Stabilization of Finite Automata with Application to Hybrid Systems Control

Abstract

This paper discusses the state feedback stabilization problem of a deterministic finite automaton (DFA), and its application to stabilizing model predictive control (MPC) of hybrid systems. In the modeling of a DFA, a linear state equation representation recently proposed by the authors is used. First, this representation is briefly explained. Next, after the notion of equilibrium points and stabilizability of the DFA are defined, a necessary and sufficient condition for the DFA to be stabilizable is derived. Then a characterization of all stabilizing state feedback controllers is presented. Third, a simple example is given to show how to follow the proposed procedure. Finally, control Lyapunov functions for hybrid systems are introduced based on the above results, and the MPC law is proposed. The effectiveness of this method is shown by a numerical example.

Appendix: Derivation procedure of state equation expressing deterministic finite automata

The procedure deriving a state equation from a given DFA is as follows. This is a more sophisticated version of our approach derived in Kobayashi and Imura (2007).

1.1 Procedure of deriving a state equation

Step 1 :

For a given deterministic finite automata ${\cal A}$ with m nodes and n ( ≥ m) arcs, let ${\cal I}_a$ denote the set of combinations of (i,j) such that the arc from node i to node j exists, and assign a binary variable δ _l to the arc l. Furthermore, set $\xi(k) := [~ \delta_{1}(k) ~~ \delta_{2}(k) ~~ \cdots ~~ \delta_{n}(k) ~]^T \in \{ 0,1 \}^n$. Then the input-output relation of δ _l (k) on each node gives the implicit system of

$$ \Sigma_{I}: \left\{ \begin{array}{l} E \xi(k+1) = F \xi(k), \\[2.5pt] \xi(k) \in \{ 0,1 \}^n, ~~ \xi(0) \in \Xi_0 \left(\delta^{M}_0\right). \end{array} \right. $$

where E, F ∈ {0, 1}^{m ×n},

$$ \Xi_0\le\left(\delta^{M}_0\right) := \left\{~ \eta \in \{ 0,1 \}^{n} ~|~ e^{T}_n \eta = 1, ~ E \eta = \delta^{M}_0 ~\right\} $$

and $\delta^{M}_0\in \{0,1\}^{m}$ denotes a given initial mode satisfying $e^{T}_m \delta^{M}_0=1$.

Step 2 :

Derive a permutation matrix P satisfying $E P = [~ I_m ~~ \tilde{E} ~]$, where $\tilde{E} \in \{ 0,1 \}^{m \times (n-m)}$ is some matrix. Then by using

$$ \hat{V} = \left[ \begin{array}{cc} I_m & \tilde{E} \\ 0_{(n-m) \times m} & I_{n-m} \end{array} \right] P^{-1}, $$

compute $E \hat{V}^{-1} = [~ I_m ~~ 0_{m \times (n-m)} ~]$ and

$$ F \hat{V}^{-1} = \left[~ \tilde{F}_1 ~~ -\tilde{F}_1 \tilde{E} + \tilde{F}_2 ~\right] =: \left[~ \hat{A} ~~ \hat{B} ~\right] $$

where $[~ \tilde{F}_1 ~~ \tilde{F}_2 ~] := FP$. Thus letting $[~ {x}^T(k) ~~ \hat{u}^T(k) ~]^T:=\hat{V} \xi(k)$, the state equation with inequality constraints is obtained as

$$ \left\{ \begin{array}{l} {x}(k+1) = \hat{A} {x}(k) + \hat{B} \hat{u}(k), \\[1.5pt] -x(k) + \tilde{E} \hat{u}(k) \leq 0, \\[1.5pt] {x}(k) \in {\bf R}^m, ~~ \hat{u}(k) \in \{ 0,1 \}^{n-m}, \\[1.5pt] {x}(0) = x_0 \in {\cal X}_0 := \{~ \zeta \in \{ 0,1 \}^m ~|~ e_m^T \zeta = 1 ~\}. \end{array} \right. $$

(32)

If $\hat{B}$ is full row rank, then Eq. 32 with $u(k) := \hat{u}(k)$ is the state-equation-based model to be found. Otherwise, go to Step 3.

Step 3 :

Reduce the matrix $\hat{B}$ to

$$ \hat{B} = P_B \left[ \begin{array}{cc} I_{\hat{\alpha}} & 0 \\ \tilde{B} & 0 \end{array} \right] T_B $$

(33)

where $\hat{\alpha} := {\rm rank} \hat{B}$, P _B is a permutation matrix, T _B is a nonsingular matrix, and $\tilde{B}$ is some matrix. Next, define

$$ \left[ \begin{array}{c} \tilde{u}(k) \\ \tilde{u}_e(k) \end{array} \right] := T_B \hat{u}(k) $$

(34)

where $\tilde{u}_e(k)$ denotes redundant input variables. Then applying the input transformation

$$ \tilde{u}(k) = \hat{A}_u {x}(k) + u(k) $$

(35)

where $\hat{A}_u := -[~ I_{\hat{\alpha}} ~~ 0_{\hat{\alpha} \times (m-\hat{\alpha})} ~] P_B^{-1} \tilde{F}_1$ and $u(k) \in \{ 0,1 \}^{\hat{\alpha}}$ is the binary input vector, to Eq. 32 yields

$$ \left\{ \begin{array}{l} x(k+1) = A x(k) + B u(k), \\[1.5pt] C x(k) + D u(k) \leq G, \\[1.5pt] x(k) \in {\bf R}^{m}, ~~ u(k) \in \{ 0,1 \}^{\hat{\alpha}}, \\[1.5pt] x(0) = x_0 \in {\cal X}_0 \end{array} \right. $$

(36)

where

$$\begin{array}{rll} && A := P_B \left[ \begin{array}{cc} 0 & 0 \\ -\tilde{B} & I_{m-\hat{\alpha}} \end{array} \right] P_B^{-1} \hat{A},~~ B := P_B \left[ \begin{array}{c} I_{\hat{\alpha}} \\ \tilde{B} \end{array} \right], \\ && C := \left[ \begin{array}{c} I_m - \Phi A \\ e_m^T A \\ -e_m^T A \end{array} \right] , ~~ D := \left[ \begin{array}{c} - \Phi B \\ e_m^T B \\ -e_m^T B \end{array} \right] , ~~ G := \left[ \begin{array}{c} 0_{\hat{\alpha} \times 1} \\ 1 \\ -1 \end{array} \right], \end{array} $$

and Φ is the adjacency matrix of a given finite automaton.

In Step 1, noting that n ≥ m holds, for a given $\delta^M_0$, ξ(0) is not uniquely determined. So we consider the set $\Xi_0(\delta^M_0)$.

In Step 2, the state equation (32) includes the inequality $-x(k) + \tilde{E} \hat{u}(k) \leq 0$. To explain this inequality, we show a very simple example. Consider the linear scalar system x(k + 1) = x(k) − u ₁(k) − u ₂(k), where x(k + 1), x(k), u ₁(k), u ₂(k) ∈ {0, 1}. To satisfy the binary property of x(k + 1), the constraint must be considered for u ₁(k), u ₂(k). If x(k) = 0, then u ₁(k) = u ₂(k) = 0 must hold. If x(k) = 1, then we must consider only two cases: (i) u ₁(k) = 1, u ₂(k) = 0 and (ii) u ₁(k) = 0, u ₂(k) = 1. From the above discussion, the inequality constraint − x(k) + u ₁(k) + u ₂(k) ≤ 0 must be imposed. This inequality corresponds to $-x(k) + \tilde{E} \hat{u}(k) \leq 0$ in Eq. 32. Furthermore, the state x implies a dependent variable, which can be determined from m equations in Σ _I . The input $\hat{u}$ implies an independent (free) variable.

In Step 3, by substituting Eq. 35 into Eq. 32 and replacing the inequality of Eq. 32 to the inequality using the adjacency matrix Φ, we obtain Eq. 36. The input transformation of Eq. 35 guarantees the binary property of the input vectors because $\tilde{u}$ itself does not always take binary values due to some transformation (Eq. 34).

In addition, although the matrices P, P _B , T _B in the above procedure are not unique, P, P _B , T _B satisfying the conditions can be derived by elementary transformations of matrices, which can be easily implemented by a suitable software such as MATLAB. Note here that the dimension of u does not depend on selection of P, P _B , T _B . Note here that the computation cost of the above procedure is very small, since there does not exist iteration in all steps of the proposed procedure.