(New page: ==Linearity== The System takes the input and gives the output as follows: <math> e^{2jt} \to System \to te^{-2jt}\!</math> <math> e^{-2jt} \to System \to te^{2jt}\!</math> now,we know ...) |
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thus <math>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)</math> | thus <math>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)</math> | ||
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| + | thus | ||
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| + | <math>cos(2t) \to system \to tcos(2t)</math> | ||
Revision as of 12:17, 18 September 2008
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) $
