(New page: <math>x(t)=e^{2jt} \to sys \to y(t)=te^{-2jt}</math> <math>x(t)=e^{-2jt} \to sys \to y(t)=te^{2jt}</math> This implies that if <math>x(t)=cos(2t)</math> then <math>y(t)=tcos(-2t)=tcos(2t...) |
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<math>x(t)=e^{-2jt} \to sys \to y(t)=te^{2jt}</math> | <math>x(t)=e^{-2jt} \to sys \to y(t)=te^{2jt}</math> | ||
| − | + | We want to know the output associated with the input <math>x(t)=cos(2t)</math>. If you expand <math>cos(2t)</math> into two exponentials you will get <math>\frac{e^{2jt} + e^{-2jt}}{2}</math>. Now you can use linearity to solve the problem. Linearity implies that if <math>x(t)=cos(2t)</math> then <math>y(t)=tcos(-2t)=tcos(2t)</math>. | |
Revision as of 11:24, 18 September 2008
$ x(t)=e^{2jt} \to sys \to y(t)=te^{-2jt} $
$ x(t)=e^{-2jt} \to sys \to y(t)=te^{2jt} $
We want to know the output associated with the input $ x(t)=cos(2t) $. If you expand $ cos(2t) $ into two exponentials you will get $ \frac{e^{2jt} + e^{-2jt}}{2} $. Now you can use linearity to solve the problem. Linearity implies that if $ x(t)=cos(2t) $ then $ y(t)=tcos(-2t)=tcos(2t) $.
