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<math>\int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \;</math> | <math>\int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \;</math> | ||
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| + | <math>=\int_{0}^{\infty}te^{-6t-j\omega t}dt \;</math> | ||
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| + | <math>=\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \;</math> | ||
| + | |||
| + | Do integration by parts | ||
Revision as of 18:41, 7 October 2008
Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ x(t)=(t-1)e^{-6t+6}u(t-1) \,\ $
$ X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \; $
$ x(t) \,\ $looks like $ te^{-6t}u(t) \,\ $ so we evaluate that
the F.T of $ te^{-6t}u(t) \,\ $ is
$ \int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-6t-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \; $
Do integration by parts
