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==Part B: The basics of linearity== | ==Part B: The basics of linearity== | ||
| − | <math>e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} </math> | + | <math>x_1(t) = e^{2jt} \rightarrow linear-system \rightarrow y_1(t) = te^{-2jt} </math> |
| − | <math>e^{-2jt} \rightarrow linear-system \rightarrow te^{2jt} </math> | + | <math>x_2(t) = e^{-2jt} \rightarrow linear-system \rightarrow y_2(t) = te^{2jt} </math> |
The input, cos(2t) is equal to <math>\frac{1}{2}(e^{j2t} + e^{-j2t})</math> | The input, cos(2t) is equal to <math>\frac{1}{2}(e^{j2t} + e^{-j2t})</math> | ||
| − | From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear system \rightarrow ay_1(t) + by_2(t)</math> | + | From the properties of a linear system <math>ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t)</math>, |
| + | |||
| + | The response to cos(2t) is <math>\frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math> | ||
Revision as of 08:08, 18 September 2008
Part B: The basics of linearity
$ 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} $
The input, cos(2t) is equal to $ \frac{1}{2}(e^{j2t} + e^{-j2t}) $
From the properties of a linear system $ ax_1(t) + bx_2(t) \rightarrow linear-system \rightarrow ay_1(t) + by_2(t) $,
The response to cos(2t) is $ \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $
