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thus now from the above equations we can see that <math>e^{2jt}</math> converts to <math>te^{-2jt}</math> and <math>e^{-2jt}</math> converts to <math> te^{2jt}</math> | thus now from the above equations we can see that <math>e^{2jt}</math> converts to <math>te^{-2jt}</math> and <math>e^{-2jt}</math> converts to <math> te^{2jt}</math> | ||
| − | thus <math>cos(2t)=\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} \to | + | 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 | thus | ||
| − | <math>cos(2t) \to | + | <math>cos(2t) \to System \to tcos(2t)</math> |
Latest revision as of 12:18, 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) $
