Energy of Sinc function - Signal Processing Stack Exchange most recent 30 from dsp.stackexchange.com 2019-11-18T16:56:12Z https://dsp.stackexchange.com/feeds/question/48962 https://creativecommons.org/licenses/by-sa/4.0/rdf https://dsp.stackexchange.com/q/48962 0 Energy of Sinc function user5045 https://dsp.stackexchange.com/users/5045 2018-05-03T10:26:34Z 2018-05-07T21:46:03Z <p>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?</p> https://dsp.stackexchange.com/questions/48962/-/48994#48994 1 Answer by Matt L. for Energy of Sinc function Matt L. https://dsp.stackexchange.com/users/4298 2018-05-04T19:32:02Z 2018-05-04T19:32:02Z <p>The energy of $x(t)$ is given by</p> <p>$$E_x=\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\frac{\sin^2(\pi t)}{(\pi t)^2}dt\tag{1}$$</p> <p>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 $t=0$:</p> <p>$$E_x=(x\star x)(0)=\int_{-\infty}^{\infty}\frac{\sin(\pi\tau)}{\pi\tau}\frac{\sin(\pi(t-\tau))}{\pi(t-\tau)}d\tau{\huge|}_{t=0}\tag{2}$$</p> <p>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 $x(t)$:</p> <p>$$(x\star x)(t)=x(t)\tag{3}$$</p> <p>And, consequently,</p> <p>$$E_x=x(0)=1\tag{4}$$</p> <p>(Note that the value of $x(t)$ at $t=0$ is defined as a limit.)</p> https://dsp.stackexchange.com/questions/48962/-/49049#49049 1 Answer by Atul Ingle for Energy of Sinc function Atul Ingle https://dsp.stackexchange.com/users/1825 2018-05-07T21:46:03Z 2018-05-07T21:46:03Z <p><a href="https://dsp.stackexchange.com/a/48994/1825">Matt L.'s answer</a> is great because it uses an insight from signal processing.</p> <p>Here's a purely "turn the crank" method that uses no signal processing:</p> <p>\begin{align} \int_{-\infty}^\infty \frac{\sin^2(\pi x)}{(\pi x)^2} dx &amp;= \frac{1}{\pi}\int_{-\infty}^\infty \frac{\sin^2 x}{x^2} dx \tag{1} \\ &amp;= \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}\\ &amp;= \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}\\ &amp;= 0 + \frac{1}{\pi} \int_{-\infty}^\infty \frac{\sin x}{x} dx \tag{4}\\ &amp;= \frac{1}{\pi} \cdot \pi \tag{5}\\ &amp;= 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 <a href="https://math.stackexchange.com/a/891822/36936">this math.SE answer</a>.</p>