(New page: == System Response == <math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>. We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^...)
 
(System Response)
 
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== System Response ==
 
== System Response ==
<math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>.
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Based on the Euler Formula, <math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>.
  
 
We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^{-2jt}\,</math> is <math>te^{2jt}\,</math>.
 
We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^{-2jt}\,</math> is <math>te^{2jt}\,</math>.
  
Assuming the system is a LTI system, we can substitute the response of <math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math> with the values above.
+
Since the system is a LTI system, we have
  
Thus the system will produce the following response as an output :
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Output = Response of <math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>
  
<math>\frac{te^{2jt}+te^{-2jt}}{2}\,</math>
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      = Response of <math>\frac{e^{2jt}}{2}\,</math> + Response of <math>\frac{e^{-2jt}}{2}\,</math>
  
<math>=t\frac{e^{2jt}+e^{-2jt}}{2}\,</math>
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      <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math>  
  
<math>=t\cos(2t)\,</math>
+
(since the SYSTEM is linear, so we can do addition and division by constant to the output according to the different inputs)
 +
 
 +
      <math>=t\frac{e^{2jt}+e^{-2jt}}{2}\,</math>
 +
 
 +
      <math>=t\cos(2t)\,</math>

Latest revision as of 08:32, 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 $ \frac{e^{2jt}}{2}\, $ + Response of $ \frac{e^{-2jt}}{2}\, $
     $ =\frac{te^{2jt}+te^{-2jt}}{2}\, $ 

(since the SYSTEM is linear, so we can do addition and division by constant to the output according to the different inputs)

     $ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
     $ =t\cos(2t)\, $

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