- #1
thrillhouse86
- 80
- 0
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
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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