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Answer:
  
<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>.
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<math>cos(2t)\!</math> can also be written as <math>(e^{2jt} + e^{-2jt})/2\!</math> which can also be written as <math>1/2*[(e^{2jt} + e^{-2jt})]\!</math> so therefore the linear system's response  is <math>t/2*[(e^{-2jt} + e^{2jt})]\!</math> which equals <math>t*cos(2t)\!</math>.
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(Note:  The star in this case is the multiplication operator, not the convolution operator)

Latest revision as of 17:18, 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\! $ which can also be written as $ 1/2*[(e^{2jt} + e^{-2jt})]\! $ so therefore the linear system's response is $ t/2*[(e^{-2jt} + e^{2jt})]\! $ which equals $ t*cos(2t)\! $.

(Note: The star in this case is the multiplication operator, not the convolution operator)

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett