Question

Show that if (12.2) is written with the factor $\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi }}}$multiplying each integral, then the corresponding form of Parseval’s (12.24) theorem is $\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{\left|}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right|}}^{\mathbf{2}}\mathit{d}\mathit{x}\mathbf{=}\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^2\mathit{d}\mathit{\alpha }$.

Short answer

Start from equation 12.20 and change the constant from $\frac{1}{2\pi }$to $\frac{1}{\sqrt{2\pi }}$. Then $\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$.

Step-by-step solution

Given Information.

The given function is $\overline{g_1}\left(\alpha \right)=\underset{-\infty }{\overset{\infty }{\int }}\overline{f_1}\left(x\right)e^{i\alpha x}dx$.It is to verify that the corresponding form of Parseval theorem is $\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$after multiplying each integral with$\frac{1}{\sqrt{2\pi }}$.

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

$\overline{g_1}\left(\alpha \right)=\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}\overline{f_1}\left(x\right)e^{i\alpha x}dx$

Multiply both sides by $g_2\left(\alpha \right)$and integrate over $\alpha$from $-\infty$to $\infty$.

$\begin{array}{c}\underset{-\infty }{\overset{\infty }{\int }}\overline{g_1}\left(\alpha \right)g_2\left(\alpha \right)d\alpha =\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}\left[\underset{-\infty }{\overset{\infty }{\int }}\overline{f_1}\left(x\right)e^{i\alpha x}dx\right]g_2\left(\alpha \right)d\alpha \\ =\underset{-\infty }{\overset{\infty }{\int }}{\overline{f}}_1\left(x\right)\left[\frac{1}{\sqrt{2\pi }}\underset{-\infty }{\overset{\infty }{\int }}{e}^{i\alpha x}{g}_2\left(\alpha \right)d\alpha \right]dx\\ =\underset{-\infty }{\overset{\infty }{\int }}\overline{f_1}\left(x\right)f_2\left(x\right)dx\end{array}$

If $f_1=f_2=f$and $g_1=g_2=g$then

$\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$

Thus, Start from equation 12.20 and change the constant from $\frac{1}{2\pi }$to localid="1664271226752" $\frac{1}{\sqrt{2\pi }}$. Then $\underset{-\infty }{\overset{\infty }{\int }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\underset{-\infty }{\overset{\infty }{\int }}{\left|f\left(x\right)\right|}^{2}dx$.