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Fourier Series

  • Thread starter Mindscrape
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  • #1
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I am expanding the function [tex]f(t) = e^{i \omega t}[/tex] from (-π,π) as a complex fourier series where w is not an integer. I am stuck figuring out how the series expands with n.

[tex] c_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i \omega t} e^{-int} dt [/tex]

Join exponentials

[tex] c_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{i t(\omega - n) [/tex]

Then integrate the function and evaluate the limits

[tex] c_n = \frac{1}{2 \pi} \frac{1}{i(\omega - n)} \frac{dw}{dt} (e^{i \pi (\omega - n)} - e^{ -i \pi(\omega - n)}) [/tex]

Use the identity for sinx

[tex] c_n = \frac{1}{\pi} \frac{1}{(\omega - n)} \frac{dw}{dt} sin(\pi \omega - \pi n)[/tex]

This is the point when I am not quite sure if I am done, or I need to do something more. Should I just put this into back into the expression for the complex fourier series?

[tex] f(x) = \sum_{n=-\infty}^{n=+\infty} c_n e^{int}[/tex]
 
Last edited:

Answers and Replies

  • #2
Dr Transport
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So far so good.....
 
  • #3
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So then the function as a fourier series would just be:

[tex] f(x) = \sum_{n=-\infty}^{n=+\infty} \frac{1}{\pi} \frac{1}{(\omega - n)} \frac{dw}{dt} sin(\pi \omega - \pi n) e^{int} [/tex]
 

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