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== Signal == | == Signal == | ||
| − | <math>x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}* | + | <math>x(t) = e^{3jt}*(u(t+5) - u(t-5)) + e^{-2t}*u(t)\,</math> |
| Line 8: | Line 8: | ||
<math>X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\,</math> | <math>X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\,</math> | ||
| − | <math> = \int_{-\infty}^{\infty}e^{3jt}*(u(t+5) - u(t-5))e^{-j\omega t}dt + \int_{-\infty}^{\infty}e^{-2t} | + | <math> = \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\,</math> |
Revision as of 12:18, 7 October 2008
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\, $
