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

Revision as of 10:23, 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

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood