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Solved reachability and controllability exercises, from basic to exam level

Six exercises with solutions: Kalman's test, systems with two inputs, reachability with a parameter, companion form, minimum energy of the double integrator, a controllable but not reachable system.

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Exercises on reachability and controllability of the kind set in Systems Theory and Automatic Control courses, at three levels: basic, for the direct use of Kalman’s test; intermediate, with parameters and canonical forms; exam, with the Gramian and discrete-time systems. The result is given at once; the full solution opens with a click. The proofs are in the lesson and in the in-depth page linked at the top.

Basic level

Exercise 1. Kalman’s test on a second-order system

The system \dot x = Ax + Bu has A = \begin{pmatrix} 1 & 2 \\ 0 & -1 \end{pmatrix} and B = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. Is it reachable?

Result: \det\mathcal{R} = -4 \ne 0: yes, the system is reachable.

Solution

Compute AB = \begin{pmatrix} 1 + 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} and place the columns side by side:

    \[ \mathcal{R} = \begin{pmatrix} B & AB \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 1 & -1 \end{pmatrix},\qquad \det\mathcal{R} = 1\cdot(-1) - 3\cdot 1 = -4 \]

The determinant is nonzero, so the rank is 2 and the system is reachable. (The exact value of the determinant does not matter: all that matters is that it is not zero.)

Exercise 2. Two inputs

With A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} and B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} (two independent integrators, each with its own input), is the system reachable? And if the second input is removed?

Result: With two inputs yes (rank 2); with the first input only, no (rank 1).

Solution

With two inputs \mathcal{R} = \begin{pmatrix} B & AB \end{pmatrix} is a 2\times 4 matrix that already contains B = I: rank 2, reachable. Each input moves its own state.

With the first input only, B = (1\ \ 0)^T and AB = 0: \mathcal{R} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} has rank 1. The second integrator receives nothing and stays where it is: the state x_2 is not reachable.

Intermediate level

Exercise 3. Reachability with a parameter

For which values of a is the system with A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 1 \\ a \end{pmatrix} not reachable? Interpret the result.

Result: Not reachable for a = -1 and for a = -2, when B is aligned with an eigenvector of A.

Solution

AB = \begin{pmatrix} a \\ -2 - 3a \end{pmatrix} and

    \[ \det\mathcal{R} = \det\begin{pmatrix} 1 & a \\ a & -2 - 3a \end{pmatrix} = -2 - 3a - a^2 = -(a + 1)(a + 2) \]

The system is not reachable for a = -1 and a = -2. The eigenvalues of A are -1 and -2, with eigenvectors (1\ \ -1)^T and (1\ \ -2)^T. For a = -1 the vector B coincides with the eigenvector of -1: the input excites only that mode, and the state stays on the eigenvector line. The same holds for a = -2 with the other mode. In general: an input aligned with an eigenvector is not enough to move the whole state.

Exercise 4. The companion form

Check that the system with A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3 \end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} is reachable.

Result: \det\mathcal{R} = -1: rank 3, reachable.

Solution

AB = (0\ \ 1\ \ -3)^T, A^2B = A\,(0\ \ 1\ \ -3)^T = (1\ \ -3\ \ 7)^T. Hence

    \[ \mathcal{R} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & -3 \\ 1 & -3 & 7 \end{pmatrix},\qquad \det\mathcal{R} = 1\cdot(0\cdot(-3) - 1\cdot 1) = -1 \]

Rank 3. It is no coincidence: the reachability matrix of a companion form (controllable canonical form) is always triangular with respect to the antidiagonal, with ones on the antidiagonal, so its determinant is \pm 1 whatever the coefficients. This is why every reachable single-input system can be brought into this form, which makes the design of the feedback immediate.

Exam level

Exercise 5. The cost of being quick

A 1 kg platform (the double integrator) must move by 2 m and stop in 2 s. Find the peak force and the minimum energy. How do they change if the time drops to 1 s?

Result: With T = 2 s: 3 N and 6 N²·s. With T = 1 s: 12 N and 48 N²·s.

Solution

From the formulas derived above, with d = 2 and T = 2: peak force u(0) = 6d/T^2 = 3 N, energy E = 12d^2/T^3 = 6 N²·s. With T = 1 s: u(0) = 12 N (four times as much), E = 48 N²·s (eight times as much). If the actuator cannot exceed 5 N, the minimum time with this profile is T = \sqrt{6d/5} \approx 1.55 s. With a bounded force, the absolute minimum time is obtained instead with a “full forward, full reverse” profile (bang-bang control), not with the minimum-energy one.

Exercise 6. Controllable but not reachable

In discrete time, with A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} and B = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, study reachability and controllability.

Result: Not reachable (rank 1), but controllable: every state goes to zero in two steps.

Solution

AB = (0\ \ 0)^T: \mathcal{R} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} has rank 1, the system is not reachable (the second component cannot be made nonzero). However A^2 = 0: with zero input, x(2) = A^2x(0) = 0 for every initial state. Every state is driven to zero in two steps, so the system is controllable. The difference comes from the zero eigenvalue of A; in continuous time it cannot happen, because e^{At} is always invertible.

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