(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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With an input of <math>cos(2t)</math>, which is <math>\frac{1}{2}(e^{j2t}+e^{-j2t})</math> according to Euler's Forumla. | With an input of <math>cos(2t)</math>, which is <math>\frac{1}{2}(e^{j2t}+e^{-j2t})</math> according to Euler's Forumla. | ||
| − | Using the property of linearity, the response is: <math>t\frac{1}{2}(e^{-j2t}+e^{j2t})</math> which is equal to <math>tcos(2t)</math> | + | Using the property of linearity, the response is: <math>t\frac{1}{2}(e^{-j2t}+e^{j2t})</math> which is equal to <math>t*cos(2t)</math> |
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| + | <math>x(t) = cos(2t) \rightarrow </math> Linear System <math> \rightarrow y(t) = tcos(2t) </math> | ||
Latest revision as of 13:41, 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 $ t*cos(2t) $
$ x(t) = cos(2t) \rightarrow $ Linear System $ \rightarrow y(t) = tcos(2t) $
