topsquark
Senior Member
- Joined
- Aug 27, 2012
- Messages
- 2,307
How is taking the Fourier transform of [imath]e^{i \omega \nu } F( \nu )[/imath] where [imath]F( \nu )[/imath] is the inverse Fourier transform of [imath]f( \omega )[/imath] violating anything? You simply put [imath]e^{i \omega \nu } F( \nu )[/imath] into the formula for the Fourier transform, just like you did on the line below. It's a substitution, not anything against the rules.Are you telling me to violate the definition and force it to include [imath]e^{i\omega \nu} F(\omega)[/imath]?
(Parentheses added for emphasis.)Or this?
[imath]\int_{-\infty}^{\infty} \left ( e^{i\omega \nu} F(\omega) \right ) e^{-i\omega x} \ d\omega[/imath]
If you want to learn something then try to do the integral above. It's not that hard if you apply the definition of a Fourier transform. It's a one-liner. (Hint: Combine the exponentials. What does this look like?)I am not telling you to give me the answer for no effort. I have already tried my best but failed. By giving me the answer, I will be able to solve any problem with shifting. If you don't want to help me, just say it. It seems that you are right, I should have not bothered you and searched in somewhere else.
But you keep demanding that we just give you the fish...Don't give me the fish. Teach me how to fish.
-Dan