Duobinary error; entropy; capacity as B → ∞

Minimizing error in duo-binary coding

In duobinary signalling the received sample is $c_k = a_k + a_{k-1}$, so the decoder normally finds the present bit by subtracting the previous decision: $\hat a_k = c_k - \hat a_{k-1}$. One wrong decision then spoils all the following decisions. This is called error propagation. It is reduced or removed by:

• Precoding: before the duobinary filter, form $d_k = b_k \oplus d_{k-1}$ (modulo-2 sum). The receiver can then decide each bit from the present sample alone: if $|c_k| < 1$ (i.e. $c_k = 0$) then $\hat b_k = 1$; if $|c_k| > 1$ ($c_k = \pm 2$) then $\hat b_k = 0$. An error stays in one bit only.
• Using the correct decision thresholds ($\pm 1$) for the three-level signal and good receive filtering to keep noise low.

Information and entropy

• Information: the information carried by a message $m_k$ of probability $p_k$ is $I_k = \log_2 \frac{1}{p_k}$ bits. A less likely message carries more information; a certain message ($p_k = 1$) carries none.
• Entropy: the average information per symbol of a source with $M$ symbols:

$$H = \sum_{k=1}^{M} p_k \log_2 \frac{1}{p_k}\ \text{bits/symbol}, \qquad 0 \le H \le \log_2 M$$

Upper limit of channel capacity as $B \to \infty$

By the Shannon–Hartley theorem, with signal power $S$ and white noise of one-sided PSD $N_0$ (noise power $N = N_0 B$):

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

Multiply and divide by $S/N_0$ and put $x = \dfrac{S}{N_0 B}$:

$$\begin{aligned} C &= \frac{S}{N_0}\cdot\frac{N_0 B}{S}\log_2\left(1+\frac{S}{N_0B}\right) = \frac{S}{N_0}\cdot\frac{1}{x}\log_2(1+x) \\ &= \frac{S}{N_0}\log_2 (1+x)^{1/x} \end{aligned}$$

As $B \to \infty$, $x \to 0$ and $\lim_{x\to 0}(1+x)^{1/x} = e$. Therefore

$$C_\infty = \lim_{B\to\infty} C = \frac{S}{N_0}\log_2 e = 1.44\,\frac{S}{N_0}\ \text{bits/s}$$

So increasing bandwidth does not give unlimited capacity, because noise power also grows with $B$. Capacity saturates at $1.44\,S/N_0$.

Putting $S = E_b C$ at the limit gives $\dfrac{E_b}{N_0} = \ln 2 = 0.693$, i.e. −1.6 dB (the Shannon limit): no system can communicate error-free below this $E_b/N_0$.

Answer: $C_\infty = 1.44\,S/N_0$ bits/s.