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<math>X(\omega)=\int_{-\infty}^{\infty}t^2 u(t) e^{-j\omega t}dt \; = \int_{0}^{\infty}t^2 e^{-j\omega t}dt</math> | <math>X(\omega)=\int_{-\infty}^{\infty}t^2 u(t) e^{-j\omega t}dt \; = \int_{0}^{\infty}t^2 e^{-j\omega t}dt</math> | ||
| + | |||
| + | Integration by Parts | ||
| + | |||
| + | <math>u=t^2 \; \; \; \; \; dv = e^(-j \omega t)</math> | ||
| + | |||
| + | <math>du=2t \; \; \; \; \; dv = \frac{1}{-j\omega}e^(-j \omega t)</math> | ||
Revision as of 09:40, 3 October 2008
Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ x(t)=t^2 u(t) $
$ X(\omega)=\int_{-\infty}^{\infty}t^2 u(t) e^{-j\omega t}dt \; = \int_{0}^{\infty}t^2 e^{-j\omega t}dt $
Integration by Parts
$ u=t^2 \; \; \; \; \; dv = e^(-j \omega t) $
$ du=2t \; \; \; \; \; dv = \frac{1}{-j\omega}e^(-j \omega t) $
