5 Fourier series for functions of general period

This is a straightforward extension of the period $2\pi$ case that we have already discussed.

Using $x$ (instead of $t$ ) temporarily as the variable. We have seen that a $2\pi$ periodic function $f\left(x\right)$ has a Fourier series

$f\left(x\right)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left(a_{n}cosnx+b_{n}sinnx\right)$

with

$a_n=\frac{1}{\pi }{\int \; <!--nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)cosnxdxn=0,1,2,\dots b_n=\frac{1}{\pi }{\int \; <!--nolimits\; -->}_{-\pi }^{\pi }f\left(x\right)sinnxdxn=1,2,\dots$

Suppose we now change the variable to $t$ where $x=\frac{2\pi }{T}t.$

Thus $x=\pi$  corresponds to   $t=T∕2$ and    $x=-\pi$ $$  corresponds to $t=-T∕2.$

Hence regarded as a function of $t$ , we have a function with period $T$ .

Making the substitution $x=\frac{2\pi }{T}t$ , and hence $dx=\frac{2\pi }{T}dt$ , in the expressions for $a_n$ and $b_n$ we obtain

These integrals give the Fourier coefficients for a function of period $T$ whose Fourier series is

$f\left(t\right)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left\{a_{n}cos\left(\frac{2n\pi t}{T}\right)+b_{n}sin\left(\frac{2n\pi t}{T}\right)\right\}$

Various other notations are commonly used in this case e.g. it is sometimes convenient to write the period $T=2\ell$ . (This is particularly useful when Fourier series arise in the solution of partial differential equations.) Another alternative is to use the angular frequency $\omega$ and put $T=2\pi ∕\omega$ .

Task!

Write down the form of the Fourier series and expressions for the coefficients if  

1. $T=2\ell$
2. $T=2\pi ∕\omega$ .

Answer

You should note that, as usual, any convenient integration range of length $T$ (or $2\ell$ or $\frac{2\pi }{\omega }$ ) can be used in evaluating $a_n$ and $b_n$ .

Example 1

Find the Fourier series of the function shown in Figure 11 which is a saw-tooth wave with alternate portions removed.

Figure 11

Solution

Here the period $T=2\ell =4$ so $\ell =2$ . The Fourier series will have the form

$f\left(t\right)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left\{a_{n}cos\left(\frac{n\pi t}{2}\right)+b_{n}sin\left(\frac{n\pi t}{2}\right)\right\}$

The coefficients $a_n$ are given by

$a_n=\frac{1}{2}{\int \; <!--nolimits\; -->}_{-2}^{2}f\left(t\right)cos\left(\frac{n\pi t}{2}\right)dt$

where

$f\left(t\right)=\left\{\begin{array}{ccc}\hfill 0& \hfill & \hfill -2<t<0\\ \hfill t& \hfill & \hfill 0<t<2\end{array}\right.$ $f\left(t+4\right)=f\left(t\right)$

Hence $a_n=\frac{1}{2}{\int \; <!--nolimits\; -->}_0^{2}tcos\left(\frac{n\pi t}{2}\right)dt.$

The integration is readily performed using integration by parts:

Hence, since $a_n=\frac{1}{2}{\int \; <!--nolimits\; -->}_0^{2}tcos\left(\frac{n\pi t}{2}\right)dt$

$a_n=\left\{\begin{array}{ccc}\hfill 0\hfill & \hfill \hfill & \hfill n=2,4,6,\dots \hfill \\ \hfill \hfill \\ \hfill -\frac{4}{n^2{\pi }^2}\hfill & \hfill \hfill & \hfill n=1,3,5,\dots \hfill \end{array}\right.$

The constant term is $\frac{a_0}{2}$ where $a_0=\frac{1}{2}{\int \; <!--nolimits\; -->}_0^{2}tdt=1.$

Similarly

$b_n=\frac{1}{2}{\int \; <!--nolimits\; -->}_0^{2}tsin\left(\frac{n\pi t}{2}\right)dt$

where

${\int \; <!--nolimits\; -->}_0^{2}tsin\left(\frac{n\pi t}{2}\right)dt={\left[-t\frac{2}{n\pi }cos\left(\frac{n\pi t}{2}\right)\right]}_0^2+\frac{2}{n\pi }{\int \; <!--nolimits\; -->}_0^{2}cos\left(\frac{n\pi t}{2}\right)dt.$

The second integral gives zero. Hence

$b_n=-\frac{2}{n\pi }cosn\pi =\left\{\begin{array}{ccc}\hfill -\frac{2}{n\pi }\hfill & \hfill \hfill & \hfill n=2,4,6,\dots \hfill \\ \hfill \hfill \\ \hfill +\frac{2}{n\pi }\hfill & \hfill \hfill & \hfill n=1,3,5,\dots \hfill \end{array}\right.$

Hence, using all these results for the Fourier coefficients, the required Fourier series is

Notice that because the Fourier coefficients depend on $\frac{1}{n^2}$ (rather than $\frac{1}{n}$ as was the case for the square wave) the sinusoidal components in the Fourier series have quite rapidly decreasing amplitudes. We would therefore expect to be able to approximate the original saw-tooth function using only a quite small number of terms in the series.

Task!

Obtain the Fourier series of the function

First write out the form of the Fourier series in this case:

Answer

Now write out integral expressions for $a_n$ and $b_n$ . Will there be a constant term in the Fourier series?

Answer

Answer

Exercises

For each of the following periodic signals

• sketch the given function over a few periods
• find the trigonometric Fourier coefficients
• write out the first few terms of the Fourier series.

1. $f\left(t\right)=\left\{\begin{array}{cc}\hfill 1\hfill & \hfill 0<t<\pi ∕2\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill \pi ∕2<t<2\pi \hfill \\ \hfill \hfill \end{array}\right.f\left(t+2\pi \right)=f\left(t\right)\text{ square wave}$
2. $f\left(t\right)=t^2-1<t<1f\left(t+2\right)=f\left(t\right)$
3. $f\left(t\right)=\left\{\begin{array}{cc}\hfill -1\hfill & \hfill -T∕2<t<0\hfill \\ \hfill \hfill \\ \hfill 1\hfill & \hfill 0<t<T∕2\hfill \\ \hfill \hfill \end{array}\right.f\left(t+T\right)=f\left(t\right)\text{ square wave}$
4. $f\left(t\right)=\left\{\begin{array}{cc}\hfill 0\hfill & \hfill -\pi <t<0\hfill \\ \hfill \hfill \\ \hfill t^2\hfill & \hfill 0<t<\pi \hfill \\ \hfill \hfill \end{array}\right.f\left(t+2\pi \right)=f\left(t\right)$
5. $f\left(t\right)=\left\{\begin{array}{cc}\hfill 0\hfill & \hfill -T∕2<t<0\hfill \\ \hfill \hfill \\ \hfill A\text{ sin}\frac{2\pi t}{T}\hfill & \hfill 0<t<T∕2\hfill \\ \hfill \hfill \end{array}\right.f\left(t+T\right)=f\left(t\right)$  half-wave rectifier

Answer