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A voice signal band limited to 3.4 kHz is sampled at 8 kHz and pulse code modulated using 64 quantization levels...

 

31) A voice signal band limited to 3.4 kHz is sampled at 8 kHz and pulse code modulated using 64 quantization levels. Ten such signals are time division multiplexed using an 5-bit synchronizing word. The minimum channel band width will be

  • (1) 64 kHz
  • (2) 128 kHz
  • (3) 320 kHz
  • (4) 520 kHz
Answer: Option D

Solution

Step 1: Calculate bits per sample (n)
The number of quantization levels (L) is 64.
Formula: n = log2(L)
n = log2(64) = 6 bits per sample

Step 2: Calculate total bits per TDM frame
A frame contains one sample from each of the 10 signals plus the sync bits.
Total bits = (Number of signals × bits per sample) + synchronization bits
Total bits = (10 × 6) + 5 = 65 bits per frame

Step 3: Calculate the bit rate (Rb)
The sampling rate is given as 8 kHz (8,000 samples per second).
Rb = Sampling Rate × Bits per frame
Rb = 8,000 × 65 = 520,000 bits/sec (520 kbps)

Step 4: Determine the minimum channel bandwidth
For PCM transmission, the minimum bandwidth is typically taken as equal to the bit rate (Rb).
Minimum Bandwidth = 520 kHz

Correct Option: (4) 520 kHz

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