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<math>du=1 \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math>
 
<math>du=1 \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}</math>
  
<math>X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{0}^{\infty} + \frac{2}{j \omega}[\frac{tj}{\omega}e^{-j\omega t}|_{0}^{\infty}+\int_{0}^{\infty}t^2 e^{-j\omega t}dt]</math>
+
<math>X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{0}^{\infty} + \frac{2}{j \omega}[\frac{tj}{\omega}e^{-j\omega t}|_{0}^{\infty}+\frac{1}{j \omega}\int_{0}^{\infty}e^{-j\omega t}dt]</math>

Revision as of 08:53, 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}|_{0}^{\infty} + \frac{2}{j \omega}\int_{0}^{\infty}t^2 e^{-j\omega t}dt $

Integration by Parts

$ u=t \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t} $

$ du=1 \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t} $

$ X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{0}^{\infty} + \frac{2}{j \omega}[\frac{tj}{\omega}e^{-j\omega t}|_{0}^{\infty}+\frac{1}{j \omega}\int_{0}^{\infty}e^{-j\omega t}dt] $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn