(New page: == System Response == <math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>. We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^...) |
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== System Response == | == System Response == | ||
| − | <math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>. | + | Based on the Euler Formula, <math>\cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\,</math>. |
We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^{-2jt}\,</math> is <math>te^{2jt}\,</math>. | We already had the response of <math>e^{2jt}\,</math> is <math>te^{-2jt}\,</math> and the response of <math>e^{-2jt}\,</math> is <math>te^{2jt}\,</math>. | ||
| − | + | Since the system is a LTI system, we have | |
| − | + | Output = Response(<math>\frac{e^{2jt}+e^{-2jt}}{2}\,</math>) | |
| − | <math>\frac{te^{2jt}+te^{-2jt}}{2}\,</math> | + | <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> |
| − | <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 07:50, 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}\, $)
$ =\frac{te^{2jt}+te^{-2jt}}{2}\, $
$ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
$ =t\cos(2t)\, $
