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</font>
  
<math>x(t)\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{-j\omega t}d\omega</math>
+
<math>x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{-j\omega t}d\omega</math>
 +
 
 +
<math>=\frac{2-3j}{2}\int_{-\infty}^{\infty}delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}delta(\omega + 4\pi)e^{j\omega t}d\omega </math>

Revision as of 10:22, 3 October 2008

Inverse Fourier Transform

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

$ X(\omega) = \pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j) $

$ x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}[\pi\delta(\omega - 4\pi)(2-3j) + \pi\delta(\omega + 4\pi)(2+3j)]e^{-j\omega t}d\omega $

$ =\frac{2-3j}{2}\int_{-\infty}^{\infty}delta(\omega - 4\pi)e^{j\omega t}d\omega + \frac{2+3j}{2}\int_{-\infty}^{\infty}delta(\omega + 4\pi)e^{j\omega t}d\omega $

Alumni Liaison

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

Buyue Zhang