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Consider the audio signal with spectral components limited to the frequency band of 300 to 3300 Hz. A PCM signal is generated with a sampling rate of 8000 samples/sec. The required output signal to quantizing noise ratio is 30 dB. The minimum number of bits per sample needed is :
  • a)
    4
  • b)
    5
  • c)
    6
  • d)
    7
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider the audio signal with spectral components limited to the freq...
(SNR) in dB = 1.76 + 20 log L
Where L is the number of levels

So minimum number of levels are 26, and minimum number of bits are 5.
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Most Upvoted Answer
Consider the audio signal with spectral components limited to the freq...
Given parameters:
Frequency band: 300 to 3300 Hz
Sampling rate: 8000 samples/sec
Signal to quantizing noise ratio (SQNR): 30 dB

To determine the minimum number of bits per sample needed for PCM signal, we can follow the below steps:

Step 1: Determine the bandwidth of the signal
Bandwidth = 3300 - 300 = 3000 Hz

Step 2: Apply Nyquist theorem to determine the minimum sampling rate required
According to Nyquist theorem, the minimum sampling rate required is twice the bandwidth of the signal.
Minimum sampling rate = 2 x 3000 = 6000 samples/sec

Since the given sampling rate of 8000 samples/sec is greater than the minimum required sampling rate, it is sufficient for this signal.

Step 3: Calculate the maximum signal amplitude
The maximum signal amplitude can be determined using the SQNR formula:
SQNR = 6.02 x b + 1.76 dB
where b is the number of bits per sample.

Substituting the given SQNR of 30 dB, we get:
30 = 6.02 x b + 1.76
28.24 = 6.02 x b
b = 4.69

We can round up the value of b to the nearest integer, which is 5.

Therefore, the minimum number of bits per sample needed for PCM signal is 5 (option B).
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