(New page: <math>\cos{2t} = \frac{e^{2jt} - e^{-2jt}}{2}</math> Since we know, <math>e^{2jt} \Longrightarrow System \Longrightarrow te^{-2jt}</math> and <math>e^{-2jt} \Longrightarrow System \Lon...) |
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− | <math>\cos{2t} = \frac{e^{2jt} | + | <math>\cos{2t} = \frac{e^{2jt} + e^{-2jt}}{2}</math> |
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− | <math>\cos{2t} \Longrightarrow System \Longrightarrow \frac{ | + | <math>\cos{2t} \Longrightarrow System \Longrightarrow t(\frac{e^{-2jt} + e^{2jt}}{2}) = t\cos{2t}</math> |
Latest revision as of 15:06, 18 September 2008
$ \cos{2t} = \frac{e^{2jt} + e^{-2jt}}{2} $
Since we know,
$ e^{2jt} \Longrightarrow System \Longrightarrow te^{-2jt} $
and
$ e^{-2jt} \Longrightarrow System \Longrightarrow te^{2jt} $
then
$ \cos{2t} \Longrightarrow System \Longrightarrow t(\frac{e^{-2jt} + e^{2jt}}{2}) = t\cos{2t} $