Shannon reconstruction formula: who can prove alternative form?

Question

I am familiar with the standard formulation of the Shannon-Whittaker reconstruction formula which is

$$s(t) = \sum\nolimits_{n = - \infty }^\infty {s(nT){\mathop{\rm sinc}\nolimits} (B(t - nT))}$$

where B = 1/T.

However, in the standard book "Simulation of Communication Systems, 2nd Ed" by Michel C. Jeruchim et al., the following formula is given for the reconstruction of $s(t-\tau)$, where $\tau$ is a delay parameter (real number) in Chapter 9, eq. 9.1.30:

$$s(t-\tau) = \sum\nolimits_{n = - \infty }^\infty {s(t-nT){\mathop{\rm sinc}\nolimits} (B(\tau - nT))}$$

How does s(.) in the summation become a function of $t$ (in addition to the sample timing offsets $nT$) and the $\mathop{\rm sinc(.)}$ devoid of the continuous variable $t$? So in essence, instead of samples of $s(t)$, samples of $\mathop{\rm sinc()}$ are being used to reconstruct the (delayed) signal ??

In the case $\tau$ = 0, the formula reduces to $$s(t) = \sum\nolimits_{n = - \infty }^\infty {s(t-nT){\mathop{\rm sinc}\nolimits} (B(-nT))} = \sum\nolimits_{n = - \infty }^\infty {s(t-nT){\mathop{\rm sinc}\nolimits} (B(nT))}$$ since $\mathop{\rm sinc()}$ is an even function.

Thanks for any help clarifying this!

Answer 1

First note that in the formula (9.1.30), $t$ is a fixed value parameter, whereas $\tau$ is the independent variable (taking arbitrary values). However, I will consider below the common notation where $t$ is the independent time variable and $\tau$ is a (fixed) delay. You can translate the results by simply switching the notation.

From the standard sampling theorem expansion, it holds

$$ \tag{1} s(t)=\sum_n s(nT) \text{sinc}(B(t-nT)) , t\in \mathbb{R}. $$

Now, consider another signal $y(t)=s(t+\tau)$, which is also bandlimited. Applying the sampling theorem to $y(t)$ gives a more general version of (1) as

$$ \tag{2} s(t+\tau)=\sum_n s(\tau+nT) \text{sinc}(B(t-nT)),t\in \mathbb{R}, \tau\in \mathbb{R}. $$

From (2), we have

$$ \begin{align} s(-t+\tau)&=\sum_n s(\tau+nT) \text{sinc}(B(-t-nT))\\ &\stackrel{(a)}=\sum_m s(\tau-mT) \text{sinc}(B(-t+mT))\\ &\stackrel{(b)}=\sum_m s(\tau-mT) \text{sinc}(B(t-mT)) \end{align} $$

where $(a)$ follows by changing the summation index variable to $m=-n$ and $(b)$ follows by noting that $\text{sinc}(\cdot)$ is a symmetric function, i.e., $\text{sinc}(x)=\text{sinc}(-x)$.

This is formula (9.1.30) with the notations $t$-$\tau$ exchanged.