Question

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

31. $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$as in figure 12.1. Hint: Integrate by parts and use (12.18) to evaluate$\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{\left|}\mathbf{g}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{\right|}}^{\mathbf{2}}\mathit{d}\mathit{\alpha }$.

Short answer

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

Step-by-step solution

Given Information.

The given function is$\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha$.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 the function and its Fourier transform are

$\begin{array}{l}f\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{sin\alpha cos\left(\alpha x\right)}{\alpha }d\alpha \\ g\left(\alpha \right)=\frac{sin\alpha }{\alpha \pi }\end{array}$

Thus,$\begin{array}{c}\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx=\frac{1}{2\pi }\underset{-1}{\overset{1}{\int }}dx\\ =\frac{1}{\pi }\end{array}$

And

$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{{\pi }^2}\underset{-\infty }{\overset{\infty }{\int }}\frac{si{n}^2\alpha }{{\alpha }^2}d\alpha \\ =\frac{1}{{\pi }^2}\underset{-\infty }{\overset{\infty }{\int }}\frac{d}{d\alpha }(-\frac{si{n}^2\alpha }{\alpha })d\alpha +\frac{1}{{\pi }^2}\underset{-\infty }{\overset{\infty }{\int }}\frac{d}{d\alpha }\left(\frac{2sin\alpha cos\alpha }{\alpha }\right)d\alpha \\ =0+\frac{4}{{\pi }^2}\underset{0}{\overset{\infty }{\int }}\frac{sin\alpha cos\alpha }{\alpha }d\alpha \end{array}$

It is known that $\underset{0}{\overset{\infty }{\int }}\frac{sin\alpha cos\alpha }{\alpha }d\alpha =\frac{\pi }{4}$

Therefore

$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{4}{{\pi }^2}\underset{0}{\overset{\infty }{\int }}\frac{sin\alpha cos\alpha }{\alpha }d\alpha \\ =\frac{4}{{\pi }^2}\frac{\pi }{4}\\ =\frac{1}{\pi }\end{array}$

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