0
$\begingroup$

Given a nonlinear dynamical system like

$$ \begin{split} \dot{x}_1 &= f_1(\mathbf{x}) \\ \dot{x}_2 &= f_2(\mathbf{x}) + g(\mathbf{x})u \,. \end{split} $$

Now assume I have a found a stabilizing nonlinear controller $u$ that contains $\dot{x}_1$ like (just an illustrative example):

$$ u = -k_1 x_1 - k_2 x_2 - f_1(\mathbf{x}) \,. $$

Since the control law $u$ contains the nonlinear function $f_1(\mathbf{x})$, it is a nonlinear control law.

However, what happens if I can actually measure $\dot{x}_1 = f_1(\mathbf{x})$? Then it would be just a linear combination of (measured) signals, is this still (technically) considered a nonlinear controller then? Or if I estimate $\dot{x}_1$ using a numerical derivative like

$$ D(s) = \frac{s}{T s + 1} $$

with $T$ small enough... $u$ still a nonlinear controller? Or a linear one?

$\endgroup$
0
$\begingroup$

The system $A(x, u)$ is linear with respect to the controller $u$ iff it satisfies the linearity arguments:

1) $y=A(x, u) \rightarrow \alpha y= A(x, \alpha u)$ and

2) $y_1=A(x, u_1), \ y_2=A(x, u_2) \rightarrow y_1+y_2=A(x,u_1+u_2) $

$\endgroup$
  • $\begingroup$ I don't mean the system itself but just the controller. With using for example state feedback linearization, some nonlinear systems can be exactly rendered linear - the controller itself might be nonlinear nevertheless. $\endgroup$ – SampleTime Jun 25 '17 at 15:46
  • $\begingroup$ Well, you design the control law that contains a nonlinear function. The control is thus nonlinear. Then you approximate the nonlin. f-n with a linear one and the control becomes linear. There is nothing inherently linear or nonlinear in control, it depends on the way you write it. $\endgroup$ – Dmitry Jun 26 '17 at 8:14
  • $\begingroup$ I don't think this is linear approximation of the nonlinear controller. By making $T$ arbitrary small, the approximation error gets arbitrary small (assuming no measurement noise). Therefore, if anything, it should be a nonlinear approximation of the nonlinear controller? $\endgroup$ – SampleTime Jun 27 '17 at 17:45
  • 1
    $\begingroup$ You can say that you approximate (using $f(x)=\dot{x} \sim D(s)x$) or reformulate (using $\dot{x}=f(x)$) the nonlinear static control by a linear dynamic one. The feedback control becomes dynamic as you either use the derivative of the state or transform the state by a dynamic block. $\endgroup$ – Dmitry Jun 29 '17 at 9:45

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.