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