Consider a SISO non-linear system $$\dot{x} = F(x,u)$$ in which $\vec{0}$ is an equilibrium point. In the process of determining that it is indeed an equilibrium point, the input did not matter at all. It was completely independent, due to the term $x_4 \cdot u$

  • $u(t) \in \mathbb{R}$ is the input
  • $x_4$ is a state variable

I want to linearize the system about $\vec{0}$. But I haven't completely understood the theory behind it, and I would appreciate some help.

1. Can I completely ignore the term $x_4 \cdot u$, i.e. consider it as a higher order term, as we would do for example with 2 state variables $x_1 \cdot x_2$?

If so problem solved.

2. According to the theory, when I want to bring the system to the linear form $$\dot{x} = A x + B u$$ then $$A = \frac{\partial{F}}{\partial{\vec{x}}} \text{ calculated at (0,u*) i.e. at the equilibrium point}$$

However, in the system I am examining we cannot extract any information about u*. What would even u* mean in this system?

Thanks for your time. If you can provide an answer to the 1st question I think I would be ok.

EDIT: The system equations are $$\dot{x_1} = x_2$$ $$\dot{x_2} = \frac{2x_2 + \sin(x_1) + x_3 x_2^2}{1 + x_1^2} + x_3 $$ $$\dot{x_3} = x_4$$ $$\dot{x_4} = 3x_4 + u \cdot x_4 + 2x_3^2 +x_1^2 x_3^2 + \sin(x_2)$$

  • $\begingroup$ Could you please write the system equations ? How does the term $x_4 u$ turn up ? $\endgroup$
    – drC1Ron
    Commented Jun 19, 2022 at 15:51
  • $\begingroup$ @RonnyLandsverk sorry, now I fixed it. $\endgroup$
    – Anonymous
    Commented Jun 19, 2022 at 17:11
  • 1
    $\begingroup$ Note that this system is not controllable in the operating point $x_4 = 0$ since you are multiplying the control with 0, it has no effect. The $B$ matrix will be zero. $\endgroup$ Commented Jun 20, 2022 at 9:09
  • $\begingroup$ Yes you are right, thanks for noting, I didn't have a problem with matrix B, since u "disappears". Thus I can substitute $x_4$ with $0$. The problem is with element $A_{44}$ in which I don't know how to substitute u, is $u*$ meaningful ? Can I throw away the term $u \cdot x_4$ ? $\endgroup$
    – Anonymous
    Commented Jun 20, 2022 at 9:19
  • 1
    $\begingroup$ The $A_{44}$ element will be equal to $3+u$. $\endgroup$
    – drC1Ron
    Commented Jun 20, 2022 at 13:00

1 Answer 1


One way to solve this is to create a new input variable $v$ and set it equal to to $u\ x_4$. $(v = u\ x_4)$

This can be used to obtain a linear system with $x_i$ as states and $v$ as input, and will result in the following $A$ and $B$ matrices.

$$\begin{array}{cc} \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ \end{array} \right) & \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right) \\ \end{array}$$

When you do a control simulation, you will have to feed in $u$ to the system which is $v/x_4$.

The Mathematica code I used to compute the linearization.

 x1'[t] == x2[t], 
 x2'[t] == (2 x2[t] + Sin[x1[t]] + x3[t] x2[t]^2)/(1 + x1[t]^2) + x3[t], 
 x3'[t] == x4[t], 
 x4'[t] == 3 x4[t] + v[t] + 2 x3[t]^2 + x1[t]^2 x3[t]^2 + Sin[x2[t]]
{x1[t], x2[t], x3[t], x4[t]}, 

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