Line 19: | Line 19: | ||
<math>e^{-2jt} = t*(cos(2t)+jsin(2t)) | <math>e^{-2jt} = t*(cos(2t)+jsin(2t)) | ||
</math> | </math> | ||
+ | |||
+ | using Euler's formula | ||
+ | |||
+ | <math>e^{ix} = cosx + isinx</math> | ||
+ | |||
input is | input is | ||
<math>x(t)=\cos(2t)</math> | <math>x(t)=\cos(2t)</math> |
Revision as of 05:35, 18 September 2008
first,
$ e^{2jt}=cos(2t) + jsin(2t) $
$ e^{-2jt}=cos(2t) - jsin(2t) $
then we put into system, we got..
$ e^{2jt}\rightarrow system\rightarrow t*e^{-2jt} $
$ e^{-2jt}\rightarrow system\rightarrow t*e^{2jt} $
it means
$ e^{2jt} = t*(cos(2t)-jsin(2t)) $
$ e^{-2jt} = t*(cos(2t)+jsin(2t)) $
using Euler's formula
$ e^{ix} = cosx + isinx $
input is
$ x(t)=\cos(2t) $