1 Parseval’s theorem

Recall from HELM booklet  23.2 on Fourier series that for a periodic signal $f_T\left(t\right)$ with complex Fourier coefficients $c_n\left(n=0,\pm 1,\pm 2,\dots \right)$ Parseval’s theorem holds:

$\frac{1}{T}{\int \; <!--nolimits\; -->}_{-\frac{T}{2}}^{+\frac{T}{2}}{f}_T^2\left(t\right)dt={\sum }_{n=-\infty }^{\infty }{\left|c_n\right|}^2,$

where the left-hand side is the mean square value of the function (signal) over one period.

For a non-periodic real signal $f\left(t\right)$ with Fourier transform $F\left(\omega \right)$ the corresponding result is

${\int \; <!--nolimits\; -->}_{-\infty }^{\infty }{f}^2\left(t\right)dt=\frac{1}{2\pi }{\int \; <!--nolimits\; -->}_{-\infty }^{\infty }{\left|F\left(\omega \right)\right|}^{2}d\omega .$

This result is particularly significant in filter theory. For reasons that we do not have space to go into, the left-hand side integral is often referred to as the total energy of the signal. The integrand on the right-hand side

$\frac{1}{2\pi }{\left|F\left(\omega \right)\right|}^2$

is then referred to as the energy density (because it is the frequency domain quantity that has to be integrated to obtain the total energy).

Task!

Verify Parseval’s theorem using the one-sided exponential function

$f\left(t\right)=e^{-t}u\left(t\right).$

Firstly evaluate the integral on the left-hand side:

Answer

Now obtain the Fourier transform $F\left(\omega \right)$ and evaluate the right-hand side integral:

Answer