Feedback Linearization

Feedback Linearization Peter Al Hokayem and Eduardo Gallestey May 14, 2015 1 Introduction Consider a class o single-input-single-output (SISO) nonlinear systems o the orm ẋ = (x) + g(x)u (1) y = h(x) (2) where x D R n, u R 1, : D R n, g : D R n, and the domain D contains the origin 1 In this lecture, we shall answer the ollowing two (tightly related) questions: [ ] η 1 Does there exist a nonlinear change o variables z = = T (x), and a ξ control input u = α(x)+β(x)v that would transorm the system (1)-(2) into the ollowing partially linear orm? η = 0 (η, ξ) ξ = Aξ + Bv y = Cξ 2 Does there exist a nonlinear change o variables z = T (x) and a control input u = α(x) + β(x)v that would transorm the system (1) into the ollowing ully linear orm? ż = Ãz + Bv 1 This condition is not a limitation to the theory o eedback linearization and the results can be extended to the multiple-input-multiple-output (MIMO) cases Feedback Linearization 1 o 14

I the answer to question 2 is positive, then we say that the system (1) is eedback linearizable I the answer to question 1 is positive, then we say that the system (1)-(2) is input-output linearizable Both scenarios are very attractive rom the control design point o view, since in either case we can rely on linear design techniques to render the closed-loop system stable Since we cannot expect every nonlinear system to possess such properties o eedback linearization, it is interesting to understand the structural properties that the nonlinear system should possess that render it eedback linearizable Beore we dive into the theoretical developments o eedback linearization, let us irst look into two simple examples v α(x) + β(x)v u ẋ = (x) + g(x)u y = h(x) y x Figure 1: General idea o eedback linearization Example 1 Consider the example o the pendulum that we have seen in previous lectures The dynamics are given by I we choose the control ẋ 1 = x 2 ẋ 2 = a [sin (x 1 + δ) sin δ] bx 2 + cu u = a c [sin (x 1 + δ) sin δ] + v c we can cancel the nonlinear term a [sin (x 1 + δ) sin δ] The resulting linear system is given by ẋ 1 = x 2 ẋ 2 = bx 2 + v As such, the stabilization problem or the nonlinear system has been reduced to a stabilization problem or a controllable linear system We can proceed to design a stabilizing linear state eedback control v = k 1 x 1 k 2 x 2 Feedback Linearization 2 o 14

that renders the closed-loop system ẋ 1 = x 2 ẋ 2 = k 1 x 1 (k 2 + b) x 2 asymptotically stable The overall state eedback control law comprises linear and nonlinear parts u = Example 2 Consider the system ( a c ) [sin (x 1 + δ) sin δ] 1 c (k 1x 1 + k 2 x 2 ) ẋ 1 = a sin x 2 ẋ 2 = x 2 1 + u We cannot simply choose u to cancel the nonlinear term a sin x 2 However, let us irst apply the ollowing change o variables z 1 = x 1 z 2 = a sin x 2 = ẋ 1 Then, the new variables z 1 and z 2 satisy ż 1 = z 2 ( ż 2 = a cos x 2 ẋ 2 = a cos sin 1 z 2 a and the nonlinearities can be canceled by the control u = x 2 1 + 1 a cos x 2 v ) (u ) z 2 1 which is well deined or π < x < π The state equation in the new coordinates can be ound by inverting the transormation to express (x 1, x 2 ) in the 2 2 terms o (z 1, z 2 ), that is, x 1 = z 1 x 2 = sin 1 ( z2 a ) Feedback Linearization 3 o 14

which is well deined or a < z 2 < a The transormed state equation is given by ż 1 = z 2 ( ż 2 = a cos sin 1 ( z2 a )) ( z 2 1 + u ) which is in the required orm to use state eedback Finally, the control input that we would use is o the ollowing orm u = x 2 1 + 1 ( k 1 z 1 k 2 z 2 ) = x 2 1 1 + ( k 1 x 1 k 2 a sin(x 2 )) a cos x 2 a cos x 2 2 Input-Output Linearization Consider the single-input-single-output system (1)-(2), where the vector ields : D R n and g : D R n and the unction h : D R are suiciently smooth Deinition 1 is called smooth i C, that is, is continuous and all its derivatives o all orders are continuous The goal is to derive conditions under which the input-output map can be rendered linear The idea is to take a suicient number o derivatives o the output, until the input appears We proceed by taking the derivative o y, which is given by ẏ = h x [ (x) + g (x) u] = L + L g u where L h (x) is called the Lie Derivative o h with respect to x (along) I L g = 0, then ẏ = L, independent o u and we repeat the dierentiation process again Calculating the second derivative o y, denoted by y (2), we obtain y (2) = (L h) x [ (x) + g (x) u] = L 2 + L g L u Once again, i L g L = 0, then y (2) = L 2 is independent o u and we repeat the process Actually, we repeat the processes o taking derivatives o the output until we see that satisies L g L i 1 = 0, i = 1, 2,, ρ 1; L g L ρ 1 0 Feedback Linearization 4 o 14

Thereore, u does not appear in the expressions o y, ẏ,,y (ρ 1) and appears in the expression o y (ρ) with a nonzero coeicient, ie, y (ρ) = L ρ + L gl ρ 1 u (3) We can clearly see rom (3), that the system is input-output linearizable, since the state eedback control 1 ( u = L ρ L g L ρ 1 + v) (4) reduces the input-output map (in some domain D 0 D) to y (ρ) = v which is a chain o ρ integrators In this case, the integer ρ is called the relative degree o the system Deinition 2 The nonlinear system (1)-(2) is said to have a relative degree ρ, 1 ρ n, in the region D 0 D i { Lg L i h(x) = 0, i = 1,, ρ 2 L g L ρ 1 (5) h(x) 0 or all x D 0 Example 3 Consider the controlled van der Pol equation ẋ 1 = x 2 ẋ 2 = x 1 + ε ( 1 x 2 1) x2 + u with output y = x 1 Calculating the derivatives o the output, we obtain ẏ = ẋ 1 = x 2 ÿ = ẋ 2 = x 1 + ε (1 x 2 1) x 2 + u Hence, the system has relative degree two in R 2 For the output y = x 1 + x 2 2, we have that [ ẏ = x 2 + 2x 2 x1 + ε ( ) 1 x 2 1 x2 + u ] and the system has relative degree one in D 0 = {x R 2 x 2 0} As such, we can see that the procedure o input-output linearization is dependent on the choice o the output map h(x) Feedback Linearization 5 o 14

Continuing with our derivations, we now let T 1 (x) z = T (x) = T 2 (x) φ 1 (x) φ n ρ (x) L ρ 1 where φ 1 (x) to φ n ρ (x) are chosen such that φ (x) ψ (x) η ξ (6) φ i g (x) = 0, 1 i n ρ (7) x which ensures that the η-dynamics η = φ φ [ (x) + g (x) u] = x x (x) x=t 1 (z) are independent o u Note also that our deinition o T 2 (x) in (6) results in y y (1) ξ = y (ρ 1) O course, it is crucial to know at this point i such unctions φ exist in order to deine the transormation T ; the next result shows that Deinition 3 A continuously dierentiable transormation T with a continuously dierential inverse is a called a dieomorphism Theorem 1 Consider the system (1)-(2), and suppose that it has a relative degree ρ n in D I ρ = n, then or every x 0 D, a neighborhood N o x 0 exists such that the map T (x) = L h(x) L n 1 Feedback Linearization 6 o 14

restricted to N is a dieomorphism on N I ρ < n, then or every x 0 D, a neighborhood N o x 0 and smooth maps φ 1 (x),, φ n ρ (x) exist such that the map such that the map T (x) = φ 1 (x) φ n ρ (x) L ρ 1 restricted to N, is a dieomorphism on N We can now apply the change o variables (see the Appendix) z = T (x) to transorm the system (1)-(2) into η= 0 (η, ξ) (8) ξ=a c ξ + B c γ (x) [u α (x)] (9) y=c c ξ (10) where ξ R ρ, η R n ρ, (A C, B C, C C ) is a canonical orm representation o a chain o ρ integrators, ie, 0 1 0 0 0 0 1 0 A c =, B c = 0 0 0 1 0, C c = [ 1 0 0 ] 1 0 0 0 and 0 (η, ξ) = φ x (x) x=t 1 (z) (11) γ (x) = L g L ρ 1 (12) α (x) = Lρ L g L ρ 1 (13) Feedback Linearization 7 o 14

v = ξ ξρ 1 = ξ ρ ξ ξ ρ 2 = ξ ρ 1 1 = ξ 2 ρ y = ξ 1 ξ η = 0 (η, ξ) Figure 2: Transormed system (8)-(10) and the input deined as in (4) We have kept α and γ expressed in the original coordinates These two unctions are uniquely determined in terms o, g, and h and are independent o the choice o φ They can be expressed in the new coordinates by setting α 0 (η, ξ) = α ( T 1 (z) ), γ 0 (η, ξ) = γ ( T 1 (z) ) (14) but now they are dependent on the choice o the unctions φ Regarding the deinition o the equilibrium point or the transormed system, assume that x is the open-loop equilibrium o the system (1), then η = φ( x), ξ = [ h( x) 0 0 ] T (15) The transormed system (8)-(10) is said to be in the normal orm This orm decomposes the system into an external part ξ and an internal part η The external part is linearized by the state eedback control u = α (x) + β (x) v (16) where β (x) = γ 1 (x), while the internal part is made unobservable by the same control (see Figure 2) Setting ξ = 0 in the internal dynamics (8), results in η = 0 (η, 0) (17) which is called the zero dynamics o the system I the zero dynamics o the system are (globally) asymptotically stable, the system is called minimum phase Finally, the linearized system may then be stabilized by the choice o an appropriate state eedback: v = Kξ Feedback Linearization 8 o 14

21 Relationship to Linear Systems The notions o relative degree and minimum phase can also be ound in linear systems Consider a linear system represented by the transer unction G (s) = b ms m + b m 1 s m 1 + + b 0 s n + a n 1 s n 1 + + a 0 (18) where m < n and b m 0 We can realize the transer unction in (18) in state-space orm as ẋ = Ax + Bu, y = Cx (19) where 0 1 0 0 0 0 1 0 0 A =, B = 0 0 0 1 0 1 a 0 a 1 a n 1 C = [ b 0 b 1 b m 0 0 ] This linear space model is a special case o ẋ = (x) + g (x) u, y = where (x) = Ax, g = B, and = Cx We know that the relative degree o the transer unction is given by n m However, let us veriy that this is the case We take the derivative o the output, ie, ẏ = Cẋ = CAx + CBu (20) Now, i m = n 1, then CB = b n 1 0 and the system has relative degree one In general we have that and we have the conditions that y (n m) = CA (n m) x + CA (n m 1) Bu (21) CA (i 1) B = 0, i = 1, 2,, n m 1, and CA (n m 1) B = b m 0 (22) Feedback Linearization 9 o 14

and the relative degree o the system is n m, ie, the dierence between the order o the denominator and the numerator o the transer unction G(s) Moreover, we know that a minimum phase system means that the zeros o the transer unction G(s) lie in the open let-hal complex plane We can show as well that there is a correspondence o the zeros o H(s) and the normal orm zero dynamics 3 Full State Feedback Linearization We know rom the previous section, that the system (1) is eedback linearizable i we can ind a suiciently smooth unction h(x) in a domain D such that the system (1)-(2) has relative degree n within some region D 0 D This is because the normal orm would have no zero dynamics In act, we can show that this result holds as an i and only i statement As such, it remains to show that such a unction h(x) exists We begin with some deinitions For any two vector ields and g on D R n, the Lie bracket [, g] is a third vector ield that is deined as [, g](x) g (x) x x g(x) = L g(x) L g (x) (23) Taking Lie brackets can be repeated, as such we deine the ollowing notation: ad 0 g(x) = g(x) ad g(x) = [, g](x) ad k g(x) = [, ad k 1 g](x), k 1 Example 4 Consider the two vector ields (x) = Ax and g is constant Then, ad g(x) = [, g](x) = Ag, ad 2 g(x) = [, ad g](x) = A( Ag) = A 2 g, and ad k g = ( 1)k A k g Deinition 4 (Distribution) Consider the vector ields 1,, k on a domain D R n The distribution, denoted by = { 1, 2,, k } (24) is the collection o all vectors spaces (x) or x D, where (x) = span{ 1 (x), 2 (x),, k (x)} is the subspace o R n spanned by the vectors 1 (x), k (x) Feedback Linearization 10 o 14

Deinition 5 A distribution is called involutive, i g 1, and g 2 [g 1, g 2 ] And now we are ready to state the main result o this section Theorem 2 (Feedback Linearization) The SISO system (1) is eedback linearizable i and only i 1 the matrix M(x) = [g(x), ad g(x),, ad n 1 g(x)] has ull rank (= n) or all x D 0, and 2 the distribution = span{g, ad g,, ad n 2 g} is involutive on D 0 Example 5 Consider the system [ ] a sin(x2 ) ẋ = (x) + gu = + We compute the Lie bracket Accordingly, the matrix x 2 1 ad g = [, g](x) = x g = M(x) = [g, ad g] = [ ] 0 u 1 [ ] a cos(x2 ) 0 [ ] 0 a cos(x2 ) 1 0 has ull rank or any x such that cos(x 2 ) 0 Moreover, the distribution = span{g} is involutive Thereore, the conditions o Theorem (2) are satisied and the system is eedback linearizable in the domain D 0 = {x R 2 cos(x 2 ) 0} Let us now ind the unction h(x) or which the system is eedback linearizable h(x) should satisy the ollowing conditions Now, h x g = h x 2 h x g = 0, (L h) g 0, h(0) = 0 x = 0 which implies that h(x) is independent o x 2 Thereore, = h x 1 a cos(x 2 ) 0 is satisied in D 0 or any choice o h that satisies h x 1 0 There are many such choices or h, or example, h(x) = x 1 or h(x) = x 1 + x 3 1 L h(x) = h x 1 a sin(x 2 ) The condition (L h) x g = (L h) x 2 Feedback Linearization 11 o 14

4 Stability Consider again the input-output linearized system in normal orm (8)-(10), and assume that we have designed the eedback control law u = α(x) + β(x)v where β(x) = γ 1 (x), and v = Kξ, such that (A c B c K) is Hurwitz The resulting system reduces to the ollowing dynamics η = 0 (η, ξ) (25) ξ = (A c B c K)ξ (26) Theorem 3 The origin o the system (25)-(25) is asymptotically stable, i the origin o the zero dynamics η = 0 (η, 0) is asymptotically stable (minimum phase) Proo: The idea is to construct a special Lyapunov unction Since the zero dynamics are asymptotically stable, there exists (by converse Lyapunov theorem) a continuously dierentiable unction V η (η) such that V η (η, 0) < α ( η ) η in some neighborhood o η = 0, where α is a strictly increasing continuous unction with α(0) = 0 Let P = P T > 0 be the solution o the Lyapunov equation P (A c B c K) + (A c B c K) T P = I Consider the unction The derivative o V is given by V = V 1 η 0(η, ξ) + V (η, ξ) = V η (η) + k ξ T P ξ, k > 0 (27) k 2 ξ T P ξ ξt [P (A C B C K) + (A C B C K)P ] ξ }{{} = V 1 η (η, 0) + V 1 η ( k 0(η, ξ) 0 (η, 0)) 2 ξ T P ξ ξt ξ α 3 ( ξ ) + k 1 ( ξ ) kk 2 ( ξ ) I Feedback Linearization 12 o 14

or k 1, k 2 > 0 2 By choosing k large enough, we guarantee V < 0, and the result ollows Remark 1 The result in the last theorem is local and does not extend to the global setup, even i the zero dynamics are globally asymptotically stable In order to make the result global, we have to impose more restrictive requirements regarding the zero dynamics, namely the notion o input-tostate stability o the zero dynamics (which is beyond the current scope o the course) 5 Robustness Feedback linearization is based on exact cancellation o the nonlinearities in the system dynamics, which is practically very diicult to achieve In a realistic setup, we would have only approximations ˆα, ˆβ and ˆT (x) o the true α, β, and T (x) The eedback control law has then the orm u = ˆα(x) + ˆβ(x)v = ˆα(x) ˆβ(x)Kξ = ˆα(x) ˆβ(x)K ˆT 2 (x) Accordingly, the closed-loop system o the normal orm is given by η = (η, ξ) (28) [ ξ = Aξ + Bγ (x) (ˆα(x) α(x)) ˆβ(x)K ˆT ] 2 (x) (29) Adding and subtracting BKξ to the ξ equation we obtain where δ (z) = γ(x) η = (η, ξ) (30) ξ = (A BK) ξ + Bδ (z) (31) [ ( ) (ˆα(x) α(x)) ˆβ(x) β(x) KT ˆβ(x)K ( )] ˆT2 (x) T (x) x=t 1(z) Hence, the local closed loop system diers rom the nominal one by an additive perturbation Thus, we can show that locally the closed-loop system remains stable, despite small uncertainties in the eedback linearization maps For more details on eedback linearization concepts and extensions, see [1] and [2, Ch13] 2 k 1 > 0 ollows rom local continuity and dierentiability o the unction 0, and k 2 > 0 ollows rom P Feedback Linearization 13 o 14

Reerences [1] A Isidori Nonlinear control systems Springer-Verlag, 3rd edition, 1995 [2] H K Khalil Nonlinear systems Prentice hall, 3rd edition, 2002 Appendix The η-dynamics can be derived as ollows η = φ(x) = φ xẋ = φ ((x) + g(x)u) x = φ φ (x) = x x (x) 0 (η, ξ) x=t 1 (x) The ξ-dynamics can be derived as ollows ξ 1 y y (1) ξ 2 ξ 2 ξ = = d y (1) dt = y (2) = ξ ρ ξ ρ y (ρ 1) y (ρ) y (ρ) [ = A c ξ + B c y (ρ) = A c ξ + B c (L g L ρ 1 h(x)) we can now deine (u γ(x) L g L ρ 1 h(x), α(x) Lρ h(x) L g L ρ 1 h(x) Finally, our choice o the transormation T 2 (x) yields y = h(x) = ξ 1 = C c ξ ( Lρ h(x) L g L ρ 1 h(x) )] Feedback Linearization 14 o 14