A communication system uses digitization of audio signals with BW 15 kHz followed by PCM encoding and real time transmission. Assuming that uniform quantization with 1024 levels is used and there are no overheads in PCM encoding, what is the minimum permissible bit rate?

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  1. 150 kbps
  2. 300 kbps
  3. 750 kbps
  4. 600 kbps

Answer (Detailed Solution Below)

Option 2 : 300 kbps
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Detailed Solution

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Explanation:

Minimum Permissible Bit Rate Calculation:

Given Data:

  • Bandwidth (BW) of the audio signal = 15 kHz
  • Number of quantization levels = 1024
  • No overheads in PCM encoding

Step 1: Determine the number of bits per sample

The number of quantization levels (L) is given as 1024. The number of bits required to represent each quantization level can be calculated using the formula:

Number of bits per sample = log2(L)

Here, log2(1024) = 10. Therefore, each sample requires 10 bits to represent the quantized value.

Step 2: Apply Nyquist Theorem to determine the sampling rate

According to Nyquist theorem, the minimum sampling rate (fs) should be at least twice the bandwidth of the signal:

fs = 2 × BW

Given BW = 15 kHz:

fs = 2 × 15 kHz = 30 kHz

Step 3: Calculate the minimum permissible bit rate

The bit rate (R) is given by:

R = Sampling Rate × Number of bits per sample

Substituting the values:

R = 30 kHz × 10 bits/sample = 300 kbps

Therefore, the minimum permissible bit rate for the given communication system is 300 kbps.

Correct Option Analysis:

The correct option is:

Option 2) 300 kbps

This option correctly represents the minimum permissible bit rate for the given communication system using PCM encoding, considering the bandwidth, quantization levels, and sampling rate.

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