Contents CAS, Mines-ParisTech 2008
Contents Contents 1, Linear Case Introductory Example: Linear Motor with Appended Mass General Solution (Linear Case)
Contents Contents 1, Linear Case Introductory Example: Linear Motor with Appended Mass General Solution (Linear Case) 2 Example of of Implicit Systems Flatness Necessary and Sufficient Conditions
Introductory Example General Solution Plan 1, Linear Case Introductory Example: Linear Motor with Appended Mass General Solution (Linear Case) 2 Example of of Implicit Systems Flatness Necessary and Sufficient Conditions
Introductory Example General Solution Introductory Example: Linear Motor with Appended Mass Model: Mẍ = F k(x z) r(ẋ ż) m z = k(x z) + r(ẋ ż) Aim: Fast and high-precision rest-to-rest displacements. Measurements: Motor position and velocity z not measured. rail linear motor flexible beam mass bumper Experiment realized with the help of Micro-Controle.
Introductory Example General Solution Flat Output Computation (J.L. et D.V. Nguyen, S&CL, 2003) We want to express x, z and F as Thus: d x = a 0 y + a 1 ẏ +... = (a 0 + a 1 dt +...)y = P x( d dt )y d z = b 0 y + b 1 ẏ + = (b 0 + b 1 dt +...)y = P z( d dt )y d F = c 0 y + c 1 ẏ +... = (c 0 + c 1 dt +...)y = P F( d dt )y ( ) M d2 + r d dt 2 dt + k P x y ( r d dt + k) P z y = P F y ( ( ) r d dt + k) P x y + m d2 + r d dt 2 dt + k P z y = 0 Solution: P x = 1 k (m d2 dt 2 + r d ) dt + k, P z = 1 (r ddt ) k + k
Introductory Example General Solution Flat Output: ) y = (1 r2 mk x + r2 z r mk k ż x = y + r k ẏ + m k ÿ, z = y + r ( k ẏ F = (M + m) ÿ + r ) Mm k y(3) + (M + m)k y(4)
Introductory Example General Solution General Solution (Linear Case) Consider the linear controllable system A( d dt )x = Bu. We are looking for P and Q such that x = Py, u = Qy with y: flat output to be determined. Let C be s.t. C T B = 0. We thus have to solve C T AP = 0, AP = BQ General Solution Smith Decomposition of C T A( d dt ): U, V unimodular such that VC T AU ( = ) ( 0). Thus: I P = U P 0 0 ( I Indeed: C T AP = C T AU 0 ) P 0 = V 1 ( 0) A flat output y is deduced by left inversion of P. ( I 0 ) P 0 = 0
Plan, Linear Case 1, Linear Case Introductory Example: Linear Motor with Appended Mass General Solution (Linear Case) 2 Example of of Implicit Systems Flatness Necessary and Sufficient Conditions
Example of ẋ = u cos θ ẏ = u sin θ θ = u l tan ϕ After elimination of the input variables u and ϕ: ẏ ẋ tan θ = 0 Implicit representation invariant by dynamic extension. Variational Equation In Polynomial Form : ( tan θ d dt dẏ tan θ dẋ d dt ẋ cos 2 θ dθ = 0 ẋ ) cos 2 θ dx dy dθ = 0.
If (z 1, z 2 ) is a flat output, then (dz 1, dz 2 ) is a flat output of the variational system and we must have ( dx = 2 j 0 i=1 P x,i,jdz (j) i = P x,1 dz 1 + P x,2 dz 2 dy = 2 j 0 i=1 P y,i,jdz (j) i = P y,1 dz 1 + P y,2 dz 2 dθ = 2 j 0 i=1 P θ,i,jdz (j) i = P θ,1 dz 1 + P θ,2 dz 2 tan θ d dt d dt ẋ ) cos 2 θ P x,1 P x,2 P y,1 P y,2 P θ,1 P θ,2 = 0
Smith Decomposition ( tan θ d dt d dt ẋ ) 0 0 1 0 1 0 cos 2 θ cos2 θ cos 2 θ d ẋ ẋ dt = ( 1 0 0 ) sin θ cos θ ẋ d dt One can verify that P is given by 0 1 P = 1 0 P 0, (P 0 arbitrary). cos 2 θ d sin θ cos θ d ẋ dt ẋ dt and thus dx 0 1 ( ) dy = 1 0 cos dθ 2 θ d sin θ cos θ d P dz1 0 dz 2 ẋ dt ẋ dt or dx = dz 2, dy = dz 1. Therefore: x = z 2, y = z 1.
of Nonlinear Systems Consider the explicit system ẋ = f (x, u) with rank ( f u ) = m locally on X R m. A representation invariant by endogenous dynamic extension is obtained by elimination of the input u = ν(x, ẋ), yielding the (n m)-dimensional implicit system: F(x, ẋ) = 0 with rank ( F ẋ ) = n m. We introduce the global coordinates x = (x, ẋ, ẍ,...) on the manifold X R n endowed with the trivial Cartan field τ n. The implicit representation is thus given by the triple (X R n, τ n, F).
of Implicit Systems Consider two implicit systems (X R n, τ n, F) and (Y R n, τ n, G). They are said Lie-Bäcklund equivalent iff there exists a locally C mapping Φ : Y R n X R n, with locally C inverse Ψ s.t. (i) Φ τ n = τ n and Ψ τ n = τ n ; (ii) for every y s.t. Lτ k n G(y) = 0 for all k 0, then x = Φ(y) satisfies Lτ k n F(x) = 0 for all k 0 and conversely. The system (X R n, τ n, F) is flat iff it is Lie-Bäcklund equivalent to (R m, τ m, 0). Variational Property The system (X R n, τ n, F) is flat iff there exists a locally C and invertible mapping Φ : X R n R m such that Φ df = 0.
We have: df = F F dx + x ẋ dẋ = ( F x + F ) d dx = P(F)dx ẋ dt and with Φ df = P(F) P(Φ 0 )dy P(Φ 0 ) = j 0 Φ 0 d j y (j) dt j We thus have to find a polynomial matrix P(Φ 0 ) solution to P(F) P(Φ 0 ) = 0. This solution is deduced from the Smith decomposition of P(F).
Notations: K: field of meromorphic functions from X R n to R and K[ d dt ] principal ideal ring of polynomials of d dt = L τ n with coefficients in K. M p,q [ d dt ]: module of the p q matrices over K[ d dt ], with p and q arbitrary integers. U p [ d dt ]: group of unimodular matrices of M p,p[ d dt ]. Smith Decomposition: If P(F) M n m,n [ d dt ], there exist V U n m [ d dt ] and U U n[ d dt ] such that VP(F)U = (, 0 n m,m ). We note V L Smith (P(F)) and U R Smith (P(F)). Moreover, if the linear tangent system around an arbitrary trajectory is controllable, one can prove that = I n m.
Flatness Necessary and Sufficient Conditions (Lévine (2004), Lévine (2006)) Theorem The system (X R n, τ n, F), assumed first-order controllable, ) is flat iff for all U R Smith (P(F)) and Q L Smith (Û, with ( ) 0n m,m Û = U, there exists an m m matrix µ such that for all I m κ Λ p (X) m and all p N, µκ Λ p+1 (X), and a matrix M U m [ d dt ] such that, if we denote by ω = (I m, 0 m,n m ) Q(x)dx, we have: dω = µω, d (µ) = µ 2, d (M) = Mµ with d extension of the exterior derivative d to polynomial matrices with coefficients in Λ(X).
Comments, Linear Case The last conditions generalize the moving frame structure equations of Cartan to manifolds of jets of infinite order. They are equivalent to the existence of M s.t. d(mω) = 0, or equivalently to the existence of y s.t. dy = Mω (flat output). Their validity may be checked by computer algebra.