Question 4 of 6: Controllability, observability and BIBO stability of a parametrised system

Question 4: Controllability, observability and BIBO stability of a parametrised system (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 realisation whose $(2,1)$ entry carries a free parameter $\alpha$; the state matrix is lower triangular, so its eigenvalues can be read off the diagonal.

Find. (a) the values of $\alpha$ for which the pair $(A,B)$ is controllable and the pair $(A,C)$ observable; (b) whether a bounded input produces a bounded output; (c) the limiting output of the zero-input response from the stated initial condition.

Approach. Build the two rank test matrices and evaluate their determinants symbolically in $\alpha$; then obtain the transfer function to see which modes actually survive from input to output, which is what BIBO stability tests; finally integrate the zero-input response, which the triangular structure makes trivial.

1. Part (a) — build the controllability matrix. With $B=\begin{pmatrix}1&1&0\end{pmatrix}^{\mathsf{T}}$,$$AB=\begin{pmatrix}-1\\ \alpha-3\\ 1\end{pmatrix},\qquad A^{2}B=\begin{pmatrix}1\\ 7-3\alpha\\ \alpha-3\end{pmatrix},\qquad \mathcal{C}=\begin{pmatrix}1&-1&1\\ 1&\alpha-3&7-3\alpha\\ 0&1&\alpha-3\end{pmatrix}$$
2. Part (a) — expand its determinant. Expanding along the first column,$$\det\mathcal{C}=\left[\left(\alpha-3\right)^{2}-\left(7-3\alpha\right)\right]-\left[-\left(\alpha-3\right)-1\right]=\left(\alpha^{2}-3\alpha+2\right)+\left(\alpha-2\right)$$$$\boxed{\det\mathcal{C}=\alpha^{2}-2\alpha=\alpha\left(\alpha-2\right)}$$so the system is controllable for every $\alpha$ except $\alpha=0$ and $\alpha=2$. Both exceptions have a clean interpretation: at $\alpha=2$ the two routes from $u$ into $x_{2}$ cancel one mode, and at $\alpha=0$ the integrator state $x_{3}$ becomes unreachable.
3. Part (a) — build the observability matrix. With $C=\begin{pmatrix}0&0&1\end{pmatrix}$, the products $CA$ and $CA^{2}$ simply pick out rows of $A$:$$CA=\begin{pmatrix}0&1&0\end{pmatrix},\qquad CA^{2}=\begin{pmatrix}\alpha-1&-2&0\end{pmatrix},\qquad \mathcal{O}=\begin{pmatrix}0&0&1\\ 0&1&0\\ \alpha-1&-2&0\end{pmatrix}$$$$\boxed{\det\mathcal{O}=-\left(\alpha-1\right)}$$so the system is observable for every $\alpha$ except $\alpha=1$. At $\alpha=1$ the coupling from $x_{1}$ into $x_{2}$ vanishes, $x_{1}$ decays on its own and never reaches the output, and that mode is hidden.
4. Part (a) — state the answer including the exam’s own case. For the generic parameter the system is both controllable and observable; it is a minimal realisation. The exceptions are $\alpha\in\{0,2\}$ (uncontrollable) and $\alpha=1$ (unobservable). In particular at the value $\alpha=3$ used in part (c) both determinants are non-zero ($\det\mathcal{C}=3$, $\det\mathcal{O}=-2$), so that case is minimal.
5. Part (b) — find the eigenvalues, then resist the obvious conclusion. $A$ is lower triangular, so its eigenvalues are its diagonal entries $-1$, $-2$ and $0$ for every $\alpha$. The pole at the origin makes the system not asymptotically stable in the Lyapunov sense, but BIBO stability is a property of the transfer function, and a mode that is uncontrollable or unobservable never appears there. The question is therefore whether the mode at the origin survives the cancellation.
6. Part (b) — compute the transfer function. Solving the state equations in turn,$$X_{1}=\frac{U}{s+1},\qquad X_{2}=\frac{\left(\alpha-1\right)X_{1}+U}{s+2}=\frac{U\left[\left(\alpha-1\right)+\left(s+1\right)\right]}{\left(s+1\right)\left(s+2\right)}=\frac{U\left(s+\alpha\right)}{\left(s+1\right)\left(s+2\right)},\qquad X_{3}=\frac{X_{2}}{s}$$and since $y=x_{3}$,$$\boxed{G(s)=\frac{s+\alpha}{s\left(s+1\right)\left(s+2\right)}}$$
7. Part (b) — conclude. The numerator zero sits at $s=-\alpha$, so it cancels the pole at the origin if and only if $\alpha=0$:$$\alpha=0:\quad G(s)=\frac{s}{s\left(s+1\right)\left(s+2\right)}=\frac{1}{\left(s+1\right)\left(s+2\right)}\quad\text{(BIBO stable)}$$$$\alpha\neq0:\quad G(s)\ \text{retains a pole at }s=0\quad \text{(not BIBO stable)}$$So the system is BIBO stable only for $\alpha=0$, and in particular not BIBO stable for the $\alpha=3$ of part (c): a bounded input such as a unit step produces an output that grows without bound, because the free integrator $x_{3}$ accumulates it. Note that $\alpha=0$ is precisely one of the two values that made the system uncontrollable, which is not a coincidence — the pole is removed from $G(s)$ exactly because that mode is decoupled from the input. Checking the other two special values makes the point: at $\alpha=1$ the zero cancels the stable pole at $-1$ and at $\alpha=2$ it cancels the stable pole at $-2$, but in both cases $s=0$ survives and the system is still not BIBO stable.
8. Part (c) — integrate the zero-input response. With $u\equiv0$ and $\alpha=3$ the state equations decouple downward:$$\dot{x}_{1}=-x_{1},\ x_{1}(0)=0\;\Rightarrow\;x_{1}(t)\equiv0$$$$\dot{x}_{2}=2x_{1}-2x_{2}=-2x_{2},\ x_{2}(0)=0\;\Rightarrow\;x_{2}(t)\equiv0$$$$\dot{x}_{3}=x_{2}=0\;\Rightarrow\;x_{3}(t)=x_{3}(0)=9\ \text{for all }t$$
9. Part (c) — read the output. Since $y=x_{3}$,$$\boxed{y(t)=9\ \text{for all }t\ge0,\qquad y_{ss}=9}$$The output does not decay, and that is the observable signature of the pole at the origin: the initial condition is stored in the integrator state and there is nothing in the unforced dynamics to bleed it away. Marginal stability of this kind is consistent with parts (a) and (b) — the system is minimal at $\alpha=3$, so the mode at the origin is fully visible, and it is precisely that visible mode which destroys BIBO stability.