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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?


    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
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