Communication Engineering Set 8

Computer Science Engineering (CSE)

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This set of Communication Engineering Multiple Choice Questions & Answers (MCQs) focuses on Communication Engineering Set 8

Q1 | Gaussian filter has zero crossings at adjacent symbol peaks.
  • true
  • false
Answer: false
Explaination:
nyquist filters have zero crossings at adjacent symbol peaks and a truncated transfer function. gaussian filter does not follow nyquist criterion and has a smooth transfer function with no zero crossings.
Q2 | Which of the following is true for a Gaussian filter?
  • large bandwidth
  • minimum isi
  • high overshoot
  • sharp cut off
Answer: sharp cut off
Explaination:
the gaussian filter has a narrow absolute bandwidth, and has a sharp cut off, low overshoot and pulse area preservation properties. this makes it attractive for use in mobile communication that uses nonlinear rf amplifiers.
Q3 | Gaussian pulse shaping filter reduces the spectral occupancy and ISI.
  • true
  • false
Answer: false
Explaination:
gaussian pulse shaping does
Q4 | Gaussian pulses are used when cost and power efficiency are major factors.
  • true
  • false
Answer: true
Explaination:
gaussian pulses are used when cost and power efficiency are major factors. but the bit error rates due to isi are deemed to be lower than what is nominally required. thus, there is a trade-off between desired rf bandwidth and irreducible error due to isi.
Q5 | The method in which small amount of controlled ISI is introduced into the data stream rather than trying to eliminate it completely is called as
  • correlative coding
  • duobinary signalling
  • partial response signalling
  • all of the mentioned
Answer: all of the mentioned
Explaination:
the interference at the detector can be cancelled out using these methods in which some controlled amount of isi is introduced into the data stream.
Q6 | From digital filter we will get the output pulse as the                of the current and the previous pulse.
  • summation
  • difference
  • product
  • ratio
Answer: summation
Explaination:
the digital filter incorporates
Q7 | In duobinary signalling method, for M-ary transmission, the number of output obtained is
  • 2m
  • 2m+1
  • 2m-1
  • m2
Answer: 2m-1
Explaination:
in duobinary coding, the number of output obtained for m-ary transmission is 2m-1.
Q8 | The method using which the error propagation in dubinary signalling can be avoided is
  • filtering
  • precoding
  • postcoding
  • none of the mentioned
Answer: precoding
Explaination:
in duobinary signalling method if one error occurs it repeats everywhere through out the next steps. to avoid this precoding method can be used.
Q9 | In precoding technique, the binary sequence is            with the previous precoded bit.
  • and-ed
  • or-ed
  • exor-ed
  • added
Answer: exor-ed
Explaination:
to avoid error propogation precoding method is used. in this each bit is encoded individually without having any effect due to its prior bit or decisions.
Q10 | The duobinary filter, He (f) is called as
  • sine filter
  • cosine filter
  • raised cosine filter
  • none of the mentioned
Answer: cosine filter
Explaination:
the transfer function is 2t cos(πft) which is called as cosine filter.
Q11 | The method which has greater bandwidth efficiency is called as
  • duobinary signalling
  • polybinary signalling
  • correlative coding
  • all of the mentioned
Answer: polybinary signalling
Explaination:
if more than three levels are introduced in duobinary signalling technique the bandwidth efficiency increases this method is called as polybinary signalling.
Q12 | In polybinary signalling method the present bit of binary sequence is algebraically added with              number of previous bits.
  • j
  • 2j
  • j+2
  • j-2
Answer: j-2
Explaination:
in polybinary signalling method the present binary digit of the sequence is formed from the modulo-2 addition of the j-2 preceding digits of the sequence and the present digit.
Q13 | The primary advantage of this method is
  • redistribution of spectral density
  • to favor low frequencies
  • redistribution of spectral density & to favor low frequencies
  • none of the mentioned
Answer: redistribution of spectral density & to favor low frequencies
Explaination:
each bit can be independently detected in-spite of strong correlation and this provides redistribution of spectral density and also favors low frequencies.
Q14 | Source encoding procedure does
  • sampling
  • quantization
  • compression
  • all of the mentioned
Answer: all of the mentioned
Explaination:
source encoding includes a sampling of continuous time signals, quantization of continuous valued signals and compression of those sources.
Q15 | The range of amplitude difference gives the value of
  • width
  • distortion
  • timing jitter
  • noise margin
Answer: distortion
Explaination:
in the eye pattern, the amplitude difference gives the value of distortion caused by isi.
Q16 | As the eye opens, ISI                
  • increases
  • decreases
  • remains the same
  • none of the mentioned
Answer: decreases
Explaination:
as the eye closes, isi increases and as the eye opens isi decreases.
Q17 | The index value n, in transversal filter can be used as.
  • time offset
  • filter coefficient identifier
  • time offset & filter coefficient identifier
  • none of the mentioned
Answer: time offset & filter coefficient identifier
Explaination:
the index n can be used as both time offset and the filter coefficient identifier, which is the address in the filter.
Q18 | The over-determined set of equations can be solved using
  • zero forcing
  • minimum mean square error
  • zero forcing & minimum mean square error
  • none of the mentioned
Answer: zero forcing & minimum mean square error
Explaination:
the matrix x in transversal equalizer if non square with dimensions 4n+1 and 2n+1. such equations are called as over-determined set. this can be solved by two methods called as zero forcing method and minimum mean square error method.
Q19 | If the filter’s tap weight remains fixed during transmission of data, then the equalization is called as
  • preset equalization
  • adaptive equalization
  • fixed equalization
  • none of the mentioned
Answer: preset equalization
Explaination:
if the weight remains fixed during transmission of data then the equalization is called as preset equalization. it is a simple method which consists of setting the tap weight according to some average knowledge of the channel.
Q20 | Equalization method which is done by tracking a slowly time varying channel response is
  • preset equalization
  • adaptive equalization
  • variable equalization
  • none of the mentioned
Answer: adaptive equalization
Explaination:
this method is implemented to perform tap weight adjustment periodically or continually. equalization is done by tracking a slowly varying channel response.
Q21 | Preamble is used for
  • detect start of transmission
  • to set automatic gain control
  • to align internal clocks
  • all of the mentioned
Answer: all of the mentioned
Explaination:
the receiver uses preamble for detecting the start of transmission, to set automatic gain control, and to align internal clocks and local oscillator with the received signal.
Q22 | The disadvantage of preset equalizer is that
  • it doesnot requires initial training pulse
  • time varying channel degrades the performance of the system
  • all of the mentioned
  • none of the mentioned
Answer: time varying channel degrades the performance of the system
Explaination:
the disadvantage of preset equalization is that it requires an initial training period that must be invoked at the start of any new transmission. also time varying channel can degrade system performance due to isi, since the tap weights are fixed.
Q23 | For AWGN, the noise variance is
  • n0
  • n0/2
  • 2n0
  • n0/4
Answer: n0/2
Explaination:
the noise variance out of the correlator for awgn is n0/2.
Q24 | A Gaussian distribution into the non linear envelope detector yields
  • rayleigh distribution
  • normal distribution
  • poisson distribution
  • binary distribution
Answer: rayleigh distribution
Explaination:
the two output signals of gaussian distribution yields rayleigh and rician distribution.
Q25 | The DPSK needs                  Eb/N0 than BPSK.
  • 1db more
  • 1db less
  • 3db more
  • 3db less
Answer: 1db more
Explaination:
the dpsk system is easier to implement than psk and it needs 1db more eb/n0 than bpsk.
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