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<math>e^{ix} = cosx + isinx</math>
 
<math>e^{ix} = cosx + isinx</math>
 
  
 
input is  
 
input is  
  
 
<math>x(t)=\cos(2t)</math>
 
<math>x(t)=\cos(2t)</math>
 +
 +
output will be..
 +
 +
<math>\frac{1}{2}*tcos(2t) + \frac{1}{2}tcos(2t) = tcos(2t)</math>
 +
 +
so,
 +
 +
<math>cos(2t)\rightarrow system \rightarrow tcos(2)</math>

Latest revision as of 05:40, 18 September 2008

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)) $

using Euler's formula

$ e^{ix} = cosx + isinx $

input is

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

output will be..

$ \frac{1}{2}*tcos(2t) + \frac{1}{2}tcos(2t) = tcos(2t) $

so,

$ cos(2t)\rightarrow system \rightarrow tcos(2) $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009