Channel Capacity — Nyquist and Shannon: 25-Comp-B5 Computer Communications, May 2015, Q2

Question 2: Channel Capacity — Nyquist and Shannon (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Bandwidth $B$	5.6 kHz
Signalling	Binary (2 levels)
Signal-to-noise ratio	30 dB

Find. (1) The noise-free maximum data rate; (2) the maximum data rate with the stated SNR.

Approach. Use the Nyquist noise-free capacity formula for a binary (2-level) signal, then the Shannon–Hartley theorem for the noisy channel.

Noise-free capacity (Nyquist). For $M=2$ signal levels (binary), $C = 2B\log_2 M$: $$C_{\text{Nyquist}} = 2(5600)\log_2(2) = \boxed{11{,}200\text{ bps} = 11.2\text{ kbps}}$$
Convert the SNR from dB to a ratio. $$\text{SNR (dB)} = 10\log_{10}(\text{SNR}) \;\Rightarrow\; \text{SNR} = 10^{30/10} = 10^{3} = 1000$$
Noisy capacity (Shannon–Hartley). $$C_{\text{Shannon}} = B\log_2(1+\text{SNR}) = 5600 \times \log_2(1001) = 5600 \times 9.9672$$ $$C_{\text{Shannon}} = \boxed{55{,}816\text{ bps} \approx 55.82\text{ kbps}}$$ Since $C_{\text{Shannon}} < 2C_{\text{Nyquist}}$ here would require $M=\lceil 2^{C/2B}\rceil$ signal levels far beyond binary, this Shannon limit represents the theoretical ceiling reachable only with sufficiently many signal levels/advanced coding, not the binary-signalling result of Step 1.

Case	Maximum data rate
(1) Noise-free, binary	11,200 bps (11.2 kbps)
(2) With SNR = 30 dB	≈ 55,816 bps (55.82 kbps)

Topic: Channel capacity — Nyquist bandwidth-limited and Shannon noisy-channel formulas.

Key relations: Nyquist $C = 2B\log_2 M$ (noise-free, $M$ signal levels); Shannon–Hartley $C = B\log_2(1+\text{SNR})$ (noisy, theoretical upper bound); $\text{SNR(dB)} = 10\log_{10}(\text{SNR})$.

Why this works: Nyquist bounds the rate purely from the signal's bandwidth and the number of distinguishable levels, assuming no noise; Shannon instead bounds the rate from bandwidth AND how much noise corrupts each level, giving the true information-theoretic ceiling — no coding scheme, however clever, can exceed it.

Common pitfall: forgetting to convert dB to a linear ratio before using it in $\log_2(1+\text{SNR})$ (using "30" directly instead of $10^{3}=1000$ gives a badly wrong, much smaller answer), or applying the Nyquist $M=2$ formula to the noisy case (Shannon's formula already accounts for noise and needs no separate $M$).

Source: Stallings, Data and Computer Communications, 10th ed., Ch.3 (Data Transmission) §3.4, Channel Capacity.

Question 2 of 7: Channel Capacity — Nyquist and Shannon

Question 2: Channel Capacity — Nyquist and Shannon (20 marks)