ISI; Nyquist criteria; duobinary with example

ISI

Intersymbol interference (ISI) is the overlap of a pulse with the neighbouring symbol intervals. A band-limited channel (or filter) spreads each pulse in time, so the tails of nearby pulses add to the sample of the current symbol:

$$y(iT_b) = \mu a_i + \mu \sum_{k \ne i} a_k\, p\big((i-k)T_b\big) + n(iT_b)$$

The summation term is the ISI. It closes the eye diagram and raises the bit error rate.

Nyquist criterion for zero ISI

If $p(t)$ is the overall pulse (transmit filter, channel and receive filter), ISI is zero when

$$p(nT_b) = \begin{cases} 1, & n = 0 \\ 0, & n \ne 0 \end{cases}$$

In the frequency domain this is

$$\sum_{k=-\infty}^{\infty} P\!\left(f - \frac{k}{T_b}\right) = T_b \quad \text{(constant)}$$

That is, copies of the pulse spectrum shifted by multiples of $R_b = 1/T_b$ must add up to a flat spectrum.

• Ideal Nyquist pulse: $p(t) = \text{sinc}(R_b t)$ with bandwidth $B_0 = R_b/2$ (the minimum possible). It cannot be built, and its slow tails are sensitive to timing error.
• Raised-cosine pulse: bandwidth $B_T = \frac{R_b}{2}(1+\alpha)$, $0 \le \alpha \le 1$. It still gives zero ISI, its tails decay fast, and it can be built.

Duobinary encoding

Duobinary signalling (a correlative-level coding scheme) adds each binary symbol to the one before it. This brings in a controlled amount of ISI that the receiver knows and can remove, and in return the data can be sent at $R_b$ b/s in the minimum Nyquist bandwidth $R_b/2$ with realisable filters.

Encoder: with $a_k = \pm 1$,

$$c_k = a_k + a_{k-1}, \qquad c_k \in \{-2, 0, +2\}$$

This is equal to passing $a_k$ through $H(f) = 1 + e^{-j2\pi f T_b}$. With the ideal Nyquist filter, the overall response becomes

$$H_I(f) = 2\cos(\pi f T_b)\, e^{-j\pi f T_b}, \quad |f| \le \frac{1}{2T_b}$$

This is a smooth half-cosine, which is easy to approximate with a real filter.

 b_k->[Precoder]->[2d-1]-a_k-+---->(+)-c_k->[LPF]-> out
      d_k=b_k XOR d_k-1     |      ^
                            +-[Tb]-+

Decoder (without precoding): $\hat a_k = c_k - \hat a_{k-1}$. One wrong decision then spreads to the bits that follow (error propagation).

Precoding removes this. Use $d_k = b_k \oplus d_{k-1}$ and send $a_k = 2d_k - 1$. Then the receiver decides each bit on its own: $|c_k| = 0 \Rightarrow b_k = 1$, and $|c_k| = 2 \Rightarrow b_k = 0$.

Example: input $b_k$ = 1011001, with initial reference $d_{-1} = 0$ (so $a_{-1} = -1$).

The decoded sequence 1011001 equals the input. Each bit is decided on its own, so there is no error propagation.