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= Response of <math>\frac{e^{2jt}}{2}\,</math> + Response of <math>\frac{e^{-2jt}}{2}\,</math> | = Response of <math>\frac{e^{2jt}}{2}\,</math> + Response of <math>\frac{e^{-2jt}}{2}\,</math> | ||
| − | <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> ( | + | <math>=\frac{te^{2jt}+te^{-2jt}}{2}\,</math> |
| + | |||
| + | (since the SYSTEM is linear, so we can do addition and division by constant to the output according to the different inputs) | ||
<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> | ||
Latest revision as of 09:32, 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 of $ \frac{e^{2jt}+e^{-2jt}}{2}\, $
= Response of $ \frac{e^{2jt}}{2}\, $ + Response of $ \frac{e^{-2jt}}{2}\, $
$ =\frac{te^{2jt}+te^{-2jt}}{2}\, $
(since the SYSTEM is linear, so we can do addition and division by constant to the output according to the different inputs)
$ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
$ =t\cos(2t)\, $
