Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to get fourier transform from fourier series

  1. Mar 31, 2012 #1
    How you get Fourier transform from Fourier series? Do Fourier series becomes Fourier transform as L --> infinity?

    http://mathworld.wolfram.com/FourierTransform.html

    I don't understand where discrete A sub n becomes continuous F(k)dk ( where F(k) is exactly like A sub n in fourier series)?
    I also have a general question which is;
    How to transform a discrete variable to a continuous variable in order to convert a summation to integral?
     
  2. jcsd
  3. Apr 4, 2012 #2
    I think I know the answer now.

    " For a function periodic in [-L/2,L/2], Fourier series is
    [tex] f(x) = \sum_{n=-\infty}^{\infty }A_{n}e^{i(2\pi nx{/}L)}

    \\A_{n} = 1{/}L \int_{-L{/}2}^{L{/}2}f(x)e^{-i(2\pi nx{/}L)}dx. [/tex]"

    The part that was bothering me was " The Fourier transform is a generalization of the complex Fourier series in the limit as L->infty. Replace the discrete [tex]A_{n} [/tex] with the continuous F(k)dk while letting n/L->k. Then change the sum to an integral, and the equations become
    [tex] f(x) = \int_{-\infty}^{\infty} F(k)e^{2\pi ikx}dk

    \\ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi ikx}dx [/tex] "
    My question was how [tex]A_{n}[/tex] becomes [tex]F(k) d(k) [/tex]. Especially where the dk comes from?
    [tex] k=n{/}L [/tex]; then [tex]\Delta k = (n+1){/}L -n{/}L =1{/}L [/tex] , when [tex]L [/tex] goes to infinity [tex]\Delta k [/tex]becomes dk.
    So when [tex] L \rightarrow \infty; A_{n} =F(k)dk [/tex]. Indeed! I am a happy man now :).
     
    Last edited: Apr 4, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to get fourier transform from fourier series
  1. Fourier transform (Replies: 1)

Loading...