(New page: I have no idea if i am doing this right because i see a whole mess of people having done the same thing. I looked at the problem and saw that <math>\,\ y(t) = tx(-t)</math> so for <math...) |
(No difference)
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Revision as of 18:44, 18 September 2008
I have no idea if i am doing this right because i see a whole mess of people having done the same thing. I looked at the problem and saw that
$ \,\ y(t) = tx(-t) $
so for $ \,\ x(t)=cos(2t) $ we will get
$ \,\ y(t) = tcos(-2t) $ which is the same as what everybody else has because $ \,\ cos(-2t) = cos(2t) $
