非线性系统理论

Introduction of Lyapunov-Based Control1An Example of Nonlinear SystemsLinear System ˙x=Ax+Buy=Cx(1) it has the superposition property.Besides,the stability of the linear system completely depends on its parameters.Nonlinear System ˙x=f(x,u)y=g(x)(2)superposition does not hold for nonlinear systems,and the stability of a nonlinear system depends on both system parameters and initial conditions.Example:The dynamic model for a2-DOF overhead crane system(see Figure??)can be presented as followsM(q)¨q+V m(q,˙q)˙q+G(q)=u(3)q=[x(t)θ(t)]T(4) where x(t)∈R1denotes the gantry position,θ(t)∈R1denotes the payload angle with respect to the vertical,and M(q)∈R2×2,V m(q,˙q)∈R2×2,G(q)∈R2,and u(t)∈R2are defined as followsM(q)= m c+m p−m p L cosθ−m p L cosθm p L2 ,V m(q,˙q)= 0m p L sinθ˙θ00 ,G(q)= 0m p gL sinθ T,u(t)= F0 T,(5)where m c,m p∈R1represent the gantry mass and the payload mass,respectively,L∈R1represents the length of the rod to the payload,g∈R1represents the gravity coefficient,and F(t)∈R1 represents the control force input acting on the gantry(see Figure??).2Common Nonlinear Systems Behaviors2.1Multiple Equilibrium PointsFor the system˙x=f(x)(6)1Figure1:To obtain equilibrium points,solve the following equationf(x s)=0if it has multiple solutions,then the system has multiple equilibrium points.Example:˙x=−x+x2(7)−x+x2=0=⇒x s=1or x s=0has too equilibrium points:x s=1and x s=0.Question:Which one is stable?Why?Rewrite the equation as follows˙x=x(x−1)then if x>1,˙x>0,x increases with time.For the case of x<1,(x−1)<0,thus˙x<0,for1>x>0˙x>0,for x<0therefore,x s=1is unstable and x s=0is stable.We can solve the system equation to obtain the response as followsx(t)=x0e−t1−x0+x0e−t=1−1−x01−x0+x0e−tFrom this formula,it can be seen that if x0<1,the denominator1−x0+x0e−t>0,thuslimt→∞x(t)=0.The response of x0=0.99and x0=1.0001is demonstrated in Figure??.2Figure2:2.2Limit CycleLimit Cycle:Oscillation offixed amplitude andfixed frequency without external excitation.Example:¨x+(x2−1)˙x+x=0if|x|>1,x2−1>0,the damper consumes energy.If|x|<1,x2−1<0,the damper produces energy.Therefore,the system has a tendency to make x oscillate around some region.Response ofx0=5,˙x0=5andx0=−1,˙x0=−3is shown in Figure32.3ChaosThe system output is extremely sensitive to initial conditions.Example:¨x+0.1˙x+x5=6sin(t)Figure4shows the response of the system to two almost identical initial conditionsx0=2,˙x0=3andx0=2.01,˙x0=3.01.Due to the high nonlinearity in x5,the response become apparently different after a certain time.3Figure3:42.4Other BehaviorsSaturation,dead zone,hysteresis,and so on.3Why Nonlinear ControlAll physical systems are nonlinear in nature.Linear model is only an approximation.Conven-tionally,the nonlinear system is linearized around its operating point and a linear controller is then designed for the obtained simplified model.In some cases,this cannot achieve satisfactory performance.1.To improve system performance;2.Linerization cannot provide correct solutionExample:for the same system˙x=−x+x2Linearizing it yields˙x=−xit has only1stable equilibrium point at x s=0??3.To deal with model uncertainties:Example:˙x=f(x,u)+g(x,u)where f(x,u)denotes the modeled dynamics,while g(x,u)represents unmodeled uncertainties including noise,disturbance,etc.4General Nonlinear Control4.1Phase Plane Analysis:A graphical method mainly for2nd-order systems.4.2Describing System AnalysisAn extended version of the frequency response method,it can be used to approximately analyze and predict nonlinear behavior.Main Use:Prediction of limit cycles in nonlinear systems,unfortunately only an approximate way.4.3Lyanpunov MethodIt isfirst introduced to judge the stability of a nonlinear system.Stable:state not blow up,how to judge?Linear System:Route Criteria,Nyquist CriteriaExample for Stability Analysis:˙x=12sin(x)−xIs this system stable?5ChooseV=1 2 x2then V∈L∞(V remains bounded)implies x∈L∞.Taking the time derivative of V yields˙V=x˙x=x 12sin(x)−x≤−x2+12|x sin(x)|≤−x2+12x2≤−12x2therefore,˙V≤−VandV(t)≤V0e−tx(t)≤x0e−12tQuestion:If V cannot be solved out,how to judge the stability from the differential equation or inequalities?Lyapunov Theorem,Babalat’s Lemma,and so on.Besides stability analysis,the Lyapunov method is also a powerful tool to design nonlinear controllers.Example:Suppose we have the following system˙x=f(x)+g(x)uwhere f(x),g(x)are known functions and u is the control input.Besides,g(x)N g0with g0being positive constant.How to design a controller to regulate x?ChooseV=1 2 x2then we take its time derivative and substitute into the system dynamics to obtain˙V=x˙x=x(f(x)+g(x)u)Can we choose a suitable controller u to make˙V≤0or further,˙V≤−x2?Makeu=−f(x)g(x)−xg(x)with k representing a positive control gain.Then˙V=−x2Similarly as the example above,x goes to zero exponentially fast.Question1.Why do we need the assumption of g(x)N g0?To make the controller free of singularity,or make the system controllable.Question2.There are two terms within the controller u,which one is feedback,which one is feedforwad?What are they for?6Advantage:Implement the controller design and stability analysis simultaneously;Backbone of the existing nonlinear controllers;Heart and soul of model control.Disadvantage:hard to construct a suitable Lyapunov function for a give complex dynamic system,conservative method:the conclusion you made for a system based on the Lyapunov analysis can be weaker than its actual situation.We will show this in the future class.5Differences between Linear Control and Lyapunov Con-trola)Model-Free Control vs Model-Based Controlb)Feedback Control vs Feedforward Control6HomeworkUse Matlab/Simulink to simulate the following system¨x+0.1˙x+x5=6sin(t)for two set of initial conditionsx0=2,˙x0=3andx0=2.01,˙x0=3.017。