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Output = Response(<math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>) | Output = Response(<math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>) | ||
| − | = Response(<math>\frac{e^{2jt}</math>)+ | + | = Response of(<math>\frac{e^{2jt}\,</math>)+ Response of(<math>\frac{e^{-2jt}\,</math>) |
| − | <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> | + | <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> (derive from given conditions) |
<math>=t\frac{e^{2jt}+e^{-2jt}}{2}\,</math> | <math>=t\frac{e^{2jt}+e^{-2jt}}{2}\,</math> | ||
<math>=t\cos(2t)\,</math> | <math>=t\cos(2t)\,</math> | ||
Revision as of 09:20, 18 September 2008
System Response
Based on the Euler Formula, $ \cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\, $.
We already had the response of $ e^{2jt}\, $ is $ te^{-2jt}\, $ and the response of $ e^{-2jt}\, $ is $ te^{2jt}\, $.
Since the system is a LTI system, we have
Output = Response($ \frac{e^{2jt}+e^{-2jt}}{2}\, $)
= Response of($ \frac{e^{2jt}\, $)+ Response of($ \frac{e^{-2jt}\, $)
$ =\frac{te^{2jt}+te^{-2jt}}{2}\, $ (derive from given conditions)
$ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
$ =t\cos(2t)\, $
