Minimum attention control for linear systems

Abstract

In this paper, we present a novel solution to the minimum attention control problem for linear systems. In minimum attention control, the objective is to minimise the ‘attention’ that a control task requires, given certain performance requirements. Here, we interpret ‘attention’ as the inverse of the interexecution time, i.e., the inverse of the time between two consecutive executions. Instrumental for our approach is a particular extension of the notion of a control Lyapunov function and the fact that we allow for only a finite number of possible interexecution times. By choosing this extended control Lyapunov function to be an ∞-norm-based function, the minimum attention control problem can be formulated as a linear program, which can be solved efficiently online. Furthermore, we provide a technique to construct a suitable ∞-norm-based (extended) control Lyapunov function. Finally, we illustrate the theory using a numerical example, which shows that minimum attention control outperforms an alternative ‘attention-aware’ control law available in the literature.

Appendix: Proofs of Theorems and Lemmas

Proof

(Lemma 2) Since Eq. 11 holds and since the solutions to Eq. 1 with Eq. 2 satisfy

$$ x(t_k+\hbar_l) = e^{A \hbar_l} x(t_k)+\int_0^{\hbar_l}e^{As}B \mathrm{d}s \, \hat{u}_k, $$

(31)

we have that

$$ V(x(t_k+\hbar_l)) \leqslant e^{-\alpha q (t_k+\hbar_l)} V(x(0)). \label{eq5:stab_cond_1} $$

(32)

for all l ∈ {0, ..., L − 1} and for all t _k , k ∈ ℕ, with $\hbar_0=0$. Now using Eq. 10, we have that Eq. 32 implies

$$ \| x(t_k+\hbar_l) \| \leqslant \sqrt[q]{\tfrac{\overline{a}}{\underline{a}}} e^{-\alpha (t_k+\hbar_l)} \|x(0)\|, \label{eq5:stab_cond_2} $$

(33)

for all l ∈ {0, ..., L − 1} and for all t _k , k ∈ ℕ, with $\hbar_0=0$. Moreover, because it holds that $\| \hat{u}_k \| \leqslant \beta \| x(t_k) \|$, the solutions to Eq. 1 with Eq. 2 also satisfy

$$ \begin{array}{rll} \| x(t) \| &\leqslant& \| e^{A (t-t_k-\hbar_l)} \| \, \| x(t_k+\hbar_l) \| +\int_{t_k+\hbar_l}^t \| e^{A(t-s)} \| \mathrm{d}s\, \| B \| \| \hat{u}_k \| \\ &\leqslant& e^{\|A\| \Delta_\hbar} \, \|x(t_k+\hbar_l) \| + \beta \int_0^{\Delta_\hbar} e^{\|A\|s} \mathrm{d}s\, \|B\|\, \| x(t_k) \|, \label{eq5:stab_cond_3} \end{array} $$

(34)

for all $t\in[t_k+\hbar_l,t_k+\hbar_{l+1})$, k ∈ ℕ, l ∈ {0, ..., L − 1}, with $\Delta_\hbar$ as defined in the hypothesis of the lemma. Substituting Eq. 33 into this expression (twice) yields

$$ \| x(t) \| \leqslant \sqrt[q]{\tfrac{\overline{a}}{\underline{a}}} \Bigg( e^{\|A\| \Delta_\hbar} \, e^{-\alpha (t_k+\hbar_l)} + \beta\, \int_0^{\Delta_\hbar} e^{\|A\|s} \mathrm{d}s \, \|B\| \, e^{-\alpha t_k} \Bigg) \|x(0)\|, \label{eq5:stab_cond_4} $$

(35)

for all $t\in[t_k+\hbar_l,t_k+\hbar_{l+1})$, k ∈ ℕ, l ∈ {0, ..., L − 1}. Now realising that for all $t\in[t_k+\hbar_l,t_k+\hbar_{l+1})$, k ∈ ℕ, l ∈ {0, ..., L − 1}, it holds that $e^{-\alpha (t_k+\hbar_l)} < e^{-\alpha t + \alpha\Delta_\hbar}$ and that $e^{-\alpha t_k} < e^{-\alpha t + \alpha \hbar_L }$ we have Eq. 4 with c as given in the hypothesis of Lemma 2.□

Proof

(Theorem 1) Using the arguments given in Section 3.3, we have that the hypotheses of the theorem guarantee that F _MAC(x) ≠ ∅ for all $x\in\mathbb{R}^{n_x}$. By following a similar reasoning as done in the proof of Lemma 2, we can show that the MAC law guarantees that Eq. 35 holds for all $t\in[t_k+\hbar_l,t_k+\hbar_{l+1})$, k ∈ ℕ, $l\in\{0,\ldots,\bar{L}^\star(x(t_k))-1\}$, with $\hbar_0=0$, and all $x\in\mathbb{R}^{n_x}$. Again realising that for all $t\in[t_k+\hbar_l,t_k+\hbar_{l+1})$, k ∈ ℕ, $l\in\{0,\ldots,\bar{L}^\star(x(t_k))-1\}$, it holds that $e^{-\alpha (t_k+\hbar_l)} < e^{-\alpha t + \alpha\Delta_\hbar}$ and that $e^{-\alpha t_k} < e^{-\alpha t + \alpha \hbar_{\bar{L}^\star(x(t_k))}} \leqslant e^{-\alpha t + \alpha \hbar_L}$ yields Eq. 4 with gain $c=\bar{c}(\alpha,\beta,\Delta_\hbar,\hbar_L)$ as in Eq. 13.□

Proof

(Lemma 3) The proof follows the same line of reasoning as in Kiendl et al. (1992) and Polański (1995). GES of Eq. 1 with Eq. 19 with convergence rate $\hat\alpha$ and gain $\hat{c} = \overline{a}/\underline{a}$ is implied by the existence of a positive definite function, satisfying Eq. 10 and

$$ \lim\limits_{s\downarrow0} \tfrac{1}{s} \big( V(x(t+s)) - V(x(t)) \big) \leqslant -\hat\alpha V(x(t)), \label{eq5:interevent2} $$

(36)

for all t ∈ ℝ₊, which follows from the Comparison Lemma, see, e.g., Khalil (1996). Now using the fact that the solutions to Eq. 1 with Eq. 19 satisfy $\tfrac{\mathrm{d}}{\mathrm{d}t}x = (A+BK) x$, and using Eq. 17, we obtain that Eq. 36 is implied by

$$ \lim\limits_{s\downarrow0} \tfrac{1}{s} ( \| P (I + s (A+BK)) x(t) \|_\infty - \| P x(t) \|_\infty ) \leqslant -\hat\alpha \| P x(t) \|_\infty,\label{eq5:interevent5} $$

(37)

for all t ∈ ℝ₊. Using Eq. 20a, we have that, for all t ∈ ℝ₊, Eq. 37 implied by

$$ \lim\limits_{s\downarrow0} \tfrac{1}{s} ( \| (I + s Q ) \|_\infty - 1 ) \| P x(t) \|_\infty \leqslant -\hat\alpha \| P x(t) \|_\infty, $$

(39)

which is, due to positivity of ||Px||_∞ for all x ≠ 0, equivalent to $\lim_{s\downarrow0} \tfrac{1}{s} ( \| (I + s Q ) \|_\infty - 1 ) \leqslant -\hat\alpha$, which is implied by Eq. 20b. This completes the proof.□

Proof

(Lemma 4) The proof is based on showing that the Lyapunov function obtained using Lemma 3 also guarantees Eqs. 1 and 2, with Eq. 21 and t _k + 1 = t _k  + h, k ∈ ℕ, to be GES with convergence rate α and gain $c:=\bar{c}(\alpha,\beta,h)$, where $\bar{c}(\alpha,\beta,h)$ as in Eq. 9, for all h < h _max(α) as in Eq. 22. To do so, observe that the solutions of Eqs. 1 and 2, with Eq. 21 and t _k + 1 = t _k  + h, k ∈ ℕ, satisfy

$$ x(t) = \Bigg( e^{A (t-t_k)} + \int_0^{t-t_k}e^{As} BK \mathrm{d}s \Bigg) x(t_k), \label{eq5:interevent1a} $$

(39)

for all t ∈ [t _k ,t _k  + h), k ∈ ℕ. Now by following the ideas used in the proof of Lemma 2, and the candidate Lyapunov function of the form of Eq. 17, we have that GES with convergence rate α and gain c of Eqs. 1 and 2, with Eq. 22 and t _k + 1 = t _k  + h, k ∈ ℕ, is implied by requiring that

$$ \| P x(t_k+h) \|_\infty - e^{-\alpha h} \| P x(t_k) \|_\infty \leqslant 0, \label{eq5:interevent1} $$

(40)

for all t _k , k ∈ ℕ, and some well-chosen h > 0. Substituting Eq. 39 and defining $\hat{x} := P x$, yielding $x = (P^\top P)^{-1}P^\top \hat{x}$, yields that Eq. 40 is implied by

$$ \Bigg( \Big\| P ( e^{A h}+\int_0^he^{As}BK \mathrm{d}s ) (P^{\top}P)^{-1}P^{\top}\Big\|_\infty - e^{-\alpha h}\Bigg) \| \hat{x}(t_k) \|_\infty \leqslant 0, $$

(41)

for all $\hat{x}(t_k)\in\mathbb{R}^m$, which holds for all h > 0, satisfying h < h _max(α), as given in the hypothesis of the lemma, meaning that Eq. 40 holds for all $\hat{x}(t_k)\in\mathbb{R}^m$ and for all h > 0, satisfying h < h _max(α). This completes the proof.□

Proof

(Theorem 2) As a result of Lemma 4, we have that the control input given by Eq. 21 renders the plant with ZOH, Eqs. 1, 2, GES with convergence rate α and gain $c:=\bar{c}(\alpha,\|K\|_\infty,h)$ as in Eq. 9, for any interexecution time h < h _max(α) as in Eq. 22. To obtain a well-defined control law, we need that F _MAC(x) ≠ ∅, for all $x\in\mathbb{R}^{n_x}$, which is guaranteed if and only if Eq. 14 satisfies F ₁(x) ≠ ∅ for all $x\in\mathbb{R}^{n_x}$, as argued in Section 3.3. This can be achieved by choosing $\beta\geqslant\|K\|_\infty$ and choosing the set $\mathcal{H}:=\{\hbar_1,\ldots,\hbar_L\}$, L ∈ ℕ, such that $\hbar_1<h_{\max}(\alpha)$, as this yields that $F_1(x) \supseteq \{ K x \} \neq \emptyset$, if V is chosen as in Eq. 17. GES with the convergence rate α and the gain $c\geqslant\bar{c}(\alpha,\beta,\Delta_\hbar,\hbar_L)$ of Eqs. 1, 2 and 3, with Eqs. 8, 14, 15, 16 and 17, follows directly from Theorem 1. This completes the proof.□