Match the following : Codes :List – I List – II a. Power efficient transmission i. SSB-SC b. Most bandwidth efficient transmission of voice signal ii. VSB c. Simplest receiver iii. FM d. Bandwidth efficient transmission of signals with significant d.c. component iv. AM
a-iii, b-i, c-iv, d-ii
Each scheme is matched to the single respect in which it is best: a-iii, b-i, c-iv, d-ii — option 2.
| Requirement | Scheme | Why |
|---|---|---|
| a. Power efficient | iii. FM | Constant envelope; noise traded for bandwidth |
| b. Bandwidth efficient (voice) | i. SSB-SC | Half the bandwidth, no carrier |
| c. Simplest receiver | iv. AM | A diode envelope detector suffices |
| d. Signals with DC content | ii. VSB | Retains a vestige near DC |
a — FM's power efficiency. Its envelope is constant, so the transmitter's output stage can be run in class C or saturated switching — 70 % efficient or better — whereas a linear AM stage must stay linear and wastes power doing so. FM also converts extra bandwidth into signal-to-noise improvement, so a given quality is achieved with less transmitted power.
b — SSB for voice. The two sidebands of an AM signal carry identical information, so one is pure redundancy. Removing it halves the occupied bandwidth to \(f_{m}\) and removing the carrier saves the two-thirds of power AM spends on it. Speech tolerates this well because it has no significant energy near DC — the audio band starts around 300 Hz — leaving room for the filter to cut cleanly between the sidebands. This is why long-distance HF voice communication is almost entirely SSB.
d — why VSB exists, and why it is not simply SSB. This is the pairing that explains the whole question. When a signal does have energy down to DC — television video is the standard case — SSB becomes impossible: the two sidebands meet at the carrier with no gap, and no filter can separate them without destroying the low-frequency content that carries the picture's average brightness. VSB is the compromise: transmit one sideband fully, plus a small vestige of the other, so the filter's gradual roll-off is symmetrical about the carrier and the low frequencies are recovered correctly. It costs a little more bandwidth than SSB but far less than DSB.
c — AM's one advantage. A diode, a capacitor and a resistor recover the envelope. That receiver simplicity is the entire justification for transmitting a carrier that carries no information, and it is why AM broadcasting outlived its inefficiency: transmitter power is spent once, while receivers are bought by millions.
Hence, the correct code is a-iii, b-i, c-iv, d-ii.
Match the following lists :
| List – I | List – II |
| a. PCM | i. Slope overload distortion |
| b. DM | ii. Constant carrier frequency |
| c. AM | iii. Encoding |
| d. TDM | iv. Commutator |
The correct match is :
Match the following :
| List – I | List – II |
| a. DSB-SC modulation | i. envelope detection |
| b. SSB-modulation | ii. Foster Seeley |
| c. AM-demodulation | iii. Weaver's method |
| d. Phase-shift detection | iv. Balanced modulator |
Codes :
Match the following lists :
| List - I | List - II |
a) ![]() | i) Frequency modulated detector |
b) ![]() | ii) Pre-emphasis |
c) ![]() | iii) Amplitude modulator |
d) ![]() | iv) Amplitude detector |
Choose the correct answer from the codes given below:
Match the following lists :
| List -I | List - II |
| a. Power efficient transmission of signals | i. AM |
| b. Bandwidth efficient transmission of signals with significant dc signals | ii. VSB |
| c. Most bandwidth efficient transmission of voice signals | iii. SSB-SC |
| d. Simplest Receiver Circuit | iv. FM |
Correct Codes are :
Arrange in order of the increasing Bandwidth value for the below mentioned systems :
(a) DSB-SC (b) VSB (c) FM (d) SSB
Codes :
Match List - I with List - II.
| List - I | List - II |
| (A) White noise | (I) Reduces antenna height |
| (B) RMS noise voltage | (II) \(P_{C}\left(\dfrac{m^{2}}{2}\right)\) |
| (C) Modulation | (III) Inversely proportional to \(\sqrt{T}\) |
| (D) PDSBSC | (IV) Johnson Noise |
Choose the correct answer from the options given below :
What is the duty cycle of a radar with pulse width 0.3 µs and pulse repetation rate of 1 ms ?
In an AM with m=1, the power in one sideband is:
Match the following lists :
| List – I | List – II |
| a. PCM | i. Slope overload distortion |
| b. DM | ii. Constant carrier frequency |
| c. AM | iii. Encoding |
| d. TDM | iv. Commutator |
The correct match is :
Match the following :
| List – I | List – II |
| a. DSB-SC modulation | i. envelope detection |
| b. SSB-modulation | ii. Foster Seeley |
| c. AM-demodulation | iii. Weaver's method |
| d. Phase-shift detection | iv. Balanced modulator |
Codes :