(New page: ==Linearity== The System is defined as follows <math> e^{2jt} \to System\to te^{-2jt}\!</math> <math> e^{-2jt} \to System \to te^{2jt}\!</math> Now if a system is Linear then it is als...) |
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Latest revision as of 14:17, 18 September 2008
Linearity
The System is defined as follows
$ e^{2jt} \to System\to te^{-2jt}\! $
$ e^{-2jt} \to System \to te^{2jt}\! $
Now if a system is Linear then it is also additive and homogeneous
Thus $ \frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} \to System\to \frac{1}{2}te^{-2jt}+ \frac{1}{2}te^{2jt}\! $
This can also be written as
$ \frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} \to System\to \frac{1}{2}te^{-2jt}+ \frac{1}{2}te^{2jt}\! $= $ t[\frac{1}{2}e^{-2jt}+ \frac{1}{2}e^{2jt}] $
Now, $ cos(2t)=\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} $
Thus
$ cos(2t)=\frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} \to System \to \frac{1}{2}te^{-2jt}+ \frac{1}{2}te^{2jt}=t[\frac{1}{2}e^{-2jt}+ \frac{1}{2}e^{2jt}]=tcos(2t) $
So we see that for input $ cos(2t) $
$ cos(2t) \to System \to tcos(2t) $
