Line 8: Line 8:
 
=<math>\frac{1}{2}te^{-2jt} +\frac{1}{2}te^{2jt}</math><br>
 
=<math>\frac{1}{2}te^{-2jt} +\frac{1}{2}te^{2jt}</math><br>
 
=<math>\frac{1}{2}t(e^{-2jt} +e^{2jt})</math><br>
 
=<math>\frac{1}{2}t(e^{-2jt} +e^{2jt})</math><br>
=<math>\frac{1}{2}t\cos(2t)</math><br>
+
=<math>t\cos(2t)</math><br>

Latest revision as of 15:52, 18 September 2008

$ e^{2jt} --> system --> te^{-2jt} $
$ e^{-2jt} --> system --> te^{2jt} $

=>$ \cos(2t) $ --> system =$ \frac{e^{2jt}+e^{-2jt}}{2} $ --> system
=$ \frac{1}{2}(e^{2jt}+e^{-2jt}) $ -->system
=$ \frac{1}{2}e^{2jt}-->system +\frac{1}{2}e^{-2jt} $-->system
=$ \frac{1}{2}te^{-2jt} +\frac{1}{2}te^{2jt} $
=$ \frac{1}{2}t(e^{-2jt} +e^{2jt}) $
=$ t\cos(2t) $

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

Landis Huffman