(AM modulation)
(AM modulation)
Line 15: Line 15:
  
 
[[Image:Modulation_ECE301Fall2008mboutin.gif]]
 
[[Image:Modulation_ECE301Fall2008mboutin.gif]]
 +
 
[[Image:hw92_ECE301Fall2008mboutin.jpg]]
 
[[Image:hw92_ECE301Fall2008mboutin.jpg]]
  

Revision as of 16:44, 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

Hw92 ECE301Fall2008mboutin.jpg

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

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva