How Do You Test if a System is Linear?

Take a look at the definition you gave for a linear function. That's what it means to say "whether x is linear". That is, whether the operator "H" is linear. So the question of whether y is linear is the same as asking whether H is linear. In summary, the conversation discusses the determination of whether a system is linear or not. The professor provided a "recipe" for testing a system and the question asks for guidance in understanding the solution for a specific example. The answer defines a linear system and explains that the input in this case is t and y acts upon it. It is clarified that the input x is a function, not a number, and the linearity of y is dependent upon the
  • #1
FrogPad
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I basically need to know how to test a system to see if it is linear or not. The professor gave us a "recipe" to check a system. I can kinda follow the recipe, but I would like to understand it, so I have reached out to other sources. The other sources deal with operators from the beginning... thus, I am trying to understand them. The following simple example is throwing me for a loop.

Question:
Determine if the following system is linear.

[tex] y(t) = x(t-2) + x(2-t) [/tex]


Answer:
A system is linear if:
[tex] H \{ \alpha x_1 +\beta x_2 \} = \alpha H\{ x_1 \} + \alpha H\{ x_2 \} [/tex]

I have the solution (the answer is yes - linear), however I do not understand it. What is [tex] H\{ \} [/tex] in this case?

It makes sense to me in examples such as:
[tex] y(t) = \sin(t)x(t) [/tex] so [itex] \sin(t) [/itex] is acting on the input [itex] x(t) [/itex]. But... what is acting on the input here (the x(t-2) + x(2-t) example)?

In words I imagine [itex] H [/itex] to be something that yields an ouput by adding a shifted input to a shifted input. I'm kinda lost... guidance would be helpful.

thanks !
 
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  • #2
FrogPad said:
I basically need to know how to test a system to see if it is linear or not. The professor gave us a "recipe" to check a system. I can kinda follow the recipe, but I would like to understand it, so I have reached out to other sources. The other sources deal with operators from the beginning... thus, I am trying to understand them. The following simple example is throwing me for a loop.

Question:
Determine if the following system is linear.

[tex] y(t) = x(t-2) + x(2-t) [/tex]


Answer:
A system is linear if:
[tex] H \{ \alpha x_1 +\beta x_2 \} = \alpha H\{ x_1 \} + \alpha H\{ x_2 \} [/tex]

I have the solution (the answer is yes - linear), however I do not understand it. What is [tex] H\{ \} [/tex] in this case?
Well, what was "H" in your definition- it was the given system. Here H is y(t). Now, I have a question- is your "x" in the definition of y(t) a number (so that "x(t-2)" means x times t-2) or is it a function (so that x(t-2) mean x applied to t-2)? I'm going to assume that it is a number. In that case, y(t)= x(t-2)+ x(2-t)= xt- 2x+ 2x- xt= 0 for all x!

Okay, just apply your definition with y in place of H:
[itex]y(\{ \alpha t_1 +\beta t_2 \} = 0[/itex] while [itex]\alpha y(t_1)= \alpha(0)= 0[/itex] and [itex]\beta y(t_2)= \beta(0)= 0[/itex].

It makes sense to me in examples such as:
[tex] y(t) = \sin(t)x(t) [/tex] so [itex] \sin(t) [/itex] is acting on the input [itex] x(t) [/itex]. But... what is acting on the input here (the x(t-2) + x(2-t) example)?
No, x(t-2)+ x(2-t) is NOT the "input"- it's the definition of the function. Because the problem says "y(t)= ", t is the "input" and y is acting on it.

In words I imagine [itex] H [/itex] to be something that yields an ouput by adding a shifted input to a shifted input. I'm kinda lost... guidance would be helpful.

thanks !
Once again, the "input" is t and y is the function. A rather trivial function if my guess about x being a number is correct. If, on the other hand, x is a function then whether y is linear depends upon whether x is linear.
 
  • #3
I'm sorry I wasn't specific with my notation. I should have been.

The condition for a system to be linear would be more appropriately written as:
[tex] H \{ \alpha x_1(t) +\beta x_2(t) \} = \alpha H\{ x_1(t) \} + \alpha H\{ x_2(t) \} [/tex]

[tex] \alpha, \,\,\, \beta [/tex] are scalars
[tex] x_1(t) , \,\,\, x_2(t) [/tex] are functions with independent variables [itex] t [/itex]

Thus,
[tex] y(t) = x(t-2) + x(2-t) [/tex]

If we defined an operator such that [itex] S_\lambda x(t) = x(t - \lambda) [/itex]. (By the way, I don't really know how to write this properly.)

Then we could write the above system as:
[tex] y(t) = S_2 x(t) + S_{-2}x(t) [/tex]

I hope this explains it better. Thanks for the help HallsofIvy!
 
  • #4
The reason I assumed that you were just multiplying numbers x1 and x2 is that if they are functions then whether or not y is a linear function is entirely dependent upon whether x1 and x2 are. In fact, y is linear if and only if x1 and x2 are linear.
 

FAQ: How Do You Test if a System is Linear?

1. What is a linear system?

A linear system is a set of equations that can be represented by a system of linear equations, where the variables are raised to the first power and there are no products or powers of variables. In other words, it is a system of equations that can be graphed as a straight line.

2. What is the purpose of solving a linear system?

The purpose of solving a linear system is to find the values of the variables that satisfy all of the equations in the system. This allows us to find the intersection point(s) of the lines represented by the equations, which can provide valuable information in many fields such as engineering, economics, and physics.

3. How do you solve a linear system?

There are several methods for solving a linear system, including substitution, elimination, and graphing. Substitution involves solving for one variable in terms of the other and then substituting that expression into the other equations. Elimination involves adding or subtracting equations to eliminate one variable at a time. Graphing involves graphing each equation and finding the intersection point(s) of the lines.

4. Can a linear system have more than one solution?

Yes, a linear system can have one, none, or infinitely many solutions. It all depends on the relationship between the equations. If the equations are consistent and intersect at one point, there is one solution. If the equations are inconsistent and do not intersect, there is no solution. If the equations are dependent and represent the same line, there are infinitely many solutions.

5. What are some real-life applications of linear systems?

Linear systems have many real-life applications, such as determining the optimal production levels for a company, calculating the break-even point for a business, predicting population growth, and analyzing the motion of objects in physics. They can also be used in computer algorithms for image processing and pattern recognition.

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