Sinc Impulse

The preceding Fourier pair can be used to show that

$$\zbox {\lim_{\tau\to\infty} \tau\,\mbox{sinc}(f\tau) = \delta(f).}$$

Proof: The inverse Fourier transform of $\tau\,$sinc$(f\tau)$ is

$$\begin{eqnarray*} p_\tau(t) &=& \ensuremath{\int_{-\infty}^{\infty}}\tau\,\mbo... ...eq 1/2 \\ [5pt] 0, & \mbox{otherwise}. \\ \end{array} \right. \end{eqnarray*}$$

In particular, in the middle of the rectangular pulse at $t=0$, we have

$$p_\tau(0)=\ensuremath{\int_{-\infty}^{\infty}}\tau\,$$sinc$$(f\tau) df = 1, \quad \forall \tau>0.$$

This establishes that the algebraic area under $\tau\,$sinc$(\tau f)$ is 1 for every $\tau>0$. Every delta function (impulse) must have this property.

We now show that $\tau\,$sinc$(f\tau)$ also satisfies the sifting property in the limit as $\tau\to\infty$. This property fully establishes the limit as a valid impulse. That is, an impulse $\delta(t)$ is any function having the property that

$$\ensuremath{\int_{-\infty}^{\infty}}g(t)\delta(t)dt = \left<g,\delta\right> = g(0)$$

for every continuous function $g(t)$. In the present case, we need to show, specifically, that

$$\lim_{\tau\to\infty}\ensuremath{\int_{-\infty}^{\infty}}G(f)\tau\,$$sinc$$(\tau f)\,df = G(0).$$

Define $P_\tau(f)\isdef \tau\,$sinc$(f\tau)$. Then by the power theorem (§2.4.8),

$$\left<G,P_\tau\right> = \left<g,p_\tau\right> = \ensuremath{\int_{-\infty}^{\infty}}g(t) p_\tau(t)\,dt = \int_{-\tau/2}^{\tau/2} g(t)\,dt.$$

Then as $\tau\to\infty$, the limit converges to the algebraic area under $g$, which is $G(0)$ as desired:

$$\lim_{\tau\to\infty}\int_{-\tau/2}^{\tau/2} g(t)\,dt = \ensurema... ...int_{-\infty}^{\infty}}e^{-j\omega t} g(t)\,dt \right\vert _{\omega=0} = G(0).$$

We have thus established that

$${\lim_{\tau\to\infty}\tau\,\mbox{sinc}(f\tau) = \delta(f),}$$

where

   sinc$$(f)\isdef \frac{\sin(\pi f)}{\pi f}.$$

For related discussion, see [31, p. 127].