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Causal/Noncausal systems

  1. Sep 13, 2005 #1
    I'm working on homework, and I think I can explain the difference between causal and noncausal systems, but I don't know if I'm accurately recognizing them mathematically.

    Here are my homework questions and my answers:

    Are the systems described by the following equations with input x(t) and output y(t) causal or noncausal?

    A) y(t)=x(t-2) : CAUSAL; involves a delay of the signal in real time.

    B) y(t)=x(-t) : NONCAUSAL; causal signals do not exist before t=0 (I think?).

    C) y(t)=x(at), a>1 : CAUSAL; a is never negative.

    D) y(t)=x(at), a<1 : CAUSAL; a may be any real number between 0 and 1.

    Can anyone verify or correct me on any of these?
     
  2. jcsd
  3. Sep 14, 2005 #2
    Nobody can help me on this?
     
  4. Sep 14, 2005 #3

    es

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    I think some of the answers you gave are wrong. As I understand it, and it has been a while since I took this class, a casual system is one whose current value only depends on the present and past values of the input.

    So A is correct. The present value of y only depends on past values of x, or more specifically, x's that occurred 2 time units ago. i.e. for t=2, y(2)=x(0). Using this technique perhaps you can find the problems in your other answers.

    Also, usually when I solve systems I don't restrict myself to t>=0. In general I will set the signal value to 0 when t<0 but this is not the same as saying the signal does not exist before t=0 and it does not mean t cannot be negative. Your instructor may have confined your range to t>=0 though.
     
  5. Sep 14, 2005 #4
    Thank you, es. My answer for D was incorrect.
     
  6. Sep 15, 2005 #5

    es

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    My pleasure however I think you may still have some errors hiding.
    What are your opinions of the following points?
    Equation C with a=5 and t=1
    Equation D with a=-1
    Equation D with a=1/2 and t=2
     
  7. Sep 16, 2005 #6
    Well, I'm still learning, but here goes...

    Equation C remains causal with a=5 and t=1 because the output still exists after t=0.

    Equation D with a= -1 is noncausal because it places the signal before t=0.

    Equation D with a=1/2 and t=2 becomes causal because once again, it places the resultant signal after t=0. But since the parameter given was a<1, the system is noncausal because it must hold true for all values of a<1.

    Or something. :)
     
  8. Sep 16, 2005 #7

    es

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    Ah. I understand where the confusion is now.

    Causality does not reflect how events relate to t=0. Causailty relates times to each other. It means the output signal only depends on values that are occuring right now or occured in the past.

    Reusing our examples:

    1: y(t)=x(t-2) is casual
    the output, y, only depends on x's which occured two time units ago, aka
    in the past. for t=10, y(10)=x(8) which means the value of the output at
    time 10 is the same what the output was at time 8. notice both values are
    greater than t=0

    2: y(t)=x(t) is casual
    i just wanted to throw this in to show that the output can depend on what
    is happening right now and still be casual

    3: y(t)=x(-t) is non-casual
    when t<0, say -2, y(-2)=x(2). This means the value of the output at time
    t=-2 is equal to the value of the input at time t=2. However t=2 occurs
    after t=-2. This means the output depends on something which hasn't
    happened yet, a future value. This is what makes equation three non-
    casual.

    4: y(t)=x(at), a>1 is non-casual
    let's pick an example, a=5 t=2, here y(2)=x(10). Again, the output
    depends on a values which occurs in the future, t=10 occurs after t=2.
    All the t's here exist after zero but it is still non-casual.

    Note, there are t's which make equation 3 casual. For instance t=0. However if only one point in the range is non-casual then the entire system is non-casual (at least this is how I always do it). If you require the system to become casual then you could try restricting the range. In our example you could say t is only allowed to be zero.
     
    Last edited: Sep 16, 2005
  9. Feb 3, 2009 #8
    Are you allowed to take t<0. I dont think that would be physically correct or physically make sense.
    For that matter for e.g. 4 that you have if we pick a = 5 and t = -2 (t<0) then we have y(-2) = x(-10) which means that the output at -2 depends on a value that occured earlier -10 thus would make the system causal. What is the difference can you explain? Plus what would you think if a was constrained to only a<1?
     
  10. Feb 3, 2009 #9

    stewartcs

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    Science Advisor

    FYI...this thread is 4 years old. :wink:

    CS
     
  11. Feb 3, 2009 #10
    oops sorry i didnt even notice. thank you ;)
     
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