(System Response)
(System Response)
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Output = Response of <math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>
 
Output = Response of <math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>
  
       = Response of <math>\frac{e^{2jt}\,</math> + Response of <math>\frac{e^{-2jt}\,</math>
+
       = Response of <math>\e^{2jt}\,</math> + Response of <math>\e^{-2jt}\,</math>
  
 
       <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> (derive from given conditions)
 
       <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> (derive from given conditions)

Revision as of 08:22, 18 September 2008

System Response

Based on the Euler Formula, $ \cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\, $.

We already had the response of $ e^{2jt}\, $ is $ te^{-2jt}\, $ and the response of $ e^{-2jt}\, $ is $ te^{2jt}\, $.

Since the system is a LTI system, we have

Output = Response of $ \frac{e^{2jt}+e^{-2jt}}{2}\, $

      = Response of $ \e^{2jt}\, $ + Response of $ \e^{-2jt}\, $
     $ =\frac{te^{2jt}+te^{-2jt}}{2}\, $ (derive from given conditions)
     $ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
     $ =t\cos(2t)\, $

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Ryne Rayburn