I would like to understand if it is possible to demonstrate the stability of this closed-loop nonlinear system

$$a b -K_1 u = a K_2 \dot{a}$$

where $a$ is the variable I am trying to control, $u$ is the control law, and the nonlinearity is represented by the fact that $b=f(a)$, and by the multiplication of $a$ with $\dot{a}$ on the right hand side of the equation.

The control law I am using is

$$ u= \frac{[ k_p(a^*-a) + k_i\psi+b] a}{K_1} \\ \dot{\psi}=a^*-a$$

where $k_p$ and $k_i$ are the coefficients of a proportional-integral controller.

I am not a control expert, so I apologise in advance for any imprecision or mistake I might have made in formulating the question.

  • $\begingroup$ Are $K_1$ and $K_2$ constants and known and is the reference $a^*$ constant? $\endgroup$ Mar 3 at 23:40
  • $\begingroup$ What characteristics does $b=f(a)$ have? $\endgroup$
    – Cesareo
    Mar 4 at 1:07
  • $\begingroup$ And shouldn't your $b$ term in your expression for $u$ be divided by $K_1$? $\endgroup$ Mar 4 at 1:38
  • $\begingroup$ @KwinvanderVeen, yes K1 and K2 are constant, and you are right $u$ is divided by K1 (edited just now) $\endgroup$
    – Lello
    Mar 4 at 3:17
  • $\begingroup$ @KwinvanderVeen, $a^*$ is constant $\endgroup$
    – Lello
    Mar 4 at 3:50

The proposed control law makes the resulting dynamics linear. Namely, when substituting the control law in the dynamics one can factor out $a$ which gives

$$ \dot{a} = -\frac{k_p}{K_2}(a^*-a) - \frac{k_i}{K_2} \psi. $$

By using that $\dot{\psi}=a^*-a$ and that $a^*$ is constant, it follows that the second derivative of $\psi$ with respect to time would be $\ddot{\psi}=-\dot{a}$. Substituting the expression for $\dot{a}$ in yields

$$ \ddot{\psi} = \frac{k_p}{K_2} \dot{\psi} + \frac{k_i}{K_2} \psi, $$

which is linear and can be shown to be stable if $\frac{k_p}{K_2},\frac{k_i}{K_2}<0$.

  • $\begingroup$ Is your analysis valid when $a$ crosses zero? $\endgroup$
    – Arastas
    Mar 4 at 14:39
  • $\begingroup$ @Arastas The control law itself is well defined at $a=0$. The main question is whether $\dot{a}$ is. Plugging in the solution to linear second order differential equation should satisfy the initial implicit first order dynamics of $a$. The only question is whether this is the only valid solution and I am not 100% sure. $\endgroup$ Mar 4 at 15:52
  • $\begingroup$ As to me, $\dot{a}$ is not properly defined in the system. I would define the new variable $z=\frac{1}{2}a^2$. $\endgroup$
    – Arastas
    Mar 5 at 8:43
  • $\begingroup$ In my problem, I consider $a$>0, always $\endgroup$
    – Lello
    Mar 5 at 22:04

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