| Line 21: | Line 21: | ||
<math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{t(3-j\omega)}\,dt + e^{6} \int_{2}^{\infty} e^{-t(3+j\omega)} \,dt </math> | <math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{t(3-j\omega)}\,dt + e^{6} \int_{2}^{\infty} e^{-t(3+j\omega)} \,dt </math> | ||
| − | <math> \mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{2}_{-\infty} } \frac{1}{e^{6}} | + | <math> \mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{2}_{-\infty} } \frac{1}{e^{6}} + {\left. -\frac{e^{-t(3+j\omega)}}{3+j\omega} \right]^{\infty}_2 } e^{6}\,</math> |
| − | <math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \frac{e^{6-2j\omega}}{3-j\omega} + e^{6} \frac{e^{-6- | + | <math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \frac{e^{6-2j\omega}}{3-j\omega} + e^{6} \frac{e^{-6-j\omega}}{3+j\omega} </math> |
Revision as of 16:53, 8 October 2008
$ x(t) = e^{-3|t-2|} $
Noticing that there is an absolute value, we can proceed to divide in tow cases.
When
$ t-2 < 0 \rightarrow x_1(t) = e^{3t-6} $
and when,
$ t-2 >0 \rightarrow x_2(t) = e^{-3t+6} $
So, we can then compute the Fourier series by adding the integrals of each diferent case.
$ \ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \int_{-\infty}^{2} e^{3t-6}e^{-j\omega t}\,dt + \int_{2}^{\infty} e^{-3t+6}e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{3t-j\omega t}\,dt + e^{6} \int_{2}^{\infty} e^{-3t-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{t(3-j\omega)}\,dt + e^{6} \int_{2}^{\infty} e^{-t(3+j\omega)} \,dt $
$ \mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{2}_{-\infty} } \frac{1}{e^{6}} + {\left. -\frac{e^{-t(3+j\omega)}}{3+j\omega} \right]^{\infty}_2 } e^{6}\, $
$ \mathcal{X}(\omega) = \frac{1}{e^{6}} \frac{e^{6-2j\omega}}{3-j\omega} + e^{6} \frac{e^{-6-j\omega}}{3+j\omega} $
