Chap. 3. Controlled Systems,...
Transcript of Chap. 3. Controlled Systems,...
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Chap. 3. Controlled Systems, Controllability
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1. Controllability of Linear Systems1.1. Kalman’s CriterionConsider the linear system
x = Ax + Bu
wherex ∈ Rn : state vector andu ∈ Rm : input vector.A : of sizen× n andB : of sizen×m.
Definition The pair(A, B) is controllable if, given a durationT > 0and two arbitrary pointsx0, xT ∈ Rn, there exists a piecewise conti-nuous functiont 7→ u(t) from [0, T ] to Rm, such that the integralcurvex(t) generated byu with x(0) = x0, satisfiesx(T ) = xT .
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In other words
eATx0 +
∫ T
0eA(T−t)Bu(t)dt = xT .
This property depends only onA andB :
Theorem (Kalman) A necessary and sufficient condition for(A, B)to be controllable is
rankC = rank(B|AB| · · · |An−1B
)= n.
C is calledKalman’s controllability matrix (of sizen× nm).
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ProofWithout loss of generality, we can consider thatx0 = 0 by changingxT in yT = xT − eATx0 since∫ T
0eA(T−t)Bu(t)dt = yT = xT − eATx0.
Consider the matrices
C(t) = eA(T−t)B, G =
∫ T
0C(t)C ′(t)dt
whereC ′ : transposed matrix ofC.We prove 2 lemmas.
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Lemma 1 A necessary and sufficient condition for(A, B) to becontrollable is thatG is invertible.AssumeG invertible. It suffices to set :
u(t) = B′eA′(T−t)G−1yT .
Thus∫ T
0eA(T−t)Bu(t)dt =
∫ T
0eA(T−t)BB′eA′(T−t)G−1yTdt
= GG−1yT = yT
andu generates the required trajectory.The converse is immediate.
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Lemma 2 Invertibility of G is equivalent torankC = n.By contradiction, ifG isn’t invertible, ∃v ∈ Rn, v 6= 0, such thatv′G = 0. Thus
v′Gv =
∫ T
0v′C(t)C ′(t)vdt = 0.
Sincev 7→ v′C(t)C ′(t)v : quadratic form≥ 0,∫ T0 v′C(t)C ′(t)vdt = 0 impliesv′C(t)C ′(t)v = 0 ∀t ∈ [0, T ].
Thus :v′C(t) = 0 ∀t ∈ [0; T ]. But
v′C(t) = v′eA(T−t)B = v′
I +∑i≥1
Ai(T − t)i
i!
B = 0
implies thatv′B = 0, v′AiB = 0 pour touti ≥ 1 or v′C = 0 withv 6= 0, thus :
rankC < n.
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Conversely, if rankC < n, ∃v 6= 0 such that
v′C = 0.
Thus, according to what precedes,v′AiB = 0 for i = 0, . . . , n− 1.By Cayley-Hamilton’ Theorem,An+i, i ≥ 0, is a linear combinationof theAj ’s, j = 1, . . . , n− 1.Thusv′AiB = 0, ∀i ≥ 0 or v′G = 0, which proves the Theorem.
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Examplesx1 = x2 , x2 = u
is controllable :B =
(01
), A =
(0 10 0
),
C =
(0 11 0
)and rankC = 2.
x1 = u , x2 = u
isn’t controllable :B =
(11
), A =
(0 00 0
),
C =
(1 01 0
)and rankC = 1.
Note : x1 − x2 : 0, or x1 − x2 = Cste : relation between statesindependent ofu.
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1.2. Controllability Canonical Form
Definition Two systems
x = Ax + Bu , z = Fz + Gv
are said equivalent by change of coordinates and feedback (we note(A, B) ∼ (F, G)) iff there exist invertible matricesM andL and amatrixK such that{
x = Ax + Buz = Mx, v = Kx + Lu
=⇒ z = Fz + Gv
and conversely.
M : change of coordinates, invertible of sizen × n, andK andL :feedback gains, withL invertible of sizem×m andK of sizem×n.
Changes of coordinates preserve the state dimension and (non dege-nerate) feedbacks preserve the input dimension.
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The relation∼ is an equivalence relation :– reflexive :(A, B) ∼ (A, B) (M = In, K = 0 andL = Im)– symmetric :x = M−1z andu = −L−1KM−1z + L−1v– transitive : if (A, B) ∼ (F, G) and(F, G) ∼ (H, J), z = Mx, v =
Kx + Lu imply z = Fz + Gv ands = Tz, w = Nz + Pv implys = Hs + Jw. Thuss = TMx, w = (NM + PK)x + PLu.
Note that
F = M(A−BL−1K)M−1 , G = MBL−1.
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The Single Input Case
x = Ax + bu
with (A, b) controllable :
rankC = rank(b Ab . . . An−1b
)= n.
One can construct the matricesM , K andL that transform the systemin its canonical form,
z = Fz + gv
F =
0 1 0 . . . 00 0 1 0... ... ...0 0 0 10 0 0 . . . 0
, g =
00...01
or
z1 = z2, . . . , zn−1 = zn, zn = v.
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The Multi Input Case (m > 1) :
x = Ax + Bu , B =(b1 . . . bn
).
The controllability matrix
C =(b1 Ab1 . . . An−1b1 . . . bm Abm . . . An−1bm
)has rankn and one can construct a sequence of integersn1, . . . , nm
calledcontrollability indices such that
n1 + . . . + nm = n
with
C =(b1 Ab1 . . . An1−1b1 . . . bm Abm . . . Anm−1bm
)of sizen× n, invertible.The controllability indicesn1, . . . , nm exist for all controllable linearsystems, are defined up to permutation and are invariant by changeof coordinates and feedback.
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Theorem (Brunovsky) : Every linear system withn states andminputs is equivalent by change of coordinates and feedback to thecanonical form
F = diag{F1, . . . , Fm} , G = diag{g1, . . . , gm}
where each pairFi, gi is given by
Fi =
0 1 0 . . . 00 0 1 0... ... ...0 0 0 10 0 0 . . . 0
, gi =
00...01
, i = 1, . . . ,m
with Fi of sizeni × ni andgi of sizeni × 1.
Consequences : trajectory planning, feedback design.
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1.3. Trajectory Planning
x = Ax + bu
n states, 1 input, controllable.System equivalent toz = Fz + gv with z = Mx, v = Kx + Lu.We want to start from x(0) = x0 at t = 0 with u(0) = u0, andarrive at x(T ) = xT at t = T with u(T ) = uT .We translate these conditions onz andv :
z(0) = Mx0, v(0) = Kx0 + Lu0z(T ) = MxT , v(T ) = KxT + LuT
Settingy = z1, we have
y(i) = zi+1, i = 0, . . . , n− 1, y(n) = v.
The initial and final conditions are interpreted as conditions on thesuccessive derivatives ofy up to ordern at times 0 andT .
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Given a curvet ∈ [0, T ] 7→ yref (t) ∈ R, of classCn , satisfying theinitial and final conditions.
All the other system variables may be obtained by differentiation ofyref , andwithout integrating the system’s equations.
We havevref = y(n)ref anduref = −L−1KM−1zref + L−1vref , with
zref = (yref , yref , . . . , y(n−1)ref ).
Accordinglyxref = M−1zref .
The inputuref exactly generatesxref = Axref + buref .
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The previousy-trajectory may be obtained by polynomial interpola-tion :
yref (t) =
2n+1∑i=0
ai
(t
T
)i
.
with a0, . . . , a2n+1 computed from the successive derivatives ofyrefat times 0 andT :
y(k)ref (t) =
1
T k
2n+1∑i=k
i(i− 1) · · · (i− k + 1)ai
(t
T
)i−k
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At t = 0 :yref (0) = a0
y(k)ref (0) =
k!
T kak, k = 1, . . . , n− 1,
vref (0) =n!
Tnan
and att = T :
yref (T ) =
2n+1∑i=0
ai,
y(k)ref (T ) =
1
T k
2n+1∑i=k
i!
(i− k)!ai, k = 1, . . . , n− 1,
vref (T ) =1
Tn
2n+1∑i=n
i!
(i− n)!ai
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Thus
a0 = yref (0) , ak =T k
k!y
(k)ref (0) , k = 1, . . . , n− 1, an =
Tn
n!vref (0).
1 1 . . . 1n + 1 n + 2 2n + 1
(n + 1)n (n + 2)(n + 1) (2n + 1)2n... ...
(n + 1)!(n+2)!
2 . . .(2n+1)!(n+1)!
an+1
...a2n+1
=
yref (T )−
∑ni=0 ai
...
T ky(k)ref (T )−
∑ni=k
i!(i−k)!
ai...
Tnvref (T )− n!an
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1.4. Trajectory Tracking, Pole PlacementAssume that the statex is measured at every time.We want to follow the reference trajectoryyref , the system beingperturbed by non modelled disturbances. Notee = y − yref the de-viation between the measured trajectory and its reference. We havee(n) = v − vref .
Notev − vref = −∑n−1
i=0 Kie(i), the gainsKi being arbitrary. Thus
e(n) = −n−1∑i=0
Kie(i)
or e...
e(n)
=
0 1 0 . . . 00 0 1 0... ... ...0 0 0 1
−K0 −K1 −K2 . . . −Kn−1
ee...
e(n−1)
.
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the gainsKi are the coefficients of the characteristic polynomial ofthe closed-loop matrixA + BK.TheoremIf the systemx = Ax+Bu is controllable, the eigenvalues ofA+BKmay be placed arbitrarily in the complex plane by a suitable feedbacku = Kx.
CorollaryA controllable linear system is stabilizable and, by state feedback, allits characteristic exponents can be arbitrarily chosen.
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2. First Order Controllability of NonlinearSystemsConsider the nonlinear system
x = f (x, u)
with x ∈ X, n-dimensional manifold, andu ∈ Rm.Its tangent linear system around the equilibrium point(x, u) is givenby
ξ = Aξ + Bv
with A = ∂f∂x(x, u), B = ∂f
∂u(x, u).
Definition We say that a nonlinear system isfirst order control-lablearound an equilibrium point(x, u) if its tangent linear system at(x, u) is controllable, i.e. iffrankC = n, withC =
(B|AB| · · · |An−1B
).
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Definition We say that a nonlinear system islocally controllablearound an equilibrium point(x, u) if :for all ε > 0, there existsη > 0 such that for every pair of points(x0, x1) ∈ Rn×Rn satisfying‖x0− x‖ < η and‖x1− x‖ < η, thereexists a piecewise continuous controlu on [0, ε] such that‖u(t)‖ < ε∀t ∈ [0, ε] andXε(x0, u) = x1, whereXε(x0, u) is the integral curveat timeε, generated fromx0 at time 0 with the controlu.
Theorem If a nonlinear system is first-order controllable at the equi-librium point (x, u), it is locally controllable at(x, u).
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RemarkThe scalar system :x = u3 is locally controllable but not first-ordercontrollable.To join x0 andx1 in durationT = ε, x0 andx1 arbitrarily chosenclose to 0, one can use the motion planning approach :
x(t) = x0 + (x1 − x0)
(t
ε
)2 (3− 2
t
ε
).
Thusu(t) =(x(t)
)13 =
(6(
x1−x0ε
) ( tε
) (1− t
ε
))13.
One easily checks that if|x0| < η and |x1| < η with η < ε4
3 then|u(t)| < ε, which proves the local controllability. On the contrary, atthe equilibrium point(0, 0), the tangent linear system isx = 0, and isindeed not first-order controllable.
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3. Local Controllability and Lie Brackets
For simplicity’s sake, the system is assumed affine in the control, i.e.
x = f0(x) +
m∑i=1
uifi(x)
with f0(0) = 0, ((x, u) = (0, 0) is an equilibrium point).From the vector fieldsf0, . . . , fm, we construct the sequence of dis-tributions :
D0 = span{f1, . . . , fm} , Di+1 = [f0, Di] + Di , i ≥ 1
whereDi is the involutive closure of the distributionDi.
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Proposition 3.1 The sequence of distributionsDi is non decreasing,i.e. Di ⊂ Di+1 for all i, and there exists an integerk∗ and an invo-lutive distributionD∗ such thatDk∗ = Dk∗+r = D∗ for all r ≥ 0.Moreover,D∗ enjoys the two following properties :
(i) span{f1, . . . , fm} ⊂ D∗
(ii) [f0, D∗] ⊂ D∗.
Proof Di ⊂ [f0, Di] + Di = Di+1 ⊂ Di+1.Thus, there exists a largestD∗, with D
∗= D∗. But :
rankDi+1 ≥ rankDi + 1 for smalli = 0, 1, . . ..thus, there existsk∗ ≤ n such thatDk∗ = Dk∗+r = D∗.Moreover,D0 ⊂ D∗ : (i),andD∗ = [f0, D
∗] + D∗ implies [f0, D∗] ⊂ D∗ : (ii).
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Proposition 3.2 Let D be involutive with constant rank equal tok inan openU , satisfying (ii). There exists a diffeomorphismϕ such that,if we note : {
ξi = ϕi(x), i = 1, . . . , kζj = ϕk+j(x), j = 1, . . . , n− k
we have :
ϕ∗f0(ξ, ζ) =
k∑i=1
γi(ξ, ζ)∂
∂ξi+
n−k∑i=1
γk+i(ζ)∂
∂ζiwhere theγi’s areC∞ functions.
Proof We haveϕ∗f0(ξ, ζ) =
k∑i=1
γi(ξ, ζ)∂
∂ξi+
n−k∑i=1
γk+i(ξ, ζ)∂
∂ζi.
But, by Frobenius’ Theorem,D = span{ ∂∂ξ1
, . . . , ∂∂ξk}.
By (ii), we have[ϕ∗f0,∂
∂ξi] ∈ D for all i = 1, . . . , k. But :
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[ϕ∗f0,∂
∂ξi] =
k∑j=1
(γj[
∂
∂ξj,
∂
∂ξi]−
∂γj
∂ξi
∂
∂ξj
)
+
n−k∑j=1
(γk+j[
∂
∂ζj,
∂
∂ξi]−
∂γk+j
∂ξi
∂
∂ζj
)
= −k∑
j=1
∂γj
∂ξi
∂
∂ξj−
n−k∑j=1
∂γk+j
∂ξi
∂
∂ζj∈ D
Thus∂γk+j
∂ξi= 0 for all i, j, or γk+j depends only ofζ, which proves
the Proposition.
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Theorem Let D∗ be defined as in Proposition 3.1, satisfying (i) and(ii). A necessary condition for the system to be locally controllablearound the origin is that
rankD∗(x) = n , ∀x ∈ U
whereU is a neighborhood of the origin.
Proof By contradiction. Assume thatD∗ satisfies (i), (ii) and
rankD∗(x) = k < n , ∀x ∈ U
Using (i), the image byϕ of fi, i = 1, . . . ,m, is
ϕ∗fi =
k∑j=1
ηi,j∂
∂ξj.
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Therefore,
ϕ∗
f0 +
m∑i=1
uifi
= (ϕ∗f0) +
m∑i=1
ui (ϕ∗fi)
=
k∑j=1
γj(ξ, ζ) +
m∑i=1
ui ηi,j(ξ, ζ)
∂
∂ξj+
n−k∑j=1
γk+j(ζ)∂
∂ζj.
In other words, in these coordinates, the system reads ξj = γj(ξ, ζ) +
m∑i=1
ui ηi,j(ξ, ζ), j = 1, . . . , k
ζj = γk+j(ζ), j = 1, . . . , n− k
and theζ part is not controllable, which achieves the proof.
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Denote by Lie{f0, . . . , fm} the Lie algebra generated by the linearcombinations off0, . . . , fm and all their Lie brackets.Lie{f0, . . . , fm}(x) is the vector space generated by the vectors ofLie{f0, . . . , fm} at the pointx.By construction,
D∗ ⊂ Lie{f0, . . . , fm}but the equality doesn’t hold true in general.
Theorem Assume that them+1 vector fieldsf0, . . . , fm are analytic.Local controllability atx = 0, u = 0, implies :
Lie{f0, . . . , fm}(x) = TxRn, ∀x ∈ X.
Remark If f0 ≡ 0, we haveD∗ = Lie{f1, . . . , fm}. Thus, iff1, . . . , fm
are analytic, local controllability is equivalent torankD∗ = n
in some open set.
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Examplex1 = x2
2, x2 = u
Equilibrium point(x1, 0), x1 arbitrary.D0 = span{f1} = span{ ∂
∂x2} andD0 = D0 ;
D1 = span{f1, [f0, f1]} = span{ ∂∂x2
, x2∂
∂x1}, has rank 2 and is invo-
lutive if x2 6= 0.But [f1, [f0, f1]] = 2 ∂
∂x1∈ D1(x1, 0), thus
D∗ = D1(x1, 0) = span{ ∂∂x2
, ∂∂x1
}.
However, ifx1 > 0, one cannot reach points such thatx1 < 0. Thustherank condition is necessary but not sufficient for local control-lability !
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Remark For general systems :
x = f (x, u)
one can use the previous formalism by setting
ui = vi, i = 1, . . . ,m
sincex = f (x, u)u1 = v1...um = vm
has the required affine form.
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Remark The rank condition for linear systems :
x = Ax +
m∑i=1
uibi
Setf0(x) = Ax =
∑ni=1
(∑nj=1 Ai,jxj
)∂
∂xiand
fi(x) = bi =∑n
j=1 bi,j∂
∂xj, i = 1, . . . ,m.
D0 = span{b1, . . . , bm}. D0 = D0 since[bi, bj] = 0 for all i, j.
[Ax, bi] = −n∑
j=1
bi,j∂
∂xj
n∑k=1
n∑l=1
Ak,lxl∂
∂xk
= −
n∑j,k=1
bi,jAk,j∂
∂xk−
n∑j=1
(Abi)j∂
∂xj= Abi.
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Thus
D1 = [Ax, D0] + D0 = span{b1, . . . , bm, Ab1, . . . , Abm}.
andD1 involutive (made of constant vector fields).We have
Dk = span{b1, . . . , bm, Ab1, . . . , Abm, . . . , Akb1, . . . , Akbm}
for all k ≥ 1.There existsk∗ < n such that
Dk∗ = D∗
and rankD∗(x) = rankC for all x ∈ U .
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4. Some Extensions using Module Theory
4.1. Recalls on ModulesConsider :•K a principal ideal ring (not necessarily commutative)•M a group• and an external productK×M → M, i.e. satisfying
αm ∈ M for all α ∈ K andm ∈ M.
M is a module if and only if the external product satisfies :
(αβ)m = α(βm) and(α + β)m = αm + βm
for all α, β ∈ K and allm ∈ M.
Remark If K is a field, thenM is a vector-space.
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Examples•K = R[ d
dt] the set of polynomials ofddt with coefficients inR, andlet {x1, . . . , xn} be a basis ofRn. Let M be made of all vectors ofthe form
n∑i=1
ki∑j=1
ai,jdjxi
dtj=
n∑i=1
ki∑j=1
ai,jx(j)i
with k1, . . . , kn arbitrary integers.M is afinitely generatedK-module.
• Let K be the field of meromorphic functions oft, K = K[ ddt] and let
{x1, . . . , xn} be a basis ofRn. Let M be made of all vectors of theform
n∑i=1
ki∑j=1
ai,j(t)x(j)i
with k1, . . . , kn arbitrary integers and theai,j ’s meromorphic func-tions. ThenM is afinitely generatedK-module.
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Let M be aK-module. An elementm ∈ M, m 6= 0, is said to betorsion if there existsα ∈ K, α 6= 0, such thatαm = 0.We denote byT the set of all torsion elements ofM.It is obviously a submodule ofM, called thetorsion submodule.
We say that aK-moduleF is free if and only if for everym ∈ F,m 6= 0, αm = 0, with α ∈ K, impliesα = 0.
We have
Proposition 1LetK = K[ ddt] whereK is a field.
(i) A finitely generated moduleM can be uniquely decomposed intoa torsion sub-moduleT and a free sub-moduleF : M = T ⊕ F.
(ii) It is free if and only ifT = {0}.
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4.2. Linear SystemsLet K = K[ d
dt], whereK is a field, andM a finitely generatedK-module.Let A(τ ) be a(n−m)× n polynomial matrix ofτ = d
dt with coeffi-cients inK. We consider the linear system inM
A(τ )x = 0.
The quotient ofM by the lines of this system is again a finitely gene-ratedK-module.
Therefore, the data of a linear system is equivalent to the one of afinitely generatedK-module.
Definition. (Fliess, 1990)Given a finitely generatedK-moduleM,we say that the associated linear system iscontrollable if and only ifM is free.
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ExampleConsider then-dimensional time-varying linear system
x = F (t)x + G(t)u
with u ∈ Rm, F andG meromorphic w.r.t.t in a given open intervalof R, and rankG(t) = m for everyt in this interval.It reads( d
dtI − F (t))x = G(t)u.LetC be a(n−m)×n matrix such thatC(t)B(t) = 0 with rankC(t) =n−m for everyt.We have
A(d
dt)x
def= C(t)(
d
dtI − F (t))x = 0
thusA( ddt) = C(t)( d
dtI−F (t)) is a(n−m)×n matrix with coefficients
in K = K[ ddt], with K the field of meromorphic functions oft.
The linear system is equivalent to the set of linear equations
A(d
dt)x = 0.
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We denote byM the finitely generatedK-module generated by a ba-sis of Rn andM0 its quotient submodule generated by the lines ofA( d
dt)x = 0.Assume that the pair(F, G) is not controllable. By Kalman’s decom-position, there exists ak-dimensional subspace, described by the setof n− k independent equations
K(t)x = 0
such that then− k-dimensional quotient subsystem is controllable.Clearly,K ∈ K and the set{x|Kx = 0} is the torsion submodule ofM0 which is thus not free.Consequently,M0 is free if the pair(F, G) is controllable in the usualsense.The converse follows the same lines.
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Remarks.• This definition is independent of the system variables description.• For 1st order controllable nonlinear systems, the tangent linear ap-
proximation at any point constitutes a linear time-varying systemwhose module is free.