Revision as of 05:28, 18 September 2008 by Han45 (Talk)

first,

$ e^{2jt}=cos(2t) + jsin(2t) $

$ e^{-2jt}=cos(2t) - jsin(2t) $

then we put into system, we got..


$ e^{2jt}\rightarrow system\rightarrow t*e^{-2jt} $

$ e^{-2jt}\rightarrow system\rightarrow t*e^{2jt} $

it means

$ e^{2jt} = t*(cos(2t)-jsin(2t)) $

$ e^{-2jt} = t*(cos(2t)+jsin(2t)) $

input is

$ x(t)=\cos(2t) $

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