ISI in baseband systems; ideal and practical solutions

Intersymbol interference (ISI)

ISI is the spreading of each transmitted pulse beyond its own symbol interval because the channel's bandwidth is limited. Tails of earlier and later pulses then add to the sample of the current symbol, which can cause decision errors.

Derivation. The baseband PAM signal at the receiver filter output is

$$y(t) = \mu \sum_{k} a_k \, p(t - kT_b) + n(t)$$

where $a_k$ are the symbols, $p(t)$ is the overall pulse shape (normalised so $p(0) = 1$) and $n(t)$ is noise. Sampling at $t_i = iT_b$:

$$y(t_i) = \underbrace{\mu a_i}_{\text{wanted}} + \underbrace{\mu \sum_{k \ne i} a_k \, p\big((i-k)T_b\big)}_{\text{ISI}} + n(t_i)$$

The middle term is the ISI. It is zero only when $p(nT_b) = 0$ for all $n \ne 0$.

  pulse k-1     pulse k      pulse k+1
     /\           /\           /\
 ___/  \___...___/  \___...___/  \___
          tails overlap at sampling
          instants -> ISI

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 solution: Nyquist (sinc) pulse

The ideal solution uses an ideal rectangular spectrum of bandwidth $B_0 = R_b/2$:

$$P(f) = \frac{1}{2B_0}\,\text{rect}\!\left(\frac{f}{2B_0}\right) \;\Rightarrow\; p(t) = \text{sinc}(2B_0 t) = \frac{\sin 2\pi B_0 t}{2\pi B_0 t}$$

This pulse is zero at every $t = nT_b$ ($n \ne 0$), so ISI is zero. It needs the minimum bandwidth $R_b/2$ (the Nyquist bandwidth).

It is not practical because:

• the brick-wall filter cannot be built, and
• the sinc tails decay only as $1/t$, so a small timing error causes large ISI.

Practical solution: raised-cosine spectrum

The spectrum is given a gradual cosine roll-off from $f_1$ to $2B_0 - f_1$, with roll-off factor $\alpha = 1 - f_1/B_0$ ($0 \le \alpha \le 1$):

$$P(f) = \begin{cases} \dfrac{1}{2B_0}, & |f| < f_1 \\[4pt] \dfrac{1}{4B_0}\left[1 + \cos\dfrac{\pi(|f| - f_1)}{2B_0 - 2f_1}\right], & f_1 \le |f| < 2B_0 - f_1 \\[4pt] 0, & |f| \ge 2B_0 - f_1 \end{cases}$$ $$p(t) = \text{sinc}(2B_0 t)\,\frac{\cos 2\pi\alpha B_0 t}{1 - 16\alpha^2 B_0^2 t^2}, \qquad B_T = B_0(1 + \alpha) = \frac{R_b}{2}(1+\alpha)$$

• It still has zeros at all $nT_b$, so ISI is zero.
• The tails decay as $1/t^3$, so it tolerates timing error well.
• The filter can be built. The cost is extra bandwidth: $\alpha = 0.5$ needs $0.75R_b$, and $\alpha = 1$ needs $R_b$.

Other practical remedies are equalisers (e.g. a tapped-delay-line zero-forcing equaliser) to correct channel distortion, and correlative (duobinary) coding, which allows controlled ISI at the Nyquist bandwidth.