ASN4 fixing

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Parseval's Theorem

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

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

and also that

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

Therefore

$(|s(t)|{)}^2=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}S(f)e^{j2\pi ft}S(f)e^{-j2\pi ft}dfdf$

Note that

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

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