Second Shifting Theorem for Fourier Transforms ?

[thrillhouse86]

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:
$$\mathcal{L}\{f(t-a)U(t-a) \} = e^{as}F(s)$$

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 ?

Thanks

 

[Ben Niehoff]

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 $s \rightarrow i\omega$. Also, for Fourier transforms, you don't need the step function.