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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}\,d\omega \,</math>

Revision as of 10:51, 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}\,d\omega \, $

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang