(The basics of linearity)
(The basics of linearity)
 
(20 intermediate revisions by the same user not shown)
Line 8: Line 8:
  
 
<math>\cos 2t = \mathrm{Re}\{e^{jt}\} ={e^{2jt} + e^{-2jt} \over 2}</math>
 
<math>\cos 2t = \mathrm{Re}\{e^{jt}\} ={e^{2jt} + e^{-2jt} \over 2}</math>
 +
 +
<math> cos 2t = {e^{2jt} \over 2} + {e^{-2jt} \over 2} </math>
 +
 +
<math>1/2 e^{(2jt)}</math> --->[system]---><math> 1/2 te^{(-2jt)}</math>
 +
 +
<math>1/2 e^{(-2jt)}</math> --->[system]---><math> 1/2 te^{(2jt)}</math>
 +
 +
<math> 1/2 te^{(-2jt)} + 1/2 te^{(2jt)}  = {te^{2jt} + te^{-2jt} \over 2} = t{{e^{2jt} + e^{-2jt}} \over 2} = tcos(2t)</math>
 +
 +
thus, the system's response to cos(2t) is tcos(2t).

Latest revision as of 05:24, 19 September 2008

The basics of linearity

$ e^{(2jt)} $ --->[system]--->$ te^{(-2jt)} $

$ e^{(-2jt)} $ --->[system]--->$ te^{(2jt)} $

$ \cos x = \mathrm{Re}\{e^{ix}\} ={e^{ix} + e^{-ix} \over 2} $

$ \cos 2t = \mathrm{Re}\{e^{jt}\} ={e^{2jt} + e^{-2jt} \over 2} $

$ cos 2t = {e^{2jt} \over 2} + {e^{-2jt} \over 2} $

$ 1/2 e^{(2jt)} $ --->[system]--->$ 1/2 te^{(-2jt)} $

$ 1/2 e^{(-2jt)} $ --->[system]--->$ 1/2 te^{(2jt)} $

$ 1/2 te^{(-2jt)} + 1/2 te^{(2jt)} = {te^{2jt} + te^{-2jt} \over 2} = t{{e^{2jt} + e^{-2jt}} \over 2} = tcos(2t) $

thus, the system's response to cos(2t) is tcos(2t).

Alumni Liaison

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

Buyue Zhang