(New page: == Chosen Signal to Transform == The signal we will transform here will be <math>x(t)=e^{2jt}*(u(t+4)-u(t-4))</math> ==Transform by integral== <math> = \int_{-\infty}^{\infty}e^{2jt}*(u...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
| + | |||
== Chosen Signal to Transform == | == Chosen Signal to Transform == | ||
The signal we will transform here will be | The signal we will transform here will be | ||
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<math> = \frac{2sin(8 - 4\omega )}{2-\omega }\,</math> | <math> = \frac{2sin(8 - 4\omega )}{2-\omega }\,</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:35, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Chosen Signal to Transform
The signal we will transform here will be $ x(t)=e^{2jt}*(u(t+4)-u(t-4)) $
Transform by integral
$ = \int_{-\infty}^{\infty}e^{2jt}*(u(t+4)-u(t-4))e^{-j\omega t}dt\, $
$ = \int_{-4}^{4}e^{2jt}e^{-j\omega t}dt \, $
$ = \int_{-4}^{4}e^{2jt -j\omega t}dt\, $
$ = \int_{-4}^{4}e^{t*(2j -j\omega )}dt \, $
$ = \frac{e^{2jt - j\omega t}}{2j-j\omega}]_{-4}^{4} \, $
$ = \frac{e^{8j - 4j\omega} - e^{-8j + 4j\omega}}{2j-j\omega} \, $
$ = \frac{e^{j(8 - 4\omega )} - e^{-j(8 - 4\omega )}}{j(2-\omega )} \, $
$ = \frac{2sin(8 - 4\omega )}{2-\omega }\, $
