4 Parseval’s theorem

This is essentially a mathematical theorem but has, as we shall see, an important engineering interpretation particularly in electrical engineering. Parseval’s theorem states that if $f\left(t\right)$ is a periodic function with period $T$ and if $c_n$ $\left(n=0,\pm 1,\pm 2,\dots \right)$ denote the complex Fourier coefficients of $f\left(t\right)$ , then

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

In words the theorem states that the mean square value of the signal $f\left(t\right)$ over one period equals the sum of the squared magnitudes of all the complex Fourier coefficients.

4.1 Proof of Parseval’s theorem.

Assume $f\left(t\right)$ has a complex Fourier series of the usual form:

$f\left(t\right)={\sum }_{n=-\infty }^{\infty }{c}_n{e}^{\text{ i}n{\omega }_{0}t}\left({\omega }_0=\frac{2\pi }{T}\right)$

where

$c_n=\frac{1}{T}{\int \; <!--nolimits\; -->}_{-\frac{T}{2}}^{\frac{T}{2}}f\left(t\right)e^{-\text{ i}n{\omega }_{0}t}dt$

Then

$f^2\left(t\right)=f\left(t\right)f\left(t\right)=f\left(t\right)\sum c_n{e}^{\text{ i}n{\omega }_{0}t}=\sum c_{n}f\left(t\right)e^{\text{ i}n{\omega }_{0}t}$

Hence

which completes the proof.

Parseval’s theorem can also be written in terms of the Fourier coefficients $a_n,b_n$ of the trigonometric Fourier series. Recall that

$\begin{array}{c}{c}_0=\frac{a_0}{2}{c}_n=\frac{a_n-\text{ i}{b}_n}{2}n=1,2,3,\dots c_n=\frac{a_n+\text{ i}{b}_n}{2}n=-1,-2,-3,\dots \hfill \end{array}$

so

${\left|c_n\right|}^2=\frac{a_n^2+b_n^2}{4}n=\pm 1,\pm 2,\pm 3,\dots$

so

${\sum }_{n=-\infty }^{\infty }{\left|c_n\right|}^2=\frac{a_0^2}{4}+2{\sum }_{n=1}^{\infty }\frac{a_n^2+b_n^2}{4}$

and hence Parseval’s theorem becomes

$\frac{1}{T}{\int \; <!--nolimits\; -->}_{-\frac{T}{2}}^{\frac{T}{2}}{f}^2\left(t\right)dt=\frac{a_0^2}{4}+\frac{1}{2}{\sum }_{n=1}^{\infty }\left(a_n^2+b_n^2\right)$ (7)

The engineering interpretation of this theorem is as follows. Suppose $f\left(t\right)$ denotes an electrical signal (current or voltage), then from elementary circuit theory $f^2\left(t\right)$ is the instantaneous power (in a 1 ohm resistor) so that

$\frac{1}{T}{\int \; <!--nolimits\; -->}_{-\frac{T}{2}}^{\frac{T}{2}}{f}^2\left(t\right)dt$

is the energy dissipated in the resistor during one period.

Now a sinusoid wave of the form

$Acos\omega t\left(\text{ or}Asin\omega t\right)$

has a mean square value $\frac{A^2}{2}$ so a purely sinusoidal signal would dissipate a power $\frac{A^2}{2}$ in a 1 ohm resistor. Hence Parseval’s theorem in the form (7) states that the average power dissipated over 1 period equals the sum of the powers of the constant (or d.c.) components and of all the sinusoidal (or alternating) components.

Task!

The triangular signal shown below has trigonometric Fourier series

$f\left(t\right)=\frac{\pi }{2}-\frac{4}{\pi }{\sum }_{\genfrac{}{}{0ex}{}{n=1}{\left(\text{ odd}n\right)}}^{\infty }\frac{cosnt}{n^2}.$

[This was deduced in the Task in Section 23.3, page 39.]

Use Parseval’s theorem to show that ${\sum }_{\genfrac{}{}{0ex}{}{n=1}{\left(n\text{ odd}\right)}}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{96}.$

First, identify $a_0$ , $a_n$ and $b_n$ for this situation and write down the definition of $f\left(t\right)$ for this case:

Answer

Now evaluate the integral on the left hand side of Parseval’s theorem and hence complete the problem:

Answer

Exercises

Obtain the complex Fourier series for each of the following functions of period $2\pi$ .

1. $f\left(t\right)=t-\pi \le t\le \pi$
2. $f\left(t\right)=t0\le t\le 2\pi$
3. $f\left(t\right)={\text{e}}^t-\pi \le t\le \pi$

Answer