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Fourier Series

  1. Mar 5, 2007 #1
    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: Mar 5, 2007
  2. jcsd
  3. Mar 5, 2007 #2

    Dr Transport

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    So far so good.....
  4. Mar 5, 2007 #3
    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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