(→The basics of linearity) |
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<math>1/2 e^{(-2jt)}</math> --->[system]---><math> 1/2 te^{(2jt)}</math> | <math>1/2 e^{(-2jt)}</math> --->[system]---><math> 1/2 te^{(2jt)}</math> | ||
| − | <math> 1/2 te^{(-2jt)} + 1/2 te^{(2jt)} = { | + | <math> 1/2 te^{(-2jt)} + 1/2 te^{(2jt)} = {te^{2jt} + te^{-2jt} \over 2} = t{{e^{2jt} + e^{-2jt}} \over 2} = tcos(2t)</math> |
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| + | thus, the system's response to cos(2t) is tcos(2t). | ||
Latest revision as of 05:24, 19 September 2008
The basics of linearity
$ e^{(2jt)} $ --->[system]--->$ te^{(-2jt)} $
$ e^{(-2jt)} $ --->[system]--->$ te^{(2jt)} $
$ \cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2} $
$ \cos 2t = \mathrm{Re}\{e^{jt}\} ={e^{2jt} + e^{-2jt} \over 2} $
$ cos 2t = {e^{2jt} \over 2} + {e^{-2jt} \over 2} $
$ 1/2 e^{(2jt)} $ --->[system]--->$ 1/2 te^{(-2jt)} $
$ 1/2 e^{(-2jt)} $ --->[system]--->$ 1/2 te^{(2jt)} $
$ 1/2 te^{(-2jt)} + 1/2 te^{(2jt)} = {te^{2jt} + te^{-2jt} \over 2} = t{{e^{2jt} + e^{-2jt}} \over 2} = tcos(2t) $
thus, the system's response to cos(2t) is tcos(2t).
