# State space and linearization

I have a question about state space representation. How can I represent an equation in which I have only the second and first derivatives? For example

where $u$ is the control input.

If I put $x_1=x$ and $x_2=\dot{x}$ I will not have $x_1$ in my state space representation, and when finding equilibrium points or Jacobian to check controllability, I will obtain zero in the partial derivatives corresponding to $x_1$. Is there a way to overcome this with a more suitable state representation? I'm not sure how to solve this problem as I have to linearize the system around an operating point using SS. Thanks in advance.

• What is the problem with the zeros? Oct 30 '14 at 2:43
• Well it's just I don't know what to say about the stability of the system in this case or how to find out its equilibrium points, nor how to design a feedback controller for this system, control is not my background and is a little bit difficult to me to get this kind of tricky questions, as I can solve "normal" nonlinear systems. So I'm thinking I could be wrong in my SS representation, or there is a way I can solve the rest of my exercise coming from this. Oct 30 '14 at 3:13
• You need to provide more info. about what you are trying to do and what you are trying to control/design. Is your state $x,\dot{x}$ or just $\dot{x}$, does $u$ have some nominal value, etc, etc... Oct 30 '14 at 3:19
• Pitifully I do not have much more info than this. The equation is intended to rule the motion of a plane, what I call x is its attitude angle, and I have to put the equation in state space form, use linearization to analyze the stability of the system and develop feedback control for the system to track a reference angle, so x is what I want to control. Oct 30 '14 at 3:38
• Well, if you need to control $x$ then you have no choice but to include it in the state space representation. However, I am still not clear about why you think this is a problem with the representation. Oct 30 '14 at 3:41

By selecting $x_1=x$ and $x_2=\dot{x}$ you can write
\begin{align*} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= k_1 x_1 (1 - |x_1|) + k_2 u \end{align*}
Now the fixed points are $(0, 0), (1, 0), (-1, 0)$. The Jacobian around $(1, 0)$ (as an example) is
$$\begin{bmatrix} 0 & 1 \\ k_1(1 - 2 x_1) & 0 \end{bmatrix} |_{x_1=1} = \begin{bmatrix} 0 & 1 \\ -k_1 & 0 \end{bmatrix}$$