# Inverse fourier transform

• May 15th 2010, 02:25 PM
rebghb
Inverse fourier transform
Hello everyone!

I'm trying to IFT this fraction after I couldn't find it in a Fourier couples table.

$F(j\omega)=\frac{250\pi \delta(\omega)}{30+125\omega}$.
When I integrated the latter, I got: $\frac{1}{2\pi}\int_{0-}^{0+} \frac{250\pi \delta(\omega)}{30+125\omega}e^{-j\omega t}d\omega$ $\ =\frac{1}{2\pi}\int_{0-}^{0+} \frac{250\pi \delta(\omega)}{30}d\omega$ $=\frac{25}{6} u(t)$??

Is there ever a $u(t)$ in IFT ??
Thanks for any help!
• May 15th 2010, 11:55 PM
CaptainBlack
Quote:

Originally Posted by rebghb
Hello everyone!

I'm trying to IFT this fraction after I couldn't find it in a Fourier couples table.

$F(j\omega)=\frac{250\pi \delta(\omega)}{30+125\omega}$.
When I integrated the latter, I got: $\frac{1}{2\pi}\int_{0-}^{0+} \frac{250\pi \delta(\omega)}{30+125\omega}e^{-j\omega t}d\omega$ $\ =\frac{1}{2\pi}\int_{0-}^{0+} \frac{250\pi \delta(\omega)}{30}d\omega$ $=\frac{25}{6} u(t)$??

Is there ever a $u(t)$ in IFT ??
Thanks for any help!

For any function continuous in a neighbourhood of $0$ we have:

$\int_A f(\omega) \delta(\omega)\;d\omega=f(0)$

For any interval $A$ containing $0$ as an interior point.

CB
• May 16th 2010, 12:19 AM
rebghb
Isn't it that $\int_A f(\omega)\delta(\omega) d\omega$ $=\int_A f(0) \delta(\omega) d\omega = f(0)u(t)$
At least that's what I know from circuit analysis...