EL2620 2010 EL2620 2010 Output Feedback and State Feedback x˙ = f(x, u) EL2620 Nonlinear Control y = h(x) • Output feedback: Find u = k(y) such that the closed-loop Lecture 10 system x˙ = f x, k h(x) • Exact feedback linearization has nice properties. • Input-output linearization • State feedback: Find u = ℓ(x) such that • Lyapunov-based control design methods x˙ = f(x,ℓ(x)) has nice properties. k and ℓ may include dynamics. Lecture 10 1 Lecture 10 2 EL2620 2010 EL2620 2010 Nonlinear Observers What if x is not measurable? x˙ = f(x, u), y = h(x) Nonlinear Controllers Simplest observer ˙ • Nonlinear dynamical controller: z˙ = a(z,y), u = c(z) x = f(x, u) Feedback correction, as in linearb case,b • Linear dynamics, static nonlinearity: z˙ = Az + By, u = c(z) x˙ = f(x, u) + K(y − h(x)) • Linear controller: z˙ = Az + By, u = Cz Choices of K b b b • Linearize f at x0, find K for the linearization • Linearize f at x(t), find K = K(x) for the linearization Second case is called Extended Kalman Filter b b Lecture 10 3 Lecture 10 4 EL2620 2010 EL2620 2010 Exact Feedback Linearization Consider the nonlinear control-affine system x˙ = f(x) + g(x)u Some state feedback control approaches idea: use a state-feedback controller u(x) to make the system linear • Exact feedback linearization Example 1: 3 • Input-output linearization x˙ = cos x − x + u • Lyapunov-based design - backstepping control The state-feedback controller u(x) = − cos x + x3 − kx + v yields the linear system x˙ = −kx + v Lecture 10 5 Lecture 10 6 EL2620 2010 EL2620 2010 Another Example Diffeomorphisms A nonlinear state transformation z = T (x) with x˙ 1 = a sin x2 2 • T invertible for x in the domain of interest x˙ 2 = −x + u 1 − • T and T 1 continuously differentiable How do we cancel the term sin x2? is called a diffeomorphism Perform transformation of states into linearizable form: Definition: A nonlinear system z1 = x1, z2 =x ˙ 1 = a sin x2 x˙ = f(x) + g(x)u yields is feedback linearizable if there exist a diffeomorphism T whose 2 z˙1 = z2, z˙2 = a cos x2(−x1 + u) domain contains the origin and transforms the system into the form and the linearizing control becomes x˙ = Ax + Bγ(x)(u − α(x)) v u(x) = x2 + , x ∈ [−π/2, π/2] 1 a cos x 2 with (A,B) controllable and γ(x) nonsingular for all x in the domain 2 of interest. Lecture 10 7 Lecture 10 8 EL2620 2010 EL2620 2010 Exact vs. Input-Output Linearization The example again, but now with an output Input-Output Linearization 2 x˙ 1 = a sin x2, x˙ 2 = −x1 + u, y = x2 Use state feedback u(x) to make the control-affine system 2 • The control law u = x1 + v/a cos x2 yields x˙ = f(x) + g(x)u −1 z˙1 = z2 ;z ˙2 = v ; y = sin (z2/a) y = h(x) which is nonlinear in the output. linear from the input v to the output y • If we want a linear input-output relationship we could instead use • The general idea: differentiate the output, y = h(x), p times 2 untill the control u appears explicitly in y(p), and then determine u u = x1 + v so that to obtain y(p) = v x˙ 1 = a sin x2, x˙ 2 = v, y = x2 i.e., G(s) = 1/sp which is linear from v to y (but what about the unobservable state x1?) Lecture 10 9 Lecture 10 10 EL2620 2010 EL2620 2010 Lie Derivatives Example: controlled van der Pol equation Consider the nonlinear SISO system x˙ = x n 1 2 x˙ = f(x) + g(x)u ; x ∈ R , u ∈ R1 2 x˙ 2 = −x1 + ǫ(1 − x1)x2 + u y = h(x), y ∈ R y = x1 The derivative of the output Differentiate the output dh dh y˙ = x˙ = (f(x) + g(x)u) , Lf h(x) + Lgh(x)u y˙ =x ˙ 1 = x2 dx dx 2 where L h(x) and L h(x) are the Lie derivatives (L h is the y¨ =x ˙ 2 = −x1 + ǫ(1 − x1)x2 + u f g f derivative of h along the vector field of x˙ = f(x)) The state feedback controller Repeated derivatives 2 − − ⇒ k−1 u = x1 ǫ(1 x1)x2 + v y¨ = v d(L h) d(L h) Lkh(x) = f f(x), L L h(x) = f g(x) f dx g f dx Lecture 10 11 Lecture 10 12 EL2620 2010 EL2620 2010 Lie derivatives and relative degree Example • The relative degree p of a system is defined as the number of The controlled van der Pol equation integrators between the input and the output (the number of times y must be differentiated for the input u to appear) x˙ 1 = x2 2 • A linear system x˙ 2 = −x1 + ǫ(1 − x1)x2 + u m y = x1 Y (s) b0s + . + bm = n n−1 U(s) s + a1s + . + an Differentiating the output has relative degree p = n − m y˙ =x ˙ 1 = x2 • A nonlinear system has relative degree p if 2 y¨ =x ˙ 2 = −x1 + ǫ(1 − x1)x2 + u i−1 p−1 LgLf h(x) = 0, i = 1,...,p−1 ; LgLf h(x) =6 0 ∀x ∈ D Thus, the system has relative degree p = 2 Lecture 10 13 Lecture 10 14 EL2620 2010 EL2620 2010 The input-output linearizing control Consider a nth order SISO system with relative degree p and hence the state-feedback controller x˙ = f(x) + g(x)u 1 y = h(x) p u = p−1 −Lf h(x) + v LgLf h(x) Differentiating the output results in the linear input-output system =0 dh (p) y˙ = x˙ = Lf h(x) + Lgh(x)u y = v dx z }| { . (p) p p−1 y = Lf h(x) + LgLf h(x)u Lecture 10 15 Lecture 10 16 EL2620 2010 EL2620 2010 Zero Dynamics van der Pol again • Note that the order of the linearized system is , corresponding to p x˙ 1 = x2 the relative degree of the system 2 x˙ 2 = −x1 + ǫ(1 − x1)x2 + u • Thus, if p<n then n − p states are unobservable in y. • With y = x1 the relative degree p = n = 2 and there are no • The dynamics of the n − p states not observable in the linearized zero dynamics, thus we can transform the system into y¨ = v. Try dynamics of y are called the zero dynamics. Corresponds to the it yourself! dynamics of the system when y is forced to be zero for all times. • With y = x2 the relative degree p = 1 <n and the zero • A system with unstable zero dynamics is called non-minimum dynamics are given by x˙ 1 = 0, which is not asymptotically stable phase (and should not be input-output linearized!) (but bounded) Lecture 10 17 Lecture 10 18 EL2620 2010 EL2620 2010 A simple introductory example Lyapunov-Based Control Design Methods Consider x˙ = cos x − x3 + u x˙ = f(x, u) Apply the linearizing control • Find stabilizing state feedback u = u(x) u = − cos x + x3 − kx • Verify stability through Control Lyapunov function Choose the Lyapunov candidate V (x) = x2/2 • Methods depend on structure of f V (x) > 0 , V˙ = −kx2 < 0 Here we limit discussion to Back-stepping control design, which require certain f discussed later. Thus, the system is globally asymptotically stable But, the term x3 in the control law may require large control moves! Lecture 10 19 Lecture 10 20 EL2620 2010 EL2620 2010 Simulating the two controllers Simulation with x(0) = 10 The same example x˙ = cos x − x3 + u State trajectory Control input 10 linearizing 1000 Now try the control law non−linearizing linearizing 9 non−linearizing 800 u = − cos x − kx 8 7 600 Choose the same Lyapunov candidate V (x) = x2/2 6 5 400 State x ˙ 4 2 4 Input u V (x) > 0 , V = −x − kx < 0 200 3 2 Thus, also globally asymptotically stable (and more negative V˙ ) 0 1 0 −200 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time [s] time [s] The linearizing control is slower and uses excessive input Lecture 10 21 Lecture 10 22 EL2620 2010 EL2620 2010 Back-Stepping Control Design We want to design a state feedback u = u(x) that stabilizes Suppose the partial system x˙ 1 = f(x1) + g(x1)x2 (1) x˙ 1 = f(x1) + g(x1)¯v x˙ 2 = u can be stabilized by v¯ = φ(x1) and there exists Lyapunov fcn at x = 0 with f(0) = 0. V1 = V1(x1) such that Idea: See the system as a cascade connection. Design controller first for the inner loop and then for the outer. dV1 V˙1(x1) = f(x1) + g(x1)φ(x1) ≤−W (x1) u x2 x1 dx1 g(x1) for some positive definite function W . R R This is a critical assumption in backstepping control! f() Lecture 10 23 Lecture 10 24 EL2620 2010 EL2620 2010 ˙ The Trick Introduce new state ζ = x2 − φ(x1) and control v = u − φ: x˙ 1 = f(x1) + g(x1)φ(x1) + g(x1)ζ Equation (1) can be rewritten as ζ˙ = v x˙ 1 = f(x1) + g(x1)φ(x1) + g(x1)[x2 − φ(x1)] x˙ = u where 2 dφ dφ φ˙(x ) = x˙ = f(x ) + g(x )x x2 1 1 1 1 2 u x1 dx1 dx1 g(x1) u ζ x1 R R g(x1) −φ(x1) f + gφ R R −φ˙(x1) f + gφ Lecture 10 25 Lecture 10 26 EL2620 2010 EL2620 2010 2 Back-Stepping Lemma Consider V2(x1,x2) = V1(x1) + ζ /2.