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To find the response of the system above we first note that <math>e^{j2t} = cos(2t) + jsin(2t)
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To find the response of the system above we first note that <math>e^{j2t} = cos(2t) + jsin(2t)\!

Revision as of 09:10, 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)\! $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn