(Basics of Linearity)
(Basics of Linearity)
 
(2 intermediate revisions by the same user not shown)
Line 5: Line 5:
 
: The Signal is Linear
 
: The Signal is Linear
  
 +
:Since the system is linear you can split the signal in two parts
  
:<math>\cos x = \dfrac{e^{i x}+e^{-i x}}{2}</math>
 
 
:<math>\cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2}</math>
 
:<math>\cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2}</math>
  
Line 18: Line 18:
 
:so the response is equal to
 
:so the response is equal to
  
:<math>\t cos 2t \, </math>
+
:<math>t\cos 2t \, </math>

Latest revision as of 07:35, 18 September 2008

Basics of Linearity

Given

$ e^{2 x i}=t e^{-2 x i}\, $
$ e^{-2 x i}=t e^{2 x i}\, $
The Signal is Linear
Since the system is linear you can split the signal in two parts
$ \cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2} $

The Systems response to $ \cos 2x $ is $ \ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} $ but

$ e^{2 x i}=\cos 2x + i \sin 2x \, $ and
$ e^{-2 x i}=\cos 2x - i \sin 2x \, $
so the response is equal to
$ t\cos 2t \, $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood