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Quick question about Causal LTI system

  1. Feb 12, 2013 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    I am reading a little bit on Causal LTI systems. I know causality means that a system cannot depend on the future, but why does that mean an impulse function, say

    h(t) = 0, for all t<0

    I mean, if t is negative, then it's not really depending on the future, but the past right? Or am I visualizing this wrong?
     
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  3. Feb 12, 2013 #2

    Dick

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    h(t)=0 for all t<0 isn't really depending on the past either. It doesn't depend on anything. You just said what it is. There's no evolution equation to specify any time evolution.
     
  4. Feb 13, 2013 #3

    ElijahRockers

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    It doesn't depend on anything? But I thought being being a function of t by definition means it depends on t. Sorry if that sounds obtuse, I just can't relate anything you said to anything I've learned..., apparently I am struggling with this.

    I just dont see why the system, being causal, means that h(t) has to be 0 for all t<0....

    assuming I have some impulse h[n] = h1[n]+h1[n-1] and I know h[n] = 2 for n>0

    then at n=0, 2 = h1[0] + h1[-1]. since the system is causal, i could say h1[0] = 2, because h1[-1] = 0. what i don't get is WHY i can say h1[-1]=0...

    I hope that makes sense.. it's hard for me to give an example of something I dont really understand in the first place
     
  5. Feb 13, 2013 #4

    Dick

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    I'm just saying you haven't presented the system you are evolving. You need to look at that. Say,
    [tex]y(t)=\int_{-\infty}^{\infty} x(t-\tau)h(\tau) d\tau[/tex]
    where x is the input, y in the output and h is the response function. If you want to calculate ##y(0)## then that doesn't depend, for example, on ##x(1)## because the contribution of ##x(1)## to the integral happens when ##\tau=(-1)## and being causal says ##h(-1)=0##.
     
  6. Feb 13, 2013 #5

    ElijahRockers

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    It was explained to me like this, earlier today:

    [itex] y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k][/itex]

    [itex] y[n] = \sum_{k=-\infty}^{n}x[k]h[n-k] + \sum_{k=n+1}^{\infty}x[k]h[n-k][/itex]



    then let h[n] = 0 for n<0

    => h[n-k] = 0 for k>n

    => [itex]\sum_{k=n+1}^{\infty}x[k]h[n-k] = 0[/itex]

    So now x[k] only depends on values from -infinity to n, which is the past and present.
     
    Last edited: Feb 13, 2013
  7. Feb 13, 2013 #6

    Dick

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    Sure, same thing basically. That's the discrete case.
     
  8. Feb 14, 2013 #7

    ElijahRockers

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    Ya, we are doing convolutions with discrete time signals so this is the only kind I've really had exposure to, other than a tiny bit in diff eq. Makes sense now, thanks for your help!
     
  9. Feb 14, 2013 #8

    vela

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    Another way to think of it is that h(t) is the response of the system to a unit impulse at t=0. A causal system can't have a non-zero response before t=0, otherwise it wouldn't be causal.
     
  10. Feb 14, 2013 #9

    ElijahRockers

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    See that's what is tripping me up....

    When it's explained simply like that, I just can't picture it. Seeing the sums broken up helps, but I feel there is something crucial about impulses that I am not understanding.
     
  11. Feb 14, 2013 #10

    vela

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    You want h(t) to be the response only to the impulse, so the system, by assumption, needs to be at rest before the impulse occurs. At t=0, an impulse is imparted to the system. If h(t) isn't zero when t<0, that means either the system wasn't at rest, which by assumption isn't true, or the system began responding to the impulse before it was imparted, which means the system is not causal.
     
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