# How to re-write systems of two linear second order ODEs in a state space form

I have recently been working on how to convert ODES into state space form, and I came across this problem that has me confused. Should B in the state space be [1; -2] or [1 0; 0 -2]?

I think the second one should be correct, but when I try to calculate the value of a when the system is controllable. controllability matrix is not a square matrix, which leads me not to be able to use determinants to determine. What other way is there to determine whether the matrix is controllable?

Can people help me with this?

• Are there two dots above $z_i$ on the left hand side of the equations? Or is it a typographical error? If, so please correct either the error or add more states to make the equations first order differential equations.
– AJN
Commented Apr 9, 2023 at 16:40
• Please use edit option below the question to add details to the question or to correct errors in the question.
– AJN
Commented Apr 9, 2023 at 16:41

State Space form:

Assuming the double dots are a mistake, and they're supposed to be single dots (please clarify):

$$\begin{bmatrix} \dot{z_1}\\ \dot{z_2} \end{bmatrix} = \begin{bmatrix} -1 & a \\ a & -1 \end{bmatrix} \begin{bmatrix} z_1\\ z_2 \end{bmatrix}+ \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}$$

Controllability:

The controllability matrix does not need to be square: You determine if its RANK is equal to the amount of states in your system. The controllability matrix is (with $$n$$ the amount of states):

$$\mathcal{C} = \begin{bmatrix} B, AB, A^2B, \dots, A^{(n-1)}B \end{bmatrix} = \begin{bmatrix} B, AB \end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 & -2a\\ 0 & -2 & a & 2 \end{bmatrix}$$

We easily see that the rank of $$\mathcal{C}$$ is 2 for any $$a$$ (because of the left most square $$2 \times 2$$ matrix in $$\mathcal{C}$$). Since the rank of $$\mathcal{C}$$ is equal to the amount of states, we see that the system is controllable.