Applied Mathematics/Parseval's Theorem

Parseval's theorem

 * $$\int_{-\infty}^\infty | x(t) |^2 \, dt  =  \int_{-\infty}^\infty | X(f) |^2 \, df  $$

where $$X(f) = \mathcal{F} \{ x(t) \}$$ 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 $$\bar{X}(f)$$ be the complex conjugation of $$X(f)$$.


 * $$X(-f)=\int_{-\infty}^\infty x(-t) e^{-ift}$$
 * $$=\int_{-\infty}^\infty x(t) e^{ift}$$
 * $$=\bar{X}(f)$$


 * $$\int_{-\infty}^\infty | X(f) |^2 \, df $$

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
 * $$ \int_{-\infty}^\infty \bar{X}(f) X(f) \, df  $$
 * $$=\int_{-\infty}^\infty \left( \frac{1}{\sqrt 2 \pi }\int_{-\infty}^\infty x(t) e^{ift} dt \right) \left( \frac{1}{\sqrt 2 \pi}\int_{-\infty}^\infty x(t') e^{-ift'} dt' \right)df$$
 * $$=\int_{-\infty}^\infty x(t)x(t') \left(\frac{1}{ 2 \pi} e^{-if(t-t')} df \right)dtdt'$$
 * $$=\int_{-\infty}^\infty \int_{-\infty}^\infty x(t)x(t') \delta (t-t') dtdt'$$
 * $$=\int_{-\infty}^\infty |x(t)|^2 dt $$

Hence
 * $$\int_{-\infty}^\infty | x(t) |^2 \, dt  =   \int_{-\infty}^\infty | X(f) |^2 \, df  $$