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<math> \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 </math> | <math> \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 </math> | ||
| − | <math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{3t-j\omega t}\,dt + \frac{ | + | <math> \mathcal{X}(\omega) = \frac{1}{e^{6}} \int_{-\infty}^{2} e^{3t-j\omega t}\,dt + \frac{1}{e^{6}} \int_{2}^{\infty} e^{-3t-j\omega t} \,dt </math> |
<math> | <math> | ||
Revision as of 16:34, 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 + \frac{1}{e^{6}} \int_{2}^{\infty} e^{-3t-j\omega t} \,dt $
