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Contents

Chapter 6. Differentially Flat Systems

Jean LEVINE

CAS, Mines-ParisTech

2008

Jean LEVINE Chapter 6. Differentially Flat Systems
http://find/http://goback/

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Contents

Contents

1 Flatness NSC, Linear Case

Introductory Example: Linear Motor with Appended Mass

General Solution (Linear Case)


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Contents

Contents




2 Flatness NSC, General Case

Example of Non Holonomic Vehicle

Implicit Representation

Lie-Backlund Equivalence of Implicit SystemsFlatness Necessary and Sufficient Conditions


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Flatness NSC, Linear CaseFlatness NSC, General Case

Introductory ExampleGeneral Solution

Plan









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Model:

Mx = F k(x z) r(x z)mz = k(x z) + r(x z)

Aim:

Fast and high-precision rest-to-rest

displacements.

Measurements:Motor position and velocity

z not measured.

mass

flexible beam

bumper

linear motorrail

Experiment realized with the help of

Micro-Controle.


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Flat Output Computation (J.L. et D.V. Nguyen, S&CL, 2003)

We want to express x, z and F as

x = a0y + a1 y + . . . = (a0 + a1ddt

+ . . .)y = Px(ddt

)y

z = b0y + b1 y + = (b0 + b1ddt

+ . . .)y = Pz(ddt

)y

F = c0y + c1 y + . . . = (c0 + c1ddt

+ . . .)y = PF(ddt

)y

Thus: Md

2

dt2+ rd

dt+ k

Pxy

rd

dt+ k

Pzy = PFy

rddt

+ kPxy + md2

dt2+ rd

dt+ kPzy = 0

Solution:

Px =1

k

m

d2

dt2+ r

d

dt+ k

, Pz =

1

k

r

d

dt+ k


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Flat Output:

y =r2

mk

x +1 r2

mkz

r

k

z

x = y +r

ky +

m

ky, z = y +

r

ky

F = (M+ m)

y +

r

ky(3) +

Mm

(M+ m)ky(4)


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Consider the linear controllable system A( ddt)x = Bu. We are lookingfor P and Q such that x = Py, u = Qy with y: flat output to bedetermined.

Let C be s.t. CTB = 0. We thus have to solveCTAP = 0, AP = BQ

General Solution

Smith Decomposition ofCTA( ddt

):U, V unimodular such that VCTAU = (|0). Thus:

P = U I

0

P0

Indeed: CTAP = CTAU

I

0

P0 = V

1 (|0)

I

0

P0 = 0

A flat output y is deduced by left inversion ofP.


Non Holonomic Vehicle

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Non Holonomic VehicleImplicit RepresentationLie-Backlund EquivalenceFlatness NSC

Plan










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Non Holonomic VehicleImplicit RepresentationLie-Backlund EquivalenceFlatness NSC


x = u cos y = u sin

=u

ltan

After elimination of the input variables u and :

y x tan = 0Implicit representation invariant by dynamic extension.

Variational Equation

dy tan dx

x

cos2 d = 0In Polynomial Form :

tan d

dt

d

dt

x

cos2

dx

dy

d

= 0.



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Implicit RepresentationLie-Backlund EquivalenceFlatness NSC

If(z1,z2) is a flat output, then (dz1, dz2) is a flat output of thevariational system

dx =

j0

2i=1 Px,i,jdz

(j)i = Px,1dz1 + Px,2dz2

dy = j02i=1 Py,i,jdz(j)i = Py,1dz1 + Py,2dz2d =

j0

2i=1 P,i,jdz

(j)i = P,1dz1 + P,2dz2

and we must have

tan

d

dt

d

dt x

cos2 Px,1 Px,2

Py,1 Py,2P,1 P,2

= 0



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Smith Decomposition

tan

d

dt

d

dt

x

cos2

0 0 10 1 0 cos

2 x

cos2 x

ddt

sin cos xddt

=

1 0 0

One can verify that P is given by

P =

0 11 0

cos2 x

ddt

sin cos xddt

P0, (P0 arbitrary).

and thus dxdy

d

=

0 1

1 0

cos2

x

d

dt

sin cos

x

d

dt

P0

dz1dz2

or dx = dz2, dy = dz

1. Therefore: x = z

2, y = z

1.



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Implicit Representation of Nonlinear Systems

Consider the explicit system x = f(x, u) with rank f

u

= m locallyon X Rm.A representation invariant by endogenous dynamic extension is

obtained by elimination of the input u = (x, x), yielding the(n m)-dimensional implicit system:

F(x, x) = 0

with rankFx

= n m.

We introduce the global coordinates

x = (x, x,x, . . .)

on the manifold X Rn endowed with the trivial Cartan field n.The implicit representation is thus given by the triple

(XR

n

, n, F).Jean LEVINE Chapter 6. Differentially Flat Systems


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Lie-Backlund Equivalence of Implicit Systems

Consider two implicit systems (X Rn, n, F) and (Y R

n

, n , G).They are said Lie-Backlund equivalent iff there exists a locally C

mapping : Y Rn

X Rn, with locally C

inverse s.t.

(i) n = n and n = n ;

(ii) for every y s.t. LknG(y) = 0 for all k 0, then

x = (y) satisfies Lkn F(x) = 0 for all k 0 andconversely.

The system (XRn, n, F) is flat iff it is Lie-Backlund equivalent to(Rm, m, 0).

Variational Property

The system (XRn, n, F) is flat iff there exists a locally C and

invertible mapping : X Rn Rm such that

dF = 0.


Fl NSC Li CNon Holonomic VehicleI li i R i

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We have:

dF =F

xdx +

F

xdx =

F

x+

F

x

d

dt

dx

= P(F)dx

and

dF = P(F) P(0)dy

with

P(0) =j0

0

y(j)dj

dtj

We thus have to find a polynomial matrix P(0) solution to

P(F) P(0) = 0.

This solution is deduced from the Smith decomposition ofP(F).


Fl t NSC Li CNon Holonomic VehicleI li it R t ti

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Notations:

K: field of meromorphic functions from XRn

to R and K[d

dt]principal ideal ring of polynomials of d

dt= Ln with coefficients

in K.

Mp,q[ddt

]: module of the p q matrices over K[ ddt

], with p and qarbitrary integers.

Up[ddt

]: group ofunimodular matrices ofMp,p[ddt

].

Smith Decomposition: IfP(F) Mnm,n[ddt

], there exist

V Unm[ddt

] and U Un[ddt

] such that

VP(F)U = (, 0nm,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 = Inm.


Flatness NSC Linear CaseNon Holonomic VehicleImplicit Representation

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Flatness Necessary and Sufficient Conditions

(Levine (2004), Levine (2006))

Theorem

The system (X Rn, n, F), assumed first-order controllable, is flat

iff for all U R Smith (P(F)) andQ L SmithU, withU = U

0nm,m

Im

, there exists an m m matrix such that for all

p(X)m and all p N, p+1(X), and a matrix M Um[ddt

]such that, if we denote by

= (Im, 0m,nm) Q(x)dx ,we have:

d = , d () = 2, d (M) = Mwith d extension of the exterior derivative d to polynomial matrices

with coefficients in (X).


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Comments

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(M) = 0, orequivalently to the existence ofy s.t. dy = M (flat output).

Their validity may be checked by computer algebra.


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