Revision as of 06:28, 18 September 2008 by Han45 (Talk)

first,

$ e^{2jt}=cos(2t) + jsin(2t) $

$ e^{-2jt}=cos(2t) - jsin(2t) $

then we put into system, we got..


$ e^{2jt}\rightarrow system\rightarrow t*e^{-2jt} $

$ e^{-2jt}\rightarrow system\rightarrow t*e^{2jt} $

it means

$ e^{2jt} = t*(cos(2t)-jsin(2t)) $

$ e^{-2jt} = t*(cos(2t)+jsin(2t)) $

input is

$ x(t)=\cos(2t) $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang