Inverse fourier transform

**rebghb** · May 15th 2010, 01:25 PM

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!

**CaptainBlack** · May 15th 2010, 10:55 PM

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

**rebghb** · May 15th 2010, 11:19 PM

For any function continuous in a neighbourhood of

we have:

For any interval

containing

as an interior point.

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...

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