EL2620 Nonlinear Control Lecture notes Karl Henrik Johansson, Bo Wahlberg and Elling W. Jacobsen This revision December 2012 Automatic Control KTH, Stockholm, Sweden Preface Many people have contributed to these lecture notes in nonlinear control. Originally it was developed by Bo Bernhardsson and Karl Henrik Johansson, and later revised by Bo Wahlberg and myself. Contributions and comments by Mikael Johansson, Ola Markusson, Ragnar Wallin, Henning Schmidt, Krister Jacobsson, Björn Johansson and Torbjörn Nordling are gratefully acknowledged. Elling W. Jacobsen Stockholm, December 2012 Lecture 1 EL2620 • Practical information • Course outline • Linear vs Nonlinear Systems • Nonlinear differential equations Lecture 1 EL2620 Nonlinear Control [email protected] STEX (entrance floor, Osquldasv. 10), course material [email protected] Hanna Holmqvist, course administration Farhad Farokhi, Martin Andreasson, teaching assistants [email protected], [email protected] [email protected] Elling W. Jacobsen, lectures and course responsible • Instructors 28h lectures, 28h exercises, 3 home-works 7.5 credits, lp 2 Automatic Control Lab, KTH EL2620 Nonlinear Control • Disposition Lecture 1 EL2620 3 2012 1 2012 Course Goal Today’s Goal Lecture 1 • Derive equilibrium points • Transform differential equations to first-order form • Describe distinctive phenomena in nonlinear dynamic systems • Define a nonlinear dynamic system You should be able to EL2620 Lecture 1 • use some practical nonlinear control design methods • apply the most powerful nonlinear analysis methods • understand common nonlinear control phenomena You should after the course be able to 4 2012 2 2012 To provide participants with a solid theoretical foundation of nonlinear control systems combined with a good engineering understanding EL2620 Course Information Course Outline Lecture 1 • Summary (L14) fuzzy control (L11-L13) 7 2012 5 2012 • Alternatives: gain scheduling, optimal control, neural networks, methods (L7-L10) • Control design: compensation, high-gain design, Lyapunov functions (L3-L6) • Feedback analysis: linearization, stability theory, describing simulation (L1-L2) • Introduction: nonlinear models and phenomena, computer EL2620 Lecture 1 • Written 5h exam on January 18 2014 (we are strict on this!) • Homeworks are compulsory and have to be handed in on time Social) • All info and handouts are available at course homepage (KTH EL2620 Material Software: Matlab / Simulink Homeworks: 3 computer exercises to hand in Exercises: Class room and home exercises Lecture notes: Copies of transparencies (from previous year) Linear Systems superposition scaling 0 Lecture 1 Notice the importance to have zero initial conditions Y (s) = G(s)U (s) 8 2012 : M → M is ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t), x(0) = 0 Z t g(τ )u(t − τ )dτ y(t) = g(t) ⋆ u(t) = Example: Linear time-invariant systems S(αu) = αS(u) S(u + v) = S(u) + S(v) Definition: Let M be a signal space. The system S linear if for all u, v ∈ M and α ∈ R EL2620 Lecture 1 Two course compendia sold by STEX (also available at course homepage) Only references to Khalil will be given. 6 2012 Alternative textbooks (decreasing mathematical rigour): Sastry, Nonlinear Systems: Analysis, Stability and Control; Vidyasagar, Nonlinear Systems Analysis; Slotine & Li, Applied Nonlinear Control; Glad & Ljung, Reglerteori, flervariabla och olinjära metoder. • • • • 2002. Optional but highly recommended. • Textbook: Khalil, Nonlinear Systems, Prentice Hall, 3rd ed., EL2620 Linear Systems Have Nice Properties Linear model valid only if sin φ ≈φ Unrealistic answer. Clearly outside linear region! 0.1 φ0 ≈ = 2 rad = 114◦ 0.05 Lecture 1 Must consider nonlinear model. Possibly also include other nonlinearities such as centripetal force, saturation, friction etc. so that = φ0 ) gives t2 x(t) ≈ 10 φ0 ≈ 0.05φ0 2 Linear model (step response with φ Can the ball move 0.1 meter in 0.1 seconds from steady state? EL2620 Lecture 1 Frequency analysis possible Sinusoidal inputs give sinusoidal outputs: Y (iω) = G(iω)U (iω) Superposition Enough to know a step (or impulse) response 11 2012 9 2012 Local stability=global stability Stability if all eigenvalues of A (or poles of G(s)) are in the left half-plane EL2620 Lecture 1 2012 Linear Models are not Rich Enough 12 2012 10 mẍ(t) = mg sin φ(t), Linear model: ẍ(t) = gφ(t) x Linear models can not describe many phenomena seen in nonlinear systems EL2620 Lecture 1 Nonlinear model: φ Linear Models may be too Crude Approximations Example:Positioning of a ball on a beam EL2620 Σ 1 s Motor − 1 x2 x2 1 + f (1 − x1 ) − ǫf (x2 − xc ) 0 0 50 100 150 200 0 10 15 50 50 100 time [s] Output 100 Input 150 150 200 200 15 2012 0 0 5 0 0 10 2 0 0 4 0.2 0.4 5 5 5 10 10 10 15 Time t 15 Time t 15 Time t 20 20 20 25 25 25 S TEP R ESPONSES 30 30 30 2012 Stable Periodic Solutions Motor: 1 s(1+5s) Sum K=5 G(s) = Constant 0 P-controller 5 -1 Motor 5s 2+s 1 Backlash Example: Position control of motor with back-lash EL2620 Lecture 1 y 16 2012 14 (The linearized gain of the valve increases with increasing amplitude) Stability depends on amplitude of the reference signal! Lecture 1 0.7, ǫ = 0.4 − 20 y r = 1.72 r = 1.68 r = 0.2 Lecture 1 = f x1 exp 1 (s+1)(s+1) y Controller: = ẋ2 −x1 exp -1 2 u 1 (s+1)2 Process = u2 EL2620 Existence of multiple stable equilibria for the same input gives hysteresis effect = ẋ1 s 1 −1 Valve Multiple Equilibria Example: chemical reactor EL2620 Lecture 1 Simulink block diagram: r Example: Control system with valve characteristic f (u) 13 2012 Stability Can Depend on Reference Signal c 2 PSfr EL2620 Coolant temp x Temperature x Output y Output y Output y Back-lash induces an oscillation 0 10 10 10 20 20 20 Time t Time t Time t 30 30 30 Lecture 1 0 10 -2 10 0 10 -2 10 40 40 40 0 10 Frequency (Hz) a=2 0 10 Frequency (Hz) a=1 a sin t 0 -2 -1 0 0 1 2 -2 -1 0 1 2 Saturation 1 1 2 3 Time t 2 3 Time t y(t) = Harmonic Distortion How predict and avoid oscillations? -0.5 0 0 0.5 -0.5 0 0 0.5 -0.5 0 0.5 Example: Sinusoidal response of saturation EL2620 Lecture 1 Output y Output y Output y 50 50 50 k=1 4 4 5 5 P∞ Period and amplitude independent of initial conditions: EL2620 Amplitude y Amplitude y 19 Ak sin(kt) 2012 17 2012 Automatic Tuning of PID Controllers 0 y u Relay 5 A T u −1 Process 10 y Time = Energy in tone k Energy in tone 1 k=2 Lecture 1 Trade-off between effectivity and linearity 20 2012 Introduces spectrum leakage, which is a problem in cellular systems Effective amplifiers work in nonlinear region Example: Electrical amplifiers Total Harmonic Distortion P∞ Nonlinearities such as rectifiers, switched electronics, and transformers give rise to harmonic distortion Example: Electrical power distribution EL2620 Lecture 1 −1 0 1 Σ PID 18 2012 Relay induces a desired oscillation whose frequency and amplitude are used to choose PID parameters EL2620 0 10 10 Time t 15 15 Time t Lecture 1 20 20 25 25 30 30 tf = 1 x0 1 0≤t< x0 Integrate ⇒ ẋ = x2 ⇒ dx = dt x2 −1 −1 x0 − = t ⇒ x(t) = x(t) x(0) 1 − x0 t Recall the trick: Solution interval depends on initial condition! Solution not defined for x0 , has solution x(t) = 1 − x0 t = x2 , x(0) = x0 Existence Problems 5 5 + ẏ + y − y 3 = a sin(ωt) Subharmonics Example: The differential equation ẋ EL2620 Lecture 1 -1 -0.5 0 0.5 1 0 0 0.5 a sin ωt -0.5 y Example: Duffing’s equation ÿ EL2620 23 2012 21 2012 Nonlinear Differential Equations Lecture 1 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 2 Time t 3 Finite escape time of dx/dt = x2 4 Finite Escape Time 1 (1) 2012 5 24 2012 22 : [0, T ] → Rn such that (1) ẋ = Ax, x(0) = x0 , gives x(t) = exp(At)x0 Simulation for various initial conditions x0 EL2620 Lecture 1 Example: • When is the solution unique? • When does there exists a solution? over an interval [0, T ] is a C1 function x is fulfilled. ẋ(t) = f (x(t)), x(0) = x0 Definition: A solution to EL2620 x(t) ẋ = 0 1 2 Time t 3 4 5 Lipschitz Continuity -1 -0.5 0 0.5 1 1.5 2 x, x(0) = 0, has many solutions: (t − C)2 /4 t > C x(t) = 0 t≤C √ Uniqueness Problems Lecture 1 kxk2 = x21 + · · · + x2n Euclidean norm is given by Slope L kf (x) − f (y)k ≤ Lkx − yk Definition: f : Rn → Rn is Lipschitz continuous if there exists L, r > 0 such that for all x, y ∈ Br (x0 ) = {z ∈ Rn : kz − x0 k < r}, EL2620 Lecture 1 Example: EL2620 x(t) 27 2012 25 2012 Physical Interpretation x(T ) = 0 x(0) = x0 Lecture 1 • f being C 1 implies Lipschitz continuity (L = maxx∈Br (x0 ) f ′ (x)) • f being C 0 is not sufficient (cf., tank example) • δ = δ(r, L) Remarks Appendix C.1. Based on the contraction mapping theorem has a unique solution in Br (x0 ) over [0, δ]. Proof: See Khalil, ẋ(t) = f (x(t)), > 0 such that Local Existence and Uniqueness dx dx =− ds dt = T − t ⇒ ds = −dt and thus Theorem: If f is Lipschitz continuous, then there exists δ EL2620 Lecture 1 Hint: Reverse time s 28 2012 26 2012 where x is the water level. It is then impossible to know at what time t < T the level was x(t) = x0 > 0. √ ẋ = − x, Consider the reverse example, i.e., the water tank lab process with EL2620 State-Space Models ẋ = f (x) + g(x)u, ẋ = Ax + Bu, Affine in u: Linear: dn y Transformation to First-Order System y = Cx y = h(x) Lecture 1 x = θ θ̇ T 2 = y dy dt ... dn−1 y dtn−1 ẋ2 = − k g x2 − sin x1 2 MR R ẋ1 = x2 gives M R θ̈ + k θ̇ + M gR sin θ = 0 Example: Pendulum express the equation in x T Given a differential equation in y with highest derivative dtn , EL2620 Lecture 1 ẋ = f (x, u), Explicit: y = h(x) f (x, u, y, ẋ, u̇, ẏ, . . .) = 0 General: State x, input u, output y EL2620 31 2012 29 2012 ẋ = f (x, t) Equilibria 2012 30 2012 f (x∗ , u∗ , y ∗ , 0, 0, . . .) = 0 0 = f (x∗ , u∗ ), y ∗ = h(x∗ ) 0 = f (x∗ ) + g(x∗ )u∗ , y ∗ = h(x∗ ) 0 = Ax∗ + Bu∗ , y ∗ = Cx∗ Lecture 1 Often the equilibrium is defined only through the state x∗ Linear: Affine in u: Explicit: General: Corresponds to putting all derivatives to zero: 32 Definition: A point (x∗ , u∗ , y ∗ ) is an equilibrium, if a solution starting in (x∗ , u∗ , y ∗ ) stays there forever. EL2620 Lecture 1 ẋ = f (x, xn+1 ) ẋn+1 = 1 is always possible to transform to an autonomous system by introducing xn+1 = t: A nonautonomous system Transformation to Autonomous System EL2620 Multiple Equilibria k g sin x1 x − 2 M R2 R Lecture 1 • When we want to push performance to the limit Lecture 1 • Phase plane analysis • When the range of operation is large Backlash Dead Zone Math Function eu Coulomb & Viscous Friction Look-Up Table Saturation Next Lecture • When distinctive nonlinear phenomena are relevant • Linearization EL2620 Lecture 1 Relay Sign Abs |u| Some Common Nonlinearities in Control Systems EL2620 • Simulation in Matlab 35 2012 33 2012 • When the system is strongly nonlinear, or not linearizable When do we need Nonlinear Analysis & Design? EL2620 Lecture 1 ẋ1 = ẋ2 = 0 gives x∗2 = 0 and sin(x∗1 ) = 0 ẋ2 = − ẋ1 = x2 Alternatively in first-order form: θ̈ = θ̇ = 0 gives sin θ = 0 and thus θ∗ = kπ M R2 θ̈ + k θ̇ + M gR sin θ = 0 Example: Pendulum EL2620 36 2012 34 2012 Lecture 2 EL2620 Nonlinear Control Analysis Through Simulation = f (t, x, u) =0 Lecture 2 • Spice, EMTP, ADAMS, gPROMS Special purpose simulation tools • Omsim, Dymola, Modelica http://www.modelica.org DAE’s F (t, ẋ, x, u) • ACSL, Simnon, Simulink ODE’s ẋ 3 2012 1 2012 • Wrap-up of Lecture 1: Nonlinear systems and phenomena • Modeling and simulation in Simulink • Phase-plane analysis Simulation tools: EL2620 Lecture 2 EL2620 Today’s Goal Lecture 2 > matlab >> simulink EL2620 Lecture 2 Simulink • Do phase-plane analysis using pplane (or other tool) • Linearize using Simulink • Model and simulate in Simulink You should be able to EL2620 4 2012 2 2012 An Example in Simulink Transfer Fcn Scope To Workspace2 u Transfer Fcn s+1 1 To Workspace To Workspace1 y Lecture 2 >> plot(t,y) 7 2012 5 2012 Check “Save format” of output blocks (“Array” instead of “Structure”) Step Clock t Save Results to Workspace stepmodel.mdl EL2620 Lecture 2 Step s+1 1 File -> New -> Model Double click on Continous Transfer Fcn Step (in Sources) Scope (in Sinks) Connect (mouse-left) Simulation -> Parameters EL2620 Choose Simulation Parameters How To Get Better Accuracy 2 3 4 5 6 7 8 9 10 -0.8 Lecture 2 -1 -0.6 -0.8 -1 -0.4 -0.6 0 -0.4 0.2 0 -0.2 0.4 0.2 -0.2 0.6 1 1 0.8 0.4 0 Refine = 1 0.6 0.8 1 Refine adds interpolation points: 0 1 2 3 4 5 Refine = 10 6 7 8 9 10 8 2012 6 2012 Modify Refine, Absolute and Relative Tolerances, Integration method EL2620 Lecture 2 Don’t forget “Apply” EL2620 Use Scripts to Document Simulations 1 s Lecture 2 1 In In 1 Sum h Subsystem In q Gain 1/A h q Subsystem2 In s Integrator 1 ḣ = (u − q)/A p √ q = a 2g h f(u) Fcn 1 Out h 2 q 1 Example: Two-Tank System The system consists of two identical tank models: EL2620 11 2012 -1 u2 (s+1)(s+1) 1 1 (s+1)2 Process y Linearization in Simulink s 1 −1 Valve y = u2 2012 10 2012 Lecture 2 12 >> A=2.7e-3;a=7e-6,g=9.8; >> [x0,u0,y0]=trim(’twotank’,[0.1 0.1]’,[],0.1) x0 = 0.1000 0.1000 u0 = 9.7995e-006 y0 = 0.1000 >> [aa,bb,cc,dd]=linmod(’twotank’,x0,u0); >> sys=ss(aa,bb,cc,dd); >> bode(sys) Linearize about equilibrium (x0 , u0 , y0 ): EL2620 Lecture 2 Σ Motor Lecture 2 r Example: Control system with valve characteristic f (u) Nonlinear Control System Simulink block diagram: PSfr EL2620 open_system(’stepmodel’) set_param(’stepmodel’,’RelTol’,’1e-3’) set_param(’stepmodel’,’AbsTol’,’1e-6’) set_param(’stepmodel’,’Refine’,’1’) tic sim(’stepmodel’,6) toc subplot(2,1,1),plot(t,y),title(’y’) subplot(2,1,2),plot(t,u),title(’u’) 9 2012 If the block-diagram is saved to stepmodel.mdl, the following Script-file simstepmodel.m simulates the system: EL2620 Differential Equation Editor deedemo1 deedemo3 Homework 1 deedemo2 DEE Differential Equation Editor deedemo4 Lecture 2 Write in English. • The report should be short and include only necessary plots. • See the course homepage for a report example report. 15 2012 13 2012 • Follow instructions in Exercise Compendium on how to write the • Use your favorite phase-plane analysis tool EL2620 Lecture 2 Run the demonstrations >> dee dee is a Simulink-based differential equation editor EL2620 Phase-Plane Analysis Lecture 2 This was the preferred tool last year! http://math.rice.edu/˜dfield • Down load DFIELD and PPLANE from http://www.control.lth.se/˜ictools • Download ICTools from EL2620 14 2012 Local Stability • Stability definitions • Linearization • Phase-plane analysis • Periodic solutions Lecture 3 EL2620 Nonlinear Control x(t) ⇒ δ ǫ kx(t)k < ǫ, ∀t ≥ 0 = 0 is not stable it is called unstable. Lecture 3 If x∗ kx(0)k < δ = f (x) with f (0) = 0 Definition: The equilibrium x∗ = 0 is stable if for all ǫ > 0 there exists δ = δ(ǫ) > 0 such that Consider ẋ EL2620 Lecture 3 EL2620 3 2012 1 2012 Asymptotic Stability ⇒ lim x(t) = 0 t→∞ Lecture 3 4 2012 2 2012 The equilibrium is globally asymptotically stable if it is stable and limt→∞ x(t) = 0 for all x(0). kx(0)k < δ Definition: The equilibrium x = 0 is asymptotically stable if it is stable and δ can be chosen such that EL2620 Lecture 3 • Analyze stability of periodic solutions through Poincaré maps center points • Classify equilibria into nodes, focuses, saddle points, and • Sketch phase portraits for two-dimensional systems • Linearize around equilibria and trajectories Today’s Goal • Explain local and global stability You should be able to EL2620 Linearization Around a Trajectory m(t) Lecture 3 2012 5 2012 7 = (h0 (t), v0 (t), m0 (t))T , u0 (t) ≡ u0 > 0, be a solution. ḣ(t) = v(t) v̇(t) = −g + ve u(t)/m(t) ṁ(t) = −u(t) Example 0 1 0 0 e u0 ˙ ve ũ(t) x̃(t) = 0 0 − (m0v−u 2 x̃(t) + m0 −u0 t 0 t) 1 0 0 0 Let x0 (t) Then, h(t) EL2620 Lecture 3 (x0 (t) + x̃(t), u0 (t) + ũ(t)) (x0 (t), u0 (t)) ẋ(t) = f (x0 (t) + x̃(t), u0 (t) + ũ(t)) ∂f (x0 (t), u0 (t))x̃(t) = f (x0 (t), u0 (t)) + ∂x ∂f + (x0 (t), u0 (t))ũ(t) + O(kx̃, ũk2 ) ∂u Let (x0 (t), u0 (t)) denote a solution to ẋ = f (x, u) and consider another solution (x(t), u(t)) = (x0 (t) + x̃(t), u0 (t) + ũ(t)): EL2620 ∂f (x0 (t), u0 (t)) ∂x ∂f B(x0 (t), u0 (t)) = (x0 (t), u0 (t)) ∂u A(x0 (t), u0 (t)) = −1 + α cos2 t 1 − α sin t cos t −1 − α sin t cos t −1 + α sin2 t Lecture 3 is unbounded solution for α > 1. 8 2012 6 2012 , α>0 √ α − 2 ± α2 − 4 λ(t) = λ = 2 which are stable for 0 < α < 2. However, (α−1)t e cos t e−t sin t x(t) = x(0), −e(α−1)t sin t e−t cos t Pointwise eigenvalues are given by A(t) = Pointwise Left Half-Plane Eigenvalues of A(t) Do Not Impose Stability EL2620 Lecture 3 Note that A and B are time dependent. However, if (x0 (t), u0 (t)) ≡ (x0 , u0 ) then A and B are constant. where ˙ x̃(t) = A(x0 (t), u0 (t))x̃(t) + B(x0 (t), u0 (t))ũ(t) Hence, for small (x̃, ũ), approximately EL2620 = ∂f (x0 ) ∂x and α(A) = 0 needs further investigation. x(t) = c1 e λ1 t v1 + c 2 e λ2 t v2 , = λ1 v1 etc). Lecture 3 where the constants c1 and c2 are given by the initial conditions This implies that where v1 , v2 are the eigenvectors of A (Av1 −1 e λ1 t 0 eAt = V eΛt V −1 = v1 v2 v v 1 2 0 e λ2 t If A is diagonalizable, then x(t) = eAt x(0), t ≥ 0. d x1 x =A 1 x2 dt x2 Linear Systems Revival Analytic solution: EL2620 Lecture 3 A proof is given next lecture. The theorem is also called Lyapunov’s Indirect Method. The case α(A) The fundamental result for linear systems theory! • If α(A) > 0, then x0 is unstable • If α(A) < 0, then x0 is asymptotically stable Denote A 1 = f (x) with f ∈ C . = max Re(λ(A)). Lyapunov’s Linearization Method Theorem: Let x0 be an equilibrium of ẋ EL2620 11 2012 9 2012 x(t) = c1 eλ1 t v1 + c2 eλ2 t v2 Lecture 3 Fast eigenvalue/vector: x(t) ≈ c1 eλ1 t v1 + c2 v2 for small t. Moves along the fast eigenvector for small t Slow eigenvalue/vector: x(t) ≈ c2 eλ2 t v2 for large t. Moves along the slow eigenvector towards x = 0 for large t Solution: Given the eigenvalues λ1 < λ2 < 0, with corresponding eigenvectors v1 and v2 , respectively. Example: Two real negative eigenvalues EL2620 Lecture 3 x0 is thus an asymptotically stable equilibrium for the nonlinear system. ẋ2 = cos x2 − x31 − 5x2 ẋ1 = −x21 + x1 + sin x2 Example = (1, 0)T is given by −1 1 A= , λ(A) = {−2, −4} −3 −5 at the equilibrium x0 The linearization of EL2620 12 2012 10 2012 center point Example—Unstable Focus unstable focus Re λi = 0 λ1 < 0 < λ2 saddle point 2012 Lecture 3 θ̇ = ω ṙ = σr 15 σ −ω ẋ = x, σ, ω > 0, λ1,2 = σ ± iω ω σ −1 σt iωt 1 1 1 1 e e 0 x(0) x(t) = eAt x(0) = −i i 0 eσt e−iωt −i i p In polar coordinates r = x21 + x22 , θ = arctan x2 /x1 (x1 = r cos θ , x2 = r sin θ ): EL2620 Lecture 3 stable focus Re λi > 0 Im λi 6= 0 : Re λi < 0 unstable node λ1 , λ2 > 0 stable node Im λi = 0 : λ1 , λ2 < 0 Six cases: The location of the eigenvalues λ(A) determines the characteristics of the trajectories. 13 2012 Phase-Plane Analysis for Linear Systems EL2620 Lecture 3 EL2620 x1 λ1,2 = 1 ± i Re λ Im λ x2 λ1,2 = 0.3 ± i Equilibrium Points for Linear Systems Lecture 3 EL2620 16 2012 14 2012 (λ1 , λ2 ) = (−1, −2) and v1 v2 −1 1 ẋ = x 0 −2 Lecture 3 Hint: Two cases; only one linear independent eigenvector or all vectors are eigenvectors = λ2 ? 1 −1 = 0 1 5 minute exercise: What is the phase portrait if λ1 EL2620 Lecture 3 Example—Stable Node v1 is the slow direction and v2 is the fast. EL2620 19 2012 17 2012 + c3 2012 18 2012 ẋ = f (x) = Ax + g(x), Lecture 3 Remark: If the linearized system has a center, then the nonlinear system has either a center or a focus. with limkxk→0 kg(x)k/kxk = 0. If ż = Az has a focus, node, or saddle point, then ẋ = f (x) has the same type of equilibrium at the origin. Theorem: Assume Close to equilibrium points “nonlinear system” ≈ “linear system” 20 Phase-Plane Analysis for Nonlinear Systems EL2620 Lecture 3 Fast: x2 = −x1 Slow: x2 = 0 EL2620 How to Draw Phase Portraits Lecture 3 + nπ, 0) since x2 (t) + KT −1 sin(θin − x1 (t)) ẋ1 = 0 ⇒ x2 = 0 ẋ2 = 0 ⇒ sin(θin − x1 ) = 0 ⇒ x1 = θin + nπ Equilibria are (θin ẋ2 (t) = −T −1 ẋ1 (t) = x2 (t) = (θout , θ̇out ), K, T > 0, and θin (t) ≡ θin . Phase-Plane Analysis of PLL dee or pplane Let (x1 , x2 ) EL2620 Lecture 3 1. Matlab: By computer: 5. Guess solutions 4. Try to find possible periodic orbits dx2 ẋ1 = dx1 ẋ2 3. Sketch (ẋ1 , ẋ2 ) for some other points. Notice that 2. Sketch local behavior around equilibria 1. Find equilibria By hand: EL2620 23 2012 21 2012 sin(·) K θ̇out 1 + sT Filter 1 s VCO Classification of Equilibria − e Phase Detector λ2 + T −1 λ + KT −1 = 0 Lecture 3 >0 λ2 + T −1 λ − KT −1 = 0 Saddle points for all K, T n odd: K > (4T )−1 gives stable focus 0 < K < (4T )−1 gives stable node n even: Linearization gives the following characteristic equations: EL2620 Lecture 3 θin sin θout “θout ” = A sin[ωt + θin (t)]. Phase-Locked Loop A PLL tracks phase θin (t) of a signal sin (t) EL2620 24 2012 22 2012 Phase-Plane for PLL Lecture 3 Only r gives ṙ θ̇ 1 = r r cos θ r sin θ − sin θ cos θ θ̇ = 1 ṙ = r(1 − r2 ) ẋ2 = r(1 − r ) sin θ + r cos θ 2 ẋ1 ẋ2 ẋ2 = sin θṙ + r cos θθ̇ ẋ1 = cos θṙ − r sin θθ̇ ẋ1 = r(1 − r2 ) cos θ − r sin θ ⇒ ⇒ = 1 is a stable equilibrium! Now, from (1) x2 = r sin θ x1 = r cos θ Periodic solution: Polar coordinates. implies EL2620 Lecture 3 (K, T ) = (1/2, 1): focuses 2kπ, 0 , saddle points (2k + 1)π, 0 EL2620 27 2012 25 2012 Periodic Solutions ∀t ≥ 0 >0 Lecture 3 Note that x(t) ≡ const is by convention not regarded periodic • When is it stable? • When does there exist a periodic solution? A periodic orbit is the image of x in the phase portrait. x(t + T ) = x(t), A system has a periodic solution if for some T EL2620 Lecture 3 ẋ2 = x1 + x2 − x2 (x21 + x22 ) ẋ1 = x1 − x2 − x1 (x21 + x22 ) Example of an asymptotically stable periodic solution: EL2620 28 2012 26 (1) 2012 φt (x0 ) = f (x) is sometimes denoted Flow P (x∗ ) = x∗ = x∗ corresponds to a periodic orbit. x∗ is called a fixed point of P . Lecture 3 ∈ Rn Existence of Periodic Orbits A point x∗ such that P (x∗ ) EL2620 Lecture 3 φt (·) is called the flow. to emphasize the dependence on the initial point x0 The solution of ẋ EL2620 31 2012 29 2012 Poincaré Map x Lecture 3 32 2012 30 2012 1 minute exercise: What does a fixed point of P k corresponds to? EL2620 Lecture 3 P (x) Σ φt (x) P (x) = φτ (x) (x) where τ (x) is the time of first return. Let Σ ⊂ Rn be an n − 1-dim hyperplane transverse to f at x0 . Definition: The Poincaré map P : Σ → Σ is Assume φt (x0 ) is a periodic solution with period T . EL2620 Stable Periodic Orbit e−4π 0 0 1 Lecture 3 ⇒ Stable periodic orbit (as we already knew for this example) ! dP W = (1, 2πk) = d(r0 , θ0 ) The periodic solution that corresponds to (r(t), θ(t)) asymptotically stable because = (1, t) is [1 + (r0−2 − 1)e−2·2π ]−1/2 θ0 + 2π (r0 , θ0 ) = (1, 2πk) is a fixed point. P (r0 , θ0 ) = The Poincaré map is EL2620 Lecture 3 • If |λi (W )| > 1 for some i, then the periodic orbit is unstable. orbit is asymptotically stable • If |λi (W )| < 1 for all i 6= j , then the corresponding periodic • λj (W ) = 1 for some j if x is close to x∗ . P (x) ≈ W x The linearization of P around x∗ gives a matrix W such that EL2620 35 2012 33 2012 [1 + (r0−2 Lecture 3 −2t −1/2 τ (r0 , θ0 ) = 2π. ∈ Σ is − 1)e First return time from any point (r0 , θ0 ) φt (r0 , θ0 ) = = {(r, θ) : r > 0, θ = 2πk}. The solution is Choose Σ θ̇ = 1 ṙ = r(1 − r2 ) ] , t + θ0 Example—Stable Unit Circle Rewrite (1) in polar coordinates: EL2620 34 2012 • Lyapunov methods for stability analysis Lecture 4 EL2620 Nonlinear Control Lecture 4 If the total energy is dissipated, the system must be stable. Formalized the idea: Doctoral thesis “The general problem of the stability of motion,” 1892. Master thesis “On the stability of ellipsoidal forms of equilibrium of rotating fluids,” St. Petersburg University, 1884. 3 2012 1 2012 Alexandr Mihailovich Lyapunov (1857–1918) EL2620 Lecture 4 EL2620 spring + Fspring ds V (0, 0) = 0 0 Rx 4 2012 2 2012 d V (x, ẋ) = mẋẍ + k0 xẋ + k1 x3 ẋ = −b|ẋ|3 < 0, ẋ 6= 0 dt • Change in energy along any solution x(t) Lecture 4 mv 2 2 b, k0 , k1 > 0 V (x, ẋ) = mẋ2 /2 + k0 x2 /2 + k1 x4 /4 > 0, • Total energy = kinetic + potential energy: V = damping mẍ = − bẋ|ẋ| − k0 x − k1 x3 , | {z } | {z } x m A Motivating Example • Balance of forces yields EL2620 Lecture 4 LaSalle’s invariant set theorem • Prove stability of a set (e.g., a periodic orbit) using Lyapunov’s method Today’s Goal • Prove local and global stability of equilibria using You should be able to EL2620 ⇒ kx(t)k < ǫ, ∀t ≥ 0 ⇒ lim x(t) = 0 t→∞ Lyapunov Function Lecture 4 V̇ (x) = x1 V x2 ∂V ∂V d V (x) = ẋ = f (x) ≤ 0 dt ∂x ∂x Condition (3) means that V is non-increasing along all trajectories in Ω: A function V that fulfills (1)–(3) is called a Lyapunov function. EL2620 Lecture 4 Globally asymptotically stable, if asymptotically ∀x(0) kx(0)k < δ ∈ Rn . > 0 there exists δ > 0 such that Locally asymptotically stable, if locally stable and kx(0)k < δ Locally stable, if for every ǫ = 0 of ẋ = f (x) is Stability Definitions Recall from Lecture 3 that an equilibrium x EL2620 7 2012 5 2012 ∂V f (x) ∂x ∂V ∂x to the level surface = 0, i.e. the vector field 2012 6 2012 ∂V f (x) ∂x ≤ 0, i.e. the vector field to the level surface V (x) = c make an V̇ (x) = Lecture 4 Example: Total energy of a mechanical system with damping or total fluid in a system that leaks. f (x) and the normal obtuse angle. ∂V ∂x Dissipation of energy: 8 Example: Total energy of a lossless mechanical system or total fluid in a closed system. f (x) is everywhere orthogonal to the normal V (x) = c. V̇ (x) = Conservation and Dissipation Conservation of energy: EL2620 Lecture 4 The result is called Lyapunov’s Direct Method = 0 is locally asymptotically stable. V̇ (x) < 0 for all x ∈ Ω, x 6= 0 then x (4) = 0 is locally stable. Furthermore, if V̇ (x) ≤ 0 for all x ∈ Ω (3) then x V (x) > 0 for all x ∈ Ω, x 6= 0 V (0) = 0 (2) (1) ∈ Ω ⊂ Rn . If there Lyapunov Stability Theorem Theorem: Let ẋ = f (x), f (0) = 0, and 0 exists a C1 function V : Ω → R such that EL2620 V (x) =constant f (x) 0 10 1 2 3 4 x1 0 -10 -2 0 x2 2 V (x) = (1 − cos x1 )g/ℓ + x22 /2 g ẋ1 = x2 , ẋ2 = − sin x1 ℓ Lyapunov function candidate: Lecture 4 x(t) Example—Pendulum Is the origin stable for a mathematical pendulum? EL2620 Lecture 4 Trajectories can only go to lower values of V (x) dV dx Geometric interpretation Vector field points into sublevel sets EL2620 11 2012 9 2012 0 t V̇ (x(τ ))dτ ≤ V (x(0)) Conservation of energy! Lecture 4 12 2012 10 2012 0 is not asymptotically stable, so, of course, (4) is not 6< 0, ∀x 6= 0. = 0 is locally stable. Note that x = fulfilled: V̇ (x) Hence, x V̇ (x) = gℓ ẋ1 sin x1 + x2 ẋ2 = 0, ∀x V (x) > 0 for −2π < x1 < 2π and (x1 , x2 ) 6= 0 (2) (3) V (0) = 0 (1) EL2620 Lecture 4 For stability it is thus important that the sublevel sets {z | V (z) ≤ c)} are locally bounded. {z | V (z) ≤ V (x(0))} which means that the whole trajectory lies in the set Z Boundedness: V ((x(t)) = V (x(0)) + For an trajectory x(t) EL2620 Radial Unboundedness is Necessary 2012 13 Lecture 4 -2 -10 -1.5 -1 -0.5 0 0.5 1 1.5 2 -8 Example: Assume V (x) -6 -4 -2 0 x1 2 4 6 8 10 15 = x21 /(1 + x21 ) + x22 is a Lyapunov function for some system. Then might x(t) → ∞ even if V̇ (x) < 0, as shown by the contour plot of V (x): If (4) is not fulfilled, then global stability cannot be guaranteed. EL2620 Lecture 4 V (x) > 0 for all x 6= 0 V̇ (x) ≤ −αV (x) for all x (2) (3) Lecture 4 then x ≤ c} are bounded for all c ≥ 0 = 0 is globally asymptotically stable. (4) The sublevel sets {x|V (x) V (0) = 0 (1) 2012 14 16 = 0. If there exists a C1 function Somewhat Stronger Assumptions = 0 is globally asymptotically stable. Theorem: Let ẋ = f (x) and f (0) V : Rn → R such that EL2620 Lecture 4 then x V (x) → ∞ as kxk → ∞ V̇ (x) < 0 for all x 6= 0 (3) V (x) = xT x (4) V (x) > 0 for all x 6= 0 (2) Can you say something about global stability of the equilibrium? V (0) = 0 2012 = 0. If there exists a C1 function Lyapunov Theorem for Global Asymptotic Stability Theorem: Let ẋ = f (x) and f (0) V : Rn → R such that EL2620 (1) 2012 For what values of ǫ is the steady-state (0, 0) locally stable? Hint: try the ”standard” Lyapunov function ẋ2 = −x1 (t) − ǫx21 (t)x2 (t) ẋ1 = x2 (t) 5 minute exercise: Consider Example 2 from Lecture 3: EL2620 x2 2012 → 0, t → ∞. Positive Definite Matrices → 0, t → ∞. 2012 17 Lecture 4 19 Note that if M = M T is positive definite, then the Lyapunov function candidate V (x) = xT M x fulfills V (0) = 0 and V (x) > 0, ∀x 6= 0. • M = M T is positive semidefinite ⇐⇒ λi (M ) ≥ 0, ∀i • M = M T is positive definite ⇐⇒ λi (M ) > 0, ∀i Lemma: Definition: A matrix M is positive definite if xT M x > 0 for all x 6= 0. It is positive semidefinite if xT M x ≥ 0 for all x. EL2620 Lecture 4 Using the properties of V it follows that x(t) Hence, V (x(t)) log V (x(t)) − log V (x(0) ≤ −αt ⇒ V (x(t)) ≤ e−αt V (x(0)) V̇ (x) ≤ −α V (x) 6= 0 (otherwise we have x(τ ) = 0 for all τ > t). Then Proof Idea Integrating from 0 to t gives Assume x(t) EL2620 Converse Lyapunov theorems = M T . Then Symmetric Matrices Lecture 4 20 2012 18 2012 Hint: Use the factorization M = U ΛU T , where U is an orthogonal matrix (U U T = I ) and Λ = diag(λ1 , . . . , λn ). λmin (M )kxk2 ≤ xT M x ≤ λmax (M )kxk2 Assume that M EL2620 Lecture 4 There exist also Lyapunov instability theorems! then there is a Lyapunov function that proves that it is globally asymptotically stable. ||x(t)|| ≤ M e−βt ||x(0)||, M > 0, β > 0 Example: If the system is globally exponentially stable EL2620 P A + AT P = −Q < 0 ∀t if there exist a Q = QT > 0 such that Q t Lecture 4 23 Lecture 4 Theorem: Let x0 be an equilibrium of ẋ = f (x) with f ∂f Denote A = ∂x (x0 ) and α(A) = max Re(λ(A)). > 0 then x0 is unstable Qe dt z = z P z At (2) If α(A) e AT t Recall from Lecture 3: ∈ C1 . Lyapunov’s Linearization Method < 0 then x0 is asymptotically stable 0 ∞ EL2620 Lecture 4 24 2012 22 R∞ T = 0 eA t QeAt dt. Then Z ∞ T AT t At AT t At T A e Qe + e Qe A dt A P + PA = 0 Z ∞ h T i∞ d AT t At A t At = e Qe = −Q dt = e Qe dt 0 0 Proof: Choose P P A + AT P = −Q (1) If α(A) 0 = Ax, x(0) = z . Then Z ∞ T t x (t)Qx(t)dt = z 2012 21 Converse Theorem for Linear Systems 2012 If Re λi (A) < 0, then for every symmetric positive definite Q there exist a symmetric positive definite matrix P such that EL2620 Thus v(z) = z T P z is the cost-to-go from the point z (no input) with integral quadratic cost function using weighting matrix Q. Z Assume ẋ EL2620 Interpretation Global Asymptotic Stability: If Q is positive definite, then the Lyapunov Stability Theorem implies global asymptotic stability, and hence the eigenvalues of A must satisfy Re λi (A) < 0 for all i Lecture 4 2012 V̇ (x) = xT P ẋ + ẋT P x = xT (P A + AT P ) x = −xT Qx | {z } Thus, V̇ ⇒ V (x) = xT P x, P = P T > 0 Lyapunov equation: Consider the quadratic function ẋ = Ax Lyapunov Stability for Linear Systems Linear system: EL2620 T T xT Qx ≥ λmin (Q)kxk2 = −xT Qx + 2xT P g(x) = x (P A + A P )x + 2x P g(x) T = xT P [Ax + g(x)] + [xT AT + g T (x)]P x V (x) > 0 for all x 6= 0 V̇ (x) ≤ 0 for all x (2) (3) Lecture 4 then x = f (x) such that V̇ (x) = 0 is x(t) = 0 27 = 0. If there exists a C1 function = 0 is globally asymptotically stable. (5) The only solution of ẋ for all t V (x) → ∞ as kxk → ∞ V (0) = 0 (1) Theorem: Let ẋ = f (x) and f (0) V : Rn → R such that (4) 2012 25 2012 LaSalle’s Theorem for Global Asymptotic Stability EL2620 Lecture 4 • we need to show that k2xT P g(x)k < λmin (Q)kxk2 where T V̇ (x) = x P f (x) + f (x)P x T Let f (x) = Ax + g(x) where limkxk→0 kg(x)k/kxk = 0. The Lyapunov function candidate V (x) = xT P x satisfies V (0) = 0, V (x) > 0 for x 6= 0, and Proof of (1) in Lyapunov’s Linearization EL2620 ∀kxk < r V̇ < −λmin (Q)kxk2 + 2γλmax (P )kxk2 kg(x)k < γkxk, > 0 there exists r > 0 such that Lecture 4 Hence, x(t) = 0 is the only possible trajectory for which V̇ (x) and the LaSalle theorem gives global asymptotic stability. which means that ẋ(t) can not stay constant. 28 2012 26 2012 = 0, = 0, x(t) 6= 0. Then d k0 k1 ẋ(t) = − x(t) − x3 (t) 6= 0, dt m m Assume that there is a trajectory with ẋ(t) V̇ (x) = −b|ẋ| 3 V (x) = (2mẋ2 + 2k0 x2 + k1 x4 )/4 > 0, V (0, 0) = 0 A Motivating Example (cont’d) 1 λmin (Q) 2 λmax (P ) mẍ = −bẋ|ẋ| − k0 x − k1 x3 EL2620 Lecture 4 γ< which becomes strictly negative if we choose Thus, For all γ EL2620 Invariant Sets M = f (x), if 2012 Lecture 4 V̇ (x) = 31 d ∂V f (x) = 2(x21 + x22 − 1) (x21 + x22 − 1) ∂x dt 2 2 = −2(x1 + x2 − 1)2 (x21 + x22 ) ≤ 0, ∀x ∈ Ω It is possible to show that Ω = {kxk ≤ R} is invariant for sufficiently large R > 0. Let V (x) = (x21 + x22 − 1)2 . ẋ2 = x1 + x2 − x2 (x21 + x22 ) ẋ1 = x1 − x2 − x1 (x21 + x22 ) : kxk = 1} ∪ {0} for Example—Periodic Orbit Show that x(t) approaches {x EL2620 Lecture 4 29 2012 Definition: x(t) approaches a set M as t → ∞, if for each ǫ > 0 there is a T > 0 such that dist(x(t), M ) < ǫ for all t > T . Here dist(p, M ) = inf x∈M kp − xk. x(0) x(t) Definition: A set M is invariant with respect to ẋ x(0) ∈ M implies that x(t) ∈ M for all t ≥ 0. EL2620 LaSalle’s Invariant Set Theorem 0 E M M V̇ (x) x 2012 Lecture 4 32 d 2 (x + x22 − 1) = −2(x21 + x22 − 1)(x21 + x22 ) = 0 for x ∈ M dt 1 = E because E = {x ∈ Ω : V̇ (x) = 0} = {x : kxk = 1} ∪ {0} The largest invariant set of E is M EL2620 Lecture 4 Note that V does not have to be a positive definite function. Ω E 30 2012 for x ∈ Ω. Let E be the set of points in Ω where V̇ (x) = 0. If M is the largest invariant set in E , then every solution with x(0) ∈ Ω approaches M as t → ∞. Theorem: Let Ω ⊂ Rn be a compact set invariant with respect to ẋ = f (x). Let V : Rn → R be a C 1 function such that V̇ (x) ≤ EL2620 − f (·) G(s) y S History u y • Input–output stability Lecture 5 EL2620 Nonlinear Control Lecture 5 • Solution by Popov (1960) • Kalman’s conjecture (1957) (False!) (False!) y f (y) (Led to the Circle Criterion) • Aizerman’s conjecture (1949) • Luré and Postnikov’s problem (1944) For what G(s) and f (·) is the closed-loop system stable? r EL2620 Lecture 5 EL2620 3 2012 1 2012 2012 S y Lecture 5 Question: How should we measure the size of u and y ? 4 Here S can be a constant, a matrix, a linear time-invariant system, etc The gain γ of S is the largest amplification from u to y u 2 2012 Idea: Generalize the concept of gain to nonlinear dynamical systems EL2620 Lecture 5 – Passivity – Circle Criterion – Small Gain Theorem • analyze stability using Gain Today’s Goal • derive the gain of a system You should be able to EL2620 and + Norms x21 + ··· + Lecture 5 Why? What amplification is described by the eigenvalues? is not a gain (nor a norm). ρ(M ) = max |λi (M )| i Eigenvalues are not gains kxk = max{|x1 |, . . . , |xn |} kxk = p The spectral radius of a matrix M EL2620 Lecture 5 Max norm: Euclidean norm: Examples: • kαxk = |α| · kxk, for all α ∈ R x2n kxk = 0 ⇔ x = 0 : Ω → R , such that for all x, y ∈ Ω • kx + yk ≤ kxk + kyk • kxk ≥ 0 Definition: A norm is a function k · k A norm k · k measures size EL2620 7 2012 5 2012 Σ = diag {σ1 , . . . , σn } ; U ∗ U = I ; V ∗ V = I M = U ΣV ∗ ∈ Cn×n has a singular value decomposition Gain of a Matrix kM xk kxk Lecture 5 sup-norm: kxk2 = 0 qR ∞ kxk∞ = supt∈R+ |x(t)| 2-norm (energy norm): Examples: |x(t)|2 dt A signal norm k · kk is a norm on the space of signals x. : R+ → R. Signal Norms A signal x is a function x EL2620 Lecture 5 where k · k is the Euclidean norm. x∈Rn σmax (M ) = σ1 = sup The “gain” of M is the largest singular value of M : σi - the singular values where Every matrix M EL2620 8 2012 6 2012 Parseval’s Theorem Z 0 ∞ = Z 0 ∞ Z −∞ ∞ 1 |x(t)| dt = 2π 2 1 y(t)x(t)dt = 2π Z −∞ ∞ |X(iω)|2 dω. Y ∗ (iω)X(iω)dω. 0 Lecture 5 u S2 y ≤ γ(S1 )γ(S2 ). S1 2 minute exercise: Show that γ(S1 S2 ) EL2620 Lecture 5 11 2012 The power calculated in the time domain equals the power calculated in the frequency domain kxk22 In particular, then 0 ∈ L2 have the Fourier transforms Z ∞ Z ∞ −iωt e−iωt y(t)dt, e x(t)dt, Y (iω) = X(iω) = Theorem: If x, y 9 2012 L2 denotes the space of signals with bounded energy: kxk2 < ∞ EL2620 u∈L2 γ(S) = sup S f (·) u∈L2 y(t) x∗ Kx x f (x) ≤ K|x| and Gain of a Static Nonlinearity Lecture 5 R∞ R∞ kyk22 = 0 f 2 u(t) dt ≤ 0 K 2 u2 (t)dt = K 2 kuk22 , where u(t) = x∗ , t ∈ (0, 1), gives equality, so γ(f ) = sup kyk2 kuk2 = K Proof: u(t) = αu(t) is |α|kuk2 kαuk2 = sup = |α| kuk2 u∈L2 kuk2 Lemma: A static nonlinearity f such that |f (x)| f (x∗ ) = Kx∗ has gain γ(f ) = K . EL2620 Lecture 5 u∈L2 γ(α) = sup 12 2012 10 2012 kyk2 kS(u)k2 = sup kuk2 u∈L2 kuk2 y Example: The gain of a scalar static system y(t) The gain of S is defined as u y = S(u). System Gain A system S is a map from L2 to L2 : EL2620 γ(G) Z ∞ 1 = |Y (iω)|2 dω 2π −∞ Z ∞ 1 = |G(iω)|2 |U (iω)|2 dω ≤ K 2 kuk22 2π −∞ r1 e1 S2 S1 e2 r2 The Small Gain Theorem Lecture 5 13 15 2012 Theorem: Assume S1 and S2 are BIBO stable. If γ(S1 )γ(S2 ) < 1, then the closed-loop system is BIBO stable from (r1 , r2 ) to (e1 , e2 ) EL2620 Lecture 5 1 10 2012 (0, ∞) and |G(iω ∗ )| = K 0 10 |G(iω)| Arbitrary close to equality by choosing u(t) close to sin ω ∗ t. kyk22 Proof: Assume |G(iω)| ≤ K for ω ∈ for some ω ∗ . Parseval’s theorem gives 10 -1 10 -2 -1 10 0 10 10 1 Gain of a Stable Linear System kGuk2 γ G = sup = sup |G(iω)| u∈L2 kuk2 ω∈(0,∞) Lemma: EL2620 BIBO Stability S y f (·) 0≤ Lecture 5 Ky y f (y) ∈ (0, 1/2). 16 2012 14 2012 f (y) ≤ K, ∀y 6= 0, f (0) = 0 y Small Gain Theorem gives BIBO stability for K γ(G) = 2 and γ(f ) ≤ K . 2 , (s + 1)2 − G(s) y Example—Static Nonlinear Feedback G(s) = r EL2620 Lecture 5 < ∞. kS(u)k2 γ S = sup kuk2 u∈L2 = Ax is asymptotically stable then G(s) = C(sI − A)−1 B + D is BIBO stable. Example: If ẋ u Definition: S is bounded-input bounded-output (BIBO) stable if γ(S) EL2620 ke1 k2 ≤ kr1 k2 + γ(S2 )kr2 k2 1 − γ(S2 )γ(S1 ) ke1 k2 ≤ kr1 k2 + γ(S2 )[kr2 k2 + γ(S1 )ke1 k2 ] “Proof” of the Small Gain Theorem kr2 k2 + γ(S1 )kr1 k2 ke2 k2 ≤ 1 − γ(S1 )γ(S2 ) 2012 17 2012 Lecture 5 r − f (·) G(s) y -1 -0.5 -1 0 0.5 1 -0.5 0 0.5 1 1.5 G(iω) 2 19 Let f (y) = Ky in the previous example. Then the Nyquist Theorem proves stability for all K ∈ [0, ∞), while the Small Gain Theorem only proves stability for K ∈ (0, 1/2). Small Gain Theorem can be Conservative EL2620 Lecture 5 Note: Formal proof requires k · k2e , see Khalil so also e2 is bounded. Similarly we get γ(S2 )γ(S1 ) < 1, kr1 k2 < ∞, kr2 k2 < ∞ give ke1 k2 < ∞. gives EL2620 − G(s) -1 -1 -0.5 0 0.5 1 -0.5 The Nyquist Theorem 0 From: U(1) 0.5 1 − f (·) G(s) y y k1 y k2 y f (y) − k11 The Circle Criterion − k12 G(iω) 2012 18 2012 f (y) ≤ k2 , ∀y 6= 0, y f (0) = 0 Lecture 5 If the Nyquist curve of G(s) does not encircle or intersect the circle defined by the points −1/k1 and −1/k2 , then the closed-loop system is BIBO stable. 0 ≤ k1 ≤ 20 Theorem: Assume that G(s) has no poles in the right half plane, and r EL2620 Lecture 5 Theorem: If G has no poles in the right half plane and the Nyquist curve G(iω), ω ∈ [0, ∞), does not encircle −1, then the closed-loop system is stable. EL2620 To: Y(1) 2012 y f (y) -1 -0.5 -0.5 1 − K -1 0 0.5 1 0 0.5 1 2 −fe(·) e G(s) r2 R −k 1 G(iω) Lecture 5 e |G(iω)| 1 = 1 +k >R G(iω) 23 2012 21 ∈ (0, 4). < 1, where G e= is stable (This has to be checked later). Hence, G 1 + kG e Small Gain Theorem gives stability if |G(iω)|R re1 EL2620 Lecture 5 the Circle Criterion gives that the system is BIBO stable if K = −1/K . 1.5 G(iω) The “circle” is defined by −1/k1 = −∞ and −1/k2 Since min Re G(iω) = −1/4 Ky Example—Static Nonlinear Feedback (cont’d) EL2620 y2 e1 −f (·) G(s) e2 y1 r2 re1 1 G(iω) − k11 − k12 Lecture 5 r2 y1 G(iω) G is stable since −1/k is inside the circle. 1 + kG R − k1 Note that G(s) may have poles on the imaginary axis, e.g., integrators are allowed Note that −k2 −fe(·) G −k fe(0) = 0 2012 22 2012 24 ∈ C : |z + k| > R} mapped e G fe(y) k2 − k1 =: R, ∀y 6= 0, ≤ y 2 = (k1 + k2 )/2, fe(y) = f (y) − ky , and re1 = r1 − kr2 : Proof of the Circle Criterion The curve G−1 (iω) and the circle {z through z 7→ 1/z gives the result: EL2620 Lecture 5 r1 Let k EL2620 Passivity and BIBO Stability u S y Passive System > 0 such that Lecture 5 Warning: There exist many other definitions for strictly passive hy, uiT ≥ ǫ(|y|2T + |u|2T ), for all T > 0 and all u and strictly passive if there exists ǫ 27 2012 Definition: Consider signals u, y : [0, T ] → Rm . The system S is passive if hy, uiT ≥ 0, for all T > 0 and all u EL2620 Lecture 5 25 2012 The main result: Feedback interconnections of passive systems are passive, and BIBO stable (under some additional mild criteria) EL2620 y T (t)u(t)dt S y cos φ = hy, uiT =0 |y|T |u|T u = sin t and y = cos t are orthogonal if T = kπ , hy, uiT ≤ |y|T |u|T Lecture 5 2 minute exercise: Is the pure delay system y(t) = u(t − θ) passive? Consider for instance the input u(t) = sin ((π/θ)t). EL2620 Lecture 5 Example: because Cauchy-Schwarz Inequality: 28 2012 26 2012 = |y|T |u|T cos φ p |y|T = hy, yiT - length of y , φ - angle between u and y 0 T If u and y are interpreted as vectors then hy, uiT hy, uiT = Z u Scalar Product Scalar product for signals y and u EL2620 − ǫ) ≥ 0, Lecture 5 passive. 1 is strictly passive, G(s) = s+1 1 is passive but not strictly G(s) = s Example: Re G(iω -0.4 -0.2 0 0.2 0.4 0.6 0 0.2 0.4 0.6 G(iω) 0.8 1 > 0 such that ∀ω > 0 It is strictly passive if and only if there exists ǫ 31 2012 du Cu2 (T ) dt = ≥0 dt 2 di Li2 (T ) L i(t)dt = ≥0 dt 2 u(t)C Ri2 (t)dt ≥ Rhi, iiT ≥ 0 Passivity of Linear Systems T T 0 T Theorem: An asymptotically stable linear system G(s) is passive if and only if Re G(iω) ≥ 0, ∀ω > 0 EL2620 Lecture 5 Z di u = L : hu, iiT = dt 0 0 Z du : hu, iiT = i=C dt u(t) = Ri(t) : hu, iiT = Z 29 2012 Example—Passive Electrical Components EL2620 y2 − e1 S2 S1 e2 y1 r2 = hy1 , r1 iT + hy2 , r2 iT = hy, riT ≥ 0 if hy1 , e1 iT ≥ 0 and hy2 , e2 iT ≥ 0. S = supu∈L2 y kyk2 kuk2 < ∞. Lecture 5 Hence, ǫkyk22 1 kyk2 ≤ kuk2 ǫ ≤ kyk2 · kuk2 , so ǫ(|y|2∞ + |u|2∞ ) ≤ hy, ui∞ ≤ |y|∞ · |u|∞ = kyk2 · kuk2 Proof: Lemma: If S is strictly passive then γ(S) u 32 2012 A Strictly Passive System Has Finite Gain EL2620 Lecture 5 Hence, hy, riT hy, eiT = hy1 , e1 iT + hy2 , e2 iT = hy1 , r1 − y2 iT + hy2 , r2 + y1 iT Proof: 30 2012 Lemma: If S1 and S2 are passive then the closed-loop system from (r1 , r2 ) to (y1 , y2 ) is also passive. r1 Feedback of Passive Systems is Passive EL2620 r1 y2 − e1 S2 S1 e2 y1 r2 The Passivity Theorem y2 − e1 S2 S1 e2 y1 r2 φ1 The Passivity Theorem is a “Small Phase Theorem” φ2 Lecture 5 S2 passive ⇒ cos φ2 ≥ 0 ⇒ |φ2 | ≤ π/2 S1 strictly passive ⇒ cos φ1 > 0 ⇒ |φ1 | < π/2 r1 EL2620 Lecture 5 Theorem: If S1 is strictly passive and S2 is passive, then the closed-loop system is BIBO stable from r to y . EL2620 35 2012 33 2012 Proof 1 ≤ 2hy2 , r2 iT + hy, riT ≤ ǫ − G(s) → ∞ and the result follows. |y|2T 1 2+ |y|T |r|T ǫ 1 |y1 |2T + |y2 |2T − 2hy2 , r2 iT + |r1 |2T ≤ hy, riT ǫ 1 |y1 |2T + hr1 − y2 , r1 − y2 iT ≤ hy, riT ǫ 2012 34 2012 Lecture 5 36 2 minute exercise: Apply the Passivity Theorem and compare it with the Nyquist Theorem. What about conservativeness? [Compare the discussion on the Small Gain Theorem.] EL2620 Lecture 5 Let T Hence or Therefore S1 strictly passive and S2 passive give ǫ |y1 |2T + |e1 |2T ≤ hy1 , e1 iT + hy2 , e2 iT = hy, riT EL2620 Example—Gain Adaptation ∗ u − c s G(s) − θ θ (θ − θ∗ )u S ym − y Gain Adaptation is BIBO Stable c > 0. ym y Lecture 5 S is passive (see exercises). If G(s) is strictly passive, the closed-loop system is BIBO stable EL2620 Lecture 5 G(s) θ(t) Model G(s) θ∗ Process dθ = −cu(t)[ym (t) − y(t)], dt Adaptation law: u Applications in telecommunication channel estimation and in noise cancellation etc. EL2620 39 2012 37 2012 Lecture 5 Let G(s) EL2620 Lecture 5 u G(s) G(s) ym − y − c θ s = 0 0 0 0.5 1 1.5 -2 0 2 5 5 10 10 θ y , ym 15 15 20 20 1 , c = 1, u = sin t, and θ(0) = 0. s+1 Simulation of Gain Adaptation θ(t) θ∗ Gain Adaptation—Closed-Loop System replacements EL2620 40 2012 38 2012 ∀x, u ∀x 6= 0 : R → R such that n Lyapunov vs. Passivity Lecture 5 V̇ ≤ uT y Passivity idea: “Increase in stored energy ≤ Added energy” V̇ ≤ 0 Lyapunov idea: “Energy is decreasing” Storage function is a generalization of Lyapunov function EL2620 Lecture 5 absorbed energy 43 2012 41 2012 • V (T ) represents the stored energy in the system Z T y(t)u(t)dt + V (x(0)) , ∀T > 0 • V (x(T )) ≤ | {z } | {z } 0 | {z } stored energy at t = 0 stored energy at t = T Remark: • V̇ (x) ≤ uT y , • V (0) = 0 and V (x) ≥ 0, A storage function is a C function V 1 ẋ = f (x, u), y = h(x) Storage Function Consider the nonlinear control system EL2620 Storage Function and Passivity hy, uiT = Z 0 T Example—KYP Lemma ẋ = Ax + Bu, y = Cx BT P = C Lecture 5 and hence the system is strictly passive. This fact is part of the Kalman-Yakubovich-Popov lemma. = −0.5xT Qx + uy < uy, x 6= 0 42 44 2012 V̇ = 0.5(ẋT P x + xT P ẋ) = 0.5xT (AT P + P A)x + uB T P x = 0.5xT P x. Then AT P + P A = −Q, Assume there exists positive definite matrices P, Q such that Consider V 2012 y(t)u(t)dt ≥ V (x(T ))−V (x(0)) = V (x(T )) ≥ 0 > 0, Consider an asymptotically stable linear system EL2620 Lecture 5 y = h(x) = 0, then the system is passive. Proof: For all T with x(0) ẋ = f (x, u), Lemma: If there exists a storage function V for a system EL2620 Today’s Goal Lecture 5 – Passivity – Circle Criterion – Small Gain Theorem • analyze stability using • derive the gain of a system You should be able to EL2620 45 2012 − u G(s) Lecture 6 -2 -1 0 0 1 2 5 u 10 y 4 and u = sat e give a stable oscillation. s(s + 1)2 e y Motivating Example • Describing function analysis Lecture 6 EL2620 Nonlinear Control • How can the oscillation be predicted? G(s) = r EL2620 Lecture 6 EL2620 15 20 3 2012 1 2012 Today’s Goal A Frequency Response Approach Lecture 6 But, can we talk about the frequency response, in terms of gain and phase lag, of a static nonlinearity? 4 2012 A (linear) feedback system will have sustained oscillations (center) if the loop-gain is 1 at the frequency where the phase lag is −180o Nyquist / Bode: EL2620 Lecture 6 function analysis 2 2012 • Analyze existence and stability of periodic solutions by describing • Derive describing functions for static nonlinearities You should be able to EL2620 r n=1 p N.L. u G(s) y a2n + b2n sin[nωt + arctan(an /bn )] − e Key Idea → φ/ω Lecture 6 That is, we assume all higher harmonics are filtered out by G 5 7 2012 ≪ |G(iω)| for n ≥ 2, then n = 1 suffices, so that q y(t) ≈ |G(iω)| a21 + b21 sin[ωt + arctan(a1 /b1 ) + arg G(iω)] If |G(inω)| u(t) = ∞ X e(t) = A sin ωt gives EL2620 Lecture 6 Note: Sometimes we make the change of variable t = 2π/T and Z Z 2 T 2 T u(t) cos nωt dt, bn (ω) = u(t) sin nωt dt an (ω) = T 0 T 0 where ω a0 (an cos nωt + bn sin nωt) + 2 n=1 ∞ 2012 = u(t + T ) has a Fourier series expansion Fourier Series a0 X p 2 + = an + b2n sin[nωt + arctan(an /bn )] 2 n=1 u(t) = ∞ X A periodic function u(t) EL2620 Z 0 N.L. 2 u(t) − u bk (t) dt Lecture 6 Amplitude dependent gain and phase shift! can be used instead of u(t) to analyze the system. N (A, ω) u b1 (t) u b1 (t) = |N (A, ω)|A sin[ωt + arg N (A, ω)] e(t) b1 (ω) + ia1 (ω) A = 0, then u(t) N (A, ω) = If G is low pass and a0 e(t) T Definition of Describing Function 2 min û T The describing function is EL2620 Lecture 6 solves a0 X (an cos nωt + bn sin nωt) u bk (t) = + 2 n=1 k The Fourier Coefficients are Optimal The finite expansion EL2620 8 2012 6 2012 Z e -1 -0.5 0 0 0.5 1 − 3 4 5 6 a1 = e 4H b1 (ω) + ia1 (ω) = A πA f (·) u G(s) y A −1/N (A) G(iω) Existence of Periodic Solutions N (A) = ⇔ G(iω) = −1/N (A) Lecture 6 The intersections of the curves G(iω) and −1/N (A) give ω and A for a possible periodic solution. G(iω)N (A) = −1 11 2012 9 2012 Proposal: sustained oscillations if loop-gain 1 and phase-lag −180o 0 replacements EL2620 Lecture 6 2 u φ = 2πt/T 1 e 1 2π u(φ) cos φ dφ = 0 π 0 Z Z 1 2π 2 π 4H b1 = u(φ) sin φ dφ = H sin φ dφ = π 0 π 0 π −H H u Describing Function for a Relay The describing function for a relay is thus EL2620 Odd Static Nonlinearities − e G(s) = u G(s) y -1 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 -0.8 -0.6 -0.4 -0.2 G(iω) −1/N (A) 0 Periodic Solutions in Relay System 2012 10 2012 Lecture 6 12 3 with feedback u = −sgn y (s + 1)3 √ No phase lag in f (·), arg G(iω) = −π for ω = 3 = 1.7 √ G(i 3) = −3/8 = −1/N (A) = −πA/4 ⇒ A = 12/8π ≈ 0.48 r EL2620 Lecture 6 • Nf +g (A) = Nf (A) + Ng (A) • Nαf (A) = αNf (A) • Nf (A, ω) = Nf (A) • Im Nf (A, ω) = 0 Assume f (·) and g(·) are odd (i.e. f (−e) = −f (e)) static nonlinearities with describing functions Nf and Ng . Then, EL2620 0 2 y 4 u 6 8 10 Lecture 6 EL2620 Lecture 6 = = b1 = a1 = Z 1 2π u(φ) cos φ dφ = 0 π 0 Z Z 1 2π 4 π/2 u(φ) sin φ dφ = u(φ) sin φ dφ π 0 π 0 Z Z 4A φ0 2 4D π/2 sin φ dφ + sin φ dφ π 0 π φ0 A 2φ0 + sin 2φ0 π Note that G filters out almost all higher-order harmonics. -1 -0.5 0 0.5 1 The prediction via the describing function agrees very well with the true oscillations: EL2620 15 2012 13 2012 D e -1 -0.5 0 0 0.5 1 1 2 u 3 φ e 4 5 6 = arcsin D/A. Lecture 6 0.1 0 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 2 4 6 8 N (A) for H = D = 1 10 1 Hence, if H = D , then N (A) = 2φ0 + sin 2φ0 . π If H 6= D , then the rule Nαf (A) = αNf (A) gives H 2φ0 + sin 2φ0 N (A) = Dπ EL2620 Lecture 6 where φ0 = A sin ωt = A sin φ. First set H = D. Then for φ ∈ (0, π) A sin φ, φ ∈ (0, φ0 ) ∪ (π − φ0 , π) u(φ) = D, φ ∈ (φ0 , π − φ0 ) −H −D H u Describing Function for a Saturation Let e(t) EL2620 16 2012 14 2012 Ω Lecture 6 oscillation amplitude is decreasing. • If G(Ω) does not encircle the point −1/N (A), then the amplitude is increasing. • If G(Ω) encircles the point −1/N (A), then the oscillation −1/N (A) G(Ω) Stability of Periodic Solutions Assume that G(s) is stable. EL2620 Lecture 6 19 2012 17 2012 5 minute exercise: What oscillation amplitude and frequency do the describing function analysis predict for the “Motivating Example”? EL2620 − KG(s) −1/K G(iω) The Nyquist Theorem 2012 −1/N (A) G(Ω) An Unstable Periodic Solution Lecture 6 An intersection with amplitude A0 is unstable if A < A0 leads to decreasing amplitude and A > A0 leads to increasing. EL2620 Lecture 6 system is stable (damped oscillations) 20 2012 18 • If G(iω) does not encircle the point −1/K , then the closed-loop is unstable (growing amplitude oscillations). • If G(iω) encircles the point −1/K , then the closed-loop system displays sustained oscillations • If G(iω) goes through the point −1/K the closed-loop system Assume that G is stable, and K is a positive gain. EL2620 − u G(s) y -0.2 -5 -0.15 -0.1 -0.05 0 0.05 0.1 -4 -3 -2 -1 −1/N (A) G(iω) (s + 10)2 with feedback u = −sgn y G(s) = (s + 1)3 e 1 u D e 1 -1 -0.8 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 0.8 1 2 u 3 e 4 = arcsin D/A. = A sin ωt = A sin φ. Then for φ ∈ (0, π) 0, φ ∈ (0, φ0 ) u(φ) = 1, φ ∈ (φ0 , π − φ0 ) where φ0 Lecture 6 0 5 6 Describing Function for a Quantizer Let e(t) EL2620 Lecture 6 gives one stable and one unstable limit cycle. The left most intersection corresponds to the stable one. r 0.15 0.2 Stable Periodic Solution in Relay System EL2620 23 2012 21 2012 Automatic Tuning of PID Controller Lecture 6 EL2620 Lecture 6 0 y u Relay 5 A T u −1 Process 10 y Time Z 1 2π u(φ) cos φ dφ = 0 a1 = π 0 Z Z 1 2π 4 π/2 b1 = u(φ) sin φdφ = sin φdφ π 0 π φ0 4 4p = cos φ0 = 1 − D2 /A2 π π ( 0, A<D N (A) = 4 p 1 − D2 /A2 , A ≥ D πA −1 0 1 Σ PID Period and amplitude of relay feedback limit cycle can be used for autotuning: EL2620 24 2012 22 2012 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 6 8 10 _ C = sgn y : _ F G Friction y Lecture 6 The oscillation depends on the zero at s = z. s(s − z) G = 3 with feedback u = −sgn y 1 + GC s + 2s2 + 2s + 1 Corresponds to yref Control loop with friction F 27 2012 25 2012 Accuracy of Describing Function Analysis EL2620 Lecture 6 4 ≈ 1.3/A for large amplitudes 2 N (A) for D = 1 Plot of Describing Function Quantizer Notice that N (A) EL2620 Describing Function Pitfalls 0 15 15 20 20 25 25 30 30 0 0.2 0.4 0.6 0.8 Lecture 6 0.5 Accurate results only if y is close to sinusoidal! (17.3, 0.23), respectively. 0 = (11.4, 1.00) and z = 1/3 -0.5 z = 4/3 DF gives period times and amplitudes (T, A) 0 -0.4 -1 10 z = 4/3 10 1 1.2 -0.4 5 5 z = 1/3 -0.2 y y -0.2 0 0.2 0.4 -1 0 1 EL2620 Lecture 6 and can be far from the true values. 28 2012 26 2012 • The predicted amplitude and frequency are only approximations • A limit cycle may exist even if the DF does not predict it. • A DF may predict a limit cycle even if one does not exist. Describing function analysis can give erroneous results. EL2620 =x ? 2 Analysis of Oscillations—A Summary Lecture 6 • Powerful graphical methods • Approximate results • Describing function analysis Frequency-domain: • Hard to use for large problems • Rigorous results but only for simple examples • Poincaré maps and Lyapunov functions Time-domain: EL2620 Lecture 6 2 minute exercise: What is N (A) for f (x) EL2620 31 2012 29 2012 e(t) = A sin ωt f (·) u(t) Harmonic Balance 2012 30 2012 Lecture 6 function analysis 32 • Analyze existence and stability of periodic solutions by describing Today’s Goal • Derive describing functions for static nonlinearities You should be able to EL2620 Lecture 6 Example: f (x) = x2 gives u(t) = (1 − cos 2ωt)/2. Hence by considering a0 = 1 and a2 = 1/2 we get the exact result. may give much better result. Describing function corresponds to k = 1 and a0 = 0. k a0 X u bk (t) = (an cos nωt + bn sin nωt) + 2 n=1 A few more Fourier coefficients in the truncation EL2620 PSfr r - e PID v u G y The Problem with Saturating Actuator • Compensation for friction • Friction models • Compensation for saturation (anti-windup) Lecture 7 EL2620 Nonlinear Control Lecture 7 1 + Td s Recall: CPID (s) = K 1 + Ti s – Example: I-part in PID 3 2012 1 2012 • Leads to problem when system and/or the controller are unstable behavior! • The feedback path is broken when u saturates ⇒ Open loop EL2620 Lecture 7 EL2620 Today’s Goal −0.1 0 0.1 0 1 2 0 0 20 20 40 Time 40 60 60 80 80 Example—Windup in PID Controller Lecture 7 PID controller without (dashed) and with (solid) anti-windup EL2620 Lecture 7 • Compensation for friction • Anti-windup for PID and state-space controllers You should be able to analyze and design EL2620 y u 4 2012 2 2012 Anti-Windup for PID Controller y K Ti ∑ K KTd s ∑ K KTd s 1 Tt 1 s 1 Tt 1 s ∑ v v − − ∑ es + u ∑ u Actuator es + Actuator Actuator model ∑ − xˆ Σ L State feedback Actuator x̂˙ = Ax̂ + B sat v + K(y − C x̂) v = L(xm − x̂) Observer xm Lecture 7 sat v Anti-Windup for Observer-Based State Feedback Controller −y K Ti −y x̂ is estimate of process state, xm desired (model) state Need actuator model if sat v is not measurable EL2620 Lecture 7 e (b) e (a) Anti-windup (a) with actuator output available and (b) without EL2620 7 2012 5 2012 Anti-Windup is Based on Tracking Z Ke(τ ) es (τ ) 1 t es (τ ) dτ + dτ ≈ Ti Tt Tt 0 2012 Lecture 7 (a) y SA u (b) y SB u 6 8 2012 Idea: Rewrite representation of control law from (a) to (b) with the same input–output relation, but where the unstable SA is replaced by a stable SB . If u saturates, then (b) behaves better than (a). Windup possible if F unstable and u saturates ẋc = F xc + Gy u = Cxc + Dy Anti-Windup for General State-Space Controller 0 State-space controller: EL2620 Lecture 7 I(t) = Z t Remark: If 0 < Tt ≪ Ti , then the integrator state becomes sensitive to the instances when es 6= 0: • Tt = Ti √ • Tt = Ti Td Common choices of Tt : The tracking time Tt is the design parameter of the anti-windup When the control signal saturates, the integration state in the controller tracks the proper state EL2620 Controllers with ”Stable” Zeros = G − KD. ⇒ 2012 9 2012 Lecture 7 Note that this implies G − KD = 0 in the figure on the previous slide, and we thus obtain P-feedback with gain D under saturation. 11 and the eigenvalues of the ”observer” based controller becomes equal to the zeros of F (s) when u saturates K = G/D F − KC = F − GC/D ⇒u=0 Thus, choose ”observer” gain ẋc = F xc + Gy u = Cxc + Dy z }| { ẋc = (F − GC/D) xc y = −C/Dxc zero dynamics Most controllers are minimum phase, i.e. have zeros strictly in LHP EL2620 Lecture 7 where G0 ẋc = F0 xc + G0 y + Ku u = sat(Cxc + Dy) Choose K such that F0 = F − KC has desired (stable) eigenvalues. Then use controller ẋc = F xc + Gy + K(u − Cxc − Dy) = (F − KC)xc + (G − KD)y + Ku Mimic the observer-based controller: EL2620 ∑ ∑ K s −1 F − KC s −1 xc xc C D C ∑ ∑ v u sat u y + − Σ 1 − 1 D F (s) D sat u Lecture 7 If the transfer function (1/F (s) − 1/D) in the feedback loop is stable (stable zeros) ⇒ No stability problems in case of saturation 10 12 2012 = lims→∞ F (s) and consider the feedback implementation Controller F (s) with ”Stable” Zeros G − KD G 2012 It is easy to show that transfer function from y to u with no saturation equals F (s)! Let D EL2620 Lecture 7 y y F D State-space controller without and with anti-windup: EL2620 Internal Model Control (IMC) − Q(s) u b G(s) G(s) − yb y − Lecture 7 Choose Q r ≈ G−1 . u v b G G − y IMC with Static Nonlinearity Q − yb. ≈ G and choose Q stable with Q ≈ G−1 . Include nonlinearity in model EL2620 Lecture 7 b Design: assume G Assume G stable. Note: feedback from the model error y r C(s) IMC: apply feedback only when system G and model Ĝ differ! EL2620 15 2012 13 2012 Lecture 7 T1 s + 1 umax y± τs + 1 τs + 1 An alternative way to implement anti-windup! No integration. u=− > umax (v = ±umax ): If |u| −(T1 s + 1) y τs 1 T1 s + 1 y+ v τs + 1 τs + 1 < umax (v = u): PI controller u = b =− u = −Q(y − Gv) = 0 and abuse of Laplace transform notation Example (cont’d) if |u| Assume r EL2620 Lecture 7 PI-controller! ⇒ b 1 − QG T1 1 T1 s + 1 = 1+ F = τs τ T1 s Q T1 s + 1 , τ < T1 τs + 1 1 T1 s + 1 Example b G(s) = Q= F = Gives the controller Choose EL2620 16 2012 14 2012 Other Anti-Windup Solutions Friction Lecture 7 • How to detect friction in control loops? • How to compensate for friction? • How to model friction? Problems: – Earthquakes • Sometimes too small: – Friction in brakes • Sometimes good: – Friction in valves and other actuators • Often bad: Friction is present almost everywhere EL2620 Lecture 7 • Apply optimal control theory (Lecture 12) • Conditionally reset integration state to zero • Don’t update controller states at saturation • Tune controller to avoid saturation Other solutions include: Solutions above are all based on tracking. EL2620 19 2012 17 2012 Bumpless Transfer 1 Tr 1 Tm ∑ uc y ∑ 1 Tr 1 s PD 1 s ∑ A − ∑ M + − ∑ + u Lecture 7 Ff EL2620 Lecture 7 y Fp x 0 0 1 0 0 0.2 0.4 0 0 1 2 10 Ff Fp 10 10 Stick-Slip Motion Note the incremental form of the manual control mode (u̇ e uc 1 Tr PID with anti-windup and bumpless transfer: 20 20 20 ẏ 20 Time Time Time y x 2012 18 2012 ≈ uc /Tm ) Another application of the tracking idea is in the switching between automatic and manual control modes. EL2620 Lecture 7 EL2620 xr 1 0 0 −1 0 1 −0.2 0.2 0 0 0.5 u 20 20 20 Σ 40 40 40 1 ms 60 60 60 Friction v 80 80 80 1 Friction Modeling PID − F s 100 Time 100 Time 100 Time x Position Control of Servo with Friction Lecture 7 EL2620 23 2012 21 2012 Stribeck Effect Lecture 7 -0.04 -200 -150 -100 -50 0 50 100 150 200 Stribeck (1902) -0.03 -0.02 -0.01 0 0.01 Velocity [rad/sec] Steady state friction: Joint 1 0.02 0.03 0.04 Friction increases with decreasing velocity (for low velocities) EL2620 Lecture 7 5 minute exercise: Which are the signals in the previous plots? EL2620 Friction [Nm] 24 2012 22 2012 Classical Friction Models Integral Action Lecture 7 xr _ PID u _ F 1 ms Friction v 1 s • If friction force changes quickly, then large integral action (small Ti ) necessary. May lead to stability problem • Works if friction force changes slowly (v(t) ≈ const) • Integral action compensates for any external disturbance EL2620 Lecture 7 Advanced models capture various friction phenomena better EL2620 x 27 2012 25 2012 Friction Compensation K Ti η e(t) ê(τ )dτ where ê(t) Rt Lecture 7 Disadvantage: Error won’t go to zero Advantage: Avoid that small static error introduces oscillation = Modified Integral Action Modify the integral part to I EL2620 Lecture 7 • The Knocker • Adaptive friction compensation • Model-based friction compensation • Dither signal • Integral action • Lubrication EL2620 28 2012 26 2012 Dither Signal _ PID u F F̂ + + Friction estimator 1 ms Friction 1 ms F̂ = â sgn v v ż = kuPID sgn v â = z − km|v| F = a sgn v uPID − Friction estimator: Lecture 7 _ Friction v 1 s x 1 s x Adaptive Friction Compensation Coulomb friction model: EL2620 Lecture 7 Cf., mechanical maze puzzle (labyrintspel) xr F Avoid sticking at v = 0 (where there is high friction) by adding high-frequency mechanical vibration (dither ) replacements EL2620 31 2012 29 2012 u = uPID + F̂ Lecture 7 = 0. = −k sgn v(F − F̂ ) = −k(a − â) = −ke dâ dz d de =− = − + km |v| dt dt dt dt = −kuPID sgn v + kmv̇ sgn v = −k sgn v(uPID − mv̇) e = a − â → 0 as t → ∞ d Remark: Careful with dt |v| at v Proof: Adaptation converges: EL2620 Lecture 7 • u and F does apply at the same point • An estimate F̂ ≈ F is available Possible if: where uPID is the regular control signal and F̂ an estimate of F . use control signal mẍ = u − F Model-Based Friction Compensation For process with friction F : EL2620 32 2012 30 2012 0 0 2 2 3 u 3 6 6 5 5 7 7 8 8 0 -10 -5 0 0 5 10 -1 -0.5 0 1 1 2 2 3 3 4 4 0 -10 -5 0 0 5 10 -1 -0.5 0 0.5 1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 5 5 PI-controller P-controller with adaptive friction compensation 4 4 1 0.5 6 6 7 7 8 8 Detection of Friction in Control Loops 1 1 v P-controller vref 50 50 100 100 150 150 200 time 200 Lecture 7 Horch: PhD thesis (2000) and patent 10.5 11 11.5 12 98 99 100 101 102 250 250 300 300 350 350 • Idea: Monitor loops automatically and estimate friction • Q: When should valves be maintained? • Friction is due to wear and increases with time EL2620 Lecture 7 -10 -5 0 5 10 -1 -0.5 0 0.5 1 EL2620 y u 35 2012 33 2012 The Knocker 0 1 2 3 4 5 6 7 8 Today’s Goal Lecture 7 • Compensation for friction 9 10 • Anti-windup for PID, state-space, and polynomial controllers You should be able to analyze and design EL2620 Lecture 7 Hägglund: Patent and Innovation Cup winner -0.5 0 0.5 1 1.5 2 2.5 Coulomb friction compensation and square wave dither Typical control signal u EL2620 36 2012 34 2012 Lecture 7 • Quantization • Backlash EL2620 Next Lecture 37 2012 Lecture 8 Fin✲ EL2620 Lecture 8 EL2620 Tin θin ✛ D✲ xin 2D ✛ D✲ xout θout Linear and Angular Backlash • Backlash • Quantization Lecture 8 EL2620 Nonlinear Control Tout Fout ✲ 3 2012 1 2012 Today’s Goal Lecture 8 • sometimes inducing oscillations • bad for control performance • necessary for a gearbox to work in high temperature • increasing with wear Backlash • present in most mechanical and hydraulic systems Backlash (glapp) is EL2620 Lecture 8 • Quantization • Backlash You should be able to analyze and design for EL2620 4 2012 2 2012 otherwise in contact D xout θout D xin θin − xin | = D and ẋin (xin − xout ) > 0. ẋout ẋin , = 0, Backlash Model Effects of Backlash Lecture 8 θref − e K u 1 1 + sT θ̇in 1 s θin P-control of motor angle with gearbox having backlash with D EL2620 Lecture 8 backlash is a dynamic phenomena. θout 7 2012 5 2012 = 0.2 • Multivalued output; current output depends on history. Thus, “in contact” if |xout EL2620 5 10 15 20 No backlash K 25 30 35 = 0.25, 1, 4 40 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 5 10 15 Backlash K 20 25 30 35 = 0.25, 1, 4 xin − xout (θin − θout ) 40 Lecture 8 but never vanish • Note that the amplitude will decrease with decreasing D, 8 2012 6 2012 • Oscillations for K = 4 but not for K = 0.25 or K = 1. Why? 0 0 0.5 1 1.5 EL2620 Lecture 8 D (Torque) Force Alternative Model Not equivalent to “Backlash Model” EL2620 Describing Function for Backlash K=4 -2 Real Axis 0 K = 0.25 K=1 −1/N (A) Nyquist Diagrams 4 0 0 5 10 15 Lecture 8 Describing function predicts oscillation well = 0.33, ω = 1.24 Simulation: A = 0.33, ω = 2π/5.0 = 1.26 DF analysis: Intersection at A 2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20 25 30 35 Input and output of backlash Describing Function Analysis K = 4, D = 0.2: -4 -4 -3 -2 -1 0 1 2 3 4 EL2620 Re N (A) = 1 π + arcsin(1 − 2D/A) π 2 s D D + 2(1 − 2D/A) 1− A A D 4D 1− Im N (A) = − πA A < D then N (A) = 0 else Lecture 8 If A EL2620 Imaginary Axis 40 11 2012 9 2012 -3.5 -2.5 -2 -1.5 -1 Re -1/N(A) -0.5 0 0.5 1 Stability Proof for Backlash System − e K u 1 1 + sT θ̇in 1 s Lecture 8 Q: What about θ̇in and θ̇out ? Note that θin and θout will not converge to zero θref θin Do there exist conditions that guarantee stability (of the steady-state)? θout 12 2012 10 2012 → −1 as A → ∞ (physical interpretation?) -3 The describing function method is only approximate. replacements EL2620 Lecture 8 Note that −1/N (A) -16 -4 -14 -12 -10 -8 -6 -4 -2 0 −1/N (A) for D = 0.2: EL2620 Im -1/N(A) G(s) Lecture 8 • Backlash inverse • Linear controller design • Deadzone G(s) BL θ̇in in contact 0 otherwise θ̇in θ̇out Backlash Compensation • Mechanical solutions EL2620 Lecture 8 θout θ̇out = The block “BL” satisfies θin Rewrite the system: EL2620 15 2012 13 2012 Homework 2 G(s) BL Lecture 8 θref − e 1+sT2 K 1+sT 1 u 1 1 + sT θ̇in 1 s θin Linear Controller Design Introduce phase lead compensation: EL2620 Lecture 8 Circle Criterion BL contained in sector [0, 1] =1 θ̇out Small Gain Theorem BL has gain γ(BL) Passivity Theorem BL is passive θ̇in Analyze this backlash system with input–output stability results: EL2620 θout 16 2012 14 2012 -5 without filter Real Axis 0 Nyquist Diagrams 10 -2 -1 0 0 1 5 u, with/without filter 5 y, with/without filter 10 10 15 15 20 20 b if u(t) > u(t−) u+D b if u(t) < u(t−) xin (t) = u−D x (t−) otherwise in Oscillation removed! 5 with filter 2 0 0 0.5 1 1.5 Lecture 8 b > D then over-compensation (may give oscillation) • If D b < D then under-compensation (decreased backlash) • If D b = D then perfect compensation (xout = u) • If D EL2620 Lecture 8 -10 -10 -8 -6 -4 -2 0 2 4 6 8 10 1+sT2 F (s) = K 1+sT with T1 = 0.5, T2 = 2.0: 1 EL2620 Imaginary Axis 19 2012 17 2012 Backlash Inverse ✲ ✻ ✲ ✲ xout D D Example—Perfect Compensation ✻ ✲ xin ✲ Lecture 8 -2 0 2 4 6 0 0.4 0.8 1.2 0 0 2 2 4 4 6 6 8 8 10 10 Motor with backlash on input in feedback with PD-controller EL2620 Lecture 8 u Idea: Let xin jump ±2D when ẋout should change sign xout EL2620 xin 20 2012 18 2012 40 60 80 21 Lecture 8 2012 EL2620 -2 0 1 2 2 3 3 4 4 5 5 2012 22 u G y A/D Lecture 8 • Roundoff, overflow, underflow in computations • Quantization of parameters • Quantization in A/D and D/A converters D/A ∆ ∆/2 • What precision is needed in computations? (8–64 bits?) 23 y e fe de = 2 Z u −∆/2 ∆/2 + e y ∆2 e2 de = ∆ 12 Lecture 8 24 • But, added noise can never affect stability while quantization can! Q −∞ ∞ • May be reasonable if ∆ is small compared to the variations in u u Var(e) = Z Model quantization error as a uniformly distributed stochastic signal e independent of u with 20 1 • What precision is needed in A/D and D/A converters? (8–14 bits?) 0 0 2 4 6 0 2012 Linear Model of Quantization 0 2 4 80 0 60 0 40 0.4 0.4 20 0.8 0.8 Example—Over-Compensation 1.2 0 EL2620 1.2 Example—Under-Compensation 2012 Quantization EL2620 Lecture 8 EL2620 0 1 0 ( -1 -0.8 -0.6 -0.4 -0.2 0 0 0.2 0.4 0.6 1 2 u 3 e 4 5 2 A/Delta 4 0, A<D 4 p 1 − D2 /A2 , A > D πA e 1 0.8 6 2012 25 2012 Lecture 8 • Reducing ∆ reduces only the oscillation amplitude • Controller with gain margin > 1/0.79 = 1.27 avoids oscillation 27 • Predicts oscillation if Nyquist curve intersects negative real axis to the left of −π/4 ≈ −0.79 • The maximum value is 4/π ≈ 1.27 attained at A ≈ 0.71∆. EL2620 Lecture 8 D N (A) = Lecture 6 ⇒ 1 u Describing Function for Deadzone Relay EL2620 N (A) u Q y ∆ ∆/2 y Describing Function for Quantizer u Lecture 8 r − K u G(s) y Q > 2/1.27 = 1.57 with Q. 1−e−s s Stable oscillation predicted for K Nyquist of linear part (K & ZOH & G(s)) intersects at −0.5K : Stability for K < 2 without Q. = 0.2 Example—Motor with P-controller. Quantization of process output with ∆ EL2620 Lecture 8 A < ∆2 0, s 2 n N (A) = 2i − 1 4∆ P 1− ∆<A< ∆ , 2n−1 2 πA i=1 2A EL2620 28 2012 26 2n+1 ∆ 2 2012 (b) (c) Quantization ∆ EL2620 Lecture 8 (a) 0 0 0 Time K = 1.6 K = 1.2 K = 0.8 31 −0.05 0 0.05 −0.05 0 0.05 0 1 0 0 0 50 50 50 Time 100 100 100 Lecture 8 150 −2 D/A Real axis = 0.01 in D/A converter: F −4 Lecture 8 100 150 150 Quantization ∆ EL2620 Lecture 8 −6 Au = 0.005 and T = 39 Simulation: Au = 0.005 and T = 39 Time 100 100 2012 29 A/D −2 0 Describing function: 50 50 50 50 50 50 2 0 G 150 150 150 Example—1/s2 & 2nd-Order Controller EL2620 Ay = 0.01 and T = 39 Simulation: Ay = 0.01 and T = 28 0 0 0 Describing function: −0.05 0 0.05 0.98 1.02 0 1 = 0.02 in A/D converter: 0 1 0 1 0 1 2012 Imaginary axis EL2620 Output Output Output Output Output Input Output Unquantized Input 32 2012 30 2012 Quantization Compensation Lecture 8 pling, and digital lowpass filter to improve accuracy of A/D converter • Use analog dither, oversam- anti-windup to improve D/A converter • Use the tracking idea from + A/D controller Digital filter • Avoid unstable controller and gain margins < 1.3 • Improve accuracy (larger word length) EL2620 Analog decim. +− D/A 33 2012 Today’s Goal Lecture 8 • Quantization • Backlash You should now be able to analyze and design for EL2620 34 2012 k f (·) Feedback amplifiers were the solution! f (·) − Lecture 9 Black, Bode, and Nyquist at Bell Labs 1920–1950. f (·) r High signal amplification with low distortion was needed. y History of the Feedback Amplifier • Nonlinear control design based on high-gain control Lecture 9 EL2620 Nonlinear Control New York–San Francisco communication link 1914. EL2620 Lecture 9 EL2620 3 2012 1 2012 Today’s Goal α2 e f (e) Lecture 9 K f (·) y ⇒ ≫ 1/α1 , yields f (e) ≤ α2 e − choose K α1 ≤ r y≈ 1 r K α1 α2 r ≤ y ≤ r 1 + α1 K 1 + α2 K e α1 e 4 2012 2 2012 Linearization Through High Gain Feedback EL2620 Lecture 9 • Sliding mode controllers • High-gain control systems You should be able to analyze and design EL2620 A Word of Caution − f (·) k s u̇ = k v − f (u) u Lecture 9 > 0 large and df /du > 0, then u̇ → 0 and 0 = k v − f (u) ⇔ f (u) = v ⇔ If k v f (u) u = f −1 (v) u 7 2012 Remark: How to Obtain f −1 from f using Feedback EL2620 Lecture 9 5 2012 Nyquist: high loop-gain may induce oscillations (due to dynamics)! EL2620 Inverting Nonlinearities fˆ−1 (·) f (·) G(s) 2012 6 2012 − e K u Lecture 9 The case K 0 0 0.2 0.4 0.6 0.8 0.2 = u2 through feedback. f (·) y 0.4 0.6 y(r) = 100 is shown in the plot: y(r) ≈ r. Linearization of f (u) r 1 0.8 f (u) 1 8 Example—Linearization of Static Nonlinearity EL2620 Lecture 9 Should be combined with feedback as in the figure! − F (s) Controller Compensation of static nonlinearity through inversion: EL2620 ⇒ F G = 0.1 K G Lecture 9 Gtot = (Gcl )100 ≈ 10100 dGtot = (1 + 10−3 )100 − 1 ≈ 0.1 Gtot and total distortion y S = (1 + GK)−1 ≈ 0.01. Then − dG dGcl =S ≈ 0.001 Gcl G ⇒ r 100 feedback amplifiers in series give total amplification Choose K = 0.1 y 9 11 2012 1 dG dG dGcl = =S Gcl 1 + GF G G − Example—Distortion Reduction Let G = 1000, distortion dG/G EL2620 Lecture 9 2012 = (1 + GF )−1 S is the closed-loop sensitivity to open-loop perturbations. dGcl 1 = dG (1 + GF )2 Small perturbations dG in G gives G Gcl = 1 + GF The closed-loop system is r The Sensitivity Function S EL2620 K f (·) − K 2012 10 2012 Lecture 9 1–4 16 480 1938 1941 1 1914 1923 Channels Year 30000 1000 150–400 60 Loss (dB) 600 40 6–20 3–6 No amp’s The feedback amplifier was patented by Black 1937. 12 Transcontinental Communication Revolution EL2620 Lecture 9 − Several links give distortion-free high gain. f (·) Distortion Reduction via Feedback The feedback reduces distortion in each link. EL2620 −1 r GF (iω) r − y |S(iω)| ≤ r−1 f (·) GF Sensitivity and the Circle Criterion ⇔ On–Off Control − e u G(s) y Lecture 9 The relay corresponds to infinitely high gain at the switching point. r Common in temperature control, level control etc. 15 2012 13 2012 1 f (y) 1 ≤ ≤ 1+r y 1−r On–off control is the simplest control strategy. EL2620 Lecture 9 Then, the Circle Criterion gives stability if |1 + GF (iω)| > r Consider a circle C := {z ∈ C : |z + 1| = r}, r ∈ (0, 1). GF (iω) stays outside C if EL2620 y k1 y ẋ = Ax + Bu, u ∈ [−1, 1] = xT P x, P = P T > 0, represents the energy of A Control Design Idea 2012 Lecture 9 ẋ = Ax + B ẋ = Ax − B B T QP x = 0 16 is minimized if u = − sgn(B T P x) (Notice that V̇ = a + bu, i.e. just a segment of line in u, −1 < u < 1. Hence the lowest value is at an endpoint, depending on the sign of the slope b. ) V̇ = xT (AT P + P A)x + 2B T P xu Choose u such that V decays as fast as possible: Assume V (x) EL2620 Lecture 9 1 k1 = 1+r 1 k2 = 1−r Hence, |S(iω)| small implies low sensitivity to nonlinearities. k2 y f (y) This corresponds to a large sector for f (·). If |S(iω)| is small, we can choose r large (close to one). 14 2012 Small Sensitivity Allows Large Uncertainty EL2620 σ(x) > 0 σ(x) < 0 f+ Sliding Modes f− σ(x) < 0 σ(x) > 0 Lecture 9 x1 = −1 This implies the following behavior x1 = 1 x2 < 0 x2 > 0 ẋ2 (t) ≈ x1 + 1, dx2 ≈ 1 + x1 dx1 dx2 ≈ 1 − x1 ẋ2 (t) ≈ x1 − 1, dx1 For small x2 we have EL2620 Lecture 9 f+ x2 = 0 x2 < 0 x2 > 0 αf + + (1 − α)f − = {x : σ(x) = 0}. f− The sliding surface is S The sliding mode is ẋ = αf + + (1 − α)f − , where α satisfies αfn+ + (1 − α)fn− = 0 for the normal projections of f + , f − + f (x), ẋ = f − (x), EL2620 19 2012 17 2012 Ax − B, Ax + B, x2 > 0 x2 < 0 Lecture 9 ueq is called the equivalent control. Sliding Mode Dynamics ẋ = The dynamics along the sliding surface S is obtained by setting u = ueq ∈ [−1, 1] such that x(t) stays on S . EL2620 Lecture 9 Example 0 −1 1 ẋ = x+ u = Ax + Bu 1 −1 1 u = − sgn σ(x) = − sgn x2 = − sgn(Cx) is equivalent to EL2620 20 2012 18 2012 =0 ẋ = Ax + Bu u = − sgn σ(x) = − sgn(Cx) = −CAx/CB . Lecture 9 because σ(x) = x2 = 0. (Same result as before.) ueq = −CAx/CB = − 1 −1 x = −x1 , Example (cont’d): For the example: gives ueq > 0. The sliding surface S = {x : Cx = 0} so dσ 0 = σ̇(x) = f (x) + g(x)u = C Ax + Bueq dx Assume CB EL2620 = {x : x2 = 0}. ⇒ ueq = −x1 Equivalent Control for Linear System gives the dynamics on the sliding surface S =0 ẋ1 = − x2 + ueq = −x1 |{z} Insert this in the equation for ẋ1 : Lecture 9 2012 23 2012 21 = ueq such that σ̇(x) = ẋ2 = 0 on σ(x) = x2 = 0 gives Example (cont’d) 0 = ẋ2 = x1 − x2 + ueq = x1 + ueq |{z} Finding u EL2620 if the inverse exists. dσ f (x) dx Lecture 9 Remark: The condition that Cx = 0 corresponds to the zero at s = 0, and thus this dynamic disappears on S = {x : Cx = 0}. under the constraint Cx = 0, where the eigenvalues of (I − BC/CB)A are equal to the zeros of sG(s) = sC(sI − A)−1 B . = {x : Cx = 0} is given by 1 BC Ax, ẋ = Ax + Bueq = I − CB The dynamics on S EL2620 −1 Sliding Dynamics dσ ueq = − g(x) dx The equivalent control is thus given by Lecture 9 2012 24 2012 22 = {x : σ(x) = 0}. Then, for x ∈ S , dσ dx dσ · = 0 = σ̇(x) = f (x) + g(x)u dx dt dx ẋ = f (x) + g(x)u u = − sgn σ(x) Deriving the Equivalent Control has a stable sliding surface S Assume EL2620 Proof 2012 = σ 2 (x)/2 with σ(x) = pT x. Then, Closed-Loop Stability = 0. + ··· + pn−1 x(1) n + pn x(0) n 2012 Lecture 9 27 where x(k) denote time derivative. Now p corresponds to a stable differential equation, and xn → 0 exponentially as t → ∞ . The state relations xk−1 = ẋk now give x → 0 exponentially as t → ∞. = p1 x(n−1) n 0 = σ(x) = p1 x1 + · · · + pn−1 xn−1 + pn xn so x tend to σ(x) V̇ = −µσ(x) sgn σ(x) < 0 With the chosen control law, we get V̇ = σ T (x)σ̇(x) = xT p pT f (x) + pT g(x)u Consider V (x) EL2620 > 0: Lecture 9 Note that ts → 0 as µ → ∞. ts = = 0) is σ0 <∞ µ Hence, the time to the first switch (σ(x) σ̇(x) = −µ it follows that as long as σ(x) σ(x)σ̇(x) = −µσ(x) sgn σ(x) = σ(x0 ) > 0. Since Time to Switch Consider an initial point x0 such that σ0 EL2620 Lecture 9 Lecture 9 µ pT f (x) − T sgn σ(x), T p g(x) p g(x) > 0 is a design parameter, σ(x) = pT x, and pT = p1 . . . pn are the coefficients of a stable polynomial. u=− Choose control law where µ 25 Design of Sliding Mode Controller 2012 28 2012 26 Idea: Design a control law that forces the state to σ(x) = 0. Choose σ(x) such that the sliding mode tends to the origin. Assume x1 f1 (x) + g1 (x)u x1 d x2 .. = = f (x) + g(x)u .. dt . . xn xn−1 EL2620 but this transfer function is also 1/(sG(s)) Hence, the eigenvalues of (I − BC/CB)A are equal to the zeros of sG(s). −1 1 1 −1 + CA(sI − ((I − BC)A))−1 B CB CB CB CB Hence, the transfer function from ẏ to u equals 1 1 CAx − ẏ ⇒ y = Cx ⇒ ẏ = CAx + CBu ⇒ u = CB CB 1 1 ẋ = I − BC Ax − B ẏ CB CB ẋ = Ax + Bu EL2620 Example—Sliding Mode Controller T . σ0 =3 µ Lecture 9 ẏ = −y u x1 Time Plots x2 µ pT Ax − T sgn σ(x) u=− T p B p B = 2x1 − µ sgn(x1 + x2 ) and sliding dynamics ts = 2012 31 2012 29 = s + 1 so that σ(x) = x1 + x2 . The controller is Simulation agrees well with time to switch x(0) = 1.5 0 Initial condition EL2620 Lecture 9 Choose p1 s + p2 given by 1 0 1 ẋ = x+ u 1 0 0 y= 0 1 x Design state-feedback controller for EL2620 −1 0 1 2 x1 The Sliding Mode Controller is Robust −2 = sgn(pT gb) and µ > 0 is sufficiently large. Lecture 9 (High gain control with stable open loop zeros) The closed-loop system is thus robust against model errors! if sgn(pT g) pT g pT (f gbT − fbg T )p − µ T sgn σ(x) < 0 V̇ = σ(x) pT gb p gb 32 2012 30 2012 = 0.5. Note the sliding surface σ(x) = x1 + x2 . Phase Portrait Assume that only a model ẋ = fb(x) + g b(x)u of the true system ẋ = f (x) + g(x)u is known. Still, however, EL2620 Lecture 9 0 1 −1 x2 Simulation with µ EL2620 Comments on Sliding Mode Control Next Lecture Lecture 9 • Exact feedback linearization • Lyapunov design methods EL2620 Lecture 9 • Compare puls-width modulated control signals • Applications in robotics and vehicle control • Smooth version through low pass filter or boundary layer • Often impossible to implement infinite fast switching • Efficient handling of model uncertainties EL2620 35 2012 33 2012 Today’s Goal Lecture 9 • Sliding mode controllers • High-gain control systems You should be able to analyze and design EL2620 34 2012 • Lyapunov-based control design methods • Input-output linearization • Exact feedback linearization Lecture 10 EL2620 Nonlinear Control Lecture 10 • Linear controller: ż = Az + By , u = Cz 3 2012 1 2012 • Linear dynamics, static nonlinearity: ż = Az + By , u = c(z) • Nonlinear dynamical controller: ż = a(z, y), u = c(z) Nonlinear Output Feedback Controllers EL2620 Lecture 10 EL2620 Nonlinear Observers x ḃ = f (b x, u) Lecture 10 Second case is called Extended Kalman Filter • Linearize f at x b(t), find K = K(b x) for the linearization • Linearize f at x0 , find K for the linearization Choices of K x ḃ = f (b x, u) + K(y − h(b x)) Feedback correction, as in linear case, Simplest observer ẋ = f (x, u), y = h(x) What if x is not measurable? EL2620 Lecture 10 k and ℓ may include dynamics. has nice properties. ẋ = f (x, ℓ(x)) • State feedback: Find u = ℓ(x) such that • Output feedback: Find u = k(y) such that the closed-loop system ẋ = f x, k h(x) ẋ = f (x, u) y = h(x) Output Feedback and State Feedback has nice properties. EL2620 4 2012 2 2012 ẋ2 = −x21 + u ẋ1 = a sin x2 Another Example ż1 = z2 , ż2 = a cos x2 (−x21 + u) Lecture 10 v , x2 ∈ [−π/2, π/2] u(x) = x21 + a cos x2 and the linearizing control becomes yields z1 = x1 , z2 = ẋ1 = a sin x2 Perform transformation of states into linearizable form: How do we cancel the term sin x2 ? EL2620 Lecture 10 • Lyapunov-based design - backstepping control • Input-output linearization • Exact feedback linearization Some state feedback control approaches EL2620 7 2012 5 2012 Exact Feedback Linearization ẋ = cos x − x3 + u Lecture 10 with (A, B) controllable and γ(x) nonsingular for all x in the domain of interest. ẋ = Ax + Bγ(x) (u − α(x)) 8 2012 is feedback linearizable if there exist a diffeomorphism T whose domain contains the origin and transforms the system into the form ẋ = f (x) + g(x)u Definition: A nonlinear system is called a diffeomorphism • T and T −1 continuously differentiable • T invertible for x in the domain of interest = T (x) with Diffeomorphisms ẋ = −kx + v A nonlinear state transformation z EL2620 Lecture 10 yields the linear system u(x) = − cos x + x3 − kx + v The state-feedback controller Example 1: 6 2012 idea: use a state-feedback controller u(x) to make the system linear ẋ = f (x) + g(x)u Consider the nonlinear control-affine system EL2620 Exact vs. Input-Output Linearization + u, y = x2 Lecture 10 u = x1 − ǫ(1 − The state feedback controller x21 )x2 +v ⇒ ÿ = v ÿ = ẋ2 = −x1 + ǫ(1 − x21 )x2 + u ẏ = ẋ1 = x2 Differentiate the output ẋ2 = −x1 + ǫ(1 − x21 )x2 + u y = x1 ẋ1 = x2 Example: controlled van der Pol equation EL2620 Lecture 10 Caution: the control has rendered the state x1 unobservable ẋ1 = a sin x2 , ẋ2 = v, y = x2 which is linear from v to y to obtain u = x21 + v 11 2012 9 2012 • If we want a linear input-output relationship we could instead use which is nonlinear in the output. ż1 = z2 ; ż2 = v ; y = sin−1 (z2 /a) • The control law u = x21 + v/a cos x2 yields ẋ1 = a sin x2 , ẋ2 = −x21 The example again, but now with an output EL2620 Input-Output Linearization 2012 = 1/sp Lie Derivatives 2012 10 Lecture 10 Lkf h(x) d(Lfk−1 h) d(Lf h) f (x), Lg Lf h(x) = g(x) = dx dx Repeated derivatives 12 dh dh ẋ = (f (x) + g(x)u) , Lf h(x) + Lg h(x)u dx dx where Lf h(x) and Lg h(x) are Lie derivatives (Lf h is the derivative of h along the vector field of ẋ = f (x)) ẏ = The derivative of the output ẋ = f (x) + g(x)u ; x ∈ Rn , u ∈ R1 y = h(x), y ∈ R Consider the nonlinear SISO system EL2620 Lecture 10 i.e., G(s) y (p) = v 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 linear from the input v to the output y ẋ = f (x) + g(x)u y = h(x) Use state feedback u(x) to make the control-affine system EL2620 Lie derivatives and relative degree 2012 =n−m The input-output linearizing control Lecture 10 z }| { dh ẋ = Lf h(x) + Lg h(x)u dx =0 y (p) = Lpf h(x) + Lg Lfp−1 h(x)u .. . ẏ = Differentiating the output repeatedly ẋ = f (x) + g(x)u y = h(x) Consider a nth order SISO system with relative degree p EL2620 Lecture 10 15 2012 13 Lg Lfi−1 h(x) = 0, i = 1, . . . , p−1 ; Lg Lfp−1 h(x) 6= 0 ∀x ∈ D • A nonlinear system has relative degree p if has relative degree p b0 sm + . . . + bm Y (s) = n U (s) s + a1 sn−1 + . . . + an • A linear system integrators between the input and the output (the number of times y must be differentiated for the input u to appear) • The relative degree p of a system is defined as the number of EL2620 Example Lg Lfp−1 h(x) 1 Lecture 10 =2 −Lpf h(x) + v y (p) = v results in the linear input-output system u= and hence the state-feedback controller EL2620 Lecture 10 Thus, the system has relative degree p ÿ = ẋ2 = −x1 + ǫ(1 − x21 )x2 + u ẏ = ẋ1 = x2 Differentiating the output ẋ2 = −x1 + ǫ(1 − x21 )x2 + u y = x1 ẋ1 = x2 The controlled van der Pol equation EL2620 16 2012 14 2012 Zero Dynamics 2012 Lecture 10 Here we limit discussion to Back-stepping control design, which require certain f discussed later. • Methods depend on structure of f • Verify stability through Control Lyapunov function • Find stabilizing state feedback u = u(x) ẋ = f (x, u) 17 19 2012 Lyapunov-Based Control Design Methods EL2620 Lecture 10 phase (and should not be input-output linearized!) • A system with unstable zero dynamics is called non-minimum • 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. • Thus, if p < n then n − p states are unobservable in y . the relative degree of the system • Note that the order of the linearized system is p, corresponding to EL2620 ẋ2 = −x1 + ǫ(1 − x21 )x2 + u ẋ1 = x2 van der Pol again 2012 ẋ = cos x − x3 + u A simple introductory example = x2 /2 Lecture 10 18 20 2012 But, the term x3 in the control law may require large control moves! Thus, the system is globally asymptotically stable V (x) > 0 , V̇ = −kx2 < 0 Choose the Lyapunov candidate V (x) u = − cos x + x3 − kx Apply the linearizing control Consider EL2620 Lecture 10 (but bounded) • With y = x2 the relative degree p = 1 < n and the zero dynamics are given by ẋ1 = 0, which is not asymptotically stable • With y = x1 the relative degree p = n = 2 and there are no zero dynamics, thus we can transform the system into ÿ = v . EL2620 u = − cos x − kx = 0 with f (0) = 0. R (1) 2012 21 Lecture 10 u R x2 g(x1 ) f () x1 23 Idea: See the system as a cascade connection. Design controller first for the inner loop and then for the outer. at x ẋ1 = f (x1 ) + g(x1 )x2 ẋ2 = u = u(x) that stabilizes Back-Stepping Control Design We want to design a state feedback u EL2620 Lecture 10 Thus, also globally asymptotically stable (and more negative V̇ ) V (x) > 0 , V̇ = −x4 − kx2 < 0 = x2 /2 ẋ = cos x − x3 + u Choose the same Lyapunov candidate V (x) Now try the control law The same example 2012 0.5 1 1.5 2 2.5 time [s] 3 3.5 State trajectory 4 4.5 linearizing non-linearizing 5 -200 0 0 200 400 600 800 1000 0.5 1 1.5 2 2.5 time [s] 3 Control input 3.5 φ(x1 ) and there exists Lyapunov fcn Lecture 10 This is a critical assumption in backstepping control! for some positive definite function W . dV1 V̇1 (x1 ) = f (x1 ) + g(x1 )φ(x1 ) ≤ −W (x1 ) dx1 can be stabilized by v̄ = V1 = V1 (x1 ) such that ẋ1 = f (x1 ) + g(x1 )v̄ Suppose the partial system EL2620 Lecture 10 The linearizing control is slower and uses excessive input. Thus, linearization can have a significant cost! 0 0 1 2 3 4 5 6 7 8 9 10 = 10 Simulating the two controllers Simulation with x(0) EL2620 State x EL2620 Input u 4 24 2012 22 4.5 linearizing non-linearizing 2012 5 The Trick R −φ(x1 ) x2 g(x1 ) f + gφ R x1 dV1 g(x1 ) − kζ, dx1 R ζ g(x1 ) R f + gφ x1 ≤ −W ). Then, Lecture 10 Lecture 10 2012 26 2012 28 = 0 with V (z) + (xk − φ(z))2 /2 being a Lyapunov fcn. dV dφ f (z) + g(z)xk − g(z) − (xk − φ(z)) u= dz dz stable, and V (z) a Lyapunov fcn (with V̇ ż = f (z) + g(z)φ(z) = 0, f (0) = 0, ż = f (z) + g(z)xk ẋk = u = (x1 , . . . , xk−1 )T and Back-Stepping Lemma −φ̇(x1 ) Assume φ(0) stabilizes x 2 k>0 V̇2 (x1 , x2 ) ≤ −W (x1 ) − kζ v=− u ζ̇ = v dφ dφ ẋ1 = f (x1 ) + g(x1 )x2 φ̇(x1 ) = dx1 dx1 Lemma: Let z EL2620 Lecture 10 where = x2 − φ(x1 ) and control v = u − φ̇: ẋ1 = f (x1 ) + g(x1 )φ(x1 ) + g(x1 )ζ Introduce new state ζ EL2620 = 0 is asymptotically stable for (1) with control law u(x) = φ̇(x) + v(x). If V1 radially unbounded, then global stability. Hence, x gives Choosing Consider V2 (x1 , x2 ) 27 2012 25 2012 = V1 (x1 ) + ζ 2 /2. Then, dV1 dV1 f (x1 ) + g(x1 )φ(x1 ) + g(x1 )ζ + ζv V̇2 (x1 , x2 ) = dx1 dx1 dV1 ≤ −W (x1 ) + g(x1 )ζ + ζv dx1 EL2620 Lecture 10 u ẋ1 = f (x1 ) + g(x1 )φ(x1 ) + g(x1 )[x2 − φ(x1 )] ẋ2 = u Equation (1) can be rewritten as EL2620 Strict Feedback Systems Back-Stepping Example 32 Lecture 10 Lecture 10 u = φn (x1 , . . . , xn ) V2 = x21 /2 + (x2 + x21 + x1 )2 /2 2012 30 2012 u1 = (−2x1 − 1)(x21 + x2 ) − x1 − (x2 + x21 + x1 ) = φ2 (x1 , x2 ) = −x21 − x1 stabilizes the first equation. With V1 (x1 ) = x21 /2, Back-Stepping Lemma gives where φ1 (x1 ) ẋ1 = x21 + φ1 (x1 ), ẋ2 = u1 Step 0 Verify strict feedback form Step 1 Consider first subsystem ẋ1 = x21 + x2 , ẋ2 = x3 , ẋ3 = u Design back-stepping controller for EL2620 Lecture 10 ẋ = Ax + Bu on strict feedback form. 2 minute exercise: Give an example of a linear system EL2620 Back-stepping results in the final state feedback 31 2012 29 2012 by “stepping back” from x1 to u (see Khalil pp. 593–594 for details). Vk (x1 , . . . , xk ) = Vk−1 (x1 , . . . , xk−1 ) + [xk − φk−1 ]2 /2 Back-stepping generates stabilizing feedbacks φk (x1 , . . . , xk ) (equal to u in Back-Stepping Lemma) and Lyapunov functions on strict feedback form. ẋ = f (x) + g(x)u Back-Stepping Lemma can be applied recursively to a system EL2620 6= 0 x1 , . . . , xk do not depend on xk+2 , . . . , xn . Lecture 10 Note: where gk ẋn = fn (x1 , . . . , xn ) + gn (x1 , . . . , xn )u .. . ẋ1 = f1 (x1 ) + g1 (x1 )x2 ẋ2 = f2 (x1 , x2 ) + g2 (x1 , x2 )x3 ẋ3 = f3 (x1 , x2 , x3 ) + g3 (x1 , x2 , x3 )x4 Back-stepping Lemma can be applied to stabilize systems on strict feedback form: EL2620 Lecture 10 which globally stabilizes the system. 33 2012 dφ2 dV2 g(z) − (xn − φ2 (z)) f (z) + g(z)xn − u = u2 = dz dz ∂φ2 2 ∂φ2 ∂V2 = (x1 + x2 ) + x3 − − (x3 − φ2 (x1 , x2 )) ∂x1 ∂x2 ∂x2 gives ẋ1 = x21 + x2 ẋ2 = x3 ẋ3 = u Step 2 Applying Back-Stepping Lemma on EL2620 0 1 ẋ = f (x, u) Controllability • Gain scheduling • Nonlinear controllability Lecture 11 EL2620 Nonlinear Control Lecture 11 is controllable if for any x , x there exists T > 0 and u : [0, T ] → R such that x(0) = x0 and x(T ) = x1 . Definition: EL2620 Lecture 11 EL2620 3 2012 1 2012 Today’s Goal Lecture 11 Is there a corresponding result for nonlinear systems? has full rank. Wn = B AB . . . An−1 B ẋ = Ax + Bu Linear Systems is controllable if and only if Lemma: EL2620 Lecture 11 • Apply gain scheduling to simple examples • Determine if a nonlinear system is controllable You should be able to EL2620 4 2012 2 2012 2012 0 0 B1 = , 0 1 = 0 and ż = Az + B1 u1 + B2 u2 = u2 = 0 gives Lecture 11 5 7 2012 Linearization does not capture the controllability good enough linearization is not controllable. Still the car is controllable! cos(ϕ0 + θ0 ) sin(ϕ0 + θ0 ) B2 = sin(θ0 ) 0 rank Wn = rank B AB . . . An−1 B = 2 < 4, so the with A Linearization for u1 EL2620 Lecture 11 system is not controllable • A nonlinear system can be controllable, even if the linearized • Hence, if rank Wn = n then there is an ǫ > 0 such that for every x1 ∈ Bǫ (0) there exists u : [0, T ] → R so that x(T ) = x1 Remark: at x = 0 with f (0) = 0. If the linear system is controllable then the nonlinear system is controllable in a neighborhood of the origin. ẋ = f (x) + g(x)u ż = Az + Bu Controllable Linearization be the linearization of Lemma: Let EL2620 (x, y) y θ ϕ x Input: velocity u1 steering wheel velocity, u2 forward Car Example 2012 ∂f ∂g f− g ∂x ∂x Lecture 11 x1 cos x2 , g= f= 1 x1 0 − sin x2 x1 1 0 cos x2 − [f, g] = 1 0 1 0 0 x1 cos x2 + sin x2 = −x1 Example: [f, g] = : Rn → Rn is a vector field Lie Brackets Lie bracket between vector fields f, g defined by EL2620 Lecture 11 8 2012 6 x 0 cos(ϕ + θ) d y 0 sin(ϕ + θ) = u1 + u2 = g1 (z)u1 + g2 (z)u2 dt ϕ 0 sin(θ) θ 1 0 EL2620 4. Finally, for t Lecture 11 t ∈ [0, ǫ) t ∈ [ǫ, 2ǫ) t ∈ [2ǫ, 3ǫ) t ∈ [3ǫ, 4ǫ) ∈ [2ǫ, 3ǫ] 2 x(4ǫ) = x0 + ǫ ∈ [3ǫ, 4ǫ] 2 dg1 dg2 g1 − g2 dx dx dg1 1 dg2 dg2 g1 − g2 + g2 dx dx 2 dx Proof, continued x(3ǫ) = x0 + ǫg2 + ǫ 3. Similarily, for t EL2620 Lecture 11 (1, 0), (0, 1), (−1, 0), (0, −1), x(4ǫ) = x(0) + ǫ2 [g1 , g2 ] + O(ǫ3 ) (u1 , u2 ) = ẋ = g1 (x)u1 + g2 (x)u2 Lie Bracket Direction The system can move in the [g1 , g2 ] direction! gives motion the control For the system EL2620 11 2012 9 2012 Proof = g2 (x0 ) + dg2 dx ǫg1 (x0 ) Car Example (Cont’d) Lecture 11 ∂g2 ∂g1 g3 := [g1 , g2 ] = g1 − g2 ∂x ∂x 0 0 − sin(ϕ + θ) − sin(ϕ + θ) 0 cos(ϕ + θ) 0 0 0 cos(ϕ + θ) = − 0 0 cos(θ) 0 0 0 0 0 0 0 1 − sin(ϕ + θ) cos(ϕ + θ) = cos(θ) 0 EL2620 Lecture 11 12 2012 10 x(2ǫ) = x0 + ǫ(g1 (x0 ) + g2 (x0 ))+ dg2 1 dg2 2 1 dg1 (x0 )g1 (x0 ) + (x0 )g1 (x0 ) + (x0 )g2 (x0 ) ǫ 2 dx dx 2 dx and with x(ǫ) from (1), and g2 (x(ǫ)) 1 dg2 g2 (x(ǫ))ǫ2 2 dx 1 dg1 g1 (x0 )ǫ2 + O(ǫ3 ) 2 dx x(2ǫ) = x(ǫ) + g2 (x(ǫ))ǫ + ∈ [ǫ, 2ǫ] x(ǫ) = x0 + g1 (x0 )ǫ + (1) 2012 ∈ [0, ǫ], assuming ǫ small and x(0) = x0 , Taylor series yields 2. Similarily, for t 1. For t EL2620 2012 ϕ x Parking Theorem θ Lecture 11 Wriggle, Drive, −Wriggle, −Drive 13 15 2012 You can get out of any parking lot that is ǫ > 0 bigger than your car by applying control corresponding to g4 , that is, by applying the control sequence EL2620 Lecture 11 (x, y) y (u1 , u2 ) = {(1, 0), (0, 1), (−1, 0), (0, −1)} We can hence move the car in the g3 direction (“wriggle”) by applying the control sequence EL2620 2012 14 2012 Lecture 11 16 2 minute exercise: What does the direction [g1 , g2 ] correspond to for a linear system ẋ = g1 (x)u1 + g2 (x)u2 = B1 u1 + B2 u2 ? EL2620 Lecture 11 sideways movement g4 direction corresponds to (−sin(ϕ), cos(ϕ)) − sin(ϕ + 2θ) cos(ϕ + 2θ) ∂g3 ∂g2 g3 − g2 = . . . = g4 := [g3 , g2 ] = ∂x ∂x 0 0 The car can also move in the direction EL2620 [g1 , [g2 , [g1 , g2 ]]] Lecture 11 Controllable because {g1 , g2 , [g1 , g2 ]} spans R3 cos θ 0 sin θ g1 = sin θ , g2 = 0 , [g1 , g2 ] = − cos θ 0 1 0 2012 19 2012 17 [g2 , [g2 , [g1 , g2 ]]] [g2 , [g1 , g2 ]] θ (x1 , x2 ) Example—Unicycle [g2 , [g1 , [g1 , g2 ]]] [g1 , g2 ] The Lie Bracket Tree x1 cos θ 0 d x2 = sin θ u1 + 0 u2 dt θ 0 1 EL2620 Lecture 11 [g1 , [g1 , [g1 , g2 ]]] [g1 , [g1 , g2 ]] EL2620 2012 Lecture 11 • Is the linearization of the unicycle controllable? • Show that {g1 , g2 , [g1 , g2 ]} spans R3 for the unicycle 2 minute exercise: EL2620 Lecture 11 20 2012 18 • The system can be steered in any direction of the Lie bracket tree Remark: is controllable if the Lie bracket tree (together with g1 and g2 ) spans Rn for all x ẋ = g1 (x)u1 + g2 (x)u2 Controllability Theorem Theorem:The system EL2620 φ θ Gain Scheduling Controller Process Gain schedule Output Operating condition 2012 21 Lecture 11 23 Examples of scheduling variable are production rate, machine speed, Mach number, flow rate = K(α), where α is the scheduling Control signal Example: PID controller with K variable. Command signal Controller parameters Control parameters depend on operating conditions: EL2620 Lecture 11 2012 0 cos θ x1 d x2 sin θ 0 u1 + u2 = dt θ 0 1 0 1 φ Example—Rolling Penny Controllable because {g1 , g2 , [g1 , g2 ], [g2 , [g1 , g2 ]]} spans R4 (x1 , x2 ) EL2620 2012 Lecture 11 EL2620 Lecture 11 v u α ẋ = f x T Flow Linear Quickopening Equalpercentage Position Valve Characteristics β z A: The answer requires Lie brackets and further concepts from differential geometry (see Khalil and PhD course) 24 2012 22 Q: When can we transform ẋ = f (x) + g(x)u into ż = Az + bv by means of feedback u = α(x) + β(x)v and change of variables z = T (x) (see previous lecture)? When is Feedback Linearization Possible? EL2620 uc Σ 0 Lecture 11 5.0 5.1 1.0 1.1 0.2 0.3 0 0 0 uc y uc uc y y With gain scheduling: EL2620 Lecture 11 0 10 Valve characteristics EL2620 20 20 20 PI 0.5 c 40 40 40 −1 fˆ −1 1 u 60 60 60 1.5 v fˆ(u) f Process Nonlinear Valve 80 80 80 2 f (u) G 0 (s) 100 Time 100 Time 100 Time y 27 2012 25 2012 0 0 0 Lecture 11 θ uc uc y uc V α q = θ˙ Pitch dynamics EL2620 Lecture 11 5.0 5.2 1.0 1.1 0.2 0.3 y y 20 20 20 Nz Flight Control 10 10 10 Without gain scheduling: EL2620 30 30 30 δe 40 Time 40 Time 40 Time 28 2012 26 2012 0 1.6 0.8 1.2 Mach number 2 4 Today’s Goal 0.4 1 3 Flight Control 2.0 Lecture 11 • Apply gain scheduling to simple examples • Determine if a nonlinear system is controllable You should be able to EL2620 Lecture 11 0 20 40 60 80 Operating conditions: EL2620 Altitude (x1000 ft) 2.4 31 2012 29 2012 Lecture 11 EL2620 Filter Pitch rate Filter Acceleration Filter Position Pitch stick A/D A/D A/D M T2 s 1+ T2 s 1 1+ T3 s T1s 1+ T1s H K DSE VIAS H M VIAS K QD K Q1 M Σ K SG H Gear VIAS H M H K NZ Filter − Σ Σ D/A D/A Σ Σ To servos Σ The Pitch Control Channel 30 2012 Optimal Control Problems • Optimal control Lecture 12 EL2620 Nonlinear Control 2012 1 2012 Lecture 12 - determining the optimal controller can be hard - difficult to formulate control objectives as a single objective function + can deal with constraints + applicable to nonlinear problems + provides a systematic design framework u(t) min J(x, u, t), ẋ = f (t, x, u) 3 Idea: formulate the control design problem as an optimization problem EL2620 Lecture 12 EL2620 Today’s Goal Lecture 12 v(x2 ) ẋ1 (t) = v(x2 ) + u1 (t) ẋ2 (t) = u2 (t) x1 (0) = x2 (0) = 0 max x1 (tf ) u:[0,tf ]→U u(t) ∈ U = {(u1 , u2 ) : u21 + u22 = 1} Rudder angle control: =1 x2 Sail as far as possible in x1 direction Example—Boat in Stream Speed of water v(x2 ) with dv/dx2 EL2620 Lecture 12 • Design controllers based on optimal control theory You should be able to EL2620 x1 4 2012 2 2012 min 0 tf L(x(t), u(t)) dt + φ(x(tf )) ẋ(t) = f (x(t), u(t)), x(0) = x0 u:[0,tf ]→U Z Lecture 12 • Final time tf fixed (free later) • Constraints on x from the dynamics • Infinite dimensional optimization problem: Optimization over functions u : [0, tf ] → U • U ⊂ Rm set of admissible control Remarks: [1 − u(t)]x(t)dt ẋ(t) = γu(t)x(t) x(0) = x0 > 0 0 tf > 0) Optimal Control Problem Standard form: EL2620 Lecture 12 max u:[0,tf ]→[0,1] Z amount of stored profit change of production rate (γ portion of x stored portion of x reinvested production rate Maximization of stored profit Example—Resource Allocation x(t) ∈ [0, ∞) u(t) ∈ [0, 1] 1 − u(t) γu(t)x(t) [1 − u(t)]x(t) EL2620 7 2012 5 2012 Example—Minimal Curve Length 2012 p a Pontryagin’s Maximum Principle 1 + u2 (t)dt tf Lecture 12 u∈U u∗ (t) = arg min H(x∗ (t), u, λ(t)) Moreover, the optimal control is given by ∂H T ∗ ∂φT ∗ ∗ λ̇(t) = − (x (t), u (t), λ(t)), λ(tf ) = (x (tf )) ∂x ∂x where λ(t) solves the adjoint equation u∈U 6 8 2012 t min H(x∗ (t), u, λ(t)) = H(x∗ (t), u∗ (t), λ(t)), ∀t ∈ [0, tf ] Suppose the optimal control problem above has the solution u∗ : [0, tf ] → U and x∗ : [0, tf ] → Rn . Then, H(x, u, λ) = L(x, u) + λT f (x, u) Theorem: Introduce the Hamiltonian function EL2620 Lecture 12 0 tf ẋ(t) = u(t) x(0) = a u:[0,tf ]→R min Z (t, x(t)) with x(0) = a Line: Vertical through (tf , 0) Curve: x(t) Find the curve with minimal length between a given point and a line EL2620 Remarks 2012 = 0) λ1 (t) u1 +u2 =1 λ21 (t) + λ22 (t) , u2 (t) = − p λ2 (t) + u1 ) + λ2 (t)u2 1 tf − t , u2 (t) = p 1 + (t − tf )2 1 + (t − tf )2 λ21 (t) + λ22 (t) u1 (t) = p Lecture 12 λ1 (t)(v(x∗2 (t)) λ1 (t)u1 + λ2 (t)u2 = arg 2min 2 u1 (t) = − p Hence, or u1 +u2 =1 u (t) = arg 2min 2 ∗ Optimal control EL2620 Lecture 12 • “maximum” is due to Pontryagin’s original formulation 9 11 2012 • Solution involves 2n ODE’s with boundary conditions x(0) = x0 and λ(tf ) = ∂φT /∂x(x∗ (tf )). Often hard to solve explicitly. • The Maximum Principle provides all possible candidates. there may exist many or none solutions (cf., minu:[0,1]→R x(1), ẋ = u, x(0) • Pontryagin’s Maximum Principle provides necessary condition: increase the criterium. Then perform a clever Taylor expansion. • See textbook, e.g., Glad and Ljung, for proof. The outline is simply to note that every change of u(t) from the optimal u∗ (t) must EL2620 λ2 (tf ) = 0 λ1 (tf ) = −1 λ1 (t) = −1, λ2 (t) = t − tf λ̇2 (t) = −λ1 (t), λ̇1 (t) = 0, φ(x) = −x1 v(x2 ) + u1 λ2 u2 x(0) = x0 [u(t) − 1]x(t)dt λ̇(t) = 1 − u∗ (t) − λ(t)γu∗ (t), Adjoint equation λ(tf ) = 0 H = L + λT f = (u − 1)x + λγux Hamiltonian satisfies Lecture 12 0 tf ẋ(t) = γu(t)x(t), min u:[0,tf ]→[0,1] Z Example—Resource Allocation (cont’d) EL2620 Lecture 12 have solution Adjoint equations ∂H = 0 λ1 , ∂x H = λ f = λ1 T Example—Boat in Stream (cont’d) Hamiltonian satisfies EL2620 12 2012 10 2012 = arg min u(1 + λ(t)γ), u∈[0,1] 0, λ(t) ≥ −1/γ = 1, λ(t) < −1/γ (x (t) > 0) ∗ < tf − 1/γ , we have u∗ (t) = 1 and thus λ̇(t) = −γλ(t). Lecture 12 min 0 tf p ẋ(t) = u(t), u:[0,tf ]→R Z x(0) = a 1 + u2 (t)dt 5 minute exercise: Find the curve with minimal length by solving EL2620 ∗ ≈ tf , we have u∗ (t) = 0 (why?) and thus λ̇(t) = 1. Lecture 12 For t For t u∈[0,1] ∗ u (t) = arg min (u − 1)x (t) + λ(t)γux (t) ∗ Optimal control EL2620 15 2012 13 2012 0 0.4 0.4 0.6 0.6 0.8 0.8 1 1 u (t) ∗ 1.2 1.2 1.4 1.4 1.6 1.6 1.8 1.8 2 2 Lecture 12 0 1 u4 dt + x(1) ẋ = −x + u x(0) = 0 min Z 5 minute exercise II: Solve the optimal control problem EL2620 Lecture 12 t ∈ [0, tf − 1/γ] t ∈ (tf − 1/γ, tf ] 0.2 0.2 λ(t) • It is optimal to reinvest in the beginning 0 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 1, ∗ • u (t) = 0, EL2620 16 2012 14 2012 History—Calculus of Variations A B Lecture 12 maxu h(tf ) 0 ≤ u ≤ umax and m(tf ) = m1 (empty) Optimization criterion: Constraints: u motor force, D = D(v, h) air resistance (v(0), h(0), m(0)) = (0, 0, m0 ), g, γ > 0 h m How to send a rocket as high up in the air as possible? Goddard’s Rocket Problem (1910) u − D v − g d m h = v dt m −γu EL2620 Lecture 12 • Find the curve enclosing largest area (Euler) Solved by John and James Bernoulli, Newton, l’Hospital J(y) = Z p 1 + y ′ (x)2 p dx 2gy(x) p p 1 + y ′ (x)2 dx2 + dy 2 ds dt = = = p dx v v 2gy(x) Minimize time 19 2012 17 2012 • Brachistochrone (shortest time) problem (1696): Find the (frictionless) curve that takes a particle from A to B in shortest EL2620 History—Optimal Control Z 0 tf L(x(t), u(t)) dt + φ(x(tf )) Lecture 12 • Final state is constrained: ψ(x(tf )) = x3 (tf ) − m1 = 0 • End time tf is free Note the diffences compared to standard form: ẋ(t) = f (x(t), u(t)), x(0) = x0 ψ(x(tf )) = 0 u:[0,tf ]→U min Generalized form: EL2620 Lecture 12 – Finance—portfolio theory – Physics—Snell’s law, conservation laws – Aeronautics—satellite orbits – Robotics—trajectory generation • Huge influence on engineering and other sciences: • Bellman’s Dynamic Programming (1957) • Pontryagin’s Maximum Principle (1956) • The space race (Sputnik, 1957) EL2620 20 2012 18 2012 Solution to Goddard’s Problem H(x∗ (tf ), u∗ (tf ), λ(tf ), n0 ) ∂ψ ∂φ (tf , x∗ (tf )) = −n0 (tf , x∗ (tf )) − µT ∂t ∂t ∂H T ∗ (x (t), u∗ (t), λ(t), n0 ) λ̇(t) = − ∂x ∂φ ∂ψ λT (tf ) = n0 (tf , x∗ (tf )) + µT (tf , x∗ (tf )) ∂x ∂x Lecture 12 • ψ defines end point constraints • With fixed tf : H(x∗ (tf ), u∗ (tf ), λ(tf ), n0 ) = 0 • tf may be a free variable Remarks: EL2620 Lecture 12 • Took 50 years before a complete solution was presented resistance is high, in order to burn fuel at higher level • Hard: e.g., it can be optimal to have low speed when air D(v, h) 6≡ 0: • Burn fuel as fast as possible, because it costs energy to lift it • Easy: let u(t) = umax until m(t) = m1 D(v, h) ≡ 0: 23 2012 21 2012 x = (v, h, m)T , L ≡ 0, φ(x) = −x2 , ψ(x) = x3 − m1 Goddard’s problem is on generalized form with EL2620 L(x(t), u(t)) dt + φ(tf , x(tf )) ≥ 0, µ ∈ Rn such that (n0 , µT ) 6= 0 and 2012 Example—Minimum Time Control H(x, u, λ, n0 ) = n0 L(x, u) + λT f (x, u) 2012 22 min 0 tf 1 dt = tf ψ(x(tf )) = (x1 (tf ), x2 (tf ))T = (0, 0)T u∗ (t) = arg min 1 + λ1 (t)x∗2 (t) + λ2 (t)u u∈[−1,1] 1, λ2 (t) < 0 = −1, λ2 (t) ≥ 0 Optimal control is the bang-bang control Lecture 12 min u:[0,tf ]→[−1,1] ẋ1 (t) = x2 (t), ẋ2 (t) = u(t) u:[0,tf ]→[−1,1] Z 24 Bring the states of the double integrator to the origin as fast as possible EL2620 Lecture 12 where min H(x∗ (t), u, λ(t), n0 ) = H(x∗ (t), u∗ (t), λ(t), n0 ), t ∈ [0, tf ] Then, there exists n0 u∈U 0 tf : [0, tf ] → U and x∗ : [0, tf ] → Rn are ẋ(t) = f (x(t), u(t)), x(0) = x0 ψ(tf , x(tf )) = 0 u:[0,tf ]→U min Z Theorem: Suppose u∗ solutions to General Pontryagin’s Maximum Principle EL2620 x1 (t) = x1 (0) + x2 (0)t + ζt2 /2 x2 (t) = x2 (0) + ζt = ζ = ±1, we have min u:[0,∞)→Rm Lecture 12 where L 0 ∞ (xT Qx + uT Ru) dt SA + AT S + Q − SBR−1 B T S = 0 = R−1 B T S and S > 0 is the solution to u = −Lx ẋ = Ax + Bu Z Linear Quadratic Control has optimal solution with EL2620 Lecture 12 These define the switch curve, where the optimal control switch x1 (t) ± x2 (t)2 /2 = const Eliminating t gives curves With u(t) = 0, λ̇2 (t) = −λ1 (t) gives λ1 (t) = c1 , λ2 (t) = c2 − c1 t Adjoint equations λ̇1 (t) EL2620 27 2012 25 2012 2012 Properties of LQ Control Lecture 12 1978) • But, then system may have arbitrarily poor robustness! (Doyle, If x is not measurable, then one may use a Kalman filter; leads to linear quadratic Gaussian (LQG) control. • Phase margin 60 degrees • Closed-loop system stable with u = −α(t)Lx for α(t) ∈ [1/2, ∞) (infinite gain margin) • Stabilizing EL2620 Lecture 12 • Efficient design method for nonlinear problems reference value to a linear regulator, which keeps the system close to the wanted trajectory • We may use the optimal open-loop solution u∗ (t) as the 28 2012 26 • Optimal control problem makes no distinction between open-loop control u∗ (t) and closed-loop control u∗ (t, x). Reference Generation using Optimal Control EL2620 Move milk in minimum time without spilling Tetra Pak Milk Race Pros & Cons for Optimal Control Lecture 12 − Hard to solve the equations that give optimal controller − Hard to find suitable criteria + Captures limitations (as optimization constraints) + Applicable to nonlinear control problems + Systematic design procedure EL2620 Lecture 12 [Grundelius & Bernhardsson,1999] EL2620 31 2012 29 2012 15 Acceleration 0 0 0.05 0.05 0.1 0.1 0.15 0.15 0.2 0.2 Slosh 0.25 0.25 0.3 0.3 0.35 0.35 0.4 0.4 SF2852 Optimal Control Theory 2012 30 2012 Lecture 12 Applications: Aeronautics, Robotics, Process Control, Bioengineering, Economics, Logistics 32 Numerical Methods: Numerical solution of optimal control problems Pontryagin’s Maximum principle: Main results; Special cases such as time optimal control and LQ control Dynamic Programming: Discrete & continuous; Principle of optimality; Hamilton-Jacobi-Bellman equation http://www.math.kth.se/optsyst/ • Optimization and Systems Theory • Period 3, 7.5 credits EL2620 Lecture 12 Optimal time = 375 ms, TetraPak = 540ms -1 -0.5 0 0.5 1 -15 -10 -5 0 5 10 Given dynamics of system and maximum slosh φ = 0.63, solve Rt minu:[0,tf ]→[−10,10] 0 f 1 dt, where u is the acceleration. EL2620 Today’s Goal Lecture 12 • Understand possibilities and limitations of optimal control – Generalized form – Standard form • Design controllers based on optimal control theory for You should be able to EL2620 33 2012 • Fuzzy logic and fuzzy control • Artificial neural networks Lecture 13 EL2620 Nonlinear Control Fuzzy Control Lecture 13 IF Speed is High AND Traffic is Heavy THEN Reduce Gas A Bit Example of a rule: • Implement as rules (instead of as differential equations) • Model operator’s control actions (instead of the plant) Idea: • Transfer process knowledge to control algorithm is difficult • Many plants are manually controlled by experienced operators EL2620 Lecture 13 Some slides copied from K.-E. Årzén and M. Johansson EL2620 3 2012 1 2012 Today’s Goal Model Controller Instead of Plant Lecture 13 r − e C u P Fuzzy control design Model manual control → Implement control rules y Conventional control design Model plant P → Analyze feedback → Synthesize controller C Implement control algorithm EL2620 Lecture 13 • understand simple neural networks • understand the basics of fuzzy logic and fuzzy controllers You should EL2620 → 4 2012 2 2012 Fuzzy Set Theory x ∈ A to a certain degree µA (x) x ∈ A or x 6∈ A 0 Fuzzy Logic AND Y OR Z x Lecture 13 Mimic human linguistic (approximate) reasoning [Zadeh,1965] Defines logic calculations as X Fuzzy logic: AND: µA∩B (x) = min(µA (x), µB (x)) OR: µA∪B (x) = max(µA (x), µB (x)) NOT: µA′ (x) = 1 − µA (x) Conventional logic: AND: A ∩ B OR: A ∪ B NOT: A′ How to calculate with fuzzy sets (A, µA )? EL2620 Lecture 13 25 Warm Cold 10 µW µC A fuzzy set is defined as (A, µA ) 0 1 Membership function: µA : Ω → [0, 1] expresses the degree x belongs to A Fuzzy set theory: Conventional set theory: Specify how well an object satisfies a (vague) description EL2620 7 2012 5 2012 Example 0 Cold µC 10 0 10 Cold Lecture 13 0 1 25 µC∩W Warm Q1: Is it cold AND warm? EL2620 Lecture 13 0 1 x x 0 1 0 10 Cold µC∪W 25 Warm Q2: Is it cold OR warm? 25 Warm µW = 1/3. Example 15 Q2: Is x = 15 warm? A2: It is not really warm since µW (15) Q1: Is the temperature x = 15 cold? A1: It is quite cold since µC (15) = 2/3. EL2620 x 8 2012 6 2012 r Fuzzy Controller u Plant Fuzzy Control System y Fuzzifier Lecture 13 0 1 Example y = 15: µC (15) 0 15 25 Warm Cold 10 µW µC = 2/3 and µW (15) = 1/3 Fuzzy set evaluation of input y EL2620 Lecture 13 y Fuzzy controller is a nonlinear mapping from y (and r ) to u r, y, u : [0, ∞) 7→ R are conventional signals EL2620 11 2012 9 2012 y Fuzzifier Fuzzy Controller Fuzzy Inference u Fuzzy Inference Lecture 13 Rule 2: | {z } 2. 1. 2. IF y is Warm THEN u | is{zLow} | {z } 1. | {z } Examples of fuzzy rules: Rule 1: IF y is Cold THEN u is High 3. Aggregate rule outputs 2. Calculate fuzzy output of each rule 1. Calculate degree of fulfillment for each rule Fuzzy Inference: EL2620 Lecture 13 Fuzzifier and defuzzifier act as interfaces to the crisp signals Defuzzifier Fuzzy Controller Fuzzifier: Fuzzy set evaluation of y (and r ) Fuzzy Inference: Fuzzy set calculations Defuzzifier: Map fuzzy set to u EL2620 12 2012 10 2012 Lecture 13 EL2620 Lecture 13 3. Aggregate rule outputs 15 2012 13 2012 1. Calculate degree of fulfillment for rules EL2620 2. Calculate fuzzy output of each rule Lecture 13 EL2620 Lecture 13 Defuzzifier 16 2012 14 2012 Note that ”mu” is standard fuzzy-logic nomenclature for ”truth value”: EL2620 y Fuzzifier Fuzzy Controller Fuzzy Inference Defuzzifier Lecture 13 EL2620 19 2012 Lecture 13 EL2620 Lecture 13 Lecture 13 20 2012 18 2012 Example—Fuzzy Control of Steam Engine EL2620 http://isc.faqs.org/docs/air/ttfuzzy.html 17 2012 Defuzzifier: Map fuzzy set to u 3. Aggregate rule outputs 2. Calculate fuzzy output of each rule 1. Calculate degree of fulfillment for each rule Fuzzy Inference: Fuzzy set calculations u Fuzzy Controller—Summary Fuzzifier: Fuzzy set evaluation of y (and r ) EL2620 Lecture 13 EL2620 Lecture 13 EL2620 Rule-Based View of Fuzzy Control 23 2012 21 2012 Lecture 13 EL2620 Lecture 13 EL2620 Nonlinear View of Fuzzy Control 24 2012 22 2012 Pros and Cons of Fuzzy Control 2012 Lecture 13 EL2620 Lecture 13 Brain neuron xn x1 x2 wn Σ b φ(·) Artificial neuron w1 w2 Neurons y 27 2012 25 Fuzzy control is a way to obtain a class of nonlinear controllers • Not obvious how to include dynamics in controller • Sometimes hard to combine with classical control • Limited analysis and synthesis Disadvantages • Intuitive for non-experts in conventional control • Explicit representation of operator (process) knowledge • User-friendly way to design nonlinear controllers Advantages EL2620 Lecture 13 xn w2 x2 wn w1 x1 Σ b φ(·) y =φ b+ Model of a Neuron Inputs: x1 , x2 , . . . , xn Weights: w1 , w2 , . . . , wn Bias: b Nonlinearity: φ(·) Output: y EL2620 Lecture 13 • A network of computing components (neurons) Neural Networks • How does the brain work? EL2620 y i=1 wi xi Pn 28 2012 26 2012 A Simple Neural Network Lecture 13 y Success Stories Lecture 13 – Pattern recognition (e.g., speech, vision), data classification • Applications took off in mid 80’s 31 2012 29 2012 • Complex problems with unknown and highly nonlinear structure • McCulloch & Pitts (1943), Minsky (1951) Artificial neural networks: – Cement kilns, washing machines, vacuum cleaners • Applications took off in mid 70’s • Complex problems but with possible linguistic controls • Zadeh (1965) Fuzzy controls: EL2620 Output Layer Represents a nonlinear mapping from inputs to outputs u3 u2 u1 Input Layer Hidden Layer Neural network consisting of six neurons: EL2620 Neural Network Design Today’s Goal Lecture 13 • understand simple neural networks • understand the basics of fuzzy logic and fuzzy controllers You should EL2620 Lecture 13 The choice of weights are often done adaptively through learning 3. How to choose the weights? 2. How many neurons in each layer? 1. How many hidden layers? EL2620 32 2012 30 2012 Next Lecture Lecture 13 • PhD thesis projects • Master thesis projects • Spring courses in control • EL2620 Nonlinear Control revisited EL2620 33 2012 What’s on the exam? Question 1 • Master thesis projects • Spring courses in control • Summary and repetition Lecture 14 EL2620 Nonlinear Control Question 2 Lecture 14 Lecture 14 • Is the system generically nonlinear? E.g, ẋ = xu – saturations, valves etc • Optimal control • Nonlinear controllability • Some nonlinearities to compensate for? – analyze and simulate with nonlinear model – varying operating conditions? – strong nonlinearities (under feedback!)? • Evaluate: – linear methods (loop shaping, state feedback, . . . ) • Start with the simplest: The answer is highly problem dependent. Possible (learning) approach: What design method should I use in practice? EL2620 Lecture 14 • See homepage for old exams TEFYMA or BETA (No other material: textbooks, exercises, calculators etc. Any other basic control book must be approved by me before the exam.). • You may bring lecture notes, Glad & Ljung “Reglerteknik”, and • Sign up on course homepage Exam • Regular written exam (in English) with five problems EL2620 • Exact feedback linearization, input-output linearization, zero dynamics • Lyapunov based design: back-stepping • Sliding modes, equivalent controls • Describing functions • Compensating static nonlinearities • Circle Criterion, Small Gain Theorem, Passivity Theorem • Lyapunov stability (local and global), LaSalle 3 2012 1 2012 • Nonlinear models: equilibria, phase portaits, linearization and stability EL2620 Lecture 14 EL2620 4 2012 2 2012 Question 3 y k1 y − k11 − k12 The Circle Criterion k2 y f (y) G(iω) Lecture 14 If the Nyquist curve of G(s) stays on the correct side of the circle defined by the points −1/k1 and −1/k2 , then the closed-loop system is BIBO stable. k1 ≤ f (y)/y ≤ k2 . Theorem Consider a feedback loop with y = Gu and u = −f (y). Assume G(s) is stable and that EL2620 Lecture 14 of them can be used to prove instability. 7 2012 • Since they do not provide necessary conditions for stability, none • But, if one method does not prove stability, another one may. Criterion all provide only sufficient conditions for stability • No, the Small Gain Theorem, Passivity Theorem and Circle 5 2012 Can a system be proven stable with the Small Gain Theorem and unstable with the Circle Criterion? EL2620 Question 4 k1 < 0 < k2 : Stay inside the circle Lecture 14 Only Case 1 and 2 studied in lectures. Only G stable studied. 8 2012 6 2012 < 0 < k2 ? 0 = k1 < k2 : Stay to the right of the line Re s = −1/k2 0 < k1 < k2 : Stay outside circle The different cases Other cases: Multiply f and G with −1. 3. 2. 1. Stable system G EL2620 Lecture 14 Can you review the circle criterion? What about k1 EL2620 Please repeat antiwindup Question 5 G − KD G Lecture 14 Choose K such that F y y K s −1 F − KC s −1 xc xc C D C ∑ ∑ v u sat − KC has stable eigenvalues. ∑ ∑ F D u Antiwindup—General State-Space Model EL2620 Lecture 14 EL2620 11 2012 9 2012 Lecture 14 EL2620 Lecture 14 EL2620 e (b) e (a) K Ti ∑ K 1 Tt 1 s 1 Tt 1 s ∑ v v − − ∑ es + u ∑ u Actuator es + Actuator Actuator model ∑ Question 6 KTd s ∑ K KTd s Please repeat Lyapunov theory −y K Ti −y Tracking PID 12 2012 10 2012 ⇒ kx(t)k < R, t ≥ 0 > 0 there exists r > 0, such that ⇒ lim x(t) = 0 t→∞ Lecture 14 then x = 0 is globally asymptotically stable. • V (x) → ∞ as kxk → ∞ • V̇ (x) < 0 for all x 6= 0 • V (x) > 0, for all x 6= 0 • V (0) = 0 15 2012 13 2012 = 0. Assume that V : R → R n Lyapunov Theorem for Global Stability Theorem Let ẋ = f (x) and f (0) is a C 1 function. If EL2620 Lecture 14 globally asymptotically stable, if asymptotically stable for all x(0) ∈ Rn . kx(0)k < r locally asymptotically stable, if locally stable and kx(0)k < r locally stable, if for every R = 0 of ẋ = f (x) is Stability Definitions An equilibrium point x EL2620 = 0 is locally stable. Furthermore, if = 0 is locally asymptotically stable. V̇ (x) ≤ 0 for all x (3) Lecture 14 then x = f (x) such that V̇ (x) = 0 is x(t) = 0 16 = 0. If there exists a C1 function = 0 is globally asymptotically stable. (5) The only solution of ẋ for all t V (x) → ∞ as kxk → ∞ V (x) > 0 for all x 6= 0 (2) (4) V (0) = 0 (1) Theorem: Let ẋ = f (x) and f (0) V : Rn → R such that 14 2012 LaSalle’s Theorem for Global Asymptotic Stability EL2620 Lecture 14 then x • V̇ (x) < 0 for all x ∈ Ω, x 6= 0 then x • V̇ (x) ≤ 0 along all trajectories in Ω • V (x) > 0, for all x ∈ Ω, x 6= 0 • V (0) = 0 2012 ∈ Ω ⊂ Rn . Assume that Lyapunov Theorem for Local Stability Theorem Let ẋ = f (x), f (0) = 0, and 0 V : Ω → R is a C 1 function. If EL2620 2012 g k ẋ1 = x2 , ẋ2 = − sin x1 − x2 l m k 1 g V (x) = (1 − cos x1 ) + x22 ⇒ V̇ = − x22 l 2 m Example: Pendulum with friction 2012 17 Lecture 14 • The domain is compact if we consider Ω = {(x1 , x2 ) ∈ R2 |V (x) ≤ c} M = {(x1 , x2 )|x1 = kπ, x2 = 0} 19 • The invariant points in E are given by ẋ1 = x2 = 0 and ẋ2 = 0. Thus, the largest invariant set in E is • The set E = {(x1 , x2 )|V̇ = 0} is E = {(x1 , x2 )|x2 = 0} • We can not prove global asymptotic stability; why? EL2620 Lecture 14 and V is a positive definite function Ω = {x ∈ Rn |V (x) ≤ c} Remark : a compact set (bounded and closed) is obtained if we e.g., consider Let V : Rn → R be a C 1 function such that V̇ (x) ≤ 0 for x ∈ Ω. Let E be the set of points in Ω where V̇ (x) = 0. If M is the largest invariant set in E , then every solution with x(0) ∈ Ω approaches M as t → ∞ ẋ = f (x). ∈ Rn be a bounded and closed set that is invariant LaSalle’s Invariant Set Theorem Theorem Let Ω with respect to EL2620 2012 Lecture 14 asymptotic stability of the origin. • If we e.g., consider Ω : x21 + x22 ≤ 1 then M = {(x1 , x2 )|x1 = 0, x2 = 0} and we have proven EL2620 Lecture 14 In particular, if the compact region does not contain any fixed point then the ω -limit set is a limit cycle 20 2012 18 Poincare-Bendixson Any orbit of a continuous 2nd order system that stays in a compact region of the phase plane approaches its ω -limit set, which is either a fixed point, a periodic orbit, or several fixed points connected through homoclinic or heteroclinic orbits Relation to Poincare-Bendixson Theorem EL2620 Question 7 ẋ1 = −ax1 (t) Lecture 14 Choose large a: fast convergence along sliding manifold σ(x) = ax1 + x2 = 0 ⇒ ẋ1 = x2 (t) ẋ2 = x1 (t)x2 (t) + u(t) Example Choose S for desired behavior, e.g., EL2620 Lecture 14 3. The equivalent control 2. The sliding control 1. The sliding manifold There are 3 essential parts you need to understand: 23 2012 21 2012 Please repeat the most important facts about sliding modes. EL2620 Step 1. The Sliding Manifold S S of 2012 Step 2. The Sliding Controller Lecture 14 Example: f (x) u=− 2012 u = −x1 x2 − x2 − sgn(x1 + x2 ) 24 f (x) + x2 + sgn(σ) g(x) = x1 x2 , g(x) = 1, σ = x1 + x2 , yields ⇐ = x2 , ẋ2 = f (x) + g(x)u and = 0.5σ 2 yields V̇ = σ σ̇ V̇ = σ (x2 + f (x) + g(x)u) < 0 For 2nd order system ẋ1 σ = x1 + x2 we get Lyapunov function V (x) Use Lyapunov ideas to design u(x) such that S is an attracting invariant set EL2620 22 ∈ R2 then S is R1 , i.e., a curve in the state-plane (phase plane). Lecture 14 If x and make S invariant S = {x ∈ Rn |σ(x) = 0} σ ∈ R1 Idea: use u to force the system onto a sliding manifold dimension n − 1 in finite time ẋ = f (x) + g(x)u, x ∈ Rn , u ∈ R1 Aim: we want to stabilize the equilibrium of the dynamic system EL2620 2012 ⇒ ueq = −x2 − x1 x2 Question 8 Lecture 14 Can you repeat backstepping? EL2620 Lecture 14 27 2012 25 Thus, the sliding controller will take the system to the sliding manifold S in finite time, and the equivalent control will keep it on S . σ̇ = ẋ1 + ẋ2 = x2 + x1 x2 + ueq = 0 Example: However, an equivalent control ueq (t) that keeps x(t) on S can be computed from σ̇ = 0 when σ = 0 ∈ S , then u will chatter Step 3. The Equivalent Control When trajectory reaches sliding mode, i.e., x (high frequency switching). EL2620 Note! Backstepping Design Lecture 14 In this course we only consider f (x, u) with a special structure, namely strict feedback structure V (x) > 0 , V̇ (x) < 0 ∀x ∈ Rn 28 2012 26 2012 General Lyapunov control design: determine a Control Lyapunov function V (x, u) and determine u(x) so that ẋ = f (x, u) We are concerned with finding a stabilizing control u(x) for the system EL2620 Lecture 14 This is OK, but is not completely general (see example) u = −sgn(σ) Previous years it has often been assumed that the sliding mode control always is on the form EL2620 The Backstepping Result 2012 29 2012 Lecture 14 31 dφ dV u(x) = f (x1 )+g(x1 )x2 − g(x1 )−(xk −φ(x1 ))−f2 (x1 , x2 ) dx1 dx1 with corresponding controller ẋ1 = f1 (x1 ) + g1 (x1 )x2 ẋ2 = f2 (x1 , x2 ) + u − φ(x1 ))2 /2 is a Control = φ(x1 ). Then V2 (x1 , x2 ) = V1 (x1 ) + (x2 Lyapunov Function for the system with corresponding controller u ẋ1 = f1 (x1 ) + g1 (x1 )u Let V1 (x1 ) be a Control Lyapunov Function for the system EL2620 6= 0 ẋn = fn (x1 , . . . , xn ) + gn (x1 , . . . , xn )u .. . ẋ1 = f1 (x1 ) + g1 (x1 )x2 ẋ2 = f2 (x1 , x2 ) + g2 (x1 , x2 )x3 ẋ3 = f3 (x1 , x2 , x3 ) + g3 (x1 , x2 , x3 )x4 Strict Feedback Systems x1 , . . . , xk do not depend on xk+2 , . . . , xn . Lecture 14 Note: where gk EL2620 The Backstepping Idea Lecture 14 EL2620 Lecture 14 Repeat backlash compensation Question 9 x˙1 = f1 (x1 ) + g1 (x1 )x2 x˙2 = f2 (x1 , x2 ) + u find a Control Lyapunov function V2 (x1 , x2 ), with corresponding control u = φ2 (x1 , x2 ), for the system ẋ1 = f1 (x1 ) + g1 (x1 )u Given a Control Lyapunov Function V1 (x1 ), with corresponding control u = φ1 (x1 ), for the system EL2620 32 2012 30 2012 Backlash Compensation − e 1+sT2 K 1+sT 1 1 1 + sT θ̇in 1 s Question 10 u θin Lecture 14 Can you repeat linearization through high gain feedback? EL2620 Lecture 14 θref Linear controller design: Phase lead compensation • Backlash inverse • Linear controller design • Deadzone EL2620 33 35 2012 θout 2012 -5 without filter 10 -2 -1 0 0 1 5 u, with/without filter 5 y, with/without filter Oscillation removed! 5 with filter 2 0 0 0.5 1 1.5 10 10 Inverting Nonlinearities Real Axis 0 Nyquist Diagrams fˆ−1 (·) f (·) Lecture 14 Should be combined with feedback as in the figure! − F (s) Controller 15 15 G(s) Compensation of static nonlinearity through inversion: EL2620 Lecture 14 -10 -10 -8 -6 -4 -2 0 2 4 6 8 10 1+sT2 F (s) = K 1+sT with T1 = 0.5, T2 = 2.0: 1 describing function is removed 20 20 36 2012 34 2012 • Choose compensation F (s) such that the intersection with the EL2620 Imaginary Axis − ê f (·) k s ê = v − f (u) u Lecture 14 EL2620 Lecture 14 What about describing functions? Question 12 > 0 large and df /du > 0, then ê → 0 and 0 = v − f (u) ⇔ f (u) = v ⇔ If k v f (u) 39 2012 u = f −1 (v) u 37 2012 Remark: How to Obtain f −1 from f using Feedback EL2620 What should we know about input–output stability? Question 11 kyk2 kuk2 n=1 N.L. u G(s) y a2n + b2n sin[nωt + arctan(an /bn )] − e 2012 38 2012 Lecture 14 40 ≪ |G(iω)| for n ≥ 2, then n = 1 suffices, so that q y(t) ≈ |G(iω)| a21 + b21 sin[ωt + arctan(a1 /b1 ) + arg G(iω)] If |G(inω)| u(t) = ∞ X p e(t) = A sin ωt gives r Idea Behind Describing Function Method EL2620 Lecture 14 • Passivity Theorem • Circle Criterion • Small Gain Theorem • BIBO stability • System gain γ(S) = supu∈L2 You should understand and be able to derive/apply EL2620 N.L. e(t) N (A, ω) u b1 (t) More Courses in Control Lecture 14 • EL2421 Project Course in Automatic Control, per 2 • EL1820 Modelling of Dynamic Systems, per 1 • EL2520 Control Theory and Practice, Advanced Course, per 4 • EL2745 Principles of Wireless Sensor Networks, per 3 • EL2450 Hybrid and Embedded Control Systems, per 3 EL2620 Lecture 14 = 0 then u(t) u b1 (t) = |N (A, ω)|A sin[ωt + arg N (A, ω)] ≈ u(t) If G is low pass and a0 e(t) N (A, ω) = b1 (ω) + ia1 (ω) A Definition of Describing Function The describing function is EL2620 43 2012 41 2012 y ⇒ A G(iω) = − G(iω) −1/N (A) y = G(iω)u = −G(iω)N (A)y e f (·) u G(s) − Existence of Periodic Solutions EL2520 Control Theory and Practice, Advanced Course 1 N (A) Lecture 14 Contact: Mikael Johansson [email protected] • Lectures, exercises, labs, computer exercises – Real time optimization: Model Predictive Control (MPC) – Design of multivariable controllers: LQG, H∞ -optimization – Robustness and performance – Linear multivariable systems • Multivariable control: • Period 4, 7.5 p 44 2012 42 2012 Aim: provide an introduction to principles and methods in advanced control, especially multivariable feedback systems. EL2620 Lecture 14 The intersections of the curves G(iω) and −1/N (A) give ω and A for a possible periodic solution. 0 replacements EL2620 EL2450 Hybrid and Embedded Control Systems EL1820 Modelling of Dynamic Systems Lecture 14 Contact: Håkan Hjalmarsson, [email protected] • Lectures, exercises, labs, computer exercises • Computer tools for modeling, identification, and simulation – experiments: parametric identification, frequency response – physics: lagrangian mechanics, electrical circuits etc • Model dynamical systems from • Period 1, 6 p Aim: teach how to systematically build mathematical models of technical systems from physical laws and from measured signals. EL2620 Lecture 14 Contact: Dimos Dimarogonas [email protected] • Lectures, exercises, homework, computer exercises – Control over communication networks – Scheduling of real-time software – Computer-implementation of control algorithms • How is control implemented in reality: • Period 3, 7.5 p Aim:course on analysis, design and implementation of control algorithms in networked and embedded systems. EL2620 47 2012 45 2012 EL2745 Principles of Wireless Sensor Networks – Research topics in WSNs – Design of practical WSNs Lecture 14 Lecture 14 Contact: Jonas Mårtensson [email protected] • No regular lectures or labs • Project management (lecturers from industry) • Preparation for Master thesis project • Team work • “From start to goal...”: apply the theory from other courses • Period 4, 12 p 46 48 2012 Aim: provide practical knowledge about modeling, analysis, design, and implementation of control systems. Give some experience in project management and presentation. EL2421 Project Course in Control Contact: Carlo Fischione [email protected] EL2620 2012 – Essential tools within communication, control, optimization and signal processing needed to cope with WSN • THE INTERNET OF THINGS • Period 3, 7.5 cr Aim: provide the participants with a basic knowledge of wireless sensor networks (WSN) EL2620 Lecture 14 • Check old projects • Discuss with professors, lecturers, PhD and MS students • The topic and the results of your thesis are up to you Hints: ◦ Get insight in research and development ◦ Collaboration with leading industry and universities ◦ The research edge ◦ Cross-disciplinary ◦ Theory and practice 49 2012 Doing Master Thesis Project at Automatic Control Lab EL2620 Doing PhD Thesis Project at Automatic Control Lecture 14 • 4-5 yr’s to PhD (lic after 2-3 yr’s) teaching (20%), fun (100%) • Research (50%), courses (30%), • World-wide job market • Competitive • International collaborations and travel • Get paid for studying • Intellectual stimuli EL2620 50 2012