Question 2 of 6: Controllability, observability and stability of a parameter-dependent state model

Question 2: Controllability, observability and stability of a parameter-dependent state model (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A third-order single-input single-output state model whose $(1,2)$ entry and input vector both depend on one real parameter $\alpha$.

Find. The values of $\alpha$ for which the pair $(A,B)$ is controllable and the pair $(A,C)$ observable, together with the poles of the system and the stability conclusion that follows.

Approach. Build the two Kalman rank matrices, reduce each determinant to a single factor in $\alpha$, then read the poles from the characteristic polynomial and classify stability case by case.

1. (a) Build the controllability matrix. Propagating $B$ through $A$ twice,$$AB=\begin{bmatrix}0\\ \alpha\\ 0\end{bmatrix},\qquad A^{2}B=\begin{bmatrix}\alpha(1-2\alpha)\\ 0\\ -2\alpha\end{bmatrix},\qquad \mathcal{W}_{c}=\begin{bmatrix}B & AB & A^{2}B\end{bmatrix}=\begin{bmatrix}\alpha & 0 & \alpha(1-2\alpha)\\ 0 & \alpha & 0\\ 1 & 0 & -2\alpha\end{bmatrix}.$$The condition for controllability is the Kalman rank condition: $\operatorname{rank}\mathcal{W}_{c}=n=3$, equivalently $\det\mathcal{W}_{c}\neq 0$.
2. Evaluate the determinant. Expanding along the second row, whose only non-zero entry is $\alpha$ in position $(2,2)$,$$\det\mathcal{W}_{c}=\alpha\begin{vmatrix}\alpha & \alpha(1-2\alpha)\\ 1 & -2\alpha\end{vmatrix}=\alpha\bigl[-2\alpha^{2}-\alpha(1-2\alpha)\bigr]=-\alpha^{2}$$$$\boxed{\,\det\mathcal{W}_{c}=-\alpha^{2}\;\Longrightarrow\;\text{controllable for every }\alpha\neq 0\,}$$At $\alpha=0$ the input vector collapses to $B=\begin{bmatrix}0&0&1\end{bmatrix}^{\mathsf T}$ and $AB=0$, so the reachable subspace is the single direction $B$ and the rank falls to one.
3. (b) Build the observability matrix. Propagating $C$ through $A$ twice,$$CA=\begin{bmatrix}0 & -(1+2\alpha) & 0\end{bmatrix},\qquad CA^{2}=\begin{bmatrix}-(1+2\alpha) & 0 & 0\end{bmatrix},$$$$\mathcal{W}_{o}=\begin{bmatrix}C\\ CA\\ CA^{2}\end{bmatrix}=\begin{bmatrix}1 & 0 & 1\\ 0 & -(1+2\alpha) & 0\\ -(1+2\alpha) & 0 & 0\end{bmatrix}.$$Note how the $(1,2)$ entry of $A$ and the fixed $-2$ in its third row combine: $C A$ picks up $(1-2\alpha)+(-2)=-(1+2\alpha)$.
4. Evaluate the determinant. Writing $\beta=-(1+2\alpha)$ and expanding along the first row,$$\det\mathcal{W}_{o}=1\cdot(\beta\cdot 0-0)-0+1\cdot(0-\beta^{2})=-\beta^{2}=-(1+2\alpha)^{2}$$$$\boxed{\,\det\mathcal{W}_{o}=-(1+2\alpha)^{2}\;\Longrightarrow\;\text{observable for every }\alpha\neq -\tfrac12\,}$$At $\alpha=-\tfrac12$ the second and third rows vanish and the only information the sensor returns is the combination $x_{1}+x_{3}$.
5. (c) Characteristic polynomial. Because the third column of $A$ is zero, the determinant expands cleanly along the third column of $sI-A$:$$\det(sI-A)=\begin{vmatrix}s & -(1-2\alpha) & 0\\ -1 & s & 0\\ 0 & 2 & s\end{vmatrix}=s\bigl[s^{2}-(1-2\alpha)\bigr]=s^{3}-(1-2\alpha)s$$$$\boxed{\,s_{1}=0,\qquad s_{2,3}=\pm\sqrt{1-2\alpha}\,}$$One pole is pinned at the origin for every $\alpha$; the other two are a symmetric pair about the origin.
6. Classify the stability. Three cases follow directly from the sign of $1-2\alpha$:
7. Case by case. For $\alpha\lt\tfrac12$ the pair is real and symmetric, so one pole sits at $+\sqrt{1-2\alpha}$ in the right half plane and the system is unstable. For $\alpha=\tfrac12$ all three poles coalesce at the origin, a triple integrator, whose response grows like $t^{2}$ — also unstable. For $\alpha\gt\tfrac12$ the pair becomes $\pm j\sqrt{2\alpha-1}$, so the poles are $0$ and a simple imaginary pair: the system is marginally stable, not asymptotically stable, since the response neither decays nor grows.$$\boxed{\,\text{Never asymptotically stable: unstable for }\alpha\le\tfrac12,\ \text{marginally stable for }\alpha\gt\tfrac12\,}$$
8. Tie the three parts together. The two singular values of $\alpha$ both fall inside the unstable range, and in each case the lost mode is an unstable one. At $\alpha=0$ the poles are $0,\pm 1$ and the uncontrollable subspace contains $s=+1$, so the pair is not stabilisable. At $\alpha=-\tfrac12$ the poles are $0,\pm\sqrt{2}$ and the eigenvector of $s=+\sqrt{2}$, namely $\begin{bmatrix}\sqrt{2} & 1 & -\sqrt{2}\end{bmatrix}^{\mathsf T}$, satisfies $x_{1}+x_{3}=0$ and therefore lies in the unobservable subspace, so the pair is not detectable. Away from those two values the system is both controllable and observable, hence a state-feedback plus observer design can place all three closed-loop poles wherever the designer wishes.