Nyquist channel capacity theorem

Nyquist channel capacity theorem: for a noiseless channel of bandwidth $B$ Hz, the maximum rate at which independent pulses (symbols) can be sent without ISI is $2B$ symbols per second. If each symbol takes one of $M$ distinct levels, the maximum bit rate (capacity) is

$$C = 2B\log_2 M\ \text{bits/s}$$

Explanation

1. Signalling rate limit. A channel of bandwidth $B$ can pass the sinc pulse $p(t) = \text{sinc}(2Bt)$, whose zeros occur every $\frac{1}{2B}$ s. Pulses spaced $T_s = \frac{1}{2B}$ apart therefore do not interfere at the sampling instants (Nyquist zero-ISI criterion). Hence the maximum symbol rate is

$$r_{max} = \frac{1}{T_s} = 2B\ \text{symbols/s (bauds)}$$

This rate $2B$ is called the Nyquist rate and $B = r/2$ the Nyquist bandwidth. Sending faster than $2B$ causes ISI that cannot be removed.

2. Bits per symbol. With $M$ equally likely levels, each symbol carries $\log_2 M$ bits. With binary signalling ($M = 2$), $C = 2B$ bits/s.

3. Capacity. Multiplying, $C = 2B\log_2 M$.

Examples

For a telephone channel with $B = 3$ kHz:

$$\begin{aligned} M = 4: \quad C &= 2(3000)\log_2 4 = 12\ 000\ \text{bits/s} \\ M = 16: \quad C &= 2(3000)\log_2 16 = 24\ 000\ \text{bits/s} \end{aligned}$$

Points to note

• The theorem assumes a noiseless channel. In theory, $C$ grows without limit as $M$ increases.
• In practice, noise limits $M$: closely spaced levels cannot be told apart. The noise-limited capacity is given by Shannon–Hartley, $C = B\log_2(1 + S/N)$. Comparing the two, the useful number of levels is about $M \approx \sqrt{1 + S/N}$.
• The ideal pulse needed for $2B$ symbols/s (brick-wall filter) is not realizable; with a raised cosine pulse of roll-off $\alpha$, the rate is $\frac{2B}{1+\alpha}$ symbols/s.