McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 51. |
Frequency modulation index defines the relationship between the and bandwidth of transmitted signal. |
| A. | frequency of message signal |
| B. | amplitude of message signal |
| C. | amplitude of carrier signal |
| D. | frequency of carrier signal |
| Answer» B. amplitude of message signal | |
| Explanation: the frequency modulation | |
| 52. |
Which of the following are two methods for generating FM signal? |
| A. | coherent method, noncoherent method |
| B. | product detector, envelope detector |
| C. | direct method, indirect method |
| D. | slope detector, zero crossing detector |
| Answer» C. direct method, indirect method | |
| Explanation: direct method and indirect method are the methods used for generating fm signals. these methods are differentiated by the variation of the carrier frequency. | |
| 53. |
In indirect method, the carrier frequency is directly varied in accordance with the input modulating signal. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the above is the case for direct method. in the indirect method, a narrowband fm signal is generated using a balanced | |
| 54. |
Which of the following is used to vary the frequency of the carrier frequency in accordance with the baseband signal amplitude variations in direct method of FM generation? |
| A. | integrator |
| B. | envelope detector |
| C. | multivibrator |
| D. | voltage controlled oscillators |
| Answer» D. voltage controlled oscillators | |
| Explanation: in direct method, vcos are used to vary the frequency of the carrier signal in accordance with the baseband signal amplitude variations. these oscillators use devices with reactance that can be varied by the application of a voltage. | |
| 55. |
Frequency demodulator is a frequency to amplitude converter circuit. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: frequency demodulator produces an output voltage with instantaneous amplitude that is directly proportional to the instantaneous frequency of the input fm signal. thus, frequency demodulator is a frequency to amplitude converter circuit. | |
| 56. |
Which of the following is not a technique for FM demodulation? |
| A. | slope detection |
| B. | zero crossing detection |
| C. | product detector |
| D. | phase locked discriminator |
| Answer» C. product detector | |
| Explanation: various techniques such as slope detection, zero crossing detection, phase locked discrimination and quadrature | |
| 57. |
Which of the following FM demodulator is sometimes known as pulse averaging discriminator? |
| A. | slope detection |
| B. | zero crossing detection |
| C. | quadrature detection |
| D. | phase locked discriminator |
| Answer» B. zero crossing detection | |
| Explanation: zero crossing detector is sometimes known as pulse averaging discriminator. the rationale behind this technique is to use the output of the zero crossing detector to generate a pulse train with an average value that is proportional to frequency of the input signal. | |
| 58. |
PLL in FM detection stands for |
| A. | phase locked loop |
| B. | programmable logic loop |
| C. | phase locked logic |
| D. | programmable locked loop |
| Answer» A. phase locked loop | |
| Explanation: pll stands for phase locked loop. the pll is a closed loop control system which can track the variations in the received signal phase and frequency. | |
| 59. |
In angle modulation, signal to noise ratio before detection is a function of |
| A. | modulation index |
| B. | input signal to noise ratio |
| C. | maximum frequency of the message |
| D. | if filter bandwidth |
| Answer» D. if filter bandwidth | |
| Explanation: in angle modulation systems, the signal to noise ratio before detection is the function of the receiver if filter bandwidth, received carrier power, and received interference. however, signal to noise ratio after detection is a function of maximum | |
| 60. |
FM can improve the receiver performance through adjustment of transmitted power. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: fm can improve receiver performance through adjustment of the modulation index at the transmitter, and not the transmitted power. this is not the case in am since linear modulation techniques do not trade bandwidth for snr. | |
| 61. |
De-emphasis circuit is used |
| A. | before detection |
| B. | after detection |
| C. | before encoding |
| D. | after encoding |
| Answer» B. after detection | |
| Explanation: de-emphasis means attenuation of those frequencies by the amount by which they are boosted. it is done at the receiver end | |
| 62. |
Why frequency fogging is used in a carrier system? |
| A. | to reduce noise |
| B. | to reduce cross talk |
| C. | to converge frequencies |
| D. | to reduce distortion |
| Answer» B. to reduce cross talk | |
| Explanation: the interchanging of the frequencies of carrier channels to accomplish specific purposes. it is used to prevent feedback and oscillation. it is also used to reduce cross-talk and also to correct for a high frequency response slope in the transmission line. | |
| 63. |
For a phase modulated signal, the frequency deviation is proportional to |
| A. | frequency only |
| B. | amplitude only |
| C. | only width |
| D. | phase only |
| Answer» B. amplitude only | |
| Explanation: for a phase modulated system, it is amplitude which is directly proportional to the deviation. | |
| 64. |
The frequency deviation is proportional to frequency in phase modulated signal. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in phase modulation, the phase of the carrier signal is varied with respect to the amplitude of the message signal. for a phase modulated signal, the frequency deviation is proportional to amplitude. | |
| 65. |
Which of the following is not necessarily an advantage of FM over AM? |
| A. | less modulating power is required |
| B. | better noise immunity is provided |
| C. | higher bandwidth is required |
| D. | carrier is of any shape |
| Answer» C. higher bandwidth is required | |
| Explanation: in frequency modulation, frequency of carrier gets varied with respect to the wave being propagated. fm has many advantages over am but requirement of higher bandwidth is not an efficient condition. | |
| 66. |
What is number of possible outputs if there is 7 line digital input? |
| A. | 64 |
| B. | 32 |
| C. | 16 |
| D. | 128 |
| Answer» D. 128 | |
| Explanation: total possible outputs will be 27 which is equal to 128. | |
| 67. |
What is the frequency of the stereo sub carrier signal in FM broadcasting? |
| A. | 19 khz |
| B. | 45 khz |
| C. | 55 khz |
| D. | 38 khz |
| Answer» D. 38 khz | |
| Explanation: stereo broadcasting is made possible by using a subcarrier on fm radio stations, which takes the left channel and “subtracts” the right channel from it. a | |
| 68. |
What is the bandwidth required in SSB signal? |
| A. | fm |
| B. | 2fm |
| C. | > 2fm |
| D. | < 2fm |
| Answer» A. fm | |
| Explanation: in an am modulated system, total bandwidth required is from fc + fm to fc | |
| 69. |
One of the advantage of using a high frequency carrier wave is that it dissipates very small power. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the main advantage of using high frequency signals is that the signal gets transmitted over very long distances and thus dissipates very less power. the antenna height required for transmission also gets reduced at high frequencies. and also it allows less noise interference and enables multiplexing. this is the reason for sending the audio signals at high frequency carrier signals for communication purpose. | |
| 70. |
The antenna current is 10A. Find the percentage of modulation when the antenna current increases to 10.4A? |
| A. | 50% |
| B. | 30% |
| C. | 28.5% |
| D. | 23% |
| Answer» C. 28.5% | |
| 71. |
Find the total power, if the carrier of an AM transmitter is 800W and it is modulated to 50%? |
| A. | 100w |
| B. | 800w |
| C. | 500w |
| D. | 900w |
| Answer» D. 900w | |
| Explanation: p =p (1+u2 ), according to | |
| 72. |
Spread spectrum is used for |
| A. | encrypting signal |
| B. | hiding signal |
| C. | encrypting & hiding signal |
| D. | none of the mentioned |
| Answer» C. encrypting & hiding signal | |
| Explanation: spread spectrum is used for hiding and encrypting signals. | |
| 73. |
Which is a quantization process? |
| A. | rounding |
| B. | truncation |
| C. | rounding & truncation |
| D. | none of the mentioned |
| Answer» C. rounding & truncation | |
| Explanation: rounding and truncation are examples of quantization process. | |
| 74. |
Quantization is a process. |
| A. | few to few mapping |
| B. | few to many mapping |
| C. | many to few mapping |
| D. | many to many mapping |
| Answer» C. many to few mapping | |
| Explanation: quantization is a many to few mapping process. | |
| 75. |
Which conveys more information? |
| A. | high probability event |
| B. | low probability event |
| C. | high & low probability event |
| D. | none of the mentioned |
| Answer» B. low probability event | |
| Explanation: high probability event conveys less information than a low probability event. | |
| 76. |
If the channel is noiseless information conveyed is and if it is useless channel information conveyed is |
| A. | 0,0 |
| B. | 1,1 |
| C. | 0,1 |
| D. | 1,0 |
| Answer» D. 1,0 | |
| Explanation: if the channel is noiseless information conveyed is 1 and if it is useless channel information conveyed is 0. | |
| 77. |
The mutual information between a pair of events is |
| A. | positive |
| B. | negative |
| C. | zero |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the mutual information between a pair of events can be positive negative or zero. | |
| 78. |
2 QUANTIZATION |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the output of the source encoder is a sequence of binary digits. the conversion of source output to digital form is done here in source encoder. | |
| 79. |
The output of an information source is |
| A. | random |
| B. | deterministic |
| C. | random & deterministic |
| D. | none of the mentioned |
| Answer» A. random | |
| Explanation: the output of any information source is random. | |
| 80. |
When the base of the logarithm is e, the unit of measure of information is |
| A. | bits |
| B. | bytes |
| C. | nats |
| D. | none of the mentioned |
| Answer» C. nats | |
| Explanation: the unit of measure of information is determined based on the base of logarithm. if the base is e then the unit is nats( natural unit). | |
| 81. |
Flat top sampling of low pass signals |
| A. | gives rise to aperture effect |
| B. | implies over sampling |
| C. | leads to aliasing |
| D. | introduces delay distortion |
| Answer» A. gives rise to aperture effect | |
| Explanation: flat top sampling of low pass signals gives rise to aperture effect. | |
| 82. |
In a delta modulation system, granular noise occurs when the |
| A. | modulating signal increases rapidly |
| B. | pulse rate decreases |
| C. | pulse amplitude decreases |
| D. | modulating signal remains constant |
| Answer» D. modulating signal remains constant | |
| Explanation: in a delta modulation system, granular noise occurs when the modulating signal remains constant. | |
| 83. |
A PAM signal can be detected using |
| A. | low pass filter |
| B. | high pass filter |
| C. | band pass filter |
| D. | all pass filter |
| Answer» A. low pass filter | |
| Explanation: a pam signal can be detected by using low pass filter. | |
| 84. |
Coherent demodulation of FSK signal can be performed using |
| A. | matched filter |
| B. | bpf and envelope detectors |
| C. | discriminator |
| D. | none of the mentioned |
| Answer» A. matched filter | |
| Explanation: coherent demodulation of fsk signal can be performed using matched filter. | |
| 85. |
The use of non uniform quantization leads to |
| A. | reduction in transmission bandwidth |
| B. | increase in maximum snr |
| C. | increase in snr for low level signals |
| D. | simplification of quantization process |
| Answer» C. increase in snr for low level signals | |
| Explanation: the use of non uniform quantization leads to increase in snr for low level signals. | |
| 86. |
A PWM signal can be generated by |
| A. | an astable multi vibrator |
| B. | a monostable multi vibrator |
| C. | integrating a ppm signal |
| D. | differentiating a ppm signal |
| Answer» B. a monostable multi vibrator | |
| Explanation: a pwm signal can be generated by a mono stable multi vibrator. | |
| 87. |
TDM is less immune to cross-talk in channel than FDM. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: false because different message signals are not applied to the channel simultaneously. | |
| 88. |
In an ideal TDM system, the cross correlation between two users of the system is |
| A. | 1 |
| B. | 0 |
| C. | infinity |
| D. | -1 |
| Answer» B. 0 | |
| Explanation: in an ideal tdm system, the cross correlation between two users of the system is 0. | |
| 89. |
TDM requires |
| A. | constant data transmission |
| B. | transmission of data samples |
| C. | transmission of data at random |
| D. | transmission of data of only one measured |
| Answer» B. transmission of data samples | |
| Explanation: tdm requires transmission of data samples. | |
| 90. |
Which waveforms are also called as line codes? |
| A. | pcm |
| B. | pam |
| C. | fm |
| D. | am |
| Answer» A. pcm | |
| Explanation: when pulse modulation is applied to binary symbol we obtain pulse code modulated waveforms. these waveforms are also called as line codes. | |
| 91. |
When pulse code modulation is applied to non binary symbols we obtain waveform called as |
| A. | pcm |
| B. | pam |
| C. | m-ary |
| D. | line codes |
| Answer» C. m-ary | |
| Explanation: when pulse code modulation is applied to binary symbols we get pcm waveforms and when it is applied to non binary symbols we obtain m-ary waveforms. | |
| 92. |
Examples of PCM waveforms are |
| A. | non return to zero |
| B. | phase encoded |
| C. | multilevel binary |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: some of the examples or classification of pulse code modulated signals are non return to zero, return to zero, phase encoded, multilevel binary etc. | |
| 93. |
Which type is used and preferred in digital logic circuits? |
| A. | nrz-l |
| B. | nrz-m |
| C. | nrz-s |
| D. | none of the mentioned |
| Answer» A. nrz-l | |
| Explanation: nrz-l is extensively used in | |
| 94. |
Which method is called as differential encoding? |
| A. | nrz-l |
| B. | nrz-m |
| C. | nrz-s |
| D. | none of the mentioned |
| Answer» B. nrz-m | |
| Explanation: in nrz-m, logic 1 is represented by a change in voltage level and logic 0 is represented by no change in level. this is called as differential encoding. | |
| 95. |
Which method is preferred in magnetic tape recording? |
| A. | nrz-l |
| B. | nrz-m |
| C. | nrz-s |
| D. | none of the mentioned |
| Answer» B. nrz-m | |
| Explanation: nrz-m is also called as differential encoding and it is most preferred in magentic tape recording. | |
| 96. |
NRZ-S is complement of |
| A. | nrz-l |
| B. | nrz-m |
| C. | nrz-l & nrz-m |
| D. | none of the mentioned |
| Answer» B. nrz-m | |
| Explanation: nrz-s is a complement of nrz-m. logic 0 is represented by a change in voltage level and logic 1 is represented as no change in voltage level. | |
| 97. |
The return to zero waveform consists of |
| A. | unipolar rz |
| B. | bipolar rz |
| C. | rz-ami |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: different types of return to zero waveforms are unipolar rz, bipolar rz, rz- ami. these are used in baseband transmission and in magnetic recording. | |
| 98. |
Phase encoded group consists of |
| A. | manchester coding |
| B. | bi-phase-mark |
| C. | miller coding |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: different types of phase encoded waveform consists of manchester coding, bi-phase-mark, bi-phase-space, delay modulation. | |
| 99. |
In which waveform logic 1 is represented by half bit wide pulse and logic 0 is represented by absence of pulse? |
| A. | unipolar rz |
| B. | bipolar rz |
| C. | rz-ami |
| D. | manchester coding |
| Answer» A. unipolar rz | |
| Explanation: in unipolar rz waveform, logic 1 is represented by half bit wide pulse and logic 0 is represented by the absence of a pulse. | |
| 100. |
In which waveform logic 1 and logic 0 are represented by opposite one half bit wide pulses? |
| A. | unipolar rz |
| B. | bipolar rz |
| C. | rz-ami |
| D. | manchester coding |
| Answer» B. bipolar rz | |
| Explanation: in bipolar return to zero waveform ones and zeroes are represented by opposite level pulses one half bit wide pulses. | |
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