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<math>X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt</math> | <math>X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt</math> | ||
| − | <math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e{-4(t-1)})e^{-jwt}dt</math> | + | <math>=\int_{-\infty}^{\infty}(e^{-5t}u(t) + e^{-4(t-1)}u(t-3))e^{-jwt}dt</math> |
<math>=\int_{0}^{\infty}e^{-5t} e^{-jwt}dt + \int_{3}^{\infty}e^{-4(t-1)}e^{-jwt}dt</math> | <math>=\int_{0}^{\infty}e^{-5t} e^{-jwt}dt + \int_{3}^{\infty}e^{-4(t-1)}e^{-jwt}dt</math> | ||
Revision as of 12:39, 24 October 2008
Question: Compute the Fourier transform of the signal
$ x(t) = e^{-5t}u(t) + e^{-4(t-1)}u(t-3)\! $
Answer:
$ X(w) = \int_{-\infty}^{\infty}x(t)e^{-jwt}dt $
$ =\int_{-\infty}^{\infty}(e^{-5t}u(t) + e^{-4(t-1)}u(t-3))e^{-jwt}dt $
$ =\int_{0}^{\infty}e^{-5t} e^{-jwt}dt + \int_{3}^{\infty}e^{-4(t-1)}e^{-jwt}dt $
$ =\frac{e^{-(5+jw)t}}{-(5+jw)}\bigg]^{\infty}_{0}+ \int_{3}^{\infty}e^{-4t}e^{4}e^{-jwt}dt $
$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4*\int_{\infty}^{3}e^{-(4+jw)t}dt $
$ =\frac{e^{-5t}e^{-jwt}}{-(5+jw)}\bigg]^{infty}_{0} + e^4 * \frac{e^{-(4+jw)t}}{-(4+jw)}\bigg]_3^\infty $
$ =\frac{1}{(5+jw)}+ e^4 * \frac{e^{-(12+3jw)}}{(4+jw)} $
$ =\frac{1}{(5+jw)} + \frac{e^{-(8+3jw)}}{(4+jw)} $
