Line 19: Line 19:
 
<math>e^{-2jt} = t*(cos(2t)+jsin(2t))
 
<math>e^{-2jt} = t*(cos(2t)+jsin(2t))
 
</math>
 
</math>
 +
 +
using Euler's formula
 +
 +
<math>e^{ix} = cosx + isinx</math>
 +
  
 
input is  
 
input is  
  
 
<math>x(t)=\cos(2t)</math>
 
<math>x(t)=\cos(2t)</math>

Revision as of 05:35, 18 September 2008

first,

$ e^{2jt}=cos(2t) + jsin(2t) $

$ e^{-2jt}=cos(2t) - jsin(2t) $

then we put into system, we got..


$ e^{2jt}\rightarrow system\rightarrow t*e^{-2jt} $

$ e^{-2jt}\rightarrow system\rightarrow t*e^{2jt} $

it means

$ e^{2jt} = t*(cos(2t)-jsin(2t)) $

$ e^{-2jt} = t*(cos(2t)+jsin(2t)) $

using Euler's formula

$ e^{ix} = cosx + isinx $


input is

$ x(t)=\cos(2t) $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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