1
$\begingroup$

Given the nonlinear system

$$ \begin{align} \dot{x}_1 &= -4x_1 + 10x_2 + u \\ \dot{x}_2 &= -x_1 - 2x_2 - \log(1 + x_1^2) \\ y &= x_1 + x_2 \end{align} $$

Assume the system should be forced to stay at the operating point $x^*_1 = 5$, $y^* = 10$. From that follows that $x^*_2 = 5$.

However, by choosing an input $u^*$ for the operating point, I can only make $\dot{x}_1 = 0$, but not $\dot{x}_2 = 0$: Assume $u^* = -30$, then

$$ \begin{align} \dot{x}^*_1 &= 0 \\ \dot{x}^*_2 &= -18.2581 \\ \end{align} $$

So it seems there is no way to make this operating point an equilibrium. However, the linearization at $(x^*_1,x^*_2,u^*) = (5,5,-30)$ is stable, controllable and observable...

Question: I would like to use linearization at the required operating point, but how to deal with non-existing equilibrium at this point?

Edit: To give another example, take the normalized pendulum:

$$ \begin{align} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= \sin(x_1) - \frac{1}{2}x_2 + u \end{align} $$

Usually, the task is to balance the pendulum at the top position s.t. $x_1 = x_2 = 0$.

But what if the task is to rotate the pendulum at a constant rate of change of its angle, such that $\dot{\phi} = \dot{x}_1 = c \neq 0$? To achieve this, $x_2 = c$ must hold, and then I can choose $u$ such that $\dot{x}_2$ gets zero, but what to do whith the non-zero $\dot{x}_1$?

$\endgroup$
1
$\begingroup$

If an operating point is not an equilibrium point then by definition it is not stable, since at the operating point the time derivatives are not zero and thus the system will always move away from it. So in short performing a linearization at an operating point that is not an equilibrium point is not very helpful, since you will always move away from it after which your dynamics might change and the linearization is not a valid approximation anymore.

For the second example you are not considering an operating point, but an operating curve. Namely $x_1^*(t)=a+c\,t$, $x_2^*(t)=c$ and $u^*(t)=\frac{1}{2}c-\sin(a+c\,t)$. The dynamics of the error defined as $\delta x_1=x_1-x_1^*$ and $\delta x_2=x_2-x_2^*$ can be written as

\begin{align} \delta \dot{x}_1 &= x_2 - c \\ &= \delta x_2 \\ \delta \dot{x}_1 &= \sin(x_1) - \frac{1}{2} x_2 + u \\ &= \sin(x_1) -\sin(a+c\,t) - \frac{1}{2} \delta x_2 + \delta u \end{align}

with $\delta u = u - u^*$. Under the assumption that $x_1\approx x_1^*$ you can simplify the two sine terms with

$$ \sin(x_1) -\sin(a+c\,t) \approx \cos(a+c\,t) \delta x_1 $$

The resulting error dynamics is linear time variant and not linear time invariant, so there is not one linearization that would approximate well for all time.

$\endgroup$
  • $\begingroup$ So there is just nothing one can do about this? $\endgroup$ – SampleTime Oct 20 '18 at 9:37
  • $\begingroup$ @SampleTime You can change the operating point, such that the derivatives are zero. $\endgroup$ – Kwin van der Veen Oct 20 '18 at 11:00
  • $\begingroup$ But what if this is the only operating point of interest? I edited my question to give another example, how could one deal with this case? $\endgroup$ – SampleTime Oct 20 '18 at 13:04
  • $\begingroup$ @SampleTime I have updated my answer, in short you can't use linearization to find a LTI system that approximates it well, but you can find a LTV system that does. But you do require that the operating point/curve satisfies the differential equation. $\endgroup$ – Kwin van der Veen Oct 20 '18 at 14:02
  • 1
    $\begingroup$ @SampleTime I assume you mean $|c\,t| \ll \pi$, but then you might as well consider $x_2^*=0$ and $x_1^*=a$. $\endgroup$ – Kwin van der Veen Oct 20 '18 at 19:05

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.