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A Fourier transform

  1. May 23, 2008 #1
    Hi!

    I want to find the Fourier transform of

    [tex] \int_{-\infty}^t f(s-t)g(s) ds [/tex].

    The FT

    [tex] \int_{-\infty}^t h(s) ds \rightarrow H(\omega)/i\omega + \pi H(0) \delta(\omega)[/tex]

    is found in lots of textbooks. So if I let h(s) = f(s-t)g(s), I need to find the FT of h(s)

    [tex] H(\omega) = \int_{-\infty}^{\infty} h(s) \frac{e^{-i\omega s}}{\sqrt(2\pi)}ds = \int_{-\infty}^{\infty} f(s-t)g(s) \frac{e^{-i\omega s}}{\sqrt(2\pi)} ds [/tex].

    But there I'm stumped, what can I do? I can do FT of products like f(s)g(s), but this isn't exactly like that.
     
    Last edited: May 23, 2008
  2. jcsd
  3. May 23, 2008 #2

    Defennder

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    Homework Helper

    Did you try the Fourier transform of a convolution?
     
  4. May 23, 2008 #3

    rbj

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    yeah, but the OP needs to reverse the "time" argument in f(.).

    WarnK, where did you get that icky [itex]1/\sqrt{2 \pi}[/itex] definition for the F.T.?

    i really recommend this definition:

    [tex] X(f) = \int_{-\infty}^{+\infty} x(t) e^{-i 2 \pi f t} dt [/tex]

    with inverse

    [tex] x(t) = \int_{-\infty}^{+\infty} X(f) e^{i 2 \pi f t} df [/tex]

    while remembering that [itex] \omega \equiv 2 \pi f [/itex], and when comparing to the double-sided Laplace Transform to substitute [itex] i \omega \leftarrow s [/itex].

    it will make your life much easier.
     
  5. May 23, 2008 #4

    Defennder

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    I'm rather new to this myself. So the answer differs by a minus sign?
     
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