Passive control. Carles Batlle. II EURON/GEOPLEX Summer School on Modeling and Control of Complex Dynamical Systems Bertinoro, Italy, July

Transcription

1 Passive control theory II Carles Batlle II EURON/GEOPLEX Summer School on Modeling and Control of Complex Dynamical Systems Bertinoro, Italy, July

2 Contents of this lecture Interconnection and Damping Assignment Passivity Based Control (IDA-PBC) Magnetic levitation system Boost converter DC motor How to solve quasilinear PDEs

3 IDA-PBC Control-as-interconnection has some problems: Nonlinear PDE for the Casimir functions Dissipation obstacle Both problems can be somehow overcomed by considering state-modulated interconnection feedback and controllers with energy function not bounded from below However, some intuition is lost in the process, so it may be better to go for a more radical approach, which allows much more flexibility, at the expense of immediate physical intuition

4 Idea: try to find a feedback control such that the closed-loop system is ẋ =(J d (x) R d (x)) H d x (x) J T d = J d R T d = R d 0 Interconnection assignment instead of just Damping assignment ẋ =(J(x) R(x)) H d x (x) with H d with a global minimum at the desired regulation point x To do that, one just matches the original dynamics to the desired one (J(x) R(x)) x H(x)+ g(x)β(x) =(J d (x) R d (x)) x H d (x)) closed-loop control u Matching equation

5 The formal result is as follows Find a (vector) function K(x), a function β(x), a skew-symmetric matrix J a (x), and a symmetric, semipositive definite matrix R a (x) suchthat (J(x)+J a (x) R(x) R a (x))k(x) = (J a (x) R a (x)) H x (x)+g(x)β(x) with K the gradient of an scalar, K(x) = x H a (x). Then the closed-loop dynamics with u = β(x) isaphdswith H d = H + H a, J d = J + J a and R d = R + R a with everything else fixed, this is a PDE for H a (x) However, we can try to select J a and R a to make its solution easier

6 Magnetic levitation system u i φ = Ri + u ẏ = v m v = F m + mg F m = y W c (i, y) W c = 1 2 L(y)i2 m g y ẋ = L(y) = k a + y x 1 = φ, x 2 = y, x 3 = mv = p R H x u H(x) = 1 2k (a + x 2)x m x2 3 mgx 2 magnetic co-energy expresed in energy variables (coincides with energy due to the linearity φ = L(y)i)

7 ( H) T = a+x 2 k x 1 1 2k x2 1 mg x 3 m x = Given a desired equilibrium point y 2kmg y 0 u = R k x 1(a + x 2) Set first J a =0,R a =0 (J(x)+J a (x) R(x) R a (x))k(x) = (J a (x) R a (x)) H x (x)+g(x)β(x) (J R)K(x) =gβ(x) RK 1 (x) = β(x) K 3 (x) = 0 K 2 (x) = 0 H a (x) =H a (x 1 )

8 Unfortunately 2 H d x 2 (x) = 1 k (a + x 2)+Ha 00 (x 1 ) x 1 x 1 k 0 k m has at least one negative eigenvalue no matter which H a we choose no minimum at x Let us try something different and put R a =0but J a = 0 0 α α 0 0

9 (J(x)+J a (x) R(x) R a (x))k(x) = (J a (x) R a (x)) H x (x)+g(x)β(x) αk 3 RK 1 (x) = α m x 3 + β(x) K 3 (x) = 0 H a = H a (x 1,x 3 ) αk 1 (x) K 2 (x) = α k (a + x 2)x 1 u = β(x) =RK 1 α x 3 m α H a x 1 H a = α x 1(a + x 2 ) x 2 k This is a quasilinear PDE for H a and we have to solve it

10 Method of characteristics Equations of the form a(x, y, u)u x + b(x, y, u)u y = c(x, y, u) where u x = x u(x, y), u y = y u(x, y) are called quasilinear because the derivatives of u appear linearly. The method of characteristics works as follows. Construct the following system of ODE for x(τ), y(τ), u(τ) x 0 (τ) = a(x(τ),y(τ),u(τ)) y 0 (τ) = b(x(τ),y(τ),u(τ)) u 0 (τ) = c(x(τ),y(τ),u(τ)) the solutions are called characteristic curves and their projections on u =0 are simply called characteristics

11 x 0 (τ) = a(x(τ),y(τ),u(τ)) y 0 (τ) = b(x(τ),y(τ),u(τ)) u 0 (τ) = c(x(τ),y(τ),u(τ)) We introduce next a curve of initial conditions (x(0,s),y(0,s),u(0,s)) parameterized by s If the curve of initial conditions does not lie on a characteristic curve, their evolution will generate a two dimensional manifold in R 3 Finally, from (x(τ,s),y(τ,s),u(τ,s)) x = x(τ,s) we can eliminate τ and s in terms of x and y and obtain the solution u(x, y) tothepde y = y(τ,s) u = u(τ,s) The solution depends on arbitrary functions specifying the curve of initial conditions

12 As an example, consider 3u x +5u y = u with an initial curve (s, 0,f(s)) where f is arbitrary. x 0 = 3 y 0 = 5 u 0 = u x = 3τ + s y = 5τ u = f(s)e τ We get τ = y/5 ands = x 3y/5 andthen u(x, y) =f(x 3y 5 )e y 5 Exercise. Solve the PDE for H a (x 1,x 2 ) for the levitating system. It is better to give the initial condition curve in the form (s, 0,f(s).

13 Boost converter Consider the averaged model of the boost converter, where we set u =1 S: J (u) = µ 0 u u 0, R = µ 1/R H(x 1,x 2 )= 1 2C x L x2 2 µ 0, g = 1 The control goal is to regulate the load voltage (resistor) at a desired value V d input voltage x =(CV d, LV d 2 RE ) u = E 1 V d output (load) voltage

14 One can get a controller by setting J a =0,R a =0, but a better one can be obtained if J a =0but R a = µ 1/R 0 0 r a so that R d = µ r a with r a > 0. The IDA-PBC equation is now µ µ 0 u 1 H a u r a 2 H a = µ 1 R 0 0 r a µ x1 C x 2 L + µ 0 1 E The standard trick when the control appears in both equations is to isolate the two partial derivatives x 1 1 H a = r a RC u r a 2 L 2 H a = 1 x 1 RC u x 2 u E u

15 x 1 1 H a = r a RC u r a 2 L 2 H a = 1 x 1 RC u x 2 u E u since 2 1 H a = 1 2 H a, we get, with α =1 r a RC/L, µ 2r a x 1 + r arc L ux 2 + ERCu 2 u x 1 u 1 u + αu 2 =0 This is a PDE of the kind we know how to solve. However, if we look for solutions of the form u = u(x 1 ) x 1 1 u = αu with solution, satisfying the appropiate fixed point limit, u(x 1,x 2 )= E µ α x1 V d x which makes sense provided that α 0. 1 It remains to be checked that one can obtain an H d with a minimum at the desired point.

16 DC motor Consider a DC motor for which we do not consider the field coil dynamics (or it has just a permanent magnet). The system is then 2-dimensional, with port Hamiltonian structure ẋ =(J R) x H + g + g u u H(λ,p)= 1 2L λ J p2 J = µ 0 K K 0 R = µ r 0 0 b g = µ 0 τ L g u = µ 1 0 Assume the control objective is regulation of the mechanical speed to ω = ω d. i = 1 K (bω d + τ L ) u = ri + Kω d

17 Here we adopt a new completely different approach to solve the IDA-PBC equation. We impose an desired Hamiltonian with the correct minimum andthendeterminej a and R a so that the equation is satisfied. H d = 1 2L (λ λ ) J (p p ) 2 Infact,inthiscaseitisbettertoworkwiththeMatchingEquation instead of the IDA-PBC one (they are the same) µ We go for a completely arbitrary rd j J d R d = d j d b d The second row of the matching equation imposes j d (i i ) b d (ω ω d )=Ki bω τ L Setting b d = b, and using the equilibrium point expression, j d = K.

18 Finally,using this and going to the first row of the matching equation, u = r d (i i ) ri + Kω d wherewestillhaver d > 0freetotunethecontroller. Notice that this controller depends on i,whichinturn depends on the mechanical torque τ L. This makes this controller useless as a regulating speed controller, since it will not be able to do the job if the external torque changes. We know from PID theory that this can be solved by adding an integrator: Z u i = r d (i i ) ri + Kω d (ω ω d )dt This new controller has been obtained outside the Hamiltonian framework. work in progress

19 Controller comparison. ω d =250,andτ L is changed at t = Mechanical speed (w) w [rad/s] IDA PBC IDA PBC integral time [s]

20 Conclusions Energy based control is well suited for energy based modeling. Physical structure can be incorporated. Open problems and extensions: Robustness. Non regulation problems. Distributed systems: boundary and bulk control. Other ideas.