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Question

Assertion (A) : As compared to PWM, PAM and PPM, PCM is much better for noise immunity.

Reason (R) : PCM depends only on the presence or absence of pulses at any given time, not on any characteristics of the pulses which could be distorted.

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This question was previously asked in
UGC NET 2015 Paper 1 Question Paper (27-Dec-2015)
The correct answer is

Both (A) and (R) are true and (R) is the correct explanation of (A).

The distinction is between analogue and digital pulse modulation, and it decides everything.

PAM, PWM and PPM are all analogue pulse schemes: the information lives in a continuously variable property of the pulse.

SchemeInformation carried inVulnerable to
PAMPulse amplitudeAny additive noise — worst of the three
PWMPulse widthEdge jitter and threshold shift
PPMPulse positionTiming jitter
PCMPresence or absence of a pulseOnly noise large enough to flip a bit

Because a corrupted amplitude, width or position is a corrupted sample value, noise in an analogue pulse scheme translates directly into error at the output, and it accumulates at every repeater along the route.

PCM breaks that chain. The receiver has to answer only a yes/no question: was a pulse present? Provided the noise stays below half the decision threshold, the answer is correct and the regenerated pulse is perfect — the noise picked up on that hop is discarded rather than passed on. A chain of regenerative repeaters therefore has essentially the noise performance of a single hop, which is exactly why long-haul and undersea links are digital. So (R) states precisely the mechanism behind (A), and the code is 1.

What PCM pays for the immunity. Sampling at \(f_{s}\ge2f_{m}\) and coding each sample into n bits needs a bandwidth of roughly

\(B_{PCM}\approx n f_{s}\)

far more than the message bandwidth. It also introduces quantisation noise, which for a uniform quantiser of step \(\Delta\) has power \(\Delta^{2}/12\) and gives

\(\left(\dfrac{S}{N}\right)_{dB}\approx6n+1.76\)

— the familiar 6 dB per bit rule. Quantisation noise is unavoidable, but it is fixed at the encoder and, unlike channel noise, does not grow along the route.

The threshold effect is the other side of the coin: PCM performance is excellent above a certain carrier-to-noise ratio and degrades abruptly below it, whereas analogue schemes fade gracefully.

Hence, both (A) and (R) are true and (R) is the correct explanation of (A).

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Similar Questions

  1. Consider the following :

    (i) Pulse – position – modulation
    (ii) Pulse – Code – modulation
    (iii) Pulse – Width modulation
    (iv) Delta Modulation

    The output of which of these in not digital ?

  2. A signal m(t) of bandwidth B=4 kHz it transmitted using a binary companded PCM with µ = 100 and L = 64. The total number of bits for encoding and output SNR are respectively

  3. Which of the following statements are correct ?

    (a) PAM can be detected by a band pass filter.

    (b) The PWM needs more power than PPM.

    (c) Noise can be reduced by increasing sampling rate.

    (d) Pulse code modulation is a digital modulation technique.

  4. A channel with bit rate Rb = 36 Kbps is available for PCM voice transmission. Find the sampling rate fs, assuming fm = 3.2 kHz.

  5. Which of the following statements are incorrect :

    (a) In a frequency modulation system, the frequency of the carrier wave is varied in accordance to the frequency of the modulating signal.

    (b) In PAM, there is a greater peak power compared to conventional amplitude modulation systems.

    (c) In a PCM, the sampled amplitude is directly taken for conversion to a code of pulses.

    (d) The noise in a PCM system is reduced by increasing the number of quantising levels.


Important Questions from Pulse Modulation

  1. PDM is generated by employing __________

  2. The correct sequence of PCM modulation

  3. Compared to PPM, the PDM has the disadvantage that it requires

  4. Comparing Delta Modulation (DM) with PCM system: DM requires 

    A. Lower Sampling Rate 

    B. Higher Sampling Rate 

    C. Large Bandwidth 

    D. Simpler Hardware 

    E. Complex Hardware 

    Choose the correct answer from the options given below:

  5. Arrange PCM system starting from transmission to reception of signal 

    A. Sampling and Quantization 

    B. Quantization and decoding 

    C. Holding circuit 

    D. Encoding 

    E. Low pass filter (LPF) 

    Choose the correct answer from the options given below:

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