(New page: Hmm... this space seems empty...) |
|||
| Line 1: | Line 1: | ||
| − | + | We are told that a system is linear and given inputs | |
| + | |||
| + | <math>\,x_1(t)=e^{2jt}\,</math> yields <math>\,y_1(t)=te^{-2jt}\,</math> | ||
| + | |||
| + | <math>\,x_2(t)=e^{-2jt}\,</math> yields <math>\,y_2(t)=te^{2jt}\,</math> | ||
| + | |||
| + | |||
| + | The input | ||
| + | |||
| + | <math>\,x(t)=cos(2t)\,</math> | ||
| + | |||
| + | can be rewritten as | ||
| + | |||
| + | <math>\,x(t)=\frac{e^{2jt}+e^{-2jt}}{2}\,</math> | ||
| + | |||
| + | <math>\,x(t)=\frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt}\,</math> | ||
Revision as of 16:35, 17 September 2008
We are told that a system is linear and given inputs
$ \,x_1(t)=e^{2jt}\, $ yields $ \,y_1(t)=te^{-2jt}\, $
$ \,x_2(t)=e^{-2jt}\, $ yields $ \,y_2(t)=te^{2jt}\, $
The input
$ \,x(t)=cos(2t)\, $
can be rewritten as
$ \,x(t)=\frac{e^{2jt}+e^{-2jt}}{2}\, $
$ \,x(t)=\frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt}\, $
