(Non-Linear System Example)
 
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Consider the system  
 
Consider the system  
<math> y(t)=x(n)^2 \,</math>  
+
<math> y(t)=x(t)^2 \,</math>  
  
  
 
let     
 
let     
  
<math> \mathbf{a} = \begin{bmatrix}2 & 2 \end{bmatrix} </math>
+
<math> x(t_0)=2\,</math>
 
+
 
+
<math> \mathbf{b} = \begin{bmatrix}4 & 1 \end{bmatrix} </math>
+
 
+
 
+
<math> \mathbf{M} = \begin{bmatrix}1 & 2 \\ 3 & 4 \\ \end{bmatrix} </math>
+
 
+
<math> k=3\,</math>
+
  
 +
<math>  x(t_1)=-2\,</math>
  
  
Line 86: Line 79:
  
  
<math>y[\mathbf{a}+\mathbf{b}]=y[\mathbf{a}]+y[\mathbf{b}] \,</math>
+
<math>y(t_0+t_1)=y(t_0)+y(t_1) \,</math>
 
+
thus the system is linear.
+
 
+
 
+
 
+
<math>y[k\mathbf{a}]=ky[\mathbf{a}] \,</math>
+
 
+
 
+
  
Here is the proof that the first prop holds:
 
  
<math> y[a] = \begin{bmatrix}8 & 12 \end{bmatrix} </math>
+
<math>y(kt)=ky(t) \,</math>
  
  
<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> 
 
  
 +
To see if the system is linear test the first property:
  
<math>y[a]+y[b]= \begin{bmatrix}8 & 12 \end{bmatrix} +\begin{bmatrix}16 & 6 \end{bmatrix} = \begin{bmatrix}24 & 18 \end{bmatrix}</math>
+
<math> y(t_0)+y(t_1)= 4 +4 =8 \,</math>
  
  
And the second:
+
<math> y(t_0+t_1)= x(t_0+t_1)^2= 0\, </math> 
  
<math>ky[\mathbf{a}] =\begin{bmatrix} 24 &36 \end{bmatrix} \,</math>
 
  
<math>y[k\mathbf{a}] =\begin{bmatrix} 6 &6 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} \,</math>
+
Since <math>0\neq 8</math>, the system isn't linear.
<math>              =\begin{bmatrix} 24 &36 \end{bmatrix} </math>
+

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

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett