McqMate
| 1. |
If the peak message signal amplitude is half the peak amplitude of the carrier signal, the signal is modulated. |
| A. | 100% |
| B. | 2% |
| C. | 50% |
| D. | 70% |
| Answer» C. 50% | |
| Explanation: the modulation is also expressed in percentage. it is also called percentage modulation. the signal is said to be 50% modulated if the peak message signal | |
| 2. |
A percentage of modulation greater than |
| A. | 10% |
| B. | 25% |
| C. | 50% |
| D. | 100% |
| Answer» D. 100% | |
| Explanation: a percentage of modulation greater than 100% will distort the message signal if detected by an envelope detector. in this case the lower excursion of the signal will drive the carrier amplitude below zero, making it negative (and hence changing its phase). | |
| 3. |
Single sideband AM systems occupy same bandwidth as of conventional AM systems. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: single sideband (ssb) am systems transmit only one of the sidebands (either upper or lower) about the carrier. | |
| 4. |
How is the performance of SSB AM systems in fading channels? |
| A. | poor |
| B. | best |
| C. | good |
| D. | average |
| Answer» A. poor | |
| Explanation: ssb systems have the advantage of being very bandwidth efficient. but their performance in fading channels is very poor. for proper detection, the frequency of the oscillator at the product detector mixer in the receiver must be same as that of the incoming carrier frequency. | |
| 5. |
Which of the following is a disadvantage of tone-in-band SSB system? |
| A. | high bandwidth |
| B. | bad adjacent channel protection |
| C. | effects of multipath |
| D. | generation and reception of signal is complicated |
| Answer» D. generation and reception of signal is complicated | |
| Explanation: tone-in-band ssb systems has the advantage of maintaining the low bandwidth property of the ssb signals, while at the same time providing good adjacent | |
| 6. |
FFSR in AM systems stands for |
| A. | feedforward signal regeneration |
| B. | feedbackward signal regeneration |
| C. | feedbackward system restoration |
| D. | feedforward system restoration |
| Answer» A. feedforward signal regeneration | |
| Explanation: ffsr stands for feedforward signal regeneration. if the pilot tone and the information bearing signal undergo correlated fading, it is possible at the receiver to counteract the effects of fading through signal processing based on tracking of pilot tone. | |
| 7. |
AM demodulation technique can be divided into and demodulation. |
| A. | direct, indirect |
| B. | slope detector, zero crossing |
| C. | coherent, noncoherent |
| D. | quadrature detection, coherent detection |
| Answer» C. coherent, noncoherent | |
| Explanation: am demodulation techniques may be broadly divide into two main categories. they are called coherent and noncoherent demodulation. they are differentiated by the knowledge of transmitted carrier frequency and phase at the receiver. | |
| 8. |
Non coherent detection requires the knowledge of transmitted carrier frequency and phase at the receiver. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: non coherent detection does require the knowledge of phase information. however, coherent detection requires knowledge of the transmitted carrier frequency and phase at the receiver. | |
| 9. |
A product detector in AM systems is also called |
| A. | envelope detector |
| B. | differentiator |
| C. | integrator |
| D. | phase detector |
| Answer» D. phase detector | |
| Explanation: a product detector is also called a phase detector. it forms a coherent demodulator for am signals. it is a down converter circuit which converts the input bandpass signal to a baseband signal. | |
| 10. |
AM system use only product detector for demodulation. They never use envelope detectors. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: am systems can use either product detector or envelope detector for demodulation. as a rule, envelope detectors are useful when input signal power is at least 10db greater than noise power, whereas product detectors are able to process the am signals with input signal to noise ratios well below 0 db. | |
| 11. |
LCD uses |
| A. | sematic crystals |
| B. | twisted nematic crystals |
| C. | nematic crystals |
| D. | cholesteric crystals |
| Answer» B. twisted nematic crystals | |
| Explanation: lcd uses liquid crystal display. it uses twisted nematic crystals which are a type of liquid crystal, consisting of a substance called the nematic. the nematic liquid crystal is placed between two plates of polarized glass. | |
| 12. |
Which of the following stage is present in FM receiver but not in AM receiver? |
| A. | amplitude limiter |
| B. | demodulator |
| C. | am amplifier |
| D. | mixer |
| Answer» A. amplitude limiter | |
| Explanation: amplitude limiter circuit is used in fm receiver to remove the noise or any variation in amplitude present in the received signal. thus, the output of the amplitude limiter has a constant amplitude. so it is only used in frequency modulation and not in amplitude modulation. | |
| 13. |
Function of duplexer in a RADAR is to permit the use of same antenna for transmission and reception. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: a duplexer is being an electronic unit, allows bi-directional communication over the same path. the transmitter and receiver can communicate simultaneously. in radar, the duplexer isolates the receiver from the transmitter while allowing them to share a common antenna. | |
| 14. |
Single Sideband Modulation (SSB) is generally reserved for point-to-point communication. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: a point-to-point communication refers to bidirectional communication between only one transmitter and one receiver. in ssb-sc modulation technique, the carrier is suppressed and only one of the two side-bands are transmitted. | |
| 15. |
For an AM transmitter, class C amplifier can be used after the modulation stage. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in an am transmitter, the required transmission power is obtained from class c amplifier, as it is a power amplifier, for low-level or high-level modulation. so it is not used after the modulation stage. | |
| 16. |
For which of the modulated system, the linear amplified modulated stage is used? |
| A. | low level amplitude modulated system |
| B. | high level amplitude modulated system |
| C. | high level frequency modulated system |
| D. | low level frequency modulated system |
| Answer» A. low level amplitude modulated system | |
| Explanation: in low-level modulation, the generation of amplitude modulated signal takes place at low power levels. the generated am signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. thus, linear amplified modulated stage is used in low level amplitude modulated system. | |
| 17. |
When noise is passed through a narrow band filter, the output of filter should be? |
| A. | triangular |
| B. | square |
| C. | parabolic |
| D. | sinusoidal |
| Answer» D. sinusoidal | |
| Explanation: narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. it is a combination of band pass and band reject filter. when noise gets passed through it, the output of it should be sinusoidal. | |
| 18. |
A narrow band noise can exist in |
| A. | am only |
| B. | pcm only |
| C. | fm only |
| D. | am and fm both |
| Answer» D. am and fm both | |
| Explanation: narrow band filter is used to isolate a narrow band of frequencies from a wider bandwidth signal. it is a combination of band pass and band reject filter. so it can be used in both am and fm to pass a band of frequencies or to attenuate a band of frequencies. | |
| 19. |
The upper and lower sideband frequencies for 5KHz amplitude modulation with a 30KHz carrier frequency will be? |
| A. | 35khz and 25khz |
| B. | 34khz and 24khz |
| C. | 25khz and 35khz |
| D. | 0.35khz and 0.25khz |
| Answer» A. 35khz and 25khz | |
| Explanation: upper sideband frequency will be (30 + 5) = 35 khz and lower sideband frequency will be (30 – 5) = 25 khz. | |
| 20. |
Phase array radar can track many targets together. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: a phased array radar is an array of radiating elements, with each having a phase-shifter. the phase of the signal being emitted from the radiating element is changed to produce beams, thereby producing constructive or destructive interference for steering the beams in the required direction. | |
| 21. |
A duplex arrangement use separate frequencies for transmission. |
| A. | true |
| B. | false |
| C. | topic 1.3 ssbsc |
| Answer» A. true | |
| Explanation: in duplex communication, two- way interaction is favourable simultaneously. thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units. | |
| 22. |
VSB modulation is used in televisions because it avoids phase distortion at low frequencies. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: vestigial sideband modulation (vsb) is a type of amplitude modulation in which the carrier and only one sideband is completely transmitted and the other sideband is partly transmitted. thus, television production is done using vsb modulation as it reduces bandwidth to half. | |
| 23. |
A cordless telephone that uses separate frequencies for transmission in base and portable units is called |
| A. | half duplex |
| B. | duplex |
| C. | simplex |
| D. | one-way communication |
| Answer» B. duplex | |
| Explanation: in duplex communication, two- way interaction is favourable simultaneously. thus, a cordless telephone is duplex which uses separate frequencies for transmission in base and portable units. | |
| 24. |
Which polarization is used to reduce the depolarization effect on received waves? |
| A. | circular polarization |
| B. | linear polarization |
| C. | atomic polarization |
| D. | dipolar polarization |
| Answer» A. circular polarization | |
| Explanation: in circular polarization at each point the electric field of electromagnetic wave has a constant magnitude but its | |
| 25. |
Circular polarization involves critical alignment between transmitting and receiving antenna. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in circular polarization at each point the electric field of the electromagnetic wave has a constant magnitude but its direction changes as it rotates with time at a steady rate, in a plane perpendicular to the direction of propagation of the wave. it is used to reduce depolarization effect on received waves. it does not involve alignment between transmitting and receiving antenna. | |
| 26. |
It is only the reflected color that decided the color of an object. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: color of any object is decided by the reflected color for opaque object and wavelength transmitted through it for transparent object, while both reflector color and wavelength transmitted are considered for a translucent object. | |
| 27. |
What do you understand by the term “carrier” in modulation? |
| A. | voltage to be transmitted |
| B. | resultant wave |
| C. | voltage for which amplitude, phase or frequency can be varied |
| D. | voltage for which amplitude, phase or frequency remains constant |
| Answer» C. voltage for which amplitude, phase or frequency can be varied | |
| Explanation: carrier wave is the wave with frequency higher than the message signal, | |
| 28. |
Carrier wave in modulation is a resultant wave. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: carrier wave is the wave with frequency higher than the message signal, whose certain characteristics like amplitude, phase or frequency are varied with respect to the instantaneous amplitude of the message signal. thus forming the modulated wave which is the wave to be transmitted. | |
| 29. |
For a low level AM system, amplifier modulated stage must have |
| A. | harmonic devices |
| B. | linear devices |
| C. | non-linear devices |
| D. | class a amplifiers |
| Answer» B. linear devices | |
| Explanation: in low-level modulation, the generation of amplitude modulated signal takes place at low power levels. the generated am signal is then amplified using a chain of linear amplifiers, which are required to avoid waveform distortion. thus, linear devices are used in low level amplitude modulated system. | |
| 30. |
Which is the greatest disadvantage of Pulse Code Modulation? |
| A. | highly prone to noise |
| B. | cannot travel long distances |
| C. | its inability to handle analog signals |
| D. | large bandwidth is required for it |
| Answer» D. large bandwidth is required for it | |
| Explanation: pulse code modulation (pcm) is a digital form of communication. for demodulation of pcm, it is necessary to convert it into pam. quantization noise occurs in pcm only. its greatest disadvantage is its requirement for large bandwidth. | |
| 31. |
Inductance and capacitance of a line is 0.8 |
| A. | 158 |
| B. | 166 |
| C. | 143 |
| D. | 127 |
| Answer» A. 158 | |
| 32. |
Quantization noise occurs in |
| A. | frequency division multiplexing |
| B. | time division multiplexing |
| C. | delta modulation |
| D. | amplitude modulation |
| Answer» D. amplitude modulation | |
| Explanation: quantisation is the process | |
| 33. |
What the main advantage of PCM? |
| A. | can travel small distances |
| B. | higher bandwidth |
| C. | lower noise |
| D. | good reception |
| Answer» C. lower noise | |
| Explanation: pulse code modulation is a technique in which the amplitude of an analogue signal is converted to a binary value represented as a series of pulses. it is less prone to noise and can travel through long distances without loss of data. | |
| 34. |
In AM pilot carrier, transmission has |
| A. | carrier and part of one side band |
| B. | two side bands and a carrier |
| C. | two side bands |
| D. | carrier, one side band and part of other side band |
| Answer» B. two side bands and a carrier | |
| Explanation: in amplitude modulated wave, the transmitted wave has two side bands and a carrier. thus it’s bandwidth is twice the maximum modulating frequency. | |
| 35. |
Quantization noise depends upon both sampling rate and number of quantization levels. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: quantization noise in pulse code modulation (pcm) depends upon only on number of quantization levels. | |
| 36. |
Which of the following frequency is not transmitted in AM transmission? |
| A. | upper side band frequency |
| B. | carrier frequency |
| C. | lower side band frequency |
| D. | audio frequency |
| Answer» D. audio frequency | |
| Explanation: audio frequency is the frequency that is not transmitted in am transmission. | |
| 37. |
Companding is used in PCM transmitters to allow amplitude limiting in the receivers. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: companding is the process through which the signal to noise ratio of a wave is reduced by compressing and expanding the signal. it decreases the number of bits required to record the strongest signal. companding also improves signal to noise ratio. | |
| 38. |
What is the use of Companding? |
| A. | in pcm transmitters to allow amplitude limiting in the receivers |
| B. | in pcm receiver to overcome impulse noise |
| C. | to overcome quantizing noise in pcm |
| D. | to protect small signals in pcm from quantizing distortion |
| Answer» D. to protect small signals in pcm from quantizing distortion | |
| Explanation: companding is the process through which the signal to noise ratio of a wave is reduced by compressing and expanding the signal. it decreases the number of bits required to record the strongest signal. companding also improves signal to noise ratio. it is mainly used to protect small signals in pcm from quantizing distortion. | |
| 39. |
5 PSD MODULATORS AND DEMODULATORS |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: modern mobile communication systems use digital modulation techniques. | |
| 40. |
Which of the following is not an advantage of digital modulation? |
| A. | greater noise immunity |
| B. | greater security |
| C. | easier multiplexing |
| D. | less bandwidth requirement |
| Answer» D. less bandwidth requirement | |
| Explanation: digital modulation offer many advantages over analog modulation. some advantages include greater noise immunity and robustness. they provide easier multiplexing of various forms of information and greater security. | |
| 41. |
The performance of modulation scheme is not measured in terms of |
| A. | power efficiency |
| B. | bandwidth efficiency |
| C. | cost and complexity |
| D. | transmitted power |
| Answer» D. transmitted power | |
| Explanation: the performance of modulation scheme is often measured in terms of its power efficiency and bandwidth efficiency. | |
| 42. |
In digital communication system, in order to increase noise immunity, it is necessary to increase |
| A. | signal power |
| B. | signal amplitude |
| C. | signal frequency |
| D. | signal magnitude |
| Answer» A. signal power | |
| Explanation: in digital communication system, in order to increase noise immunity, it is necessary to increase signal power. | |
| 43. |
Which of the following is the ratio of signal energy per bit to noise power spectral density? |
| A. | bandwidth efficiency |
| B. | spectral density |
| C. | power efficiency |
| D. | power density |
| Answer» C. power efficiency | |
| Explanation: power efficiency is often expressed as the ratio of signal energy per bit to noise power spectral density required at the receiver input for a certain probability of error. power efficiency is a measure of how favourably the trade-off between fidelity and signal power is made. | |
| 44. |
Increasing the data rate implies the increase in pulse width of digital symbol. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: there is an unavoidable relationship between data rate and bandwidth occupancy. increasing the data rate implies decreasing the pulse width of a digital symbol, which increases the bandwidth of the signal. | |
| 45. |
Which of the following is the ratio of the throughput data rate per Hertz? |
| A. | bandwidth efficiency |
| B. | spectral density |
| C. | power efficiency |
| D. | power density |
| Answer» A. bandwidth efficiency | |
| Explanation: bandwidth efficiency reflects how efficiently the allocated bandwidth is utilized. it is defined as the ratio of throughput data rate per hertz in a given bandwidth. it describes the ability of a modulation scheme to accommodate data within a limited bandwidth. | |
| 46. |
Which of the following is defined as the range of frequencies over which the signal has a non zero power spectral density? |
| A. | null to null bandwidth |
| B. | half power bandwidth |
| C. | 3 db bandwidth |
| D. | absolute bandwidth |
| Answer» D. absolute bandwidth | |
| Explanation: the absolute bandwidth is defined as the range of frequencies over which the signal has a non-zero power spectral density. for symbols represented as rectangular baseband pulses, the psd profile extends over an infinite range of frequencies, and has an absolute bandwidth of infinity. | |
| 47. |
is equal to width of main spectral lobe. |
| A. | null to null bandwidth |
| B. | half power bandwidth |
| C. | 3 db bandwidth |
| D. | absolute bandwidth |
| Answer» A. null to null bandwidth | |
| Explanation: null to null bandwidth is a simpler and more widely accepted measure of bandwidth. it is equal to the width of main spectral lobe. | |
| 48. |
Half power bandwidth is also called |
| A. | absolute bandwidth |
| B. | null to null bandwidth |
| C. | 3 db bandwidth |
| D. | zero db bandwidth |
| Answer» C. 3 db bandwidth | |
| Explanation: half power bandwidth is also called the 3 db bandwidth. it is defined as the interval between frequencies at which the psd has dropped to half power, or 3 db below the peak value. | |
| 49. |
FM is a part of general class of modulation known as |
| A. | angle modulation |
| B. | phase modulation |
| C. | amplitude modulation |
| D. | frequency modulation |
| Answer» A. angle modulation | |
| Explanation: fm is a part of general class of modulation known as angle modulation. | |
| 50. |
Frequency modulated signal is regarded as the phase modulated signal in which the modulating wave is differentiated before modulation. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: frequency modulated signal is regarded as the phase modulated signal in which the modulating wave is integrated before modulation. this means that an fm signal can be generated by first integrating the message signal and then using the result as an input to a phase modulator. | |
| 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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