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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]] | ||
| + | ---- | ||
| − | <math>X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> | + | <math>X(t)=e^{-5t}cos{(2t)}u(t)dt</math> |
| + | |||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\int_{0}^{+\infty}e^{-5t}cos{(2t)}e^{-j\omega t}\,dt,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-5t}e^{2jt}e^{-j\omega t}\,dt + 1/2\int_{0}^{+\infty}e^{-5t}e^{-2jt}e^{-j\omega t}\,dt\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-t(5-2j+j\omega)}\,dt + 1/2\int_{0}^{+\infty}e^{-t(5+2j+j\omega)}\,dt\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\left. 1/2\frac{e^{-t(5-2j+j\omega)}}{-(5-2j+j\omega)}\right]_{0}^{+\infty} + \left. 1/2\frac{e^{-t(5+2j+j\omega)}}{-(5+2j+j\omega)}\right]_{0}^{+\infty}</math> | ||
| + | |||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\frac{1}{2(5-2j+j\omega)} + \frac{1}{2(5+2j+j\omega)}</math> | ||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:22, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
$ X(t)=e^{-5t}cos{(2t)}u(t)dt $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{0}^{+\infty}e^{-5t}cos{(2t)}e^{-j\omega t}\,dt, $
$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-5t}e^{2jt}e^{-j\omega t}\,dt + 1/2\int_{0}^{+\infty}e^{-5t}e^{-2jt}e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-t(5-2j+j\omega)}\,dt + 1/2\int_{0}^{+\infty}e^{-t(5+2j+j\omega)}\,dt\, $
$ \,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\left. 1/2\frac{e^{-t(5-2j+j\omega)}}{-(5-2j+j\omega)}\right]_{0}^{+\infty} + \left. 1/2\frac{e^{-t(5+2j+j\omega)}}{-(5+2j+j\omega)}\right]_{0}^{+\infty} $
$ \,\mathcal{X}(\omega)=\frac{1}{2(5-2j+j\omega)} + \frac{1}{2(5+2j+j\omega)} $
