# A $w$ system to stabilize.

I have the following system to be stabilized: \begin{equation} \begin{aligned}\dot{w}=Aw+Bv \\& A=\left( \begin{array}{ccc} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 2 & 0 \\ \end{array} \right), B=\begin{pmatrix}-1 \\ 1 \\ 1 \\ \end{pmatrix} \end{aligned} \end{equation} where $w\in \mathbb{R}^3$, and $v\in \mathbb{R}$ is the input.

The characteristic polynomial of $A$ is $p(\lambda) = p(\lambda)=-\lambda^3 +3\lambda^2 +7\lambda-3$

Now I want to transform this system into a stabilized system, i.e into a new system of $\bar{w}=\bar{A}\bar{w}+\bar{B}v$, where the eigenvalues of $\bar{A}$ have a negative real part. How do I do it? I forgot, and I can't find my notes from class.

• A simple linear transform $\bar{w}=Pw$ with $P$ nonsingular does not change the system eigenvalues. Or maybe by "transform" you mean taking some feedback control law of the form $v=Kw+v_0$? – RTJ Apr 22 '15 at 18:27
• I meant the second option. I think I need to use here Cayley-Hamilton on the above polynomial with $p(A)=0$, but I am not sure how to proceed. – MathematicalPhysicist Apr 22 '15 at 18:40
Since $(A,B)$ is controllable and the system has only one input you can use Ackermann's formula. Particularly you can take $$v=Kw+v_0$$ with $$K=-e_3^T \mathcal{C}^{-1}\alpha_d(A)$$ where $e_3$ is the third column of the identity matrix of dimension 3, $$\mathcal{C}:=\left[\matrix{B & AB & A^2B}\right]$$ the controllability matrix and $$a_d(s)=s^3+d_2s^2+d_1s +d_0$$ the desired polynomial whose roots you want to assign to $A+BK$. Note that $\alpha_d(A)$ is a $3\times 3$ matrix given by $$a_d(A)=A^3+d_2A^2+d_1 A +d_0\mathbb{I}_3$$