A 4 kHz noiseless channel is sampled once every 125 microseconds (i.e., 8000 samples per second) to carry digital signals. If delta modulation is used (1 bit per sample), how many bits per second are actually sent?

Introduction / Context:
Delta modulation is a simple form of analog-to-digital conversion in which each sample conveys only the sign of the change relative to the previous sample. Because each sample is encoded as a single bit, the bit rate equals the sampling rate. This question ties together channel sampling and the delta modulation bit budget.

Given Data / Assumptions:

• Channel bandwidth mentioned: 4 kHz (contextual; sampling rate is explicitly provided).
• Sampling interval: every 125 microseconds → sampling rate = 8000 samples per second.
• Delta modulation uses 1 bit per sample.

Concept / Approach:
Bit rate in delta modulation = samples per second * bits per sample. With 8000 samples per second and 1 bit each, the transmitted bit rate is 8000 bits per second, i.e., 8 kbps (using kbps as 1000 bps in telecom shorthand). No additional framing or coding overhead is considered in the basic calculation.

Step-by-Step Solution:

Verification / Alternative check:
By comparison, PCM with 8 bits per sample at 8000 samples per second yields 64 kbps; delta modulation reduces this by a factor of 8 by using 1-bit coding per sample under ideal tracking conditions.

Why Other Options Are Wrong:

• 32 kbps / 64 kbps / 128 kbps: correspond to multi-bit PCM scenarios (e.g., 4-, 8-, or 16-bit PCM), not delta modulation.
• 16 kbps: could match 2 bits/sample at 8 kHz (not delta modulation in its basic 1-bit form).

Common Pitfalls:
Confusing “bandwidth” with “bit rate” directly; forgetting the given sampling interval; mixing PCM and delta modulation bit budgets.

Final Answer:
8 kbps