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<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> | ||
| + | |||
| + | output will be.. | ||
| + | |||
| + | <math>\frac{1}{2}*tcos(2t) + \frac{1}{2}tcos(2t) = tcos(2t)</math> | ||
| + | |||
| + | so, | ||
| + | |||
| + | <math>cos(2t)\rightarrow system \rightarrow tcos(2)</math> | ||
Latest revision as of 05:40, 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) $
output will be..
$ \frac{1}{2}*tcos(2t) + \frac{1}{2}tcos(2t) = tcos(2t) $
so,
$ cos(2t)\rightarrow system \rightarrow tcos(2) $
