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<math>\,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \delta(\omega - 3\pi) e^{(j\omega - 1)t}\,d\omega \,</math>
 
<math>\,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \delta(\omega - 3\pi) e^{(j\omega - 1)t}\,d\omega \,</math>
  
<math>\,x(t)=\frac{1}{2\pi} e^{(j(-3\pi) - 1)t}\,d\omega \,</math>
+
<math>\,x(t)=\frac{1}{2\pi} e^{(j(-3\pi) - 1)t}\,</math>

Revision as of 09:51, 6 October 2008

Compute the inverse fourier transform of the fourier transform below:

$ \,\mathcal{X}(\omega)= \delta(\omega - 3\pi) e^{-t}\, $


$ \,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{j\omega t}\,d\omega \, $

$ \,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \delta(\omega - 3\pi) e^{-t} e^{j\omega t}\,d\omega \, $

$ \,x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \delta(\omega - 3\pi) e^{(j\omega - 1)t}\,d\omega \, $

$ \,x(t)=\frac{1}{2\pi} e^{(j(-3\pi) - 1)t}\, $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett