Revision as of 15:04, 12 September 2008 by Dbarjum (Talk)

LINEARITY

For a system to be called Linear the following two scenarios must yield output signals that are equal to each other.


1) Signals $ X_1 $ and $ Y_1 $ are first multiplied by constants $ C_1 \in \mathbb{C} $ and $ C_2\in \mathbb{C} $ respectively, then added together and passed through a system that yields a signal $ Z(t) $.

and

2) Signals $ X_1 $ and $ Y_1 $ each pass through a system, their results are multiplied by constants $ C_1 \in \mathbb{C} $ and $ C_2\in \mathbb{C} $ respectively, and then added together yielding a signal $ W(t) $.

For this system to be linear, signals $ Z(t) $ and $ W(t) $ must be equal to each other.

$ Z(t) = W(t) $

LINEAR SYSTEM

NON-LINEAR SYSTEM

$ X(t) \to Y(t)^3 $


PROOF


let $ a \in \mathbb{{C}} $ and $ b \in \mathbb{{C}} $,


$ X_1(t) \Rightarrow Y_1(t) = X_1(t)^3, a*X_1(t)^3 \downarrow $

........................................................................$ \bigoplus \to Z(t) = a*X_1(t)^3 + b*X_2(t)^3 $

$ X_2(t) \Rightarrow Y_2(t) = X_2(t)^3, b*X_2(t)^3 \uparrow $



$ a*X_1(t) \downarrow $

..................$ \bigoplus \to a*X_1(t) + b*X_2(t) \Rightarrow W(t)^3 = (a*X_1(t) + b*X_2(t))^3 $

$ a*X_2(t) \uparrow $

$ Z(t) \not\equiv W(t) \Rightarrow $ Non-Linear System

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood