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Tyler, i think you got your expansion wrong. <math>e^{2jt} = cos(2t) + jsin(2t)</math> and then <math>e^{-2jt} = cos(-2t) + jsin(-2t) = cos(2t) - jsin(2t)</math> Try that and apply it and you should get the answer you thought it was. Dont really know how you got tcos(2t) from what you typed out, but i might have read it wrong. -Steve Anderson | Tyler, i think you got your expansion wrong. <math>e^{2jt} = cos(2t) + jsin(2t)</math> and then <math>e^{-2jt} = cos(-2t) + jsin(-2t) = cos(2t) - jsin(2t)</math> Try that and apply it and you should get the answer you thought it was. Dont really know how you got tcos(2t) from what you typed out, but i might have read it wrong. -Steve Anderson | ||
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| + | Good job on the corrections. Using the expansion worked well for me too. | ||
Revision as of 06:02, 19 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
Tyler, i think you got your expansion wrong. $ e^{2jt} = cos(2t) + jsin(2t) $ and then $ e^{-2jt} = cos(-2t) + jsin(-2t) = cos(2t) - jsin(2t) $ Try that and apply it and you should get the answer you thought it was. Dont really know how you got tcos(2t) from what you typed out, but i might have read it wrong. -Steve Anderson
Good job on the corrections. Using the expansion worked well for me too.
