CAD.3:4.1 - Construct an invariant error set

For the linear interface of CAD.2, write

e^+=Fe+Hz+Gv+w.

Choose a proposed command set S of pairs (z,v), a bounded mismatch set W, and an error set E containing the admitted initial error. A sufficient one-step condition is

FE\ \oplus\ \{Hz+Gv:(z,v)\in S\}\ \oplus\ W\ \subseteq E.

Here \oplus is addition of sets. This is a property to establish, not a consequence of naming E an error tube. Substitution and induction then keep the error in E for every permitted command sequence. Preserve correlations between z and v when they materially reduce conservatism; replacing S by a Cartesian product can enlarge the forcing unnecessarily.

One usable construction takes a positive-definite M and the norm \|e\|_M=\sqrt{e^\mathsf{T}Me}. Verify

q=\|M^{1/2}FM^{-1/2}\|_2<1,\qquad d\geq\sup_{(z,v)\in S,\ w\in W}\|Hz+Gv+w\|_M.

The triangle inequality gives \|e^+\|_M\leq q\|e\|_M+d. Therefore

E=\{e:\|e\|_M\leq\varepsilon\},\qquad \varepsilon\geq\max\{\|e_0\|_M,d/(1-q)\}

is invariant. For a stable fixed F, a positive-definite solution of a discrete Lyapunov equation is a candidate M; verify the displayed norm inequality for the actual values used. Failure of this sufficient construction does not prove that no controller or invariant set exists.

For multiple timing modes, one sufficient condition for arbitrary admissible switching is a common set E containing the initial error and preserved by every admitted mode and reset map. This includes resets caused by a changed reference or error coordinate. Each step then starts in the same E, so induction covers any admitted sequence.

Separate invariant sets for the frozen modes are not sufficient by themselves. With mode-dependent or within-period phase sets E_i, prove that every allowed transition from mode or phase i to j, including its reset, sends the admitted part of E_i into E_j for all admitted commands and disturbances. Include initialization and the evolution within each phase. A boundary-map proof can use this construction, but a bound only at period boundaries leaves the intermediate phases uncovered.