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A sinusoidal massage signal m(t) is transmitted by binary PCM without compression. If the signal to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be
  • a)
    8
  • b)
    10
  • c)
    12
  • d)
    14
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A sinusoidal massage signal m(t) is transmitted by binary PCM without ...
3(L2)/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 28, So 8 bits per sample is required.
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A sinusoidal massage signal m(t) is transmitted by binary PCM without ...
Question Analysis:

  • A sinusoidal massage signal m(t) is transmitted by binary PCM without compression.

  • The signal to-quantization-noise ratio is required to be at least 48 dB.

  • The minimum number of bits per sample is to be determined.



Solution:

  • PCM stands for Pulse Code Modulation.

  • The signal to quantization noise ratio is given by the formula, SNR = 6.02N + 1.76 dB. Here, N is the number of bits per sample.

  • Given, SNR = 48 dB.

  • Substituting the above values in the formula, we get,

  • 48 = 6.02N + 1.76

  • 46.24 = 6.02N

  • N = 7.68 ≈ 8 (Rounding off to the nearest integer)

  • Therefore, the minimum number of bits per sample is 8.



Therefore, the correct answer is option 'A' (8).
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