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<math>e^{2jt}\!</math> through a system produces <math>e^{-2jt}\!</math>, and <math>e^{-2jt}\!</math> produces <math>e^{2jt}\!</math>. what is the output of <math>cos(2t)\!</math>
 
<math>e^{2jt}\!</math> through a system produces <math>e^{-2jt}\!</math>, and <math>e^{-2jt}\!</math> produces <math>e^{2jt}\!</math>. what is the output of <math>cos(2t)\!</math>
  
 +
Answer:
  
<math>cos(2t)\!</math> can also be written as <math>(e^{2jt} + e^{-2jt})/2\!</math>
+
<math>cos(2t)\!</math> can also be written as <math>(e^{2jt} + e^{-2jt})/2\!</math> so therefore the linear system's response to <math>cos(2t)\!</math> is <math>(e^{-2jt} + e^{2jt})/2\!</math> which would just be <math>cos(2t)\!</math>.

Revision as of 17:03, 18 September 2008

Basics of Linearity

Given:

$ e^{2jt}\! $ through a system produces $ e^{-2jt}\! $, and $ e^{-2jt}\! $ produces $ e^{2jt}\! $. what is the output of $ cos(2t)\! $

Answer:

$ cos(2t)\! $ can also be written as $ (e^{2jt} + e^{-2jt})/2\! $ so therefore the linear system's response to $ cos(2t)\! $ is $ (e^{-2jt} + e^{2jt})/2\! $ which would just be $ cos(2t)\! $.

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

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