(New page: The two functions given to us happen to be part of the breakdown of cos(2t). <math>cos(2t)=e^{2jt}+e^{-2jt}</math> Referenced: Max Paganini) |
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The two functions given to us happen to be part of the breakdown of cos(2t). | The two functions given to us happen to be part of the breakdown of cos(2t). | ||
| − | <math>cos(2t)=e^{2jt}+e^{-2jt}</math> | + | <math>cos(2t)=\frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt}</math> |
| + | If the response to <math>e^{2jt}=te^{-2jt}</math> and <math>e^{-2jt}=te^{2jt}</math> | ||
| + | then, | ||
| + | cos(2t) -> System -> <math>t\frac{1}{2}e^{-2jt} + t\frac{1}{2}e^{2jt} = tcos(2t)</math> | ||
Latest revision as of 05:58, 19 September 2008
The two functions given to us happen to be part of the breakdown of cos(2t).
$ cos(2t)=\frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt} $
If the response to $ e^{2jt}=te^{-2jt} $ and $ e^{-2jt}=te^{2jt} $ then, cos(2t) -> System -> $ t\frac{1}{2}e^{-2jt} + t\frac{1}{2}e^{2jt} = tcos(2t) $
Referenced: Max Paganini
