The article introduces the exponential Fourier series by transforming the traditional trigonometric Fourier series into its exponential form using Euler’s formulas. It explains the derivation process, key formulas, and provides a solved example using a rectangular wave to demonstrate the application.
Replacing the sinusoidal terms in the trigonometric Fourier series by the exponential equivalents,
$\cos (n{{\omega }_{o}}t)=\frac{1}{2}({{e}^{jn{{\omega }_{o}}t}}+{{e}^{-jn{{\omega }_{o}}t}})$
And
$\sin (n{{\omega }_{o}}t)=\frac{1}{j2}({{e}^{jn{{\omega }_{o}}t}}-{{e}^{-jn{{\omega }_{o}}t}})$
Now, let us put the above exponential equivalents in the trigonometric Fourier series and get the Exponential Fourier Series expression:
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The trigonometric Fourier series can be represented as:
\[f(t)=\frac{{{a}_{o}}}{2}+\sum\limits_{n=1}^{\infty }{(}{{a}_{n}}\cos (n{{\omega }_{o}}t)+{{b}_{n}}\sin (n{{\omega }_{o}}t))\text{ }\cdots \text{ (1)}\]
Where
$\begin{matrix} {{a}_{n}}=\frac{2}{T}\int\limits_{{{t}_{o}}}^{{{t}_{o}}+T}{f(t)cos(n{{\omega }_{o}}t)}dt, & \text{n=0,1,2,}\cdots & {} \\ {} & {} & (2) \\ {{b}_{n}}=\frac{2}{T}\int\limits_{{{t}_{o}}}^{{{t}_{o}}+T}{f(t)\sin (n{{\omega }_{o}}t)}dt, & \text{n=1,2,3,}\cdots & {} \\\end{matrix}\text{ }$
Let us replace the sinusoidal terms in (1)
$f(t)=\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{\infty }{\frac{{{a}_{n}}}{2}({{e}^{jn{{\omega }_{o}}t}}+{{e}^{-jn{{\omega }_{o}}t}})+}\frac{{{b}_{n}}}{2}({{e}^{jn{{\omega }_{o}}t}}-{{e}^{-jn{{\omega }_{o}}t}})$
$f(t)=\frac{{{a}_{0}}}{2}+\sum\limits_{n=1}^{\infty }{\left[ \left( \frac{{{a}_{n}}}{2}-\frac{j{{b}_{n}}}{2} \right){{e}^{jn{{\omega }_{o}}t}}+\left( \frac{{{a}_{n}}}{2}+\frac{j{{b}_{n}}}{2} \right){{e}^{-jn{{\omega }_{o}}t}} \right]}\text{ }\cdots \text{ (3)}$
If we define a new coefficient cn by
${{c}_{n}}=\frac{{{a}_{n}}-j{{b}_{n}}}{2}$
And then substitute for an and bn from (2), with to=-T/2, we have
${{c}_{n}}=\frac{1}{2}\left[ \frac{2}{T}\int\limits_{-T/2}^{T/2}{f(t)(cos(n{{\omega }_{o}}t)dt}-\frac{j2}{T}\int\limits_{-T/2}^{T/2}{f(t)(sin(n{{\omega }_{o}}t)dt} \right]$
${{c}_{n}}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{f(t)(cos(n{{\omega }_{o}}t)-j\sin (n{{\omega }_{o}}t))}dt$
So, By Euler Formula
${{e}^{-j\theta }}=\cos \theta -j\sin \theta $
We can simply write,
${{c}_{n}}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{f(t){{e}^{-jn{{\omega }_{o}}t}}}dt\text{ }\cdots \text{ (4)}$
We also observe that the conjugate of cn is given by
$c_{n}^{*}=\frac{{{a}_{n}}+j{{b}_{n}}}{2}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{f(t)(cos(n{{\omega }_{o}}t)+j\sin (n{{\omega }_{o}}t))}dt$
Which is evidently c-n (cn with n replaced by -n). That is,
${{c}_{-n}}=\frac{{{a}_{n}}+j{{b}_{n}}}{2}\text{ }\cdots \text{ (5)}$
Finally, let us observe that
$\frac{{{a}_{0}}}{2}=\frac{1}{T}\int\limits_{-T/2}^{T/2}{f(t)}dt\text{ }$
Which by (4) is
$\frac{{{a}_{0}}}{2}={{c}_{0}}\text{ }\cdots \text{ (6)}$
Summing up, (4), (5) and (6) enable us to write (3) in the form
$\begin{align} & \text{f(t)=}{{c}_{0}}+\sum\limits_{n=1}^{\infty }{{{c}_{n}}}{{e}^{jn{{\omega }_{o}}t}}+\sum\limits_{n=1}^{\infty }{{{c}_{-n}}}{{e}^{-jn{{\omega }_{o}}t}} \\ & =\sum\limits_{n=0}^{\infty }{{{c}_{n}}}{{e}^{jn{{\omega }_{o}}t}}+\sum\limits_{n=-1}^{-\infty }{{{c}_{n}}}{{e}^{jn{{\omega }_{o}}t}} \\\end{align}$
We have combined co with the first summation and replaced the dummy summation index n by –n in the second summation. The result is more compactly written as
[stextbox id=”info” caption=”Exponential Fourier Series”]\[f(t)=\sum\limits_{n=-\infty }^{\infty }{{{c}_{n}}}{{e}^{jn{{\omega }_{o}}t}}\text{ }\cdots \text{ (7)}\][/stextbox]
Where cn is given by (4). This version of the Fourier series is called the exponential Fourier series and is generally easier to obtain because only one set of coefficients needs to be evaluated.
Example of Rectangular Wave
As an example, let us find the exponential series for the following rectangular wave, given by
$\begin{matrix} \begin{matrix} f(t)=4, \\ =-4, \\ f(t+2)=f(t) \\\end{matrix} & \begin{matrix} 0<t<1 \\ 1<t<2 \\ {} \\\end{matrix} \\\end{matrix}$
With T=2. We have ωo=2π/T= π, and thus by (4)
${{c}_{n}}=\frac{1}{2}\int\limits_{-1}^{1}{f(t){{e}^{-jn\pi t}}}dt\text{ }$
For n≠0 this is
${{c}_{n}}=\frac{1}{2}\int\limits_{-1}^{0}{(-4){{e}^{-jn\pi t}}}dt+\frac{1}{2}\int\limits_{0}^{1}{4{{e}^{-jn\pi t}}}dt\text{ =}\frac{4}{jn\pi }\left[ 1-{{(-1)}^{n}} \right]\text{ }$
Also, we have
\[\begin{align} & {{c}_{n}}=\frac{1}{2}\int\limits_{-1}^{1}{f(t)}dt \\ & =\frac{1}{2}\int\limits_{-1}^{0}{4}dt-\frac{1}{2}\int\limits_{0}^{1}{4}dt=0\text{ } \\\end{align}\]
Since cn=0 for n even and cn=8/jnπ for n odd, we may write the exponential series in the form
[stextbox id=”info”]\[f(t)=\frac{8}{j\pi }\sum\limits_{n=-\infty }^{\infty }{\frac{1}{2n-1}{{e}^{j(2n-1)\pi t}}}\][/stextbox]
This expression covers the function for both even and odd values.
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Exponential Fourier Series Key Takeaways
The exponential Fourier series offers a compact and efficient representation of periodic signals using complex exponentials instead of sines and cosines. This transformation simplifies mathematical analysis, especially in the context of signal processing, communications, and systems engineering. By consolidating the Fourier coefficients into a single complex term, it becomes easier to manipulate, differentiate, integrate, and analyze signals in both time and frequency domains. This representation is especially powerful in applications such as modulation, filtering, and spectral analysis, where working with exponential functions aligns naturally with the behavior of linear time-invariant (LTI) systems. The example of the rectangular wave illustrates how even complex-looking periodic signals can be broken down into manageable exponential components, making the exponential Fourier series an indispensable tool in engineering and applied mathematics.