(New page: <math>\,cos(2t)</math> can be written as <math>\ {e^{-2jt} + e^{2jt} \over 2}</math> based on Euler's forumla: :<math>\cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2}</math> Sin...)
 
 
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:<math>\frac{e^{2t}}{2} \to t*\frac{e^{-2t}}{2}</math> and <math>\frac{e^{-2t}}{2} \to t*\frac{e^{2t}}{2}</math>
+
:<math>\frac{e^{2t}}{2} \to t\cdot\frac{e^{-2t}}{2}</math> and <math>\frac{e^{-2t}}{2} \to t\cdot\frac{e^{2t}}{2}</math>
  
  
:<math>t*\frac{e^{-2jt}}{2} + t*\frac{e^{2jt}}{2}</math>
+
:<math>t\cdot\frac{e^{-2jt}}{2} + t\cdot\frac{e^{2jt}}{2}</math>
  
  
:<math>    t*\frac{e^{-2jt}+e^{2jt}}{2}      </math>
+
:<math>    t\cdot\frac{e^{-2jt}+e^{2jt}}{2}      </math>
  
  
<math>\therefore \ t * cos(2t)</math>
+
<math>\therefore \ t \cdot cos(2t)</math>

Latest revision as of 17:31, 19 September 2008

$ \,cos(2t) $ can be written as $ \ {e^{-2jt} + e^{2jt} \over 2} $ based on Euler's forumla:

$ \cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2} $

Since the system is linear, and one of the properties of linear system is that:


Input $ \,ax_1(t)+bx_2(t) $ equals to the output $ \, ay_1(t)+by_2(t) $


and


$ \frac{e^{2t}}{2} \to t\cdot\frac{e^{-2t}}{2} $ and $ \frac{e^{-2t}}{2} \to t\cdot\frac{e^{2t}}{2} $


$ t\cdot\frac{e^{-2jt}}{2} + t\cdot\frac{e^{2jt}}{2} $


$ t\cdot\frac{e^{-2jt}+e^{2jt}}{2} $


$ \therefore \ t \cdot cos(2t) $

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