Chap. 3. Controlled Systems, Controllability

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 .

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 .

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 .

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

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

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

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)

.

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.

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

).

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).

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.

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.

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).

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 :

[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.

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.

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.

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.

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 !

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.

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.

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 .

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.

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.

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}.

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.

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.

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.

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.