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Exponential Fourier Series with Solved Example

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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:

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

 

Exponential Fourier Series
\[f(t)=\sum\limits_{n=-\infty }^{\infty }{{{c}_{n}}}{{e}^{jn{{\omega }_{o}}t}}\text{      }\cdots \text{    (7)}\]

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

 

\[f(t)=\frac{8}{j\pi }\sum\limits_{n=-\infty }^{\infty }{\frac{1}{2n-1}{{e}^{j(2n-1)\pi t}}}\]

This expression covers the function for both even and odd values.

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About Ahmad Faizan

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.

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