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Good job on the corrections. Using the expansion worked well for me too. | Good job on the corrections. Using the expansion worked well for me too. | ||
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| + | It's very unclear how you went about your derivation. You shouldn't write as if you were teaching yourself. That'll get you more points on exams as I've found that over clarifying answers usually let's you stumble across some partial credit, and it will make your answers more useful here. -Allen Humphreys | ||
Latest revision as of 15:59, 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.
It's very unclear how you went about your derivation. You shouldn't write as if you were teaching yourself. That'll get you more points on exams as I've found that over clarifying answers usually let's you stumble across some partial credit, and it will make your answers more useful here. -Allen Humphreys
