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| − | http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t | + | http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t --- Adam Frey |
Revision as of 07:27, 23 July 2009
$ e^{j 2 \pi t} = \left( e^{j 2 \pi} \right)^t= \left( cos{2 \pi } + j sin{2 \pi } \right)^t= 1^t =1 $ This is incorrect (as one sentence). But,
It is correct that:
$ \left( {e^{j 2 \pi }}\right) ^t = 1^t = 1 $
However
$ \left( e^{j 2 \pi t} \right) = \left( cos{2 \pi t} + j sin{2 \pi t } \right) $
So there error would be that (2 Pi t) is the theta, so it must stay in the theta when converted into sin and cos.
Therefore the "t" may not be separated into the exponent in that case.
But if "t" starts off in the exponent, then the result will equal 1.
A great source would be this web site:
http://www00.wolframalpha.com/input/?i=(e^(2+i+pi+))^t --- Adam Frey
