Revision as of 08:03, 18 September 2008 by Cztan (Talk)

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}) $

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