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To find the response of the system above we first note that | To find the response of the system above we first note that | ||
| − | <math>e^{j2t} = cos(2t) + jsin(2t)\!</math> | + | <math>e^{j2t} = cos(2t) + jsin(2t)\!</math> |
<math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | <math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | ||
Revision as of 10:11, 19 September 2008
Given:
For a linear system we have:
$ e^{j2t} \rightarrow [system] \rightarrow te^{-j2t}\! $
$ e^{-j2t} \rightarrow [system] \rightarrow te^{j2t}\! $
To find the response of the system above we first note that
$ e^{j2t} = cos(2t) + jsin(2t)\! $
$ e^{-j2t} = cos(2t) - jsin(2t)\! $
