91Ó°ÊÓ

The given function is $\stackrel{¯}{g_1}\left(\mathrm{Î\pm }\right)=\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\stackrel{¯}{f_1}\left(x\right)e^{i\mathrm{Î\pm }x}dx$.It is to verify that the corresponding form of Parseval theorem is $\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|g\left(\mathrm{Î\pm }\right)\right|}^{2}d\mathrm{Î\pm }=\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|f\left(x\right)\right|}^{2}dx$after multiplying each integral with$\frac{1}{\sqrt{2\mathrm{Ï€}}}$.

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

$\stackrel{¯}{g_1}\left(\mathrm{Î\pm }\right)=\frac{1}{\sqrt{2\mathrm{Ï€}}}\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\stackrel{¯}{f_1}\left(x\right)e^{i\mathrm{Î\pm }x}dx$

Multiply both sides by $g_2\left(\mathrm{Î\pm }\right)$and integrate over $\mathrm{Î\pm }$from $-\mathrm{∞}$to $\mathrm{∞}$.

$\begin{array}{c}\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\stackrel{¯}{g_1}\left(\mathrm{Î\pm }\right)g_2\left(\mathrm{Î\pm }\right)d\mathrm{Î\pm }=\frac{1}{\sqrt{2\mathrm{Ï€}}}\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\left[\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\stackrel{¯}{f_1}\left(x\right)e^{i\mathrm{Î\pm }x}dx\right]g_2\left(\mathrm{Î\pm }\right)d\mathrm{Î\pm }\\ =\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\stackrel{¯}{f}}_1\left(x\right)\left[\frac{1}{\sqrt{2\mathrm{Ï€}}}\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{e}^{i\mathrm{Î\pm }x}{g}_2\left(\mathrm{Î\pm }\right)d\mathrm{Î\pm }\right]dx\\ =\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\stackrel{¯}{f_1}\left(x\right)f_2\left(x\right)dx\end{array}$

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

$\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|g\left(\mathrm{Î\pm }\right)\right|}^{2}d\mathrm{Î\pm }=\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|f\left(x\right)\right|}^{2}dx$

Thus, Start from equation 12.20 and change the constant from $\frac{1}{2\mathrm{Ï€}}$to localid="1664271226752" $\frac{1}{\sqrt{2\mathrm{Ï€}}}$. Then $\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|g\left(\mathrm{Î\pm }\right)\right|}^{2}d\mathrm{Î\pm }=\underset{-\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}{\left|f\left(x\right)\right|}^{2}dx$.