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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

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# Second Shifting Theorem for Fourier Transforms ?

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