(AM Demodulation)
 
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== AM Demodulation ==
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== AM modulation ==
  
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Now we know that <br>
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<math> x(t)</math> &rArr; <math>X(\omega)</math><br>
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Now suppose the input signal was multiplied by a cosine wave
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[[Image:aa_ECE301Fall2008mboutin.jpg]]
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then the fourier transform of the wave would look as follows
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<math>x(t)*cos(\frac{\pi t}{4})</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})
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+ X(e^{j(\theta + \pi/4)}) ]</math>.<br>
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In short we are getting two side bands which look something like this
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[[Image:Modulation_ECE301Fall2008mboutin.gif]]
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== AM Demodulation ==
  
 
[[Image:Hw9_ECE301Fall2008mboutin.JPG]]
 
[[Image:Hw9_ECE301Fall2008mboutin.JPG]]
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<math>r(n)= x(n)cos^2(n \theta)= \frac{1}{2} x(n) + \frac{1}{2}x(n)cos(2n\theta)</math><br>
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<math>Y(e^{j\theta})= R(e^{j\theta})H(e^{j\theta})</math><br>
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<math> = [\frac{1}{2}X(e^{j\theta})+\frac{1}{4}X(e^{j\theta-2\phi})+\frac{1}{4}X(e^{j\theta+2\phi})]H(e^{j\theta})</math><br>
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<math> = X(e^{j\theta}) </math>
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[[Image:hw91_ECE301Fall2008mboutin.jpg]]
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[[Image:hw92_ECE301Fall2008mboutin.jpg]]

Latest revision as of 16:48, 17 November 2008

AM modulation

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

Now suppose the input signal was multiplied by a cosine wave Aa ECE301Fall2008mboutin.jpg

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

AM Demodulation

Hw9 ECE301Fall2008mboutin.JPG

$ r(n)= x(n)cos^2(n \theta)= \frac{1}{2} x(n) + \frac{1}{2}x(n)cos(2n\theta) $

$ Y(e^{j\theta})= R(e^{j\theta})H(e^{j\theta}) $
$ = [\frac{1}{2}X(e^{j\theta})+\frac{1}{4}X(e^{j\theta-2\phi})+\frac{1}{4}X(e^{j\theta+2\phi})]H(e^{j\theta}) $
$ = X(e^{j\theta}) $

Hw91 ECE301Fall2008mboutin.jpg Hw92 ECE301Fall2008mboutin.jpg

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BSEE 2004, current Ph.D. student researching signal and image processing.

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