How to use the iircomb filter in matlab
Precision delay is, in my opinion, best understood in terms of the Nyquist-Shannon sampling and reconstruction theorem.

If a continuous-time (a.k.a. "analog", but that is somewhat imprecise) signal $x(t)$ is bandlimited to below the Nyquist frequency $\frac{1}{2T}$ (where $T$ is the sampling period and $f_\mathrm{s} \triangleq \frac1T$ is the sample rate), then the original continuous-time signal $x(t)$ can be reconstructed from the discrete samples $x[n]$ as:

$$\begin{align}
x(t) &amp;= \sum_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\left(\tfrac{t - nT}{T}\right) \\
&amp;= \sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}\left(\tfrac{t - nT}{T}\right) \\
\end{align}$$

where

$$ \operatorname{sinc}(u) \triangleq \begin{cases} 
\frac{\sin(\pi u)}{\pi u},  &amp; \text{if } u \ne 0 \\
1, &amp; \text{if } u = 0 \\
\end{cases} $$

All terms of the summation are bandlimited to a maximum frequency of $\frac{1}{2T}$, so the summation is bandlimited to the same bandlimit as the original $x(t)$.  The samples are

$$ x[n] \triangleq x(t) \Bigg|_{t = nT} \triangleq x(nT)$$

Because $x(t)$ can be evaluated at any $t$, not only multiples of $T$, this is the basis in which we interpolate to get a precision delay.

Suppose we wanna delay by an amount of time $\tau$, then the output of the delay is

$$\begin{align}
y(t) &amp;= x(t-\tau) \\
&amp;= \sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}\left(\tfrac{t-\tau - nT}{T}\right) \\
\end{align}$$

So that is explicitly where you get the delayed signal of $y(t) = x(t-\tau)$ from the samples $x[n]$.  The first problem we have is that we cannot add up an infinite number of terms.  So our first practical approximation is truncating terms.  But that is the same as applying a rectangular window (usually considered to be the worst one), so instead, we'll apply a better window.

$$\begin{align}
y(t) &amp;= x(t-\tau) \\
&amp;= \sum_{n=-\infty}^{\infty} x[n] \, \operatorname{sinc}\left(\tfrac{t-\tau - nT}{T}\right) \\
&amp;= \sum_{n=-\infty}^{\infty} x[n-n_\tau] \, \operatorname{sinc}\left(\tfrac{t-\tau - (n-n_\tau)T}{T}\right) \\
&amp;= \sum_{n=-\infty}^{\infty} x[n-n_\tau] \, \operatorname{sinc}\left(\tfrac{t-(\tau-n_\tau T)}{T}-n\right) \\
\end{align}$$

... for any integer $n_\tau$.  Let's choose it as the integer part of the dimensionless $\frac{\tau}{T}$, which is the delay in sample units.

$$ n_\tau \triangleq \left\lfloor \frac{\tau}{T} \right\rfloor $$

The fractional part of the delay (in sample units) is

$$ \frac{\tau}{T} - \left\lfloor \frac{\tau}{T} \right\rfloor $$

(i gotta return to this a little bit later.)