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Fourier Transform of delta functions
 
Fourier Transform of delta functions
  
<math> x(t) = \delta (t+1) + \delta (t-1)
+
<math> x(t) = \delta (t+1) + \delta (t-1) </math>
  
X(\omega) = \int_{-\infty}^{\infty} \delta (t+1)e^{-j \omega t} + \int_{-\infty}^{\infty} \delta (t-1)e^{-j \omega t} dt  
+
<math> X(\omega) = \int_{-\infty}^{\infty} \delta (t+1)e^{-j \omega t} + \int_{-\infty}^{\infty} \delta (t-1)e^{-j \omega t} dt  
  
 
  X(\omega) = e^{j \omega}+ e^{-j \omega} = \frac{1}{2} (e^ {j \omega} + e^ {-j \omega})^2
 
  X(\omega) = e^{j \omega}+ e^{-j \omega} = \frac{1}{2} (e^ {j \omega} + e^ {-j \omega})^2
  
 
<math> X(\omega) = 2cos(\omega) </math>
 
<math> X(\omega) = 2cos(\omega) </math>

Revision as of 18:12, 24 October 2008

Fourier Transform of delta functions

$ x(t) = \delta (t+1) + \delta (t-1) $

$ X(\omega) = \int_{-\infty}^{\infty} \delta (t+1)e^{-j \omega t} + \int_{-\infty}^{\infty} \delta (t-1)e^{-j \omega t} dt X(\omega) = e^{j \omega}+ e^{-j \omega} = \frac{1}{2} (e^ {j \omega} + e^ {-j \omega})^2 <math> X(\omega) = 2cos(\omega) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood