Exponential Fourier Series with Solved Example

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 c_n by

${{c}_{n}}=\frac{{{a}_{n}}-j{{b}_{n}}}{2}$

And then substitute for a_n and b_n from (2), with t_o=-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 c_n 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 (c_n 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 c_o 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 c_n 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 ω_o=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 c_n=0 for n even and c_n=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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