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Response of a 2nd order system

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    b85072.png

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I understand how they got the answer and the calculations they did but I have 3 questions about this screenshot.

    1) Why the box in red is the transfer function? Is there a way to tell this from the Y(s) = ... expression?
    2) Why is the second term the free response (green box) and the first term the forced response (blue box)?
    3) Why does the 3e^(-t) - e^(-3t) confirm that the system is stable?

    Thanks
     
  2. jcsd
  3. Nov 23, 2016 #2

    FactChecker

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    That is the part that relates how U is transferred to Y. It is the coefficient of U in the equation.
    The free response is not "forced" by the input U. The "forced" response is forced by the input U.
    As time, t, increases in the positive direction, the exponentials disappear. If the exponents were positive, the any tiny disturbance would grow exponentially.
     
  4. Nov 23, 2016 #3
    I understand that but how did this then lead to the conclusion which of the terms is which?

    Thanks
     
  5. Nov 23, 2016 #4
    Generally we've Y(s) = G(s)U(s) but in this case it's Y(s)=G(s)U(s) + another term. Is there a reason why we don't have the usual Y(s) = G(s)U(s) ?
     
  6. Nov 23, 2016 #5

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    The other terms are coming from the initial conditions of Y, Y' and Y''. U is not involved in driving those. The current Y, Y' and Y'' are called state variables. Since some of them have nonzero initial values, their effect is independent of U and is added in.
     
  7. Nov 28, 2016 #6
    That makes sense. Thanks
     
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