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Causal Systems

  1. Sep 23, 2007 #1
    We have been going over causal systems and I am still having trouble determining what defines a system to be causal.

    I was told that if the input is anything besides x(a*t) where a=1 then the system is non causal. I can kind of see this, but it is still a bit blurry for me. I also was wondering if that would still apply if you removed t directly from the input equation...

    say like if you had [tex] y(t) = \int_{-\infty}^{t}x(5{\tau}) d\tau [/tex]

    then is this automatically not causal because of the the 5 coefficient on the inside of x()
     
  2. jcsd
  3. Sep 24, 2007 #2
    That's not true.

    The system defined by [tex]y(t) = x(t-1)[/tex] is causal although [tex]x(t-1)[/tex] is something else than [tex]x(t)[/tex].

    The general definition for a causal system (linear or non-linear, time-invariant or time-variant) is:

    Given 2 input signals [tex]x_1(t)[/tex] and [tex]x_2(t)[/tex] such that [tex]x_1(t) = x_2(t)[/tex] for any [tex]t < t_0[/tex], the system is causal if the output signals [tex]y_1(t) = y_2(t)[/tex] for any [tex]t < t_0[/tex]

    If the system is linear then if we apply a signal [tex]x(t) = x_1(t) - x_2(t)[/tex] the output should be [tex]y(t) = y_1(t) - y_2(t)[/tex], so the condition for the system to be causal (in the case of linear systems) reduces to:

    if [tex]x(t) = 0[/tex] for [tex]t<t_0[/tex] then [tex]y(t) = 0[/tex] for [tex]t<t_0[/tex]

    If the system is linear and time invariant, the condition for causality reduces to:

    [tex]h(t)=0[/tex] for [tex]t<0[/tex]

    So depending on the kind of system and your known data you should check one of these conditions.

    In the case of [tex] y(t) = \int_{-\infty}^{t}x(5{\tau}) d\tau [/tex] I know that it's linear because it's defined by an integral which is a linear operation so I will check the second condition.
    I pick an instant [tex]t_0[/tex] at which the output will be [tex] y(t_0) = \int_{-\infty}^{t_0}x(5{\tau}) d\tau [/tex]

    So we see that the output depends on values of [tex]x(t)[/tex] till [tex]5t_0[/tex] but
    we know that [tex]x(t) = 0[/tex] only for [tex]t<t_0[/tex] and thus the output will not be 0 for any [tex]t<t_0[/tex] which means that the system is not causal.
     
    Last edited: Sep 25, 2007
  4. Sep 24, 2007 #3

    rbj

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    i sorta like the Wikipedia definition of causal system (i had a hand in it before they kicked me out of Wikipedia):


    i think you can conclude that for an LTI system, causality is equivalent to the impulse response h(t) being zero for all t < 0. t0 is not a parameter of the impulse response. the impulse response is the LTI to a unit impulse applied at t=0.
     
    Last edited: Sep 24, 2007
  5. Sep 25, 2007 #4
    You're right of course. I just copied the latex expression above it and forgot to replace t0 with 0 as well. I'll edit it now. Thanks!
     
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