(New page: ==The basics of linearity== <math>cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! </math> We are given that the system's output for <math>e^{2jt} \!</math> is <math>t*e^{-2jt} \!</math>, and ...) |
(No difference)
|
Latest revision as of 13:43, 18 September 2008
The basics of linearity
$ cos(2t)=\frac{1}{2}(e^{2jt}+e^{-2jt}) \! $
We are given that the system's output for $ e^{2jt} \! $ is $ t*e^{-2jt} \! $, and the output for $ e^{-2jt} \! $ is $ t*e^{2jt} \! $. Since the system is linear, we can say that $ \frac{1}{2}(te^{-2jt}+te^{2jt})=tcos(2t) $.
It is rather straightforward.
