| Line 4: | Line 4: | ||
<font "size"=4> | <font "size"=4> | ||
| − | <math>x(t)= | + | <math>x(t)=(t-1)e^{-6t+6}u(t-1) \,\ </math> |
</font> | </font> | ||
| − | <math>X(\omega)=\int_{-\infty}^{\infty}t^ | + | <math>X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \;</math> |
| + | |||
| + | |||
| + | <math>x(t) \,\ </math>looks like <math>te^{-6t}u(t) \,\ </math> so we evaluate that | ||
| + | |||
| + | the F.T of <math>te^{-6t}u(t) \,\ </math> is | ||
| + | |||
| + | <math>\int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \;</math> | ||
Revision as of 18:37, 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 \; $
