(AM Demodulation)
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== AM Demodulation ==
 
== AM Demodulation ==
  
<math>r(n)= x(n)*cos^2(n \theta)= \frac{1}{2} x(n) + \frac{1}{2}x(n)*cos(2n\theta)</math><br>
+
<math>r(n)= x(n)cos^2(n \theta)= \frac{1}{2} x(n) + \frac{1}{2}x(n)cos(2n\theta)</math><br>
  
<math>X(e^{j\theta))
+
<math>Y(e^{j\theta})= R(e^{j\theta})H(e^{j\theta})</math><br>
+ X(e^{j(\theta)</math>
+
<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>
 +
<math> = X(e^{j\theta}) </math>
  
 
[[Image:Hw9_ECE301Fall2008mboutin.JPG]]
 
[[Image:Hw9_ECE301Fall2008mboutin.JPG]]

Revision as of 16:33, 17 November 2008

AM Demodulation

$ 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}) $

Hw9 ECE301Fall2008mboutin.JPG

Alumni Liaison

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

Buyue Zhang