How to find energy of sinc function $x(t) = \frac{\sin \pi t}{\pi t}$ with out the help of Fourier transform or Parseval's theorem?

  • $\begingroup$ Take the Fourier Transform and integrate the magnitude squared. What problem are you encountering? $\endgroup$
    – user28715
    May 3, 2018 at 13:26
  • 2
    $\begingroup$ This has an answer on math.SE: math.stackexchange.com/questions/891812/… $\endgroup$
    – Atul Ingle
    May 3, 2018 at 15:33
  • 1
    $\begingroup$ @AtulIngle: That answer shows the computation of the integral of a sinc function, not of a squared sinc. $\endgroup$
    – Matt L.
    May 5, 2018 at 12:41

2 Answers 2


The energy of $x(t)$ is given by

$$E_x=\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\frac{\sin^2(\pi t)}{(\pi t)^2}dt\tag{1}$$

If we may assume that we know that $x(t)$ is the impulse response of an ideal low pass filter, the integral $(1)$ can be computed without using the Fourier transform and Parseval's theorem by noticing that it can be represented as a convolution evaluated at $t=0$:

$$E_x=(x\star x)(0)=\int_{-\infty}^{\infty}\frac{\sin(\pi\tau)}{\pi\tau}\frac{\sin(\pi(t-\tau))}{\pi(t-\tau)}d\tau{\huge|}_{t=0}\tag{2}$$

The convolution in $(2)$ can be seen as filtering an ideal low-pass signal with an ideal low-pass filter with the same cut-off frequency as the signal. Clearly, the filtering does not affect the signal, and the result of the convolution equals $x(t)$:

$$(x\star x)(t)=x(t)\tag{3}$$

And, consequently,


(Note that the value of $x(t)$ at $t=0$ is defined as a limit.)


Matt L.'s answer is great because it uses an insight from signal processing.

Here's a purely "turn the crank" method that uses no signal processing:

\begin{align} \int_{-\infty}^\infty \frac{\sin^2(\pi x)}{(\pi x)^2} dx &= \frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx \tag{1} \\ &= \frac{1}{\pi}\left[ \sin^2x \int \frac{1}{x^2} dx\right]_{-\infty}^\infty -\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{d(\sin^2 x)}{dx} \int \left(\frac{1}{x^2} dx\right) dx \tag{2}\\ &= \frac{1}{\pi}\left[ -\frac{\sin^2 x}{x}\right]_{-\infty}^\infty + \frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin 2x}{x} dx \tag{3}\\ &= 0 + \frac{1}{\pi} \int_{-\infty}^\infty \frac{\sin x}{x} dx \tag{4}\\ &= \frac{1}{\pi} \cdot \pi \tag{5}\\ &= 1 \end{align} where (1) is by a change of variable $\pi x$ to $x$, (2) is by integrating by parts, (4) is by a change of variable $2x$ to $x$ and finally (5) follows from this math.SE answer.


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