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I have the matrix A and B and I computed the controllability matrix and found the rank is lower than n. I know I can't apply pole-placement , how else can I solve the system?

n is the order of A

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Controlability means that a state-space feedback is able to assign closed-loop eigenvalues of a linear system to anywhere in the complex plane. However, the stabilizability of a system is a weaker notion, requiring only that the closed-loop eigenvalues all end up on the left half-plane.

Consider \$A\$ is of order \$n\$. For all \$\lambda \in \mathbb{C}\$ that are eigenvalues of \$A\$ with positive real parts (\$\Re (\lambda )\geq 0\$):

\$\text{rank}\begin{bmatrix}\lambda I-A&B\end{bmatrix} = n\$ \$\iff\$ the system is stabilizable

Example: $$ A = \begin{bmatrix} 1 & 1 \\ 4 & -2\end{bmatrix} \qquad B = \begin{bmatrix}1\\1\end{bmatrix} $$

This system is not controllable. Its controllability matrix, with rank 1, is $$ Co = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix}1 & 2 \\ 1 & 2\end{bmatrix} $$ However, this system is stabilizable. There are 2 eigenvalues of \$A\$: \$\lambda=-3\$ and \$\lambda=2\$. $$ \text{rank}\begin{bmatrix}2I-A&B\end{bmatrix} = \text{rank}\begin{bmatrix}1&-1&1\\-4& 4 & 1\end{bmatrix} = 2 $$

One way of stabilizing this system, using state-feedback with pole placement ideas, is to keep the stable (yet uncontrollable) eigenvalue \$\lambda=-3\$ in its place for the closed-loop system. If one wishes for the closed-loop eigenvalues of \$\lambda=-3\$ and \$\lambda=-5\$, then it follows the characteristic equation: $$(s+3)(s+5) = s^2+8s+15$$

Gain calculation: $$K = \begin{bmatrix}k_1 & k_2\end{bmatrix}$$ $$|sI-(A-BK)| = \text{det}\begin{bmatrix}s-1+k_1&-1+k_2\\-4+k_1&s+2+k_2\end{bmatrix}=$$ $$ = s^2+s(k_1+k_2+1)+3k_1+3k_2-6$$ $$\begin{cases} k_1+k_2+1 = 8\\ 3k_1+3k_2-6 = 15 \end{cases} $$ This system has infinite possible solutions. Assuming \$k_1 = 0\$, then \$K=\begin{bmatrix}0&7\end{bmatrix}\$ achieves the desired closed-loop eigenvalues.

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