1010/

11010

/14 7EC

to get the functionality get find its truth table

1 2A B T T Y

0 0 OFF ON 0

0 1 ON OFF A 0

1 0 OFF ON 0

1 1 ON OFF A 1

So, Y satisfy AND gate table.

18. The minimum number of 2-input NAND gates required to implement a 2-input XOR gate is

(A) 4 (B) 5 (C) 6 (D) 7

Key: (A)

Exp:

19. The block diagram of a feedback control system is shown in the figure. The overall closed-loop gain

G of the system is

(A) 1 2

1 1

G GG

1 G H

(B) 1 2

1 2 1 1

G GG

1 G G G H

(C) 1 2

1 2 1

G GG

1 G G H

(D) 1 2

1 2 1 2 1

G GG

1 G G G G H

Key: (B)

Exp:

2G 1G

1H

1

1 1

G

1 G H

A

B

Y AB AB A B




11 1111/

11111

/14 7EC

1 2

1 1 1 2

G GY

x 1 G H G G

20. For the unity feedback control system shown in the figure, the open-loop transfer function G(s) is

given as

2G s

s s 1

The steady state error ess due to a unit step input is

(A) 0 (B) 0.5 (C) 1.0 (D) ∞

Key: (A)

Exp: For unit step input p

1ess

1 k

.

ps 0 s 0

ss

2k limG(s) lim

s(s 1)

1So, e 0

1

21. For a superheterodyne receiver, the intermediate frequency is 15 MHz and the local oscillator frequency

is 3.5 GHz. If the frequency of the received signal is greater than the local oscillator frequency, then

the image frequency (in MHz) is .

Key: 3485

Exp: imagef = ?

S L0 Si L0f = f (f - f )

Sif = Image freq

L0f = Signal freq

S L0 Si L0f - f f + f

Si L0 Sf 2f f

S L0 1Ff f f

Si L0 1F L0f 2f f f L0 1Ff f 3500 15 =3485 mHz




12 1212/

11212

/14 7EC

22. An analog baseband signal, band limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and

p2 = p3. The information rate (bits/sec) of the message source is

Key: 360 to 363

23. A binary baseband digital communication system employs the signal

S

S

1, 0 t T

Tp t

0, otherwise

for transmission of bits. The graphical representation of the matched filter output y(t) for this signal will

be

(A) (B)

(C) (D)

Key: (C)

Exp: P(t) h(t) y(t)

Sh(t) = P(t -t)

ST 0

S

1

T

t

S2T 0

1




13 1313/

11313

/14 7EC

24. If a right-handed circularly polarized wave is incident normally on a plane perfect conductor, then the

reflected wave will be

(A) right-handed circularly polarized (B) left-handed circularly polarized

(C) elliptically polarized with a tilt angle of 450

(D) horizontally polarized

Key: (B)

Exp: If incident wave is right handed polarized then the reflected wave is left handed polarized.

25. Faraday‟s law of electromagnetic induction is mathematically described by which one of the following

equations?

(A) .B 0

(B) V.D

(C)

BE

t

(D) D

H Et

Key: (C)

Q. No. 26 – 55 carry Two Marks Each

26. The particular solution of the initial value problem given below is

2

2

x 0

d y dy dy12 36y 0 with y 0 3 and 36

dx dx dx

(A) 6x3 18x e (B) 6x3 25x e (C) 6x3 20x e (D) 6x3 12x e

Key: (A)

Exp: 2D 12D 36 0

6x

1 2

6x

1 2

1

6x 6x

2 1 2

D 6, 6

C.F C C x e

y C C x e

y 0 3

3 C

dyC e C C x e 6

dx

x 0

2 2

6x

dy36

dx

36 C 3 0 .1 6 C 18

y 3 18x e

27. If the vectors e1 = (1, 0, 2), e2 = (0, 1, 0) and e3 = (−2, 0, 1) form an orthogonal basis of the three-

dimensional real space R3, then the vector 3u 4,3, 3 R can be expressed as

(A) 1 2 3

2 11u e 3e e

5 5 (B)

1 2 3

2 11u e 3e e

5 5

(C) 1 2 3

2 11u e 3e e

5 5 (D)

1 2 3

2 11u e 3e e

5 5




14 1414/

11414

/14 7EC

Key: (D)

Exp: From option (D),

1 2 3

2 11u e 3e e

3 5

2 111,0,2 3 0,1,0 2,0,1

5 5

4,3, 3

28. A triangle in the xy-plane is bounded by the straight lines 2x = 3y, y = 0 and x = 3. The volume

above the triangle and under the plane x + y + z = 6 is

Key: 10

Exp: Volume = R

zdxdy

3 2x

3

y 0x 0

2x3 2 3

x 0 y 0

32

32 2

33

2 3

0

6 x y dy dx

y6 x y .dx

z

2x 1 46 x x dx

3 2 9

2 24x x x dx

3 9

8 x 82x 2 9 3 10cubic units

9 3 27

29. The values of the integral

z

c

1 edz

2 j z 2 along a closed contour c in anti-clockwise direction for

(i) The point z0 = 2 inside the contour c, and

(ii) The point z0 = 2 outside the contour c,

Respectively, are

(A) (i) 2.72, (ii) 0 (B) (i) 7.39, (ii) 0 (C) (i) 0, (ii) 2.72 (D) (i) 0, (ii) 7.39

Key: (B)

Exp: (i) 2

21 e 12 jf 2 e 7.39

2 j z 2 2 j

(ii) 0

30. A signal 2

2cos t cos t3

is the input to an LTI system with the transfer function

s sH s e e .

If Ck denotes the kth coefficient in the exponential Fourier series of the output signal, then C3 is equal to

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

O

y

x

B

A

Rx b

2x 3y

x limits : 0 to 3

2y limits: 0 to x

3




15 1515/

11515

/14 7EC

Key: (B)

Exp: 0

nd rd

2output 2cos t 2cos t

3 3

2 harmonic 3 harmonic

In E.F.S 3rd

harmonic j 3t j 3t

3 32cos t e e

The coefficient of j 3t

3

3e is1So C 1.

31. The ROC (region of convergence) of the z-transform of a discrete-time signal is represented by the shaded

region in the z-plane. If the signal n

x n 2.0 , n , then the ROC of its z-transform is

represented by

(A) (B)

(C) (D)




16 1616/

11616

/14 7EC

Key: (D)

Exp:

n

n n

n

n

x n 2 , n

x n 2 u n 2 u n 1

2 u n ; ROC : z 2

2 u n 1 ; ROC : z 0.5

Thus combined ROC does not exist for x[n]

32. Assume that the circuit in the figure has reached the steady state before time t = 0 when the 3

resistor suddenly burns out, resulting in an open circuit. The current i(t) (in ampere) at t = 0+ is ______.

Key: 1

Exp: At t 0, the circuit is on steady state i.e. the capacitor is open circuited so the circuit will be

3F 2 3 3

2F 3

V V V V 4V

V V 6V

at t = 0+ when is open circuited, the capacitors will have an ideal voltage source of values 4V and

6V so the circuit will be

So the current through 2 resistor at t = 0+ should be

41A

2 2

2A

1

2

3

3F

12V

2F

2




17 1717/

11717

/14 7EC

33. In the figure shown, the current i (in ampere) is ______.

Key: -1

Exp: Nodal equation at V

V 8 V V 8 V

01 1 1 1

4V 16

V 4V

By using KCL at node „a‟.

1 1

8 41 i 0 i 5A

1

KCL at b

1

4i i 0 4 5 i 0

1

i 1A

1A

1

51i

a

v1

b0V

i1

8V

8V

12

2

6V

4V

12

Burn out

1

2

2

6V

4V

at t 0

12




18 1818/

11818

/14 7EC

34. The z-parameter matrix 11 12

21 22

z z

z z

for the two-port network shown in

(A) 2 2

2 2

(B) 2 2

2 2

(C) 9 3

6 9

(D) 9 3

6 9

Key: (A)

Exp: Since the given network is symmetric

and reciprocal 11 22

12 21

Z Z

Z Z

2

111

1 I 0

V 3 6Z 2

I 3 6

2

221

1 I 0

VZ

I

We know 2 1 21V V Z 2

So 11 12

21 22

Z Z 2 2

Z Z 2 2

35. A continuous-time speech signal xa(t) is sampled at a rate of 8 kHz and the samples are subsequently

grouped in blocks, each of size N. The DFT of each block is to be computed in real time using the radix-

2 decimation-in-frequency FFT algorithm. If the processor performs all operations sequentially, and takes

20 s for computing each complex multiplication (including multiplications by 1 and −1) and the time

required for addition/subtraction is negligible, then the maximum value of N is _____

Key: 4096

36. The direct form structure of an FIR (finite impulse response) filter is shown in the figure.

The filter can be used to approximate a

2V

LI 0

3

1I

1V

input

output

6

3




19 1919/

11919

/14 7EC

(A) low-pass filter (B) high-pass filter (C) band-pass filter (D) band-stop filter

Key: (C)

Exp:

j 2 j

j

j

y n 5x n 5x n 2

H e 5 1 e

At 0, H e 0

At ; H e 0

Thus the filter is band pass filter

37. The injected excess electron concentration profile in the base region of an npn BJT, biased in the

active region, is linear, as shown in the figure. If the area of the emitter-base junction is 0.001 cm2,

2

n 800cm V s in the base region and depletion layer widths are negligible, then the collector

current Ic (in mA) at room temperature is __________.

(Given: thermal voltage VT = 26 mV at room temperature, electronic charge19q 1.6 10 C )

Key: 6.65

Exp: c n n t

dn dnI qAD qA V

dx dx

1419 3

4

10 01.6 10 0.001 800 26 10

0.5 10

cI 6.65mA




20 2020/

12020

/14 7EC

38. Figures I and II show two MOS capacitors of unit area. The capacitor in Figure I has insulator materials

X (of thickness t1 = 1 nm and dielectric constant 1 4 ) and Y (of thickness t2 = 3 nm and dielectric

constant 2 20 ). The capacitor in Figure II has only insulator material X of thickness tEq. If the

capacitors are of equal capacitance, then the value of tEq (in nm) is ___________.

Key: 1.6

Exp:

1 2

1 2 1 2I

1 2 1 2 2 1

1 2

E E.

t t E E 4 20C 2.5

E E E t E t (4 2) (20 1)

t t

1II

Eq Eq

E 4C

t t 2.5 Eqt 1.6nm

39. The I-V characteristics of the zener diodes D1 and D2 are shown in Figure I. These diodes are used in

the circuit given in Figure II. If the supply voltage is varied from 0 to 100 V, then breakdown occurs in

(A) D1 only (B) D2 only (C) both D1and D2 (D) none of D1 and D2

Key: (A)




21 2121/

12121

/14 7EC

40. For the circuit shown in the figure, 1 2 3R R R 1 , L 1 H and C 1 F. If the input

6

inV cos 10 t , then the overall voltage gain out inV V of the circuit is ______.

Key: -1

Exp: 11 6 6

R 1A 1 1 2

j L 10 10

32

2 C6 6

R 1 1 1A

1R X 1 1 21

10 10

The overall voltage gain outv 1 2

in

VA A A

V

out

in

V 12 1

V 2

41. In the circuit shown in the figure, the channel length modulation of all transistors is non-zero

0 . Also, all transistors operate in saturation and have negligible body effect. The ac small

signal voltage gain 0 inV V of the circuit is

(A) m1 01 02 03g r || r || r (B) m1 01 03

m3

1g r || || r

g

(C) m1 01 02 03

m2

1g r || || r || r

g

(D) m1 01 03 02

m3

1g r || || r || r

g

Key: (C)




22 2222/

12222

/14 7EC

42. In the circuit shown in the figure, transistor M1 is in saturation and has transconductance gm = 0.01

siemens. Ignoring internal parasitic capacitances and assuming the channel length modulation to be

zero, the small signal input pole frequency (in kHz) is .

Key: 57.9

Exp: in mC 50PF 1 g R 550PF

m

m

in

in

in in

g 0.015

R 1k

R 1k

1 g R 11

R 5k

1f 57.9 kHz

2 R C

43. Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sum-

of-product (SOP) expression for the function is

(A) P QSX P QSX Q R SX QRSX (B) QSX QSX

(C) QSX QSX (D) QS QS

Key: (B)




23 2323/

12323

/14 7EC

44. For the circuit shown in the figure, the delays of NOR gates, multiplexers and inverters are 2 ns, 1.5

ns and 1 ns, respectively. If all the inputs P, Q, R, S and T are applied at the same time instant, the

maximum propagation delay (in ns) of the circuit is .

Key: (6)

Exp: Case (i) When T = 0

Ttotal = delay of NOR + delay of 1st MUX + delay of 2

nd MUX= 2+1.5+1.5 = 5ns

Case (ii) When T = 1

Ttotal = delay of 1st NOT-gate + delay of 1

st MUX + delay of 2

nd NOR-gate + delay of 2

nd MUX

= 1+1.5+2+1.5 = 6 ns

So, the maximum delay = 6 ns.

45. For the circuit shown in the figure, the delay of the bubbled NAND gate is 2 ns and that of the

counter is assumed to be zero.

If the clock (Clk) frequency is 1 GHz, then the counter behaves as a

(A) mod-5 counter (B) mod-6 counter (C) mod-7 counter (D) mod-8 counter

Key: (D)

Exp: The time period of clock is 1nsec and that of nand gate is 2n sec.

The time period of clock 1n sec and that of nand gate is 2n.sec.




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If the nand gate would have 0 delay then the counter must be MOD 6 because on 6th CP it will be

rejected.

But as the delay is 2nsec, the effect of 6th CP will be observed on reset in after 2nsec, but in this

additional 2n.sec, 2 more clock cycle would passed in total 8 cycles are needed to rest the counter so

MOD8 counter.

46. The first two rows in the Routh table for the characteristic equation of a certain closed-loop control

system are given as

The range of K for which the system is stable is

(A) −2.0 < K < 0.5 (B) 0 < K < 0.5 (C) 0 K (D) 0.5 K

Key: (D)

Exp: 3

2

1 2k 3S

2k 4S

From the table we can find characteristic equation

3 2s 2ks 2k 3 s 4 0

For stability

2

2k 2k 3 4

4k 6k 4 0

1k k 2 0

2

So the conditions are 1

k2

and k 2 combiningly k > – 2

47. A second-order linear time-invariant system is described by the following state equations

1 1

2 2

dx t 2x t 3u t

dt

dx t x t u t

dt

where x1(t) and x2(t) are the two state variables and u(t) denotes the input. If the output c(t) = x1(t), then

the system is

(A) controllable but not observable

(B) observable but not controllable

(C) both controllable and observable

(D) neither controllable nor observable

Key: (A)

Exp: The set of equation of the system are




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1i1 1 2

dx t2x t 3u t x t 2x t 0x t 3u t

dt

2

2

2 1 2

1 2

dx tx t u t x t 0 x t x t u t

dt

c t x t 0x t

we can frame the state space of the system as

1 1

22

1

2

xx 2 0 34

x0 1 1x

xy 1 0

x

2 0A matrix is

0 1

3Bmatrix is

1

C matrix is 1 0

for controllability determinant of B AB 0

3 6

3 6 3 01 1

so controllable

for observability determinant of C

0Ca

1 0

02 0

so not observable

final controllable but not observable

48. The forward-path transfer function and the feedback-path transfer function of a single loop negative

feedback control system are given as

2

K s 2G s and H s 1,

s 2s 2

Respectively, If the variable parameter K is real positive, then the location of the breakaway point on the

root locus diagram of the system is .

Key: -3.414

Exp: To find break point, from characteristic equation we need to arrange k as function of s, then the root of

dk0

ds gives break point.

Characteristic equation is given by




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2s 2s 2 k 2k 0

2

2

k s 2 s 2s 2

s 2s 2k

s 2

2 2

2

d ds 2 s 2s 2 s 2s s 2

dk ds ds

ds s 2

2

2

dk s 2 2s 2 s 2s 2

ds s 2

dk0

ds

2 2

2

2s 2s 4s 4 s 2s 2 0

s 4s 2 0

s 0.58 and 3.414

To find the valid break point we need to find that lies on root locus

– 3.414 lies on root locus

So break point – 3.414.

49. A wide sense stationary random process X(t) passes through the LTI system shown in the figure. If the

autocorrelation function of X(t) is XR , then the autocorrelation function YR of the output Y(t) is

equal to

(A) X X 0 X 02R R T R T (B) X X 0 X 02R R T R T

(C) X X 02R 2R 2T (D) X X 02R 2R 2T

Key: (B)

2

1

j

j




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Exp: YYR (τ) = E[Y(t)Y(t - τ)]

0 0E[{X(t)-X(t - T )}{X(t-τ)-X(t -τ T )}]

X X 0 X 0 X= R (τ) - R (-τ-T )-R (-T+T )+R (-τ)

YY X X 0 X 0R (τ) = 2R (τ) - R (τ + T )-R (τ-T )

50. A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0

kHz and two-sided noise power spectral density 52.5 10

2

Watt per Hz. If information at the rate

of 52 kbps is to be transmitted over this channel with arbitrarily small bit error rate, then the minimum

bit-energy Eb (in mJ/bit) necessary is .

Key: 31.5

Exp: Information rate

R C Channel capacity

For Error free transmission

2

S C = B log 1

N

B = 4KHz

b b b bS = E / T E R

η

N = .2B = ηB2

b 2

SR B log 1

N

b bb 2

E RR B log 1+

ηB

bE 31.5mJ/bit

bminE 31.5mJ/bit

51. The bit error probability of a memoryless binary symmetric channel is 10−5

. If 105

bits are sent over this

channel, then the probability that not more than one bit will be in error is _________.

Key: 0.7 to 0.75




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52. Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. At 10

GHz operating frequency, the value of the propagation constant (per meter) of the corresponding

propagating mode is _________.

Key: 157

53. Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. The

increasing order of the cut-off frequencies for different modes is

(A) TE01 < TE10 < TE11 < TE20 (B) TE20 < TE11 < TE10 < TE01

(C) TE10 < TE20 < TE01 < TE11 (D) TE10 < TE11 < TE20 < TE01

Key: (C)

Exp:

2 2

c

01 c 2

C m nf

2 a b

C 1 Cfor TE f 98.4

2 1.016 10 2

for TE10 c 2

C 1 Cf 43.74

2 2.286 10 2

For TE11 c

Cf 107.7

2

For TE20 c 2

C 2 Cf 87.49

2 2.286 10 2

TE10 < TE20 < TE01 < TE11

54. A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a

gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having

radar cross-section of 3 m2. If it transmits 100 kW, then the received power in W is

__________.

Key: 0.012

55. Consider the charge profile shown in the figure. The resultant potential distribution is best described by




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(A) (B)

(C) (D)

Key: (D)

Exp: Electrical 0 d no d po

q qN x N x

Potential 0(x) 2

0

xx

2w

a

1P

b

1P

P(x)

1P

a

b

V(x)

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