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Second Shifting Theorem for Fourier Transforms ?

  1. Jun 15, 2010 #1

    I know from my the t shifting theorem that if I take the laplace transform of a function which is multiplied by a step function:
    \mathcal{L}\{f(t-a)U(t-a) \} = e^{as}F(s)

    Does this same rule apply for Fourier Transforms ? i.e.
    \mathcal{F}\{f(t-a)U(t-a) \} = e^{as}F(\omega)

    EDIT - a simple perusal of Fourier Transform tables has shown me that this is not the case, is there a different rule for Fourier Transforms ?

    Last edited: Jun 15, 2010
  2. jcsd
  3. Jun 15, 2010 #2

    Ben Niehoff

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    You can derive the appropriate rule easily by doing a change of variable in the Fourier integral. I think all you need to do is put [itex]s \rightarrow i\omega[/itex]. Also, for Fourier transforms, you don't need the step function.
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