Question 2 of 6: State-space model and pole placement for a satellite attitude loop

Question 2: State-space model and pole placement for a satellite attitude loop (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. The open-loop dynamics of a satellite attitude control system, with $\theta$ the attitude angle, $\omega$ the body rate and $h$ the momentum-wheel state:

Find. The state-space quadruple, the controllability and observability of the realisation, the free response $\omega(t)$ from the stated initial conditions, and a state-feedback gain that places the three closed-loop poles as specified.

Approach. Read the matrices directly off the three first-order equations, test the rank of the controllability and observability matrices, integrate the decoupled $\theta$–$\omega$ subsystem for the free response, and match the characteristic polynomial of $A - BK$ against the desired one coefficient by coefficient.

1. Part (a) — assemble the state-space model. The second equation is already $\dot{\omega} = -3\theta - u$ because $\ddot{\theta} = \dot{\omega}$. Stacking the three equations in the stated state order gives $$\dot{x} = Ax + Bu, \qquad y = Cx + Du,$$ $$\boxed{\;A = \begin{bmatrix} 0 & 1 & 0\\ -3 & 0 & 0\\ 0 & 0 & 0\end{bmatrix}, \quad B = \begin{bmatrix} 0\\ -1\\ 1\end{bmatrix}, \quad C = \begin{bmatrix} 0 & 1 & 0\end{bmatrix}, \quad D = 0\;}$$ The zero third row of $A$ says that the wheel momentum is a pure integrator of the control and is not fed by the attitude states; the zero third column says that $h$ acts on nothing. Both facts drive the answers to part (b).
2. Part (b) — test controllability. Building the controllability matrix column by column, $AB = [-1\ \ 0\ \ 0]^{T}$ and $A^2B = [0\ \ 3\ \ 0]^{T}$, so $$\mathcal{C} = \begin{bmatrix} B & AB & A^2B\end{bmatrix} = \begin{bmatrix} 0 & -1 & 0\\ -1 & 0 & 3\\ 1 & 0 & 0\end{bmatrix}, \qquad \det \mathcal{C} = -3 \ne 0 .$$ $$\boxed{\;\operatorname{rank}\mathcal{C} = 3 \;\Rightarrow\; \text{the system is controllable}\;}$$ A single torque input reaches all three states because it enters $\dot{\omega}$ and $\dot{h}$ directly and reaches $\theta$ through the kinematic integrator.
3. Test observability. With $C = [0\ \ 1\ \ 0]$ the successive rows are $CA = [-3\ \ 0\ \ 0]$ and $CA^2 = [0\ \ -3\ \ 0]$, so $$\mathcal{O} = \begin{bmatrix} C\\ CA\\ CA^2\end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ -3 & 0 & 0\\ 0 & -3 & 0\end{bmatrix}.$$ The third row is $-3$ times the first, and the third column is identically zero, so $$\boxed{\;\operatorname{rank}\mathcal{O} = 2 \lt 3 \;\Rightarrow\; \text{the system is NOT observable}\;}$$ The unobservable state is $h$, and this is physically obvious rather than an algebraic accident: the momentum stored in the wheel never appears in the body rate, so no record of $\omega$ — however long — can reveal it. Worse, the hidden mode is the eigenvalue at $s = 0$, which is not asymptotically stable, so the realisation is not even detectable: an observer driven by $\omega$ alone cannot be built. A practical attitude system therefore measures wheel speed directly rather than trying to infer it.
4. Part (c) — solve the free response. With $u \equiv 0$ the momentum equation gives $h(t) = h(0) = 0$ and the remaining two states decouple into $$\ddot{\theta} + 3\theta = 0, \qquad \theta(0) = 1,\ \dot{\theta}(0) = 0 .$$ This is undamped simple harmonic motion at $\sqrt{3}$ rad/s, so $\theta(t) = \cos\!\left(\sqrt{3}\,t\right)$, and differentiating, $$\boxed{\;\omega(t) = \dot{\theta}(t) = -\sqrt{3}\,\sin\!\left(\sqrt{3}\,t\right) = -1.732\sin(1.732\,t)\;}$$ The open-loop eigenvalues are $0$ and $\pm j\sqrt{3}$: the satellite simply oscillates forever, which is exactly why a controller is wanted.
5. Part (d) — write both characteristic polynomials. The assigned poles give $$\alpha_c(s) = (s+2-j)(s+2+j)(s+4) = (s^2+4s+5)(s+4) = s^3 + 8s^2 + 21s + 20 .$$ With $u = -Kx$ and $K = [\,k_1\ \ k_2\ \ k_3\,]$, forming $A - BK$ and expanding the determinant $\det\!\left(sI - A + BK\right)$ gives $$s^3 + (k_3 - k_2)\,s^2 + (3 - k_1)\,s + 3k_3 .$$ Note the open-loop polynomial $s^3 + 3s$ is recovered when $K = 0$, a useful check on the expansion before any matching is attempted.
6. Match coefficients and solve. Equating like powers, $$3k_3 = 20, \qquad 3 - k_1 = 21, \qquad k_3 - k_2 = 8,$$ which solve in one pass, bottom-up: $$\boxed{\;K = \begin{bmatrix} -18 & -\tfrac{4}{3} & \tfrac{20}{3}\end{bmatrix} = \begin{bmatrix} -18 & -1.333 & 6.667\end{bmatrix}, \qquad u = 18\,\theta + \tfrac{4}{3}\,\omega - \tfrac{20}{3}\,h\;}$$ Such a gain exists precisely because the pair $(A,B)$ was found controllable in part (b); had the rank test failed, only the controllable modes could have been moved. Substituting back, the eigenvalues of $A - BK$ are $-2 \pm j$ and $-4$ as required.
7. Read the result. The negative signs in $K$ are a consequence of the control entering $\dot{\omega}$ with a minus sign; the loop is still negative feedback. The $h$ gain deserves comment: it is the term that stops the wheel from saturating, because without it ($k_3 = 0$) the closed-loop polynomial would have no constant term and one pole would stay stuck at the origin. Momentum feedback is what converts the free integrator into a placed pole, and it is available here only because $h$ is measured — part (b) showed it cannot be estimated from $\omega$.