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Answer:
 
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/2*cos(2t)\!</math>.
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(Note:  The star in this case is the multiplication operator, not the convolution operator)

Revision as of 17:17, 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/2*cos(2t)\! $.

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

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang