Output Feedback and State Feedback. EL2620 Nonlinear Control. Nonlinear Observers. Nonlinear Controllers. ẋ = f(x,u), y = h(x)

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1 Output Feedback and State Feedback EL2620 Nonlinear Control Lecture 10 Exact feedback linearization Input-output linearization Lyapunov-based control design methods ẋ = f(x,u) y = h(x) Output feedback: Find u = k(y) such that the closed-loop system ẋ = f ( x,k ( h(x) )) has nice properties. State feedback: Find u = l(x) such that ẋ = f(x,l(x)) has nice properties. k and l may include dynamics. Lecture 10 1 Lecture 10 2 What if x is not measurable? Nonlinear Observers Nonlinear Controllers Nonlinear dynamical controller: ż = a(z,y), u = c(z) Simplest observer ẋ = f(x,u), y = h(x) x = f( x,u) Linear dynamics, static nonlinearity: ż = Az + By, u = c(z) Linear controller: ż = Az + By, u = Cz Feedback correction, as in linear case, Choices of K x = f( x,u) + K(y h( x)) Linearize f at x 0, find K for the linearization Linearize f at x(t), find K = K( x) for the linearization Second case is called Extended Kalman Filter Lecture 10 3 Lecture 10 4

2 Exact Feedback Linearization Some state feedback control approaches Exact feedback linearization Input-output linearization Lyapunov-based design - backstepping control Consider the nonlinear control-affine system idea: use a state-feedback controller u(x) to make the system linear Example 1: The state-feedback controller ẋ = cos x x 3 + u u(x) = cosx + x 3 kx + v yields the linear system ẋ = kx + v Lecture 10 5 Lecture 10 6 Another Example ẋ 1 = a sin x 2 ẋ 2 = x u How do we cancel the term sin x 2? Perform transformation of states into linearizable form: yields z 1 = x 1, z 2 = ẋ 1 = a sin x 2 ż 1 = z 2, ż 2 = a cos x 2 ( x u) and the linearizing control becomes u(x) = x v a cos x 2, x 2 [ π/2,π/2] Diffeomorphisms A nonlinear state transformation z = T(x) with T invertible for x in the domain of interest T and T 1 continuously differentiable is called a diffeomorphism Definition: A nonlinear system is feedback linearizable if there exist a diffeomorphism T whose domain contains the origin and transforms the system into the form ẋ = Ax + Bγ(x) (u α(x)) with (A,B) controllable and γ(x) nonsingular for all x in the domain of interest. Lecture 10 7 Lecture 10 8

3 Exact vs. Input-Output Linearization The example again, but now with an output ẋ 1 = a sin x 2, ẋ 2 = x u, y = x 2 The control law u = x v/a cos x 2 yields ż 1 = z 2 ; ż 2 = v ; y = sin 1 (z 2 /a) which is nonlinear in the output. If we want a linear input-output relationship we could instead use to obtain u = x v ẋ 1 = a sin x 2, ẋ 2 = v, y = x 2 which is linear from v to y (but what about the unobservable state x 1?) Input-Output Linearization Use state feedback u(x) to make the control-affine system y = h(x) linear from the input v to the output y The general idea: differentiate the output, y = h(x), p times untill the control u appears explicitly in y (p), and then determine u so that y (p) = v i.e., G(s) = 1/s p Lecture 10 9 Lecture Example: controlled van der Pol equation Differentiate the output ẋ 1 = x 2 ẋ 2 = x 1 + ǫ(1 x 2 1)x 2 + u y = x 1 ẏ = ẋ 1 = x 2 The state feedback controller ÿ = ẋ 2 = x 1 + ǫ(1 x 2 1)x 2 + u u = x 1 ǫ(1 x 2 1)x 2 + v ÿ = v Lie Derivatives Consider the nonlinear SISO system The derivative of the output ; x R n,u R 1 y = h(x), y R ẏ = dh dxẋ = dh dx (f(x) + g(x)u) L fh(x) + L g h(x)u where L f h(x) and L g h(x) are the Lie derivatives (L f h is the derivative of h along the vector field of ẋ = f(x)) Repeated derivatives L k fh(x) = d(lk 1 f h) f(x), L g L f h(x) = d(l fh) dx dx g(x) Lecture Lecture 10 12

4 Lie derivatives and relative degree The relative degree p of a system is defined as the number of integrators between the input and the output (the number of times y must be differentiated for the input u to appear) A linear system Y (s) U(s) = has relative degree p = n m b 0 s m b m s n + a 1 s n a n A nonlinear system has relative degree p if L g L i 1 f h(x) = 0,i = 1,...,p 1 ; L g L p 1 f h(x) 0 x D Example The controlled van der Pol equation ẋ 1 = x 2 Differentiating the output ẋ 2 = x 1 + ǫ(1 x 2 1)x 2 + u y = x 1 ẏ = ẋ 1 = x 2 ÿ = ẋ 2 = x 1 + ǫ(1 x 2 1)x 2 + u Thus, the system has relative degree p = 2 Lecture Lecture The input-output linearizing control Consider a nth order SISO system with relative degree p Differentiating the output y = h(x) ẏ = dh dxẋ = L fh(x) +. =0 {}}{ L g h(x)u y (p) = L p f h(x) + L gl p 1 f h(x)u and hence the state-feedback controller u = 1 ( L p L g L p 1 fh(x) + v) f h(x) results in the linear input-output system y (p) = v Lecture Lecture 10 16

5 Zero Dynamics van der Pol again Note that the order of the linearized system is p, corresponding to the relative degree of the system Thus, if p < n then n p states are unobservable in y. The dynamics of the n p states not observable in the linearized dynamics of y are called the zero dynamics. Corresponds to the dynamics of the system when y is forced to be zero for all times. A system with unstable zero dynamics is called non-minimum phase (and should not be input-output linearized!) ẋ 1 = x 2 ẋ 2 = x 1 + ǫ(1 x 2 1 )x 2 + u With y = x 1 the relative degree p = n = 2 and there are no zero dynamics, thus we can transform the system into ÿ = v. Try it yourself! With y = x 2 the relative degree p = 1 < n and the zero dynamics are given by ẋ 1 = 0, which is not asymptotically stable (but bounded) Lecture Lecture A simple introductory example Lyapunov-Based Control Design Methods ẋ = f(x,u) Find stabilizing state feedback u = u(x) Verify stability through Control Lyapunov function Methods depend on structure of f Here we limit discussion to Back-stepping control design, which require certain f discussed later. Consider Apply the linearizing control ẋ = cos x x 3 + u u = cosx + x 3 kx Choose the Lyapunov candidate V (x) = x 2 /2 V (x) > 0, V = kx 2 < 0 Thus, the system is globally asymptotically stable But, the term x 3 in the control law may require large control moves! Lecture Lecture 10 20

6 The same example Now try the control law ẋ = cos x x 3 + u u = cosx kx Choose the same Lyapunov candidate V (x) = x 2 /2 V (x) > 0, V = x 4 kx 2 < 0 Simulating the two controllers Simulation with x(0) = 10 State x State trajectory Control input 10 linearizing non linearizing Input u linearizing non linearizing Thus, also globally asymptotically stable (and more negative V ) time [s] time [s] The linearizing control is slower and uses excessive input Lecture Lecture Back-Stepping Control Design We want to design a state feedback u = u(x) that stabilizes at x = 0 with f(0) = 0. ẋ 1 = f(x 1 ) + g(x 1 )x 2 ẋ 2 = u Idea: See the system as a cascade connection. Design controller first for the inner loop and then for the outer. u x 2 x 1 g(x 1 ) f() (1) Suppose the partial system ẋ 1 = f(x 1 ) + g(x 1 ) v can be stabilized by v = φ(x 1 ) and there exists Lyapunov fcn V 1 = V 1 (x 1 ) such that V 1 (x 1 ) = dv ( ) 1 f(x 1 ) + g(x 1 )φ(x 1 ) W(x 1 ) for some positive definite function W. This is a critical assumption in backstepping control! Lecture Lecture 10 24

7 Equation (1) can be rewritten as The Trick ẋ 1 = f(x 1 ) + g(x 1 )φ(x 1 ) + g(x 1 )[x 2 φ(x 1 )] ẋ 2 = u u x 2 x 1 g(x 1 ) φ(x 1 ) f + gφ Introduce new state ζ = x 2 φ(x 1 ) and control v = u φ: where ẋ 1 = f(x 1 ) + g(x 1 )φ(x 1 ) + g(x 1 )ζ ζ = v φ(x 1 ) = dφ ẋ 1 = dφ ) (f(x 1 ) + g(x 1 )x 2 u ζ x 1 g(x 1 ) φ(x 1 ) f + gφ Lecture Lecture Consider V 2 (x 1,x 2 ) = V 1 (x 1 ) + ζ 2 /2. Then, V 2 (x 1,x 2 ) = dv ( ) 1 f(x 1 ) + g(x 1 )φ(x 1 ) + dv 1 g(x 1 )ζ + ζv Choosing gives W(x 1 ) + dv 1 g(x 1 )ζ + ζv v = dv 1 g(x 1 ) kζ, k > 0 V 2 (x 1,x 2 ) W(x 1 ) kζ 2 Hence, x = 0 is asymptotically stable for (1) with control law u(x) = φ(x) + v(x). If V 1 radially unbounded, then global stability. Back-Stepping Lemma Lemma: Let z = (x 1,...,x k 1 ) T and Assume φ(0) = 0, f(0) = 0, ż = f(z) + g(z)x k ẋ k = u ż = f(z) + g(z)φ(z) stable, and V (z) a Lyapunov fcn (with V W ). Then, u = dφ ) (f(z) + g(z)x k dv dz dz g(z) (x k φ(z)) stabilizes x = 0 with V (z) + (x k φ(z)) 2 /2 being a Lyapunov fcn. Lecture Lecture 10 28

8 Strict Feedback Systems Back-stepping Lemma can be applied to stabilize systems on strict feedback form: ẋ 1 = f 1 (x 1 ) + g 1 (x 1 )x 2 ẋ 2 = f 2 (x 1,x 2 ) + g 2 (x 1,x 2 )x 3 ẋ 3 = f 3 (x 1,x 2,x 3 ) + g 3 (x 1,x 2,x 3 )x 4. ẋ n = f n (x 1,...,x n ) + g n (x 1,...,x n )u 2 minute exercise: Give an example of a linear system ẋ = Ax + Bu on strict feedback form. where g k 0 Note: x 1,...,x k do not depend on x k+2,...,x n. Lecture Lecture Back-Stepping Back-Stepping Lemma can be applied recursively to a system on strict feedback form. Back-stepping generates stabilizing feedbacks φ k (x 1,...,x k ) (equal to u in Back-Stepping Lemma) and Lyapunov functions V k (x 1,...,x k ) = V k 1 (x 1,...,x k 1 ) + [x k φ k 1 ] 2 /2 by stepping back from x 1 to u (see Khalil pp for details). Back-stepping results in the final state feedback u = φ n (x 1,...,x n ) Design back-stepping controller for Example ẋ 1 = x x 2, ẋ 2 = x 3, ẋ 3 = u Step 0 Verify strict feedback form Step 1 Consider first subsystem ẋ 1 = x φ 1 (x 1 ), ẋ 2 = u 1 where φ 1 (x 1 ) = x 2 1 x 1 stabilizes the first equation. With V 1 (x 1 ) = x 2 1/2, Back-Stepping Lemma gives u 1 = ( 2x 1 1)(x x 2 ) x 1 (x 2 + x x 1 ) = φ 2 (x 1,x 2 ) V 2 = x 2 1/2 + (x 2 + x x 1 ) 2 /2 Lecture Lecture 10 32

9 Step 2 Applying Back-Stepping Lemma on gives u = u 2 = dφ 2 dz ẋ 1 = x x 2 ẋ 2 = x 3 ẋ 3 = u (f(z) + g(z)x n ) dv 2 dz g(z) (x n φ 2 (z)) = φ 2 x 1 (x x 2 ) + φ 2 x 2 x 3 V 2 x 2 (x 3 φ 2 (x 1,x 2 )) which globally stabilizes the system. Gain scheduling Next Lecture Nonlinear controllability (How to park a car) Lecture Lecture 10 34