| Line 6: | Line 6: | ||
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> | ||
Revision as of 09:04, 18 September 2008
Part B: The basics of linearity
$ e^{2jt} \rightarrow linear-system \rightarrow te^{-2jt} $
$ e^{-2jt} \rightarrow linear-system \rightarrow 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) $
