ASN4 fixing

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Parseval's Theorem

Parseval's Theorem says that $\int _{-\infty }^{\infty }(|s(t)|)^{2}dt$ in time transforms to $\int _{-\infty }^{\infty }(|S(f)|)^{2}df$ in frequency

Note that $(|s(t)|)^{2}=s(t)^{.}s^{*}(t)\!$

and also that

$s(t)=F^{-1}[S(f)]=\int _{-\infty }^{\infty }S(f)e^{j2\pi ft}df\!$

Therefore

$(|s(t)|)^{2}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }S(f)e^{j2\pi ft}S(f)e^{-j2\pi f't}dfdf^{'}\!$

$s(t)e^{-j2\pi ft}e^{j2\pi ft}s(t)e^{-j2\pi ft}e^{-j2\pi f't}dfdf^{'}\!$

and

$\int _{-\infty }^{\infty }(|s(t)|)^{2}dt=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }S(f)e^{j2\pi ft}S(f)e^{-j2\pi f't}dfdf^{'}dt$

$|s(t)|=F^{-1}[S(f)]=|\int _{-\infty }^{\infty }S(f)e^{j2\pi ft}df|$

Note that $|e^{j2\pi ft}|={\sqrt {cos^{2}(2\pi ft)+sin^{2}(2\pi ft)}}=1$

The above equation of $|s(t)|$ simplifies to then $|s(t)|=\int _{-\infty }^{\infty }S(f)df=|S(f)|$

Therefore,squaring the function and intergrating it in the time domain $\int _{-\infty }^{\infty }(|s(t)|)^{2}dt$ is to do the same in the frequency domain $\int _{-\infty }^{\infty }(|S(f)|)^{2}df$