1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear System - need some help

  1. Apr 23, 2007 #1
    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 !
     
    Last edited: Apr 23, 2007
  2. jcsd
  3. Apr 23, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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].

    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.

    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.
     
  4. Apr 23, 2007 #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!
     
  5. Apr 23, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear System - need some help
  1. Linear system (Replies: 1)

Loading...