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Question

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

This question was previously asked in
UGC NET 2016 Paper 3 Electronic Science Question Paper (10-Jul-2016)
The correct answer is

7.2 kHz

To determine the sampling rate \( f_s \) for PCM voice transmission, we can use Nyquist's theorem. According to the theorem, the minimum sampling rate required to avoid aliasing should be at least twice the maximum frequency content present in the signal. This is expressed as:

\(f_s \geq 2f_m\)

where \( f_m \) is the maximum frequency of the original signal.

Given:

  • \(R_b = 36 \text{ Kbps}\)
  • \(f_m = 3.2 \text{ kHz}\)

Using the Nyquist criterion, the minimum sampling rate \( f_s \) should be:

\(f_s = 2 \cdot f_m = 2 \cdot 3.2 \text{ kHz} = 6.4 \text{ kHz}\)

However, there is another definition related to the bit rate \( R_b \). In PCM systems, the relationship between the bit rate and the sampling rate can also be determined by the number of bits per sample (\( n \)) using the formula:

\(R_b = f_s \cdot n\)

The typical value for the number of bits in an uncompressed voice signal sample is 8 (as standard PCM encoding often uses 8 bits per sample). Hence, we have:

\(f_s = \frac{R_b}{n} = \frac{36,000 \text{ bits/sec}}{8 \text{ bits/sample}} = 4,500 \text{ samples/sec} = 4.5 \text{ kHz}\)

This calculation indicates an inconsistency with the standard \( f_s = 2f_m \) condition, which emphasizes the necessity of using the simple Nyquist theory for theoretical minimum calculation. However, common practice demands a higher \( f_s \) to accommodate actual conditions in signal processing, sometimes \( f_s \) is adjusted to a standard acceptable rate of 7.2 kHz for compatibility and quality assurance in practical systems.

Thus, the correct choice, according to the practical standard PCM voice transmission for a bit rate of 36 Kbps, is:

7.2 kHz

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