# stability of closed-loop nonlinear system

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.

• Are $K_1$ and $K_2$ constants and known and is the reference $a^*$ constant? Mar 3 at 23:40
• What characteristics does $b=f(a)$ have? Mar 4 at 1:07
• And shouldn't your $b$ term in your expression for $u$ be divided by $K_1$? Mar 4 at 1:38
• @KwinvanderVeen, yes K1 and K2 are constant, and you are right $u$ is divided by K1 (edited just now) Mar 4 at 3:17
• @KwinvanderVeen, $a^*$ is constant 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 $$\frac{k_p}{K_2},\frac{k_i}{K_2}<0$$.

• Is your analysis valid when $a$ crosses zero? Mar 4 at 14:39
• @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. Mar 4 at 15:52
• As to me, $\dot{a}$ is not properly defined in the system. I would define the new variable $z=\frac{1}{2}a^2$. Mar 5 at 8:43
• In my problem, I consider $a$>0, always Mar 5 at 22:04