Shannon channel capacity theorem and its implication

Statement

Shannon's channel capacity theorem (Shannon–Hartley law): for a channel of bandwidth $B$ Hz with additive white Gaussian noise of power $N = N_0 B$ and average signal power $S$, the channel capacity is

$$C = B \log_2\!\left(1 + \frac{S}{N}\right)\ \text{bits/s}$$

If the information rate $R \le C$, there is a coding scheme that sends the information with an arbitrarily small probability of error. If $R > C$, the error probability cannot be made small, whatever coding is used.

Implications in communication systems

1. Upper bound on data rate. No modem or code can beat $C$. Real designs are judged by how close they get, e.g. LDPC/turbo codes come within about 1 dB.
2. Bandwidth–power exchange. The same $C$ can come from a large $B$ with low SNR, or a small $B$ with high SNR.
   • Spread spectrum and CDMA work at very low SNR by using wide bandwidth.
   • High-order QAM (e.g. 256-QAM in cable modems) uses high SNR to pack more bits into a narrow band.
3. Error-free transmission over noisy channels is possible. Noise limits the rate, not the accuracy. This motivates channel coding (error-correcting codes).
4. Diminishing returns from power. $C$ grows only logarithmically with $S/N$. At high SNR, doubling the power adds only about 1 bit/s/Hz.
5. Limited gain from bandwidth. As $B \to \infty$, $C \to 1.44\, S/N_0$, so capacity stays finite because noise grows with bandwidth.
6. Shannon limit. The minimum $E_b/N_0$ for reliable communication is $\ln 2 = -1.6$ dB. This is the benchmark for power-efficient systems such as deep-space links.
7. Design guide. It tells designers the best spectral efficiency $C/B$ for a given SNR, e.g. 3.46 b/s/Hz at 10 dB.

Example: a telephone channel with $B = 3.1$ kHz and SNR = 30 dB has $C = 3100 \log_2(1001) \approx 30.9$ kb/s. This is close to the speed of the V.34 modem (33.6 kb/s), whose real SNR is slightly higher.