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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(\frac{\pi t}{4})</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})
 +
+ X(e^{j(\theta + \pi/4)}) ]</math>.<br>
  
==<math>x(t)*cos(t)</math> &rArr; <math>\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]</math>. ==
+
In short we are getting two side bands which look something like this
 +
 
 +
[[Image:Modulation_ECE301Fall2008mboutin.gif]]

Latest revision as of 09:24, 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(\frac{\pi t}{4}) $$ \frac{1}{2}[X(e^{j(\theta - \pi/4)}) + X(e^{j(\theta + \pi/4)}) ] $.

In short we are getting two side bands which look something like this

Modulation ECE301Fall2008mboutin.gif

Alumni Liaison

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

Buyue Zhang