Applied Mathematics/Parseval's Theorem

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Parseval's theorem[edit]

\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[edit]

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 eqaul 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