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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]] | ||
| + | ---- | ||
== Signal == | == Signal == | ||
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<math> = \frac{e^{3jt - j\omega t}}{3j-j\omega}]_{-5}^{5} + \frac{e^{-2t - j\omega t}}{-2 -j\omega}]_{0}^{\infty}\,</math> | <math> = \frac{e^{3jt - j\omega t}}{3j-j\omega}]_{-5}^{5} + \frac{e^{-2t - j\omega t}}{-2 -j\omega}]_{0}^{\infty}\,</math> | ||
| + | |||
| + | <math> = \frac{e^{15j - 5j\omega} - e^{-15j + 5j\omega}}{3j-j\omega} + \frac{e^{-2*\infty - \infty j \omega} - e^{0}}{-j -j\omega}\,</math> | ||
| + | |||
| + | <math> = \frac{e^{j(15 - 5\omega )} - e^{-j(15 - 5\omega )}}{j(3-\omega )} + \frac{1}{j(1-\omega )}\,</math> | ||
| + | |||
| + | <math> = \frac{2sin(15 - 5\omega )}{3-\omega } + \frac{1}{j(1-\omega )}\,</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:28, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Signal
$ x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}*u(t)\, $
Transformed
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\, $
$ = \int_{-\infty}^{\infty}e^{3jt}*(u(t+5) - u(t-5))e^{-j\omega t}dt + \int_{-\infty}^{\infty}e^{-2t}u(t)e^{-j\omega t}dt\, $
$ = \int_{-5}^{5}e^{3jt}e^{-j\omega t}dt + \int_{0}^{\infty}e^{-2t}e^{-j\omega t}dt\, $
$ = \int_{-5}^{5}e^{3jt -j\omega t}dt + \int_{0}^{\infty}e^{-2t -j\omega t}dt\, $
$ = \int_{-5}^{5}e^{t*(3j -j\omega )}dt + \int_{0}^{\infty}e^{t*(-2 -j\omega )}dt\, $
$ = \frac{e^{3jt - j\omega t}}{3j-j\omega}]_{-5}^{5} + \frac{e^{-2t - j\omega t}}{-2 -j\omega}]_{0}^{\infty}\, $
$ = \frac{e^{15j - 5j\omega} - e^{-15j + 5j\omega}}{3j-j\omega} + \frac{e^{-2*\infty - \infty j \omega} - e^{0}}{-j -j\omega}\, $
$ = \frac{e^{j(15 - 5\omega )} - e^{-j(15 - 5\omega )}}{j(3-\omega )} + \frac{1}{j(1-\omega )}\, $
$ = \frac{2sin(15 - 5\omega )}{3-\omega } + \frac{1}{j(1-\omega )}\, $
