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

Alumni Liaison

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

Francisco Blanco-Silva