# Inverse Fourier transform

• Jun 11th 2011, 02:35 PM
Ulysses
Inverse Fourier transform
Hi there. I have some trouble with this. I have to find the inverse fourier transform for: $\frac{e^{i 6\omega}}{\omega}$
So I'm using a table. I have that:
$F(sg(t))=\frac{2}{i \omega}$ and $F(\delta(t-t_0))=e^{-i \omega t_0}$

Then $F^{-1}(e^{i6\omega})=\delta(t+6)$ and $F^{-1}\left (\frac{1}{\omega}\right )=F^{-1}\left ( \frac{i}{2}\frac{2}{i \omega}\right )=\frac{i}{2}sg(t)$

Finally using the properties for convolutions:
$F^{-1}\left ( \frac{e^{i 6\omega}}{\omega}\right )=F^{-1}\left ( e^{i 6\omega}\right ) * F^{-1}\left ( \frac{1}{\omega}\right )=2\pi\left[ \delta (t+6)*\frac{i}{2}sg(t)\right]$ where (*) represents the convolution.

Well, everything fine till there, but when I tried to corroborate my result with mathematica I get:
http://www.physicsforums.com/attachm...1&d=1307828102

I don't know whats wrong.
• Jun 14th 2011, 06:17 PM
ojones
Re: Inverse Fourier transform
You're output from Mathematica didn't display. Can you re-post it. In any event, I think there's an easier way for this: the Fourier transform of $f(t-t_0)$ is $F(\omega)e^{-i\omega t_0}$. Hence the inverse of $\frac{e^{6i\omega}}{\omega}$ is $\frac{i}{2}\text{sgn}(t+6)$.