McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering .
1. |
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» D. both a and c are correct |
2. |
The process of converting the analog sample into discrete form is called |
A. | modulation |
B. | multiplexing |
C. | quantization |
D. | sampling |
Answer» C. quantization |
3. |
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» A. continuous in nature |
4. |
The modulation techniques used to convert analog signal into digital signal are |
A. | pulse code modulation |
B. | delta modulation |
C. | adaptive delta modulation |
D. | all of the above |
Answer» D. all of the above |
5. |
The sequence of operations in which PCM is done is |
A. | sampling, quantizing, encoding |
B. | quantizing, encoding, sampling |
C. | quantizing, sampling, encoding |
D. | none of the above |
Answer» A. sampling, quantizing, encoding |
6. |
In PCM, the parameter varied in accordance with the amplitude of the modulating signal is |
A. | amplitude |
B. | frequency |
C. | phase |
D. | none of the above |
Answer» D. none of the above |
7. |
One of the disadvantages of PCM is |
A. | it requires large bandwidth |
B. | very high noise |
C. | cannot be decoded easily |
D. | all of the above |
Answer» A. it requires large bandwidth |
8. |
The expression for bandwidth BW of a PCM system, where v is the number 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» A. bw ≥ vfm |
9. |
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» D. all of the above |
10. |
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 |
11. |
In digital transmission, the modulation technique that requires minimum bandwidth is |
A. | delta modulation |
B. | pcm |
C. | dpcm |
D. | pam |
Answer» A. delta modulation |
12. |
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 |
13. |
In Differential Pulse Code Modulation techniques, the decoding is performed by |
A. | accumulator |
B. | sampler |
C. | pll |
D. | quantizer |
Answer» A. accumulator |
14. |
DPCM suffers from |
A. | slope over load distortion |
B. | quantization noise |
C. | both a & b |
D. | none of the above |
Answer» C. both a & b |
15. |
The noise that affects PCM |
A. | transmission noise |
B. | quantizing noise |
C. | transit noise |
D. | both a and b are correct |
Answer» D. both a and b are correct |
16. |
The factors that cause quantizing error in delta modulation are |
A. | slope overload distortion |
B. | granular noise |
C. | white noise |
D. | both a and b are correct |
Answer» D. both a and b are correct |
17. |
Granular noise occurs when |
A. | step size is too small |
B. | step size is too large |
C. | there is interference from the adjacent channel |
D. | bandwidth is too large |
Answer» B. step size is too large |
18. |
The crest factor of a waveform is given as – |
A. | 2peak value/ rms value |
B. | rms value / peak value |
C. | peak value/ rms value |
D. | peak value/ 2rms value |
Answer» C. peak value/ rms value |
19. |
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 |
20. |
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» B. adaptive delta modulation |
21. |
In Adaptive Delta Modulation, the slope error reduces and |
A. | quantization error decreases |
B. | quantization error increases |
C. | quantization error remains same |
D. | none of the above |
Answer» B. quantization error increases |
22. |
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 |
23. |
The maximum data transmission rate in T1 carrier system is |
A. | 2.6 megabits per second |
B. | 1000 megabits per second |
C. | 1.544 megabits per second |
D. | 5.6 megabits per second |
Answer» C. 1.544 megabits per second |
24. |
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» A. for pcm voice transmission |
25. |
Matched filter may be optimally used only for |
A. | gaussian noise |
B. | transit time noise |
C. | flicker |
D. | all of the above |
Answer» A. gaussian noise |
26. |
Matched filters may be used |
A. | to estimate the frequency of the received signal |
B. | in parameter estimation problems |
C. | to estimate the distance of the object |
D. | all of the above |
Answer» D. all of the above |
27. |
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» A. line coding |
28. |
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 |
29. |
Regenerative repeaters are used for |
A. | eliminating noise |
B. | reconstruction of signals |
C. | transmission over long distances |
D. | all of the above |
Answer» D. all of the above |
30. |
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 |
31. |
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» D. both b and c are correct |
32. |
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» A. 1 is transmitted by a positive pulse and 0 is transmitted by negative pulse |
33. |
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» A. complete pulse duration |
34. |
The format in which the positive half interval pulse is followed by a negative 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» C. manchester format |
35. |
The maximum synchronizing capability in coding techniques is present in |
A. | manchester format |
B. | polar nrz |
C. | polar rz |
D. | polar quaternary nrz |
Answer» A. manchester format |
36. |
The advantage of using Manchester format of coding is |
A. | power saving |
B. | polarity sense at the receiver |
C. | noise immunity |
D. | none of the above |
Answer» A. power saving |
37. |
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» A. pseudo ternary coding |
38. |
In DPSK technique, the technique used to encode bits is |
A. | ami |
B. | differential code |
C. | uni polar rz format |
D. | manchester format |
Answer» B. differential code |
39. |
Overhead bits are |
A. | framing and synchronizing bits |
B. | data due to noise |
C. | encoded bits |
D. | none of the above |
Answer» A. framing and synchronizing bits |
40. |
ISI may be removed by using |
A. | differential coding |
B. | manchester coding |
C. | polar nrz |
D. | none of the above |
Answer» A. differential coding |
41. |
Timing jitter is |
A. | change in amplitude |
B. | change in frequency |
C. | deviation in location of the pulses |
D. | all of the above |
Answer» C. deviation in location of the pulses |
42. |
Probability density function defines |
A. | amplitudes of random noise |
B. | density of signal |
C. | probability of error |
D. | all of the above |
Answer» A. amplitudes of random noise |
43. |
Impulse noise is caused due to |
A. | switching transients |
B. | lightening strikes |
C. | power line load switching |
D. | all of the above |
Answer» D. all of the above |
44. |
In coherent detection of signals, |
A. | local carrier is generated |
B. | carrier of frequency and phase as same as transmitted carrier is generated |
C. | the carrier is in synchronization with modulated carrier |
D. | all of the above |
Answer» D. all of the above |
45. |
Synchronization of signals is done using |
A. | pilot clock |
B. | extracting timing information from the received signal |
C. | transmitter and receiver connected to master timing source |
D. | all of the above |
Answer» D. all of the above |
46. |
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» A. the integrated product of two different code words is zero |
47. |
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» B. 0 is encoded as no pulse and 1 is encoded as negative pulse |
48. |
Advantages of using AMI |
A. | needs least power as due to opposite polarity |
B. | prevents build-up of dc |
C. | may be used for longer distance |
D. | all of the above |
Answer» D. all of the above |
49. |
The interference caused by the adjacent pulses in digital transmission is called |
A. | inter symbol interference |
B. | white noise |
C. | image frequency interference |
D. | transit time noise |
Answer» A. inter symbol interference |
50. |
Eye pattern is |
A. | is used to study isi |
B. | may be seen on cro |
C. | resembles the shape of human eye |
D. | all of the above |
Answer» D. all of the above |
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