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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 | + | <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>. |
(Note: The star in this case is the multiplication operator, not the convolution operator) | (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)
