| 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 | |
| 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. | |
| 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. | |
| 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. | |
| 101. | In which waveform logic 1 is represented by equal amplitude alternating pulses? | |
| A. | unipolar rz | |
| B. | bipolar rz | |
| C. | rz-ami | |
| D. | manchester coding | |
| Answer» C. rz-ami | ||
| Explanation: in rz-ami logic 1 is represented by equal amplitude alternating pulses and logic 0 is represented by the absence of a pulse. | ||
| 102. | The signals which are obtained by encoding each quantized signal into a digital word is called as |
| A. | pam signal |
| B. | pcm signal |
| C. | fm signal |
| D. | sampling and quantization |
| Answer» B. pcm signal | |
| Explanation: pulse code modulation is the name for the class of signals which are obtained by encoding the quantized signals into a digital word. | |
| 103. | The length of the code-word obtained by encoding quantized sample is equal to |
| A. | l=log(to the base 2)l |
| B. | l=log(to the base 10)l |
| C. | l=2log(to the base 2)l |
| D. | l=log(to the base 2)l/2 |
| Answer» A. l=log(to the base 2)l | |
| Explanation: the quantized sample which are digitally encoded into l bit value code- word. the length l can be calculated as l=log(to the base 2)l. | |
| 104. | In PCM encoding, quantization level varies as a function of |
| A. | frequency |
| B. | amplitude |
| C. | square of frequency |
| D. | square of amplitude |
| Answer» B. amplitude | |
| Explanation: in linear pcm the quantization levels are uniform. but in normal pcm encoding the quantization level vary according to the amplitude, based of a-law of myu-law. | |
| 105. | What is bit depth? |
| A. | number of quantization level |
| B. | interval between two quantization levels |
| C. | number of possible digital values to represent each sample |
| D. | none of the mentioned |
| Answer» C. number of possible digital values to represent each sample | |
| Explanation: one of the properties of pcm signal which determines its stream fidelity is bit depth which is the number of possible digital values that can be used to represent each sample. | |
| 106. | Choosing a discrete value that is near but not exactly at the analog signal level leads to |
| A. | pcm error |
| B. | quantization error |
| C. | pam error |
| D. | sampling error |
| Answer» B. quantization error | |
| Explanation: one of the limitations of pcm is quantization error which occurs when we choose a discrete value at some near by value and not at the analog signal level. | |
| 107. | In PCM the samples are dependent on |
| A. | time |
| B. | frequency |
| C. | quanization leavel |
| D. | interval between quantization level |
| Answer» A. time | |
| Explanation: the samples depend on time,an accurate clock is required for accurate reproduction. | |
| 108. | DPCM encodes the PCM values based on |
| A. | quantization level |
| B. | difference between the current and predicted value |
| C. | interval between levels |
| D. | none of the mentioned |
| Answer» B. difference between the current and predicted value | |
| Explanation: differential pcm encodes the pcm value based on the difference between the previous sample and the present sample value. | |
| 109. | Delta modulation uses bits per sample. |
| A. | one |
| B. | two |
| C. | four |
| D. | eight |
| Answer» A. one | |
| Explanation: delta modulation is used for analog to digital conversion and vice versa. it is a simple form of dpcm. its uses 1 bit per sample. it also depends on the difference between the current and previous sample values. | |
| 110. | Sample resolution for LPCM bits per sample. |
| A. | 8 |
| B. | 16 |
| C. | 24 |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: common sampling resolution for lpcm are 8, 16, 20, 24 bits per sample. | |
| 111. | Adaptive DPCM is used to |
| A. | increase bandwidth |
| B. | decrease bandwidth |
| C. | increase snr |
| D. | none of the mentioned |
| Answer» B. decrease bandwidth | |
| Explanation: adaptive dpcm is used to decrease required bandwidth for the given snr. | |
| 112. | Vocoders analyse the speech signals at |
| A. | transmitter |
| B. | receiver |
| C. | channel |
| D. | if filter |
| Answer» A. transmitter | |
| Explanation: vocoders are a class of speech coding systems. they analyse the speech signal at the transmitter. and then transmit the parameters derived from the analysis. | |
| 113. | Vocoders the voice at the receiver. |
| A. | analyse |
| B. | synthesize |
| C. | modulate |
| D. | evaluate |
| Answer» B. synthesize | |
| Explanation: vocoders synthesize the voice at the receiver. all vocoder systems attempt to model the speech generation process as a dynamic system and try to quantify certain physical constraints of the system. | |
| 114. | 6 CHANNEL VOCODER |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: vocoders are much more complex than the waveform coders. they can achieve very high economy in transmission bit rate but are less robust. | |
| 115. | Which of the following is not a vocoding system? |
| A. | linear predictive coder |
| B. | channel vocoder |
| C. | waveform coder |
| D. | formant vocoder |
| Answer» C. waveform coder | |
| Explanation: waveform coder is not a vocoding system. lpc (linear predictive coding) is the most popular vocoding system. other vocoding systems are channel vocoder, formant vocoder, cepstrum vocoder etc. | |
| 116. | Which of the following pronunciations lead to voiced sound? |
| A. | ‘f’ |
| B. | ‘s’ |
| C. | ‘sh’ |
| D. | ‘m’ |
| Answer» D. ‘m’ | |
| Explanation: voiced sounds are ‘m’, ‘n’ and ‘v’ pronounciations. they are a result of quasiperiodic vibrations of the vocal chord. | |
| 117. | Channel vocoders are the time domain vocoders. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: channel vocoders are frequency domain vocoders. they determine the envelope of the speech signal for a number of frequency bands and then sample, encode and multiplex these samples with the encoded outputs of the other filters. | |
| 118. | is often called the formant of the speech signal. |
| A. | pitch frequency |
| B. | voice pitch |
| C. | pole frequency |
| D. | central frequency |
| Answer» C. pole frequency | |
| Explanation: the pole frequencies correspond to the resonant frequencies of the vocal tract. they are often called the formants of the speech signal. for adult speakers, the formants are centered around 500 hz, 1500 hz, 2500 hz and 3500 hz. | |
| 119. | Formant vocoders use large number of control signals. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: formant vocoders use fewer control signals. therefore, formant vocoders can operate at lower bit rates than the channel vocoder. instead of transmitting the power spectrum envelope, formant vocoders attempt to transmit the position of peaks of spectral envelope. | |
| 120. | Cepstrum vocoder uses |
| A. | wavelet transform |
| B. | inverse wavelet transform |
| C. | cosine transform |
| D. | inverse fourier transform |
| Answer» D. inverse fourier transform | |
| Explanation: cepstrum vocoders use inverse fourier transform. it separates the excitation and vocal tract spectrum by fourier transforming spectrum to produce the cepstrum of the signal. | |
| 121. | The real part of an antenna’s input impedance is due to |
| A. | swr |
| B. | radiated signal |
| C. | reflected signal |
| D. | refracted signal |
| Answer» B. radiated signal | |
| Explanation: in antenna impedance, impedance related the voltage and current at the input of the antenna. the real part of antenna impedance represents power that is either radiated away or absorbed within the antenna and the imaginary part of antenna impedance represents power that is stored in the near field of antenna. | |
| 122. | What is the other name for half-wave dipole antenna? |
| A. | helical antenna |
| B. | isotropic antenna |
| C. | hertz antenna |
| D. | maxwell antenna |
| Answer» C. hertz antenna | |
| Explanation: the hertz antenna is also known as half wave dipole antenna. it consists of two straight collinear conductors of equal length separated by a small feeding gap. | |
| 123. | Measured on the ground, the field strength of a horizontally polarized half wave dipole antenna is strongest |
| A. | in one direction |
| B. | in two directions |
| C. | depends on the number of elements |
| D. | depends on the shape of antenna |
| Answer» B. in two directions | |
| Explanation: as the name suggests, half wave dipole is half wavelength long. this antenna has the shortest resonant length that can be used for a resonant dipole. the field strength of a horizontally polarized half wave dipole antenna is strongest in two directions. | |
| 124. | When an antenna radiates more energy in one direction than in other directions, it is called |
| A. | selectivity |
| B. | directivity |
| C. | active antenna |
| D. | resonance |
| Answer» B. directivity | |
| Explanation: when an antenna radiates more energy in one direction than in other directions is called directivity. an antenna that radiates equally in all directions has effectively zero directionality, and the directivity of this type of antenna should be 1 (or 0db). | |
| 126. | The polarization of plane waves received from satellite is changed by | |
| A. | faraday rotation | |
| B. | gamma rays | |
| C. | helical rotation | |
| D. | distance travelled | |
| Answer» A. faraday rotation | ||
| Explanation: generally for satellite communication circular polarization is required. the polarization received by waves from satellite is changed by faraday rotation. | ||
| 127. | What is the input impedance to a lossless antenna, at resonance? |
| A. | infinite |
| B. | 0 |
| C. | resistive |
| D. | capacitive |
| Answer» C. resistive | |
| Explanation: in antenna impedance, impedance related the voltage and current at the input of the antenna. the real part of antenna impedance represents power that is either radiated away or absorbed within the antenna and the imaginary part of antenna impedance represents power that is stored in the near field of antenna. the input impedance of a lossless antenna is purely resistive. | |
| 128. | TDMA stands for |
| A. | time division multiple access |
| B. | time domain multiple access |
| C. | time division mutual access |
| D. | time domain mutual access |
| Answer» A. time division multiple access | |
| Explanation: tdma stands for time division multiple access. it can be seen as a channel access method for shared-medium networks. | |
| 129. | Which term is used when signals move from one line to another? |
| A. | path switching |
| B. | space switching |
| C. | line switching |
| D. | cross-point switching |
| Answer» B. space switching | |
| Explanation: space switching is the used term for signals moving from one line to another. | |
| 130. | PSK stands for Pulse Shift Keying. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: psk stands for phase shift keying. it is a modulation scheme that conveys information by changing the phase of carrier. | |
| 131. | Which term is used for moving PCM samples from one time slot to another? |
| A. | time switching |
| B. | space switching |
| C. | phase switching |
| D. | frequency switching |
| Answer» A. time switching | |
| Explanation: time switching is the used term for moving pcm samples moving from one time slot to another. | |
| 132. | Power can be coupled into or out of a waveguide with a magnetic field probe. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: a waveguide is a line through which electromagnetic waves are passed for | |
| 133. | What is the full form of LOS? |
| A. | level of signal |
| B. | line of sight |
| C. | loss of signal |
| D. | level of sight |
| Answer» B. line of sight | |
| Explanation: line of sight is a line between two points. it is a straight path between a transmitting antenna and a receiving antenna. | |
| 134. | How we can define the satisfactory performance of an analog microwave system? |
| A. | carrier to noise ratio that exceeds a given value |
| B. | carrier to noise ratio that is below a given value |
| C. | an erp value that exceeds a given value |
| D. | an erp value that is below a given value |
| Answer» A. carrier to noise ratio that exceeds a given value | |
| Explanation: we can measure performance of an analog microwave system by calculating the carrier to noise ratio that exceeds a given value. it gives the signal to noise ratio. | |
| 135. | RGB stands for |
| A. | red green brown |
| B. | red green black |
| C. | red gold blue |
| D. | red green blue |
| Answer» D. red green blue | |
| Explanation: rgb stands for red green blue. it is an additive color model in which red, green and blue light intensity and different shades are added together in various ways to reproduce a broad variety of colors. | |
| 136. | How many lines are there in an NTSC signal? |
| A. | 1024 |
| B. | 1856 |
| C. | 625 |
| D. | 525 |
| Answer» D. 525 | |
| Explanation: ntsc stands for national television system committee. in ntsc, it is standardized fixed that it has total 525 lines. | |
| 137. | Luminance refers to |
| A. | contrast |
| B. | diffusion |
| C. | brightness |
| D. | aperture |
| Answer» C. brightness | |
| Explanation: luminance refers to brightness. it is a photometric measure of luminous intensity per unit area of light travelling in a given direction. | |
| 138. | Linear modulation techniques are not bandwidth efficient. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: linear modulation techniques are bandwidth efficient. they are used in wireless communication systems when there is an increasing demand to accommodate more and more users within a limited spectrum. | |
| 139. | Which of the following is not a linear modulation technique? |
| A. | oqpsk |
| B. | π/4 qpsk |
| C. | fsk |
| D. | bpsk |
| Answer» C. fsk | |
| Explanation: oqpsk, π/4 qpsk and bpsk are the most popular linear modulation techniques. they have very good spectral efficiency. however, fsk is an non-linear modulation technique. | |
| 140. | In BPSK, the of constant amplitude carrier signal is switched between two values according to the two possible values. |
| A. | amplitude |
| B. | phase |
| C. | frequency |
| D. | angle |
| Answer» B. phase | |
| Explanation: in binary phase shift keying (bpsk), the phase of a constant amplitude carrier signal is switched between two possible values m1 and m2. these two values corresponds to binary 1 and 0 respectively. | |
| 141. | BPSK uses non-coherent demodulator. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: bpsk uses coherent or synchronous demodulation. it requires the information about the phase and frequency of the carrier be available at the receiver. | |
| 142. | DPSK uses coherent form of PSK. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: differential phase shift keying uses noncoherent form of phase shift keying. noncoherent form avoids the need for a coherent reference signal at the receiver. | |
| 143. | In DPSK system, input signal is differentially encoded and then modulated using a modulator. |
| A. | amplitude |
| B. | frequency |
| C. | bpsk |
| D. | qpsk |
| Answer» C. bpsk | |
| Explanation: in dpsk system, input binary sequence is first differentially encoded and then modulated using a bpsk modulator. the differentially encoded sequence is generated | |
| 144. | QPSK has the bandwidth efficiency of BPSK. |
| A. | twice |
| B. | same |
| C. | half |
| D. | four times |
| Answer» A. twice | |
| Explanation: quadrature phase shift keying (qpsk) has twice the bandwidth of bpsk. it is because two bits are transmitted in a single modulation symbol. the phase of the carrier takes on one of the four equally spaced values, where each value of phase corresponds to a unique pair of message bit. | |
| 145. | QPSK provides twice the bandwidth efficiency and energy efficiency as compared to BPSK. |
| A. | twice |
| B. | half |
| C. | same |
| D. | four times |
| Answer» C. same | |
| Explanation: the bit error probability of qpsk is identical to bpsk but twice as much data can be sent in the same bandwidth. thus, when compared to bpsk, qpsk provides twice the spectral efficiency with exactly the same efficiency. | |
| 146. | What is the full form of OQPSK? |
| A. | optical quadrature phase shift keying |
| B. | orthogonal quadrature pulse shift keying |
| C. | orthogonal quadrature phase shift keying |
| D. | offset quadrature phase shift keying |
| Answer» D. offset quadrature phase shift keying | |
| Explanation: oqpsk stands for offset quadrature phase shift keying. it is a modified form of qpsk which is less susceptible to deleterious effects and supports more efficient amplification. oqpsk is sometimes also called staggered qpsk. | |
| 147. | The bandwidth of OQPSK is to QPSK. |
| A. | identical |
| B. | twice |
| C. | half |
| D. | four times |
| Answer» A. identical | |
| Explanation: the spectrum of an oqpsk signal is identical to that of qpsk signal. hence, both signals occupy the same bandwidth. the staggered alignment of the even and odd bit streams in oqpsk signal does not change the nature of spectrum. | |
| 148. | QPSK signals perform better than OQPSK in the presence of phase jitter. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: oqpsk signal perform better than qpsk in the presence of phase jitter. it is due to the presence of noisy reference signal at the receiver. | |
| 149. | Which of the following is not a detection technique used for detection of π/4 QPSK signals? |
| A. | baseband differential detection |
| B. | if differential detection |
| C. | fm discriminator detection |
| D. | envelope detection |
| Answer» D. envelope detection | |
| Explanation: there are various types of detection techniques used for the detection of π/4 qpsk signals. they include baseband differential detection, if differential detection and fm discriminator detection. | |
| 151. | In an M-ary signalling scheme two or more bits are grouped together to form a | |
| A. | chip | |
| B. | symbol | |
| C. | byte | |
| D. | pattern | |
| Answer» B. symbol | ||
| Explanation: in an m-ary signalling scheme two or more bits are grouped together to form symbols. and one of the m possible signals is transmitted during each symbol period of duration ts. | ||
| 152. | M-ary signalling techniques are not sensitive to timing jitters. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: timing errors increase when smaller distances between signals in the constellation diagram are used. m-ary signalling techniques are attractive for use in bandlimited channel, but are limited in their applications due to sensitivity in timing jitter. | |
| 153. | M-ary modulation schemes have very good power efficiency. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: m-ary modulation schemes have poor power efficiency, but they have a better bandwidth efficiency. an 8-psk system requires a bandwidth that is 3 times smaller than a bpsk system, whereas its ber performance is very worse since signals are packed more closely in the signal constellation. | |
| 154. | What is the radius of the circle in M-ary PSK on which message points are equaly spaced? |
| A. | √es |
| B. | √eb |
| C. | eb |
| D. | es |
| Answer» A. √es | |
| Explanation: the m-ary message points are equally spaced on a circle of radius √es centred at the origin. here es is energy per symbol. thus, mpsk is a constant envelope signal when no pulse shaping is used. | |
| 155. | As the value of M the bandwidth efficiency |
| A. | increases, same. |
| B. | increases, decreases |
| C. | increases, increases |
| D. | decreases, same |
| Answer» C. increases, increases | |
| Explanation: the first null bandwidth of m- ary psk signals decrease as m increases while rb is held constant. therefore, as the value of m increases, the bandwidth efficiency also increases. | |
| 156. | The power efficiency of the M ary PSK decreases because of the |
| A. | freely packed constellation |
| B. | increment of bandwidth efficiency |
| C. | fixed null bandwidth |
| D. | densely packed constellation |
| Answer» D. densely packed constellation | |
| Explanation: bandwidth efficiency increases as the value of m increases. but at the same time, increasing m implies that the constellation is more densely packed. hence the power efficiency or noise tolerance is decreased. | |
| 157. | In QAM, the amplitude is and phase is |
| A. | varied, constant |
| B. | varied, varied |
| C. | constant, varied |
| D. | constant, constant |
| Answer» B. varied, varied | |
| Explanation: quadrature amplitude modulation (qam) is obtained by allowing the amplitude to also vary with the phase. thus, the constellation consists of square lattice of signal points. | |
| 158. | M-ary QAM signal have constant energy per symbol. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: m-ary qam does not have constant energy per symbol. it also does not have constant distance between possible symbol states. it reasons that particular values of m-ary qam signal will be detected with higher probability than others. | |
| 159. | In comparison to M-ary PSK, M-ary QAM bandwidth efficiency is and power efficiency is |
| A. | identical, superior |
| B. | less, superior |
| C. | identical, identical |
| D. | superior, superior |
| Answer» A. identical, superior | |
| Explanation: the power spectrum and bandwidth efficiency of qam modulator is identical to m-ary psk modulation. but, in terms of power efficiency qam is superior to m-ary psk. | |
| 160. | The bandwidth efficiency of an M-ary FSK signal with in M. |
| A. | constant, increase |
| B. | increases, increase |
| C. | decreases, increase |
| D. | decreases, decrease |
| Answer» C. decreases, increase | |
| Explanation: the bandwidth efficiency of an m-ary fsk signal decreases with increase in | |
| 161. | Power efficiency of M-ary FSK increases, since |
| A. | constellation is densely packed |
| B. | m signals are non-orthogonal |
| C. | fixed null bandwidth |
| D. | m-signals are orthogonal |
| Answer» D. m-signals are orthogonal | |
| Explanation: in m-ary fsk, all the m signals are orthogonal and there is no crowding in the signal space. hence, power efficiency of m- ary fsk increases with m. | |
| 162. | The method in which the tail of one pulse smears into adjacent symbol interval is called as |
| A. | intersymbol interference |
| B. | interbit interference |
| C. | interchannel interference |
| D. | none of the mentioned |
| Answer» A. intersymbol interference | |
| Explanation: due to the effect of system filtering the received pulse can overlap on one and another. the tail of one pulse smears | |
| 163. | If each pulse of the sequence to be detected is in shape, the pulse can be detected without ISI. |
| A. | sine |
| B. | cosine |
| C. | sinc |
| D. | none of the mentioned |
| Answer» C. sinc | |
| Explanation: the sinc shaped pulse is the ideal nyquist pulse. if each pulse in the sequence to be detected is in sinc shape the pulses can be detected without isi. | |
| 164. | What is symbol rate packing? |
| A. | maximum possible symbol transmission rate |
| B. | maximum possible symbol receiving rate |
| C. | maximum bandwidth |
| D. | maximum isi value allowed |
| Answer» A. maximum possible symbol transmission rate | |
| Explanation: a system with bandwidth rs/2 can support a maximum transmission rate of rs without isi. thus for ideal nyquist filtering the maximum possible symbol transmission rate is called as symbol rate packing and it is equal to 2 symbols/s/hz. | |
| 165. | A nyquist pulse is the one which can be represented by shaped pulse multiplied by another time function. |
| A. | sine |
| B. | cosine |
| C. | sinc |
| D. | none of the mentioned |
| Answer» C. sinc | |
| Explanation: a nyquist filter is one whose frequency transfer function can be represented by a rectangular function convolved with any real even symmetric frequency function and a nyquist pulse is one | |
| 166. | Examples of nyquist filters are |
| A. | root raised cosine filter |
| B. | raised cosine filter |
| C. | root raised & raised cosine filter |
| D. | none of the mentioned |
| Answer» C. root raised & raised cosine filter | |
| Explanation: the most popular among the class of nyquist filters are raised cosine and root raised cosine filter. | |
| 167. | The minimum nyquist bandwidth for the rectangular spectrum in raised cosine filter is |
| A. | 2t |
| B. | 1/2t |
| C. | t2 |
| D. | 2/t |
| Answer» B. 1/2t | |
| Explanation: for raised cosine spectrum the minimum nyquist bandwidth is equal to 1/2t. | |
| 168. | Roll off factor is the fraction of |
| A. | excess bandwidth and absolute bandwidth |
| B. | excess bandwidth and minimum nyquist bandwidth |
| C. | absolute bandwidth and minimum nyquist bandwidth |
| D. | none of the mentioned |
| Answer» B. excess bandwidth and minimum nyquist bandwidth | |
| Explanation: the roll off factor is defined by a fraction of excess bandwidth and the | |
| 169. | A pulse shaping filter should satisfy two requirements. They are |
| A. | should be realizable |
| B. | should have proper roll off factor |
| C. | should be realizable & have proper roll off factor |
| D. | none of the mentioned |
| Answer» C. should be realizable & have proper roll off factor | |
| Explanation: a pulse shaping filter should provide the desired roll off and should be realizable, that is the impulse response needs to be truncated to a finite length. | |
| 170. | Examples of double side band signals are |
| A. | ask |
| B. | psk |
| C. | ask & psk |
| D. | none of the mentioned |
| Answer» C. ask & psk | |
| Explanation: ask and psk needs twice the transmission bandwidth of equivalent baseband signals. thus these are called as double side band signals. | |
| 171. | ISI is by increasing channel bandwidth. |
| A. | maximized |
| B. | minimized |
| C. | zero |
| D. | infinite |
| Answer» B. minimized | |
| Explanation: increasing channel bandwidth is one of the method to minimize intersymbol interference. but mobile communication systems use minimal bandwidth, thus other methods to reduce isi are desirable. | |
| 172. | Why is pulse shaping technique used? |
| A. | to increase isi |
| B. | to increase spectral width of modulated signal |
| C. | to reduce isi |
| D. | to reduce power spectral density |
| Answer» C. to reduce isi | |
| Explanation: pulse shaping techniques reduces the intersymbol interference. they are also used to reduce the spectral width of the modulated digital signal. | |
| 173. | Raised cosine filter does not satisfy Nyquist criteria. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: raised cosine filter is the most popular pulse shaping filter used in mobile communication. it belongs to the class of filters that satisfy nyquist criterion. | |
| 174. | As the roll off factor in raised cosine rolloff filter the occupied bandwidth |
| A. | manchester |
| B. | faraday |
| C. | graham bell |
| D. | nyquist |
| Answer» D. nyquist | |
| Explanation: nyquist was the first to solve the problem of isi. he overcome the problem of isi while keeping the transmission bandwidth low. he observed that isi can be completely nullified if at every instant, the response due to all symbols except the current symbol is equal to zero. | |
| 176. | Gaussian filter has zero crossings at adjacent symbol peaks. | |
| A. | true | |
| B. | false | |
| Answer» B. false | ||
| Explanation: nyquist filters have zero crossings at adjacent symbol peaks and a truncated transfer function. gaussian filter does not follow nyquist criterion and has a smooth transfer function with no zero crossings. | ||
| 177. | Which of the following is true for a Gaussian filter? |
| A. | large bandwidth |
| B. | minimum isi |
| C. | high overshoot |
| D. | sharp cut off |
| Answer» D. sharp cut off | |
| Explanation: the gaussian filter has a narrow absolute bandwidth, and has a sharp cut off, low overshoot and pulse area preservation properties. this makes it attractive for use in mobile communication that uses nonlinear rf amplifiers. | |
| 178. | Gaussian pulse shaping filter reduces the spectral occupancy and ISI. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: gaussian pulse shaping does | |
| 179. | Gaussian pulses are used when cost and power efficiency are major factors. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: gaussian pulses are used when cost and power efficiency are major factors. but the bit error rates due to isi are deemed to be lower than what is nominally required. thus, there is a trade-off between desired rf bandwidth and irreducible error due to isi. | |
| 180. | The method in which small amount of controlled ISI is introduced into the data stream rather than trying to eliminate it completely is called as |
| A. | correlative coding |
| B. | duobinary signalling |
| C. | partial response signalling |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the interference at the detector can be cancelled out using these methods in which some controlled amount of isi is introduced into the data stream. | |
| 181. | From digital filter we will get the output pulse as the of the current and the previous pulse. |
| A. | summation |
| B. | difference |
| C. | product |
| D. | ratio |
| Answer» A. summation | |
| Explanation: the digital filter incorporates | |
| 182. | In duobinary signalling method, for M-ary transmission, the number of output obtained is |
| A. | 2m |
| B. | 2m+1 |
| C. | 2m-1 |
| D. | m2 |
| Answer» C. 2m-1 | |
| Explanation: in duobinary coding, the number of output obtained for m-ary transmission is 2m-1. | |
| 183. | The method using which the error propagation in dubinary signalling can be avoided is |
| A. | filtering |
| B. | precoding |
| C. | postcoding |
| D. | none of the mentioned |
| Answer» B. precoding | |
| Explanation: in duobinary signalling method if one error occurs it repeats everywhere through out the next steps. to avoid this precoding method can be used. | |
| 184. | In precoding technique, the binary sequence is with the previous precoded bit. |
| A. | and-ed |
| B. | or-ed |
| C. | exor-ed |
| D. | added |
| Answer» C. exor-ed | |
| Explanation: to avoid error propogation precoding method is used. in this each bit is encoded individually without having any effect due to its prior bit or decisions. | |
| 185. | The duobinary filter, He (f) is called as |
| A. | sine filter |
| B. | cosine filter |
| C. | raised cosine filter |
| D. | none of the mentioned |
| Answer» B. cosine filter | |
| Explanation: the transfer function is 2t cos(πft) which is called as cosine filter. | |
| 186. | The method which has greater bandwidth efficiency is called as |
| A. | duobinary signalling |
| B. | polybinary signalling |
| C. | correlative coding |
| D. | all of the mentioned |
| Answer» B. polybinary signalling | |
| Explanation: if more than three levels are introduced in duobinary signalling technique the bandwidth efficiency increases this method is called as polybinary signalling. | |
| 187. | In polybinary signalling method the present bit of binary sequence is algebraically added with number of previous bits. |
| A. | j |
| B. | 2j |
| C. | j+2 |
| D. | j-2 |
| Answer» D. j-2 | |
| Explanation: in polybinary signalling method the present binary digit of the sequence is formed from the modulo-2 addition of the j-2 preceding digits of the sequence and the present digit. | |
| 188. | The primary advantage of this method is |
| A. | redistribution of spectral density |
| B. | to favor low frequencies |
| C. | redistribution of spectral density & to favor low frequencies |
| D. | none of the mentioned |
| Answer» C. redistribution of spectral density & to favor low frequencies | |
| Explanation: each bit can be independently detected in-spite of strong correlation and this provides redistribution of spectral density and also favors low frequencies. | |
| 189. | Source encoding procedure does |
| A. | sampling |
| B. | quantization |
| C. | compression |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: source encoding includes a sampling of continuous time signals, quantization of continuous valued signals and compression of those sources. | |
| 190. | The range of amplitude difference gives the value of |
| A. | width |
| B. | distortion |
| C. | timing jitter |
| D. | noise margin |
| Answer» B. distortion | |
| Explanation: in the eye pattern, the amplitude difference gives the value of distortion caused by isi. | |
| 191. | As the eye opens, ISI |
| A. | increases |
| B. | decreases |
| C. | remains the same |
| D. | none of the mentioned |
| Answer» B. decreases | |
| Explanation: as the eye closes, isi increases and as the eye opens isi decreases. | |
| 192. | The index value n, in transversal filter can be used as. |
| A. | time offset |
| B. | filter coefficient identifier |
| C. | time offset & filter coefficient identifier |
| D. | none of the mentioned |
| Answer» C. time offset & filter coefficient identifier | |
| Explanation: the index n can be used as both time offset and the filter coefficient identifier, which is the address in the filter. | |
| 193. | The over-determined set of equations can be solved using |
| A. | zero forcing |
| B. | minimum mean square error |
| C. | zero forcing & minimum mean square error |
| D. | none of the mentioned |
| Answer» C. zero forcing & minimum mean square error | |
| Explanation: the matrix x in transversal equalizer if non square with dimensions 4n+1 and 2n+1. such equations are called as over-determined set. this can be solved by two methods called as zero forcing method and minimum mean square error method. | |
| 194. | If the filter’s tap weight remains fixed during transmission of data, then the equalization is called as |
| A. | preset equalization |
| B. | adaptive equalization |
| C. | fixed equalization |
| D. | none of the mentioned |
| Answer» A. preset equalization | |
| Explanation: if the weight remains fixed during transmission of data then the equalization is called as preset equalization. it is a simple method which consists of setting the tap weight according to some average knowledge of the channel. | |
| 195. | Equalization method which is done by tracking a slowly time varying channel response is |
| A. | preset equalization |
| B. | adaptive equalization |
| C. | variable equalization |
| D. | none of the mentioned |
| Answer» B. adaptive equalization | |
| Explanation: this method is implemented to perform tap weight adjustment periodically or continually. equalization is done by tracking a slowly varying channel response. | |
| 196. | Preamble is used for |
| A. | detect start of transmission |
| B. | to set automatic gain control |
| C. | to align internal clocks |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the receiver uses preamble for detecting the start of transmission, to set automatic gain control, and to align internal clocks and local oscillator with the received signal. | |
| 197. | The disadvantage of preset equalizer is that |
| A. | it doesnot requires initial training pulse |
| B. | time varying channel degrades the performance of the system |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. time varying channel degrades the performance of the system | |
| Explanation: the disadvantage of preset equalization is that it requires an initial training period that must be invoked at the start of any new transmission. also time varying channel can degrade system performance due to isi, since the tap weights are fixed. | |
| 198. | For AWGN, the noise variance is |
| A. | n0 |
| B. | n0/2 |
| C. | 2n0 |
| D. | n0/4 |
| Answer» B. n0/2 | |
| Explanation: the noise variance out of the correlator for awgn is n0/2. | |
| 199. | A Gaussian distribution into the non linear envelope detector yields |
| A. | rayleigh distribution |
| B. | normal distribution |
| C. | poisson distribution |
| D. | binary distribution |
| Answer» A. rayleigh distribution | |
| Explanation: the two output signals of gaussian distribution yields rayleigh and rician distribution. | |
| 201. | Coherent PSK and non coherent orthogonal FSK have a difference of in PB. | |
| A. | 1db | |
| B. | 3db | |
| C. | 4db | |
| D. | 6db | |
| Answer» C. 4db | ||
| Explanation: the difference of pb is approximately 4db for the best ( coherent psk ) and the worst (non coherent orthogonal fsk). | ||
| 202. | Which is easier to implement and is preferred? |
| A. | coherent system |
| B. | non coherent system |
| C. | coherent & non coherent system |
| D. | none of the mentioned |
| Answer» B. non coherent system | |
| Explanation: a non coherent system is desirable because there may be difficulty is establishing and maintaining a coherent reference. | |
| 203. | Which is the main system consideration? |
| A. | probability of error |
| B. | system complexity |
| C. | random fading channel |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the major system considerations are error probability, complexity and random fading channel. considering all this non coherent system is more desirable than coherent. | |
| 204. | Self information should be |
| A. | positive |
| B. | negative |
| C. | positive & negative |
| D. | none of the mentioned |
| Answer» A. positive | |
| Explanation: self information is always non negative. | |
| 205. | The unit of average mutual information is |
| A. | bits |
| B. | bytes |
| C. | bits per symbol |
| D. | bytes per symbol |
| Answer» A. bits | |
| Explanation: the unit of average mutual information is bits. | |
| 206. | When probability of error during transmission is 0.5, it indicates that |
| A. | channel is very noisy |
| B. | no information is received |
| C. | channel is very noisy & no information is received |
| D. | none of the mentioned |
| Answer» C. channel is very noisy & no information is received | |
| Explanation: when probability of error during transmission is 0.5 then the channel is very noisy and thus no information is received. | |
| 207. | The event with minimum probability has least number of bits. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: in binary huffman coding the event with maximum probability has least number of bits. | |
| 208. | The method of converting a word to stream of bits is called as |
| A. | binary coding |
| B. | source coding |
| C. | bit coding |
| D. | cipher coding |
| Answer» B. source coding | |
| Explanation: source coding is the method of converting a word to stream of bits that is 0’s and 1’s. | |
| 209. | When the base of the logarithm is 2, then the unit of measure of information is |
| A. | bits |
| B. | bytes |
| C. | nats |
| D. | none of the mentioned |
| Answer» A. bits | |
| Explanation: when the base of the logarithm is 2 then the unit of measure of information is bits. | |
| 210. | When X and Y are statistically independent, then I (x,y) is |
| A. | 1 |
| B. | 0 |
| C. | ln 2 |
| D. | cannot be determined |
| Answer» B. 0 | |
| Explanation: when x and y are statistically independent the measure of information i (x,y) is 0. | |
| 211. | The self information of random variable is |
| A. | 0 |
| B. | 1 |
| C. | infinite |
| D. | cannot be determined |
| Answer» C. infinite | |
| Explanation: the self information of a random variable is infinity. | |
| 212. | Entropy of a random variable is |
| A. | 0 |
| B. | 1 |
| C. | infinite |
| D. | cannot be determined |
| Answer» C. infinite | |
| Explanation: entropy of a random variable is also infinity. | |
| 213. | Which is more efficient method? |
| A. | encoding each symbol of a block |
| B. | encoding block of symbols |
| C. | encoding each symbol of a block & encoding block of symbols |
| D. | none of the mentioned |
| Answer» B. encoding block of symbols | |
| Explanation: encoding block of symbols is more efficient than encoding each symbol of a block. | |
| 214. | Lempel-Ziv algorithm is |
| A. | variable to fixed length algorithm |
| B. | fixed to variable length algorithm |
| C. | fixed to fixed length algorithm |
| D. | variable to variable length algorithm |
| Answer» A. variable to fixed length algorithm | |
| Explanation: lempel-ziv algorithm is a variable to fixed length algorithm. | |
| 215. | While recovering signal, which gets attenuated more? |
| A. | low frequency component |
| B. | high frequency component |
| C. | low & high frequency component |
| D. | none of the mentioned |
| Answer» B. high frequency component | |
| Explanation: high frequency components are attenuated more when compared to low frequency components while recovering the signals. | |
| 216. | Mutual information should be |
| A. | positive |
| B. | negative |
| C. | positive & negative |
| D. | none of the mentioned |
| Answer» C. positive & negative | |
| Explanation: mutual information can also be negative. | |
| 217. | ASCII code is a |
| A. | fixed length code |
| B. | variable length code |
| C. | fixed & variable length code |
| D. | none of the mentioned |
| Answer» A. fixed length code | |
| Explanation: ascii code is a fixed length code. it has a fixed length of 7 bits. | |
| 218. | Which reduces the size of the data? |
| A. | source coding |
| B. | channel coding |
| C. | source & channel coding |
| D. | none of the mentioned |
| Answer» A. source coding | |
| Explanation: source coding reduces the size of the data and channel coding increases the size of the data. | |
| 219. | In digital image coding which image must be smaller in size? |
| A. | input image |
| B. | output image |
| C. | input & output image |
| D. | none of the mentioned |
| Answer» B. output image | |
| Explanation: in digital image coding, output image must be smaller than the input image. | |
| 220. | Which coding method uses entropy coding? |
| A. | lossless coding |
| B. | lossy coding |
| C. | lossless & lossy coding |
| D. | none of the mentioned |
| Answer» B. lossy coding | |
| Explanation: lossy source coding uses entropy coding. | |
| 221. | Which achieves greater compression? |
| A. | lossless coding |
| B. | lossy coding |
| C. | lossless & lossy coding |
| D. | none of the mentioned |
| Answer» B. lossy coding | |
| Explanation: lossy coding achieves greater compression where as lossless coding achieves only moderate compression. | |
| 222. | Which are uniquely decodable codes? |
| A. | fixed length codes |
| B. | variable length codes |
| C. | fixed & variable length codes |
| D. | none of the mentioned |
| Answer» A. fixed length codes | |
| Explanation: fixed length codes are uniquely decodable codes where as variable length codes may or may not be uniquely decodable. | |
| 223. | A rate distortion function is a |
| A. | concave function |
| B. | convex function |
| C. | increasing function |
| D. | none of the mentioned |
| Answer» B. convex function | |
| Explanation: a rate distortion function is a monotone decreasing function and also a convex function. | |
| 224. | Which of the following algorithms is the best approach for solving Huffman codes? |
| A. | exhaustive search |
| B. | greedy algorithm |
| C. | brute force algorithm |
| D. | divide and conquer algorithm |
| Answer» B. greedy algorithm | |
| Explanation: greedy algorithm is the best approach for solving the huffman codes problem since it greedily searches for an optimal solution. | |
| 226. | Which bit is reserved as a parity bit in an ASCII set? | |
| A. | first | |
| B. | seventh | |
| C. | eighth | |
| D. | tenth | |
| Answer» C. eighth | ||
| Explanation: in an ascii character set, seven bits are reserved for character representation while the eighth bit is a parity bit. | ||
| 227. | How many bits are needed for standard encoding if the size of the character set is X? |
| A. | log x |
| B. | x+1 |
| C. | 2x |
| D. | x2 |
| Answer» A. log x | |
| Explanation: if the size of the character set is x, then [log x] bits are needed for representation in a standard encoding. | |
| 228. | The code length does not depend on the frequency of occurrence of characters. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: the code length depends on the frequency of occurrence of characters. the more frequent the character occurs, the less is the length of the code. | |
| 229. | From the following given tree, what is the code word for the character ‘a’? |
| A. | 011 |
| B. | 010 |
| C. | 100 |
| D. | 101 |
| Answer» A. 011 | |
| Explanation: by recording the path of the node from root to leaf, the code word for character ‘a’ is found to be 011. | |
| 230. | From the following given tree, what is the computed codeword for ‘c’? |
| A. | 111 |
| B. | 101 |
| C. | 110 |
| D. | 011 |
| Answer» C. 110 | |
| Explanation: by recording the path of the node from root to leaf, assigning left branch as 0 and right branch as 1, the codeword for c is 110. | |
| 231. | What will be the cost of the code if character ci is at depth di and occurs at frequency fi? |
| A. | cifi |
| B. | ∫cifi |
| C. | ∑fidi |
| D. | fidi |
| Answer» C. ∑fidi | |
| Explanation: if character ci is at depth di and occurs at frequency fi, the cost of the codeword obtained is ∑fidi. | |
| 232. | An optimal code will always be present in a full tree. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: an optimal tree will always | |
| 233. | The type of encoding where no character code is the prefix of another character code is called? |
| A. | optimal encoding |
| B. | prefix encoding |
| C. | frequency encoding |
| D. | trie encoding |
| Answer» B. prefix encoding | |
| Explanation: even if the character codes are of different lengths, the encoding where no character code is the prefix of another character code is called prefix encoding. | |
| 234. | What is the running time of the Huffman encoding algorithm? |
| A. | o(c) |
| B. | o(log c) |
| C. | o(c log c) |
| D. | o( n log c) |
| Answer» C. o(c log c) | |
| Explanation: if we maintain the trees in a priority queue, ordered by weight, then the running time is given by o(c log c). | |
| 235. | What is the running time of the Huffman algorithm, if its implementation of the priority queue is done using linked lists? |
| A. | o(c) |
| B. | o(log c) |
| C. | o(c log c) |
| D. | o(c2) |
| Answer» D. o(c2) | |
| Explanation: if the implementation of the priority queue is done using linked lists, the running time of huffman algorithm is o(c2). | |
| 236. | Notch is a |
| A. | high pass filter |
| B. | low pass filter |
| C. | band stop filter |
| D. | band pass filter |
| Answer» C. band stop filter | |
| Explanation: notch filter is a band stop filter that allows most frequencies to pass through it, except frequencies in a specific range. it is just opposite of a band-pass filter. high pass filter allows higher frequencies to pass while low pass filter allows lower frequencies to pass through it. | |
| 237. | Sin wave is |
| A. | aperiodic signal |
| B. | periodic signal |
| C. | random signal |
| D. | deterministic signal |
| Answer» B. periodic signal | |
| Explanation: periodic signal is that which repeats itself after a regular interval. sin wave is a periodic function since it’s value can be determined at any point of time, as it repeats itself at a regular interval. aperiodic signal does not repeat itself at regular interval of time. random signals are the signals which have uncertain values at any time. while deterministic signals are the signals which are constant over a period of time. | |
| 238. | What is the role of channel in communication system? |
| A. | acts as a medium to send message signals from transmitter to receiver |
| B. | converts one form of signal to other |
| C. | allows mixing of signals |
| D. | helps to extract original signal from incoming signal |
| Answer» A. acts as a medium to send message signals from transmitter to receiver | |
| Explanation: channel acts as a medium to transmit message signal from source transmitter to the destination receiver. | |
| 239. | Sum of a periodic and aperiodic signal always be an aperiodic signal. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: periodic signal is a signal which repeats itself after a regular interval. while aperiodic signal does not repeat itself at regular interval of time. | |
| 240. | Noise is added to a signal |
| A. | in the channel |
| B. | at receiving antenna |
| C. | at transmitting antenna |
| D. | during regeneration of information |
| Answer» A. in the channel | |
| Explanation: noise is an unwanted signal that gets mixed with the transmitted signal while passing through the channel. the noise interferes with the signal and provides distortion in received signal. the transmitting antenna transmits modulated message signal while the receiving antenna receives the transmitted signal. regeneration of information refers to demodulating the received signal to produce the original message signal. | |
| 241. | Agreement between communication devices are called |
| A. | transmission medium |
| B. | channel |
| C. | protocol |
| D. | modem |
| Answer» C. protocol | |
| Explanation: protocol is a set of rules that looks after data communication, by acting as an agreement between communication devices. channel is the transmission medium or the path through which information travels. modem is a device that modulates and demodulates data. | |
| 242. | What is the advantage of superheterodyning? |
| A. | high selectivity and sensitivity |
| B. | low bandwidth |
| C. | low adjacent channel rejection |
| D. | low fidelity |
| Answer» A. high selectivity and sensitivity | |
| Explanation: the main advantage of superheterodyning is that it provides high selectivity and sensitivity. it’s bandwidth remains same. it has high adjacent channel rejection and high fidelity. | |
| 243. | Low frequency noise is |
| A. | flicker noise |
| B. | shot noise |
| C. | thermal noise |
| D. | partition noise |
| Answer» A. flicker noise | |
| Explanation: flicker noise is a type of electronic noise which is generated due to fluctuations in the density of carrier. it’s also known as 1/f as it’s power spectral density increases with a decrease in frequency or increase in offset from a signal. | |
| 244. | Relationship between amplitude and frequency is represented by |
| A. | time-domain plot |
| B. | phase-domain plot |
| C. | frequency-domain plot |
| D. | amplitude-domain plot |
| Answer» C. frequency-domain plot | |
| Explanation: relationship between amplitude and frequency is represented by a frequency-domain plot. also, it represents the | |
| 245. | A function f(x) is even, when? |
| A. | f(x) = -f(x) |
| B. | f(x) = f(-x) |
| C. | f(x) = -f(x)f(-x) |
| D. | f(x) = f(x)f(-x) |
| Answer» B. f(x) = f(-x) | |
| Explanation: geometrically a function f(x) is even, if plot of the function is symmetric over y-axis. algebraically, for any function f(x) to be even, f(x) = f(-x). | |
| 246. | The minimum nyquist bandwidth needed for baseband transmission of Rs symbols per second is |
| A. | rs |
| B. | 2rs |
| C. | rs/2 |
| D. | rs2 |
| Answer» C. rs/2 | |
| Explanation: theoretical minimum nyquist bandwidth needed for the baseband transmission of rs symbols per second without isi is rs/2. | |
| 247. | The capacity relationship is given by |
| A. | c = w log2 ( 1+s/n ) |
| B. | c = 2w log2 ( 1+s/n ) |
| C. | c = w log2 ( 1-s/n ) |
| D. | c = w log10 ( 1+s/n ) |
| Answer» A. c = w log2 ( 1+s/n ) | |
| Explanation: the capacity relationship from shannon-hartley capacity theorem is given by c = w log2 ( 1+s/n ). | |
| 248. | Which parameter is called as Shannon limit? |
| A. | pb/n0 |
| B. | eb/n0 |
| C. | ebn0 |
| D. | none of the mentioned |
| Answer» B. eb/n0 | |
| Explanation: there exists a limiting value for eb/n0 below which they can be no error free communication at any information rate. this eb/n0 is called as shannon limit. | |
| 249. | Entropy is the measure of |
| A. | amount of information at the output |
| B. | amount of information that can be transmitted |
| C. | number of error bits from total number of bits |
| D. | none of the mentioned |
| Answer» A. amount of information at the output | |
| Explanation: entropy is defined as the average amount of information per source output. | |
| 251. | For a error free channel, conditional probability should be | |
| A. | zero | |
| B. | one | |
| C. | equal to joint probability | |
| D. | equal to individual probability | |
| Answer» A. zero | ||
| Explanation: for a error free channel | ||
| 252. | Average effective information is obtained by |
| A. | subtracting equivocation from entropy |
| B. | adding equivocation with entropy |
| C. | ratio of number of error bits by total number of bits |
| D. | none of the mentioned |
| Answer» A. subtracting equivocation from entropy | |
| Explanation: according to shannon the average effective information is obtained by subtracting the equivocation from the entropy of the source. | |
| 253. | Turbo codes are |
| A. | forward error correction codes |
| B. | backward error correction codes |
| C. | error detection codes |
| D. | none of the mentioned |
| Answer» A. forward error correction codes | |
| Explanation: turbo codes are a class of high performance forward error correction codes. | |
| 254. | Components used for generation of turbo codes are |
| A. | inter leavers |
| B. | punching pattern |
| C. | inter leavers & punching pattern |
| D. | none of the mentioned |
| Answer» C. inter leavers & punching pattern | |
| Explanation: there are many instances of turbo codes, using different component encoders, input/output ratios, inter leavers, punching patterns. | |
| 255. | Decoders are connected in series. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: two elementary decoders are | |
| 256. | The inter leaver connected between the two decoders is used to |
| A. | remove error bursts |
| B. | scatter error bursts |
| C. | add error bursts |
| D. | none of the mentioned |
| Answer» B. scatter error bursts | |
| Explanation: an inter leaver installed between the two decoders connected in series is used to scatter error bursts. | |
| 257. | In soft decision approach what does -127 mean? |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | very likely one |
| Answer» B. certainly zero | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here -127 represents certainly zero. | |
| 258. | In soft decision approach 100 means? |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | very likely one |
| Answer» D. very likely one | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here 100 represents very likely one. | |
| 259. | In soft decision approach 0 represents |
| A. | certainly one |
| B. | certainly zero |
| C. | very likely zero |
| D. | could be either zero or one |
| Answer» D. could be either zero or one | |
| Explanation: the decoder front end produces an integer for each bit in the data stream. this integer is the measure of how likely it is that the bit 0 or 1 and is called as soft bit. it ranges from -127 to 127. here 0 represents ‘could be either zero or one’. | |
| 260. | In layering, n layers provide service to |
| A. | n layer |
| B. | n-1 layer |
| C. | n+1 layer |
| D. | none of the mentioned |
| Answer» C. n+1 layer | |
| Explanation: in layering n layer provides service to n+1 layer and use the service provided by n-1 layer. | |
| 261. | Which can be used as an intermediate device in between transmitter entity and receiver entity? |
| A. | ip router |
| B. | microwave router |
| C. | telephone switch |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: ip router, microwave router and telephone switch can be used as an intermediate device between communication of two entities. | |
| 262. | Which has comparatively high frequency component? |
| A. | sine wave |
| B. | cosine wave |
| C. | square wave |
| D. | none of the mentioned |
| Answer» C. square wave | |
| Explanation: square wave has comparatively high frequency component in them. | |
| 263. | Which has continuous transmission? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous has continuous transmission where as asynchronous have sporadic transmission. | |
| 264. | Which requires bit transitions? |
| A. | asynchronous |
| B. | synchronous |
| C. | asynchronous & synchronous |
| D. | none of the mentioned |
| Answer» B. synchronous | |
| Explanation: synchronous transmission needs bit transition. | |
| 265. | In synchronous transmission, receiver must stay synchronous for |
| A. | 4 bits |
| B. | 8 bits |
| C. | 9 bits |
| D. | 16 bits |
| Answer» C. 9 bits | |
| Explanation: in synchronous transmission, receiver must stay synchronous for 9 bits. | |
| 266. | How error detection and correction is done? |
| A. | by passing it through equalizer |
| B. | by passing it through filter |
| C. | by amplifying it |
| D. | by adding redundancy bits |
| Answer» D. by adding redundancy bits | |
| Explanation: error can be detected and corrected by adding additional information that is by adding redundancy bits. | |
| 267. | Which is more efficient? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: cyclic redundancy check is more efficient than parity check. | |
| 268. | Which can detect two bit errors? |
| A. | parity check |
| B. | cyclic redundancy check |
| C. | parity & cyclic redundancy check |
| D. | none of the mentioned |
| Answer» B. cyclic redundancy check | |
| Explanation: crc is more powerful and it can detect various kind of errors like 2 bit errors. | |
| 269. | CRC uses |
| A. | multiplication |
| B. | binary division |
| C. | multiplication & binary division |
| D. | none of the mentioned |
| Answer» C. multiplication & binary division | |
| Explanation: crc uses more math like multiplication and binary division. | |
| 270. | Block codes can achieve a larger coding gain than convolution coding. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: convolution code can achieve a larger coding gain that can be achieved using a block coding with the same complexity. | |
| 271. | Which of the following indicates the number of input bits that the current output is dependent upon? |
| A. | constraint length |
| B. | code length |
| C. | search window |
| D. | information rate |
| Answer» A. constraint length | |
| Explanation: constraint length determines the number of input data bits that the current output is dependent upon. the constraint length determines how powerful and complex the code is. | |
| 272. | Which of the following is not a way to represent convolution code? |
| A. | state diagram |
| B. | trellis diagram |
| C. | tree diagram |
| D. | linear matrix |
| Answer» D. linear matrix | |
| Explanation: linear matrix is not a way to represent convolution code. various ways of representing convolution codes are generator matrix, generator polynomial, logic tables, state diagram, tree diagram and trellis diagram. | |
| 273. | Which of the following is not an algorithm for decoding convolution codes? |
| A. | viterbi algorithm |
| B. | stack algorithm |
| C. | fano’s sequential coding |
| D. | ant colony optimization |
| Answer» D. ant colony optimization | |
| Explanation: there are a number of techniques for decoding convolution codes. the most important of these methods is viterbi algorithm. other decoding algorithms for convolutional codes are fano’s sequential coding, stack algorithm and feedback coding. | |
| 274. | Fano’s algorithm searches all the paths of trellis diagram at same time to find the most probable path. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: fano’s algorithm searches for the most probable path through the trellis diagram by examining one path at a time. the error rate performance of fano’s algorithm is comparable to viterbi’s algorithm. | |
| 276. | In comparison to stack algorithm, Fano’s algorithm is simpler. | |
| A. | true | |
| B. | false | |
| Answer» B. false | ||
| Explanation: in comparison to fano’s | ||
| 277. | Which of the following is not an error correction and detection code? |
| A. | block code |
| B. | convolutional codes |
| C. | passive codes |
| D. | turbo codes |
| Answer» C. passive codes | |
| Explanation: there are three basic types of error correction and detection codes. they are block codes, convolutional codes and turbo codes. a channel coder operates on digital message data by encoding the source information into a code sequence. | |
| 278. | Which decoding method involves the evaluation by means of Fano’s algorithm? |
| A. | maximum likelihood decoding |
| B. | sequential decoding |
| C. | maximum a priori |
| D. | minimum mean square |
| Answer» B. sequential decoding | |
| Explanation: fano’s algorithm involves sequential decoding. it searches for the most probable path through the trellis by examining one path at a time. | |
| 279. | In Viterbi’s algorithm, the selected paths are regarded as |
| A. | survivors |
| B. | defenders |
| C. | destroyers |
| D. | carriers |
| Answer» A. survivors | |
| Explanation: in viterbi’s algorithm, the selected paths are regarded as survivors. the path thus defined is unique and corresponds to the decoded output. | |
| 280. | Pseudorandom signal predicted. |
| A. | can be |
| B. | cannot be |
| C. | maybe |
| D. | none of the mentioned |
| Answer» A. can be | |
| Explanation: random signals cannot be predicted whereas pseudorandom sequence can be predicted. | |
| 281. | The properties used for pseudorandom sequence are |
| A. | balance |
| B. | run |
| C. | correlation |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the three basic properties that can be applied for pseudorandom sequence are balance, run and correlation properties. | |
| 282. | The shift register needs to be controlled by clock pulses. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the shift register operation is controlled by clock pulses. | |
| 283. | A linear feedback shift register consists of |
| A. | feedback path |
| B. | modulo 2 adder |
| C. | four stage register |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: a linear feedback shift register | |
| 284. | If the initial pulse of 1000 is fed to shift register, after how many clock pulses does the sequence repeat? |
| A. | 15 |
| B. | 16 |
| C. | 14 |
| D. | 17 |
| Answer» A. 15 | |
| Explanation: if the initial pulse 1000 is given to shift register, the foregoing sequence repeats after 15 clock pulses. | |
| 285. | The sequences produced by shift register depends on |
| A. | number of stages |
| B. | feedback tap connections |
| C. | initial conditions |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the sequences produced by shift register depends on the number of stages, the feedback tap connections and initial conditions. | |
| 286. | For maximal length sequence, the sequence repetition clock pulses p is given by |
| A. | 2n + 1 |
| B. | 2n -1 |
| C. | 2n |
| D. | none of the mentioned |
| Answer» B. 2n -1 | |
| Explanation: for maximal length sequence, produced by n stage linear feedback shift register the sequence repetition clock pulses p is given by 2n -1 . | |
| 287. | For any cyclic shift, the auto-correlation function is equal to |
| A. | 1/p |
| B. | -1/p |
| C. | –p |
| D. | p |
| Answer» B. -1/p | |
| Explanation: for any cyclic shift the auto- correlation function is equal to -1/p. | |
| 288. | Which method is better? |
| A. | to share same bandwidth |
| B. | to share different bandwidth |
| C. | to share same & different bandwidth |
| D. | none of the mentioned |
| Answer» B. to share different bandwidth | |
| Explanation: if the jammer noise shares the same bandwidth, the result could be destructive. | |
| 289. | Pulse jammer consists of |
| A. | pulse modulated excess band noise |
| B. | pulse modulated band-limited noise |
| C. | pulse width modulated excess band noise |
| D. | pulse width modulated band-limited noise |
| Answer» B. pulse modulated band-limited noise | |
| Explanation: pulse jammer consists of pulse modulated band-limited noise. | |
| 290. | Which are the design options for anti jam communicator? |
| A. | time diversity |
| B. | frequency diversity |
| C. | special discrimination |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| Explanation: the design options for anti-jam communicator are time diversity, frequency diversity and special discrimination. | |
| 291. | The ratio (J/S)reqd gives the measure of |
| A. | vulnerability to interference |
| B. | invulnerability to interference |
| C. | all of the mentioned |
| D. | none of the mentioned |
| Answer» B. invulnerability to interference | |
| Explanation: the ratio (j/s)reqd gives the measure of how invulnerable the system is to interference. | |
| 292. | The system should have |
| A. | larger (j/s)reqd |
| B. | greater system’s noise rejection capability |
| C. | larger (j/s)reqd & greater system’s noise rejection capability |
| D. | none of the mentioned |
| Answer» C. larger (j/s)reqd & greater system’s noise rejection capability | |
| Explanation: the system will be efficient if it has greater (j/s)reqd and larger system’s noise rejection capability. | |
| 293. | The broadband jammer jams the entire |
| A. | w |
| B. | wss |
| C. | w & wss |
| D. | none of the mentioned |
| Answer» B. wss | |
| Explanation: the broadband jammer or wide-band jammer is the one which jams the entire wss with its fixed power. | |
| 294. | To increase error probability, the processing gain should be |
| A. | increased |
| B. | decreased |
| C. | exponentially increased |
| D. | exponentially decreased |
| Answer» A. increased | |
| Explanation: in a system, to increase the error probability the processing gain should be increased. | |
| 295. | Which jamming method produces greater degradation? |
| A. | broadband jamming |
| B. | partial jamming |
| C. | broadband & partial jamming |
| D. | none of the mentioned |
| Answer» B. partial jamming | |
| Explanation: greater degradation is possible more with partial jamming than broadband jamming. | |
| 296. | The jammer which monitors a communicator’s signal is known as |
| A. | frequency follower jammers |
| B. | repeat back jammers |
| C. | frequency follower & repeat back jammers |
| D. | none of the mentioned |
| Answer» C. frequency follower & repeat back jammers | |
| Explanation: the smart jammers which monitor a communicator’s signals is known as frequency follower or repeat back jammers. | |
| 297. | DS/BPSK includes |
| A. | despreading |
| B. | demodulation |
| C. | despreading & demodulation |
| D. | none of the mentioned |
| Answer» C. despreading & demodulation | |
| Explanation: ds/bpsk is a two step precess which includes despreading and demodulation. | |
| 298. | In direct sequence process which step is performed first? |
| A. | de-spreading |
| B. | demodulation |
| C. | despreading & demodulation |
| D. | none of the mentioned |
| Answer» A. de-spreading | |
| Explanation: in direct sequence process, de- spreading correlator is followed by a modulator. | |
| 299. | Chip is defined as |
| A. | shortest uninterrupted waveform |
| B. | largest uninterrupted waveform |
| C. | shortest diversion |
| D. | none of the mentioned |
| Answer» A. shortest uninterrupted waveform | |
| Explanation: a chip is defined as the shortest uninterrupted waveform in the system. | |
| 301. | Which modulation scheme is preferred for direct sequence spread spectrum process? | |
| A. | bpsk | |
| B. | qpsk | |
| C. | bpsk & qpsk | |
| D. | none of the mentioned | |
| Answer» C. bpsk & qpsk | ||
| Explanation: both the modulation scheme bpsk and qpsk can be used for direct sequence spread spectrum process. | ||
| 302. | The processing gain is given as modulation scheme. |
| A. | wss/r |
| B. | r/wss |
| C. | wss/2r |
| D. | r/2wss |
| Answer» A. wss/r | |
| Explanation: the processing gain is given by | |
| 303. | The minimum spacing between consecutive hop positions gives the |
| A. | minimum number of chips necessary |
| B. | maximum number of chips necessary |
| C. | chip rate |
| D. | none of the mentioned |
| Answer» A. minimum number of chips necessary | |
| Explanation: the minimum spacing between consecutive hop positions given the minimum number of chips necessary in the frequency word. | |
| 304. | Which system allows larger processing gain? |
| A. | direct sequence |
| B. | frequency hopping |
| C. | direct sequence & frequency hopping |
| D. | none of the mentioned |
| Answer» B. frequency hopping | |
| Explanation: frequency hopping spread spectrum system allows greater processing gain than direct sequence spread spectrum technique. | |
| 305. | In which technique is phase coherence hard to maintain? |
| A. | direct sequence |
| B. | frequency hopping |
| C. | direct sequence & frequency hopping |
| D. | none of the mentioned |
| Answer» B. frequency hopping | |
| Explanation: in frequency hopping spread spectrum phase coherence is hard to maintain from hop to hop. | |
| 306. | Which type of demodulator is used in the frequency hopping technique? |
| A. | coherent |
| B. | non coherent |
| C. | coherent & non coherent |
| D. | none of the mentioned |
| Answer» B. non coherent | |
| Explanation: as it is difficult to maintain phase coherence, non coherent demodulator is used. | |
| 307. | Robustness gives the inability of a signal to withstand the impairments. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: robustness gives the ability of a signal to withstand the impairments such as noise, jamming etc. | |
| 308. | Chips are the |
| A. | repeated symbols |
| B. | non repeated symbols |
| C. | smallest length symbols |
| D. | none of the mentioned |
| Answer» A. repeated symbols | |
| Explanation: the repeated symbols are called as chips. | |
| 309. | Slow frequency hopping is |
| A. | several hops per modulation |
| B. | several modulations per hop |
| C. | several symbols per modulation |
| D. | none of the mentioned |
| Answer» B. several modulations per hop | |
| Explanation: slow frequency hopping is several modulation per frequency hop. | |
| 310. | Fast frequency hopping is |
| A. | several modulations per hop |
| B. | several modulations per symbol |
| C. | several symbols per modulation |
| D. | none of the mentioned |
| Answer» D. none of the mentioned | |
| Explanation: fast frequency hopping is several frequency hops per modulation. | |
| 311. | Which duration is shorter? |
| A. | hop duration |
| B. | symbol duration |
| C. | chip duration |
| D. | none of the mentioned |
| Answer» A. hop duration | |
| Explanation: in frequency hopping technique hop duration is shorter than the symbol duration. | |
| 312. | Frequency division multiple access (FDMA) assigns channels to users. |
| A. | individual, individual |
| B. | many, individual |
| C. | individual, many |
| D. | many, many |
| Answer» A. individual, individual | |
| Explanation: frequency division multiple access (fdma) assigns individual channels to individual users. each user is allocated a unique frequency band or channel. these channels are assigned on demand to users who request service. | |
| 313. | If the FDMA channel is not in use, it can be used by other users. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: if an fdma channel is not in use, it sits idle and cannot be used by other users to increase or share capacity. it is essentially a wasted resource. | |
| 314. | The bandwidth of FDMA channel is |
| A. | wide |
| B. | narrow |
| C. | large |
| D. | zero |
| Answer» B. narrow | |
| Explanation: the bandwidth of fdma channels is relatively narrow as each channel supports only one circuit per carrier. that is, fdma is usually implemented in narrow band systems. | |
| 315. | Which of the following is not true for FDMA systems as compared to TDMA systems? |
| A. | low complexity |
| B. | lower cell site system cost |
| C. | tight rf filtering |
| D. | narrow bandwidth |
| Answer» B. lower cell site system cost | |
| Explanation: fdma systems have higher cell site system costs as compared to tdma systems. it is due to single channel per carrier design, and the need to use costly bandpass filters to eliminate spurious radiation at the base station. | |
| 316. | is undesired RF radiation. |
| A. | intermodulation frequency |
| B. | intermediate frequency |
| C. | instantaneous frequency |
| D. | instrumental frequency |
| Answer» A. intermodulation frequency | |
| Explanation: intermodulation (im) frequency is undesired rf radiation which can interfere with other channels in the fdma systems. the nonlinearities cause signal spreading in the frequency domain and generate im frequency. | |
| 317. | is based on FDMA/FDD. |
| A. | gsm |
| B. | w-cdma |
| C. | cordless telephone |
| D. | amps |
| Answer» D. amps | |
| Explanation: the first us analog cellular system, the advanced mobile phone system (amps) is based on fdma/fdd. a single user occupies a single channel while the call is in progress. | |
| 318. | In US AMPS, 416 channels are allocated to various operators with 10 kHz guard band and channel between them is 30 kHz. What is the spectrum allocation given to each operator? |
| A. | 12.5 khz |
| B. | 30 khz |
| C. | 12.5 mhz |
| D. | 30 mhz |
| Answer» C. 12.5 mhz | |
| Explanation: spectrum allocated to each cellular operator is 12.5 mhz. as bt = nbc + 2bguard; which is equal to 416*30*103+2(10*103) = 12.5 mhz. | |
| 319. | TDMA systems transmit in a continuous way. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: tdma systems transmit data in a buffer and burst method. thus, the transmission for any user is not continuous. | |
| 320. | are utilized to allow synchronization of the receivers between different slots and frames. |
| A. | preamble |
| B. | data |
| C. | guard bits |
| D. | trail bits |
| Answer» C. guard bits | |
| Explanation: guard times are utilized to allow synchronization of the receivers between different slots and frames. | |
| 321. | Which of the following is not true for TDMA? |
| A. | single carrier frequency for single user |
| B. | discontinuous data transmission |
| C. | no requirement of duplexers |
| D. | high transmission rates |
| Answer» A. single carrier frequency for single user | |
| Explanation: tdma share a single carrier frequency with several users, where each user makes use of non-overlapping time slots. the number of time slots per frame depends on several factors, such as modulation technique, available bandwidth etc. | |
| 322. | Because of transmissions in TDMA, the handoff process in |
| A. | continuous, complex |
| B. | continuous, simple |
| C. | discontinuous, complex |
| D. | discontinuous, simple |
| Answer» D. discontinuous, simple | |
| Explanation: because of discontinuous transmissions in tdma, the handoff process is much simpler for a subscriber unit, since it is able to listen for other base stations during idle time slots. | |
| 323. | synchronization overhead is required in TDMA due to transmission. |
| A. | high, burst |
| B. | high, continuous |
| C. | low, burst |
| D. | no, burst |
| Answer» A. high, burst | |
| Explanation: high synchronization overhead is required in tdma systems because of burst transmissions. tdma transmissions are slotted, and this requires the receivers to be synchronized for each data burst. | |
| 324. | TDMA allocates a single time per frame to different users. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: tdma has an advantage that it can allocate different numbers of time slots per frame to different users. thus, bandwidth can be supplied on demand to different users by concatenating or reassigning time slots based on priority. | |
| 326. | What is the time duration of a bit if data is transmitted at 270.833 kbps in the channel? |
| A. | 270.833 s |
| B. | 3 μs |
| C. | 3.692 μs |
| D. | 3.692 s |
| Answer» C. 3.692 μs | |
| Explanation: if data is transmitted at 270.833 kbps in the channel, the time duration of a bit will be 3.692 μs, as tb = (1/270.833 kbps) = 3.692 μs. | |
| 327. | US digital cellular system based on CDMA was standardized as |
| A. | is-54 |
| B. | is-136 |
| C. | is-95 |
| D. | is-76 |
| Answer» C. is-95 | |
| Explanation: a us digital cellular system based on cdma was standardized as interim standard 95 (is-95). it was standardized by us telecommunication industry association (tia) and promised increased capacity. | |
| 328. | IS-95 was not compatible with existing AMPS frequency band. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: like is-136, is-95 system was designed to be compatible with the existing us analog cellular system (amps) frequency band. hence, mobile and base stations can be economically produced for dual mode operation. | |
| 329. | Which of the following is used by IS-95? |
| A. | dsss |
| B. | fhss |
| C. | thss |
| D. | hybrid |
| Answer» A. dsss | |
| Explanation: is-95 uses a direct sequence spread spectrum cdma system. it allows each user within a cell to use the same radio channel, and users in adjacent cell also use the same radio channel. | |
| 330. | IS-95 uses same modulation technique for forward and reverse channel. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: is-95 uses different modulation and spreading technique for the forward and reverse links. on the forward link, the base station simultaneously transmits the user data for all mobiles in the cell by using different spreading sequence for each mobile. | |
| 331. | IS-95 is specified for reverse link operation in band. |
| A. | 869-894 mhz |
| B. | 849-894 mhz |
| C. | 849-869 mhz |
| D. | 824-849 mhz |
| Answer» D. 824-849 mhz | |
| Explanation: is-95 is specified for reverse link operation in the 824-849 mhz band and 869-894 mhz for the forward link. the pcs version of is-95 has also been designed for international use in the 1800-2000 mhz bands. | |
| 332. | User data in IS-95 is spread to a channel chip rate of |
| A. | 1.2288 mchip/s |
| B. | 9.6 mchip/s |
| C. | 12.288 mchip/s |
| D. | 0.96 mchip/s |
| Answer» A. 1.2288 mchip/s | |
| Explanation: user data is spread to a channel chip rate of 1.2288 mchip/s (a total spreading | |
| 333. | are used to resolve and combine multipath components. |
| A. | equalizer |
| B. | registers |
| C. | rake receiver |
| D. | frequency divider |
| Answer» C. rake receiver | |
| Explanation: at both the base station and the subscriber, rake receivers are used to resolve and combine multipath components, thereby reducing the degree of fading. a rake receiver exploits the multipath time delays in a channel and combines the delayed replicas of transmitted signal. | |
| 334. | CT2 was the first generation of cordless telephones. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: ct2 was the second generation of cordless telephones introduced in great britain in 1989. it is used to provide telepoint services which allow a subscriber to use ct2 handsets at a public telepoint. | |
| 335. | CT2 is analog version of first generation cordless telephones. |
| A. | true |
| B. | false |
| Answer» B. false | |
| Explanation: ct2 is a digital version of the first generation, analog, cordless telephones. when compared with analog cordless phones, ct2 offers good speech quality and is more resistant to interference. | |
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