NLControl - package for modelling, analysis and synthesis of nonlinear control systems

Transcription

1 NLControl - package for modelling, analysis and synthesis of nonlinear control systems Maris Tõnso Institute of Cybernetics Tallinn University of Technology Estonia maris@cc.ioc.ee May, 2013

2 Contents 1 Difference/differential field Sequence {H k } Problem 1: Linearization 2 Non-commutative polynomials Polynomial system Problem 2: Reduction Problem 3: Realization 2 / 15

3 The main idea Allows to address a wide range of nonlinear control problems 1 Equations One-: The nonlinear system is globally linearised and described by its infinitesimal representation. 2 Subspaces of differential one- are computed. 3 Intrinsic necessary and sufficient conditions for the existence of the solution are formulated in terms of the one-. 4 One- Eqations: Integration of one-. Integrability restrictions. 5 Final solution is found. 3 / 15

4 [E. Aranda-Bricaire, Ü. Kotta, C. Moog, 1996] x(t + 1) = f (x(t), u(t)), x(t) R n, u(t) R m K - the field of meromorphic functions in a finite number of variables from the infinite set {x(0), u j (t), j = 1,..., m, t 0} Forward-shift operator δ : K K δx i (t) = x i (t + 1) = f i ( ), i = 1,..., n δu j (t) = u j (t + 1), j = 1,..., m δ is injective (K, δ) is a difference field 4 / 15

5 Define a difference vector space E := span K {dϕ ϕ K}. The elements of E are called one-. δ : K K induces δ : E E by a i dϕ i (δa i )d(δϕ i ), a i, ϕ i K. i i δ and d are commutative The relative degree r of a one-form ω E is defined to be the least integer such that δ r ω / span K {dx(0)}. If such an integer does not exist, we set r =. 5 / 15

6 Sequence of subspaces {H k } Define a sequence of subspaces {H k } of E: H 1 = span K {dx(0)} H k+1 = {ω H k δω H k }, k 1. Each H k contains the one- whose relative degree r k {H k } is a decreasing sequence. H 1 H k H k +1 = H k +2 = =: H 6 / 15

7 Problem 1: Static state feedback linearization x(t + 1) = f (x(t), u(t)), x(t) R n, u(t) R m Theorem: The system is linearizable by static stat feedback iff All subspaces H k are integrable H = {0}. To check integrability: Frobenius Theorem: The subspace span{ω 1,..., ω k } is integrable iff dω l ω 1 ω k = 0 for all l = 1,..., k. 7 / 15

8 Non-commutative polynomial ring Noncommutative polynomials act on one- as operators allow to repersent computations in a more compact way. Introduce the left polynomial ring K[, δ]. A polynomial from K[, δ] has the form a = n a i n i, i=0 a i K polynomial indeterminate. 8 / 15

9 Non-commutative polynomial ring (continuation) Multiplication in the ring K[, δ] is not commutative since a a, a K K[, δ]. Multiplication is defined by the commutation rule: Example: a = δ(a), a K u(t) = u(t + 1) u(t) = u(t) The ring K[, δ] is an Ore ring. In the following we interprete as δ. 9 / 15

10 Polynomial system i/o equation y(t + n) = φ(y(t),..., y(t + n 1), u(t),..., u(t + s)) Differentiate the i/o equations Use definitions dy(t +j) = j dy(t), du(t +r) = r du(t) p( )dy(t) = q( )du(t) n 1 p = n φ y(t + j) j q = j=0 s r=0 φ u(t + r) r 10 / 15

11 Problem II - Reduction Problem setting y(t + n) = φ(y(t),..., y(t + n 1), u(t),..., u(t + s)) Def: A function ϕ r K is said to be an autonomous variable for a system, if there exist an integer µ 1 and a non-zero function F such that F (ϕ r, δϕ r,..., δ µ ϕ r ) = 0. The system is said to be irreducible if there does not exist any non-zero autonomous variable for it. The system is reducible if it there exists a lower order system which is transfer equivalent with the original system. 11 / 15

12 Problem II - Reduction Solution p( )dy(t) q( )du(t) = 0 Theorem: Nonlinear system is irreducible, iff p and q are relatively left prime. Let g = LeftGCD(p, q). Then p( )dy(t) q( )du(t) = g(s)[ p( )dy(t) q( )du(t)] Denote dψ := [ p( )dy(t) q( )du(t)]. Then ψ = 0 is the reduced equation. 12 / 15

13 Problem 3 - Realization Problem setting Input-output equations y(t + n) = φ(y(t),..., y(t + n 1), u(t),..., u(t + s)) State equations x(t + 1) = f (x(t), u(t)) y(t) = h(x(t)) We are looking for minimal and observable realization For arbitrary i/o equation the state space form does not necessarily exist. 13 / 15

14 Problem III - Realization Solution p( )dy(t) q( )du(t) = 0 For r = r k k + + r 1 + r 0 Define cut-and-shift opeartor δc 1 (r) := δ 1 (r k ) k δ 1 (r 2 ) + δ 1 (r 1 ) [ ] dy ω l := δc l [p, q] for l = 1,..., n du Theorem: The i/o equation is realizable if the subspace H = {ω 1,..., ω n } is integrable. s of the state coordinates dx i can be found as integrable linear combinations of ω 1,..., ω n. 14 / 15

15 More Problems Addressed by algebraic approach and implemented in NLControl Tranforming the state equations into observer form Model matching problems Flatness Input-output linearization Disturbance decoupling. 15 / 15