(New page: == Fourier transform == We are going to use the following: <math>X(\omega)=\frac{1}{(1+j\omega)(2+j\omega)}</math> == The inverse == <math>X(\omega)=\frac{1}{(1+j\omega)(2+j\omega)} = ...) |
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Revision as of 10:27, 8 October 2008
Fourier transform
We are going to use the following:
$ X(\omega)=\frac{1}{(1+j\omega)(2+j\omega)} $
The inverse
$ X(\omega)=\frac{1}{(1+j\omega)(2+j\omega)} = \frac{1}{1+j\omega} - \frac{1}{2+j\omega} $
$ f(t)= F^{-1}\frac{1}{1+j\omega} - F^{-1}\frac{1}{2+j\omega}\, $
$ = \begin{cases} 0, & t\leq 0 \\ e^{-t}+e^{-2t}, & t\geq 0 \end{cases} $
