(New page: <math>x_1(t) = e^{2jt} \rightarrow </math> Linear System <math>\rightarrow y_1(t) = te^{-2jt} </math> <math>x_2(t) = e^{-2jt} \rightarrow </math> Linear System <math> \rightarrow y_2(t) =...) |
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Revision as of 13:38, 18 September 2008
$ x_1(t) = e^{2jt} \rightarrow $ Linear System $ \rightarrow y_1(t) = te^{-2jt} $
$ x_2(t) = e^{-2jt} \rightarrow $ Linear System $ \rightarrow y_2(t) = te^{2jt} $
With an input of $ cos(2t) $, which is $ \frac{1}{2}(e^{j2t}+e^{-j2t}) $ according to Euler's Forumla.
Using the property of linearity, the response is: $ t\frac{1}{2}(e^{-j2t}+e^{j2t}) $ which is equal to $ tcos(2t) $
