Shannon capacity of an AWGN channel A communication channel with additive white Gaussian noise has bandwidth 4 kHz and signal-to-noise ratio SNR = 15 (linear). What is the channel capacity?

Introduction / Context:
The Shannon–Hartley theorem provides the maximum theoretical data rate (capacity) over a band-limited AWGN channel for a given SNR. It is central to link budgeting, modem design, and assessing the benefit of coding and bandwidth expansion.

Given Data / Assumptions:

• Bandwidth B = 4 kHz.
• SNR (linear) = 15 → 1 + SNR = 16.
• A baseband real channel interpretation.

Concept / Approach:

Shannon capacity formula: C = B * log2(1 + SNR). Insert the given values and evaluate the logarithm in base 2. Recognize that 16 is a power of 2, simplifying computation.

Step-by-Step Solution:

Verification / Alternative check:

Using natural logs: C = 4000 * ln(16)/ln(2) = 4000 * 2.7726/0.6931 ≈ 16000 bps, matching the simple power-of-two result.

Why Other Options Are Wrong:

(a) 1.6 kbps is off by a factor of 10; (c) 32 kbps assumes log2(1 + SNR) = 8, incorrect; (d) 256 kbps is far too high for 4 kHz bandwidth; (e) 64 kbps exceeds the Shannon limit for the given SNR.

Common Pitfalls:

Confusing dB with linear SNR; forgetting to use base-2 logarithms; mixing up bandwidth units (Hz vs. kHz).

Final Answer:

16 kbps