(→Comments) |
|||
| Line 7: | Line 7: | ||
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. | 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 | -Tyler Johnson | ||
| + | |||
| + | |||
| + | 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 | ||
Revision as of 07:46, 18 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
