Revision as of 11:18, 18 September 2008 by Sranka (Talk)

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Linearity

The System takes the input and gives the output as follows:

$ e^{2jt} \to System \to te^{-2jt}\! $

$ e^{-2jt} \to System \to te^{2jt}\! $

now,we know that $ cos(2t)=\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} $

thus now from the above equations we can see that $ e^{2jt} $ converts to $ te^{-2jt} $ and $ e^{-2jt} $ converts to $ te^{2jt} $

thus $ cos(2t)=\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} \to System \to \frac{1}{2}te^{-2jt}+ \frac{1}{2}te^{2jt}=t[\frac{1}{2}e^{-2jt}+ \frac{1}{2}e^{2jt}]=tcos(2t) $

thus

$ cos(2t) \to System \to tcos(2t) $

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood