(New page: == Linearity == A system is said to be linear if it satisfies the properties of scaling and superposition. Thus, the following holds true for all linear systems: :Suppose there are two in...)
 
 
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:A linear system must satisfy the condition
 
:A linear system must satisfy the condition
 
::<math>\,ay1(t) + by2(t) = C\left\{ax1(t) + bx2(t)\right\}</math>
 
::<math>\,ay1(t) + by2(t) = C\left\{ax1(t) + bx2(t)\right\}</math>
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== Example of a Linear System ==
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::<math>\,x1(t) = sin(t)</math>
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::<math>\,x2(t) = cos(t)</math>
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::<math>\,y1(t) = \pi\left\{x1(t)\right\} = \pi(sin(t))</math>
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::<math>\,y2(t) = \pi\left\{x2(t)\right\} = \pi(cos(t))</math>
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::<math>\,ay1(t) + by2(t) = a*\pi*sin(t) + b*\pi*cos(t) = \pi\left\{asin(t) + bcos(t)\right\}
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= \pi\left\{ax1(t) + bx2(t)\right\}</math>
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Thus, <math>\,y(t) = \pi(x(t))</math> is a linear system.
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== Example of a Non Linear System ==
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::<math>\,x1(t) = sin(t) </math>
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::<math>\,x2(t) = cos(t) </math>
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::<math>\,y1(t) = ln(x1(t)) = ln(sin(t))</math>
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::<math>\,y2(t) = ln(x2(t)) = ln(cos(t))</math>
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::<math>\,ay1(t) + by2(t) = a*ln(sin(t)) + b*ln(cos(t)) \neq ln(ax1(t) + bx2(t))</math>
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Thus, <math>\,y(t) = ln(x(t))</math> is not a linear system.

Latest revision as of 05:55, 10 September 2008

Linearity

A system is said to be linear if it satisfies the properties of scaling and superposition. Thus, the following holds true for all linear systems:

Suppose there are two inputs
$ \,x1(t) $
$ \,x2(t) $
with outputs
$ \,y1(t) = C\left\{x1(t)\right\} $
$ \,y2(t) = C\left\{x2(t)\right\} $
A linear system must satisfy the condition
$ \,ay1(t) + by2(t) = C\left\{ax1(t) + bx2(t)\right\} $

Example of a Linear System

$ \,x1(t) = sin(t) $
$ \,x2(t) = cos(t) $
$ \,y1(t) = \pi\left\{x1(t)\right\} = \pi(sin(t)) $
$ \,y2(t) = \pi\left\{x2(t)\right\} = \pi(cos(t)) $
$ \,ay1(t) + by2(t) = a*\pi*sin(t) + b*\pi*cos(t) = \pi\left\{asin(t) + bcos(t)\right\} = \pi\left\{ax1(t) + bx2(t)\right\} $


Thus, $ \,y(t) = \pi(x(t)) $ is a linear system.

Example of a Non Linear System

$ \,x1(t) = sin(t) $
$ \,x2(t) = cos(t) $
$ \,y1(t) = ln(x1(t)) = ln(sin(t)) $
$ \,y2(t) = ln(x2(t)) = ln(cos(t)) $
$ \,ay1(t) + by2(t) = a*ln(sin(t)) + b*ln(cos(t)) \neq ln(ax1(t) + bx2(t)) $



Thus, $ \,y(t) = ln(x(t)) $ is not a linear system.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood