Line 5: Line 5:
 
<math>e^{-2jt}=cos(2t) - jsin(2t)</math>
 
<math>e^{-2jt}=cos(2t) - jsin(2t)</math>
  
 +
then we put into system, we got..
  
  
Line 10: Line 11:
  
 
<math>e^{-2jt}\rightarrow system\rightarrow t*e^{2jt}</math>
 
<math>e^{-2jt}\rightarrow system\rightarrow t*e^{2jt}</math>
 +
 +
it means
 +
 +
<math>e^{2jt} = t*(cos(2t)-jsin(2t))
 +
</math>
 +
 +
<math>e^{-2jt} = t*(cos(2t)+jsin(2t))
 +
</math>
  
 
input is  
 
input is  
  
 
<math>x(t)=\cos(2t)</math>
 
<math>x(t)=\cos(2t)</math>

Revision as of 06:28, 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)) $

input is

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

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn