comsol

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1. The minimum sampling frequency (in samples/sec) required to reconstruct the following

signal from its samples without distortion 3 2

sin 2 1000t sin 2 1000tX t 5 7

t t

would be

(a) 32 10 (b) 34 10

(c) 36 10 (d) 38 10 [GATE-2006]

Solution:

The highest frequency signal in x(t) is 1000 * 3 = 3 kHz if expression is expanded.

Thus minimum frequency requirement is f = 2 * 3 * 10000= 36 10 Hz Hence (C) is correct option.

2. The minimum step-size required for a Delta-Modulator operating at 32 K samples/sec to track the signal (here u (t) is the unit-step function)

x t 125t u t u t 1 250 125t u t 1 u t 2 So that slope-overload is avoided, would be

(a) 102 (b) 82

(c) 62 (d) 42

[GATE-2006]

Solution:

We have

x(t) = 125t [u(t) u(t 1)] + (250 125t) [u(t 1) u(t 2)] The slope of expression x(t) is 125 and sampling frequency fs is 32 * 1000 samples/sec.

Let be the step size, then to avoid slope overload

Hence (B) is correct option.

3. A zero-mean white Gaussian noise is passed through an ideal lowpass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period tS = 0.03 msec. The

samples so obtained would be

(a) correlated (b) statistically independent

(c) uncorrelated (d) orthogonal [GATE-2006]

Solution:

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4. A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient

source encoder would have average bit rate as

(a) 6000 bits/sec (b) 4500 bits/sec

(c) 3000 bits/sec (d) 1500 bits/sec [GATE-2006]

Solution:

5. The diagonal clipping in Amplitude Demodulation (using envelope detector) can be

avoided if RC time-constant of the envelope detector satisfies the following condition,

(here W is message bandwidth and S is carrier frequency both in rad/sec)

(a) 1

RCW

(b) 1

RCW

(c) 1

RC

(d) 1

RC

[GATE-2006]

Solution:

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6. In the following figure the minimum value of the constant C, which is to be added to y1(t) such that y1 (t) and y2(t) are different, is

(a) (b) 2

(c) 2

12

(d)

L

[GATE-2006]

Solution:

7. A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with

carrier frequency fC1 = 106 Hz. The resulting signal is then passed through a Narrow-

Band Frequency Modulator with carrier frequency fC2 = 109 Hz.

The bandwidth of the output would be:

(a) 44 10 Hz (b) 62 10 Hz

(c) 92 10 Hz (d) 102 10 Hz [GATE-2006]

Solution:

Common Data for Questions 8, 9:

Let g(t)=p(t)*p(t), where * denotes convolution and x

p t u t u t 1 lim

with u (t )

being the unit step function

8. The impulse response of filter matched to the signal is s t g t t 2 *g t given as:

(a) s(1 t) (b) s(1 t) (c) s(t) (d) s(t) [GATE-2006]

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Solution:

9. An Amplitude Modulated signal is given as AM cx 100 p t 0.5g t cos t in the

interval 0 t 1 . One set of possible values of the modulating signal and modulation

index would be

(a) t, 0.5 (b) t, 1.0

(c) t, 2.0 (d) 2t , 0.5 [GATE-2006]

Solution:

Statement for Linked Answer Questions 10 & 11:

The following two questions refer to wide sense stationary stochastic processes

10. It is desired to generate a stochastic process (as voltage process) with power spectral

density 2

16S

16

, By driving a Linear-Time-Invariant system by zero mean white

noise (as voltage process) with power spectral density being constant equal to 1. The

system which can perform the desired task could be:

(a) first order low pass R-L filter (b) first order high pass R-c filter (c) tuned L-C filter (d) series R-L-C filter [GATE-2006]

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Solution:

11. The parameters of the system obtained in Q.7 would be

(a) First order R-L low pass filter would have R = 4W L = 4H

(b) First order R-C high pass filter would have R = 4W C = 0.25F

(c) Tuned L-C filter would have L = 4H C = 4F

(d) Series R-L-C low pass filter would have R = 1W, L = 4H, C = 4F [GATE-2006]

Solution:

Statement for Linked Answer Questions 12 & 13:

Consider the following Amplitude Modulated (AM) signal, where fm < B :

AM m cx t 10 1 0.5sin 2 f t cos2 f t 12. The average side band power for the AM signal given above is:

(a) 25 (b) 12.5

(c) 6.25 (d) 3.125 [GATE-2006]

Solution:

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13. The AM signal gets added to a noise with Power Spectral Density Sn(f) given in the figure below. The ratio of average sideband power to mean noise power would be:

(a) 0

25

8N B (b)

0

25

4N B

(c) 0

25

2N B (d)

0

25

N B [GATE-2006]

Solution:

14. If E denotes expectation, the variance of a random variable X is given by

(a) 2 2E x E x (b) 2 2

E x E x

(c) 2E x (d) 2

E x [ GATE-2007]

Solution:

15. A Hilbert transformer is a

(a) Non-linear system

(b) non-causal system

(c) time-varying system

(d) low-pass system [GATE-2007]

Solution:

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16. In delta modulation, the slope overload distortion can be reduced by (a) decreasing the step size

(b) decreasing the granular noise

(c) decreasing the sampling rate

(d) increasing the step size [GATE-2007]

Solution:

17. The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t ) with unity roll-off factor is given by

2 2

sin 4 Wtp t

4 Wt 1 16W t

.The value of p(t) at

1t

4W is

(a) 0.5 (b) 0 (c) 0.5 (d) [GATE-2007] Solution:

18. In the following scheme, if the spectrum M(f) of m(t ) is as shown, then the spectrum Y (f ) of y (t ) will be:

(a)

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(b)

(c)

(d)

[GATE-2007]

Solution:

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19. During transmission over a certain binary communication channel, a bit error occurs

independently with probability p. The probability of AT MOST one bit in error in a

block of n bits is given by

(a) np (b) n1 p

(c) n 1

n nnp 1 p 1 p

(d) n1 1 p [GATE-2007] Solution:

20. In a GSM system, 8 channels can co-exist in 200 KHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse

factor of (1/5) ,i.e. a five-cell repeat pattern, the maximum number of simultaneous

channels that can exist in one cell is:

(a) 200 (b) 40

(c) 25 (d) 5 [GATE-2007]

Solution:

21. In a Direct Sequence CDMA system the chip rate is 31.2288 10 chips per second. If the

processing gain is desired to be AT LEAST 100, the data rate

(a)Must be less than or equal to 312.288 10 bits per sec

(b) Must be greater than 312.288 10 bits per sec

(c) Must be exactly equal to 312.288 10 bits per sec

(d) Can take any value less than 312.288 10 bits per sec [GATE-2007]

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Solution:

Common Data for Questions 22 & 23:

Two 4-ray signal constellations are shown. It is given that 1 and 2 constitute an

orthonormal basis for the two constellations. Assume that the four symbols in both the

constellations are equiprobable. Let 0N

2 denote the power spectral density of white

Gaussian noise.

22. The ratio of the average energy of Constellation 1 to the average energy of Constellation

2 is:

(a) 24a (b) 4

(c) 2 (d) 8 [GATE-2007]

Solution:

23. If these constellations are used for digital communications over an AWGN channel, then

which of the following statements is true?

(a) Probability of symbol error for Constellation 1 is lower

(b) Probability of symbol error for Constellation 1 is higher

(c) Probability of symbol error is equal for both the constellations

(d) The value of 0 N will determine which of the two constellations has a lower

Probability of symbol error. [GATE-2007]

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Solution:

Statement for Linked Answer Questions 24 & 25:

An input to a 6-level quantizer has the probability density function f (x) as shown in the

figure. Decision boundaries of the quantizer are chosen so as t maximize the entropy of

the quantizer output. It is given that 3 consecutive decision boundaries are -1, 0 and 1.

24. The values of a and b are:

(a) a = 1/6 and b = 1/12

(b) a = 1/5 and b = 3/40

(c) a = 1/4 and b = 1/16

(d) a = 1/3 and b = 1/24 [GATE-2007]

Solution:

25. Consider the amplitude modulated (AM) signal c c m cA cos t 2cos t cos t . For b

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

(a) 2 (b) 1

(c) 0.5 (d) 0 [GATE-2008]

Solution:

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26. A memory less 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 [GATE-2008]

Solution:

27. 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 cf . The noise power at the

filter output is

(a) k (b) ckf

(c) ck f (d) [GATE-2008]

Solution:

[

28. 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) 3

1 p (d) 3 2p p 1 p [GATE-2008] Solution:

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29. 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 [GATE-2008]

Solution:

30. Consider the frequency modulated signal 10

5cos 2 10 t 5sin 2 1500t 7.5sin 2 1000t with carrier frequency of 5

10 Hz.

The modulation index is

(a) 12.5 (b) 10

(c) 7.5 (d) 5 [GATE-2008]

Solution:

31. The signal C m ccos t 0.5cos t sin t is

(a) FM only

(b) AM only

(c) both AM & FM

(d) neither AM nor FM [GATE-2008]

c

Common Data Questions 33, 34 & 35:

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

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

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

(a) 64 kHz (b) 32 kHz

(c) 8 kHz (d) 4 kHz [GATE-2008]

Solution:

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33. 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 [GATE-2008]

Solution:

34. The number of quantization levels required to reduce the quantization noise by a factor

of 4 would be

(a) 1024 (b) 512

(c) 256 (d) 64 [GATE-2008]

Solution:

35. Suppose that the modulating signal is mtm t 2cos 2 f and the carrier signal is

c c cx t A cos 2 f t which one of the following is a conventional AM signal without over-modulation?

(a) c cx t A m t cos 2 f t

(b) c cx t A 1 m t cos 2 f t

(c) cc c cA

x t A cos 2 f t m t cos 2 f t4

(d) c m c c m cx t A cos 2 f t cos 2 f t A sin 2 f t sin 2 f t [GATE-2010] Solution:

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36. Consider an angle modulated signal x(t) = 6cos[2x106t+2sin(8000t) + 4cos(8000pt)] V. The average power of x(t) is.

(a) 10W (b) 18W

(c) 20W (d) 28W [GATE-2010]

Solution:

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

(a) (b)

(c) (d)

[GATE-2010] Solution:

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38. The Nyquist sampling rate for the signal sin 500 t sin 700

s tt t

is given by

(a) 400Hz (b) 600Hz

(c) 1200Hz (d) 1400Hz

Solution:

39. 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)f 0,f0=1kHz, n any integer [GATE-2010]

Solution:

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Statement for Linked Answer Questions: 41 & 42:

Consider a base band binary PAM receiver shown below. The additive channel noise

n(t) is whit with power spectral density 200NN Ws f 10

Hz2

. The low-pass filter is

ideal with unity gain and cut-off 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, nNKP n 0.5 e

| (This has mean zero and variance 22). Assume

transmitted bits to be equiprobable and threshold z is set to6a 10 V

2 .

40. The value of the parameter (in V-1) is

(a) 1010 (b) 710

(c) 101.414 10 (d) 202 10 [GATE-2010]

Solution:

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41. The probability of bit error is

(a) 3.50.5 e (b) 50.5 e

(c) 70.5 e (d) 100.5 e [GATE-2010]

Solution:

42. Which of the following analog modulation scheme requires the minimum transmitted

power and minimum channel bandwidth?

(a) VSB (b) DSB-SC

(c) SSB (d) AM [GATE-2005]

Solution:

43. Refractive index of glass is 1.5. Find the wavelength of a beam of light with a frequency

of 1010 Hz in glass. Assume velocity of light is 83 10 m / s in vacuum.

(a) 3 m (b) 3mm

(c) 2 m (d) 1 m [GATE-2005]

Solution:

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44. A device with input x(t) and output y(t) is characterized by: 2y t x t .An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied

to this device. The bandwidth of the output signal is

(a) 370 KHz (b) 190 KHz

(c) 380 KHz (d) 95 KHz [GATE-2005]

45. A signal as shown in figure is applied to a matched filter. Which of the following does represent the output of this matched filter?

[GATE-2005] (a) (b)

(d)

(c)

Solution:

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46. Noise with uniform power spectral density of 0N W

Hz is passed through a filter

dH 2exp j t followed by an ideal low pass filter of bandwidth B Hz. The output noise power in Watts is

(a) 2N0B (b) 4N0B

(c) 8N0B (d) 16N0B [GATE-2005]

Solution:

47. A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that

phase deviation remains unchanged, the bandwidth of the PM signal is

(a) 21 kHz (b) 22 kHz

(c) 42 kHz (d) 44 kHz [GATE-2005]

Solution:

48. An output of a communication channel is a random variable n with the probability

density function as shown in figure. The mean square value of n is

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(a) 21 kHz (b) 22 kHz

(c) 42 kHz (d) 44 kHz [GATE-2005]

Solution:

Statement for Linked Answer Questions 50 & 51:

A symmetric three-level midtread quantizer is to be designed assuming equiprobable

occurrence of all quantization levels.

49. If the input probability density function is divided into three regions as shown in figure, the value of a in the figure is

(a) 1/3 (b) 2/3

(c) 1/2 (d) 1/4 [GATE-2005]

Solution:

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50. The quantization noise power for the quantization region between a and +a in the figure is

(a) 4

81 (b)

1

9

(c) 5

81 (d)

2

81 [GATE-2005]

Solution:

51) In a baseband communications link, frequencies upto 3500 Hz are used for

signaling.

Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol

interference, the maximum possible signaling rate in symbols per second is

(A) 1750 (B) 2625

(C) 4000 (D) 5250

SOL:

Option (C) is correct.

For raised cosine spectrum transmission bandwidth is given as

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(GATE 2012)

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(GATE 2012)

53) A source alphabet consists of N symbols with the probability of the first two

symbols being the same. A source encoder increases the probability of the first

symbol by a small amount and decreases that of the second by . After encoding,

the entropy of the source

(A) increases (B) remains the same

(C) increases only if N = 2 (D) decreases

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(GATE

2012)

54) An analog signal is band-limited to 4 kHz, sampled at the Nyquist rate and

the samples are quantized into 4 levels. The quantized levels are assumed to be

independent and equally probable. If we transmit two quantized samples per

second,

the information rate is ________ bits / second.

(A) 1 (B) 2

(C) 3 (D) 4

SOL : Quantized 4 level require 2 bit representation i.e. for one sample 2 bit are

required.

Since 2 sample per second are transmitted we require 4 bit to be transmitted per

second.

Hence (D) is correct option. (GATE 2011)

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55)

The Column-I lists the attributes and the Column-II lists the modulation

systems.

Match the attribute to the modulation system that best meets it

Column-I

P. Power efficient transmission of signals

Q. Most bandwidth efficient transmission of voice signals

R. Simplest receiver structure

S. Bandwidth efficient transmission of signals with Significant dc component

Column-II

1. Conventional AM

2. FM

3. VSB

4. SSB-SC

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

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

SOL:

In FM the amplitude is constant and power is efficient transmitted. No variation

in power.

There is most bandwidth efficient transmission in SSB- SC. because we

transmit

only one side band.

Simple Diode in Non linear region ( Square law ) is used in conventional AM

that

is simplest receiver structure.

In VSB dc. component exists.

Hence (B) is correct option. (GATE 2011)

56) A message signal m(t) = cos 200t + 4 cost modulates the carrier c(t) = cos 2fc t where fc = 1MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of

the detector circuit should satisfy

(A) 0.5 ms < RC < 1 ms (B) 1 s 0.5 ms

SOL : Highest frequency component in m(t) is fm = 4000/2 = 2000 Hz Carrier frequency fC = 1 MHz

For Envelope detector condition

1/fC

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Hence (B) is correct option (GATE 2011)

57) A fair dice is tossed two times. The probability that the second toss results

in a value that is higher than the first toss is

(A) 2/36 (B) 2/6 (C) 5/12 (D)1/2.

SOL : Total outcome are 36 out of which favorable outcomes are :

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);

(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.

Thus

P(E) =No of total outcomes/No of favourable outcomes=15/36=5/12

Hence (C) is correct option. (GATE 2011)

58)

Assuming high SNR and that all signals are equally probable, the additional

average transmitted signal energy required by the 8-PSK signal to achieve the

same error probability as the 4-PSK signal is

(A) 11.90 dB (B) 8.73 dB

(C) 6.79 dB (D) 5.33 dB

SOL : Here Pe for 4 PSK and 8 PSK is same because Pe depends on d . Since

Pe is same, d is same for 4 PSK and 8 PSK.

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Additional Power SNR

= (SNR)2 (SNR)1

(GATE 2011)

59)

The bit rate of a digital communication system is R kbits/s.

The modulation used is 32-QAM. The minimum bandwidth required for ISI free

transmission is

(A) R/10 Hz (B) R/10 kHz

(C) R/5 Hz (D) R/5 kHz

Answer: (B)

SOL: Bit rate given = R Kbits/second

Modulation = 32-QAM

No. of bits/symbol = 2 5 [log 32]

Symbol rate = R/5 k symbols / second

Finally we are transmitting symbols.

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BT = R/5(1+)

BT = R/5(1+1)=R/10 KHZ , The Maximum value of is 1 which is a roll factor. (GATE 2013)

60)

A band-limited signal with a maximum frequency of 5 kHz is to be sampled. According to the sampling theorem, the sampling frequency which is not valid is (A) 5 kHz (B) 12 kHz (C) 15 kHz (D) 20 kHz Answer: (A)

SOL: Given: fm 5kHz According to sampling frequency

So, only in option (a) it is less than 10KHz ie., (5KHz) (GATE 2013) 61)

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=P(Z0)=0.5=1/2,Z is standard Normal Variable (GATE 2013) 62)

(C) (F(x)-G(x)).x0 (D) (F(x)-G(x)).x0

SOL: Option(D)

F(x)=P(Xx) G(x)=P(2Xx)=P(Xx/2) For positive value of X,

(F(x)-G(x)) is always greater than zero

For negative value of x

(F(x)-G(x)) is negative but (F(x)-G(x)).x0 (GATE 2013)

63)

Let U & V are two independent and identically distributed random variables such that

P(U=1)=P(U= -1)=1/2.The entropy H(U+V) in bits is

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(GATE

2013)

64)

P(0)=P(1)=1/2

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(GATE

2013)

65)

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(GATE 2013)

66)The noise at the input to an ideal frequency detector is white. The

detector is

operating above threshold. The power spectral density of the noise at the

output is (2003)

(A) raised - cosine (B) flat (B) parabolic (D) Gaussian

Ans: (C) is correct option.

SOL: The noise at the input to an ideal frequency detector is white. The

PSD of noise at the output is parabolic

67)At a given probability of error, binary coherent FSK is inferior to binary

coherent PSK by. (2003).

(C) 6 dB (B) 3 dB (C)2 dB (D) 0 db

Ans : Hence (B) is correct option.

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SOL: We have Pe = erfc

Since Pe of Binary FSK is 3 dB inferior to binary PSK

68)Let X and Y be two statistically independent random variables uniformly

distributed in the ranges (1,1) and (2,1) respectively. Let Z=X+Y. Then the probability that (z 1) is (2003)

(A) zero (B) (C) (D)

Ans: Hence (D) is correct option.

SOL: The pdf of Z will be convolution of pdf of X and pdf of Y as

shown

below.

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69)Let X be the Gaussian random variable obtained by sampling the

process at t=ti and let

(2003)

Q () =

The probability that [ x

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E [ Y2 ] = E [ X2(2) ] =8

E [ Z2 ] = E [ X2(4) ] =8

E [YZ] = R XX (2) = 4 ( + 1)

= 6.68

E [ W2 ] = = 8 = 8 2 X 6.68 = 2.64.

71)Consider a binary digital communication system with equally likely 0s and 1s. When binary 0 is transmitted the detector input can lie between the levels 0.25 V and +0.25 V with equl probability : when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1

V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the

received signal is greater than 0.2 V, the bit is taken as 1), the average bit

error probability is (2004)

(E) 0.15 (B) 0.2 (C)0.05 (D) 0.5

Ans: Hence (A) is the correct option.

SOL: The pdf of transmission of 0 and 1 will be as shown below :

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72)A random variable X with uniform density in the interval 0 to 1 is

quantized as follows :

If (0 X 0.3), xq = 0 If (0.3 < X 1, xq = 0.7 where xq is the quantized value of X. The root-mean square value of the

quantization noise is (2004)

(F) 0.573 (B) 0.198 (C)2.205 (D) 0.266

Ans: Hence (B) is correct option.

SOL: The square mean value is:

73)A probability density function is of the form:

P (x) = K , x ( - , ). The value of K is: (2006)

(G) 0.5 (B) 1 (C)0.5 (D)

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Ans: Hence (C) is the correct option.

Sol: We know that = 1

Thus = 1

Or + = 1

i. K = .

74)A uniformly distributed random variable X with probability density

function

fx (X) = p u (x + 5) - u (x - 5)

where u(.) is the unit step function is passed through a transformation

given in the figure below. The probability density function of the

transformed random variable Y would be

(2006)

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Ans: Hence (B) is the correct option.

75)Three companies X,Y and Z supply computers to a university. The

percentage of computers supplied by them and the probability of those

being defective are tabulated below

Given that a computer is defective, the probability that was supplied by Y

is:

(2006)

(H) 0.1 (B) 0.2 (C)0.3 (D) 0.4

Ans: Hence (D) is the correct answer.

76)An examination consists of two papers, Paper 1 and Paper 2. The

probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a

student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The

probability of a student failing in both the papers is

(2007)

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(I) 0.5 (B) 0.18 (C)0.12 (D) 0.06

Ans: Hence (C) is correct answer.

SOL: Probability of failing in paper 1 is P(A)= 0.3

Possibility of failing in Paper 2 is P(B)= 0.2

Probability of failing in paper 1, when student has failed in paper 2 is P

= 0.6

77)An input to a 6-level quantizer has the probability density function f(x) as

shown in the figure. Decision boundaries of the quantizer are chosen so as to

maximize the entropy of the quantizer output. It is given that 3 consecutive

decision boundaries are 1'.'0' and '1'.

The values of a and b are (2007)

(A) a = and b = (B) a = and b =

(C) a = and b = (D) a = and b =

Ans : Hence (A) is the correct option.

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78)The probability density function (pdf) of random variable is as shown

below

The corresponding commutative distribution function CDF has the form

(2008)

Ans: Hence (A) is correct option.

SOL: CDF is the integration of PDF. Plot in option (A) is the integration

of plot given in question.

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79)Consider a Binary Symmetric Channel (BSC) with probability of error

being p. To transmit 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

(2008)

(A) P3 + 3 P 2 (1 - P) (B) P 3

(C) (1 P 3) (D) P3 + P 2 (1 - P)

Ans: Hence (D) is the correct option.

80)A fair coin is tossed 10 times. What is the probability that only the

first two tosses will yield heads?

(2009)

Ans: Hence (C) is the correct option.

SOL 1.11 Number of elements in sample space is 210. Only one element

{ H,H,T,T,T,T,T,T,T,T } is event. Thus the probability is 210.

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81)Consider two independent random variables X and Y with identical

distributions. The variables X and Y take values 0, 1 and 2 with

probabilities , and respectively. What is the conditional

probability P(X+Y=2 XY = 0)? (2009)

(A)0 (B) (C)

(D) 1

Ans: Hence (B) is correct option.

SOL:

82)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 tail shown up (2010)

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(J) (B) (C) (D)

Ans: Hence (D) is the correct option.

SOL: According to given condition head should comes 3 times or 4 times

P(Heads comes 3 times or 4times) = 4C4 ( )4 + 4C3 ( )

3 ( )

= .

83)A fair dice is tossed two times. The probability that the second toss

results in a value that is higher than the first toss is

(2011)

(K) (B) (C) (D)

Ans: Hence (C) is the correct option.

SOL: Total outcome are 36 out of which favorable outcomes are :

(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6);

(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) which are 15.

Thus P (E) = = =

84)A source alphabet consists of N symbols with the probability of the

first two

symbols being the same. A source encoder increases the probability of the

first symbol by a small amount and decreases that of the second by . After encoding, the entropy of the source

(2012)

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(L) increases (B) remains the same (M) increases only if N = 2 (D) decreases

Ans: Option (D) is correct.

SOL:

85)Two independent random variables X and Y are uniformly distributed

in the

Interval [ -1,1 ]. The probability that max (X,Y) is less than is

(2012)

(N) (B) (C) (D)

Ans: Option (B) is correct.

SOL: Probability density function of uniformly distributed variables X

and Y is shown as:

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Alternate method:

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From the given data since random variables X and Y lies in the interval

[1,1] as from the figure X, Y lies in the region of the square ABCD.

Probability for max [X,Y] < : The points for max [X,Y] < will be

inside the region of square AEFG.

So , p { max [X,Y] < } = .

= .

= .

86)A fair coin is tossed till a head appears for the first time. The

probability that the number of required tosses is odd, is

(2012)

(O) (B) (C) (D)

Ans: Option (C) is correct.

SOL: Probability of appearing a head is 1/2. If the number of required

tosses is odd, we have following sequence of events. H, TTH, TTTTH,

...........

Probability P = ( ) + ( )3 + ( )5 + ...............

P = = .

87)A and B are friends. They decide to meet between 1 PM and 2 PM on

a given day. There is a conditions that whoever arrives first will not wait

for the other for more than 15 minutes. The probability that they will

meet on that days is

(2012)

(P) (B) (C) (D)

Ans: Option (C) is correct.

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SOL: The graphical representation of their arriving time so that they met

is given as below in the figure by shaded region.

88)What is the chance that a leap year, selected at random, will contain

53 Sundays? (2013)

(Q) 2/7 (B) 3/7 (C) 1/7 (D) 5/7

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Solution: There are 52 complete weeks in a calendar year (App.) = =

852 X 7 =364days

i. Number of days in a leap year = 366

ii. Probability of 53 Saturdays = .