How to stabilise the matrix that denotes a $\frac{\pi}{2}$ turn around the z-axis?

Question

Consider the i/s/o system:

$ \left\{ \begin{array}{lll} \frac{d}{dt}x = Ax + Bu \\ y = Cx \\ u = Ky \end{array} \right.$

If $A = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} $ and $B = (0 \quad 0 \quad 0)^{\intercal}$, which means there is no feedback in this, case, we know that the system is unstable because the characteristic polynomial of $A$ is $p_{A} (t) = (t-1)(t+i)(t-i)$, which means there is an eigenvalue $t^{*} = 1 > 0$. I want to choose the vectors $B$ and $C$ in such a way that there is a $K$ such that the system becomes stable. I've tried many $B$ and $C$ before, but so far I haven't been able to find them in such a way that I can choose a $K$ that makes the system stable again.

For example, consider the example in which $B = (1 \quad 0 \quad 0)^{\intercal}$ and $C = (1 \quad 0 \quad 0)$. Then we have $y = Cx = (1 \quad 0 \quad 0)x$, so $u = Ky = (K \quad 0 \quad 0)$. This means we get $\frac{d}{dt} x = Ax + Bu = \begin{pmatrix} K & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} x = Qx $

If we compute the characteristic polynomial, we find $P_{Q} (t) = t^{3} + (-1-K)t^{2} + (1+K)t - 1 = (t-1)(t- \frac{1}{2}(K - \sqrt{k^{2} - 4}))(t - \frac{1}{2}(K + \sqrt{k^{2} - 4})) $. This still has the eigenvalue $t^{*} = 1 > 0$. The system remains unstable, for whatever value of $K$ I choose.

I would like to see an example of vectors $C$ and $B$ such that I can make the system stable again, given the matrix $A$ above. I don't care what the value of $K$ is, as long as there is some $K$ which makes the system that was once unstable, now stable again.

Can you give me an example of this?

Answer 1

Choose for example $B = \begin{bmatrix}1 & 1 & 1\end{bmatrix}^T$ and $C = \begin{bmatrix}1 & 0 & 0\end{bmatrix}$. Taking now $u = -Kx$ with $K = \begin{bmatrix}1 & 1 & 0\end{bmatrix}$ places all three eigenvalues at $-1$.

Btw. the eigenvalues of your (uncontrolled) $A$ matrix are actually $\color{red}{-1}$, $i$ and $-i$.

Answer 2

In this case you can choose $B$ and $C$, so you can view your problem as finding a state feedback law with,

$$ \dot{x} = A\,x + B\,u, \tag{1} $$

$$ u = -K\,C\,x = -\hat{K}\,x, \tag{2} $$

such that $A - \hat{K}$ is Hurwitz.

But this can only be the case if $(A,B)$ is controllable or at least stabilizable. You could check for this using the controllability matrix or the Hautus test for all eigenvalues of $A$ to test for controllability or only the eigenvalues with a positive real part to test for stabilizability.

You could also consider to use a similarity transformation and bring $A$ in to the form needed for the controllable canonical form. Namely in this form the $a_i$ terms represent all the negative coefficients (except the highest power) of the characteristic polynomial.

And finally I would like to note that your $A$ matrix is in the real Jordan form and that only the bottom right Jordan block has a positive real eigenvalue, which should be controlled to make the system stable.