(New page: A linear system’s response to <math>e^{2jt}</math> is <math>t*e^{-2jt}</math>, and its response to <math>e^{-2jt}</math> is <math>t*e^{2jt}</math>. What is the system’s response to <...) |
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− | Well, if we convert <math>\cos(2t)</math> using euler's formula, we get <math> 1/2 * e^{2jt | + | Well, if we convert <math>\cos(2t)</math> using euler's formula, we get <math> 1/2 * e^{2jt} + 1/2 * e^{-2jt} </math>. |
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+ | Since the system is linear, we can assume that with constants of 1/2, | ||
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+ | <math> 1/2 * x_1(t) + 1/2*x_2(t) => 1/2*y_1(t)+1/2*y_2(t)</math> | ||
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+ | So our result is | ||
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+ | <math> 1/2 * t * e^{-2jt} + 1/2 * t * e^{2jt} </math> |
Revision as of 17:56, 17 September 2008
A linear system’s response to $ e^{2jt} $ is $ t*e^{-2jt} $, and its response to $ e^{-2jt} $ is $ t*e^{2jt} $.
What is the system’s response to $ \cos(2t) $?
Well, if we convert $ \cos(2t) $ using euler's formula, we get $ 1/2 * e^{2jt} + 1/2 * e^{-2jt} $.
Since the system is linear, we can assume that with constants of 1/2,
$ 1/2 * x_1(t) + 1/2*x_2(t) => 1/2*y_1(t)+1/2*y_2(t) $
So our result is
$ 1/2 * t * e^{-2jt} + 1/2 * t * e^{2jt} $