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Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be (in kbps)____
Correct answer is '512'. Can you explain this answer?
Verified Answer
Eight voice signals, each limited to 4 kHz and sampled at Nyquist rat...
Nyquist rate = 4×103×2 = 8×103 samples/sec
256 levels = 8 bits.
So one sample is represented by 8 -bit. So total = 8×8×103 bits/sec
For 8 voice signal = 8×8×8×103 bits/sec
= 512 kbps
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Eight voice signals, each limited to 4 kHz and sampled at Nyquist rat...

Calculation of Bit Transmission Rate for Time-Division Multiplexed Signal

1. Given Information:
- Number of voice signals = 8
- Each signal is limited to 4 kHz
- Sampled at Nyquist rate
- Quantization levels = 256

2. Calculation of Bandwidth Required:
- Since each voice signal is limited to 4 kHz, the total bandwidth required for all 8 signals = 8 * 4 kHz = 32 kHz

3. Calculation of Bit Depth:
- With 256 quantization levels, the number of bits required to represent each sample = log2(256) = 8 bits

4. Calculation of Bit Rate:
- The bit rate for a single voice signal = 4 kHz * 8 bits = 32 kbps
- Total bit rate for all 8 signals = 8 * 32 kbps = 256 kbps

5. Calculation of Time-Division Multiplexed Signal:
- In time-division multiplexing, each signal is allotted a specific time slot
- Since there are 8 signals, each signal will be allotted 1/8th of the time
- The total bit rate for the time-division multiplexed signal = 8 * 256 kbps = 512 kbps

Therefore, the bit transmission rate for the time-division multiplexed signal will be 512 kbps.
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Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be (in kbps)____Correct answer is '512'. Can you explain this answer?
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