Pulse Modulation Techniques Questions and Answers

Q: Match List I with List II and select the correct answer using the codes given below :List IList II(a) TDM(i) Frequency detector(b) FM(ii) Envelope detector(c) AM(iii) Sensitive to propagation delay(d) FDM(iv) Non-sensitive to propagation delay

The correct code is (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv).TDM interleaves the channels in time slots, so the receiver must stay synchronised with the transmitter and the system is sensitive to propagation delay (iii). FM is demodulated by a frequency detector or discriminator (i). AM is demodulated by an envelope detector (ii). FDM separates channels by frequency, so all of them travel simultaneously and a delay in the path does not mix them up (iv).The other codes swap the detectors between AM and FM, or assign delay sensitivity to the frequency-division system.

Q: What does the sampling theorem require of the sampling rate?

The theorem requires at least twice the highest frequency in the signal, and that figure is the Nyquist rate. Sampling at the signal frequency itself loses the waveform entirely. Four times is a practical margin often used but not the theoretical requirement. And the channel's bandwidth is a separate constraint that follows from the pulse rate rather than determining it. Distinguishing a theoretical minimum from the practical rate chosen above it is the first thing to get straight about sampling.

Q: What is the Nyquist rate for a signal band-limited to 4 kHz?

Twice the highest frequency gives 8 kHz, and telephony's familiar 8 kHz sampling rate for a 3.4 kHz channel comes from exactly this calculation with a little margin for the anti-alias filter. 4 kHz is the signal's own limit and would lose everything. 2 kHz is half of it. And 16 kHz is four times, a comfortable practical rate rather than the theoretical minimum. Standard rates in real systems usually sit a little above Nyquist to leave the filter somewhere to roll off.

Q: What is aliasing?

Aliasing is a component above half the sampling rate reappearing as a lower frequency, indistinguishable from a genuine signal at that frequency, and it is irreversible once it has happened. Amplitude error from the pulse width is the aperture effect. Rounding error is quantisation noise. And channel overlap is crosstalk. Four distinct impairments, and the important property of aliasing is that no processing afterwards can undo it, which is why the filter comes first.

Q: Why is an anti-alias filter placed before the sampler?

Because aliasing cannot be corrected afterwards, the filter removes anything above half the sampling rate before the sampler can fold it down. Smoothing the recovered waveform is the reconstruction filter's job at the far end. Equalisation and amplitude limiting are separate functions. This is the same principle that governs the receiver's image response: where a stage maps two inputs onto one output, the discrimination has to happen ahead of it.

Match List I with List II and select the correct answer using the codes given below :

List I	List II
(a) TDM	(i) Frequency detector
(b) FM	(ii) Envelope detector
(c) AM	(iii) Sensitive to propagation delay
(d) FDM	(iv) Non-sensitive to propagation delay

A.(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
B.(a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
C.(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
D.(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The correct code is (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv).

TDM interleaves the channels in time slots, so the receiver must stay synchronised with the transmitter and the system is sensitive to propagation delay (iii). FM is demodulated by a frequency detector or discriminator (i). AM is demodulated by an envelope detector (ii). FDM separates channels by frequency, so all of them travel simultaneously and a delay in the path does not mix them up (iv).

The other codes swap the detectors between AM and FM, or assign delay sensitivity to the frequency-division system.

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Time-Division Multiplexing

A.can be used with PAM only.
B.combines five groups into a supergroup.
C.stacks 24 channels in adjacent frequency slots.
D.interleaves pulses belonging to different transmissions.

Exam Year(s)

UGC NET - 2012,

Answer: D

Time-division multiplexing interleaves pulses belonging to different transmissions on a common channel, each message being given a time slot in turn, so option D is correct. It can be used with PCM and other pulse modulation, not with PAM only. Combining five groups into a supergroup is a feature of frequency-division multiplexing in telephone systems, and stacking 24 channels in adjacent frequency slots also describes frequency-division multiplexing (a group), not a time-division scheme.

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Direction for [ Question No: 3 To 3 ] :

Directions : Q. Nos. 61 to 70, the following items consist of two statements, one labelled the "Assertion (A)" and the other labelled the "Reason (R)". You are to examine these two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason (R) is a correct explanation of the Assertion (A). Select your answers to these items using the codes given below and mark your answer accordingly.

Assertion (A) : TDM can be employed to transmit channels having unequal bandwidths.

Reason (R) : If sampling theorem is strictly followed, any analog signal can be reconstructed back from the samples.

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 true, but (R) is false.
D.(A) is false, but (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: B

Both statements are correct, but the reason does not explain the assertion, the second option. Time division multiplexing can carry channels of unequal bandwidth, by sampling the wider channels at a higher rate or by giving them more time slots.

The reason is also true, as the sampling theorem allows reconstruction when it is followed, but this is a property of sampling and not the cause of the ability to mix unequal bandwidths. So the first option is wrong, and neither statement is false.

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Time-Division Multiplex

A.can be used with PAM ony
B.combines five groups into a supergroup
C.stacks 24 channels in adjacent frequency slots
D.interleaves pulses in time domain belonging to different transmissions

Exam Year(s)

UGC NET - 2013,

Answer: D

Time-division multiplexing lets several transmissions share one link by interleaving narrow pulses belonging to different sources in the time domain, with the two ends synchronized, as the passage near the end of the paper also says. It is not limited to PAM, since PCM and PWM are used as well. Combining five groups into a supergroup and stacking 24 channels in adjacent frequency slots both describe frequency-division multiplexing. The answer is that it interleaves pulses in the time domain belonging to different transmissions.

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Indicate which of the following system is digital ?

A.Pulse - Position Modulation
B.Pulse - Code Modulation
C.Pulse - Width Modulation
D.Pulse - Amplitude Modulation

Exam Year(s)

UGC NET - 2013,

Answer: B

Of the modulation systems named, only pulse-code modulation is digital, because the samples are quantized and each is coded as a binary word, so the transmitted signal is a stream of bits. Pulse-position, pulse-width and pulse-amplitude modulation vary a pulse's position, width or height continuously with the signal sample, so they are analog (pulse) modulation systems although the pulses are discrete in time. The answer is pulse-code modulation.

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Direction for [ Question No: 6 To 6 ] :

Assertion-Reason type questions :

The following items consist of two statements, one labelled as ‘Assertion A’ and the other labelled the ‘Reason (R)’. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : TDM and FDM accomplish the same end by different means.
Reason (R) : FDM involves simpler instrumentation as compared to TDM.
Codes :

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: C

Assertion (A) is true, since time division and frequency division multiplexing both combine several signals on one channel, in different ways. Reason (R) is false, since FDM needs filters and many carrier oscillators and modulators, whereas TDM is simpler in its instrumentation, being based on digital switching. So (A) is true and (R) is false.

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What does the sampling theorem require of the sampling rate?

A.At least twice the highest signal frequency
B.At least equal to the highest signal frequency present
C.At least four times the highest frequency, for a margin
D.At least twice the bandwidth of the transmission channel

Answer: A

The theorem requires at least twice the highest frequency in the signal, and that figure is the Nyquist rate. Sampling at the signal frequency itself loses the waveform entirely. Four times is a practical margin often used but not the theoretical requirement. And the channel's bandwidth is a separate constraint that follows from the pulse rate rather than determining it. Distinguishing a theoretical minimum from the practical rate chosen above it is the first thing to get straight about sampling.

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What is the Nyquist rate for a signal band-limited to 4 kHz?

A.16 kHz
B.8 kHz
C.4 kHz
D.2 kHz

Answer: B

Twice the highest frequency gives 8 kHz, and telephony's familiar 8 kHz sampling rate for a 3.4 kHz channel comes from exactly this calculation with a little margin for the anti-alias filter. 4 kHz is the signal's own limit and would lose everything. 2 kHz is half of it. And 16 kHz is four times, a comfortable practical rate rather than the theoretical minimum. Standard rates in real systems usually sit a little above Nyquist to leave the filter somewhere to roll off.

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What is aliasing?

A.Distortion introduced by rounding a sample to a discrete level
B.Overlap between adjacent channels in a multiplexed pulse train
C.High frequencies appearing as lower ones after sampling
D.Loss of amplitude accuracy caused by the sampling pulse's width

Answer: C

Aliasing is a component above half the sampling rate reappearing as a lower frequency, indistinguishable from a genuine signal at that frequency, and it is irreversible once it has happened. Amplitude error from the pulse width is the aperture effect. Rounding error is quantisation noise. And channel overlap is crosstalk. Four distinct impairments, and the important property of aliasing is that no processing afterwards can undo it, which is why the filter comes first.

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10.

Why is an anti-alias filter placed before the sampler?

A.To smooth the recovered waveform at the receiving end
B.To equalise the amplitude response across the passband
C.To limit the peak amplitude the sampler has to handle
D.To remove components above half the sampling rate

Answer: D

Because aliasing cannot be corrected afterwards, the filter removes anything above half the sampling rate before the sampler can fold it down. Smoothing the recovered waveform is the reconstruction filter's job at the far end. Equalisation and amplitude limiting are separate functions. This is the same principle that governs the receiver's image response: where a stage maps two inputs onto one output, the discrimination has to happen ahead of it.

Discuss in Forum

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Pulse Modulation Techniques Questions and Answers

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