MCQOPTIONS
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MCQOPTIONS
This section includes 299 Mcqs, each offering curated multiple-choice questions to sharpen your Communication knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
To permit the selection of 1 out of 16 equiprobable events, the number of bits required is |
| A. | 2 |
| B. | log |
| C. | 16 |
| D. | 8 |
| E. | 4 |
| Answer» E. 4 | |
| 2. |
In a single tone FM discriminator (S0/N0) is |
| A. | Proportional to deviation |
| B. | Proportional to cube of deviation |
| C. | Inversely proportional to deviation |
| D. | Proportional to square of deviation |
| Answer» E. | |
| 3. |
A superheterodyne receiver with an IF of 450 kHz is tuned to a signal at 1200 kHz. The image frequency is |
| A. | 750 kHz |
| B. | 900 kHz |
| C. | 1650 kHz |
| D. | 2100 kHz |
| Answer» E. | |
| 4. |
In a broadcast superheterodyne receiver, the |
| A. | Local oscillator operates below the signal frequency |
| B. | Mixer input must be tuned to the signal frequency |
| C. | Local oscillator frequency is normally double the IF |
| D. | RF amplifier normally works at 455 kHz above the carrier frequency |
| Answer» C. Local oscillator frequency is normally double the IF | |
| 5. |
The local oscillator of a broadcast receiver is tuned to a frequency higher than the incoming frequency |
| A. | To help the image frequency rejection |
| B. | To permit easier tracking |
| C. | Because otherwise an intermediate frequency could not be produced |
| D. | To allow adequate frequency coverage without switching |
| Answer» E. | |
| 6. |
Three-point tracking is achieved with |
| A. | Variable selectivity |
| B. | The padder capacitor |
| C. | Double spotting |
| D. | Double conversion |
| Answer» C. Double spotting | |
| 7. |
A receiver has poor IF selectivity. It will therefore also have poor |
| A. | Blocking |
| B. | Double-spotting |
| C. | Diversity reception |
| D. | Sensitivity |
| Answer» B. Double-spotting | |
| 8. |
Indicate the false statement. The superheterodyne receiver replaced the TRF receiver because the latter suffered from |
| A. | Gain variation over the frequency coverage range |
| B. | Insufficient gain and sensitivity |
| C. | Inadequate selectivity at high frequencies |
| D. | Instability |
| Answer» C. Inadequate selectivity at high frequencies | |
| 9. |
One of the main functions of the RF amplifier in a superheterodyne receiver is to |
| A. | Provide improved tracking |
| B. | Permit better adjacent-channel rejection |
| C. | Increase the tuning range of the receiver |
| D. | Improve the rejection of the image frequency |
| Answer» E. | |
| 10. |
The image frequency of a superheterodyne receiver |
| A. | Is created within the receiver itself |
| B. | Is due to insufficient adjacent channel rejection |
| C. | Is not rejected by the IF tuned circuits |
| D. | Is independent of the frequency to which the receiver is tuned |
| Answer» D. Is independent of the frequency to which the receiver is tuned | |
| 11. |
The frequency generated by each decade in a direct frequency synthesizer is much higher than the frequency shown; this is done to |
| A. | Reduce the spurious frequency problem |
| B. | Increase the frequency stability of the synthesizer |
| C. | Reduce the number of decades |
| D. | Reduce the number of crystals required |
| Answer» B. Increase the frequency stability of the synthesizer | |
| 12. |
A signal x (t) = 100 cos (2 103) t is ideally sample with sampling period of 50 sec and then passed through an ideal low-pass filter with cut-off frequency of 15 kHz. Which of the following frequencies is/are present at the filter output? |
| A. | 12 kHz only |
| B. | 8 kHz only |
| C. | 12 kHz and 9 kHz |
| D. | 12 kHz and 8 kHz |
| Answer» E. | |
| 13. |
If the intermediate frequency is very high (indicate false statement) |
| A. | Image frequency rejection is very good |
| B. | The local oscillator need not be extremely stable |
| C. | The selectivity will be poor |
| D. | Tracking will be improved |
| Answer» E. | |
| 14. |
Quantization noise occurs in |
| A. | PAM |
| B. | PPM |
| C. | DM |
| D. | None of these |
| Answer» D. None of these | |
| 15. |
The most suitable method for detecting a modulated signal (2.5 + 5 cos mt) cos ct is |
| A. | Envelope detector |
| B. | Synchronous detector |
| C. | Ratio detector |
| D. | Both and |
| Answer» C. Ratio detector | |
| 16. |
If IC is unmodulated current and It is the modulated current of an AM transmitter, then It is given by (where is the modulation index) |
| A. | IC |
| B. | |
| C. | IC |
| D. | IC |
| E. | None of these |
| Answer» C. IC | |
| 17. |
In FM wave expressed in terms of Bessel functions, J0( ) determines |
| A. | The dc component |
| B. | The first pair of sidebands |
| C. | The presence of carrier |
| D. | None of these |
| Answer» D. None of these | |
| 18. |
(For the data given in above) The minimum transmission bandwidth is |
| A. | 218.16 kHz |
| B. | 468.32 kHz |
| C. | 136.32 kHz |
| D. | None of the above |
| Answer» B. 468.32 kHz | |
| 19. |
As compared to message bandwidth, the PCM bandwidth is |
| A. | much larger |
| B. | same |
| C. | much smaller |
| D. | None of these |
| Answer» B. same | |
| 20. |
For the data given in Q. 146 if X(j ) satisfies the constraints required, then the pass band gain A of the ideal low pass filter needed to recover x(t) from c(t) x(t) is |
| A. | 1 |
| B. | 2 |
| C. | 4 |
| D. | 8 |
| Answer» D. 8 | |
| 21. |
A non-linear device with a transfer characteristic given by i = (0 + 2Vi + 0 2Vi2) mA is supplied with a carrier of 1V amplitude and a sinusoidal signal of 0.5V amplitude in series. If at the output the frequency component of AM signal is considered, the depth of modulation is |
| A. | 18% |
| B. | 10% |
| C. | 20% |
| D. | 33.33% |
| Answer» C. 20% | |
| 22. |
In a single-tone FM discriminator (S0/N0) is |
| A. | proportional to deviation |
| B. | proportional to cube of deviation |
| C. | inversely proportional to deviation |
| D. | proportional to square of deviation |
| Answer» E. | |
| 23. |
An AM modulator has output xc (t) = 40 cos 400 t + 4 cos 360 t + 4 cos 440 t The AM modulation efficiency is |
| A. | 0.01 |
| B. | 0.02 |
| C. | 0.03 |
| D. | 0.04 |
| Answer» C. 0.03 | |
| 24. |
The power in normalized message signal mn (t) would be |
| A. | 0.693 |
| B. | 0.542 |
| C. | 0.254 |
| D. | None of these |
| Answer» D. None of these | |
| 25. |
The double-sided spectrum of xc(t) would be |
| A. | A |
| B. | B |
| C. | C |
| D. | None of the above |
| Answer» C. C | |
| 26. |
An AM wave is given byeAM = 10(1 + 0 4 cos 103t + 0 3 cos 104t) cos 106t The modulation index is |
| A. | 0.4 |
| B. | 0.5 |
| C. | 0.3 |
| D. | 0.9 |
| Answer» C. 0.3 | |
| 27. |
A signal m(t) band-limited to 3 kHz is sampled at a rate 331%3 higher than the Nyquist rate. The max acceptable error in the sample amplitude (max. quantization error) is 0.5% of peak amplitude Vp. The quantized samples are binary coded. Find out the minimum bandwidth required to transmit the encoded binary signal. If 24 such signals are TDM, determine the minimum bandwidth required to transmit the multiplexed signal |
| A. | 1536 kHz |
| B. | 768 kHz |
| C. | 472 kHz |
| D. | 1344 kHz |
| Answer» C. 472 kHz | |
| 28. |
The most suitable method for detecting a modulated signal (2 5 + 5 cos mt) cos mt is |
| A. | envelope circuit |
| B. | synchronous detector |
| C. | ratio detector |
| D. | Both (A) and (B) |
| Answer» C. ratio detector | |
| 29. |
A signal x(t) is multiplied by rectangular pulse train c(t) as shown below:x(t) would be recovered from the product x(t) c(t) by using an ideal LPF if X (j ) = 0 for |
| A. | > 2000 |
| B. | > 1000 |
| C. | < 1000 |
| D. | < 2000 |
| Answer» D. < 2000 | |
| 30. |
An audio signal 15 sin 2 (1500t) amplitude modulates 60 sin 2 (106t) with carrier wave. The modulation in percentage is |
| A. | 20% |
| B. | 25% |
| C. | 50% |
| D. | 75% |
| Answer» C. 50% | |
| 31. |
In an AM signal the received signal power is 10 10 W with a maximum modulation signal of 5 kHz. The noise spectral density at the receiver input is 10 18 W/Hz. If the noise power is restricted to the message signal bandwidth only, the signal-to-noise ratio at the input to the receiver is |
| A. | 43 dB |
| B. | 66 dB |
| C. | 56 dB |
| D. | 33 dB |
| Answer» B. 66 dB | |
| 32. |
The Nyquist sampling rate for the signal g(t) = 10 cos (50 t) cos2 (150 t) where t is in seconds is |
| A. | 150 samples per second |
| B. | 200 samples per second |
| C. | 300 samples per second |
| D. | 350 samples per second |
| Answer» E. | |
| 33. |
The pre-envelop and the envelop of a function f(t) = A cos ( ct) |
| A. | A e |
| B. | , A |
| C. | |
| D. | A e |
| E. | , A |
| Answer» B. , A | |
| 34. |
For the same data given in Q. 144, Nyquist rate for the function x2 (t) will be |
| A. | 100 kHz |
| B. | 150 kHz |
| C. | 250 kHz |
| D. | 400 kHz |
| Answer» E. | |
| 35. |
In respect of the block diagram shown in the figure below, the input power is 1 mW. The output power PO will be |
| A. | 2 mW |
| B. | 1mW |
| C. | 0.5 mW |
| D. | 0 |
| Answer» C. 0.5 mW | |
| 36. |
An angle-modulated signal is given by s(t) = cos 2 (2 106t + 30 sin 150t + 40 cos 150t) The maximum frequency and phase deviations of s(t) are |
| A. | 10.5 kHz, 140 rad |
| B. | 6 kHz, 80 rad |
| C. | 10.5 kHz, 100 rad |
| D. | 7.5 kHz,100 rad |
| Answer» E. | |
| 37. |
Let message signal m(t) = cos (4 103t) and carrier signal c(t) = 5 cos (2 106)t are used to generate a FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term, cos [2 (1008 103t)t] in the FM signal would be |
| A. | 5J4 (3) |
| B. | (3) |
| C. | (4) |
| D. | 5 J |
| E. | (6) |
| Answer» E. (6) | |
| 38. |
Given message signal m(t) = sin (2000 t), Kf = 100 kHz/V and Kp = 10 rad/V. The bandwidth of FM and PM signals are respectively |
| A. | 204 kHz, 44 kHz |
| B. | 202 kHz, 22 kHz |
| C. | 402 kHz, 42 kHz |
| D. | None of these |
| Answer» C. 402 kHz, 42 kHz | |
| 39. |
A superheterodyne receiver is to operate in the frequency range of 550 kHz-1650 kHz with the intermediate frequency of 450 kHz. The receiver is tuned to 700. The capacitance ratio R = Cmax/Cmin of the local oscillator would be |
| A. | 4.41 |
| B. | 2.1 |
| C. | 3 |
| D. | 9 |
| Answer» B. 2.1 | |
| 40. |
An AM modulation has output xc(t) = A cos 400 t + B cos 380 t + B cos 400 t The carrier power is 100 W and the efficiency is 40%. The value of A and B are |
| A. | 14.14, 16.33 |
| B. | 50, 10 |
| C. | 22.36,13.46 |
| D. | None of these |
| Answer» B. 50, 10 | |
| 41. |
A PCM system uses a uniform quantizer followed by a 8- bit encoder. The bit rate of the system is equal to 106bits/s. The maximum message bandwidth for which the system operates satisfactory is |
| A. | 25 MHz |
| B. | 6.25MHz |
| C. | 12.5 MHz |
| D. | 50 MHz |
| Answer» C. 12.5 MHz | |
| 42. |
The angle modulated signal is given bys(t) = cos 2 (2 106t + 30 sin 150t + 40 cos 150t)The maximum frequency deviation in radian/sec is |
| A. | 750 |
| B. | 7500 |
| C. | 75000 |
| D. | 75 |
| Answer» C. 75000 | |
| 43. |
A carrier A cos ct is modulated by a signal f(t) = 2 cos 104. 2 t + 5 cos 103. 2 t + 3 cos 104. 4 t, then the bandwidth of the FM signal will be (Given that, kf = 15 103 Hz per volt) |
| A. | 170 kHz |
| B. | 190 kHz |
| C. | 210 kHz |
| D. | 150 kHz |
| Answer» C. 210 kHz | |
| 44. |
For AM modulated signal given by i = 10 + K1 Vi + K2 Vi2 where Vi = Ac cos ct + m (t), uses square law demodulator, find given that Ac = 1V, Am = 0 5 V, K1 = 2 and K2 = 0 2 |
| A. | 0.5 |
| B. | 0.2 |
| C. | 1 |
| D. | 0.4 |
| Answer» C. 1 | |
| 45. |
Signal x(t) = 5 (1 + 2 cos 2000 t) can be recovered by (i) envelop detector (ii) synchronous detector (iii) square law |
| A. | (i) and (ii) only. |
| B. | (i) and (iii) only. |
| C. | (ii) and (iii) only. |
| D. | All. |
| Answer» D. All. | |
| 46. |
A DSB-SC wave is generated with fc = 1 MHz, V0 = a0Vi + a1Vi3 where, Vi = A c cos ( ct) m(t) then f c is |
| A. | 1 MHz |
| B. | 0 33 MHz |
| C. | 0 5 MHz |
| D. | 3 MHz |
| Answer» D. 3 MHz | |
| 47. |
The total transmitted power given by the AM signal AM(t) = 10 cos (2 106t) + 5 cos(2 106t) cos (2 103t) + 2 cos (2 106t) + cos (4 103t) |
| A. | 57.25 W |
| B. | 62.25 W |
| C. | 56.25 W |
| D. | None of these |
| Answer» B. 62.25 W | |
| 48. |
Some properties are given below: (i) A signal f (t) and its Hilbert fh(t) have the same energy density spectrum. (ii) A signal f (t) and its Hilbert transform fh(t) have the same auto correlation function. (iii) A signal f (t) and its Hilbert transform fh(t) are mutually orthogonal i.e infin; f(t) fh(t) dt = 0 (iv) If fh(t) is a Hilbert transform of f (t), then the Hilbert transform of fh(t) is f (t)i.e. If H [f(t)] = fh(t) then H [fh(t)] = f (t) The correct statements are |
| A. | (i) and (iv) only. |
| B. | (i) and (iii) only. |
| C. | (i), (ii) and (iii) only. |
| D. | All of these |
| Answer» E. | |
| 49. |
Application of Hilbert transform are: (i) generation of SSB signals. (ii) design of minimum phase type filters. (iii) representation of bandpass signals. The correct statements are |
| A. | (i) and (ii) |
| B. | (i) and (iii) |
| C. | (ii) and (iii) |
| D. | (i), (ii) and (iii) |
| Answer» E. | |
| 50. |
Which one of the following is represented by V(t) = 5[cos (106 t) sin (103 t) sin (106 t)] |
| A. | SSB upper sideband signal. |
| B. | DSB suppressed carrier signal. |
| C. | AM signal. |
| D. | Narrow band FM signal. |
| Answer» E. | |