McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
| 51. |
The time interval over which the received signal may be sampled without error may be explained by |
| A. | width of eye opening of eye pattern |
| B. | rate of closure of eye of eye pattern |
| C. | height of the eye opening of eye pattern |
| D. | all of the above |
| Answer» A. width of eye opening of eye pattern | |
| 52. |
For a noise to be white Gaussian noise, the optimum filter is known as |
| A. | low pass filter |
| B. | base band filter |
| C. | matched filter |
| D. | bessel filter |
| Answer» C. matched filter | |
| 53. |
Matched filters are used |
| A. | for maximizing signal to noise ratio |
| B. | for signal detection |
| C. | in radar |
| D. | all of the above |
| Answer» D. all of the above | |
| 54. |
The number of bits of data transmitted per second is called |
| A. | data signaling rate |
| B. | modulation rate |
| C. | coding |
| D. | none of the above |
| Answer» A. data signaling rate | |
| 55. |
Pulse shaping is done |
| A. | to control inter symbol interference |
| B. | by limiting the bandwidth of transmission |
| C. | after line coding and modulation of signal |
| D. | all of the above |
| Answer» D. all of the above | |
| 56. |
The criterion used for pulse shaping to avoid ISI is |
| A. | nyquist criterion |
| B. | quantization |
| C. | sample and hold |
| D. | pll |
| Answer» A. nyquist criterion | |
| 57. |
The filter used for pulse shaping is |
| A. | raised – cosine filter |
| B. | sinc shaped filter |
| C. | gaussian filter |
| D. | all of the above |
| Answer» D. all of the above | |
| 58. |
Roll – off factor is defined as |
| A. | the bandwidth occupied beyond the nyquist bandwidth of the filter |
| B. | the performance of the filter or device |
| C. | aliasing effect |
| D. | none of the above |
| Answer» A. the bandwidth occupied beyond the nyquist bandwidth of the filter | |
| 59. |
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» D. both a and b | |
| 60. |
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» D. both a and c | |
| 61. |
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» A. there are abrupt transitions obtained at edges of the bands | |
| 62. |
Equalization in digital communication |
| A. | reduces inter symbol interference |
| B. | removes distortion caused due to channel |
| C. | is done using linear filters |
| D. | all of the above |
| Answer» D. all of the above | |
| 63. |
Zero forced equalizers are used for |
| A. | reducing isi to zero |
| B. | sampling |
| C. | quantization |
| D. | none of the abov |
| Answer» A. reducing isi to zero | |
| 64. |
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» A. bt = 2w(1 + α) | |
| 65. |
The preferred orthogonalization process for its numerical stability is |
| A. | gram- schmidt process |
| B. | house holder transformation |
| C. | optimization |
| D. | all of the above |
| Answer» B. house holder transformation | |
| 66. |
For two vectors to be orthonormal, the vectors are also said to be orthogonal. The reverse of the same |
| A. | is true |
| B. | is not true |
| C. | is not predictable |
| D. | none of the above |
| Answer» B. is not true | |
| 67. |
Orthonormal set is a set of all vectors that are |
| A. | mutually orthonormal and are of unit length |
| B. | mutually orthonormal and of null length |
| C. | both a & b |
| D. | none of the above |
| Answer» A. mutually orthonormal and are of unit length | |
| 68. |
In On-Off keying, the carrier signal is transmitted with signal value ‘1’ and ‘0’ indicates |
| A. | no carrier |
| B. | half the carrier amplitude |
| C. | amplitude of modulating signal |
| D. | none of the above |
| Answer» A. no carrier | |
| 69. |
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» A. same as the bandwidth of baseband signal | |
| 70. |
Coherent detection of binary ASK signal requires |
| A. | phase synchronization |
| B. | timing synchronization |
| C. | amplitude synchronization |
| D. | both a and b |
| Answer» D. both a and b | |
| 71. |
The probability of error of DPSK is ______________ than that of BPSK. |
| A. | higher |
| B. | lower |
| C. | same |
| D. | not predictable |
| Answer» A. higher | |
| 72. |
In Binary Phase Shift Keying system, the binary symbols 1 and 0 are represented by carrier with phase shift of |
| A. | Π/2 |
| B. | Π |
| C. | 2Π |
| D. | 0 |
| Answer» B. Π | |
| 73. |
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» A. 1 bit/ symbol | |
| 74. |
The BPSK signal has +V volts and -V volts respectively to represent |
| A. | 1 and 0 logic levels |
| B. | 11 and 00 logic levels |
| C. | 10 and 01 logic levels |
| D. | 00 and 11 logic levels |
| Answer» A. 1 and 0 logic levels | |
| 75. |
The binary waveform used to generate BPSK signal is encoded in |
| A. | bipolar nrz format |
| B. | manchester coding |
| C. | differential coding |
| D. | none of the above |
| Answer» A. bipolar nrz format | |
| 76. |
The bandwidth of BFSK is ______________ than BPSK. |
| A. | lower |
| B. | same |
| C. | higher |
| D. | not predictable |
| Answer» C. higher | |
| 77. |
In Binary FSK, mark and space respectively represent |
| A. | 1 and 0 |
| B. | 0 and 1 |
| C. | 11 and 00 |
| D. | 00 and 11 |
| Answer» A. 1 and 0 | |
| 78. |
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» A. 50 to 1000 hz | |
| 79. |
The spectrum of BFSK may be viewed as the sum of |
| A. | two ask spectra |
| B. | two psk spectra |
| C. | two fsk spectra |
| D. | none of the above |
| Answer» A. two ask spectra | |
| 80. |
The maximum bandwidth is occupied by |
| A. | ask |
| B. | bpsk |
| C. | fsk |
| D. | none of the above |
| Answer» C. fsk | |
| 81. |
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» B. 2 bits | |
| 82. |
The data rate of QPSK is ___________ of BPSK. |
| A. | thrice |
| B. | four times |
| C. | twice |
| D. | same |
| Answer» C. twice | |
| 83. |
QPSK system uses a phase shift of |
| A. | Π |
| B. | Π/2 |
| C. | Π/4 |
| D. | 2Π |
| Answer» B. Π/2 | |
| 84. |
Minimum shift keying is similar to |
| A. | continuous phase frequency shift keying |
| B. | binary phase shift keying |
| C. | binary frequency shift keying |
| D. | qpsk |
| Answer» A. continuous phase frequency shift keying | |
| 85. |
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» B. half of the bit rate | |
| 86. |
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» C. gaussian minimum shift keying | |
| 87. |
Analog to digital conversion includes |
| A. | sampling |
| B. | quantization |
| C. | sampling & quantization |
| D. | none of above |
| Answer» C. sampling & quantization | |
| 88. |
The process of converting the analog sample into discrete form is called |
| A. | modulation |
| B. | multiplexing |
| C. | quantization |
| D. | sampling |
| Answer» C. quantization | |
| 89. |
The modulation techniques used to convert analog signal into digital signal are |
| A. | adm |
| B. | pcm |
| C. | dm |
| D. | all of above |
| Answer» D. all of above | |
| 90. |
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» D. both a and c are correct | |
| 91. |
In digital transmission, the modulation technique that requires minimum bandwidth is |
| A. | delta modulation |
| B. | pcm |
| C. | dpcm |
| D. | pam |
| Answer» A. delta modulation | |
| 92. |
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» A. n times the sampling frequency | |
| 93. |
In Differential Pulse Code Modulation techniques, the decoding is performed by |
| A. | accumulator |
| B. | sampler |
| C. | pll |
| D. | quantizer |
| Answer» A. accumulator | |
| 94. |
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» C. peak value/ rms value | |
| 95. |
The digital modulation technique in which the step size is varied according to the variation in the slope of the input is called |
| A. | delta modulation |
| B. | pcm |
| C. | adaptive delta modulation |
| D. | pam |
| Answer» C. adaptive delta modulation | |
| 96. |
Impulse sampling is also called as |
| A. | delta sampling |
| B. | ideal sampling |
| C. | natural sampling |
| D. | both a & b |
| Answer» D. both a & b | |
| 97. |
Nyquist Rate is given by |
| A. | 2w |
| B. | 2fmin |
| Answer» A. 2w | |
| 98. |
The number of voice channels that can be accommodated for transmission in T1 carrier system is |
| A. | 24 |
| B. | 32 |
| C. | 56 |
| D. | 64 |
| Answer» A. 24 | |
| 99. |
For a line code, the transmission bandwidth must be |
| A. | maximum possible |
| B. | as small as possible |
| C. | depends on the signal |
| D. | none of the above |
| Answer» B. as small as possible | |
| 100. |
Scrambling of data is |
| A. | removing long strings of 1’s and 0’s |
| B. | exchanging of data |
| C. | transmission of digital data |
| D. | all of the above |
| Answer» A. removing long strings of 1’s and 0’s | |
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