(Linear System Example)
(Non-Linear System Example)
 
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A system takes a given input and produces an output. For the system to be linear it must preserve addition and multiplication. In mathematical terms:
 
A system takes a given input and produces an output. For the system to be linear it must preserve addition and multiplication. In mathematical terms:
  
<math> x(t+t0)=x(t) + x(t0)</math>
+
<math> x(t+t_0)=x(t) + x(t_0)\,</math>
 
   
 
   
 
and
 
and
  
<math> x(k*t)=k*x(t)</math>
+
<math> x(kt)=kx(t)\,</math>
  
 
== Linear System Example ==
 
== Linear System Example ==
  
  
Consider the system <math>\mathbf{x}\mathbf{M}=\mathbf{b} </math>where <math>I^n</math> is the identity matrix and y(t) and x(t) are n x 1 vectors.
+
Consider the system  
<math>Insert formula here</math>
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<math> y[n]=x[n]\cdot\mathbf{M} </math>
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let   
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<math> \mathbf{a} = \begin{bmatrix}2 & 2 \end{bmatrix} </math>
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 +
 
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<math> \mathbf{b} = \begin{bmatrix}4 & 1 \end{bmatrix} </math>
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 +
 
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<math> \mathbf{M} = \begin{bmatrix}1 & 2 \\ 3 & 4 \\ \end{bmatrix} </math>
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 +
<math> k=3\,</math>
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 +
 
 +
 
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If the system is linear these properties hold:
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<math>y[\mathbf{a}+\mathbf{b}]=y[\mathbf{a}]+y[\mathbf{b}] \,</math>
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thus the system is linear.
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<math>y[k\mathbf{a}]=ky[\mathbf{a}] \,</math>
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Here is the proof that the first prop holds:
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<math> y[a] = \begin{bmatrix}8 & 12 \end{bmatrix} </math> 
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<math> y[a+b] = \begin{bmatrix} 6 &3 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} =\begin{bmatrix}24 & 18 \end{bmatrix} </math> 
 +
 
 +
 
 +
<math>y[a]+y[b]=  \begin{bmatrix}8 & 12 \end{bmatrix} +\begin{bmatrix}16 & 6 \end{bmatrix} = \begin{bmatrix}24 & 18 \end{bmatrix}</math>
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And the second:
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<math>ky[\mathbf{a}] =\begin{bmatrix} 24 &36 \end{bmatrix} \,</math>
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<math>y[k\mathbf{a}] =\begin{bmatrix} 6 &6 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \,</math>
 +
<math>              =\begin{bmatrix} 24 &36 \end{bmatrix} </math>
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 +
 
 +
== Non-Linear System Example ==
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 +
 
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Consider the system
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<math> y(t)=x(t)^2 \,</math>
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 +
 
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let   
 +
 
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<math>  x(t_0)=2\,</math>
 +
 
 +
<math> x(t_1)=-2\,</math>
 +
 
 +
 
 +
If the system is linear these properties hold:
 +
 
 +
 
 +
<math>y(t_0+t_1)=y(t_0)+y(t_1) \,</math>
 +
 
 +
 
 +
<math>y(kt)=ky(t) \,</math>
 +
 
 +
 
 +
 
 +
To see if the system is linear test the first property:
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 +
<math> y(t_0)+y(t_1)= 4 +4 =8 \,</math> 
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 +
 
 +
<math> y(t_0+t_1)= x(t_0+t_1)^2= 0\, </math> 
 +
 
 +
 
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Since <math>0\neq 8</math>, the system isn't linear.

Latest revision as of 06:11, 11 September 2008

Linear System Definition

A system takes a given input and produces an output. For the system to be linear it must preserve addition and multiplication. In mathematical terms:

$ x(t+t_0)=x(t) + x(t_0)\, $

and

$ x(kt)=kx(t)\, $

Linear System Example

Consider the system $ y[n]=x[n]\cdot\mathbf{M} $


let

$ \mathbf{a} = \begin{bmatrix}2 & 2 \end{bmatrix} $


$ \mathbf{b} = \begin{bmatrix}4 & 1 \end{bmatrix} $


$ \mathbf{M} = \begin{bmatrix}1 & 2 \\ 3 & 4 \\ \end{bmatrix} $

$ k=3\, $


If the system is linear these properties hold:


$ y[\mathbf{a}+\mathbf{b}]=y[\mathbf{a}]+y[\mathbf{b}] \, $

thus the system is linear.


$ y[k\mathbf{a}]=ky[\mathbf{a}] \, $


Here is the proof that the first prop holds:

$ y[a] = \begin{bmatrix}8 & 12 \end{bmatrix} $


$ y[a+b] = \begin{bmatrix} 6 &3 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} =\begin{bmatrix}24 & 18 \end{bmatrix} $


$ y[a]+y[b]= \begin{bmatrix}8 & 12 \end{bmatrix} +\begin{bmatrix}16 & 6 \end{bmatrix} = \begin{bmatrix}24 & 18 \end{bmatrix} $


And the second:

$ ky[\mathbf{a}] =\begin{bmatrix} 24 &36 \end{bmatrix} \, $

$ y[k\mathbf{a}] =\begin{bmatrix} 6 &6 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \, $ $ =\begin{bmatrix} 24 &36 \end{bmatrix} $


Non-Linear System Example

Consider the system $ y(t)=x(t)^2 \, $


let

$ x(t_0)=2\, $

$ x(t_1)=-2\, $


If the system is linear these properties hold:


$ y(t_0+t_1)=y(t_0)+y(t_1) \, $


$ y(kt)=ky(t) \, $


To see if the system is linear test the first property:

$ y(t_0)+y(t_1)= 4 +4 =8 \, $


$ y(t_0+t_1)= x(t_0+t_1)^2= 0\, $


Since $ 0\neq 8 $, the system isn't linear.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood