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Homework Help: 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

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    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

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    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.
     
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