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I'm not sure if I'm right though because when I checked it in matlab the answer I got was | I'm not sure if I'm right though because when I checked it in matlab the answer I got was | ||
| − | <pre> 4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w) < | + | <pre> 4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w) </pre> |
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[[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:32, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
$ \ x(t) = e^{-2|t|}cos(8t) $
$ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \! $
$ = \int_{-\infty}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \! $
$ = \int_{-\infty}^{0} e^{2|t|}cos(8t) e^{-j\omega t} dt \! + \int_{0}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \! $
after quite a bit of math I get the answer to be
$ \frac{1}{2}(\frac{1}{2 + j8 - jw} + \frac{1}{2 -j8 -jw} + \frac{1}{2 - j8 - jw} \frac{1}{2 + j8 + jw}) $
I'm not sure if I'm right though because when I checked it in matlab the answer I got was
4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w)
