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Lyapunov stability, mathematics vs reality

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    I have not been doing Lyapunov for a while and when doing an ordinary Lyapunov problem the other day, I ran into a funny situation.

    The correct way of doing it:
    \begin{align}
    \dot{e} &= \frac{1}{L}(u - R(e + x_{ref})) \\
    V(e) &= \frac{1}{2}Le^2 \\
    \dot{V} &= Le\dot{e} = Le \left( \frac{1}{L} \left[ u - R(e + x_{ref}) \right] \right) \\
    &= - Re^2 + e(u - Rx_{ref})
    \end{align}
    The system is a modified system, with e defined in terms of the original state x
    \begin{equation}
    e = x - x_{ref} \ \Rightarrow \ x = e + x_{ref}
    \end{equation}
    To stabilize the system, we define the feedback control law u to be
    \begin{align}
    u &= Rx_{ref} - K_pe \\
    \Rightarrow \ \dot{V} &= -Re^2 - K_pe^2 < 0
    \end{align}

    However, mathematically, one could define $u$ to be
    \begin{align}
    u &= Rx_{ref} \\
    \Rightarrow \ \dot{V} &= -Re^2 < 0
    \end{align}

    I know this will not work, u is not a constant, it is a variable.... but still, the mathematics checks out, kind of. What is the best way of explaining why this does not work?
     
    Last edited: Feb 8, 2013
  2. jcsd
  3. Feb 8, 2013 #2
    Hmm, if I think correctly, then
    \begin{equation}
    u = Rx_{ref}
    \end{equation}
    is the steady state value when
    \begin{equation}
    x=x_{ref}
    \end{equation}
    Soooo setting u to this value at all times should drive the system state to
    \begin{equation}
    x_{ref}
    \end{equation}
    It's just not a very good controller?
     
  4. Feb 8, 2013 #3
    You just add the extra Kp*e to make the solution converge to the equilibrium faster...
     
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