Applied Mathematics/Parseval's Theorem

Parseval's theorem

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|x(t){|}^{2}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|X(f){|}^{2}df$

where $X(f)=ℱ\{x(t)\}$ represents the continuous Fourier transform of x(t) and f represents the frequency component of x. The function above is called Parseval's theorem.

Derivation

Let $\overline{X}(f)$ be the complex conjugation of $X(f)$.

$X(-f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(-t)e^{-ift}$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{ift}$

$=\overline{X}(f)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|X(f){|}^{2}df$

Here, we know that $X(f)$ is equal to the expansion coefficient of $x(t)$ in fourier transforming of $x(t)$.
Hence, the integral of $|X(f){|}^2$ is

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\overline{X}(f)X(f)df$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{\sqrt{2}\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{ift}dt)(\frac{1}{\sqrt{2}\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t^{\prime })e^{-if{t}^{\prime }}d{t}^{\prime })df$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)x(t^{\prime })\left(\frac{1}{2\pi }{e}^{-if(t-t^{\prime })}df\right)dtd{t}^{\prime }$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)x(t^{\prime })\delta (t-t^{\prime })dtd{t}^{\prime }$

$=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|x(t){|}^{2}dt$

Hence

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|x(t){|}^{2}dt=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|X(f){|}^{2}df$