| Line 28: | Line 28: | ||
<math>\,y(t)=t\cos(2t)\,</math> | <math>\,y(t)=t\cos(2t)\,</math> | ||
| + | |||
| + | == Comments on Other Answers == | ||
Revision as of 16:43, 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}\, $
$ \,x(t)=\frac{1}{2}x_1(t)+\frac{1}{2}x_2(t)\, $
Because the system is linear, we can write the response as
$ \,y(t)=\frac{1}{2}y_1(t)+\frac{1}{2}y_2(t)\, $
$ \,y(t)=\frac{1}{2}te^{-2jt}+\frac{1}{2}te^{2jt}\, $
$ \,y(t)=t(\frac{e^{2jt}+e^{-2jt}}{2})\, $
$ \,y(t)=t\cos(2t)\, $
