Claude Shannon's evaluation of a $\operatorname{sinc}^2$ integral in his paper

Question

In Shannon's paper, "Communication in the Presence of noise [1], in section 3: Geometric Representation of the signals, Shannon presents an integral for a sinc based function.

Shannon represents a signal, limited by bandwidth $W$, as:

$$ f(t) = \sum_{n=1}^{2TW}x_n\frac{\sin\pi(2tW - n)}{\pi(2tW - n)}\ $$

Where $T$ is the duration of signal, $W$ is the bandwidth and $x_n$ is the amplitude of the $n$th sample.

He then evaluates the integral of $f(t)^2$:

$$ \int_{-\infty}^{\infty}f(t)^2dx = \frac{1}{2W}\sum_{}^{}x_n^2 $$

given that:

$$ \int_{-\infty}^{\infty}\frac{\sin\pi(2Wt-m)}{\pi(2Wt-m)}\frac{\sin\pi(2Wt-n)}{\pi(Wt-n)}dt =\begin{cases} 0 &m \neq n \\ \frac{1}{2W}&m=n \end{cases} $$

This last part here is where my confusion lies. Where did "$m$" come from? Where did the $2$ go in the denominator? I understand that sinc equals zero on intervals of $\pi$, but the piecewise part of this still confuses me.

[1] C.E. Shannon, Communication in the Presence of Noise, DOI: 10.1109/JRPROC.1949.232969

Answer 1

He is using different letters for the index, ie.

$$ f(t) = \sum_{n=1}^{2TW}x_n\frac{\sin\pi(2tW - n)}{\pi(2tW - n)}$$

$$ f(t) = \sum_{m=1}^{2TW}x_m\frac{\sin\pi(2tW - m)}{\pi(2tW - m)}$$

Then

$$\int_{-\infty}^{\infty}f^2(t) dt = \sum_{n=1}^{2TW}x_n \sum_{m=1}^{2TW}x_m \int_{-\infty}^{\infty}\frac{\sin\pi(2tW - n)}{\pi(2tW - n)} \frac{\sin\pi(2tW - m)}{\pi(2tW - m)} dt $$

If $m\neq n$ do the substitution $v = \pi(2tW-n)$ and use this result

$$\int_{-\infty}^{\infty}\frac{\sin\pi(2tW - n)}{\pi(2tW - n)} \frac{\sin\pi(2tW - m)}{\pi(2tW - m)} dt = \frac{1}{2\pi W} \int_{-\infty}^{\infty}\frac{\sin v}{v} \frac{\sin(v+\pi(n-m))}{(v+\pi(n-m))} dt = 0 $$

If $m=n$ do the same substitution:

$$ \int_{-\infty}^{\infty}\frac{\sin^2\pi(2tW - n)}{\left[\pi(2tW - n)\right]^2} dt = \frac{1}{2\pi W} \int_{-\infty}^{\infty} \frac{\sin^2 v}{v^2} dv = \frac{1}{\pi W} \int_{0}^{\infty} \frac{\sin v^2}{v^2}dv = \frac{1}{2W} $$

There are a lot of ways to evaluate this last integral, see here

Answer 2

$f(t)$ is a sum over $n$. For brevity let's write $f(t)=\sum_n x_nh_n(t)$. When you study $f(t)^2$, you are multiplying $f(t)$ and a copy of $f(t)$. To avoid confusion over reuse of the dummy index, you have to use a different index on the second copy: $$f(t)^2=\sum_n x_nh_n(t) \sum_m x_mh_m(t).$$ Multiplying out, this equals $$\sum_n \sum_m x_nx_m h_n(t)h_m(t).$$ Split the double sum into the case $m=n$ and the case $m\ne n$, obtaining $$\sum_n x_n^2 h_n^2(t) + \sum_{m\ne n}x_n x_m h_n(t) h_m(t).$$ Now integrate over $t$ and apply the identity ("given that"). (There is a $2$ missing from the denominator of the second factor in the integrand; looks like a misprint.)