(New page: Given: <math> e^{2jt} \rightarrow System \rightarrow te^{-2jt}</math> <math> e^{-2jt} \rightarrow System \rightarrow te^{2jt}</math> Euler's formula: <font size = 4><math>e^{iy}=cos(...) |
(No difference)
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Latest revision as of 11:09, 15 September 2008
Given:
$ e^{2jt} \rightarrow System \rightarrow te^{-2jt} $
$ e^{-2jt} \rightarrow System \rightarrow te^{2jt} $
Euler's formula: $ e^{iy}=cos(y)+isin(y) $
Use Euler's formula to get cos(2t) in the the form of the givens:
$ \frac{1}{2}e^{2jt}+\frac{1}{2}e^{-2jt}=\frac{1}{2}cos(2t)+\frac{1}{2}isin(2t)+\frac{1}{2}cos(2t)-\frac{1}{2}isin(2t)=cos(2t) $
So, the systems response to cos(2t is:
$ cos(2t) \rightarrow System \rightarrow tcos(2t) $