(New page: ==Example 1== Compute the Fourier Transform of <math>x(t)=e^{-t}u(t)</math>. <math>X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt</math> <math>=\int_{-\infty}^{\infty}e^{-t}u(t)e^{...) |
(→Example 1) |
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<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^{-t}u(t)e^{-j\omega t}dt</math> | <math>=\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt</math> | ||
| + | |||
<math>=\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt</math> | <math>=\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt</math> | ||
| + | |||
<math>=\int_{0}^{\infty}e^{-(1+j\omega )t}dt</math> | <math>=\int_{0}^{\infty}e^{-(1+j\omega )t}dt</math> | ||
| + | |||
<math>=[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty</math> | <math>=[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty</math> | ||
| − | <math> | + | |
| − | <math>=0 | + | <math>=\frac {e^{-(1+j\omega )\infty}}{-(1+j\omega)}-\frac {e^{-(1+j\omega )0}}{-(1+j\omega)}</math> |
| + | |||
| + | <math>=0+\frac {1}{(1+j\omega)}</math> | ||
| + | |||
<math>=\frac {1}{(1+j\omega)}</math> | <math>=\frac {1}{(1+j\omega)}</math> | ||
| + | ==Example 2== | ||
| + | |||
==Example 2== | ==Example 2== | ||
Revision as of 19:05, 20 October 2008
Example 1
Compute the Fourier Transform of $ x(t)=e^{-t}u(t) $.
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ =\int_{-\infty}^{\infty}e^{-t}u(t)e^{-j\omega t}dt $
$ =\int_{0}^{\infty}e^{-t}e^{-j\omega t}dt $
$ =\int_{0}^{\infty}e^{-(1+j\omega )t}dt $
$ =[\frac {e^{-(1+j\omega )t}}{-(1+j\omega)}]|_0^\infty $
$ =\frac {e^{-(1+j\omega )\infty}}{-(1+j\omega)}-\frac {e^{-(1+j\omega )0}}{-(1+j\omega)} $
$ =0+\frac {1}{(1+j\omega)} $
$ =\frac {1}{(1+j\omega)} $
