Shannon-Hartley theorem; capacity limit 1.44S/N₀

Shannon–Hartley theorem

The Shannon–Hartley theorem gives the channel capacity $C$: the highest bit rate that can be sent over a band-limited channel with additive white Gaussian noise (AWGN) with an arbitrarily small error probability.

$$C = B \log_2\!\left(1 + \frac{S}{N}\right)\ \text{b/s}, \qquad N = N_0 B$$

where $B$ is the bandwidth (Hz), $S$ the average signal power, and $N_0/2$ the two-sided noise power spectral density.

If the information rate $R \le C$, a coding scheme exists that makes the error probability as small as we like. If $R > C$, errors cannot be made arbitrarily small, whatever coding is used.

Implications

• Bandwidth–SNR trade-off: the same capacity can be reached with a wide band and low SNR (e.g. spread spectrum) or a narrow band and high SNR (e.g. M-ary QAM).
• Capacity grows linearly with B but only logarithmically with S/N. Doubling the power adds only about $B$ b/s at high SNR.
• It gives a benchmark: practical modems and codes (turbo, LDPC) are judged by how close they come to $C$.
• It proves that error-free communication is possible over noisy channels with proper coding. It does not say how to build the code.

Theoretical limits

• Ideal system: $R = C$. Writing $S = E_b C$:

$$\frac{C}{B} = \log_2\!\left(1 + \frac{E_b}{N_0}\frac{C}{B}\right) \Rightarrow \frac{E_b}{N_0} = \frac{2^{C/B} - 1}{C/B}$$

• As $C/B \to 0$ (infinite bandwidth), $E_b/N_0 \to \ln 2 = 0.693$, i.e. −1.6 dB. This is the Shannon limit: below it, no error-free transmission is possible at any bandwidth.
• Increasing bandwidth alone cannot give unlimited capacity, because noise power $N_0 B$ also grows (shown below).

Capacity with infinite bandwidth

$$C = B \log_2\!\left(1 + \frac{S}{N_0 B}\right)$$

Let $x = \dfrac{S}{N_0 B}$, so $B = \dfrac{S}{N_0 x}$. As $B \to \infty$, $x \to 0$:

$$\begin{aligned} C &= \frac{S}{N_0} \cdot \frac{1}{x} \log_2(1 + x) = \frac{S}{N_0} \log_2 (1+x)^{1/x} \\ C_\infty &= \lim_{x \to 0} \frac{S}{N_0} \log_2 (1+x)^{1/x} = \frac{S}{N_0} \log_2 e \\ &= \frac{S}{N_0} \times 1.4427 \approx 1.44 \frac{S}{N_0} \end{aligned}$$

using $\lim_{x \to 0} (1+x)^{1/x} = e$ and $\log_2 e = 1/\ln 2 = 1.4427$.

So even with infinite bandwidth, the capacity stays finite at $C_\infty \approx 1.44\, S/N_0$ b/s. Capacity can be raised only by raising the signal power or lowering the noise density.