(Non-Linear)
 
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their sum. (27Y(t) - 90) + (72Y(t)-60)) = 99Y(t)-150
 
their sum. (27Y(t) - 90) + (72Y(t)-60)) = 99Y(t)-150
  
An example of a linear system is shown below:
+
Now we will multiply the original signals by the constants, take their sum, and then send them through the system.
 +
If we end up with 99Y(t)-150, then the system must be linear.  So, (9*1X) + (6*4X) = 33x.  This gives 99Y(t) - 150.
 +
Therefore the system is linear.
 +
</pre>
  
x1(t) --> system --> y1(t)  
+
==Non-Linear==
x2(t) --> system --> y2(t)
+
<pre>
 +
SYSTEM: y = (x(t))^5
 +
a.) 1X1(t) --> SYSTEM --> (x(t))^5
 +
b.) 4X2(t) --> SYSTEM --> (4x(t))^5
 +
 
 +
To prove that this system is non-linear, we multiply output "a" by 9, then multiply output "b" by 6.  Then, we take their sum.  The calculations are as follows: 9x(t)^5 + 24x(t)^5 = 33x(t)^5.
 +
 
 +
Now we try multiplying the input in part "a" by 9 and the input of part "b" by 6.  This gives us 9x(t) + 6x(t).
 +
Now we take their sum.  This gives us 15x(t).  Now, we run 15x(t) through the system to obtain: 759375(X(t))^5.
 +
Therefore, the system is not linear.
 
</pre>
 
</pre>

Latest revision as of 18:32, 10 September 2008

Linear system

SYSTEM: y = 3x(t) - 10
a.) 1X1(t) --> SYSTEM --> 3Y1(t) - 10
b.) 4X2(t) --> SYSTEM --> 12Y2(t) - 10

We can do the following proof to show that the above system is linear.  Take two random constant numbers
such as 9 and 6.  Now multiply the output from "a" by 9.  Then multiply the output from "b" by 6.  Now take
their sum. (27Y(t) - 90) + (72Y(t)-60)) = 99Y(t)-150

Now we will multiply the original signals by the constants, take their sum, and then send them through the system.
If we end up with 99Y(t)-150, then the system must be linear.  So, (9*1X) + (6*4X) = 33x.  This gives 99Y(t) - 150.
Therefore the system is linear.

Non-Linear

SYSTEM: y = (x(t))^5
a.) 1X1(t) --> SYSTEM --> (x(t))^5
b.) 4X2(t) --> SYSTEM --> (4x(t))^5

To prove that this system is non-linear, we multiply output "a" by 9, then multiply output "b" by 6.  Then, we take their sum.  The calculations are as follows: 9x(t)^5 + 24x(t)^5 = 33x(t)^5.

Now we try multiplying the input in part "a" by 9 and the input of part "b" by 6.  This gives us 9x(t) + 6x(t).
Now we take their sum.  This gives us 15x(t).  Now, we run 15x(t) through the system to obtain: 759375(X(t))^5.
Therefore, the system is not linear.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva