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 24a.

Short answer

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

Step-by-step solution

Given Information.

The given functions are$f\left(x\right)=e^{-\left|x\right|}$and$g\left(\alpha \right)=\frac{1}{\pi (1+{\alpha }^2)}$.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

$\begin{array}{c}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}{e}^{-2x}d(-2x)\frac{-1}{2}\\ =-\frac{1}{2\pi }{\left[e^{-2x}\right]}_0^{\infty }\\ =\frac{1}{2\pi }\end{array}$

And

$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{{\pi }^2}\underset{-\infty }{\overset{\infty }{\int }}\frac{d\alpha }{{(1+{\alpha }^2)}^2}\\ =\frac{1}{{\pi }^2}{[\frac{\alpha }{2(1+{\alpha }^2)}+\frac{1}{2}arctan\alpha ]}_{-\infty }^{\infty }\\ =\frac{1}{2{\pi }^2}(\frac{\pi }{2}-(-\frac{\pi }{2}))\\ =\frac{1}{2\pi }\end{array}$

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