(New page: This problem uses the linearity. If we know that: <math>e^{2jt} \to te^{-2jt}</math> and <math>e^{-2jt} \to te^{2jt}</math> <p> Then by rewriting cos(2t) as <math>\frac{e^{2jt} + e^{-2jt}}...) |
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<math>e^{2jt} \to te^{-2jt}</math> and <math>e^{-2jt} \to te^{2jt}</math> <p> | <math>e^{2jt} \to te^{-2jt}</math> and <math>e^{-2jt} \to te^{2jt}</math> <p> | ||
Then by rewriting cos(2t) as <math>\frac{e^{2jt} + e^{-2jt}}{2}</math> then since the system in linear, | Then by rewriting cos(2t) as <math>\frac{e^{2jt} + e^{-2jt}}{2}</math> then since the system in linear, | ||
| − | take <math>\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt}</math> through the system to get <math>\frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math> which is the same as <math>t\cos(2t)</math> | + | take <math>\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt}</math> through the system to get <math>\frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt}</math> which is the same as <math>t\cos{(2t)}</math> |
Latest revision as of 10:20, 18 September 2008
This problem uses the linearity. If we know that:
$ e^{2jt} \to te^{-2jt} $ and $ e^{-2jt} \to te^{2jt} $Then by rewriting cos(2t) as $ \frac{e^{2jt} + e^{-2jt}}{2} $ then since the system in linear, take $ \frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} $ through the system to get $ \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $ which is the same as $ t\cos{(2t)} $
