# Communications Subject Wise

## Communications Subject Wise

 Question 1
If X and Y are random variables such that E[2X + Y] = 0 and E[X + 2Y] = 33, then E[X] + E[Y] = _________.
 A Fill in the Blank Type Question
Question 1 Explanation:
Given X,Y are Random variables
As per The given data
E[2x + y] = 0
E[x + 2y] = 33

⇒ 2E(X) + E(Y) = 0.........(1)
⇒ E(X) + 2 E(Y) = 33......(2)

Solving (1) and (2);
2E(x) + E(y) = 0
2E(x) + 4E(y) = 66 [equ2 mul by 2 and + to -]
------------------------------
-3E(y) = -66
E(y)=22

From (1); 2E(x) + E(y)= 0
2E(x) = -E(y)
2E(x)= -22
E(x)= -11

By solving (1) and (2)
we got E(Y) = 22
E(X) = −11
∴E(X) + E(Y) = 22 − 11 = 11
 Question 2
The baseband signal m(t) shown in the figure is phase-modulated to generate the PM signal φ(t) = cos (2πfct + k m(t)). The time t on the x-axis in the figure is in milliseconds. If the carrier frequency is fc = 50 kHz and k = 10π, then the ratio of the minimum instantaneous frequency (in kHz) to the maximum instantaneous frequency (in kHz) is _________ (rounded off to 2 decimal places).
 A Fill in the Blank Type Question
Question 2 Explanation:
 Question 3
A linear Hamming code is used to map 4-bit messages to 7-bit codewords. The encoder mapping is linear. If the message 0001 is mapped to the codeword 0000111, and the message 0011 is mapped to the codeword 100110, then the message 0010 is mapped to
 A 1111000 B 0010011 C 1111111 D 1100001
Question 3 Explanation:
As it is given that it is linear hamming code addition of two codes will produce another code.
 Question 4
A random variable X takes -1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y = X + N. The noise n is independent of X, and is uniformly distributed over the interval [-2, 2]. The receiver makes a decision

Where the threshold θ [-1, 1] is chosen so as to minimize the probability of error The minimum probability of error, rounded off to 1 decimal place, is ______________.

 A Fill in the Blank Type Question
Question 4 Explanation:
 Question 5
Let a random process Y(t) be described as Y(t) = h(t) * X(t) + Z(t), where X(t) is a white noise process with power spectral density SX(f) = 5 W/Hz. The filter h(t) has a magnitude response given by |H(f)| = 0.5 for -5 ≤ f ≤ 5, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with X(t), with power spectral density as shown in the figure. The power in Y(t), in watts, is equal to _________ W (rounded off to two decimal places).

 A Fill in the Blank Type Question
Question 5 Explanation:
 Question 6
A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal f(t) = A(1 + m(t)) cos 2πfct, where fc = 600 kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _________ Mbps.
 A Fill in the Blank Type Question
Question 6 Explanation:
 Question 7
In the circuit shown, the positive angular frequency ω (in radians per second). at which magnitude of the phase difference between the voltages V1 and V2 equals radians, is ___.
 A Fill in the Blank Type Question
Question 7 Explanation:
 Question 8
Let (X1, X2) be independent random variables. X1 has mean 0 and variance 1, while X2 has mean 1 and variance 4. The mutual information I (X1 ; X2) between X1 and X2 in bits is
 A Fill in the Blank Type Question
Question 8 Explanation:
For two independent random variable
I(X;Y) = H(X)=H(X/Y)
H(X/Y)=H(X) for independent X and Y
Therefore I(X;Y) = 0
 Question 9
Which one of the following statements about differential pulse code modulation (DPCM) is true?
 A The sum of message signal sample with its prediction is quantized B The message signal sample is directly quantized, and its prediction is not used C The difference of message signal sample and a random signal is quantized D The difference of message signal sample with its predictions is quantized
Question 9 Explanation:
In DPCM the difference of the message signal sample value and the output of prediction filter block is quantized
 Question 10
In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?
 A B C D
Question 10 Explanation:
For Inter symbol interference(ISI) free pulse, If P(t) is having spectrum P(f)
Then = constant
RS= 2.4 kHz
This condition is met by pulse given in option B.
 Question 11

In binary frequency shift keying (FSK), the given signal waveform is
U0(t)=5cos(20000πt); 0 ≤ t ≤ T, and
U1(t) =5cos(22000πt); 0 ≤ t ≤ T,
Where T is the bit-duration interval and t is in seconds. Both u0 (t)and u1 (t)are zero outside the interval 0≤t≤T. With a matched filter filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u0(t)and u1(t) uncorrelated is

 A 0.25ms B 0.5 ms C 0.75 ms D 1.0 ms
Question 11 Explanation:
 Question 12
Let X(t) be a wide sense stationary random process with the power spectral density SX(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal low pass filter with frequency response As shown in figure (b). The output of the lowpass filter is Y(t) Let E be the expectation operator and consider the following statements.
I. E(X(t)) = E(Y(t))
II. E((t)) = E((t))
III. E((t)) = 2
Select the correct option:
 A Only I is true B Only II and III are true C Only I and II are true D Only I and III are true
Question 12 Explanation:
 Question 13
A continuous time signal x(t)=4cos(200πt)+8cos(400πt), where t is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response
h(t)=
Let y(t) be the output of this filter. The maximum value of |y(t)| is ______.
 A Fill in the Blank Type Question
Question 13 Explanation:
 Question 14
A binary source generates symbols which are transmitted over a noisy channel. The probability of transmitting the both symbols is equal.Input to the threshold detector is The probability density function of the noise is shown below.
If the detection threshold is zero, then the probability of error (correct to two decimal places) is ____________.
 A Fill in the Blank Type Question
Question 14 Explanation:
 Question 15
Consider the following amplitude modulated signal :

The value of amplitude sensitivity of modulator is Ka The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is ______________
 A Fill in the Blank Type Question
Question 15 Explanation:
 Question 16
Let and be independent normal random variables with zero mean and unit variance. The probability that is the smallest among the four is ___________.
 A Fill in the Blank Type Question
Question 16 Explanation:
As X1, X2, X3, and X4 are independent normal random variables with zero mean and unit variance. Then P(X1 is the smallest ) = P(X2 is the smallest ) = P(X3 is the smallest) = P(X4 is the smallest) = 1/4 = 0.25
 Question 17
Consider a binary channel code in which each codeword has a fixed length of 5 bits. The Hamming distance between any pair of distinct codewords in this code is at least 2. The maximum number of codewords such a code can contain is ___________.
 A Fill in the Blank Type Question
Question 17 Explanation:
Given that
Fixed Length of code n=5
Minimum Hamming distance dmin
Without any constraint, 25 = 32 codewords can be formed.
Following table contains all the possible code words having minimum Hamming distance dmin = 2
Hence, total 16 such code words are possible by a 5 bit code with dmin = 2
 Question 18
A random variable takes values -0.5 and 0.5 with probabilities and respectively. The noisy observation of is where has uniform probability density over the interval (-1, 1). and are independent. If the MAP rule based detector outputs as

then the value of (accurate to two decimal places) is __________.

 A Fill in the Blank Type Question
Question 18 Explanation:
 Question 19
Consider a white Gaussian noise process with two-sided power spectral density as input to a filter with impulse response (where is in seconds) resulting in output The power in in watts is
 A 0.11 B 0.22 C 0.33 D 0.44
Question 19 Explanation:
 Question 20
Let and it is given that The signal is applied to the input of a non-linear device, whose output is related to the input as where and are positive constants. The output of the non-linear device is passed through an ideal band-pass filter with center frequency and bandwidth to produce an amplitude modulated (AM) wave. If it is desired to have the sideband power of the AM wave to be half of the carrier power, then is
 A 0.25 B 0.5 C 1 D 2
Question 20 Explanation:
 Question 21
Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memory less binary symmetric channel with crossover probability P?
 A B C D
Question 21 Explanation:
For memory less binary Symmetric channel
Channel capacity
C=1-H(p)
H(p)=
p Cross over probability
C = 1 + plog2p + (1-p)log2(1-p)
At p=0; C=1
At p=1; C=1
At p= C=0
 Question 22
Consider the random process X(t)=U+Vt, where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is _____.
 A Fill in the Blank Type Question
Question 22 Explanation:
Given
->x(t) =U+Vt
->X(2)=U+2V
->E [x (2)] =E[U+2V]
-> E[U]+2E[V]
->0+2×1=2
 Question 23
The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is _______
 A Fill in the Blank Type Question
Question 23 Explanation:
 Question 24
A modulating signal given by is fed to a phase modulator with phase deviation constant kp=5 rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is _____
 A Fill in the Blank Type Question
Question 24 Explanation:

Answer Range : 69.9 to 70.1
 Question 25
Consider a binary memory less channel characterized by the transition probability diagram shown in the figure.

The channel is
 A Lossless B Noiseless C Useless D Deterministic
Question 25 Explanation:
 Question 26
Consider binary data transmission at a rate of 56 kbps using baseband binary pulse amplitude modulation (PAM) that is designed to have a raised-cosine spectrum. The transmission bandwidth (in kHz) required for a roll-off factor of 0.25 is
 A Fill in the Blank Type Question
Question 26 Explanation:
 Question 27
A super heterodyne receiver operates in the frequency range of 58 MHz - 68 MHz. The intermediate frequency and local oscillator frequency are chosen such that. It is required that the image frequencies fall outside the 58 MHz - 68 MHz band. The minimum required (in MHz) is
 A Fill in the Blank Type Question
Question 27 Explanation:
 Question 28
The amplitude of a sinusoidal carrier is modulated by a single sinusoid to obtain the amplitude modulated signal The value of the modulation index is _______.
 A Fill in the Blank Type Question
Question 28 Explanation:
 Question 29
Consider a discrete memoryless source with alphabet and respective probabilities of occurrence. The entropy of the source (in bits) is
 A Fill in the Blank Type Question
Question 29 Explanation:
 Question 30
A digital communication system uses a repetition code for channel encoding/decoding. During transmission, each bit is repeated three times (0 is transmitted as 000, and 1 is transmitted as 111). It is assumed that the source puts out symbols independently and with equal probability. The decoder operates as follows: In a block of three received bits, if the number of zeros exceeds the number of ones, the decoder decides in favor of a 0, and if the number of one's exceeds the number of zeros, the decoder decides in favor of a 1. Assuming a binary symmetric channel with crossover probability p = 0.1, the average probability of error is
 A Fill in the Blank Type Question
Question 30 Explanation:
 Question 31
A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20, 000 sample per second and 0.1 V, respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is
 A B C D π
Question 31 Explanation:
 Question 32
Consider the signal

Where denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to . The signal s(t) is a
 A high-pass signal B low-pass signal C band-pass signal D double sideband suppressed carrier signal
Question 32 Explanation:
Given s(t) is an SSB modulated signal.
Alternate Solution:

It is the canonical representation of a bandpass signal.
 Question 33
The input X to the Binary Symmetric Channel (BSC) shown in the figure is '1' with probability 0.8. The cross-over probability is 1/7. If the received bit Y= 0, the conditional probability that '1' was transmitted is ______
 A Fill in the Blank Type Question
Question 33 Explanation:
Probability that '1' was transmitted when received bit is '0'

 Question 34
The transmitted signal in a GSM system is of 200 kHz bandwidth and 8 users share a common bandwidth using TDMA. If at a given time 12 users are talking in a cell, the total bandwidth of the signal received by the base station of the cell will be at least (in kHz) _______ .
 A Fill in the Blank Type Question
Question 34 Explanation:
It is given that GSM requires 200 kHz for 8 users and is using TDMA to transmit them. Thus for the next 4 users(9th,10th,11th,12th) will require an extra 200 kHz bandwidth. Thus, total 400 kHz bandwidth is to be used for 12 users.
 Question 35
In the system shown in Figure (a), m(t) is a low-pass signal with bandwidth W Hz. The frequency response of the band-pass filter H(f) is shown in Figure (b). If it is desired that the output signal z(t) = 10x(t). The maximum value of W (in Hz) should be strictly less than _________.
 A Fill in the Blank Type Question
Question 35 Explanation:
 Question 36
A source emits bit 0 with probability and bit 1 with probability . The emitted bits are communicate to the receiver decides for either 0 or 1 based on the received value R. It is given that the conditional density function of R are as
and
The minimum decision error probability is
 A 0 B 1/12 C 1/9 D 1/6
Question 36 Explanation:
Given the conditional density function of R as

Decision error probability that transmitted bit is 1 but receiver decides 0 is
fR/1 (r = 0) =
Again, the decision error probability that transmitted bit is 0 but receiver decides 1 is
fR/0 (r = 1) =
Hence, the minimum decision error probability is fR/1 (r = 0) =
 Question 37
A discrete memoryless source has an alphabet with corresponding probabilities. The minimum required average code world length in bits to represent 2 4 8 8 this source for error-free reconstruction is _______.
 A Fill in the Blank Type Question
Question 37 Explanation:
 Question 38
A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is______.
 A Fill in the Blank Type Question
Question 38 Explanation:
Data rate = rb= 64 kbps M = 4
Minimum bandwidth =

Therefore BWmin = 16 KHz
 Question 39
An information source generates a binary sequence {an}. an can take one of the two possible values —1 and+1 with equal probability and are statistically independent and identically distributed. This sequence is preceded to obtain another sequence , as . The sequence is used to modulate a pulse g(t) to generate the baseband signal

If there is a null at in the power spectral density of X(t), then k is _____
 A Fill in the Blank Type Question
Question 39 Explanation:
 Question 40
Consider a random process X(t)= 3V(t) - 8, where is a zero-mean stationary random process with autocorrelation . The power in is
 A Fill in the Blank Type Question
Question 40 Explanation:
 Question 41
A binary communication system makes use of the symbols "zero" and "one". There are channel errors. Consider the following events:
xo: a "zero" is transmitted
x1: a "one" is transmitted
The following probabilities are given: The information in bits that you obtain when you learn which symbol has been received (while you know that a "zero" has been transmitted) is
 A Fill in the Blank Type Question
Question 41 Explanation:
Given Binary communication channel Information content in receiving y it in given that xo is transmitted is

 Question 42
For a super heterodyne 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.
 A Fill in the Blank Type Question
Question 42 Explanation:
 Question 43
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_________
 A Fill in the Blank Type Question
Question 43 Explanation:
 Question 44
A binary baseband digital communication system employs the signal for transmission of bits. The graphical representation of the matched filter output y(t) for this signal will be
 A B C D
Question 44 Explanation:
 Question 45
A voice-grade AWGN (additive white Gaussian noise) telephone channel has a bandwidth of 4.0 kHz and two-sided noise power spectral density 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 ____
 A Fill in the Blank Type Question
Question 45 Explanation:
 Question 46
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 _____.
 A Fill in the Blank Type Question
Question 46 Explanation:
 Question 47
The modulation scheme commonly used for transmission from GSM mobile terminals is
 A 4-QAM B 16-PSK C Walsh-Hadamard orthogonal codes D Gaussian Minimum Shift Keying (GMSK)
Question 47 Explanation:

Gaussian Minimum Shift Keying (GMSK) is used for GSM mobile terminals.
 Question 48
A message signal m(t) = Am sin (2πfmt) is used to modulate the phase of a carrier Ac cos(2πfc t) to get the modulated signal y(t) = Ac cos(2πfct + m(t)). The bandwidth of y(t)
 A depends on Am but no on fm B depends on fm but not on Am C depends on both Am and fm D does not depend on Am or fm
Question 48 Explanation:

Bandwidth of PM signal is given by

So, it depends upon fm and

(mp = Am = message signal amplitude)
 Question 49
The phase margin (in degrees) of the system G(s) = is_________.
 A Fill in the Blank Type Question
Question 49 Explanation:
 Question 50
A zero mean white Gaussian noise having power spectral density is passed through an LTI filter whose impulse response h(t) is shown in the figure. The variance of the filtered noise at t = 4 is
 A B C A2N0 D
Question 50 Explanation:
 Question 51
Consider a binary, digital communication system which used pulse g(t) and –g(t) for transmitting bits over an AWGN channel. If the receiver uses a matched filter, which one of the following pulses will give the minimum probability of bit error?
 A B C D
Question 51 Explanation:
 Question 52
In a code-division multiple access (CDMA) system with N = 8 chips, the maximum number of users who can be assigned mutually orthogonal signature sequences is ________
 A 5 B 6 C 7 D 8
Question 52 Explanation:
The upper limit of the total number of user supported simultaneously is given by spreading factor. We defined the spreading factor as

Spread factor (or) process gain and determine to a certain extent the upper limit of the total number of uses supported simultaneously by a station.
Therefore the maximum number of uses. who can be assigned mutually orthogonal signature sequence is 8
 Question 53
Consider sinusoidal modulation in an AM system. Assuming no overmodulation, the modulation index (μ) when the maximum and minimum values of the envelope, respectively, are 3 V and 1 V, is ________.
 A 1 B 0.5 C 1.5 D 2
Question 53 Explanation:
 Question 54
Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms where α = 4mV Assume an AWGN channel with two-sided noise power spectral density Using an optimal receiver and the relation du the bit error probability for a data rate of 500 kbps is
 A B C D
Question 54 Explanation:
For Binary
Bit error probability
E Energy per bit [No. of symbols = No. of bits]

 Question 55
The power spectral density of a real stationary random process .X(t) is given by

The value of the expectation
 A 2 B 4 C 6 D 8
Question 55 Explanation:
 Question 56
The phase response of a passband waveform at the receiver is given by

Where fc is the centre frequency, and are positive constants. The actual signal propagation delay from the transmitter to receiver is
 A B C D
Question 56 Explanation:
Given Phase response is

∴ ty= α
Thusis actual signal propagation delay from transmitter to receiver
 Question 57
Consider an FM signal The maximum deviation of the instantaneous frequency from the carrier frequency fc is
 A B C D
Question 57 Explanation:
 Question 58
Let X(t) be a wide sense stationary (WSS) random process with power spectral density SX(f). If Y(t) is the process defined as Y(t ) = X(2t −1) , the power spectral density SY(f) is
 A B C D
Question 58 Explanation:
Given Y(t ) = X(2t −1)
Shifting in time domain does not change Power Spectral Density. Since Power Spectral Density is Fourier transform of autocorrelation function of WSS process, autocorrelation function depends on time difference.

[time scaling property of Fourier transform]

If X(t) has Power Spectral Density(PSD) SX(f) then X (2t-1) has power spectral density of 1/2 SX(f/2)
 Question 59
A real band-limited random process X(t) has two-sided power spectral density
Where f is the frequency expressed in Hz. The signal X(t)modulates a carrier cos16000 and the resultant signal is passed through an ideal band-pass filter of unity gain with centre frequency of 8 kHz and band-width of 2 kHz. The output power (in Watts) is _______.
 A 1.5 B 2 C 2.5 D 3
Question 59 Explanation:

After modulation with cos(16000)

This is obtain the power spectral density Random process y(t), we shift the given power spectral density random process x(t) to the right by fc shift it to be the left by fc and the two shifted power spectral and divide by 4.

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

Total output power is area of shaded region

 Question 60

In a PCM system, the signal V is sampled at the Nyquist rate. The samples are processed by a uniform quantizer with step size 0.75 V. The minimum data rate of the PCM system in bits per second is _____.

 A 50 B 100 C 150 D 200
Question 60 Explanation:
 Question 61
Let X be a zero-mean unit variance Gaussian random variable. E[|X|] is equal to _____
 A 0.8 B 1.34 C 5 D 6
Question 61 Explanation:
 Question 62
For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then
 A Droop rate decreases and acquisition time decreases B Droop rate decreases and acquisition time increases C Droop rate increases and acquisition time decreases D Droop rate increases and acquisition time increases
Question 62 Explanation:
 Question 63
In a double side-band (DSB) full carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power increases by a factor of_________________.
 A 2 B 4 C 6 D 8
Question 63 Explanation:
 Question 64
An M-level PSK modulation scheme is used to transmit independent binary digits over a band-pass channel with bandwidth 100 kHz. The bit rate is 200 kbps and the system characteristic is a raised-cosine spectrum with 100% excess bandwidth. The minimum value of M is ________.
 A 12 B 14 C 16 D 18
Question 64 Explanation:
 Question 65
The bit rate of a digital communication system is R kbits/s. The modulation used is 32-QAM. The minimum band width required for ISI free transmission is
 A R/10Hz B R/10 kHz C R/5Hz D R/5 kHz
Question 65 Explanation:
 Question 66
Consider two identically distributed zero-mean random variables U and V. Let the cumulative distribution functions of U and 2V be F(x) and G(x) respectively. Then, for all values of x
 A F(x) – G(x) ≤ 0 B F(x) – G(x) ≥ 0 C (F(x) – G(x)).x ≤ 0 D (F(x) – G(x)).x ≥ 0
Question 66 Explanation:
 Question 67
Let U and V be two independent and identically distributed random variables such that
. The entropy H(U+V) in bits is
 A 3/4 B 1 C 3/2 D log23
Question 67 Explanation:
 Question 68
Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the
respective received signal for both bits are as shown below

If the detection threshold is 1, the BER will be
 A B C D E None
Question 68 Explanation:
For the shown received signal, we conclude that if 0 is the transmitted signal then the received signal will be also zero as the threshold is 1 and the pdf of bit 0 is not crossing 1. Again, we can observe that there is an error when bit 1 is received as it crosses the threshold. The probability of error is given by the area enclosed by the 1 bit pdf (shown by shaded region)

P (error when bit 1 received)
Or
Since, the 1 and 0 transmission is equiprobable:
i.e.,
Hence bit error rate (BER) is
 Question 69
Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the
respective received signal for both bits are as shown below

The optimum threshold to achieve minimum bit error rate (BER) is
 A B C 1 D E None
Question 69 Explanation:
Optimum threshold is the threshold value for transmission as obtained at the intersection of two pdf. From the shown pdf. We obtain at the intersection
Transmitted = 4/5
By solving the two linear equation We can obtain the intersection
For Solving threshold we have

So Optimum Threshold to achieve Minimum bit error rate (BER) is
 Question 70
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 ε and decreases that of the second by ε. After encoding, the entropy of the source
 A increases B Remains the same C Increases only if N = 2 D Decreases
Question 70 Explanation:
 Question 71
Two independent random variable X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than 1/2 is
 A 3/4 B 9/16 C 1/4 D 2/3
Question 71 Explanation:
and
Is the entire rectangle.
The region in which maximum of {x, y} is less than 1/2 is shown below as shaded region inside this rectangle.

 Question 72
The signal m(t) as shown is applied both to a phase modulator (with kp as the phase constant) and a frequency modulator (with kf as the frequency constant) having the same earner frequency.

The ratio kp/kf (in rad/Hz) for the same maximum phase deviation is
 A 8π B 4π C 2π D π
Question 72 Explanation:
For phase modulator

maximum phase deviation is
…(1)
for frequency modulator

…..(2)
Given

 Question 73
Which one of the following specifications does not fit for a single-mode fiber?
 A The bandwidth of 1 GHz/km. B The digital communication rate is excess of 2000 Mbytes/s. C More than 100000 voice channels are available D The mode field diameter (MFD; spot size) is larger than the core diameter.
Question 73 Explanation:
It is not possible to have bandwidth of 1 GHz/Km because for a single mode fibre, the bandwidth range is from 50 GHz / Km to 100 GHZ/Km.
 Question 74
For a binary FSK signal with a mark frequency of 49 KHz, a space frequency of 51 kHz and an input bit rate of 2 kbps, the peak frequency deviation will be
 A 0.5 kHz B 1.0 kHz C 2.0 kHz D 4.0 kHz
Question 74 Explanation:
In Binary FSK signal
Mark frequency is fL = 49 kHz
Space frequency is fH = 59 kHz
Rb = 2 KbPs
Now peak frequency deviation is ΔF
2ΔF = FH – FL
2ΔF = 51 KHz – 49 KHz
∴ ΔF = 2/2 KHz
∴ ΔF = 1 KHz
 Question 75
A random process X(t) is defined as
X(t) = 2cos(2πt + Y)
Where Y is a discrete random variable with and The mean μx(1) is
 A B C D 1
Question 75 Explanation:
Random process X(t) = 2 cos(2π t + y)
X(t) = 2[cos 2πt . cos y – sin 2πt . sin y]
E [x(t)] = 2 cos 2πt E[cosy] – 2 sin 2πt E[siny]

E [x(t)] = cos2πt – sin 2πt
μx (t) = cos2πt – sin 2πt
μx (1) = 1
 Question 76
A source produces three symbols A, B and C with probabilities and The source entropy is
 A bit/symbol B 1 bit/symbol C bit/symbol D bit/symbol
Question 76 Explanation:
 Question 77
An Am wave with modulation index 0.8 has total sideband power of 4.85 kW. The carrier power and the total power radiated will be nearly
 A 12.2 kW and 20 kW B 15.2 kW and 20 kW C 12.2 kW and 25 kW D 15.2 kW and 25 Kw
Question 77 Explanation:
 Question 78
A 360 W carrier is simultaneously modulated by two audio waves with modulation percentages of 55 and 65 respectively. The effective modulation index and the total power radiated are
 A 0.85 and 490.5 W B 0.65 and 490.5 W C 0.85 and 450.5 W D 0.65 and 450.5 W
Question 78 Explanation:
Carrier power Pc = 360 W
Modulation index for 1st audio signal μ1 = 0.55
Modulation index for 2nd audio signal μ2 = 0.65

Pt = 490.5 watt
∴Total modulation index μt = 0.8514
∴Total radiated power Pt = 490.5 W
 Question 79
An amplitude modulated amplifier has a radio frequency output of 50 W at 100% modulation. The internal loss in the modulator is 10 W. The unmodulated carrier power is
 A 40 W B 50 W C 60 W D 80 W
Question 79 Explanation:
 Question 80
For an FM receiver with an input signal-to-noise ratio of 29 dB. a noise figure of 4 dB and an FM improvement factor of 16 dB, the pre-detection and post-detection signal-to-noise ratios are
 A 25 dB and 41 dB B 30 dB and 41 dB C 25 dB and 49 dB D 30 dB and 41 dB
Question 80 Explanation:
 Question 81
For Gaussian and White channel noise, the capacity of a low-pass channel with a usable bandwidth of 3000 Hz and at the channel output will be
 A 15000 bits/s B 20000 bits/s C 25000 bits/s D 30000 bits/s
Question 81 Explanation:
 Question 82
For a PM modulator with a deviation sensitivity K = 2.5 rad/V and a modulating signal vm(t) = 2cos(2π2000t), the peak phase deviation m will be
Question 82 Explanation:
 Question 83
In a PCM system, non-uniform quantization leads to
 A increased quantizer noise B simplification at the quantization process C higher average SNR D increased bandwidth
Question 83 Explanation:
In PCM system non – uniform quantization leads to higher average SNR.
• If uniform quantization is used, low amplitude signals of voice are quantized to the same value. This problem is eliminated using non-uniform quantization. If non-uniform quantization is used the average SNR will increase.
• Higher average signal to quantization noise power ratio than the uniform quantizer when the signal pdf is non uniform which is the case in many practical situation.
• Non - uniform quantization minimizes the distortion with same no of levels. Hence bandwidth requirement is same.
• Non - uniform quantizer has higher average SNR than uniform quantization
• RMS value of the quantizer noise power of a non – uniform quantizer is substantially proportional to the sampled value and hence
 Question 84
The bandwidth required in DPCM is less than that of PCM because
 A the number of bits per code is reduced resulting in a reduced bit rate B the difference signal is larger in amplitude than actual signal C more quantization levels are needed D the successive samples of signal often differ in amplitude
Question 84 Explanation:
As the number of bits required for encoding an error signal is always less than the number of bits needed to encode the actual input signal, the bit rate (signaling rate ) of DPCM is always less than that of a PCM signal
 Question 85
Which one of the following is the correct combination for a layer providing a service by means of primitives in an open systems interconnection?
 A Request, Indication, Response and Confirm B Request, Inform, Response and Service C Request, Command, Response and Action D Request, Confirm, Indication and Action
Question 85 Explanation:
Four types of primitives are:
1. Request
2. Indication
3. Response
4. Confirm
Request primitive induces an indication primitive.
If an indication primitive requires a reply, a response primitive may be issued.
This response primitive will induce a confirmation primitive.
 Question 86
A network uses a fully interconnected mesh topology to connect 10 nodes together. The number of links required will be
 A 35 B 40 C 45 D 50
Question 86 Explanation:
A fully connected mesh topology can be thought of as a complete graph in which every node is connected to every other node.
So, number of cable links with n nodes = n(n-1)/2
For a fully connected mesh set with n nodes, number of links are
 Question 87
Which of the following are the advantages of packet switching?
1) Greater link efficiency than circuit switching
2) Connections are not blocked when traffic congestion occurs
3) Direct channel established between transmitter and receiver
4) No time is taken to establish connection
Select the correct answer using the code given below.
 A 1 and 3 B 1 and 2 C 2 and 3 D 3 and 4
Question 87 Explanation:
Option 1 : is TRUE Greater quick efficiency than circuit switching because of effective resource utilization in packet switching

Option 2 : is TRUE As packet switching is implemented at network layer of model. Congestion can be handled. So, connections are not blocked.

Option 3 and 4 : False As Packet switching may be classified into connectionless packet switching, so we don’t require connection establishment.

What is Packet Switching ?
Packet switching is different from circuit switching because there is no requirement to establish a channel. The channel is available to users throughout the data network. Long messages are broken down into packets and sent individually to the network.

While packet switching may not be as suited to voice calls as circuit switching, it has a number of advantages that are hard to ignore. The main advantage that packet switching has over circuit switching is its efficiency. Packets can find their own data paths to their destination address without the need for a dedicated channel. In contrast, in-circuit switching network devices can’t use the channel until the voice communication has been terminated. Packet switching is also reliable because it helps to eliminate packet loss. With packet switching, data packets can be resent if they don’t reach their destination. This isn’t the case for circuit switching which doesn’t have the means to send lost packets. As a result, packet switching is the more reliable method of the two because it ensures that packets reach their destination. Packet switching also reduces the costs associated with running the network. Packet switching networks can transfer general network traffic and voice traffic across the network without the need for a dedicated channel. This saves you money because you don’t need to pay to have one channel available for voice communications.

The biggest limitation of packet switching is that it is unsuitable for applications that require minimal latency. In a network that uses lots of voice calls circuit switching is a necessity because it is the only setup that delivers a high-quality end call. Packet switching can only provide a voice call experience that results in choppy audio that makes it difficult for the users to understand each other. Similarly, though packet switching is able to resend lost data packets, this isn’t the case if the network becomes overwhelmed by traffic. If there is too much traffic then packets will be dropped in transit. The end result is the loss of important data. This risk is further increased by the lack of security protocols used to protect packets during data transmission. There is no IPsec to give packets that extra barrier of security against damage. Though packet switching reduces costs in a number of ways it is significantly expensive to implement. Packet switching relies on a range of complex protocols that must be managed from deployment onward.
 Question 88
A message consisting of 2400 bits is to be passed over an internet. The message is passed to the transport layer which appends a 150-bit header, followed by the network layer which uses a 120-bit header. Network layer packets are transmitted via two networks, each of which uses a 26-bit header. The destination network only accepts up to 900 bits long. The number of bits, including headers delivered to the destination network, is
 A 2706 bits B 2634 bits C 2554 bits D 2476 bits
Question 88 Explanation:
 Question 89
In a communication network, 4 T1 streams are multiplexed to form 1 T2 stream and 7 T1 streams are multiplexed to form I T3 stream, Further 6 T3 streams are multiplexed to form 1 T4 stream. If each T1 stream is of 1.544 Mbps, the data rate of 1 T4 stream should be
 A 211.8 Mbps B 232.6 Mbps C 243.4 Mbps D 274.2 Mbps
Question 89 Explanation:
Note: There is a printing mistake. 7T2 streams are multiplexed to form 1T3 stream, but given 7T1
4T1 = 1T2
7T2 = 1T3
6T3 = 1T4
T1 = 1.544 Mbps
T4 = 4 × 7 × 6 × 1.544
Therefore the T4 system has a bit rate
∴ T4≈ 274.2 mbps
 Question 90
Which of the following statements are correct regarding CDMA?
1) It is similar to GSM
2) It allows each station to transmit over the entire frequency spectrum all the time.
3) It assumes that multiple signals add linearly.
Select the correct answer using the code given below.
 A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3
Question 90 Explanation:
Option 1 is wrong : GSM uses both FDM and TDM technique whereas CDMA uses CDM technique so both are not similar.
Option 2 is correct : CDMA allows each station to transmit over entire frequency spectrum all the time because it doesn’t use FDM technique.
Option 3 is correct : CDMA consists of RAKE receiver due to which CDMA assumes that multiple signals (Multipath fading signals) add linearly.
Option (c) is correct if the 3rd statement is “multipath signals” instead of “multiple signals”.
 Question 91
Which of the following regarding cellular systems with small cells are correct?
1) Higher capacity and robustness
2) Needless transmission power and have to deal with local interference only
3) Frequency planning and infrastructure needed
4) These require both circuit switching and packet switching
Select the correct answer using the code given below
 A 1, 2 and 4 B 1, 3 and 4 C 1, 2 and 3 D 2, 3 and 4
Question 91 Explanation:
Small Cell Cellular Network
“A cellular network is a radio network distributed over land through cells where each cell includes a fixed location transceiver known as base station. These cells together provide radio coverage over a larger geographical area. Thus, principal of cellular systems is to divide a large geographic service area into cells with diameters from 2 to 50 Kms, each of which is allocated a number of radio frequency (RF) channels. Today’s cellular networks give subscribers advanced features over alternative solutions, including increased capacity, small battery power usage, a larger geographical coverage area and reduced interference from other signals. Popular cellular technologies include the Global System for Mobile (GSM) communication, General Packet Radio Service (GPRS), 3GSM and Code Division Multiple Access (CDMA).

Some of the advantages of cellular systems with small cells are brief described in following paragraphs.
1. High Capacity
2. Less transmission power
3. Local Interference only
4. Robustness

1. Infrastructure needed
2. Handover need
3. Frequency planning: to avoid interference between transmitters using some frequencies, Frequencies distributed carefully
 Question 92
A satellite is orbiting in the equatorial plane with a period from perigee to perigee of 12h. If the eccentricity = 0.002, i = 0o, K1 = 66063.17 km2, μ = 3.99 × 1014 m3/s2 and the earth’s equatorial radius = 6378.14 km, the semi-major axis will be
 A 34232 km B 30424 km C 26612 km D 22804 km
Question 92 Explanation:
From Kepler’s law

There are 43,200 seconds in 12 hours and therefore
T = 12hr = 12*hr = 12*60*60sec= 43200 sec
A = perigee distance
a = aP

aP = 26525 × 103 m
Now ap = a (1 – e)
a = semi major axis =
=
26525000/0.998
Therefore a = 26578 km
 Question 93
A single-mode optical fiber has a beat length of 8 cm at 1300 nm. The value of birefringence B, will be nearly
 A 1.6 × 10-5 B 2.7 × 10-5 C 3.2 × 10-5 D 4.9 × 10-5
Question 93 Explanation:
 Question 94
Which one of the following instruments is useful while measuring the optical power as a function of wavelength?
 A Optical power attenuator B Optical power meter C Optical spectrum analyzer D Optical return loss tester
Question 94 Explanation:
Optical spectrum analyzer is useful while measuring the optical power as function of wavelength.
The optical spectrum analyzer (OSA) is used to measure the spectral characteristics of the light. The OSA displays the optical power as a function of wavelength.
 Question 95
The optical performance monitoring involves
 A transport layer monitoring. optical signal monitoring and protocol performance monitoring B physical layer, network layer and application layer monitoring C data-link layer, presentation layer and session layer monitoring D transport layer, session layer and application layer monitoring
Question 95 Explanation:
Following 3 layers of OPM:
1) Transport monitoring,
2) Signal quality monitoring
3) Protocol monitoring.

Source : https://web.ece.ucsb.edu/publications/Blumenthal/copy.pdf
 Question 96
An earth station at sea level communicates at an elevation angle of 35 ° with GEO satellite. The vertical height of the stratiform rain is 3 km. The physical path length L through the rain will be nearly
 A 6.3 km B 5.2 km C 4.1 km D 3.0 km
Question 96 Explanation:
Angle of elevation = 35o
Stratiform rain height = 3 km
The physical pathlength, L, through the rain
is given by =

Physical path length L = 5.2 km
 Question 97
A satellite transmits a signals at a frequency of 6 GHz to a user 40000 km away. The free space path loss incurred by the signal is nearly:
 A 100 dB B 150 dB C 200 dB D 300 dB
Question 97 Explanation:
The Correct Answer Among All the Options is C
free space path loss =
d =distance in meters.
F= frequency in Hz
c= 3 m/s.
free space path loss(in db) = 10
= 200dB
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 98
The satellite communication link between two point is established with uplink carrier-to-noise ratio of 20 dB and downlink carrier to noise ratio of 14 dB. The overall C/N is close to:
 A 34 dB B 6 dB C 13 dB D 13.5 dB
Question 98 Explanation:
The Correct Answer Among All the Options is C
overall C/N ratio in satellite communication =
+ ………………………………………………………..(1)
Given,

Similarly,
=14dB

Put these values in (1)
+
=

=13dB
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 99
A system generates data at a rate of 5 Mbps. In order to provide resistance to bit errors, a rate ½ error correcting code is applied. Further, the data is mapper to a 16-QAM constellation. What is the resulting symbol rate?
 A 1.25 Msps B 2.5 Msps C 5 Msps D 10 Msps
Question 99 Explanation:
The Correct Answer Among All the Options is B

Correction code-rate = , it means 1 data bit is encoded into ‘2’ bits by adding 1 extra parity bit.
Output rate after channel adding error correction = 5 =10Mbps
Because of 16-QAM {16 =
} , symbol rate =
Refer the Topic Wise Question for Digital Communication Systems Communications
 Question 100
Considr a binary code with parity check matrix H given below.

Which of the following is a valid codeward?
 A [1 0 0 1 0 1] B [1 1 1 1 0 1] C [0 1 0 0 1 0] D [1 1 0 1 1 0]
Question 100 Explanation:
The Correct Answer Among All the Options is D
The matrix H is called parity check matrix of any code word only if C
Refer the Topic Wise Question for Information Theory Communications
 Question 101
The minimum distance of a (n,k) = (7,4) linear block code is upper bounded by:
 A 1 B 2 C 3 D 4
Question 101 Explanation:
The Correct Answer Among All the Options is C
linear block code,(7,4)=(n,k)
It means 4 bit data is encoded in 7 bit codeword.it will correct any single bit error.

t=no of error correction , here t=1.
Therefore,

Refer the Topic Wise Question for Information Theory Communications
 Question 102
Considr a 4-PSK constellation with points and a 4-PAM constellation . If all the points in the constellation occur with equal probability, the ratio of avrage energy of 4-PAM signal to that of 4-PSK signal to that of 4-PSK signal is:
 A 1 B 1.25 C 2.5 D 5
Question 102 Explanation:
The Correct Answer Among All the Options is C

Symbol energy (E) =
Average energy()= =2
average energy()=
= 5
=
=2.5
Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications
 Question 103
Pair of diffrential equations that describs motion of planet about sun using first two laws of Kepler is given as :
 A and B and C and D and
Question 103 Explanation:
The Correct Answer Among All the Options is A
Kepler's three laws of planetary motion can be described as follows:
The path of the planets about the sun is elliptical in shape, with the centre of the sun being located at one focus. (The Law of Ellipses)
An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 104
Consider identical four, 3-faced dice. When the dice are rolled, the faces of the dice appear with probabilities given below. Which distribution has the maximum entropy?
 A (1/2, 1/4, 1/4) B (1/3, 1/3, 1/3) C (1/6, 2/3, 1/6) D (1/4, 1/6, 7/12)
Question 104 Explanation:
The Correct Answer Among All the Options is B
For maximum entropy, all symbols must have equal probability.
, m=no. of symbols.
Refer the Topic Wise Question for Information Theory Communications
 Question 105
Consider waveforms and , (), to be used for non-coherent binary FSK signalling. If the symbol duration is T seconds, and is constant arbitary angle from 0 to 2, what is the minimum separation required between f1 and f2 for non-coherent, orthogonal FSK?
 A B C D
Question 105 Explanation:
The Correct Answer Among All the Options is A
For bit ‘1’
For bit ‘0’ ,
Condition for orthogonality of signals,

Therefore,
On further solving and equating the components to zero we get
Sin[2]] = Sin …………………(1)
Sin[2]= Sin……………………..(2)
Sin[2]= Sin , when
2
n =0 is not possible because it is given that ()
therefore, 2

Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications
 Question 106
A rocket with lift-off mass of m0 is launched from ground level. During flight, fuel burns at a constant rate for seconds and exhaust gases are ejecting from the bottom of the rocket at Kg/sec with speed of cm/s relative to the rocket. Ignoring air resistance and assume acceleration due to gravity, g as constant, which of the following expression represents velocity of rocket v(t).
 A B C D
Question 106 Explanation:
The Correct Answer Among All the Options is A
Initial mass = m0
Rate of decrease in mass = βt
In rocket, mass decreases exponentially and depends upon initial velocity of rocket too.
βt = m0 – m0e-u/c
e-u/c =
u = -c ln
Now applying v(t) = u + at
v(t) = -c ln - gt
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 107
Consider a mechanical system shown in figure.Masses are free to slide over frictionless horizontal surface. The equation of motion of mass m1 is
 A B C D
Question 107 Explanation:
The Correct Answer Among All the Options is C
In this question by analysing the free body diagram of m1 and m2. We can get the equation of motion.
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 108
If over the course of a day, the maximum electron density in the ionosphere varies from 1011 to 1012 m3; the critical frequency changes approximately from:
 A 2.2 MHz to 7 MHz B 2.5 MHz to 8 MHz C 2.8 MHz to 9 MHz D 3.2 MHz to 10 MHz
Question 108 Explanation:
The Correct Answer Among All the Options is C
Critical frequency of ionosphere is the limiting frequency at or below which wave component is reflected by and above which wave penetrates through ionospheric layer .it depends upon electron density of ionosphere.
,
For N=, = 2.8MHz
For N= , = 9MHz
Refer the Topic Wise Question for Satellite and Optical Communication Communications
 Question 109
Consider the system with x(t) as input and y(t) as output. The frequency domain characteristics are shown in the figure. Which combination of A and B will give Y as result?
 A B C D
Question 109 Explanation:
The Correct Answer Among All the Options is A
Here, x(t) and y(t) are given. Now, we have to determine the graph of A and B that will give us the output Y.
Now, the given x(t) can be expressed in frequency domain as:

Next, see the figure. x1(t) and 'j' are added to give x2(t). So, basically the given x(t) in question should be modified by moving the negative spectrum of the signal downwards as follows:

Next in the diagram see signal x2(t) and x(t) is getting added. So, let us do the same.

Next is block B. So, spectrum of B will be exactly same as shown in figure (a). So, answer will be option (a)
Refer the Topic Wise Question for Spectrums and Receivers Communications
 Question 110
The frequency of the output Y is
 A B C D
Question 110 Explanation:
The Correct Answer Among All the Options is A
A closed loop system is shown in the figure.
XOR gates are generally used to implement addition of binary numbers. Ex-Or gates are also regarded as addition modulo-2.
See the first block in the diagram. It can be thought of an summer. So, let us replace it by an Ex-OR gate. Next block can simply be regarded as a register.
So, input to the Ex-Or gate=32bit(Binary Number), and the other input is the output feedback, i.e 32 bit. Now, 'F' is present for both the blocks, which means both have the same synchronized frequency.
Initially, flip flop will be reset. When clock pulses begin, then let 'M' has been locked to a certain value. So, now, frequency of the output will be equal to MF/232
So, now let us assume that clock pulse has been applied.
Refer the Topic Wise Question for PSK, DPSK and ASK, FSK Communications
 Question 111
A random variable z, has a probability density function f(z) where , the probability of will be approximately
 A 0.368 B 0.135 C 0.393 D 0.865
Question 111 Explanation:
The Correct Answer Among All the Options is D
We have to find=> P(0
So, this can be written as:
dz[Since f(z)=e-z]
So on solving the integral we get: -e-2+1=>0.8
Refer the Topic Wise Question for Random Processes and Variables Communications
There are 111 questions to complete.