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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.
| 151. |
In communications, the sampling technique leads to |
| A. | Better efficiency |
| B. | Higher speed of communication |
| C. | Less costly equipment |
| D. | None of these |
| Answer» C. Less costly equipment | |
| 152. |
Carrier wave is generated by crystal oscillator because |
| A. | It provides high power output |
| B. | No buffer stage is needed |
| C. | It has good frequency stability |
| D. | None of these |
| Answer» D. None of these | |
| 153. |
Which of the following modulation is digital in nature? |
| A. | PAM |
| B. | PPM |
| C. | DM |
| D. | None of these |
| Answer» D. None of these | |
| 154. |
As compared to message bandwidth, the PCM bandwidth is |
| A. | much larger |
| B. | same |
| C. | much smaller |
| D. | None of these |
| Answer» B. same | |
| 155. |
Which of the following statements is not true for modulation? |
| A. | It allows the use of practicable antennas |
| B. | It reduces transmission over long distances |
| C. | It reduces the bandwidth |
| D. | None of these |
| Answer» D. None of these | |
| 156. |
Main disadvantage of DPSK is that |
| A. | Power margin is very low |
| B. | It requires much large power |
| C. | It is locked on a specific signalling speed. |
| D. | None of these |
| Answer» D. None of these | |
| 157. |
In differential PSK, the information is coded in terms of |
| A. | Absolute phase for each symbol. |
| B. | Both of these |
| C. | Phase changes between adjacent symbols |
| D. | None of these |
| Answer» C. Phase changes between adjacent symbols | |
| 158. |
Which of the following gives minimum probability? |
| A. | ASK |
| B. | FSK |
| C. | PSK |
| D. | DPSK |
| Answer» D. DPSK | |
| 159. |
Which of the following gives maximum probability of error? |
| A. | ASK |
| B. | FSK |
| C. | PSK |
| D. | DPSK |
| Answer» B. FSK | |
| 160. |
In PCM, if the number of quantization levels is increased from 4 to 64, then the bandwidth requirement will approximately be |
| A. | 3 times |
| B. | 4 times |
| C. | 8 times |
| D. | 16 times |
| Answer» B. 4 times | |
| 161. |
One of the following statement is wrong regarding modulation |
| A. | modulation is used to ensure that intelligence can be transmitted over long distances. |
| B. | modulation is used to ensure that different transmissions can be separated from one another. |
| C. | modulation is used to allow the use of practical antennas. |
| D. | modulation is used to reduce the signal bandwidth. |
| Answer» E. | |
| 162. |
The main advantage of high level modulation is that |
| A. | it requires a low audio driving power. |
| B. | it has a higher efficiency. |
| C. | circuitry involved is simpler. |
| D. | None of these |
| Answer» C. circuitry involved is simpler. | |
| 163. |
The main advantage of low level modulation is that |
| A. | it has maximum efficiency. |
| B. | it requires a very low audio driving power. |
| C. | circuitry involved is simpler. |
| D. | None of these. |
| Answer» C. circuitry involved is simpler. | |
| 164. |
In PDM, the bandwidth requirement is a function of |
| A. | The position of the pulse |
| B. | Maximum pulse width |
| C. | Minimum pulse width |
| D. | None of these |
| Answer» D. None of these | |
| 165. |
The PPM signal is converted into PDM with the help of a |
| A. | Monostable multivibrator |
| B. | Flip-flop |
| C. | Timer |
| D. | None of these |
| Answer» C. Timer | |
| 166. |
The sampling in PDM is |
| A. | Uniform |
| B. | Non-uniform |
| C. | Dependent on the nature of message signal |
| D. | None of these |
| Answer» C. Dependent on the nature of message signal | |
| 167. |
So far noise is concerned, as compared to direct baseband transmission, PAM is |
| A. | Better |
| B. | Similar |
| C. | Worst |
| D. | Same |
| Answer» C. Worst | |
| 168. |
Which of the following pulse modulation system has no carrier-wave equivalent? |
| A. | PPM |
| B. | PDM |
| C. | PCM |
| D. | None |
| Answer» D. None | |
| 169. |
As the sampling frequency is increased, the guard band becomes |
| A. | Smaller |
| B. | Remain same |
| C. | Larger |
| D. | None of these |
| Answer» D. None of these | |
| 170. |
The sampling rate is always between |
| A. | 0 and W |
| B. | W and 2W |
| C. | 2W and 4 W |
| D. | None of these |
| Answer» D. None of these | |
| 171. |
The minimum sampling frequency is called |
| A. | Carlson frequency |
| B. | Pulse sampling rate |
| C. | Nyquist sampling rate |
| D. | None of these |
| Answer» D. None of these | |
| 172. |
In communication, the sampling technique leads to |
| A. | Better efficiency |
| B. | Highest speed of communication |
| C. | Less costly equipment |
| D. | None of these |
| Answer» C. Less costly equipment | |
| 173. |
The PWM needs |
| A. | more power than PPM |
| B. | more samples per second than PPM |
| C. | more bandwidth than PPM |
| D. | None of these |
| Answer» B. more samples per second than PPM | |
| 174. |
Modulation index in FM depends on both maximum amplitude and frequency of the modulating signal. It is true |
| A. | yes |
| B. | no |
| C. | not necessarily true |
| D. | None of these |
| Answer» B. no | |
| 175. |
AM is less immune to noise than FM because |
| A. | in FM, intelligence is in the form of frequency variations. |
| B. | it is inherently like that. |
| C. | Noise causes amplitude variations, this spoiling the intelligence in the form of amplitude variations. |
| D. | None of the above. |
| Answer» D. None of the above. | |
| 176. |
For the data given in Q. 146 if X(j |
| A. | 1 |
| B. | 2 |
| C. | 4 |
| D. | 8 |
| Answer» D. 8 | |
| 177. |
A signal x(t) is multiplied by rectangular pulse train c(t) as shown below: |
| A. | > 2000 |
| B. | > 1000 |
| C. | < 1000 |
| D. | < 2000 |
| Answer» D. < 2000 | |
| 178. |
For the same data given in Q. 144, Nyquist rate for the function x |
| A. | 100 kHz |
| B. | 150 kHz |
| C. | 250 kHz |
| D. | 400 kHz |
| Answer» E. | |
| 179. |
Figure shows the Fourier spectra of signal x(t) the Nyquist sampling rate for the given function x(t) is |
| A. | 100 kHz |
| B. | 200 kHz |
| C. | 300 kHz |
| D. | 50 kHz |
| Answer» C. 300 kHz | |
| 180. |
In respect of the block diagram shown in the figure below, the input power is 1 mW. The output power P |
| A. | 2 mW |
| B. | 1mW |
| C. | 0.5 mW |
| D. | 0 |
| Answer» C. 0.5 mW | |
| 181. |
One of the main functions of R.F. amplifier in a superheterodyne receiver is to |
| A. | provide better tracking |
| B. | provide excellent adjacent channel selectivity. |
| C. | improve image frequency rejection. |
| D. | increase the tuning range of receiver. |
| Answer» D. increase the tuning range of receiver. | |
| 182. |
Six independent low pass signals of bandwidth 3 , , , 2 , 3 , and 2 Hz are to be time division multiplexed on a common channel using PAM. To achieve time, the minimum transmission bandwidth of the channel should be kHz |
| A. | 3 |
| B. | 6 |
| C. | 12 |
| D. | 24 |
| Answer» D. 24 | |
| 183. |
In a low-level AM system, the amplifier which follows the modulated stage must be the |
| A. | linear device |
| B. | harmonic device |
| C. | class-c amplifier |
| D. | non-linear device |
| Answer» B. harmonic device | |
| 184. |
Four voice signals, each limited to 4 kHz and sampled by Nyquist rate are converted into binary PCM using 256 quantization levels. The bit transmission rate for the time division multiplexed signal will be |
| A. | 8 kbps |
| B. | 6 kbps |
| C. | 256 kbps |
| D. | 512 kbps |
| Answer» D. 512 kbps | |
| 185. |
The AM broadcast band is given by |
| A. | 10 kHz to 30 kHz |
| B. | 500 kHz to 1500 kHz |
| C. | 2 kHz to 30 kHz |
| D. | None of these |
| Answer» C. 2 kHz to 30 kHz | |
| 186. |
The number of AM broadcast stations that can be accommodated on a 100 kHz bandwidth for the highest modulating frequency of 5 kHz will be |
| A. | 5 |
| B. | 10 |
| C. | 20 |
| D. | 50 |
| Answer» C. 20 | |
| 187. |
The modulation index of over-modulated wave is |
| A. | < 1 |
| B. | > 1 |
| C. | infinity |
| D. | 1 |
| Answer» C. infinity | |
| 188. |
The intermediate frequency of a superheterodyne receiver is 150 kHz. If the image frequency of a station is 2100 kHz, its actual frequency is |
| A. | 750 kHz |
| B. | 900 kHz |
| C. | 1200 kHz |
| D. | 1010 kHz |
| Answer» D. 1010 kHz | |
| 189. |
The Nyquist sampling rate for the signal g(t) = 10 cos (50 t) cos |
| A. | 150 samples per second |
| B. | 200 samples per second |
| C. | 300 samples per second |
| D. | 350 samples per second |
| Answer» E. | |
| 190. |
The biggest advantage of VSB system is |
| A. | that it conserves bandwidth and overcomes the problem of low video frequency attenuation |
| B. | lesser power requirement. |
| C. | Its bandwidth conservation |
| D. | simplicity of transmitter circuitry. |
| Answer» B. lesser power requirement. | |
| 191. |
In the above question, the total bandwidth required to transmit the AM wave is |
| A. | 1.5 kHz |
| B. | 3 kHz |
| C. | 100 kHz |
| D. | 10 kHz |
| Answer» C. 100 kHz | |
| 192. |
An audio signal 15 sin 2 (1500t) amplitude modulates 60 sin 2 (10 |
| A. | 20% |
| B. | 25% |
| C. | 50% |
| D. | 75% |
| Answer» C. 50% | |
| 193. |
In AM, if the modulation index is more than 100% then |
| A. | Power of the wave increases |
| B. | Efficiency of transmission increases |
| C. | The wave gets distorted |
| D. | None of these |
| Answer» D. None of these | |
| 194. |
The most commonly used system of modulation for telegraphy is |
| A. | on-off keying. |
| B. | frequency shift keying |
| C. | pulse code modulation |
| D. | differentiating the PPM signals. |
| Answer» B. frequency shift keying | |
| 195. |
A de-emphasis circuit is used |
| A. | prior to demodulation |
| B. | after demodulation |
| C. | to de-emphasise the magnitude of low frequency components. |
| D. | to boost the magnitude of high frequency. |
| Answer» C. to de-emphasise the magnitude of low frequency components. | |
| 196. |
The most commonly employed system of communication for commercial radio broadcast in india is |
| A. | FM |
| B. | PCM |
| C. | |
| D. | PM |
| E. | AM |
| Answer» E. AM | |
| 197. |
An audio signal has a bandwidth of 25 kHz. The maximum value of |m(t )| is 5V. This signal frequency modulates a carrier. If the frequency sensitivity constant of the modulator is 20 kHz/V, then according to Carson s rule, the bandwidth of the modulator output would be |
| A. | 50 kHz |
| B. | 250 kHz |
| C. | 200 kHz |
| D. | 100 kHz |
| Answer» C. 200 kHz | |
| 198. |
For 1 t 2 the phase deviation is |
| A. | 500 t rad |
| B. | 250 (1 + 2t) rad |
| C. | 20 t rad |
| D. | 0 |
| Answer» C. 20 t rad | |
| 199. |
The plot for frequency deviation is |
| A. | A |
| B. | B |
| C. | C |
| D. | D |
| Answer» C. C | |
| 200. |
Given message signal |
| 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 | |