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<math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | <math>e^{-j2t} = cos(2t) - jsin(2t)\!</math> | ||
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| + | We can use this to represent our input signal as follows: | ||
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| + | |||
| + | <math>cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)]\!</math> | ||
Revision as of 10:17, 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)\! $
We can use this to represent our input signal as follows:
$ cos(2t) = 0.5\times[cos(2t) + jsin(2t) + cos(2t) - jsin(2t)]\! $
