(x(t)*cos(t) ⇒ \frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)].)
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Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
 
Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows
  
=<math>x(t)*cos(t)</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})  
+
<math>x(t)*cos(t)</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})  
 
+ X(e^{j(\theta + \pi/4)}) ]</math>.
 
+ X(e^{j(\theta + \pi/4)}) ]</math>.

Revision as of 09:13, 24 October 2008

Now we know that
$ x(t) $$ X(\omega) $

Now suppose the input signal was multiplied by a cosine wave then the fourier transform of the wave would look as follows

$ x(t)*cos(t) $$ \frac{1}{2}[X(e^{j(\theta - \pi/4)}) + X(e^{j(\theta + \pi/4)}) ] $.

Alumni Liaison

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

Buyue Zhang