7. Feedback Linearization Nonlinear Control Systems Ant´onioPedro Aguiar [email protected] 7. Feedback Linearization IST-DEEC PhD Course http://users.isr.ist.utl.pt/%7Epedro/NCS2012/ 2012 1 7. Feedback Linearization Feedback Linearization Given a nonlinear system of the form x_ = f(x) + G(x)u y = h(x) Does exist a state feedback control law u = α(x) + β(x)v and a change of variables z = T (x) that transforms the nonlinear system into a an equivalent linear system (z_ = Az + Bv)? 2 7. Feedback Linearization Feedback Linearization Example: Consider the following system x_ = 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 x_ = 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 3 7. Feedback Linearization Feedback Linearization Example: Consider now this example: x_ 1 = a sin x2 2 x_ 2 = −x1 + u How can we do this? We cannot simply choose u to cancel the nonlinear term a sin x2! However, if we first change the variables z1 = x1 z2 = a sin x2 =x _ 1 then z_1 = z2 2 z_2 = a cos x2x_ 2 = a cos x2(−x1 + u) Therefore with 2 1 u = x1 + v; −π=2 < x2 < π=2 a cos x2 we obtain the linear system z_1 = z2 z_2 = v 4 7. Feedback Linearization Feedback Linearization • A continuously differentiable map T (x) is a diffeormorphism if T −1(x) is @T continuously differentiable. This is true if the Jacobian matrix @x is nonsingular 8x 2 D. @T n • T (x) is a global diffeormorphism if and only if @x is nonsingular 8x 2 R and T (x) is proper, that is, limkxk!1 kT (x)k = 1. Definition A nonlinear system x_ = f(x) + G(x)u (1) n n×p n where f : D ! R and G : D ! R are sufficiently smooth on a domain D ⊂ R is said to be feedback linearizable (or input-state linearizable) if there exists a n diffeomorphism T : D ! R such tat Dz = T (D) contains the origin and the change of variables z = T (x) transforms (1) into the form z_ = Az + Bγ(x)`u − α(x)´ 5 7. Feedback Linearization Feedback Linearization Suppose that we would like to solve the tracking problem for the system x_ 1 = a sin x2 2 x_ 2 = −x1 + u y = x2 If we use state feedback linearization we obtain z1 = x1 z_1 = z2 z2 = a sin x2 =x _ 1 −! z_2 = v u = x2 + 1 v y = sin−1(z =a) 1 a cos x2 2 which is not good! Linearizing the state equation does not necessarily linearize the output equation. 2 Notice however if we set u = x1 + v we obtain x_ 2 = v y = x2 There is one catch: The linearizing feedback control law made x1 unobservable from y. We have to make sure that x1 whose dynamics are given by x_ 1 = a sin x2 is well behaved. For example, if y = yd = cte −! x1(t) = x1(0) + ta sin yd. It is unbounded! 6 7. Feedback Linearization Input-Output Linearization SISO system x_ = f(x) + g(x)u y = h(x) n n where f; g; h are sufficiently smooth in a domain D ⊂ R . The mappings f : D ! R n and g : D ! R are called vector fields on D. Computing the first output derivative... @h @h y_ = x_ = [f(x) + g(x)u] =: L h(x) + Lgh(x)u @x @x f In the sequel we will use the following notation: @h L h(x) = f(x) −! Lie Derivative of h with respect to f f @x @(Lf h) LgL h(x) = g(x) f @x 0 Lf h(x) = h(x) @(L h) L2 h(x) = L L h(x) = f f(x) f f f @x @(Lk−1h) Lk h(x) = L Lk−1h(x) = f f(x) f f f @x 7 7. Feedback Linearization Input-Output Linearization y_ = Lf h(x) + Lgh(x)u If Lgh(x)u = 0 then y_ = Lf h(x) (independent of u). Computing the second derivative... (2) @(Lf h) 2 y = [f(x) + g(x)u] = L h(x) + LgL h(x)u @x f f (2) 2 If LgLf h(x)u = 0 then y_ = Lf h(x) (independent of u). Repeating this process, it follows that if i−1 LgLf h(x) = 0; i = 1; 2; : : : ; ρ − 1 ρ−1 LgLf h(x) 6= 0 then u does not appear in y; y;_ : : : ; y(ρ−1) and (ρ) ρ (ρ−1) y = Lf h(x) + LgLf h(x)u 8 7. Feedback Linearization Input-Output Linearization (ρ) ρ ρ−1 y = Lf h(x) + LgLf h(x)u Therefore, by setting 1 ρ u = ρ−1 [−Lf h(x) + v] LgLf h(x) the system is input-output linearizable and reduces to y(ρ) = v −! chain of ρ integrators Definition The nonlinear system x_ = f(x) + g(x)u y = h(x) is said to have relative degree ρ, 1 ≤ ρ ≤ n, in the region D0 ⊂ D if for all x 2 D0 i−1 LgLf h(x) = 0; i = 1; 2; : : : ; ρ − 1 ρ−1 LgLf h(x) 6= 0 9 7. Feedback Linearization Examples Example 1: Van der Pol system x_ 1 = x2 2 x_ 2 = −x1 + (1 − x1)x2 + u; > 0 1. y = x1 Calculating the derivatives... y_ =x _ 1 = x2 2 y¨ =x _ 2 = −x1 + (1 − x1)x2 + u 2 Thus the system has relative degree ρ = 2 in R . 2. y = x2 Then 2 y_ =x _ 2 = −x1 + (1 − x1)x2 + u 2 In this case the system has relative degree ρ = 1 in R . 2 3. y = x1 + x2 Then 2 y_ = x2 + 2x2(−x1 + (1 − x1)x2 + u) 2 In this case the system has relative degree ρ = 1 in D0 = fx 2 R : x2 6= 0g. 10 7. Feedback Linearization Examples Example 2: x_ 1 = x1 x_ 2 = x2 + u y = x1 Calculating the derivatives... (n) y_ =x _ 1 = x1 = y −! y = y = x1; 8n ≥ 1 The system does not have a well defined relative degree! t Why? Because the output y(t) = x1(t) = e x1(0) is independent of the input u. 11 7. Feedback Linearization Examples Example 3: b sm + b sm−1 + ··· + b H(s) = m m−1 0 n n−1 s + an−1s + ··· + a0 where m < n and bm 6= 0. A state model of the system is the following x_ = Ax + Bu y = Cx with 2 0 1 0 ··· 0 3 2 0 3 0 0 1 0 ··· : 6 . 7 6 . 7 A = . B = . C = [ b0 b1 ··· bm 0 ···0 ] 6 . 0 7 6 . 7 1×n 4 0 ··· ··· 0 1 5 4 5 −a −a ··· ··· −a 0 0 1 n−1 n×n 1 n×1 What is the relative degree ρ ? 12 7. Feedback Linearization Examples y_ = CAx + CBu If m = n − 1 −! CB = bm 6= 0 −! ρ = 1 Otherwise, CB = 0 y(2) = CA2x + CABu Note that CA is obtained by shifting the elements of C one position to the right and CAi by shifting i positions. Therefore, CAi−1B = 0; for i = 1; 2; : : : n − m − 1 n−m−1 CA B = bm 6= 0 y(n−m) = CAn−mx + CAn−m−1Bu −! 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). 13 7. Feedback Linearization Consider again the linear system given by the transfer function 8 deg D = n N(s) < H(s) = with deg N = m < n D(s) : ρ = n − m D(s) can be written as 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 1 N(s) Q(s) H(s) = = Q(s)N(s) + R(s) 1 R(s) 1 + 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 7. Feedback Linearization Note that the ρ-order transfer function 1=Q(s) has no zeros and can be realized by T ξ_ = (Ac + Bcλ )ξ + Bcbme y = Ccξ where ˆ (ρ−1)˜T ρ ξ = y y_ : : : y 2 R and (Ac;Bc;Cc) is a canonical form representation of a chain of ρ integrators: 20 1 0 ··· 03 203 60 0 1 ··· 07 607 6 7 6.7 6. .. .7 6.7 A = 6. .7 B = 6.7 C = ˆ1 0 ··· 0 0˜ c 6 7 c 6 7 c 1×ρ 6 7 6 7 6. 7 6 7 4. 0 15 405 0 ······ 0 0 ρ×ρ 1 ρ×1 20 ··· 03 6. .7 T 6. .7 ρ Bcλ = 6 7 ; λ 2 R 40 ··· 05 λT 15 7. Feedback Linearization R(s) η_ = A η + B y −! 0 0 N(s) w = C0η The eigenvalues of A0 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 η_ = A0η + B0Ccξ T ξ_ = Acξ + Bc(λ ξ − bmC0η + bmu) y = Ccξ Note that y = Ccξ and ξ_1 = ξ2 ξ_2 = ξ3 _ T ξ = Acξ + Bc(λ ξ − bmC0η + bmu) ! .