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<math>du=2t \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math> | <math>du=2t \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math> | ||
| − | <math>X(\omega)=\frac{t^2 | + | <math>X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t} + \frac{2}{j \omega}\int_{0}^{\infty}t^2 e^{-j\omega t}dt</math> |
Revision as of 09:47, 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
$ \int u \; dv = uv - \int v \; du $
$ u=t^2 \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t} $
$ du=2t \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t} $
$ X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t} + \frac{2}{j \omega}\int_{0}^{\infty}t^2 e^{-j\omega t}dt $
