Second Shifting Theorem for Fourier Transforms ?

  • #1
Hi,

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:
[tex]
\mathcal{L}\{f(t-a)U(t-a) \} = e^{as}F(s)
[/tex]

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

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 ?

Thanks
 
Last edited:

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
Gold Member
1,883
162
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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