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Signals and Systems, system properties

  1. Jul 11, 2009 #1
    I know that y(t) = t x(t) is unstable, for bounded inputs yielding unbounded outputs, but would

    y(t) = x(t)/ t also be unstable? when t is going in the negative direction? please help. test Monday.
  2. jcsd
  3. Jul 13, 2009 #2
    As t approaches zero, you get unbounded outputs from bounded inputs. So, it is unstable.

    Also, please somebody correct me if I am wrong, but I believe y(t) = t x(t) does in fact have a bounded output for a bounded input. If you don't let x go to infinity, then y will also not go to infinity, and is therefore bounded and stable.
  4. Jul 16, 2009 #3


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    No, even with a bounded x, y will grow with time, so it is unbounded.
    As, for the original question, x(t)/t is unstable for t equal zero, so no real system can have such characteristics.
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