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Checking for time invariance

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider the following input-output relationship:
    [tex] y(t) = \int_0^\infty e^{-\sigma}x(t-\sigma) d\sigma [/tex]

    A) Is the system time-invariant?
    B) Find the output y(t) when the input to the system is [tex] x(t) = \mid t \mid , -\infty < t < \infty [/tex]

    2. Relevant equations

    These are the equations to check for time invariance

    A system with an input-output transformation y(*) = T[x(*)] is time invariant if for any t and [tex] \tau [/tex]

    [tex] y(t) = T[x(t)] [/tex]

    [tex] z(t) = x(t-\tau)[/tex]

    [tex] T[z(t)] = y(t-\tau)
    [/tex]
    3. The attempt at a solution

    Edit: Ok I think it's time to get some sleep. This problem was actually pretty simple and I totally screwed it up earlier.

    [tex] z(t) = T[x(t-\tau)] = \int_0^\infty e^{-\sigma}z(t - \tau - \sigma) d\sigma [/tex]

    [tex] y(t-\tau) = \int_0^\infty e^{-\sigma}z(t - \tau - \sigma) d\sigma [/tex]

    therefore [tex] T[x(t-\tau)] = y(t-\tau) [/tex] , so it is time-invariant

    Now how do I atempt part B? I'm not seeing how the given input can be included in the given output function. My main thought is to just substitute [tex] x(t-\sigma) [/tex] with [tex] x(t) [/tex]

    My other minor thought is maybe I should use the change of variable rule from calculus to the y(t) fxn.
     
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 9, 2007 #2
    Since x(t) = |t|, then [itex]x(t - \sigma) = |t - \sigma| [/itex]. So, all you have to do is evaluate:

    [tex] y(t) = \int_0^\infty e^{-\sigma} |t - \sigma| \, d\sigma [/tex]
     
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