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Thread: Inverse Fourier transform

  1. #1
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    Inverse Fourier transform

    Hi there. I have some trouble with this. I have to find the inverse fourier transform for: $\displaystyle \frac{e^{i 6\omega}}{\omega}$
    So I'm using a table. I have that:
    $\displaystyle F(sg(t))=\frac{2}{i \omega}$ and $\displaystyle F(\delta(t-t_0))=e^{-i \omega t_0}$

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

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

    Well, everything fine till there, but when I tried to corroborate my result with mathematica I get:


    I don't know whats wrong.
    Last edited by Ulysses; Jun 11th 2011 at 02:36 PM.
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  2. #2
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    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 $\displaystyle f(t-t_0)$ is $\displaystyle F(\omega)e^{-i\omega t_0}$. Hence the inverse of $\displaystyle \frac{e^{6i\omega}}{\omega}$ is $\displaystyle \frac{i}{2}\text{sgn}(t+6)$.
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