Fourier Example: Difference between revisions

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(New page: Find the Fourier Series of the function: <br /> <center><math>\f(x)=begin{cases} 0,-pi<x<0 \end{cases} \ \ \ \ n=0,1,2,3\dots</math></center> <br /> :<math>\rho_X(x) = \begin{cases}...)
 
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Find the Fourier Series of the function:
Find the Fourier Series of the function:


<br />


<center><math>\f(x)=begin{cases}
:<math>f(x) = \begin{cases}0,& -\pi\le x<0\\
\pi,& 0 \le x \le \pi\\
\end{cases}</math>


0,-pi<x<0


\end{cases} \ \ \ \ n=0,1,2,3\dots</math></center>


<br />


== Solution ==
:<math>\rho_X(x) = \begin{cases}\frac{1}{2},& \x=0,\\

\frac{1}{2},& \x=1,\\

0,& \text{otherwise} .\end{cases}</math>

<!-- -->
Here we have




<math>a_o=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math>



<math>a_n=\int_{0}^\pi\pi cos(nx)\ dx=0, n\ge1,</math>


and


:<math>b_n = \int_{0}^\pi \pi\sin(nx)\, dx = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math>


We obtain <math>b_{2n}</math> = 0 and


<math>b_{2n+1}=\frac{2}{2n+1}</math>


Therefore, the Fourier series of f(x) is

<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>


==Solution Graph==
[[Image:Fourier.gif]]


==***BONUS*** VIDEO! (For people completely lost on Fourier Series)==


http://www.youtube.com/watch?v=nXEqrOt-nB8


==References:==
[[Fourier Series: Basic Results]]
==Readers==
[[Lau, Chris|Christopher Garrison Lau I]]

Latest revision as of 15:26, 25 January 2010

Find the Fourier Series of the function:




Solution

Here we have





and



We obtain = 0 and



Therefore, the Fourier series of f(x) is


Solution Graph

Fourier.gif


***BONUS*** VIDEO! (For people completely lost on Fourier Series)

http://www.youtube.com/watch?v=nXEqrOt-nB8


References:

Fourier Series: Basic Results

Readers

Christopher Garrison Lau I