MCQOPTIONS
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MCQOPTIONS
This section includes 125 Mcqs, each offering curated multiple-choice questions to sharpen your Analog Communications knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
A signal m(t) = 10 cos (2π 100t) is frequency-modulated. The resulting FM signal is x(t) = 20 cos {2π 106 t + 15 sin (2π 100 t)}The FM bandwidth is nearly |
| A. | 3.2 kHz |
| B. | 9.6 kHz |
| C. | 32 kHz |
| D. | 100 kHz |
| Answer» B. 9.6 kHz | |
| 52. |
A telephone signal with a cut-off frequency of 4 KHz is digitalised into 8-bit samples at Nyquist sampling rate of s = 2 W. Calculate the baseband transmission bandwidth. |
| A. | 8 KHz |
| B. | 64 KHz |
| C. | 16 KHz |
| D. | 2 KHz |
| Answer» C. 16 KHz | |
| 53. |
An FM signal can be detected by using: |
| A. | an average detector |
| B. | A discriminator |
| C. | A PLL |
| D. | An LPF |
| Answer» C. A PLL | |
| 54. |
A PAM signal can be detected by using |
| A. | An ADC |
| B. | An integrator |
| C. | A band pass filter |
| D. | A high pass filter |
| Answer» C. A band pass filter | |
| 55. |
A telephone signal band-limited to 4 kHz is to be transmitted by PCM. If the signal to quantization noise is to be at least 40 dB, the number of levels into which the signal is to be encoded will be |
| A. | 32 |
| B. | 64 |
| C. | 81 |
| D. | 128 |
| Answer» E. | |
| 56. |
In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during the time interval \(t_1 \le t\le t_2\). This indicates that during this interval. |
| A. | The input to the modulator is essentially constant |
| B. | the modulator is going through slope overload |
| C. | the accumulator is in saturation |
| D. | the speech signal is being sampled at the Nyquist rate. |
| Answer» B. the modulator is going through slope overload | |
| 57. |
Fidelity in a communication receiver is provided by |
| A. | Mixer stage |
| B. | Detector stage |
| C. | Various amplifier section |
| D. | Audio stage |
| Answer» E. | |
| 58. |
In the PCM system, if the quantization levels are increased from 2 to 8, the bandwidth requirement will |
| A. | remain same |
| B. | be doubled |
| C. | be tripled |
| D. | become four times |
| Answer» D. become four times | |
| 59. |
As compared to the A-law compander, the μ-law compander produces |
| A. | More companding at low amplitudes |
| B. | More companding at high amplitudes |
| C. | Less companding for low amplitudes |
| D. | Less companding for high amplitudes |
| Answer» B. More companding at high amplitudes | |
| 60. |
Bandwidth efficiency is expressed in units of |
| A. | Hs/bits/sec |
| B. | Hz/watts |
| C. | Bits/sec/Hz |
| D. | It has no dimension |
| Answer» D. It has no dimension | |
| 61. |
An analog voltage in the range of 0-8V is divided in eight equal intervals for conversion to 3-bit digital output. The maximum quantization error is |
| A. | 0 V |
| B. | 0.5 V |
| C. | 1.0 V |
| D. | 2.0 V |
| Answer» D. 2.0 V | |
| 62. |
In phase modulation, the frequency deviation is |
| A. | Independent of the modulating signal frequency |
| B. | Increasingly proportional to the modulating signal frequency |
| C. | Directly proportional to the modulating signal frequency |
| D. | Inversely proportional to the square root of the modulating signal frequency |
| Answer» D. Inversely proportional to the square root of the modulating signal frequency | |
| 63. |
Directions: The item consists of two statements, one labelled as the ‘Assertion (A)’ and the other as ‘Reason (R)’.You are to examine these two statements carefully and select the answers to the item using the codes given below:Assertion (A): Frequency modulation and phase modulation both produce different set of frequency bands for the same modulation depth.Reason (R): Frequency modulation and phase modulation vary the carrier phase angle or its rate. |
| A. | Both A and R individually true and R is the correct explanation of A |
| B. | Both A and R are individually true but R is not the correct explanation of A |
| C. | A is true but R is false |
| D. | A is false but R is true |
| Answer» E. | |
| 64. |
In an angle modulation system, if the carrier amplitude is AC, the power in the modulated signal is: |
| A. | 0.5 × AC2 |
| B. | 0.5 × AC4 |
| C. | 1.5 × AC2 |
| D. | 2.5 × AC2 |
| Answer» B. 0.5 × AC4 | |
| 65. |
In phase modulated signal, the frequency deviation is proportional to |
| A. | Frequency only |
| B. | Amplitude only |
| C. | Both (1) and (2) |
| D. | none of the above |
| Answer» D. none of the above | |
| 66. |
An FM signal is represented by v = 12 sin (6 × 108 t + 5 sin 1250 t). The carrier frequency fc and frequency deviation δ, respectively, are |
| A. | 191 MHz and 665 Hz |
| B. | 95.5 MHz and 995 Hz |
| C. | 191 MHz and 995 Hz |
| D. | 95.5 MHz and 665 Hz |
| Answer» C. 191 MHz and 995 Hz | |
| 67. |
In a modulation system, the modulating voltage remains the same, the modulation index is halved when the modulating frequency is doubled, the system is |
| A. | AM |
| B. | FM |
| C. | PM |
| D. | All of the above |
| Answer» C. PM | |
| 68. |
A 20 MHz carrier is modulated by a sinusoid of 1 kHz to produce a peak frequency deviation of 100 kHz. What is the approximate bandwidth? If the modulating signal is changed to 800 kHz, what is the bandwidth? |
| A. | 2 kHz, 1600 kHz |
| B. | 202 kHz, 1600 kHz |
| C. | 202 kHz, 1800 kHz |
| D. | 2 kHz, 1800 kHz |
| Answer» D. 2 kHz, 1800 kHz | |
| 69. |
For a sine wave with a frequency of 1000 Hz. the time period is _____ |
| A. | 1000 ms |
| B. | 100 ms |
| C. | 10 ms |
| D. | 1 ms |
| Answer» E. | |
| 70. |
A TDM link has 20 signal channels and each channel is sampled at 8 kHz. Each sample is represented by 7 bits and contains an additional bit for synchronization. The total bit rate for the TDM link is |
| A. | 1128 kbps |
| B. | 1180 kbps |
| C. | 1280 kbps |
| D. | 128 Mbps |
| Answer» D. 128 Mbps | |
| 71. |
In PCM, if the number of quantization levels is increased from 4 to 64, then the bandwidth requirement will approximately be: |
| A. | 4 times |
| B. | 16 times |
| C. | 8 times |
| D. | 3 times |
| Answer» E. | |
| 72. |
Consider the following:1. Pulse-position modulation.2. Pulse-code modulation.3. Pulse-width modulation.Which of the above communications are not digital? |
| A. | 1 and 2 only |
| B. | 2 and 3 only |
| C. | 1 and 3 only |
| D. | 1, 2 and 3 |
| Answer» D. 1, 2 and 3 | |
| 73. |
In frequency modulation |
| A. | Frequency of the carrier remains constant |
| B. | Carrier frequency varies in accordance with the modulating signal amplitude |
| C. | Carrier frequency varies in accordance with the modulated signal frequency |
| D. | None of the above |
| Answer» C. Carrier frequency varies in accordance with the modulated signal frequency | |
| 74. |
In DPCM, the input to the quantizer is quantized by a |
| A. | uniform scalar quantizer |
| B. | non-uniform scalar quantizer |
| C. | uniform or non-uniform scalar quantizer |
| D. | none of these. |
| Answer» C. uniform or non-uniform scalar quantizer | |
| 75. |
A high PRF would |
| A. | Increase the maximum range |
| B. | Decreases the maximum range |
| C. | Affect range ambiguity |
| D. | Increase the efficiency of radar |
| Answer» D. Increase the efficiency of radar | |
| 76. |
In delta modulation, the slope overload distortion can be reduced by. |
| A. | Decreasing the step size |
| B. | Decreasing the granular noise |
| C. | Decreasing the sampling rate |
| D. | Increasing the step size |
| Answer» E. | |
| 77. |
Digital modulation techniques are used in the satellite communication system, because |
| A. | They are easier to handle |
| B. | Large bandwidth utilization is possible |
| C. | They have a spectral efficiency |
| D. | They are less prone to interference |
| Answer» E. | |
| 78. |
PDM is generated by employing __________ |
| A. | monostable multivibrator |
| B. | Schmitt trigger |
| C. | free running multivibrator |
| D. | JK flip flop |
| Answer» B. Schmitt trigger | |
| 79. |
If transmitter bandwidth is doubled in FM, then SNR is |
| A. | also doubled |
| B. | improved four fold |
| C. | decreased by one fourth |
| D. | unaffected |
| Answer» D. unaffected | |
| 80. |
In PCM system, the encoder |
| A. | assigns quantized values |
| B. | changes quantized value to binary value |
| C. | changes quantized value to numerical value |
| D. | Changes numerical values to binary values |
| Answer» C. changes quantized value to numerical value | |
| 81. |
A carrier wave of frequency 2.5 GHz amplitude is modulated with two modulating frequencies equal to 1 kHz and 2 kHz. The modulated wave will have the total bandwidth: |
| A. | 6 kHz |
| B. | 2 kHz |
| C. | 4 kHz |
| D. | 3 kHz |
| Answer» D. 3 kHz | |
| 82. |
Directions: It consists of two statements, one labelled as the ‘Statement (I)’ and the other as ‘Statement (II). Examine these two statements carefully and select the answer using the codes given below:Statement (I): PAM, PWM and PPM modulation belong to analog modulation and the PCM modulation belongs to the digital modulation.Statement (II): PAM, PWM and PPM modulations are similar to AM, FM and PM modulation. |
| A. | Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) |
| B. | Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) |
| C. | Statement (I) is true but Statement (II) is false |
| D. | Statement (I) is false but Statement (II) is true |
| Answer» C. Statement (I) is true but Statement (II) is false | |
| 83. |
A sinusoidal signal of 2 kHz frequency is applied to a delta modulator. The sampling rate and step-size Δ of the delta modulator are 20,000 samples per second and 0.1V respectively. To prevent slope overload, the maximum amplitude of the sinusoidal signal (in Volts) is |
| A. | \(\frac{1}{{2{\rm{\pi }}}}\) |
| B. | \(\frac{1}{{\rm{\pi }}}\) |
| C. | \(\frac{2}{{\rm{\pi }}}\) |
| D. | \({\rm{\pi }}\) |
| Answer» B. \(\frac{1}{{\rm{\pi }}}\) | |
| 84. |
In the generation of modulated signal, a varactor diode can be used for |
| A. | FM generation only |
| B. | AM generation only |
| C. | PM generation only |
| D. | both (B) and (C) |
| Answer» B. AM generation only | |
| 85. |
Consider an FM signal f(t) = cos [2πfct + β1 sin 2πf1t + β2 sin 2πf2t]. The maximum deviation of the instantaneous frequency from the carrier frequency fc is |
| A. | β1f1 + β2f2 |
| B. | β1f2 + β2f1 |
| C. | β1 + β2 |
| D. | f1 + f2 |
| Answer» B. β1f2 + β2f1 | |
| 86. |
An FM detector produces a peak - to - peak output voltage of 1.2 V from an FM signal that is modulated to 10 kHz deviation by a sine wave.(a) The peak voltage is 0.6 V(b) The detector sensitivity is 6 μV/Hz(c) The peak voltage is 2.4 V(d) The detector sensitivity is 60 μV/Hz Which of the above are correct? |
| A. | (a) and (b) |
| B. | (b) and (c) |
| C. | (c) and (d) |
| D. | (a) and (d) |
| Answer» E. | |
| 87. |
An increase in modulation index leads to an increase in bandwidth in case of |
| A. | AM |
| B. | FM |
| C. | SSB |
| D. | None |
| Answer» C. SSB | |
| 88. |
Assertion (A): Vestigial side band gives rise to frequencies very close to the carrier frequency.Reason (R): It is not possible to go to the extreme and suppress one complete side band by physically realizable filters. |
| A. | Both (A) and (R) are true and (R) is the correct explanation of (A). |
| B. | Both (A) and (R) are true, but (R) is not the correct explanation of (A). |
| C. | (A) is true, but (R) is false. |
| D. | (A) is false, but (R) is true. |
| Answer» B. Both (A) and (R) are true, but (R) is not the correct explanation of (A). | |
| 89. |
Pulse Shaping in line coding is used for |
| A. | Using Optimum Power Consumption |
| B. | Using Optimum Bandwidth |
| C. | Maximize Inter symbol Interference |
| D. | Simplicity of Circuits |
| Answer» C. Maximize Inter symbol Interference | |
| 90. |
A carrier waveform 10 cos ωt and modulating signal 3 cos ωm t have fc = 100 kHz and fm = 4 kHz. Given that sensitivity of FM is 4 kHz/V and FM spectra beyond J6 is negligible, what are the channel bandwidth requirements for AM and FM, respectively? |
| A. | 12 kHz and 48 kHz |
| B. | 8 kHz and 48 kHz |
| C. | 12 kHz and 24 kHz |
| D. | 8 kHz and 24 kHz |
| Answer» C. 12 kHz and 24 kHz | |
| 91. |
For a frequency modulated signal, the modulation index is doubled. The average power of the modulated signal is |
| A. | quadrupled |
| B. | doubled |
| C. | unaltered |
| D. | none of the above |
| Answer» D. none of the above | |
| 92. |
An FM broadcasting radio station transmits signals of frequency 100 MHz with a power of 10 KW. The bandwidth of the modulation signal is from 100 Hz to 1.5 Khz. If the maximum deviation set by the FCC, (δ), is 75 KHz, the range of the modulation index is |
| A. | 100 to 750 |
| B. | 100 to 250 |
| C. | 50 to 750 |
| D. | 50 to 250 |
| Answer» D. 50 to 250 | |
| 93. |
A receiver tunes signal from 550 to 1600 KHz, with an intermediate frequency of 455 KHz. Find the minimum and maximum frequencies generated by the local oscillator. |
| A. | 1005 KHz, 2055 KHz |
| B. | 95 KHz, 1145 KHz |
| C. | 502 KHz, 1028 KHz |
| D. | 455 KHz, 2055 KHz |
| Answer» B. 95 KHz, 1145 KHz | |
| 94. |
For effective modulation, the degree of modulation should |
| A. | small |
| B. | never exceed 100% |
| C. | be large |
| D. | always more than 100% |
| Answer» C. be large | |
| 95. |
A carrier is simultaneously amplitude modulated by two sine waves with individual modulation of 30% and 40%. The overall modulation index is |
| A. | 50% |
| B. | 35% |
| C. | 70% |
| D. | 40% |
| Answer» B. 35% | |
| 96. |
Pulse communication system that is inherently highly immune to noise is |
| A. | PAM |
| B. | PWM |
| C. | PPM |
| D. | PCM |
| Answer» E. | |
| 97. |
Boosting of higher audio frequencies in FM to improve noise immunity is called |
| A. | De-Emphasis |
| B. | Pre-emphasis |
| C. | Compression |
| D. | Expansion |
| Answer» C. Compression | |
| 98. |
In amplitude modulation: |
| A. | carrier frequency is changed |
| B. | carrier amplitude is changed |
| C. | three side bands are produced |
| D. | fidelity is improved |
| Answer» C. three side bands are produced | |
| 99. |
In digital companding, the analog signal is : |
| A. | First converted to a linear PCM Code and then sampled. |
| B. | First sampled and then converted to a linear PCM Code. |
| C. | First sampled and then converted to a non - linear PCM Code. |
| D. | Never sampled. |
| Answer» C. First sampled and then converted to a non - linear PCM Code. | |
| 100. |
Match List 1 with List 2List 1List 2 (a) DPSK (i) Non coherent detector (b) AM (ii) Ratio detector (c) FM (iii) Coherent detector (d) DSB – SC (iv) Rectifier DectectorChoose the correct answer from the options given below:(1) (a) - (i), (b) - (iv), (c) - (ii), (d) - (iii)(2) (a) - (iii), (b) - (i), (c) - (iv), (d) - (ii)(3) (a) - (ii), (b) - (iii), (c) - (i), (d) - (iv)(4) (a) - (iv), (b) - (ii), (c) - (iii), (d) - (i) |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» B. 2 | |