0
$\begingroup$

I want to show a scholar example of a system such that its linearization is not controllable but the system can be stabilized with nonlinear feedback. I am thinking about this one $$ \begin{aligned} \dot{x}_1 &= x_2^3, \\ \dot{x}_2 &= u, \end{aligned} $$ where $x_1$, $x_2$, and $u$ are scalars, and the goal is to drive the system to the origin.

What is the simplest control that you would propose for this system?

$\endgroup$
  • 1
    $\begingroup$ What have you tried yourself, for example have you tried backstepping? Also wouldn't $\dot{x}=-x$ satisfy your description, since it is not controllable, but is stable? And how do you define "simplest control"? $\endgroup$ – Kwin van der Veen Nov 4 '19 at 15:02
  • $\begingroup$ I have tried myself to design this example :) My answer was $u=-x_2^3-x_2-x_1$, but as shown by @SampleTime, the term $x_2^3$ can be omitted. Normally we do not discuss controllability for autonomous systems. $\endgroup$ – Arastas Nov 4 '19 at 15:41
  • 1
    $\begingroup$ You could interpret it as $\dot{x}=-x+a\,u$ with $a=0$, otherwise another simple example would be the system $\dot{x}=x\,u$, for which $u=-x^2$ would be a "simple" stabilizing controller. $\endgroup$ – Kwin van der Veen Nov 4 '19 at 15:50
  • $\begingroup$ Thanks, $\dot{x}=xu$ is a nice example! $\endgroup$ – Arastas Nov 4 '19 at 15:53
3
$\begingroup$

Even though the linearization is not controllable, the nonlinear system can still be stabilized with linear feedback. I propose the control law

$$ u = -x_1 - x_2 \tag{1} $$

which leads to the closed loop dynamics

$$ \begin{align} \dot{x}_1 &= x_2^3 \\ \dot{x}_2 &= -x_1 - x_2 \end{align} \tag{2} $$

Take the Lyapunov function

$$ V(x) = x_1^2 + 2 x_1 x_2 + x_2^2 + \frac{1}{2} x_2^4 $$

It is easy to show that $V$ has a unique minimum at $(0, 0)$ so it is positive definite. The derivative is

$$ \dot{V}(x) = -2 (x_1 + x_2)^2 $$

which is negative semi-definite (zero along the $x_1 = -x_2$ line). If we insert that into $(2)$, we have $\dot{x}_2 = 0$ but $\dot{x}_1 = -x_1^3$, so no solution can stay in the set $\dot{V}(x) = 0$ except $x_1 = x_2 = 0$.

So, by LaSalle, the system is globally asymptotically stabilized by the linear feedback $(1)$.

This is probably also the "simplest" stabilizing control law (linear feedback with both gains being 1), but that depends on your definition of simple.

$\endgroup$
  • 1
    $\begingroup$ Thanks! Note also that it is sufficient to consider $V(x) = x_1^2 + 0.5 x_2^4$ yeilding $\dot{V} = -2x_2^2$, and then the same invariance argument works. $\endgroup$ – Arastas Nov 4 '19 at 15:38
  • 1
    $\begingroup$ Your answer destroys my example. :) I was looking for an example that cannot be stabilized by linear feedback. :) $\endgroup$ – Arastas Nov 4 '19 at 15:39
  • $\begingroup$ @Arastas Sorry about that :) I would also be interested whether such systems even exist (I guess so, but I haven't seen an example yet). $\endgroup$ – SampleTime Nov 4 '19 at 15:42
  • 2
    $\begingroup$ I do not believe there exists a linear stabilizing control law for $\dot{x}=x\,u$, namely $u=-x^2$ does stabilize it. $\endgroup$ – Kwin van der Veen Nov 4 '19 at 15:53
  • $\begingroup$ @KwinvanderVeen That is indeed a nice example! $\endgroup$ – SampleTime Nov 4 '19 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.