Match the following lists : The correct match is :List – I List – II a. PCM i. Slope overload distortion b. DM ii. Constant carrier frequency c. AM iii. Encoding d. TDM iv. Commutator
a-iii, b-i, c-ii, d-iv
Each entry in List II is the defining feature of exactly one system, so the matching is direct: a-iii, b-i, c-ii, d-iv — option 4.
| System | Feature | Why |
|---|---|---|
| a. PCM | iii. Encoding | Samples are quantised and coded as binary words |
| b. DM | i. Slope overload distortion | Fixed step size cannot follow a fast input |
| c. AM | ii. Constant carrier frequency | Amplitude varies; frequency does not |
| d. TDM | iv. Commutator | Rotating switch samples the channels in turn |
Slope overload — the characteristic defect of delta modulation. DM transmits one bit per sample, moving its staircase approximation up or down by a fixed step \(\Delta\) at each clock instant. The fastest the staircase can climb is therefore
\(\dfrac{\Delta}{T_{s}}=\Delta f_{s}\)
and if the input rises faster than this the approximation falls behind and cannot catch up until the input slows. The condition for avoiding it is
\(\Delta f_{s}\ge\left|\dfrac{dm(t)}{dt}\right|_{max}=A\omega_{m}\)
The opposite fault, granular noise, appears when the step is too large for a slowly varying input, so the staircase hunts about it. Adaptive delta modulation varies the step size to escape both — which is exactly what the fixed step of plain DM cannot do.
Encoding is what makes PCM digital. Sampling and quantising alone give PAM levels; the encoder turns each level into an n-bit word, and it is that step which buys regeneration and noise immunity.
The commutator is the classic picture of time-division multiplexing: a rotating switch that touches each channel in turn, giving every one its own time slot, with a matching de-commutator at the receiver kept in step by frame synchronisation. Contrast frequency-division multiplexing, where all channels are present at once but in different frequency bands.
Constant carrier frequency distinguishes AM from the angle-modulation family: in AM the envelope carries the message and the frequency stays put; in FM and PM the amplitude stays put and the frequency or phase carries the message.
Hence, the correct match is a-iii, b-i, c-ii, d-iv.
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 :
| 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 |
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 :
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 :
During the heterodyne process in the receiver, the modulation of the signal __________
Which of the following modulations is used in India for radio transmission?
Which of the following statements are correct?
A. DSB‐SC modulation is well suited for point to point communication involving one transmitter and one receiver.
B. VSB modulation is a linear modulation scheme.
C. SSB is a non‐linear modulation scheme.
D. FM is a linear modulation scheme.
Choose the correct answer from the options given below: