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Instead of this method, we should expand it in terms of <math>\cos(2t) =\frac{e^{2jt}+e^{-2jt}}{2}\,</math>, as we what will happen to a random signal other than the signal shown. It might produce a different random response. By expanding it we can be sure it's right. - Wei Jian Chan | Instead of this method, we should expand it in terms of <math>\cos(2t) =\frac{e^{2jt}+e^{-2jt}}{2}\,</math>, as we what will happen to a random signal other than the signal shown. It might produce a different random response. By expanding it we can be sure it's right. - Wei Jian Chan | ||
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| + | I disagree with you. If the input is expanded to be <math>e^{2jt} = cos(t) + 2j*sin(t)</math> and the output is <math>t*e^{-2jt} = t*cos(t) - 2jt*sin(t)</math> then since the system is linear, the output of cos(2t) should be <math>t*cos(2t)</math>. Does anyone disagree with this? It was just a thought. Let me know. | ||
| + | -Tyler Johnson | ||
Revision as of 05:35, 17 September 2008
Comments
Instead of this method, we should expand it in terms of $ \cos(2t) =\frac{e^{2jt}+e^{-2jt}}{2}\, $, as we what will happen to a random signal other than the signal shown. It might produce a different random response. By expanding it we can be sure it's right. - Wei Jian Chan
I disagree with you. If the input is expanded to be $ e^{2jt} = cos(t) + 2j*sin(t) $ and the output is $ t*e^{-2jt} = t*cos(t) - 2jt*sin(t) $ then since the system is linear, the output of cos(2t) should be $ t*cos(2t) $. Does anyone disagree with this? It was just a thought. Let me know.
-Tyler Johnson
