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The amplitude of a random signal is uniformly distributed between –5V and 5V. If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5dB, the step size of the quantization is approximately-
  • a)
    0.0333V
  • b)
    0.05V
  • c)
    0.0667V
  • d)
    0.10V
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The amplitude of a random signal is uniformly distributed between&ndas...
Given Parameters:
- Amplitude range of random signal = 5V to -5V
- Signal to quantization noise ratio required = 43.5dB

Calculation:
- The formula for Signal to Quantization Noise Ratio (SQNR) in dB is given by:
SQNR = 6.02 * N + 1.76 dB
where N is the number of bits used for quantization.
- Given that SQNR = 43.5dB, we can substitute this value in the formula:
43.5 = 6.02N + 1.76
41.74 = 6.02N
N ≈ 6.94
- The step size (Δ) of quantization for a uniformly distributed signal is given by:
Δ = (max amplitude - min amplitude) / (2^N)
Δ = (5 - (-5)) / (2^6.94)
Δ ≈ 0.0667V
Therefore, the step size of quantization for the given random signal is approximately 0.0667V, which is closest to option 'C'.
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Community Answer
The amplitude of a random signal is uniformly distributed between&ndas...
Given; Signal to quantization noise  = 43.5 dB
The amplitude of a random signal is uniformly distributed between –5V and 5V.

 
E[X2] = 52/3 = 25/3
E[XQE2] = δ2/12
SNR = (25/3)/ (δ2/12) = 100/ δ2
SNR (dB) = 10log(SNR)
⇒ 43.5 =  10log(SNR)
⇒  log(SNR) = 43.5/10
⇒  SNR= 104.35 
Therefore
100/ δ2 = 104.35 
⇒ δ2 = 100/104.35 
⇒ δ = 0.0667
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