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<math>e^{2jt}</math> yields <math>te^{-2jt}</math> | <math>e^{2jt}</math> yields <math>te^{-2jt}</math> | ||
| + | and | ||
| + | <math>e^{-2jt}</math> yields <math>te^{2jt}</math> | ||
| + | |||
| + | and the system is linear | ||
| + | |||
| + | since Euler's formulat states that : <math>e^{jx} = \cos x + j\sin x \!</math> | ||
| + | |||
| + | <math>e^{2jt} = \cos 2t + j\sin 2t \!</math> | ||
| + | |||
| + | and | ||
| + | |||
| + | <math>e^{-2jt} = \cos -2t + j\sin -2t \!</math> | ||
| + | <math> = \cos 2t - j\sin 2t \!</math> | ||
| + | |||
| + | |||
| + | <math>\frac{e^{2jt}+e^{-2jt}}{2}=cos2t\!</math> | ||
| + | |||
| + | Putting <math>cos2t</math> as an input will yield <math>\frac{te^{2jt}+te^{-2jt}}{2}=tcos2t=tcos-2t\!</math> | ||
Latest revision as of 16:03, 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 \! $
$ \frac{e^{2jt}+e^{-2jt}}{2}=cos2t\! $
Putting $ cos2t $ as an input will yield $ \frac{te^{2jt}+te^{-2jt}}{2}=tcos2t=tcos-2t\! $
