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Homework Help: Observability and existence and uniqueness

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    I have set up this problem for myself.

    Let P be a system of the form

    x' = Ax + Bu
    y = Cx + Du

    The definition of a "state" is:

    "x(t) is a state for a system P if knowledge of x at some initial time t_{0} and the input u(t), t \geq t_{0} is sufficient to uniquely determine y(t) for t \geq t_{0}."

    Let us consider only the free response (input 0). My understanding is that if x(t) is a state, then each selection of x_{0} will yield a unique y(t).

    Now here is one definition for observability:

    "A system P is observable if and only if the initial state x_{0} can be determined uniquely from its zero-input response over a finite time interval."

    The thing that has been bothering me is this. If a system in unobservable, then that means that we can find two different initial conditions that give the same free response. Does that mean that if a system is unobservable, the existence and uniqueness conditions are violated? And if so, does that not mean that the x(t) we chose are not even states by definition?

    2. Relevant equations

    Here is an example. Suppose i have mass spring damper system.

    x'' = -(c/m)x' - (k/m)x

    We define our state variables as:

    x_{1} = x
    x_{2} = x'


    x'_{1} = x_{2}
    x'_{2} = x'_{1} = x'' = -(c/m)x_{2} - (k/m)x_{1}

    Let my output be

    y = x_{1}+x_{2};

    Now, just for argument, suppose i define a third state variable

    x_{3} = height of a random bouncing ball far away from the mass spring damper system.

    My "state" is now x = [x_{1} x_{2} x_{3}]^{T}.

    3. The attempt at a solution

    I think it is safe to say that x_{3} will never appear in the output. The initial condition vector [1 1 10]^{T} and [1 1 20]^{T} will give the same free response. I know x_{3} is unobservable. I also think that since two different initial conditions give me the same output, therefore by definition, my choice of the state is not even a state since i do not get a unique output for two different initial conditions.

    I think i am wrong. But i don't know where.
    Last edited: Sep 11, 2010
  2. jcsd
  3. Sep 11, 2010 #2


    User Avatar

    For x_{3} to be a state variable of your system, you must write the dynamic equations describing its behavior.
    In this case, knowing the initial state and the input (0 for the ball), you can determine the output y of the system. You can also know the state (value of the 3 variables) any time in the future.
    What you can't do is to determine the initial state from observation of the output.
  4. Sep 11, 2010 #3
    My question is that if two different initial conditions give you the same free response, doesn't that violate the definition of a state?
  5. Sep 11, 2010 #4


    User Avatar

    No. Read again the definition.
    Think of a parallel RC circuit and take the voltage and current on the capacitor as state variables.
    If you charge the capacitor with any voltage, the state at t = infinity will be current and voltage both equal to zero.
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