Chap. 3. Controlled Systems, Controllability

Chap. 3. Controlled Systems, Controllability

1. Controllability of Linear Systems 1.1. Kalman s Criterion Consider the linear system ẋ = Ax + Bu where x R n : state vector and u R m : input vector. A : of size n n and B : of size n m. Definition The pair (A, B) is controllable if, given a duration T > 0 and two arbitrary points x 0, x T R n, there exists a piecewise continuous function t ū(t) from [0, T ] to R m, such that the integral curve x(t) generated by ū with x(0) = x 0, satisfies x(t ) = x T.

In other words e AT x 0 + T 0 e A(T t) Bū(t)dt = x T. This property depends only on A and B : Theorem (Kalman) A necessary and sufficient condition for (A, B) to be controllable is ( ) rank C = rank B AB A n 1 B = n. C is called Kalman s controllability matrix (of size n nm).

Proof Without loss of generality, we can consider that x 0 = 0 by changing x T in y T = x T e AT x 0 since T Consider the matrices 0 e A(T t) Bū(t)dt = y T = x T e AT x 0. C(t) = e A(T t) B, G = where C : transposed matrix of C. We prove 2 lemmas. T 0 C(t)C (t)dt

Lemma 1 A necessary and sufficient condition for (A, B) to be controllable is that G is invertible. Assume G invertible. It suffices to set : ū(t) = B e A (T t) G 1 y T. Thus T 0 e A(T t) Bū(t)dt = T 0 = GG 1 y T = y T and ū generates the required trajectory. The converse is immediate. e A(T t) BB e A (T t) G 1 y T dt

Lemma 2 Invertibility of G is equivalent to rank C = n. By contradiction, if G isn t invertible, v R n, v 0, such that v G = 0. Thus v Gv = T 0 v C(t)C (t)vdt = 0. Since v v C(t)C (t)v : quadratic form 0, T 0 v C(t)C (t)vdt = 0 implies v C(t)C (t)v = 0 t [0, T ]. Thus : v C(t) = 0 t [0; T ]. But v C(t) = v e A(T t) B = v I + i 1 A i(t t)i B = 0 i! implies that v B = 0, v A i B = 0 pour tout i 1 or v C = 0 with v 0, thus : rank C < n.

Conversely, if rank C < n, v 0 such that v C = 0. Thus, according to what precedes, v A i B = 0 for i = 0,..., n 1. By Cayley-Hamilton Theorem, A n+i, i 0, is a linear combination of the A j s, j = 1,..., n 1. Thus v A i B = 0, i 0 or v G = 0, which proves the Theorem.

Examples ẋ 1 = x 2, ( ) 0 is controllable : B =, A = ( ) 1 0 1 C = and rank C = 2. 1 0 x 2 = u ( ) 0 1, 0 0 ẋ 1 = u, ( 1 isn t controllable : B = ( ) 1 1 0 C = and rank C = 1. 1 0 ), A = x 2 = u ( ) 0 0, 0 0 Note : ẋ 1 ẋ 2 : 0, or x 1 x 2 = Cste : relation between states independent of u.

1.2. Controllability Canonical Form Définition Two systems ẋ = Ax + Bu, ż = F z + Gv are said equivalent by change of coordinates and feedback (we note (A, B) (F, G)) iff there exist invertible matrices M and L and a matrix K such that { ẋ = Ax + Bu = ż = F z + Gv z = Mx, v = Kx + Lu and conversely. M : change of coordinates, invertible of size n n, and K and L : feedback gains, with L invertible of size m m and K of size m n. Changes of coordinates preserve the state dimension and (non degenerate) feedbacks preserve the input dimension.

The relation is an equivalence relation : reflexive : (A, B) (A, B) (M = I n, K = 0 and L = I m ) symmetric : x = M 1 z and u = L 1 KM 1 z + L 1 v transitive : if (A, B) (F, G) and (F, G) (H, J), z = Mx, v = Kx + Lu imply ż = F z + Gv and s = T z, w = Nz + P v imply ṡ = Hs + Jw. Thus s = T Mx, w = (NM + P K)x + P Lu. Note that F = M(A BL 1 K)M 1, G = MBL 1.

The Single Input Case with (A, b) controllable : ẋ = Ax + bu rank C = rank ( b Ab... A n 1 b ) = n. One can construct the matrices M, K and L that transform the system in its canonical form, ż = F z + gv or F = 0 1 0... 0 0 0 1 0....... 0 0 0 1 0 0 0... 0, g = ż 1 = z 2,..., ż n 1 = z n, ż n = v. 0 0. 0 1

The Multi Input Case (m > 1) : The controllability matrix ẋ = Ax + Bu, B = ( b 1... b n ). C = ( b 1 Ab 1... A n 1 b 1... b m Ab m... A n 1 b m ) has rank n and one can construct a sequence of integers n 1,..., n m called controllability indices such that with n 1 +... + n m = n C = ( b 1 Ab 1... A n 1 1 b 1... b m Ab m... A n m 1 b m ) of size n n, invertible. The controllability indices n 1,..., n m exist for all controllable linear systems, are defined up to permutation and are invariant by change of coordinates and feedback.

Theorem (Brunovsky) : Every linear system with n states and m inputs is equivalent by change of coordinates and feedback to the canonical form F = diag{f 1,..., F m }, G = diag{g 1,..., g m } where each pair F i, g i is given by 0 1 0... 0 0 0 1 0 F i =....... 0 0 0 1, g i = 0 0 0... 0 with F i of size n i n i and g i of size n i 1. 0 0. 0 1, i = 1,..., m Consequences : trajectory planning, feedback design.

1.3. Trajectory Planning ẋ = Ax + bu n states, 1 input, controllable. System equivalent to ż = F z + gv with z = Mx, v = Kx + Lu. We want to start from x(0) = x 0 at t = 0 with u(0) = u 0, and arrive at x(t ) = x T at t = T with u(t ) = u T. We translate these conditions on z and v : Setting y = z 1, we have z(0) = Mx 0, v(0) = Kx 0 + Lu 0 z(t ) = Mx T, v(t ) = Kx T + Lu T y (i) = z i+1, i = 0,..., n 1, y (n) = v. The initial and final conditions are interpreted as conditions on the successive derivatives of y up to order n at times 0 and T.

Given a curve t [0, T ] y ref (t) R, of class C n, satisfying the initial and final conditions. All the other system variables may be obtained by differentiation of y ref, and without integrating the system s equations. We have v ref = y (n) ref and u ref = L 1 KM 1 z ref + L 1 v ref, with z ref = (y ref, ẏ ref,..., y (n 1) ref ). Accordingly x ref = M 1 z ref. The input u ref exactly generates ẋ ref = Ax ref + bu ref.

The previous y-trajectory may be obtained by polynomial interpolation : y ref (t) = 2n+1 i=0 a i ( t T ) i. with a 0,..., a 2n+1 computed from the successive derivatives of y ref at times 0 and T : y (k) ref (t) = 1 T k 2n+1 i=k i(i 1) (i k + 1)a i ( t T ) i k

At t = 0 : and at t = T : y ref (0) = a 0 y (k) k! ref (0) = T ka k, k = 1,..., n 1, v ref (0) = n! T na n y (k) ref (T ) = 1 T k y ref (T ) = 2n+1 i=k v ref (T ) = 1 T n 2n+1 i=0 a i, i! (i k)! a i, k = 1,..., n 1, 2n+1 i=n i! (i n)! a i

Thus a 0 = y ref (0), a k = T k k! y(k) ref (0), k = 1,..., n 1, a n = T n n! v ref(0). 1 1... 1 n + 1 n + 2 2n + 1 a (n + 1)n (n + 2)(n + 1) (2n + 1)2n n+1... a 2n+1 (n+2)! (2n+1)! (n + 1)! 2... (n+1)! y ref (T ) n i=0 a i. = T k y (k) ref (T ) n i=k i! (i k)! a i. T n v ref (T ) n!a n

1.4. Trajectory Tracking, Pole Placement Assume that the state x is measured at every time. We want to follow the reference trajectory y ref, the system being perturbed by non modelled disturbances. Note e = y y ref the deviation between the measured trajectory and its reference. We have e (n) = v v ref. Note v v ref = n 1 i=0 K ie (i), the gains K i being arbitrary. Thus or ė. e (n) = e (n) = n 1 i=0 K i e (i) 0 1 0... 0 0 0 1 0....... 0 0 0 1 K 0 K 1 K 2... K n 1 e ė. e (n 1).

the gains K i are the coefficients of the characteristic polynomial of the closed-loop matrix A + BK. Theorem If the system ẋ = Ax+Bu is controllable, the eigenvalues of A+BK may be placed arbitrarily in the complex plane by a suitable feedback u = Kx. Corollary A controllable linear system is stabilizable and, by state feedback, all its characteristic exponents can be arbitrarily chosen.

2. First Order Controllability of Nonlinear Systems Consider the nonlinear system ẋ = f(x, u) with x X, n-dimensional manifold, and u R m. Its tangent linear system around the equilibrium point ( x, ū) is given by with A = f f x ( x, ū), B = u ( x, ū). ξ = Aξ + Bv Definition We say that a nonlinear system is first order controllable around an equilibrium point ( x, ū) if its tangent linear system at ( x, ū) is controllable, i.e. iff rank C = n, with C = ( B AB A n 1 B ).

Definition We say that a nonlinear system is locally controllable around an equilibrium point ( x, ū) if : for all ε > 0, there exists η > 0 such that for every pair of points (x 0, x 1 ) R n R n satisfying x 0 x < η and x 1 x < η, there exists a piecewise continuous control ũ on [0, ε] such that ũ(t) < ε t [0, ε] and X ε (x 0, ũ) = x 1, where X ε (x 0, ũ) is the integral curve at time ε, generated from x 0 at time 0 with the control ũ. Theorem If a nonlinear system is first-order controllable at the equilibrium point ( x, ū), it is locally controllable at ( x, ū).

Remark The scalar system : ẋ = u 3 is locally controllable but not first-order controllable. To join x 0 and x 1 in duration T = ε, x 0 and x 1 arbitrarily chosen close to 0, one can use the motion planning approach : ( ) t 2 ( x(t) = x 0 + (x 1 x 0 ) 3 2 t ). ε ε Thus u(t) = ( ẋ(t) ) 1 ( ( ) 3 = 6 x1 x (tε ) ( 0 ε 1 t ) ) 1 3 ε. One easily checks that if x 0 < η and x 1 < η with η < ε4 3 then u(t) < ε, which proves the local controllability. On the contrary, at the equilibrium point (0, 0), the tangent linear system is ẋ = 0, and is indeed not first-order controllable.

3. Local Controllability and Lie Brackets For simplicity s sake, the system is assumed affine in the control, i.e. m ẋ = f 0 (x) + u i f i (x) with f 0 (0) = 0, (( x, ū) = (0, 0) is an equilibrium point). From the vector fields f 0,..., f m, we construct the sequence of distributions : i=1 D 0 = span{f 1,..., f m }, D i+1 = [f 0, D i ] + D i, i 1 where D i is the involutive closure of the distribution D i.

Proposition 3.1 The sequence of distributions D i is non decreasing, i.e. D i D i+1 for all i, and there exists an integer k and an involutive distribution D such that D k = D k +r = D for all r 0. Moreover, D enjoys the two following properties : (i) span{f 1,..., f m } D (ii) [f 0, D ] D. Proof D i [f 0, D i ] + D i = D i+1 D i+1. Thus, there exists a largest D, with D = D. But : rank D i+1 rank D i + 1 for small i = 0, 1,.... thus, there exists k n such that D k = D k +r = D. Moreover, D 0 D : (i), and D = [f 0, D ] + D implies [f 0, D ] D : (ii).

Proposition 3.2 Let D be involutive with constant rank equal to k in an open U, satisfying (ii). There exists a diffeomorphism ϕ such that, if we note : { ξi = ϕ i (x), i = 1,..., k we have : ϕ f 0 (ξ, ζ) = ζ j = ϕ k+j (x), k i=1 where the γ i s are C functions. Proof We have ϕ f 0 (ξ, ζ) = γ i (ξ, ζ) ξ i + k i=1 j = 1,..., n k n k i=1 γ i (ξ, ζ) ξ i + γ k+i (ζ) ζ i n k i=1 But, by Frobenius Theorem, D = span{ ξ 1,..., ξk }. By (ii), we have [ϕ f 0, ξ i ] D for all i = 1,..., k. But : γ k+i (ξ, ζ) ζ i.

[ϕ f 0, ξ i ] = + k j=1 n k j=1 = ( γ j [, ξ j ( γ k+j [, ζ j k j=1 γ j ξ i ξ j ξ i ] γ j ξ i ξ j ) ξ i ] γ k+j ξ i n k j=1 γ k+j ξ i ζ j ζ j ) D Thus γ k+j = 0 for all i, j, or γ ξ k+j depends only of ζ, which proves i 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 controllable around the origin is that rank D (x) = n, x U where U is a neighborhood of the origin. Proof By contradiction. Assume that D satisfies (i), (ii) and rank D (x) = k < n, x U Using (i), the image by ϕ of f i, i = 1,..., m, is ϕ f i = k j=1 η i,j ξ j.

Therefore, ϕ f 0 + = m u i f i = (ϕ f 0 ) + i=1 k γ j (ξ, ζ) + j=1 m u i (ϕ f i ) i=1 m u i η i,j (ξ, ζ) + ξ j i=1 n k j=1 In other words, in these coordinates, the system reads m ξ j = γ j (ξ, ζ) + u i η i,j (ξ, ζ), j = 1,..., k i=1 ζ j = γ k+j (ζ), j = 1,..., n k and the ζ part is not controllable, which achieves the proof. γ k+j (ζ) ζ j.

Denote by Lie{f 0,..., f m } the Lie algebra generated by the linear combinations of f 0,..., f m and all their Lie brackets. Lie{f 0,..., f m }(x) is the vector space generated by the vectors of Lie{f 0,..., f m } at the point x. By construction, D Lie{f 0,..., f m } but the equality doesn t hold true in general. Theorem Assume that the m+1 vector fields f 0,..., f m are analytic. Local controllability at x = 0, u = 0, implies : Lie{f 0,..., f m }(x) = T x R n, x X. Remark If f 0 0, we have D = Lie{f 1,..., f m }. Thus, if f 1,..., f m are analytic, local controllability is equivalent to rank D = n in some open set.

Example ẋ 1 = x 2 2, ẋ 2 = u Equilibrium point (x 1, 0), x 1 arbitrary. D 0 = span{f 1 } = span{ x } and D 2 0 = D 0 ; D 1 = span{f 1, [f 0, f 1 ]} = span{ x, x 2 2 x1 }, has rank 2 and is involutive if x 2 0. But [f 1, [f 0, f 1 ]] = 2 x D 1 1 (x 1, 0), thus D = D 1 (x 1, 0) = span{ x, 2 x1 }. However, if x 1 > 0, one cannot reach points such that x 1 < 0. Thus the rank condition is necessary but not sufficient for local controllability!

Remark For general systems : ẋ = f(x, u) one can use the previous formalism by setting u i = v i, i = 1,..., m since ẋ = f(x, u) u 1 = v 1. u m = v m has the required affine form.

Remark The rank condition for linear systems : m ẋ = Ax + u i b i Set and i=1 f 0 (x) = Ax = n i=1 ( nj=1 A i,j x j ) xi f i (x) = b i = n j=1 b i,j xj, i = 1,..., m. D 0 = span{b 1,..., b m }. D 0 = D 0 since [b i, b j ] = 0 for all i, j. [Ax, b i ] = = n j,k=1 n j=1 b i,j x j b i,j A k,j x k n n A k,l x l x l=1 k n (Ab i ) j k=1 j=1 x j = Ab i.

Thus D 1 = [Ax, D 0 ] + D 0 = span{b 1,..., b m, Ab 1,..., Ab m }. and D 1 involutive (made of constant vector fields). We have D k = span{b 1,..., b m, Ab 1,..., Ab m,..., A k b 1,..., A k b m } for all k 1. There exists k < n such that D k = D and rank D (x) = rank C for all x U.

4. Some Extensions using Module Theory 4.1. Recalls on Modules Consider : K a principal ideal ring (not necessarily commutative) M a group and an external product K M M, i.e. satisfying αm M for all α K and m 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 all m M. Remark If K is a field, then M is a vector-space.

Examples K = R[ dt d ] the set of polynomials of dt d with coefficients in R, and let {x 1,..., x n } be a basis of R n. Let M be made of all vectors of the form n k i d j x n a i i,j dt j = k i a i,j x (j) i i=1 j=1 with k 1,..., k n arbitrary integers. M is a finitely generated K-module. Let K be the field of meromorphic functions of t, K = K[ dt d ] and let {x 1,..., x n } be a basis of R n. Let M be made of all vectors of the form n i=1 k i j=1 i=1 j=1 a i,j (t)x (j) i with k 1,..., k n arbitrary integers and the a i,j s meromorphic functions. Then M is a finitely generated K-module.

Let M be a K-module. An element m M, m 0, is said to be torsion if there exists α K, α 0, such that αm = 0. We denote by T the set of all torsion elements of M. It is obviously a submodule of M, called the torsion submodule. We say that a K-module F is free if and only if for every m F, m 0, αm = 0, with α K, implies α = 0. We have Proposition 1 Let K = K[ dt d ] where K is a field. (i) A finitely generated module M can be uniquely decomposed into a torsion sub-module T and a free sub-module F : M = T F. (ii) It is free if and only if T = {0}.

4.2. Linear Systems Let K = K[ dt d ], where K is a field, and M a finitely generated K- module. Let A(τ) be a (n m) n polynomial matrix of τ = dt d with coefficients in K. We consider the linear system in M A(τ)x = 0. The quotient of M by the lines of this system is again a finitely generated K-module. Therefore, the data of a linear system is equivalent to the one of a finitely generated K-module. Definition. (Fliess, 1990) Given a finitely generated K-module M, we say that the associated linear system is controllable if and only if M is free.

Example Consider the n-dimensional time-varying linear system ẋ = F (t)x + G(t)u with u R m, F and G meromorphic w.r.t. t in a given open interval of R, and rank G(t) = m for every t in this interval. It reads ( dt d I F (t))x = G(t)u. Let C be a (n m) n matrix such that C(t)B(t) = 0 with rank C(t) = n m for every t. We have A( d def )x = C(t)( d dt dt I F (t))x = 0 thus A( dt d ) = C(t)( dt d I F (t)) is a (n m) n matrix with coefficients in K = K[ dt d ], with K the field of meromorphic functions of t. The linear system is equivalent to the set of linear equations A( d )x = 0. dt

We denote by M the finitely generated K-module generated by a basis of R n and M 0 its quotient submodule generated by the lines of A( dt d )x = 0. Assume that the pair (F, G) is not controllable. By Kalman s decomposition, there exists a k-dimensional subspace, described by the set of n k independent equations K(t)x = 0 such that the n k-dimensional quotient subsystem is controllable. Clearly, K K and the set {x Kx = 0} is the torsion submodule of M 0 which is thus not free. Consequently, M 0 is free if the pair (F, G) is controllable in the usual sense. 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 approximation at any point constitutes a linear time-varying system whose module is free.