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<math>e^{-2jt} = \cos -2t + j\sin -2t \!</math> | <math>e^{-2jt} = \cos -2t + j\sin -2t \!</math> | ||
| − | <math> = \cos | + | <math> = \cos 2t - j\sin 2t \!</math> |
| + | |||
| + | |||
| + | <math>{e^{2jt}+e^{-2jt}}/2=2cos2t\!</math> | ||
Revision as of 14:57, 19 September 2008
$ e^{2jt} $ yields $ te^{-2jt} $ and $ e^{-2jt} $ yields $ te^{2jt} $
and the system is linear
since Euler's formulat states that : $ e^{jx} = \cos x + j\sin x \! $
$ e^{2jt} = \cos 2t + j\sin 2t \! $
and
$ e^{-2jt} = \cos -2t + j\sin -2t \! $ $ = \cos 2t - j\sin 2t \! $
$ {e^{2jt}+e^{-2jt}}/2=2cos2t\! $
