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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>
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Integration by Parts
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<math>u=t^2 \; \; \; \; \; dv = e^(-j \omega t)</math>
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<math>du=2t \; \; \; \; \; dv = \frac{1}{-j\omega}e^(-j \omega t)</math>

Revision as of 08:40, 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) $

Alumni Liaison

EISL lab graduate

Mu Qiao