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MCQOPTIONS
This section includes 1362 Mcqs, each offering curated multiple-choice questions to sharpen your Electronics & Communication Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 351. |
When the video signal is applied to the picture tube in a TV receiver, the strength of video signal is about |
| A. | 1000 V |
| B. | 200 V |
| C. | 50 V |
| D. | 1 V |
| Answer» D. 1 V | |
| 352. |
ISDN stands for |
| A. | Integrated Services Digital Network |
| B. | Inter Services Digital Network |
| C. | Inter Subscriber Digital Network |
| D. | integrated Subscriber Digital Network |
| Answer» B. Inter Services Digital Network | |
| 353. |
In communication systems the frequency range used for voice is |
| A. | 15 Hz to 150 Hz |
| B. | 30 Hz to 1000 Hz |
| C. | 100 Hz to 1500 Hz |
| D. | 300 Hz to 3800 Hz |
| Answer» E. | |
| 354. |
Baudot code is a six bit code. |
| A. | 1 |
| B. | |
| C. | by a free running multivibrator |
| D. | by integrating the signal |
| Answer» B. | |
| 355. |
If N is the number of binary digits needed to assign individual binary code designations to the M quantization levels, the signal to quantization noise ratio S0/Nq in PCM is |
| A. | 2N |
| B. | 22N |
| C. | |
| Answer» C. | |
| 356. |
For attenuation of low frequencies we should use |
| A. | shunt capacitance |
| B. | series capacitance |
| C. | shunt inductance |
| D. | series inductance |
| Answer» C. shunt inductance | |
| 357. |
Which of the following pulse modulation is analog? |
| A. | PCM |
| B. | Differential PCM |
| C. | PWM |
| D. | Delta |
| Answer» D. Delta | |
| 358. |
If transmission bandwidth is doubled in FM, SNR is |
| A. | decreased four times |
| B. | halved |
| C. | maximum directive gain |
| D. | zero directive gain |
| Answer» D. zero directive gain | |
| 359. |
Huffman code is also known as redundancy code |
| A. | 1 |
| B. | |
| C. | in some cases |
| D. | depend upon efficiency |
| Answer» B. | |
| 360. |
Assertion (A): SSB and ISB receivers have good S/N ratioReason (R): SSB and ISB receivers have good suppression of adjacent channels. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R are correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» C. A is correct but R is wrong | |
| 361. |
In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor. |
| A. | 43624 |
| B. | 12 |
| C. | 16 |
| D. | 8 |
| Answer» D. 8 | |
| 362. |
The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is |
| A. | parabolic |
| B. | Gaussian |
| C. | BCH and Hamming code are special cases of Reed-Solomon code |
| D. | Reed-Solomon and Hamming code are special cases of BCH code |
| Answer» E. | |
| 363. |
CCIR system B employs |
| A. | 625 lines |
| B. | 625 lines with 2 : 1 interlace |
| C. | 625 lines with 4 : 1 interlace |
| D. | 500 lines with 2 : 1 interlace |
| Answer» C. 625 lines with 4 : 1 interlace | |
| 364. |
Audio amplifiers need |
| A. | Bass control |
| B. | Treble control |
| C. | Both Bass and Treble control |
| D. | Either Bass or Treble control |
| Answer» D. Either Bass or Treble control | |
| 365. |
A signal m(t) = 5 cos 2p 100 t frequency modulates a carrier. The resulting FM signal is 10 cos {(2p 105 t) + 15 sin (2p 100 t)} The approximate bandwidth of the FM would be |
| A. | 3.2 kHz |
| B. | 100 kHz |
| C. | 1, 2, 3, 4 |
| D. | 1, 2, 4 |
| Answer» E. | |
| 366. |
If F(ω) is Fourier transform of f(t), then the Fourier transform of f(t) cos ωct is |
| A. | F(ω + ωc) + F(ω - ωc) |
| B. | [F(ω + ωc) + F(ω - ωc)] |
| C. | F(ω + ωc) - F(ω - ωc) |
| D. | [F(ω + ωc) - F(ω - ωc)] |
| Answer» C. F(œâ + œâc) - F(œâ - œâc) | |
| 367. |
A single-sideband (SSB) signal contains 3 kW. The power content of the carrier is __________ kW. |
| A. | 3 |
| B. | 0 |
| C. | 1 |
| D. | |
| Answer» D. | |
| 368. |
A sinusoidal voltage of 2 kV peak value is amplitude modulated to give 20% modulation. The peak value of each side band term is |
| A. | 400 V |
| B. | 200 V |
| C. | 100 V |
| D. | 800 V |
| Answer» C. 100 V | |
| 369. |
The main advantage of TDM over FDM is |
| A. | better SNR |
| B. | required filter |
| C. | reduce the signal frequencies |
| D. | increase the signal frequencies |
| Answer» D. increase the signal frequencies | |
| 370. |
The subcarrier frequency for transmission of colour difference signals in TV systems in India is |
| A. | 2.06 MHZ |
| B. | 1.3 MHz |
| C. | good front to back ratio |
| D. | broad bandwidth |
| Answer» E. | |
| 371. |
A basic group B |
| A. | consists of five super groups |
| B. | has a frequency range of 60 to 108 kHz |
| C. | consists of erect channels only |
| D. | is formed at group translating equipment |
| Answer» C. consists of erect channels only | |
| 372. |
The pre-emphasis circuit is used |
| A. | prior to modulation |
| B. | after demodulation |
| C. | for low frequency components of the signal |
| D. | none of the above |
| Answer» B. after demodulation | |
| 373. |
In an oscillator crystal, the parallel resonant frequency is |
| A. | equal to |
| B. | higher than |
| C. | less than |
| D. | series resonant frequency |
| Answer» B. higher than | |
| 374. |
A carrier signal at frequency 100 MHz is frequency modulated with modulation index 2 by a signal at 2 kHz. At what frequencies are the sidebands produced? |
| A. | 102 MHz and 98 Mhz |
| B. | 100 MHz + 2 kHz, 100 MHz - 2 kHz |
| C. | 100 MHz ± Δfd where fd is frequency deviation |
| D. | 100 MHz ± 3n kHz, where n is 1, 2, ... S integer |
| Answer» E. | |
| 375. |
If sampling is done at the rate of 10 kHz. The bandwidth required is |
| A. | 35 kHz |
| B. | 70 kHz |
| C. | 10 kHz |
| D. | 1280 kHz |
| Answer» B. 70 kHz | |
| 376. |
For a good public address system the sound intensity at the farthest point should be at least |
| A. | 20 dB over threshold of hearing |
| B. | 40 dB over threshold of hearing |
| C. | 80 dB over threshold of hearing |
| D. | 120 dB over threshold of hearing |
| Answer» D. 120 dB over threshold of hearing | |
| 377. |
The radar cross-section of a target depends an |
| A. | frequency |
| B. | polarization of incident wave |
| C. | aspect of target |
| D. | all of the above |
| Answer» E. | |
| 378. |
Which one of the following cannot be used to remove the unwanted sideband in SSB? |
| A. | Balanced modulator |
| B. | Phase shift method |
| C. | Filter system |
| D. | Third method |
| Answer» B. Phase shift method | |
| 379. |
The efficiency of using a binary system for selection of 1 out of 13 equally probable events is |
| A. | 1 |
| B. | about 92% |
| C. | about 83% |
| D. | about 71% |
| Answer» B. about 92% | |
| 380. |
An amplitude modulated wave is e = 50 (1 + 0.4 cos 1000 t + 0.3 cos 10000 t) cos 107 t |
| A. | 0.5 |
| B. | 0.3 |
| C. | 0.7 |
| D. | 0.4 |
| Answer» B. 0.3 | |
| 381. |
Which of the following microphones is used as a standard for calibration? |
| A. | Moving coil |
| B. | Carbon |
| C. | Condenser |
| D. | Ribbon |
| Answer» D. Ribbon | |
| 382. |
In high power AM transmission modulation is done at |
| A. | IF stage |
| B. | RF power |
| C. | buffer stage |
| D. | oscillator stage |
| Answer» C. buffer stage | |
| 383. |
A typical value of insertion loss of SAW filter is |
| A. | 15 dB |
| B. | 2 dB |
| C. | 40 dB |
| D. | 60 dB |
| Answer» B. 2 dB | |
| 384. |
Consider the following statements as regards pulse repetition frequency Pulse interval should be as large as possible so that a pulse returns before next pulse is transmittedGreater the number of pulses reflected from a target, greater is the probability of distinguishing it from the noise Which of the above are correct? |
| A. | 1 only |
| B. | 2 only |
| C. | both 1 and 2 |
| D. | neither 1 nor 2 |
| Answer» D. neither 1 nor 2 | |
| 385. |
The atmospheric absorption of radio waves depends on |
| A. | frequency of wave |
| B. | the distance from transmitter |
| C. | wave polarisation |
| D. | atmospheric polarisation |
| Answer» B. the distance from transmitter | |
| 386. |
Frequency frogging is used in a carrier system to |
| A. | conserve frequencies |
| B. | reduce distortion |
| C. | reduce cross talk |
| D. | none of the above |
| Answer» D. none of the above | |
| 387. |
FM signal can be generated by |
| A. | diode reactance modulator |
| B. | saturable reactor modulator |
| C. | reactance tube modulator |
| D. | none of the above |
| Answer» E. | |
| 388. |
A sinusoidal voltage amplitude modulates a carrier of peak value 5 kV. Each side band has an amplitude of 500 V. The modulation index is |
| A. | 0.1 |
| B. | 0.2 |
| C. | 0.4 |
| D. | 0.8 |
| Answer» C. 0.4 | |
| 389. |
If e = 10 cos (2p x 108t + 20 cos 2p 103t). The bandwidth is |
| A. | 108 Hz |
| B. | 42 kHz |
| C. | 21 kHz |
| D. | 108 ± 103 Hz |
| Answer» C. 21 kHz | |
| 390. |
Consider the following statements about loop antenna It is suitable for direction finding applicationIt is a single turn or multiturn coil carrying RF currentThe loop is surrounded by a magnetic field everywhere perpendicular to the loop Which of the above statements are correct? |
| A. | 1 and 2 only |
| B. | 1 only |
| C. | 1, 2, 3 |
| D. | 2 and 3 only |
| Answer» D. 2 and 3 only | |
| 391. |
The purpose of source coding is to |
| A. | increase the information transmission rate |
| B. | decrease ratio |
| C. | decrease the information transmission rate |
| D. | decrease probability of error |
| Answer» C. decrease the information transmission rate | |
| 392. |
If k denotes Boltzmann's constant, the rms value of thermal noise voltage is proportional to |
| A. | k2 |
| B. | k |
| C. | k |
| D. | (k)1/3 |
| Answer» D. (k)1/3 | |
| 393. |
Video signals are transmitted through |
| A. | either amplitude or frequency modulation |
| B. | neither amplitude nor frequency modulation |
| C. | 3 W |
| D. | 30 W |
| Answer» B. neither amplitude nor frequency modulation | |
| 394. |
If e = 10 sin(108 t + 3 sin 104 t) then the deviation is |
| A. | 477 kHz |
| B. | 4770 kHz |
| C. | symmetrical about the carrier |
| D. | not necessarily symmetrical about the carrier |
| Answer» C. symmetrical about the carrier | |
| 395. |
Multiplex techniques are used for |
| A. | low density communications |
| B. | low density point-to-point communications |
| C. | high-density point-to-point communications |
| D. | all of the above |
| Answer» D. all of the above | |
| 396. |
In the television system in India, each channel is allowed a bandwidth of |
| A. | 20 MHz |
| B. | 12 MHz |
| C. | 7 MHz |
| D. | 3 MHz |
| Answer» D. 3 MHz | |
| 397. |
A zero mean white Gaussian noise is passed through an ideal low pass filter of bandwidth 10 kHz. The output of the samples so obtained would be |
| A. | correlated |
| B. | statistically independent |
| C. | uncorrelated |
| D. | orthogonal |
| Answer» C. uncorrelated | |
| 398. |
Assertion (A): In TV scanning is done from left to right and top to bottom. Reason (R): Interlaced scanning is useful in eliminating flicker |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R are correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» C. A is correct but R is wrong | |
| 399. |
For a FM wave carrier modulating frequency is 10 kHz and bandwidth is 2 MHz. If the modulating signal amplitude is doubled, the bandwidth will be |
| A. | 0.5 MHz |
| B. | 1 MHz |
| C. | 2 MHz |
| D. | 4 MHz |
| Answer» E. | |
| 400. |
If N bits are lumped together to get an N-bit symbol, the possible number of symbols is |
| A. | 2N |
| B. | 2N - 1 |
| C. | 2N - 2 |
| D. | 2N - 4 |
| Answer» B. 2N - 1 | |