MCQOPTIONS
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MCQOPTIONS
This section includes 152 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Communications knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Overhead bits are |
| A. | framing and synchronizing bits |
| B. | data due to noise |
| C. | encoded bits |
| D. | none of the above |
| Answer» B. data due to noise | |
| 52. |
The digital modulation scheme in which the step size is not fixed is |
| A. | delta modulation |
| B. | adaptive delta modulation |
| C. | dpcm |
| D. | pcm |
| Answer» C. dpcm | |
| 53. |
The modulation techniques used to convert analog signal into digital signal are |
| A. | adm |
| B. | pcm |
| C. | dm |
| D. | all of above |
| Answer» E. | |
| 54. |
In modulation , the baud rate is 1/4 times the bit rate |
| A. | 4-qam |
| B. | 2-psk |
| C. | 4-psk |
| D. | none of above |
| Answer» E. | |
| 55. |
The technique that may be used to reduce the side band power is |
| A. | msk |
| B. | bpsk |
| C. | gaussian minimum shift keying |
| D. | bfsk |
| Answer» D. bfsk | |
| 56. |
M-ary signaling produces error performance with orthogonal signaling and error performance with multiple phase signalling. |
| A. | degraded, improved |
| B. | improved, degraded |
| C. | improved, improved |
| D. | degraded, degraded |
| Answer» C. improved, improved | |
| 57. |
The crest factor of a waveform is given as – |
| A. | 2 x peak value/ rms value |
| B. | rms value / peak value |
| C. | peak value/ rms value |
| D. | peak value/ 2rms value |
| Answer» D. peak value/ 2rms value | |
| 58. |
Timing jitter is |
| A. | change in amplitude |
| B. | change in frequency |
| C. | deviation in location of the pulses |
| D. | all of the above |
| Answer» D. all of the above | |
| 59. |
The difficulty in achieving the Nyquist criterion for system design is |
| A. | there are abrupt transitions obtained at edges of the bands |
| B. | bandwidth criterion is not easily achieved |
| C. | filters are not available |
| D. | none of the above |
| Answer» B. bandwidth criterion is not easily achieved | |
| 60. |
In Random Process we have |
| A. | sample function |
| B. | ensemble of sample function |
| C. | sample space & sample points |
| D. | sample function & ensemble of sample function |
| Answer» E. | |
| 61. |
QPSK is a modulation scheme where each symbol consists of |
| A. | 4 bits |
| B. | 2 bits |
| C. | 1 bits |
| D. | m number of bits, depending upon the requireme |
| Answer» C. 1 bits | |
| 62. |
In Delta Modulation, the bit rate is |
| A. | n times the sampling frequency |
| B. | n times the modulating frequency |
| C. | n times the nyquist criteria |
| D. | none of the above |
| Answer» B. n times the modulating frequency | |
| 63. |
Nyquist criterion helps in |
| A. | transmitting the signal without isi |
| B. | reduction in transmission bandwidth |
| C. | increase in transmission bandwidth |
| D. | both a and b |
| Answer» E. | |
| 64. |
The polarities in NRZ format use |
| A. | complete pulse duration |
| B. | half duration |
| C. | both positive as well as negative value |
| D. | each pulse is used for twice the duration |
| Answer» B. half duration | |
| 65. |
The Nyquist theorem is |
| A. | relates the conditions in time domain and frequency domain |
| B. | helps in quantization |
| C. | limits the bandwidth requirement |
| D. | both a and c |
| Answer» E. | |
| 66. |
The processing gain of FH systems is given by ratio of |
| A. | hopping bandwidth and hopping period |
| B. | instantaneous bandwidth and hopping duration |
| C. | 3 db bandwidth and bit rate |
| D. | total hopping bandwidth and instantaneous bandwidth |
| Answer» E. | |
| 67. |
Alternate Mark Inversion (AMI) is also known as |
| A. | pseudo ternary coding |
| B. | manchester coding |
| C. | polar nrz format |
| D. | none of the above |
| Answer» B. manchester coding | |
| 68. |
The characteristics of compressor in μ-law companding are |
| A. | continuous in nature |
| B. | logarithmic in nature |
| C. | linear in nature |
| D. | discrete in nature |
| Answer» B. logarithmic in nature | |
| 69. |
TDM is |
| A. | analog multiplexing |
| B. | digital multiplexing |
| C. | a to d converter |
| D. | both a & b |
| Answer» C. a to d converter | |
| 70. |
Regenerative repeaters are used for |
| A. | eliminating noise |
| B. | reconstruction of signals |
| C. | transmission over long distances |
| D. | all of the above |
| Answer» E. | |
| 71. |
The maximum synchronizing capability in coding techniques ispresent in |
| A. | manchester format |
| B. | polar nrz |
| C. | polar rz |
| D. | polar quaternary nrz |
| Answer» B. polar nrz | |
| 72. |
In Delta modulation, |
| A. | one bit per sample is transmitted |
| B. | all the coded bits used for sampling are transmitted |
| C. | the step size is fixed |
| D. | both a and c are correct |
| Answer» E. | |
| 73. |
The process of coding multiplexer output into electrical pulses or waveforms for transmission is called |
| A. | line coding |
| B. | amplitude modulation |
| C. | fsk |
| D. | filtering |
| Answer» B. amplitude modulation | |
| 74. |
DSSS system spreads the baseband signal by the baseband pulses with a pseudo noise sequence. |
| A. | adding |
| B. | subtracting |
| C. | multiplying |
| D. | dividing |
| Answer» D. dividing | |
| 75. |
BPSK system modulates at the rate of |
| A. | 1 bit/ symbol |
| B. | 2 bit/ symbol |
| C. | 4 bit/ symbol |
| D. | none of the above |
| Answer» B. 2 bit/ symbol | |
| 76. |
The error probability of a PCM is |
| A. | calculated using noise and inter symbol interference |
| B. | gaussian noise + error component due to inter symbol interference |
| C. | calculated using power spectral density |
| D. | all of the above |
| Answer» E. | |
| 77. |
A TDMA system uses 25 MHz for the forward link, which is broken into radio channels of 200 kHz. If 8 speech channels are supported on a single radio channel, how many simultaneous users can be accommodated? |
| A. | 25 |
| B. | 200 |
| C. | 1600 |
| D. | 1000 |
| Answer» E. | |
| 78. |
The filter used for pulse shaping is |
| A. | raised – cosine filter |
| B. | sinc shaped filter |
| C. | gaussian filter |
| D. | all of the above |
| Answer» E. | |
| 79. |
The symbol time in FDMA systems is thus intersymbol interference is |
| A. | large, high |
| B. | small, low |
| C. | small, high |
| D. | large, low |
| Answer» E. | |
| 80. |
According to Parseval’s theorem the energy spectral density curve is equal to? |
| A. | area under magnitude of the signal |
| B. | area under square of the magnitude of the signal |
| C. | area under square root of magnitude of the signal |
| D. | none of the mentioned |
| Answer» C. area under square root of magnitude of the signal | |
| 81. |
T1 carrier system is used |
| A. | for pcm voice transmission |
| B. | for delta modulation |
| C. | for frequency modulated signals |
| D. | none of the above |
| Answer» B. for delta modulation | |
| 82. |
Frequency hopping involves a periodic change of transmission |
| A. | signal |
| B. | frequency |
| C. | phase |
| D. | amplitude |
| Answer» C. phase | |
| 83. |
In Differential Pulse Code Modulation techniques, the decoding isperformed by |
| A. | accumulator |
| B. | sampler |
| C. | pll |
| D. | quantizer |
| Answer» B. sampler | |
| 84. |
ASK modulated signal has the bandwidth |
| A. | same as the bandwidth of baseband signal |
| B. | half the bandwidth of baseband signal |
| C. | double the bandwidth of baseband signal |
| D. | none of the above |
| Answer» B. half the bandwidth of baseband signal | |
| 85. |
The error performance of MPSK as M or k increases. |
| A. | increases |
| B. | decreases |
| C. | stays constant |
| D. | none of the mentioned |
| Answer» C. stays constant | |
| 86. |
In DPSK technique, the technique used to encode bits is |
| A. | ami |
| B. | differential code |
| C. | uni polar rz format |
| D. | manchester format |
| Answer» C. uni polar rz format | |
| 87. |
Polar coding is a technique in which |
| A. | 1 is transmitted by a positive pulse and 0 is transmitted by negative pulse |
| B. | 1 is transmitted by a positive pulse and 0 is transmitted by zero volts |
| C. | both a & b |
| D. | none of the above |
| Answer» B. 1 is transmitted by a positive pulse and 0 is transmitted by zero volts | |
| 88. |
In MSK, the difference between the higher and lower frequency is |
| A. | same as the bit rate |
| B. | half of the bit rate |
| C. | twice of the bit rate |
| D. | four time the bit rate |
| Answer» C. twice of the bit rate | |
| 89. |
The transmission bandwidth of the raised cosine spectrum is given by |
| A. | bt = 2w(1 + α) |
| B. | bt = w(1 + α) |
| C. | bt = 2w(1 + 2α) |
| D. | bt = 2w(2 + α) |
| Answer» B. bt = w(1 + α) | |
| 90. |
The expression for bandwidth BW of a PCM system, where v is thenumber of bits per sample and fm is the modulating frequency, is given by |
| A. | bw ≥ vfm |
| B. | bw ≤ vfm |
| C. | bw ≥ 2 vfm |
| D. | bw ≥ 1/2 vfm |
| Answer» B. bw ≤ vfm | |
| 91. |
In polar RZ format for coding, symbol ‘0’ is represented by |
| A. | zero voltage |
| B. | negative voltage |
| C. | pulse is transmitted for half the duration |
| D. | both b and c are correct |
| Answer» E. | |
| 92. |
The format in which the positive half interval pulse is followed by anegative half interval pulse for transmission of ‘1’ is |
| A. | polar nrz format |
| B. | bipolar nrz format |
| C. | manchester format |
| D. | none of the above |
| Answer» D. none of the above | |
| 93. |
Matched filters are used |
| A. | for maximizing signal to noise ratio |
| B. | for signal detection |
| C. | in radar |
| D. | all of the above |
| Answer» E. | |
| 94. |
Orthogonality of two codes means |
| A. | the integrated product of two different code words is zero |
| B. | the integrated product of two different code words is one |
| C. | the integrated product of two same code words is zero |
| D. | none of the above |
| Answer» B. the integrated product of two different code words is one | |
| 95. |
In PCM, the parameter varied in accordance with the amplitude of themodulating signal is |
| A. | amplitude |
| B. | frequency |
| C. | phase |
| D. | none of the above |
| Answer» E. | |
| 96. |
In digital transmission, the modulation technique that requires minimum bandwidth is |
| A. | delta modulation |
| B. | pcm |
| C. | dpcm |
| D. | pam |
| Answer» B. pcm | |
| 97. |
In Alternate Mark Inversion (AMI) is |
| A. | 0 is encoded as positive pulse and 1 is encoded as negative pulse |
| B. | 0 is encoded as no pulse and 1 is encoded as negative pulse |
| C. | 0 is encoded as negative pulse and 1 is encoded as positive pulse |
| D. | 0 is encoded as no pulse and 1 is encoded as positive or negative pulse |
| Answer» C. 0 is encoded as negative pulse and 1 is encoded as positive pulse | |
| 98. |
The frequency shifts in the BFSK usually lies in the range |
| A. | 50 to 1000 hz |
| B. | 100 to 2000 hz |
| C. | 200 to 500 hz |
| D. | 500 to 10 hz |
| Answer» B. 100 to 2000 hz | |
| 99. |
In orthogonal signalling with symbols containing more number of bits we need power. |
| A. | more |
| B. | less |
| C. | double |
| D. | none of the mentioned |
| Answer» B. less | |
| 100. |
In uniform quantization process |
| A. | the step size remains same |
| B. | step size varies according to the values of the input signal |
| C. | the quantizer has linear characteristics |
| D. | both a and c are correct |
| Answer» E. | |