(x(t)*cos(t) ⇒ \frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)].)
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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(t)</math> &rArr; <math>\frac{\pi}{j}[\delta(\omega - \pi) - \delta(\omega + \pi)]</math>.==
+
=<math>x(t)*cos(t)</math> &rArr; <math>\frac{1}{2}[X(e^{j(\theta - \pi/4)})
 +
+ X(e^{j(\theta + \pi/4)}) ]</math>.

Revision as of 10:12, 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(t) $$ \frac{1}{2}[X(e^{j(\theta - \pi/4)}) + X(e^{j(\theta + \pi/4)}) ] $.

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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