Telephony-grade sampling with DPCM: A 4 kHz noiseless channel uses one sample every 125 microseconds (i.e., 8000 samples per second) to transmit digital signals. If Differential PCM encodes each sample using a 4-bit relative value, how many bits per second are actually sent?

Introduction / Context:
In baseband telephony and many voice codecs, the sampling frequency is typically 8000 samples per second because the voice channel bandwidth is about 4 kHz. When using Differential PCM (DPCM), each sample conveys the change relative to a predictor, often allowing fewer bits per sample than standard PCM. Calculating the resulting bit rate requires multiplying samples per second by bits per sample.

Given Data / Assumptions:

• Channel is treated as noiseless for the purpose of bit rate calculation.
• Sampling interval is 125 microseconds → 8000 samples/s.
• Each DPCM codeword carries 4 bits per sample.
• No framing or overhead is considered (payload bit rate only).

Concept / Approach:
Bit rate for a sampled system using fixed-length codewords is simply: bit_rate = samples_per_second * bits_per_sample. For DPCM here, the predictor reduces the number of bits needed to 4 per sample compared to 8 in basic PCM, halving the payload rate from 64 Kbps to 32 Kbps.

Step-by-Step Solution:

Verification / Alternative check:
Compare with standard PCM: 8000 samples/s * 8 bits/sample = 64 Kbps. Using half the bits per sample (4) yields exactly half the rate, confirming 32 Kbps.

Why Other Options Are Wrong:

Common Pitfalls:
Misreading ‘‘every 125’’ as milliseconds rather than microseconds; forgetting to convert units or to multiply by the correct samples per second.

Final Answer:
32 Kbps.