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<math> 1/2 * t * (\cos(-2t) + i\sin(-2t)) + 1/2 * t * (\cos(2t) + i\sin(2t)) </math> | <math> 1/2 * t * (\cos(-2t) + i\sin(-2t)) + 1/2 * t * (\cos(2t) + i\sin(2t)) </math> | ||
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| + | Since <math>sin(-t)=-sin(t)</math> and <math>cos(-t) = cos(t)</math> we can simplify further: | ||
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| + | <math> 1/2*t*(\cos(2t) -i\sin(2t)) + 1/2*t*(\cos(2t)+i\sin(2t)) = t\cos(2t)</math> | ||
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| + | Thus our result for the input of <math>\cos(2t)</math> is <math>t\cos(2t)</math> | ||
Latest revision as of 18:03, 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} $
Simplifying this yields
$ 1/2 * t * (\cos(-2t) + i\sin(-2t)) + 1/2 * t * (\cos(2t) + i\sin(2t)) $
Since $ sin(-t)=-sin(t) $ and $ cos(-t) = cos(t) $ we can simplify further:
$ 1/2*t*(\cos(2t) -i\sin(2t)) + 1/2*t*(\cos(2t)+i\sin(2t)) = t\cos(2t) $
Thus our result for the input of $ \cos(2t) $ is $ t\cos(2t) $
