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## Communications MCQ Set 1

1. The correlating detector is also known as
a) Maximum likelihood detector
b) Minimum likelihood detector
c) Maximum & Minimum likelihood detector
d) None of the mentioned

Answer: a [Reason:] Coherent detector considers phase as the most important parameter. This coherent detector is also known as maximum likelihood detector.

2. The minimum nyquist sampling rate is given as, fs =
a) 1/T
b) T
c) 2/T
d) 2T

Answer: a [Reason:] The input signal comprises of a prototype signal plus noise and the bandwidth is 1/2T where T is the symbol time and minimum nyquist sampling rate can be given as 2W or 1/T.

3. Phase-locked loop circuitry is used for
a) Carrier wave recovery
b) Phase estimation
c) Carrier wave recovery & Phase estimation
d) None of the mentioned

Answer: c [Reason:] The phase locked loop circuitry locks on the arriving carrier wave and estimates it phase.

4. In differential PSK the date is
a) Encoded differentially
b) Decoded differentially
c) Encoded & Decoded differentially
d) None of the mentioned

Answer: a [Reason:] In this method , the data is encode deferentially that is the presence of binary zero or one is manifested by symbol’s similarity or difference when compared to that of the previous symbol.

5. Envelope detector consists of
a) Rectifier and high pass filter
b) Rectifier and low pass filter
c) Amplifier and low pass filter
d) Amplifier and high pass filter

Answer: b [Reason:] An envelope detector consists of a rectifier and a low pass filter. The detectors are matched to the envelopes and not to the signal themselves.

6. The minimum required spacing is the difference between the
a) Center of the spectral main lobe and first zero crossing
b) First and second zero crossing
c) First and last zero crossing
d) None of the mentioned

Answer: a [Reason:] The frequency difference between the center of the spectral main lobe and the first zero crossing is called as the minimum required spacing in non coherent detection.

7. The minimum tone separation corresponds to
a) T
b) 1/T
c) 2T
d) T/2

Answer: b [Reason:] In non coherent detection the minimum tone separation corresponds to 1/T.

8. Matched filter is used for
a) Coherent detection
b) Non coherent detection
c) Coherent & Non coherent detection
d) None of the mentioned

Answer: a [Reason:] Matched filter is used for coherent detection. It cannot be used for non coherent detection because matched filter output is a function of unknown angle α.

9. In differential encoding the _________ different between two wave forms is measured.
a) Magnitude
b) Frequency
c) Phase
d) Time period

Answer: c [Reason:] In differential encoding the phase of the present signal waveform is compared with the phase of the previous signal.

10. The error probability of DPSK is ________ worse than PSK.
a) Twice
b) 3 db
c) Twice (3db)
d) None of the mentioned

Answer: c [Reason:] DPSK has twice as much as noise than in PSK. The error probability in DPSK is twice (3db) worse than PSK.

## Communications MCQ Set 2

1. Compact disc is used for
a) Digital storage
b) Reproduction of audio signals
c) Digital storage & Reproduction of audio signals
d) None of the mentioned

Answer: c [Reason:] Compact disc digital audio system is used for both digital storage and reproduction of audio signals.

2. Sources of channel errors are
a) Finger prints
b) Air bubbles
c) Unwanted particles
d) All of the mentioned

Answer: d [Reason:] Some of the sources of channel errors are some unwanted particles, air bubbles, finger prints, scratches etc.

3. The CIRC error control scheme includes
a) Correction
b) Concealment
c) Correction & Concealment
d) None of the mentioned

Answer: c [Reason:] The cross interleave reed Solomon error control scheme in the CD system involves both correction and concealment of errors.

4. Decoding step consists of
a) De-interleaving
b) Decoding
c) De-interleaving & Decoding
d) None of the mentioned

Answer: c [Reason:] The decoder step consists of both De-interleaving and decoding which are performed in the reverse order of encoding steps.

5. The ______ the error correcting capability used, the _____ will be the erasure correcting capability.
a) Larger, smaller
b) Smaller, larger
c) Smaller, smaller
d) Larger, larger

Answer: a [Reason:] There is a trade off between error correction and erasure correction, the larger the error correcting capability used the smaller will be the erasure correcting capability.

6. In interpolation
b) Unreliable samples are removed
c) Both of the mentioned
d) None of the mentioned

Answer: c [Reason:] The function of the interpolation function is to add new samples estimated from the reliable ones in place of unreliable ones.

7. The channel that exhibits multi-path fading
a) Has memory
b) Exhibits mutually dependent signal transmission impairments
c) Received signal will be distorted
d) All of the mentioned

Answer: d [Reason:] A channel that exhibits multi-path fading where the signal arrives at the receiver over two or more different paths. This channel has memory , exhibits mutually dependent signal transmission impairment.

8. Separating the symbols makes the channel memory-less.
a) True
b) False

Answer: a [Reason:] Separating the symbols in time effectively changes a channel with memory into a memory-less one.

9. The minimum end to end delay in block interleaving is
a) 2MN+2M+2
b) 2MN-2M+2
c) 2MN-2M-2
d) 2MN+2M-2

Answer: b [Reason:] The inter leaver or De-inter leaver delay is 2MN symbol times. M(N-1)+1 memory cells needs to be filled before transmission can begin. Thus the minimum end to end delay is 2MN-2M+2 symbol times, not including any channel propagation delay.

10. A concatenated code uses
a) One level of coding
b) Two levels of coding
c) Three levels of coding
d) None of the mentioned

Answer: b [Reason:] A concatenated code is the one that uses two levels of coding, inner code and outer code to achieve desired error performance.

11. A concatenated code has
a) Low error rate
b) High complexity
c) Low error rate & High complexity
d) None of the mentioned

Answer: a [Reason:] A concatenated code has low error rate with an overall implementation complexity which is less than that would be required for single level coding.

12. The performance of non binary concatenated code depends on
a) Bit errors
b) Symbols errors
c) Bit & Symbols errors
d) None of the mentioned

Answer: b [Reason:] The performance of non binary code like reed-Solomon code which is also an example for concatenated code depends only on the number of symbol errors in the block.

13. Concatenated code is an example of
a) Error detecting code
b) Error correcting code
c) Error detecting & correcting code
d) None of the mentioned

Answer: b [Reason:] Concatenated codes belongs to the class of error correcting codes.

14. Turbo code is a
a) Serial concatenation code
b) Parallel concatenation code
c) Block code
d) None of the mentioned

Answer: b [Reason:] Turbo codes implemented a parallel concatenation of two convolution codes with an inter leaver and an iterative decoder.

## Communications MCQ Set 3

1. The measure of the amount of redundancy is given by
a) Code size
b) Code weight
c) Code rate
d) Minimum distance

Answer: c [Reason:] The rate of the code gives the measure of amount of redundancy. The rate is given by the ratio of number of data bits that form an input to a block encoder to the total number of bits.

2. The number of k bit shift over which a single information bit influences the encoder output is given by
a) Code rate
b) Constraint length
c) Code length
d) Code weight

Answer: b [Reason:] The constraint length represents the number of k bit shifts over which a single information bit influences the encoder output.

3. Method used for representing convolution encoder are
a) Connection pictorial
b) State diagram
c) Tree diagram
d) All of the mentioned

Answer: d [Reason:] Several methods are used for representing convolution encoder such as connection pictorial, connection vector or polynomials, state diagram, tree diagram and trellis diagram.

4. Periodic truncation is done by
a) Appending zero to end of the input data sequence
b) Appending zero to the beginning of the input data sequence
c) Appending one’s to end of the data sequence
d) Appending one’s to beginning of the data sequence

Answer: a [Reason:] Convolution code are forced into a block structure by periodic truncation which is done by appending zero to the end of the input data sequence, for the purpose of clearing or flushing the encoding shift register of the data bits.

5. Finite state machines have
a) Past memories also
b) Only present and future memories
c) Only future memories
d) Only present memories

Answer: a [Reason:] A convolution encoder belongs to a class of device called as finite state machines which are those machines that have a memory of past signals.

6. How many number of transitions can be made at each bit time?
a) One
b) Two
c) Three
d) Half

Answer: b [Reason:] An a consequence of shifting-in-one bit at a time there are only two possible transitions that the register can make at each bit time.

7. In trellis diagram, the number of nodes ______ at successive branching.
a) Increases by 1
b) Doubles
c) Triples
d) None of the mentioned

Answer: b [Reason:] As per the algorithm, in the trellis diagram at successive branching the number of nodes doubles.

8. Example for convolution encoder state diagram is
a) Tree diagram
b) Trellis diagram
c) Tree & Trellis diagram
d) None of the mentioned

Answer: c [Reason:] Tree diagram and trellis diagram are both examples for convolution encoding state diagram.

9. In maximum likelihood decoding technique, the likelihood function compares the
a) Joint probabilities
b) Individual probabilities
c) Conditional probabilities
d) None of the mentioned

Answer: c [Reason:] If all input messages are equally likely a decoder that achieves the minimum probability of error is that which compares the conditional probabilities also called the likelihood function.

10. In maximum likelihood detector the error probability is
a) Maximum
b) Minimum
c) Zero
d) None of the mentioned

Answer: b [Reason:] The maximum likelihood detector is an optimal detector which minimizes the error probability.

11. For a binary code, the maximum number of possible sequence made up of L branch words is
a) L2
b) 2L
c) 2L
d) L/2

Answer: b [Reason:] In a binary code, all possible different code word sequences that can be made up of L branch words are 2L.

12. If the quantization level of a demodulator output is ________ then it is called as soft decision decoding.
a) Equal to 2
b) More than 2
c) Less than 2
d) None of the mentioned

Answer: b [Reason:] If the quantization level of a demodulator output is 2 it is called as hard decision decoding and if it is greater than 2 it is called as soft decision decoding.

13. Soft decision decoding requires
a) Less memory
b) More memory
c) More speed
d) None of the mentioned

Answer: b [Reason:] The price paid for soft decision decoding is an increase in required memory size at the decoder.

14. A binary symmetric channel is a
a) Discrete memoryless channel
b) Continuous memoryless channel
c) Discrete memory channel
d) Continuous memory channel

Answer: a [Reason:] A binary symmetric channel is a discrete memoryless channel that has binary input and output alphabets and symmetric transition probabilities.

15. Binary symmetric channel is hard decision channel.
a) True
b) False

Answer: a [Reason:] A binary symmetric channel is an example for hard decision channel which means that even though continuous values may be received by the demodulator the BSC only allows firm decisions.

16. Gaussian channel is a hard decision channel.
a) True
b) False

Answer: b [Reason:] The quantized Gaussian channel is referred to as the soft decision channel.

17. Branch word synchronization is the process of determining the
b) Beginning of transmitted sequence
d) End of transmitted sequence

Answer: a [Reason:] Branch word synchronization is the process of determining the beginning of a branch word in the received sequence.Such synchronization can take place without adding new information to the transmitted symbols.

## Communications MCQ Set 4

1. 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
a) Correlative coding
b) Duobinary signalling
c) Partial response signalling
d) All of the mentioned

Answer: d [Reason:] The interference at the detector can be cancelled out using these method in which some controlled amount of ISI is introduced into the data stream.

2. From digital filter we will get the output pulse as the _______ of the current and the previous pulse.
a) Summation
b) Difference
c) Product
d) Ratio

Answer: a [Reason:] The digital filter incorporates one digit delay and thus it adds the incoming pulse with the value of the previous pulse.

3. In duobinary signalling method, for M-ary transmission, the number of output obtained is
a) 2M
b) 2M+1
c) 2M-1
d) M2

Answer: c [Reason:] In duobinary coding the number of output obtained for M-ary transmission is 2M-1.

4. The method using which the error propagation in dubinary signalling can be avoided is
a) Filtering
b) Precoding
c) Postcoding
d) None of the mentioned

Answer: b [Reason:] In duobinary signalling method if one error occurs it repeats everywhere through out the next steps. To avoid this precoding method can be used.

5. In precoding technique the binary sequence is _____ with the previous precoded bit.
a) And-ed
b) Or-ed
c) EXOR-ed

Answer: c [Reason:] 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.

6. The duobinary filter, He (f) is called as
a) Sine filter
b) Cosine filter
c) Raised cosine filter
d) None of the mentioned

Answer: b [Reason:] The transfer function is 2T cos(πfT) which is called as cosine filter.

7. The method which has greater bandwidth efficiency is called as
a) Duobinary signalling
b) Polybinary signalling
c) Correlative coding
d) All of the mentioned

Answer:b [Reason:] If more than three levels are introduced in duobinary signalling technique the bandwidth efficiency increases This method is called as polybinary signalling.

8. In polybinary signalling method the present bit of binary sequence is algrebraically added with ______ number of previous bits.
a) j
b) 2j
c) j+2
d) j-2

Answer: d [Reason:] 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.

9. The primary advantage of this method is
a) redistribution of spectral density
b) to favor low frequencies
c) both of the mentioned
d) none of the mentioned

Answer: c [Reason:] Each bit can be independently detected in-spite of strong correlation and this provides redistribution of spectral density and also favors low frequencies.

10. Source encoding procedure does
a) Sampling
b) Quantization
c) Compression
d) All of the mentioned

Answer: d [Reason:] Source encoding includes sampling of continuous time signals, quantization of continuous valued signals and compression of those sources.

## Communications MCQ Set 5

1. The cyclic codes are designed using
a) Shift registers with feedback
b) Shift registers without feedback
c) Flipflops
d) None of the mentioned

Answer: a [Reason:] The cyclic codes are a subclass of linear codes. It is designed using feedback shift registers.

2. A cyclic code can be generated using
a) Generator polynomial
b) Generator matrix
c) Generator polynomial & matrix
d) None of the mentioned

Answer: a [Reason:] A cyclic code can be generated using generator polynomial and block codes can be generated using generator matrix.

3. The feedback shift register circuit is called as
a) Multiplying circuit
b) Dividing circuit
c) Feedback circuit
d) Shifting circuit

Answer: b [Reason:] The cyclic shift of a code-word polynomial and encoding involves division of one polynomial by another. Thus this feedback shift register is also called as dividing circuit.

4. In the dividing circuit, the parity polynomial is obtained by the
a) Quotient
b) Remainder
c) Dividend
d) Divisor

Answer: b [Reason:] The parity polynomial is the remainder after diving by the generator polynomial it is available in the register after n shifts through the n-k stage feedback registers.

5. The received code contains error if the syndrome vector is
a) Zero
b) Non zero
c) Infinity
d) None of the mentioned

Answer: b [Reason:] If the syndrome is an all zero vector then the received code-word is a valid code. If the syndrome is a non zero vector then the received code has errors.

6. Block codes are generated using
a) Generator polynomial
b) Generator matrix
c) Generator polynomial & matrix
d) None of the mentioned

Answer: b [Reason:] Block codes are generated using generator matrix and cyclic codes are generated using generator polynomial.

7. Extended go-lay code is formed by
a) Adding overall parity bit to perfect go-lay code
b) Ex-oaring overall parity bit with perfect go-lay code
c) Ex-oaring each bit of go-lay code
d) Dividing the overall parity bit with perfect go-lay code

Answer: a [Reason:] Extended go-lay code is formed by adding overall parity bit with the perfect bit known as the golay code.

8. Block length is the _____________ in the code word.
a) Number of elements
b) Distance between elements
c) Number of parity bits
d) None of the mentioned

Answer: a [Reason:] The block length n is the number of elements in the code word.

9. The rate of a block code is the ration of
a) Block length to message length
b) Message length to block length
c) Message weight to block length
d) None of the mentioned

Answer: b [Reason:] The rate of a block code is the ratio between its message length and the block length, R=k/n.

10. Linear codes are used for
a) Forward error correction
b) Backward error correction
c) Forward error detection
d) Backward error detection

Answer: a [Reason:] Linear codes are used in forward error correction. It allows for more efficient encoding and decoding procedures.

11. The k-bit message forms ____ distinct messages which is referred to as k-tuples.
a) 2k
b) K2
c) 2k
d) 21/k

Answer: c [Reason:] The k bit messages for 2k distinct message sequences which are referred to as k-tuples or sequence of k digits.

12. The sum of any two vectors in subset S is also in S. This is called as
b) Subset property
c) Closure property
d) Similarity property

Answer: c [Reason:] The closure property states that the sum of any two vectors in subset S is also in S.

13. To avoid corruption during transmission, the code-word should be
a) Near
b) Far apart
c) Far
d) None of the mentioned

Answer: b [Reason:] The code-words should be far apart from one and another as possible so even when the vectors experience some corruption they may still be correctly decoded.

14. In a standard matrix set code-word there are _______ cosset.
a) 2k
b) 2n+k
c) 2n-k
d) 2n

Answer: c [Reason:] Each n-tuple appears in only one location none are missing and none are replicated. There are 2n-k cosets.

15. Syndrome is calculated by
a) HT/r
b) rHT
c) rH
d) None of the mentioned

Answer: b [Reason:] The syndrome is calculated using S=rHT.

16. The _____ of the code-word is the number of non zero elements.
a) Size
b) Weight
c) Distance
d) Subspace