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

Question

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?

Answer 1

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.