Question

Verify Parseval’s theorem (12.24) for the special cases in Problems 31 to 33.

32. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$and $\mathit{g}\mathbf{\left(}\mathit{\alpha }\mathbf{\right)}$as in problem 21.

Short answer

$\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx=\frac{\sigma }{2\sqrt{\pi }}$.Thus the Parseval theorem is confirmed.

Step-by-step solution

Given Information.

The given functions are$f\left(x\right)=e^{-\frac{x^2}{2{\sigma }^2}}$and$g\left(\alpha \right)=\frac{\sigma }{\sqrt{2\pi }}{e}^{-\frac{{\sigma }^2{\alpha }^2}{2}}$.Parseval theorem is to be verified for this special case.

Definition of Parseval’s theorem

Parseval’s theorem is a theorem stating that the energy of a signal can be expressed as its frequency components’ average energy.

Verify Parseval Theorem

It is known that Gaussian integral is$\underset{-\infty }{\overset{\infty }{\int }}{e}^{-x^2}dx=\sqrt{\pi }$

Use the Gaussian integral to verify Parseval theorem

$\begin{array}{c}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{\frac{-x^2}{{\sigma }^2}}d\left(x\sigma \right)\sigma =\frac{\sigma }{2\pi }\sqrt{\pi }\\ =\frac{\sigma }{2\sqrt{\pi }}\end{array}$

And

$\begin{array}{c}\frac{{\sigma }^2}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{\sigma }^2{\alpha }^2}d\left(\alpha \sigma \right)\frac{1}{\sigma }=\frac{\sigma }{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{e}^{-x^2}dx\\ =\frac{\sigma }{2\pi }\sqrt{\pi }\\ =\frac{\sigma }{2\sqrt{\pi }}\end{array}$

$\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx=\frac{\sigma }{2\sqrt{\pi }}$.Thus the Parseval theorem is confirmed.