(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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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:inverse Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of inverse Fourier transform (CT signals) == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
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== Fourier transform == | == Fourier transform == | ||
We are going to use the following: | We are going to use the following: | ||
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e^{-t}+e^{-2t}, & t\geq 0 | e^{-t}+e^{-2t}, & t\geq 0 | ||
\end{cases}</math> | \end{cases}</math> | ||
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| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:51, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
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} $
