3 Properties of the complex Fourier coefficients

Using properties of the trigonometric Fourier coefficients $a_n$ , $b_n$ we can readily deduce the following results for the $c_n$ coefficients:

1. $c_0=\frac{a_0}{2}$ is always real.
2. Suppose the periodic function $f\left(t\right)$ is even so that all $b_n$ are zero. Then, since in the complex form the $b_n$ arise as the imaginary part of $c_n$ , it follows that for $f\left(t\right)$ even the coefficients $c_n$ ( $n=\pm 1,\pm 2,\dots$ ) are wholly real.

Task!

If $f\left(t\right)$ is odd, what can you deduce about the Fourier coefficients $c_n$ ?

Answer

3. Since

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

   then if $f\left(t\right)$ is even, $c_n$ will be real, and we have two possible methods for evaluating $c_n$ :

   1. Evaluate the integral above as it stands i.e. over the full range $\left(-\frac{T}{2},\frac{T}{2}\right)$ . Note carefully that the second term in the integrand is neither an even nor an odd function so the integrand itself is

      $\left(\text{ even function}\right)\times \left(\text{ neither even nor odd function}\right)=\text{ neither even nor odd function.}$

      Thus we cannot write $c_n=\frac{2}{T}{\int \; <!--nolimits\; -->}_0^{T∕2}f\left(t\right)e^{-\text{ i}n{\omega }_{0}t}dt$

   2. Put $e^{-\text{ i}n{\omega }_{0}t}=cosn{\omega }_{0}t-\text{ i}sinn{\omega }_{0}t$ so $$\begin{array}{rcll}f\left(t\right)e^{-\text{ i}n{\omega }_{0}t}=f\left(t\right)cosn{\omega }_{0}t-\text{ i}f\left(t\right)sinn{\omega }_{0}t& =& \left(\text{ even}\right)\left(\text{ even}\right)-\text{ i}\left(\text{ even}\right)\left(\text{ odd}\right)& \\ & =& \left(\text{ even}\right)-\text{ i}\left(\text{ odd}\right).& \end{array}$$

      Hence    $c_n=\frac{2}{T}{\int \; <!--nolimits\; -->}_0^{\frac{T}{2}}f\left(t\right)cosn{\omega }_{0}tdt=\frac{a_n}{2}.$

4. If $f\left(t+\frac{T}{2}\right)=-f\left(t\right)$ then of course only odd harmonic coefficients $c_n\left(n=\pm 1,\pm 3,\pm 5,\dots \right)$ will arise in the complex Fourier series just as with trigonometric series.

Example 4

Find the complex Fourier series of the saw-tooth wave shown in Figure 24:

Figure 24

Solution

We have

$f\left(t\right)=\frac{At}{T}0<t<Tf\left(t+T\right)=f\left(t\right)$

The period is $T$ in this case so ${\omega }_0=\frac{2\pi }{T}.$

Looking at the graph of $f\left(t\right)$ we can say immediately

1. the Fourier series will contain a constant term $c_0$
2. if we imagine shifting the horizontal axis up to $\frac{A}{2}$ the signal can be written

   $f\left(t\right)=\frac{A}{2}+g\left(t\right),$ where $g\left(t\right)$ is an odd function with complex Fourier coefficients that are purely imaginary.

Hence we expect the required complex Fourier series of $f\left(t\right)$ to contain a constant term $\frac{A}{2}$ and complex exponential terms with purely imaginary coefficients. We have, from the general theory, and using $0<t<T$ as the basic period for integrating,

$c_n=\frac{1}{T}{\int \; <!--nolimits\; -->}_0^T\frac{At}{T}{e}^{-\text{ i}n{\omega }_{0}t}dt=\frac{A}{T^2}{\int \; <!--nolimits\; -->}_0^{T}t{e}^{-\text{ i}n{\omega }_{0}t}dt$

We can evaluate the integral using parts:

But ${\omega }_0=\frac{2\pi }{T}$ so

Hence the integral becomes

$\frac{T}{-\text{ i}n{\omega }_0}-\frac{1}{{\left(\text{ i}n{\omega }_0\right)}^2}\left(e^{-\text{ i}n{\omega }_{0}T}-1\right)$

Hence

$c_n=\frac{A}{T^2}\left(\frac{T}{-\text{ i}n{\omega }_0}\right)=\frac{\text{ i}A}{2\pi n}n=\pm 1,\pm 2,\dots$

Note that

$c_{-n}=\frac{\text{ i}A}{2\pi \left(-n\right)}=\frac{-\text{ i}A}{2\pi n}=c_n^{\ast }\text{ as it must}$

Also $c_0=\frac{1}{T}{\int \; <!--nolimits\; -->}_0^T\frac{At}{T}dt=\frac{A}{2}$ as expected.

Hence the required complex Fourier series is

$f\left(t\right)=\frac{A}{2}+\frac{\text{ i}A}{2\pi }{\sum }_{\genfrac{}{}{0ex}{}{n=-\infty }{n\ne 0}}^{\infty }\frac{e^{\text{ i}n{\omega }_{0}t}}{n}$

which could be written, showing only the constant and the first two harmonics, as

$f\left(t\right)=\frac{A}{2\pi }\left\{\dots -\text{ i}\frac{e^{-\text{ i}2{\omega }_{0}t}}{2}-\text{ i}{e}^{-\text{ i}{\omega }_{0}t}+\pi +\text{ i}{e}^{\text{ i}{\omega }_{0}t}+\text{ i}\frac{e^{\text{ i}2{\omega }_{0}t}}{2}+\dots \right\}.$

The corresponding trigonometric Fourier series for the function can be readily obtained from this complex series by combining the terms in $\pm n,$ $n=1,2,3,\dots$

For example this first harmonic is

Performing similar calculations on the other harmonics we obtain the trigonometric form of the Fourier series

$f\left(t\right)=\frac{A}{2}-\frac{A}{\pi }{\sum }_{n=1}^{\infty }\frac{sinn{\omega }_{0}t}{n}.$

Task!

Find the complex Fourier series of the periodic function:

$f\left(t\right)=e^t-\pi <t<\pi$

$f\left(t+2\pi \right)=f\left(t\right)$

Firstly write down an integral expression for the Fourier coefficients $c_n$ :

Answer

Now combine the real exponential and the complex exponential as one term and carry out the integration:

Answer

Answer