Nonlinear Control Systems

Nonlinear Control Systems António Pedro Aguiar pedro@isr.ist.utl.pt 7. Feedback Linearization IST-DEEC PhD Course http://users.isr.ist.utl.pt/%7epedro/ncs1/ 1 1

Feedback Linearization Given a nonlinear system of the form ẋ = f(x) + G(x)u y = h(x) Does exist a state feedback control law and a change of variables u = α(x) + β(x)v z = T (x) that transforms the nonlinear system into a an equivalent linear system (ż = Az + Bv)?

Feedback Linearization Example: Consider the following system ẋ = Ax + Bγ(x)`u α(x) where γ(x) is nonsingular for all x in some domain D. Then, u = α(x) + β(x)v, with β(x) = γ 1 (x) yields ẋ = Ax + Bv If we would like to stabilize the system, we design v = Kx such that A BK is Hurwitz Therefore u = α(x) β(x)kx

Feedback Linearization Example: Consider now this example: ẋ 1 = a sin x ẋ = x 1 + u How can we do this? We cannot simply choose u to cancel the nonlinear term a sin x! However, if we first change the variables then ż 1 = z z 1 = x 1 z = a sin x = ẋ 1 ż = a cos x ẋ = a cos x ( x 1 + u) Therefore with u = x 1 + 1 v, π/ < x < π/ a cos x we obtain the linear system ż 1 = z ż = v 4

Feedback Linearization A continuously differentiable map T (x) is a diffeormorphism if T 1 (x) is continuously differentiable. This is true if the Jacobian matrix T is nonsingular x D. T (x) is a global diffeormorphism if and only if T T (x) is proper, that is, lim x T (x) =. is nonsingular x Rn and Definition A nonlinear system ẋ = f(x) + G(x)u (1) where f : D R n and G : D R n p are sufficiently smooth on a domain D R n is said to be feedback linearizable (or input-state linearizable) if there exists a diffeomorphism T : D R n such tat D z = T (D) contains the origin and the change of variables z = T (x) transforms (1) into the form ż = Az + Bγ(x)`u α(x) 5

Feedback Linearization Suppose that we would like to solve the tracking problem for the system ẋ 1 = a sin x ẋ = x 1 + u y = x If we use state feedback linearization we obtain z 1 = x 1 z = a sin x = ẋ 1 u = x 1 + 1 v a cos x ż 1 = z ż = v y = sin 1 (z /a) which is not good! Linearizing the state equation does not necessarily linearize the output equation. Notice however if we set u = x 1 + v we obtain ẋ = v y = x There is one catch: The linearizing feedback control law made x 1 unobservable from y. We have to make sure that x 1 whose dynamics are given by ẋ 1 = a sin x is well behaved. For example, if y = y d = cte x 1 (t) = x 1 () + ta sin y d. It is unbounded! 6

SISO system Input-Output Linearization ẋ = f(x) + g(x)u y = h(x) where f, g, h are sufficiently smooth in a domain D R n. The mappings f : D R n and g : D R n are called vector fields on D. Computing the first output derivative... ẏ = h ẋ = h [f(x) + g(x)u] =: L f h(x) + L gh(x)u In the sequel we will use the following notation: L f h(x) = h f(x) Lie Derivative of h with respect to f L gl f h(x) = (L f h) g(x) L f h(x) = h(x) L f h(x) = L f L f h(x) = (L f h) f(x) L k f h(x) = L f L k 1 f h(x) = (Lk 1 f h) f(x) 7

Input-Output Linearization ẏ = L f h(x) + L gh(x)u If L gh(x)u = then ẏ = L f h(x) (independent of u). Computing the second derivative... y () = (L f h) [f(x) + g(x)u] = L f h(x) + LgL f h(x)u If L gl f h(x)u = then ẏ () = L f h(x) (independent of u). Repeating this process, it follows that if L gl i 1 f h(x) =, i = 1,,..., ρ 1 L gl ρ 1 f h(x) then u does not appear in y, ẏ,..., y (ρ 1) and y (ρ) = L ρ f h(x) + LgL(ρ 1) f h(x)u 8

Input-Output Linearization Therefore, by setting y (ρ) = L ρ f u = h(x) + LgLρ 1 f h(x)u 1 L gl ρ 1 f h(x) [ Lρ f h(x) + v] the system is input-output linearizable and reduces to y (ρ) = v chain of ρ integrators Definition The nonlinear system ẋ = f(x) + g(x)u y = h(x) is said to have relative degree ρ, 1 ρ n, in the region D D if for all x D L gl i 1 f h(x) =, i = 1,,..., ρ 1 L gl ρ 1 f h(x) 9

Example 1: Van der Pol system Examples ẋ 1 = x ẋ = x 1 + ɛ(1 x 1 )x + u, ɛ > 1. y = x 1 Calculating the derivatives... ẏ = ẋ 1 = x ÿ = ẋ = x 1 + ɛ(1 x 1)x + u Thus the system has relative degree ρ = in R.. y = x Then ẏ = ẋ = x 1 + ɛ(1 x 1 )x + u In this case the system has relative degree ρ = 1 in R.. y = x 1 + x Then ẏ = x + x ( x 1 + ɛ(1 x 1 )x + u) In this case the system has relative degree ρ = 1 in D = {x R : x }. 1

Examples Example : ẋ 1 = x 1 ẋ = x + u y = x 1 Calculating the derivatives... ẏ = ẋ 1 = x 1 = y y (n) = y = x 1, n 1 The system does not have a well defined relative degree! Why? Because the output y(t) = x 1 (t) = e t x 1 () is independent of the input u. 11

Examples Example : where m < n and b m. H(s) = bmsm + b m 1 s m 1 + + b s n + a n 1 s n 1 + + a A state model of the system is the following ẋ = Ax + Bu y = Cx with A = 6 4 1 1 :....... 1 a a 1 a n 1 What is the relative degree ρ? 7 5 n n B = 6 7 4. 5 1 n 1 C = [ b b 1 b m ] 1 n 1

Examples ẏ = CAx + CBu If m = n 1 CB = b m ρ = 1 Otherwise, CB = y () = CA x + CABu Note that CA is obtained by shifting the elements of C one position to the right and CA i by shifting i positions. Therefore, CA i 1 B =, for i = 1,,... n m 1 CA n m 1 B = b m y (n m) = CA n m x + CA n m 1 Bu ρ = n m In this case the relative degree of the system is the difference between the degrees of the denominator and numerator polynomials of H(s). 1

Consider again the linear system given by the transfer function 8 < D(s) can be written as H(s) = N(s) D(s) with : deg D = n deg N = m < n ρ = n m D(s) = Q(s)N(s) + R(s) where the degree of the quotient deg Q = n m = ρ and the degree of the reminder deg R < m Thus H(s) = N(s) Q(s)N(s) + R(s) = 1 + 1 1 Q(s) R(s) Q(s) N(s) and therefore we can conclude that H(s) can be represented as a negative feedback connection with 1/Q(s) in the forward path and R(s)/N(s) in the feedback path. 14

Note that the ρ-order transfer function 1/Q(s) has no zeros and can be realized by where ξ = (A c + B cλ T )ξ + B cb me y = C cξ ξ = ˆy ẏ... y (ρ 1) T R ρ and (A c, B c, C c) is a canonical form representation of a chain of ρ integrators: 1 1. A c =...... 6 4 7. 15 ρ ρ B c =. 6 7 45 1 ρ 1 B cλ T = 6.. 7 4 5, λ Rρ λ T C c = ˆ1 1 ρ 15

R(s) N(s) η = A η + B y w = C η The eigenvalues of A are the zeros of the polynomial N(s), which are the zeros of the transfer function H(s). Thus, the system H(s) can be realized by the state model Note that y = C cξ and η = A η + B C cξ ξ = A cξ + B c(λ T ξ b mc η + b mu) y = C cξ ξ = A cξ + B c(λ T ξ b mc η + b mu) ξ 1 = ξ ξ = ξ. ξ ρ = λ T ξ b mc η + b mu and therefore y (ρ) = λ T ξ b mc η + b mu 16

Thus, setting results in y (ρ) = λ T ξ b mc η + b mu u = 1 b m [ λ T ξ + b mc η + v] η = A η + B C cξ Internal dynamics: It is unobservable from the output y ξ = A cξ + B cv chain of integrators y = C cξ Suppose we would like to stabilize the output y at a constant reference r, that is, ξ ξ = (r,,..., ) T. Defining ζ = ξ ξ we obtain ζ = A cζ + B cv Therefore, setting v = Kζ = K(ξ ξ ) with (A c B ck) Hurwitz we obtain the closed-loop system η = A η + B C c(ξ + ζ) ζ = (A c B ck)ζ y = C cξ where the eigenvalues of A are the zeros of H(s). If H(s) is minmum phase (zeros in the open left-half complex plan) then A is Hurwitz. 17

Feedback Linearization Can we extend this result η = A η + B C cη ξ = A cξ + B c `λt ξ b mc η + b mu y = C cx for the nonlinear system (SISO) that is find a z = T (x), where» η z = ξ ẋ = f(x) + g(x)u y = h(x) = 6 4 φ 1 (x). φ n ρ (x) h(x). L ρ 1 f. h(x) such that T (x) is a diffeomorphism on D D and φ i g(x) =, for 1 i n ρ, x D. Note that Does exist such T (x)? 7 5 η = φ i ẋ = φ i f(x) + φ i g(x)u 18

Normal form Theorem (1.1) Consider the SISO system ẋ = f(x) + g(x)u y = h(x) and suppose that it has relative degree ρ n in D. Then, for every x D, there exists a such diffeomorphism T (x) on a neighborhood of x. Using this transformation we obtain the system re-written in normal form: η = f (η, ξ) ξ = A cξ + B cγ(x)[u α(x)] y = C cξ where ξ R ρ, η R n ρ and (A c, B c, C c) is the canonical form representation of a chain of integrators, and f (η, ξ) := φ f(x) x=t 1 (z) γ(x) = L gl ρ 1 f h(x) α(x) = Lρ f h(x) L gl ρ 1 f h(x) 19

η = f (η, ξ) ξ = A cξ + B cγ(x)[u α(x)] y = C cξ The external part can be linearized by u = α(x) + β(x)v with β(x) = γ 1 (x). The internal part is described by η = f (η, ξ) Setting ξ = result η = f (η, ) This is called the zero-dynamics Note that for the linear case we have η = A η, where the eigenvalues of A are the zeros of H(s).

Definition The system is said to be minimum phase if η = f (η, ) has an asymptotically stable equilibrium point in the domain of interest. The zero dynamics can be characterized in the original coordinates by notting that y(t) =, t ξ(t) = u(t) = α(x(t)) where the first implication is due to the fact that ξ = [y, ẏ,...] T and the second due to ξ = A ξ + B γ(x)[u α(x)]. Thus, when y(t) =, the solution of the state equation is confined to the set n o Z = x D : h(x) = L f h(x) =... = L ρ 1 f h(x) = and the input that is u = u (x) := α(x) x Z ẋ = f (x) := [f(x) + g(x)α(x)] x Z In the special case that ρ = n η does not exist. In that case the system has no zero dynamics and by default is said to be minimum phase. 1

Example 1 Example ẋ 1 = x ẋ = x 1 + ɛ(1 x 1)x + u y = x It is in the normal form (ξ = y, η = x 1 ) Zero-dynamics? ẋ 1 =, which does not have an asymptotic stable equilibrium point. Hence, the system is not minimum phase. Example ẋ 1 = x 1 + + x 1 + x u ẋ = x ẋ = x x + u y = x What is the relative degree and the zero dynamics?

ẋ 1 = x 1 + + x 1 + x u Computing the time-derivative... ẋ = x ẋ = x x + u y = x ẏ = ẋ = x ÿ = ẋ = x 1 x + u Thus, the relative degree is ρ =. Analyzing the zero-dynamics we have y = ẏ = ÿ = we have x = x = and from the last we have u = x 1 x =. Therefore, ẋ 1 = x 1 and the system is minimum phase.

The single-input system Full-State Linearization ẋ = f(x) + g(x)u with f, g sufficiently smooth in a domain D R n is feedback linearizable if there exists a sufficiently smooth h : D R such that the system has relative degree n in a region D D. ẋ = f(x) + g(x)u y = h(x) This implies that the normal form reduces to Note that Thus which is equivalent to ż = A cz + B cγ(x)[u α(x)] y = C cz z = T (x) ż = T ẋ A cz + B cγ(x)[u α(x)] = T [f(x) + g(x)u] 4

Splinting in two we obtain Equation () is equivalent to T f(x) = AcT (x) Bcγ(x)α(x) () T g(x) = Bcγ(x) () T 1 f(x) = T (x) T f(x) = T (x). T n 1 f(x) = Tn(x) T n f(x) = α(x)γ(x) 5

and () is equivalent to Setting h(x) = T 1, we see that and T 1 g(x) = T g(x) =. T n 1 g(x) = T n g(x) = γ(x) T i+1 (x) = L f T i (x) = L i f h(x), i = 1,,..., n 1 L gl i 1 f h(x) =, i = 1,,..., n 1 L gl n 1 f Therefore we can conclude that if h(.) satisfies (4) the system is feedback linearizable. (4) 6

The existence of h(.) can be characterized by necessary and sufficient conditions on the vector fields f and g. First we need some terminology. Definition Given two vector fields f and g on D R n, the Lie Bracket [f, g] is the vector field [f, g](x) = g f f(x) g(x) Note that Adjoint representation [f, g] = [g, f] f = g = cte [f, g] = ad f g(x) = g(x) ad 1 f g(x) = [f, g](x) ad k f g(x) = [f, adk 1 f g](x), k 1 7

Then,» f(x) = Example 1 x, g(x) = sin x 1 x» x1 " [f, g](x) = g g1 g f 1 f(x) g(x) = 1 g g 1»»» x = 1 sin x 1 x»»» x1 = = x x 1 x 1 x 1 + x # " f1 f 1 f(x) 1 f f 1» 1 cos x 1 1 x1 # g(x) ad f g = [f, ad f g] = ad f g f f(x) ad f g(x)»»» 1 x = 1 1 1 sin x 1 x cos x 1 1» x = 1 x x 1 + x sin x 1 x 1 cos x 1» x 1 x 1 + x 8

Example : f(x) = Ax and g(x) = g is a constant vector field. Then, ad f g(x) = [f, g](x) = g f f(x) g(x) = Ag ad f g(x) = [f, ad f g](x) = ad f g f f ad f g = A( Ag) = A g ad k f g = ( 1)k A k g Definition For vector fields f 1, f,..., f k on D R n, a distribution is a collection of all vector spaces (x) = span {f 1 (x),..., f k (x)}, where for each fixed x D, (x) is the subspace of R n spanned by the vectors f 1 (x),..., f k (x). The dimension of (x) is defined by which may depend on x. dim( (x)) = rank[f 1 (x), f (x),..., f k (x)] If f 1, f,..., f k are linearly independent, then dim( (x)) = k, x D. In this case, we say that is a nonsingular distribution on D. A distribution is involutive if g 1, g [g 1, g ]. If is a nonsingular distribution on D, generated by f 1,..., f k then it is involutive if and only if [f i, f j ], 1 i, j k 9

Example Let D = R, = span {f 1, f } and f 1 = 4 x 1 5, f = 4 1 5 x dim( (x)) = rank[f 1, f ] =, x D is involutive? [f 1, f ] = f f 1 f 1 f = 4 1 5 4 x 1 5 4 5 4 1 x 5 = 4 1 5 Checking that [f 1, f ] is the same to see if [f 1, f ] can be generated by f 1, f, that is if rank[f 1 (x), f (x), [f 1, f ](x)] =, x D. But rank Hence, is not involutive. 4 x 1 1 x 1 5 =, x D

Theorem 1. Theorem The system ẋ = f(x) + g(x)u, with x R n, u R is feedback linearizable if and only if there is a domain D D such that 1. The matrix G(x) = [g(x), ad f g(x),..., ad n 1 f g] has rank n x D.. The distribution D = span{g, ad f g(x),..., ad n f g} is involutive in D. Example» a sin x ẋ = f(x) + gu, f(x) = x 1 we have seen that ad f g = [f, g] = g f f» g = a cos x x 1», g = 1» 1» a cos x =» a cos x The matrix G = [g, ad f g] = has rank G =, cos x 1. The distribution D = span {g} is involutive. Thus, we can conclude that there exists a T (x) in D = x R : cos x that allow us to do feedback linearization. 1

Now we have to find h(x) that satisfies Thus, h(.) must be independent of x. L f h(x) = h f(x) = h L f h h g = h g = ; (L f h) g ; h() = h i» h h 1 = h = 1 h 1 h a sin x 1 1 i» a sin x x 1 = h a sin x 1 i» h a sin x 1 1 = h 1 a cos x h In conclusion, and h =. 1 Examples of such h(x) include h(x) = x 1 or h(x) = x 1 + x 1. Given h(x) we can now perform input-output linearization.

Example A single link manipulator with flexible points f(x) = 6 4 x a sin x 1 b(x 1 x ) x 4 c(x 1 x ) ẋ = f(x) + gu 7 5, g = 6 4 G = [g, ad f g, ad f g, ad f g] d 7 5 a, b, c, d > ad f g = [f, g] = g f f g = 6 4 ad f g = [f, ad f g] = (ad f g) f f ad f g 1 = 6 a cos x 1 b b 7 6 4 1 5 4 c c 1 a cos x 1 b b 1 c c d 7 5 = 6 4 bd cd 7 6 5 4 7 5 d 7 5 = 6 4 d 7 5

g, h() = 4 Thus, ad f g = [f, ad f g] = ad f g f f ad f g 1 = 6 a cos x 1 b b 7 6 4 1 5 4 c c G = 6 4 d The distribution = span d bd cd bd cd n g, ad f g, ad f bd cd 7 5 = 6 4 7 5 rank G = 4, x R4 bd cd 7 5 o is involutive since g, ad f g, ad f are constant vector fields ([g, ad f g] = ). Therefore, we can conclude that there exists an h(x) : R 4 R and a T (x) that make the system full state feedback linearizable. In particular, h(x) must satisfy L i 1 f h g =, i = 1,,. L f h

For i = 1, we have h g = h 4 = and so h(x) must be independent of x 4. L f h(x) = h From f(x) = h 1 x + h [ a sin x 1 b(x 1 x )] + h x 4. L f h g = L f h 4 = h = Thus, h(x) is independent of x. Thus, L f h(x) = h x + h [ a sin x 1 b(x 1 x )] 1. L f h(x) = L f L f h(x) = (L f h) f(x) = (L f h) x + (L f h) [ a sin x 1 1 b(x 1 x )] + (L f h) x 4 + (L f h) c(x 1 x ) 4 For i =, Thus h is independent of x. L f h g = `L f h = h = 5

Hence, L L f h(x) = L f L f h = f h L f(x) = f h 1 L + f L h x 4 + f h c(x 1 x ) 4 Also, from condition L f h(x) g L f h L x + f h [ a sin x 1 b(x 1 x )] 1 L f h Therefore, let h(x) = x 1. Then, the change of variables h 1 z 1 = h(x) = x 1 z = L f h(x) = x z = L f h(x) = a sin x 1 b(x 1 x ) z 4 = L f h(x) = a cos x 1 ẋ 1 bẋ 1 + bẋ = ax cos x 1 bx + bx 4 transforms the state equation into ż 1 = z ż = z ż = z 4 ż 4 = (a cos z 1 + b + c)z + a(z c) sin z 1 + bd u 6

Example - field controlled DC motor Consider the system with Computing the f(x) = 4 ad f g = [f, g] = 4 Rank of G? a cx θx ẋ = f(x) + gu ax 1 bx + k cx 1 x θx 1 x G = [g, ad f g, ad f g] = 5, g = 5, ad f g = [f, ad f g] = 4 4 1 a cx θx 4 1 5 a (a + b) cx (b a) θx θk a (a + b) cx (b a) θx θk 5 5 7

det G = cθ( k + bx )x. Hence G has rank for x k b and x. Let s check the distribution D = span{g, ad f g} ˆg, adf g = (ad f g) g = 4 c 5 4 1 5 = 4 5 θ Hence D is involutive because [g, ad f g] D. Therefore, the conditions of Theorem 1. are satisfied in particular for the domain j D = x R : x > k ff b, x > h(.) =? 8

with f(x) = 4 ẋ = f(x) + gu ax 1 bx + k cx 1 x θx 1 x 5, g = 4 1 5 Equilibrium points: x 1 =, x = k b, x = cte. Suppose we are interested in the desired operating point x = [ k b w ] T. Then, h(x) must satisfy (n = ) and h(x ) =. h g = ; (L f h) g = ; (L f h) g that is, h must be independent of x 1. L f h(x) =... h g = h = 1 9

State feedback control Consider the system ẋ = f(x) + G(x)u and let z = T (x) = [T 1 (x), T (x)] T such that η = f (η, ξ) ξ = Aξ + Bγ(x)[u α(x)] Suppose that (A, B) is controllable, γ(x) is nonsingular x D, f (, ) = and f (η, ξ), α(x), γ(x) C 1. Goal: Design a state feedback control law to stabilize the origin z =. Setting u = α(x) + β(x)v, β(x) = γ 1 (x), we obtain the triangular system η = f (η, ξ) ξ = Aξ + Bv Let v = Kξ with (A BK) Hurwitz then we can conclude the following: 4

Lemma (1.1) State feedback control The origin z = is asymptotically stable (AS) if the origin of η = f (η, ) is AS (that is, if the system is minimum phase). Proof. By the converse theorem V 1 (η) : V 1 η f (η, ) α ( x ) in some neighborhood of η =, where α K. Let P = P T > be the solution of the Lyapunov equation P (A BK) + (A BK) T P = I. V = V 1 + k p ξ T P ξ, k > V V 1 η f (η, ξ) + k p ξ T P ξ ξt [P (A BK) + (A BK) T P ]ξ V 1 η f (η, ) + V 1 η [f (η, ξ) f (η, )] α ( η ) + k 1 ξ kk ξ, k 1, k > kξt ξ p ξ T P ξ where the last inequality follows from restricting the state to be in any bounded neighborhood of the origin and using the continuous differentiability property of V 1 and f. Thus, choosing k > k 1 /k ensures that V <, which follows that the origin is AS. 41

State feedback control Lemma (1.) The origin z = is GAS if η = f (η, ξ) is ISS. Note that if η = f (η, ) is GAS or GES does NOT imply ISS. But, if it is GES + globally Lipschitz then it is ISS. Otherwise, we have to prove ISS by further analysis. Example: η = η + η ξ ξ = v Zero dynamics: η = η η = is GES but η = η + η ξ is not ISS, e.g., if ξ(t) = 1 and η() then η(t), t, which implies that η grows unbounded. However with v = Kξ, K > we achieve AS. 4

To view this, let ν = ηξ, then ν = η ξ + ηξ = ηv ηξ + η ξ = Kηξ ηξ + η ξ = (1 + K)ν + υ = [(1 + K) ν]ν Thus, with ν() < 1 + K υ and therefore we can also conclude that T t : υ(t) 1, t T. Consider now V = 1/η. Then V = η η = η + η ξ = η (1 ηξ) = η (1 ν) <, t T Thus, η and note also that ξ = Kξ. Therefore, the control law v = Kξ can achieve semiglobal stabilization. 4

One may think that we can assign the eigenvalues of (A BK) to the left half-complex plane to make ξ = (A BK)ξ decay to zero arbitrarily fast. BUT this may have consequences: the zero-dynamics may go unstable! This is due to the peaking phenomenon. Example η = 1 (1 + ξ )η ξ 1 = ξ ξ = v» Setting v = Kξ = k ξ 1 kξ A BK = 1 k k and the eigenvalues are k, k. Note that» e (A BK)t (1 + kt)e kt te = kt k te kt (1 kt)e kt which shows that as k, ξ(t) will decay to zero arbitrary fast. However, the element (, 1) reaches a maximum value k/e at t = 1. There is a peak k of order k! Furthermore, the interaction of peaking with nonlinear growth could destabilize the system. 44

E.g., consider the initial conditions η() = η ξ 1 () = 1 ξ () = Then, and ξ (t) = k te kt In this case the solution η(t) is given by η (t) = which has a finite escape if η > 1. η = 1 (1 + ξ )η = 1 (1 k te kt )η η 1 + η [t + (1 + kt)e kt 1]. 45

Tracking η = f (η, ξ) ξ = A ξ + B γ(x)[u α(x)] y = C ξ Goal: Design u such that y asymptotically tracks a reference signal r(t). Assume that r(t), ṙ(t),..., r (ρ) are bounded and available on-line. Note that the reference signal could be the output of a pre-filter. Example: If ρ =, the pre-filter could be Then wn G(s) = s + ξw ns + w n ẏ 1 = y ẏ = w n y 1 ξw ny + w n y d r = y 1 Note that in this case ṙ = ẏ 1 = y and r = ẏ. Consider a system with relative degree ρ. Let r 6 r = 4. r (ρ 1) 7 6 5, and e = 4 ξ 1 r. ξ ρ r (ρ 1) 7 5 = ξ r 46

Then, the error system is given by η = f (η, e + r) ė = A ce + B c(γ(x)[u α(x)] r (ρ) ) Setting u = α(x) + β(x)[v r (ρ) ] with β = 1 it follows that γ(x) η = f (η, e + r) ė = A ce + B cv Thus, selecting v = Ke with (A c B ck) Hurwitz we can conclude that the states of the closed-loop system η = f (η, e + r) ė = (A B K)e are bounded if η = f (η, e + R) is ISS and that e as t. 47