Question 2 of 6: State-space model, modal response, controllability and observability

Question 2: State-space model, modal response, controllability and observability (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. Four first-order equations relating three internal variables and one output.

Find. A state-space realisation, the zero-input output response, the input-output transfer function, a verdict on bounded-input bounded-output stability, and verdicts on complete controllability and complete observability.

Approach. Every equation is already in first-order form, so the state model can be read off by inspection; the zero-input response follows from eliminating $\theta$ to leave a single second-order equation in $\gamma$; the transfer function then exposes a pole-zero cancellation whose interpretation is exactly what parts (d) and (e) ask for.

1. Part (a) — choose the state and read off the matrices. The three variables that carry initial conditions are $\theta$, $\gamma$ and $h$, so take $x = \begin{pmatrix}\theta & \gamma & h\end{pmatrix}^{T}$. Writing the three derivative equations as rows, $$\boxed{\;\dot x = \begin{pmatrix}-2 & -1 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\end{pmatrix} x + \begin{pmatrix}1\\ 0\\ 0\end{pmatrix} u, \qquad y = \begin{pmatrix}0 & 1 & 1\end{pmatrix} x, \qquad D = 0 \;}$$ The realisation is a chain of three integrators with two feedback paths into the first summing junction and an output that taps the last two integrators.
2. Record the open-loop modes. Expanding the determinant along its third column, $$\det(sI - A) = \begin{vmatrix} s+2 & 1 & 0\\ -1 & s & 0\\ 0 & -1 & s\end{vmatrix} = s\bigl[s(s+2) + 1\bigr] = s\,(s+1)^2 .$$ The system therefore has a free integrator and a repeated mode at $s = -1$; every part that follows is a statement about one of these three eigenvalues.
3. Part (b) — reduce the unforced equations to one second-order equation. With $u \equiv 0$, differentiate $\dot\gamma = \theta$ once and substitute the first equation: $$\ddot\gamma = \dot\theta = -2\theta - \gamma = -2\dot\gamma - \gamma \;\Longrightarrow\; \ddot\gamma + 2\dot\gamma + \gamma = 0 .$$ The characteristic equation is $(s+1)^2 = 0$, the critically damped case, so the solution has the form $\gamma(t) = (c_1 + c_2 t)e^{-t}$.
4. Apply the initial conditions to $\gamma$. Directly, $\gamma(0) = 0$ gives $c_1 = 0$, and $\dot\gamma(0) = \theta(0) = 1$ gives $c_2 = 1$, so $$\gamma(t) = t\,e^{-t}.$$
5. Integrate once more for $h$. Since $\dot h = \gamma$ and $h(0) = 0$, $$h(t) = \int_0^{t}\tau e^{-\tau}\,d\tau = 1 - (1+t)e^{-t}.$$
6. Add the two contributions. The output is the sum of the last two states, and the terms in $t\,e^{-t}$ cancel exactly: $$y(t) = \gamma(t) + h(t) = t e^{-t} + 1 - (1+t)e^{-t},$$ $$\boxed{\;y(t) = 1 - e^{-t}\;}$$ A third-order state trajectory produces a plain first-order output rise. That cancellation is not an accident of the initial conditions — it is the same structural feature that parts (c) and (e) uncover.
7. Part (c) — transform with zero initial conditions. The middle two equations give $\Theta = s\Gamma$ and $H = \Gamma/s$. Substituting the first into the first equation, $$s^2\Gamma = -2s\Gamma - \Gamma + U \;\Longrightarrow\; \Gamma(s) = \frac{U(s)}{(s+1)^2}.$$ Then $Y = \Gamma + H = \Gamma\left(1 + \dfrac{1}{s}\right) = \Gamma\,\dfrac{s+1}{s}$, so $$\boxed{\;\frac{Y(s)}{U(s)} = \frac{s+1}{s\,(s+1)^2} = \frac{1}{s\,(s+1)}\;}$$ One factor of $(s+1)$ has cancelled between the numerator and the denominator: the transfer function is second order although the state model is third order.
8. Part (d) — test bounded-input bounded-output stability. Bounded-input bounded-output stability is a property of the transfer function, and $1/[s(s+1)]$ retains a pole at the origin. A bounded input therefore need not produce a bounded output: the unit step, which is bounded by one, gives $$Y(s) = \frac{1}{s^2(s+1)} \;\Longrightarrow\; y(t) = t - 1 + e^{-t} \to \infty .$$ $$\boxed{\;\text{Not BIBO stable: the surviving pole at } s = 0 \text{ integrates a bounded input without limit.}\;}$$
9. Part (e)(i) — test controllability. Build the Kalman matrix from $B$, $AB$ and $A^2B$: $$AB = \begin{pmatrix}-2\\ 1\\ 0\end{pmatrix},\qquad A^2B = \begin{pmatrix}3\\ -2\\ 1\end{pmatrix},\qquad \mathcal{C} = \begin{pmatrix}1 & -2 & 3\\ 0 & 1 & -2\\ 0 & 0 & 1\end{pmatrix}.$$ The matrix is upper triangular with unit diagonal, so $\det\mathcal{C} = 1 \neq 0$ and $$\boxed{\;\text{rank}\,\mathcal{C} = 3 \;\Longrightarrow\; \text{completely controllable}\;}$$
10. Part (e)(ii) — test observability. Stacking $C$, $CA$ and $CA^2$, $$CA = \begin{pmatrix}1 & 1 & 0\end{pmatrix}, \qquad CA^2 = \begin{pmatrix}-1 & -1 & 0\end{pmatrix}, \qquad \mathcal{O} = \begin{pmatrix}0 & 1 & 1\\ 1 & 1 & 0\\ -1 & -1 & 0\end{pmatrix}.$$ The third row is the negative of the second, so $\det\mathcal{O} = 0$ and the rank is 2: $$\boxed{\;\text{rank}\,\mathcal{O} = 2 \lt 3 \;\Longrightarrow\; \text{not completely observable}\;}$$
11. Tie the three answers together. The realisation is controllable but not observable, and the missing rank is exactly the one factor of $(s+1)$ that cancelled in part (c): one of the two modes at $s = -1$ never reaches the output. Because that hidden mode is stable, the system is detectable, and the state trajectory of part (b) decays even though the output of part (d) is unbounded — the instability that spoils bounded-input bounded-output stability is the integrator at $s = 0$, which is both controllable and observable, not the hidden mode.