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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]] | ||
| + | ---- | ||
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<math>\ x(t) = e^{-2|t|}cos(8t)</math> | <math>\ x(t) = e^{-2|t|}cos(8t)</math> | ||
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| + | <math>X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt \!</math> | ||
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| + | <math> = \int_{-\infty}^{\infty} e^{-2|t|}cos(8t) e^{-j\omega t} dt \!</math> | ||
| + | |||
| + | <math> = \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 \!</math> | ||
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| + | after quite a bit of math I get the answer to be | ||
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| + | |||
| + | <math>\frac{1}{2}(\frac{1}{2 + j8 - jw} + \frac{1}{2 -j8 -jw} + \frac{1}{2 - j8 - jw} \frac{1}{2 + j8 + jw})</math> | ||
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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 | ||
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| + | <pre> 4*(68+w^2)/(68+w^2-16*w)/(68+w^2+16*w) </pre> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:32, 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)
