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While cos(2t) function can be found with this equation: | While cos(2t) function can be found with this equation: | ||
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| − | <math>\frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} = \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) = cos{(2t)}</math><br> | + | <math>\frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} =</math><br><math> \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) = cos{(2t)}</math><br> |
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Revision as of 09:35, 16 September 2008
Since
$ e^{2jt} \rightarrow system \rightarrow te^{-2jt}\! $
$ e^{-2jt} \rightarrow system \rightarrow te^{2jt}\! $
and using euler formula, we can replace exponent expressions with
Euler's formula: $ e^{iy}=cos(y)+isin(y)\, $
They will change into:
$ e^{(2jt)} = cos{(2t)} + jsin{(2t)} --> system --> t*{(cos{(2t)} - jsin{(2t)})}\, $
$ e^{(-2jt)} = cos{(2t)} - jsin{(2t)} --> system --> t*{(cos{(2t)} + jsin{(2t)})}\, $
It indicates that the system changes the expression on the middle of cos and sin.
While cos(2t) function can be found with this equation:
$ \frac{1}{2}e^{(2jt)} + \frac{1}{2}e^{(-2jt)} = $
$ \frac{1}{2}(cos{(2t)} + jsin{(2t)}) + \frac{1}{2}(cos{(2t)} - jsin{(2t)}) = cos{(2t)} $
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$ \frac{1}{2}(t*{(cos{(2t)} - jsin{(2t)})}) + \frac{1}{2}t*{(cos{(2t)} + jsin{(2t)})} = \frac{1}{2}tcos{(2t)} + \frac{1}{2}tcos{(2t)} = tcos({(2t)} $
