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51.

In M-ary FSK, as M increases error(a) Increases(b) Decreases(c) Does not get effected(d) Cannot be determinedI had been asked this question in my homework.Asked question is from M-ary Modulation and Amplifiers topic in section Modulation and Coding Trade-off of Digital Communications

Answer» CORRECT option is (b) DECREASES

The BEST I can EXPLAIN: In M-ary FSK as M increases error decreases.
Discussion
52.

In an ideal TDM system, the cross correlation between two users of the system is(a) 1(b) 0(c) Infinity(d) -1I got this question in a job interview.I'd like to ask this question from PAM,PWM and TDM in section Modulation and Coding Trade-off of Digital Communications

Answer»

The correct OPTION is (B) 0

The EXPLANATION: In an ideal TDM SYSTEM, the cross correlation between TWO users of the system is 0.
Discussion
53.

TDM is less immune to cross-talk in channel than FDM.(a) True(b) FalseI had been asked this question during an interview.Origin of the question is PAM,PWM and TDM topic in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

The correct choice is (b) FALSE

The explanation is: False because DIFFERENT MESSAGE signals are not applied to the CHANNEL simultaneously.
Discussion
54.

Coherent demodulation of FSK signal can be performed using(a) Matched filter(b) BPF and envelope detectors(c) Discriminator(d) None of the mentionedI had been asked this question by my school principal while I was bunking the class.Query is from PAM,PWM and TDM in division Modulation and Coding Trade-off of Digital Communications

Answer»

Correct choice is (a) Matched filter

To elaborate: Coherent demodulation of FSK SIGNAL can be PERFORMED USING matched filter.
Discussion
55.

A PWM signal can be generated by(a) An astable multi vibrator(b) A monostable multi vibrator(c) Integrating a PPM signal(d) Differentiating a PPM signalI had been asked this question in an international level competition.This question is from PAM,PWM and TDM in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

Right answer is (b) A monostable MULTI vibrator

To explain I WOULD say: A PWM SIGNAL can be generated by a MONO stable multi vibrator.
Discussion
56.

A PAM signal can be detected using(a) Low pass filter(b) High pass filter(c) Band pass filter(d) All pass filterI had been asked this question in unit test.My doubt is from PAM,PWM and TDM topic in section Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT option is (a) Low pass FILTER

The best explanation: A PAM SIGNAL can be detected by USING low pass filter.
Discussion
57.

In a delta modulation system, granular noise occurs when the(a) Modulating signal increases rapidly(b) Pulse rate decreases(c) Pulse amplitude decreases(d) Modulating signal remains constantThis question was addressed to me in an internship interview.I would like to ask this question from PAM,PWM and TDM in portion Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT CHOICE is (d) Modulating signal remains constant

For explanation: In a delta modulation system, granular NOISE OCCURS when the modulating signal remains constant.
Discussion
58.

The TCM decoder uses(a) Minimum likelihood detector(b) Maximum likelihood detector(c) Does not use redundant bits(d) None of the mentionedI had been asked this question in homework.This intriguing question comes from Trellis Coded Modulation topic in section Modulation and Coding Trade-off of Digital Communications

Answer»

Correct choice is (b) Maximum likelihood DETECTOR

To elaborate: The task of convolutional DECODER is to estimate the path that the MESSAGE had traversed through the encoding trellis USING maximum likelihood detector.
Discussion
59.

Which determines the error performance in TCM?(a) Reduced distance(b) Free distance(c) Bandwidth(d) None of the mentionedThis question was addressed to me in a job interview.I need to ask this question from Trellis Coded Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Right choice is (b) Free distance

To explain I would say: Free distance which is the MINIMUM distance between members of SET of allowed code sequences DETERMINES the ERROR performance.
Discussion
60.

Flat top sampling of low pass signals(a) Gives rise to aperture effect(b) Implies over sampling(c) Leads to aliasing(d) Introduces delay distortionI got this question by my college professor while I was bunking the class.Question is from PAM,PWM and TDM in division Modulation and Coding Trade-off of Digital Communications

Answer»

The correct ANSWER is (a) Gives rise to APERTURE effect

For EXPLANATION I WOULD say: Flat top sampling of low pass signals gives rise to aperture effect.
Discussion
61.

In TCM channel capacity can be increased.(a) True(b) FalseThis question was posed to me by my school principal while I was bunking the class.Question is taken from Trellis Coded Modulation in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

Right choice is (a) True

The best explanation: In trellis CODED modulation increase in CHANNEL capacity is achieved by signal SET expansion.
Discussion
62.

Transmission bandwidth depends on(a) Rate of signalling(b) Density of signal points(c) Reduced distance(d) None of the mentionedThis question was addressed to me during an internship interview.The question is from Trellis Coded Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT option is (a) Rate of signalling

To elaborate: TRANSMISSION bandwidth of non orthogonal signalling depends on rate of signalling and not on density of SIGNAL points in the CONSTELLATION.
Discussion
63.

Trellis coded modulation scheme(a) Has memory(b) Is memory-less(c) Needs more bandwidth(d) None of the mentionedThis question was addressed to me at a job interview.My question is from Trellis Coded Modulation in section Modulation and Coding Trade-off of Digital Communications

Answer»

Correct answer is (a) Has memory

The best explanation: TRELLIS coded modulation scheme has memory ( a FINITE STATE MACHINE) such as the convolutional CODE.
Discussion
64.

In finite state machines, output is predicted using(a) Past output(b) Present input(c) Past output & Present input(d) None of the mentionedI had been asked this question in an international level competition.This question is from Trellis Coded Modulation topic in section Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
65.

In TCM the signals are decoded using(a) Soft decision decoders(b) Hard decision decoders(c) Soft & Hard decision decoders(d) None of the mentionedI have been asked this question by my college professor while I was bunking the class.This is a very interesting question from Trellis Coded Modulation topic in section Modulation and Coding Trade-off of Digital Communications

Answer»

The correct OPTION is (a) Soft decision DECODERS

For EXPLANATION I would SAY: In trellis coded modulation the noisy received signals are decoded using soft decision decoders.
Discussion
66.

Choosing a closely packed signal subset from any regular lattice or array of candidate points is called as(a) Optimum signal constellation boundaries(b) Higher density lattice structures(c) Trellis coded modulation(d) None of the mentionedI got this question in an interview for job.Question is from Trellis Coded Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer» CORRECT answer is (a) OPTIMUM SIGNAL constellation BOUNDARIES

Easy explanation: Optimum signal constellation boundaries are obtained by choosing closely packed signal subset from any regular LATTICE or array of candidate points.
Discussion
67.

Adding improvement to the signal subset choice by starting with the densest possible lattice for the space is called as(a) Optimum signal constellation boundaries(b) Higher density lattice structures(c) Trellis coded modulation(d) None of the mentionedThe question was asked in an international level competition.The above asked question is from Trellis Coded Modulation topic in chapter Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT choice is (b) Higher density lattice structures

The BEST explanation: Higher density lattice structures are FORMED by ADDING improvement to the signal subset choice by starting with the densest possible lattice for the space.
Discussion
68.

Trellis coded modulation uses(a) Non binary method(b) Uses redundant bits(c) No expansion of bandwidth(d) All of the mentionedThis question was addressed to me in semester exam.My enquiry is from Trellis Coded Modulation in portion Modulation and Coding Trade-off of Digital Communications

Answer»

The correct option is (d) All of the mentioned

Best EXPLANATION: Trellis coded modulation is a achieves error performance IMPROVEMENTS WITHOUT EXPANSION of bandwidth. It uses redundant non binary modulation in COMBINATION with finite state machines.
Discussion
69.

The orbital period of communication satellites is _______ as that of earth’s rotational period.(a) Same(b) Greater(c) Lesser(d) None of the mentionedThis question was addressed to me in an interview for job.My doubt is from Evaluating Digital Communication System and Allocation of Communication Resources in division Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT answer is (a) Same

The best explanation: The communication satellite is in CIRCULAR ORBIT and has ORBITAL PERIOD same as that of the earth’s rotational period.
Discussion
70.

Channelization characterized by orthogonal spectra is called as(a) Time division multiplexing(b) Frequency division multiplexing(c) Time division & Frequency division multiplexing(d) None of the mentionedI have been asked this question in an interview for internship.I would like to ask this question from Evaluating Digital Communication System and Allocation of Communication Resources in division Modulation and Coding Trade-off of Digital Communications

Answer»

The correct option is (B) Frequency division multiplexing

For explanation I would SAY: CHANNELIZATION characterized by orthogonal wave-forms is CALLED as time division multiplexing and those characterized by orthogonal spectra is called as frequency division multiplexing.
Discussion
71.

The super group is made up of(a) Five groups(b) Sixty channels(c) Five groups & Sixty channels(d) None of the mentionedI have been asked this question in an interview.My question is based upon Evaluating Digital Communication System and Allocation of Communication Resources topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
72.

The mixing and filtering of DSB removes(a) LSB(b) USB(c) LSB & USB(d) None of the mentionedI have been asked this question during an interview.This intriguing question originated from Evaluating Digital Communication System and Allocation of Communication Resources topic in portion Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT answer is (b) USB

Easy explanation: The MIXING and filtering YIELDS frequency shifted channels and it REMOVES the USB.
Discussion
73.

In double side band spectrum which side-band is called as inverted side-band?(a) LSB(b) USB(c) LSB & USB(d) None of the mentionedThis question was addressed to me in unit test.Query is from Evaluating Digital Communication System and Allocation of Communication Resources topic in section Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
74.

In polarization division technique ______ polarization is used.(a) Orthogonal(b) Non orthogonal(c) Orthogonal & Non orthogonal(d) None of the mentionedI have been asked this question during an internship interview.This interesting question is from Evaluating Digital Communication System and Allocation of Communication Resources in section Modulation and Coding Trade-off of Digital Communications

Answer» CORRECT option is (a) Orthogonal

The explanation: In POLARIZATION DIVISION TECHNIQUE orthogonal polarization is USED to separate signals for reuse.
Discussion
75.

How can the throughput of the system be increased?(a) Increasing EIRP(b) Increasing channel bandwidth(c) Decreasing system losses(d) All of the mentionedI had been asked this question by my college director while I was bunking the class.Question is taken from Evaluating Digital Communication System and Allocation of Communication Resources in division Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT answer is (d) All of the mentioned

To explain I would SAY: The basic ways to INCREASE the throughput of the system is to increase the EIRP, by reducing the system losses, PROVIDING channel bandwidth, and efficient ALLOCATION of CR.
Discussion
76.

In which technique spot beam antennas are used to point the radio signals in different directions?(a) Code division(b) Space division(c) Frequency division(d) None of the mentionedThis question was addressed to me in an interview for job.I'm obligated to ask this question of Evaluating Digital Communication System and Allocation of Communication Resources in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

The correct answer is (b) SPACE division

To explain: In space division multiplexing, spot BEAM antennas are used to separate radio SIGNALS by POINTING in different DIRECTIONS.
Discussion
77.

For more than two quantization level, _______ is used.(a) Hard decision(b) Soft decision(c) Hard & Soft decision(d) None of the mentionedThe question was asked in quiz.This interesting question is from Evaluating Digital Communication System and Allocation of Communication Resources in section Modulation and Coding Trade-off of Digital Communications

Answer»

Correct OPTION is (b) Soft decision

The best I can explain: When the OUTPUT is quantized for more than two levels, soft decision is MADE.
Discussion
78.

The transmission rate is given as(a) R/log2M(b) log2M/R(c) R/log10M(d) log10M/RThis question was addressed to me by my college professor while I was bunking the class.My doubt is from Evaluating Digital Communication System and Allocation of Communication Resources in division Modulation and Coding Trade-off of Digital Communications

Answer»

Correct answer is (a) R/log2M

Explanation: The relationship between SYMBOL TRANSMISSION RATE and data rate R is given by RS= R/log2M.
Discussion
79.

Code’s redundancy is given as(a) n/k(b) k/n(c) nk(d) None of the mentionedI have been asked this question in an online quiz.My question is from Evaluating Digital Communication System and Allocation of Communication Resources topic in division Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
80.

Which modulation is selected for power limited system?(a) MPSK(b) MFSK(c) MPSK & MFSK(d) None of the mentionedThe question was posed to me during an interview.I want to ask this question from Evaluating Digital Communication System and Allocation of Communication Resources topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

The CORRECT OPTION is (B) MFSK

To EXPLAIN: For an uncoded system, if the channel is power limited MFSK is selected and if it is bandwidth limited MPSK is selected.
Discussion
81.

In MFSK, R/W(a) Increases with increase in M(b) Increases with decrease in M(c) Decreases with increase in M(d) Is not dependent on MI had been asked this question during an interview.I'm obligated to ask this question of Evaluating Digital Communication System and Allocation of Communication Resources topic in division Modulation and Coding Trade-off of Digital Communications

Answer»

Right ANSWER is (c) Decreases with increase in M

The EXPLANATION: In MFSK, the DENOMINATOR increases faster as the numerator when M increases. Thus R/W decreases with increase in M.
Discussion
82.

The system will be more bandwidth efficient as WTb decreases.(a) True(b) FalseThis question was posed to me by my school principal while I was bunking the class.The above asked question is from Evaluating Digital Communication System and Allocation of Communication Resources topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Correct answer is (a) True

For explanation: Any DIGITAL communication system will BECOME more bandwidth efficient as its WTb PRODUCT DECREASES.
Discussion
83.

In MPSK, R/W(a) Increases with increase in M(b) Increases with decrease in M(c) Decreases with increase in M(d) Is not dependent on MI have been asked this question in my homework.I want to ask this question from Evaluating Digital Communication System and Allocation of Communication Resources in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

Correct CHOICE is (a) Increases with INCREASE in M

The best explanation: The BANDWIDTH efficiency R/W increases as M increases in CASE of MPSK modulation.
Discussion
84.

QAM is a combination of(a) ASK and FSK(b) ASK and PSK(c) PSK and FSK(d) None of the mentionedThe question was asked by my school principal while I was bunking the class.My doubt is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in section Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT ANSWER is (b) ASK and PSK

Easy explanation: QAM is a combination of both ASK and PSK and is ALSO called as amplitude PHASE keying.
Discussion
85.

Which refers to the presence of error correction coding scheme?(a) Coded(b) Uncoded(c) Coded & Uncoded(d) None of the mentionedThe question was asked in my homework.Question is from Evaluating Digital Communication System and Allocation of Communication Resources topic in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

Right answer is (a) Coded

Easiest explanation: The term coded REFERS to the presence of ERROR correction coding scheme INVOLVING the USE of redundant bits and expanded BANDWIDTH.
Discussion
86.

Which technique can be used for bandwidth reduction?(a) BPSK(b) QPSK(c) MPSK(d) MFSKThis question was addressed to me by my college director while I was bunking the class.I want to ask this question from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in section Modulation and Coding Trade-off of Digital Communications

Answer»

Correct OPTION is (C) MPSK

To elaborate: Coherent M-ary PSK modulation is a well KNOWN technique for ACHIEVING bandwidth reduction.
Discussion
87.

QPSK amplitude modulates(a) Sine function(b) Cosine function(c) Sine & Cosine function(d) None of the mentionedI had been asked this question in exam.Query is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in division Modulation and Coding Trade-off of Digital Communications

Answer»

Correct OPTION is (C) Sine & COSINE function

To explain: QPSK modulation consists of two independent streams. ONE stream amplitude modulates cosine WAVE and the other amplitude modulates sine wave.
Discussion
88.

Which modulation spectrum has narrow main lobe?(a) QPSK(b) OQPSK(c) BPSK(d) MSKI had been asked this question in a job interview.This interesting question is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Right choice is (a) QPSK

Easiest explanation: QPSK has NARROWER main LOBE than MSK becausefor a GIVEN rate the QPSK symbol rate is half the MSK symbol rate.
Discussion
89.

Which modulation is spectrally more efficient?(a) BPSK(b) MSK(c) QPSK(d) OQPSKI have been asked this question in a job interview.Query is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in division Modulation and Coding Trade-off of Digital Communications

Answer» CORRECT option is (B) MSK

The explanation is: MSK is spectrally more efficient than the other modulation schemes. It has WIDER main lobe level and lower SIDE lobe levels.
Discussion
90.

Which modulation has lower side lobe levels?(a) QPSK(b) OQPSK(c) BPSK(d) MSKI had been asked this question in examination.My question is based upon Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

The correct option is (d) MSK

The BEST I can EXPLAIN: MSK spectrum has wider main lobe LEVEL and lower SIDE lobe levels than the other modulation spectrum.
Discussion
91.

Which modulation requires more bandwidth?(a) QPSK(b) OQPSK(c) BPSK(d) BFSKThis question was posed to me during an interview.I need to ask this question from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in division Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
92.

Which modulation scheme uses nonlinear transponders?(a) MSK(b) Offset QPSK(c) MSK & Offset QPSK(d) None of the mentionedI had been asked this question in an online quiz.I need to ask this question from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Right answer is (c) MSK & Offset QPSK

To EXPLAIN I would say: Offset QPSK and MSK are two examples of constant envelope MODULATION SCHEMES that are attractive for systems using NON linear TRANSPONDERS.
Discussion
93.

Greater the redundancy lesser is the bandwidth expansion.(a) True(b) FalseI have been asked this question during an internship interview.Query is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Right answer is (B) False

The explanation is: WITHIN the same block size, the code with greater REDUNDANCY requires more BANDWIDTH expansion.
Discussion
94.

The primary communication resource is(a) Transmitted power(b) Received power(c) Efficiency(d) None of the mentionedI have been asked this question in homework.Question is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in division Modulation and Coding Trade-off of Digital Communications

Answer»
Discussion
95.

Which modulation is the most efficient one?(a) BPSK(b) BFSK(c) QPSK(d) QAMThis question was posed to me during an interview.I'm obligated to ask this question of Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in chapter Modulation and Coding Trade-off of Digital Communications

Answer»

The CORRECT ANSWER is (d) QAM

To ELABORATE: Of all the modulation SHOWED, QAM is the most bandwidth EFFICIENT one.
Discussion
96.

The bandwidth efficiency of QFSK is _______ that of BFSK.(a) Greater than(b) Less than(c) Equal to(d) None of the mentionedThis question was addressed to me by my college director while I was bunking the class.Question is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation in chapter Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT CHOICE is (c) Equal to

The EXPLANATION: The bandwidth efficiency of QPSK is equal to that of the BFSK.
Discussion
97.

The bandwidth efficiency increases as M increases.(a) True(b) FalseThis question was posed to me in an online interview.My question is based upon Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer»

Correct ANSWER is (b) False

Easy EXPLANATION: The BANDWIDTH efficiency DECREASES as M increases.
Discussion
98.

The bandwidth efficiency of MFSK is(a) M/log2M(b) log2M/M(c) M/log10M(d) log10M/MThe question was posed to me during an online exam.Question is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in division Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT choice is (b) log2M/M

To explain: For non COHERENT orthogonal MFSK modulation the BANDWIDTH EFFICIENCY is R/W=log2M/M.
Discussion
99.

The IF transmission bandwidth of MFSK is(a) M/T(b) T/M(c) 2M/T(d) M/2The question was asked in an online quiz.Question is from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in portion Modulation and Coding Trade-off of Digital Communications

Answer» RIGHT option is (a) M/T

The best I can explain: In NON coherent ORTHOGONAL MFSK modulation, the IF transmission bandwidth is WIF = M/T.
Discussion
100.

QPSK is a composite of(a) Two BPSK(b) Three BPSK(c) Two FSK(d) Two M-ary PSKThis question was posed to me in a national level competition.This is a very interesting question from Bandwidth Efficiency Plane and Bandwidth Efficient Modulation topic in division Modulation and Coding Trade-off of Digital Communications

Answer»

The correct CHOICE is (a) Two BPSK

The best explanation: QPSK is effectively a COMPOSITE of two BPSK SIGNALS transmitted on orthogonal components of the carrier.
Discussion