FRTN10 Exercise 12. Synthesis by Convex Optimization

Transcription

1 FRTN Exercise We want to design a controller C for the stable SISO process P as shown in Figre 2. sing the Yola parametrization and convex optimization. To do this, the control loop mst first be transformed into the standard form of Figre 2.2, where z are the signals that we want to control, y are the signals available to the controller, w are the exogenos inpts and is the control signal. y Σ n x P Σ d C r Figre 2. The control loop in problem 2. z y P w C Figre 2.2 Desired form of the control loop in problem 2. The signals z and w are given by ( ) d e z =, w = n, r where the control error is e = r x. The controller C, which is a 2 transfer fnction, is the same in both figres, as is the control signal. a. What is the controller inpt y of Figre 2.2 according to Figre 2.? What is the size of the transfer matrix P? b. Find the transfer matrix of the generalized plant P so that Figre 2.2 and Figre 2. describe the same control problem. c. The Yola parametrization reslts in the closed-loop system z = Hw where the transfer fnction H is given by H = P zw + P z QP yw.

2 Exercise 2. The control objectives are (a) To make the gain H ij for all elements H ij. (b) Dring an implse distrbance experiment in d, the control signal shold satisfy (t). (c) Dring an implse distrbance experiment in d, from two seconds onward the control error shold be small: e(t).75, t 2, if the implse occrs at t =. Two transfer fnctions Q and Q 2 have been fond that satisfy objective a). Figre 2.3 shows implse responses from d to e and when sing the corresponding controllers C and C 2. Find a Q that satisfies all three objectives a), b), c). Implse Response.2.4 Amplitde e C C Time (sec) Figre 2.3 Implse responses from distrbance d to control error e (top) and control signal (bottom) for the controllers C and C 2 in problem In Problem.2 we considered a mass-spring system. If yo haven t completed that problem or don t remember it, yo shold go throgh it before solving this problem. In this problem we are going to find an (almost) optimal controller with respect to the cost fnction and constraints given in Problem.2. We are going to do this by reformlating the problem via the Yola parametrization, and then choose a finite basis for the Q parameter and a finite nmber of points where the time domain and freqency domain constraints shold be enforced. The problem is then easy to solve sing a convex optimization tool sch as sedmi or sdpt3. Since it is time-consming to interface directly with the solver, we will secvx. cvx is a Matlab-based modeling system for convex optimization. Before translating the problem to a sitable inpt to the solver,cvx checks if the problem 2

3 Exercise 2. satisfies the Disciplined Convex Programming (DCP) rle-set; if the DCP rles are satisfied the problem is garanteed to be convex, and if they are not satisfied the problem is not processed. cvx is available on the lab compters, bt it can also be downloaded at together with the solvers sdpt3 and sedmi. The files for the exercise are fond on the corse web page. The main file is spring_mass_problem.m, and the files qcvx_* help yo to set p the Yola parametrization and recover the controller. In spring_mass_problem.m, yo will be able to modify the objective fnction and the constraints. We will formlate and solve the problem in discrete time, as it gives a straightforward parametrization of Q in the time-domain. If yo are not sed to working with discrete-time systems don t worry! Jst consider the discretetime system an approximation of the continos-time system. To get a finite dimensional problem, the time-domain and freqency domain constraints can only be enforced in a finite nmber of points. For example, the constraints on the time domain signals are only enforced for the first 7 samples, which, given the sampling time of.2 seconds, amonts to the first 34 seconds. It is important that this interval covers the time-scale of the system dynamics. The mass-spring system has two poles in s = and is ths nstable. In order to get the Q optimization working we need to stabilize the plant with a nominal controller. The final controller will then be an agmented version of this stabilizing nominal controller. The fnction qcvx_q_parametrization.m designs this nominal controller and retrns the transfer fnction matrices T, T 2, and T 3, for the Q-parametrization T + T 2 QT 3, as well as state space data reqired to recover the optimal controller from Q. a. To get started yo need to rn setp_cvx and then cvx_setp. Then rn spring_mass_problem.m and stdy the otpt. A soltion satisfying the constraints of the problem has been compted. Look at the code in the script and try to nderstand it. The constraints on the nit step response from r to d are visalized in Figre 2.4. Which constraints are active? A constraint is active if the soltion toches the constraint at any point. b. Plot the Bode diagram of the controller. Can yo recognize any resemblance with any other type of controller? Can yo give some intitive explanation to any of the dips in the magnitde plot? c. The otpt from the solver looks something like this: 3

4 Exercise 2. Step Response Specifications os +kp t kp t.8 Magnitde t 5 5 t Time (s) Figre 2.4 Constraints on the closed-loop step response from the reference, r, to the position of the first mass, d. os is the maximm allowed overshoot. Calling SeDMi.34: 823 variables, 222 eqality constraints For improved efficiency, SeDMi is solving the dal problem SeDMi.34 (beta) by AdvOL, and Jos F. Strm, Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta =.25, beta =.5, eqs m = 222, order n = 285, dim = 824, blocks = 22 nnz(a) = , nnz(ada) = 4682, nnz(l) = 2452 it : b*y gap delta rate t/tp* t/td* feas cg cg prec :.E+3. : -2.8E+ 3.99E E+ 2 : -8.46E+ 2.37E E : -.42E+2.4E E-8 3 : -.42E E E Stats: Solved Optimal vale (cvx_optval): Let s not go into detail on what all colmns stand for, bt rather look only at the colmnfeas. The nmber yo end p with shold ideally be, or at least close, in order for yo to have a feasible soltion (one that satisfies all constraints). If yo end p with a vale of, then yo are certain that there is no soltion. The higher yo choose the order of yor Q filters, the more likely it is that a soltion is fond. If yo do not find a feasible soltion, even thogh the order of the Q filters is large, this tells yo that yo need to relax the constraints in order to find a controller. 4

5 Exercise 2. Take yor code and decrease NQ=n_q=n_q2 ntil yor problem is no longer feasible. What is the least ordernq yo need in order to to obtain a feasible soltion? d. Increase the order of Q and Q 2 simltaneosly (NQ=n_q=n_q2) from the vale yo obtained in the previos sbproblem and plot NQ against the cost fnction vale. Explain the shape of this plot and comment on how the max vale of the control signal changes when NQ increases. Hint: Comment ot the line where NQ is set, and rn the script spring_mass_problem.m from a new script where yo change NQ and plot the vale of optimal_cost. e. Change the weights in the objective fnction and see how this alters the soltion. Also play arond with the constraints to see if yo can achieve an extremely fast step response. Explain yor reslts. f. Take a look in the other m-files and see if yo can find the reslting closedloop transfer fnction matrix. Plot all 9 closed-loop Bode magnitde diagrams. Point ot at least one of these that yo wold have wished had a different appearance. How wold yo have preferred it to look and why? g. Plot the effect of a step distrbance on the signal p. As yo saw previosly in part b, the low-freqency gain was very low, i.e., the controller did not have integral action. Is this consistent with the distrbance response? Add a constraint on the step response from d to p and see if yo can improve the distrbance rejection. Look at the gain crve of the new controller. h. (*) Read more abot convex optimization in and more abot finding the limits of performance of linear control systems sing optimization in 5

6 Soltions 2. Soltions to Exercise 2. Synthesis by Convex Optimization 2. a. The inpts to the controller are r and y, i.e. ( ) r y =. ( ) w The inpt to P is, which contains 4 signals, and the otpt is which contains 4 signals as well. Ths P mst be 4 4. y ( ) z, y b. We know that e ( ) z = y r, Setting C =, the block diagram gives that y d ( ) w n = r. e = r x = r P (d + ), =, r = r, y = n+ P (d + ). Arranging this into matrix form gives the generalized plant model P P P =. P P c. The control objective (a) is convex in H, and H is a linear fnction of Q, so the control objective a) is convex in Q. Since it is satisfied for Q and Q 2, it is ths satisfied for any convex combination Q = wq +( w)q 2, w [, ]. We see from the implse responses that neither Q nor Q 2 satisfies (b) or (c). However, a convex combination of Q and Q 2 will give the same convex combination of the distrbance responses. Taking e.g. w =.7, The control signal satisfies (t) =.88, since (t).4 with C and (t) 2 with C 2. When t 2, the control error satisfies e(t) =.73, since e(t) with C and e(t). with C 2. Ths we can se Q =.7Q +.3Q a. See the plots in Figres and the associated captions. We see that the only inactive constraint is the one on the control signal. Also see Figre 2.3 for a Nyqist plot with a circle for the M s constraint. 6

7 Soltions 2..2 Step resonse r -> p Step response r -> Figre 2. Step responses from reference Time r to [s] mass position p and to control signal. b. bodemag(k2d) The gain crve is shown in Figre 2.4. The shape of the magnitde plot is very similar to that of a PD controller. Since the D-part of a PID controller acts to damp ot oscillations, it makes sense that we have this kind of similarity. The system has its resonance freqency at 5.8 rad/s, which is almost exactly at the same freqency as the deepest dip in the controller magnitde plot. The reason for the dip is that we do not want to amplify signals at this freqency. A PD controller does not have this kind of flexibility in its strctre to damp ot a certain freqency and is therefore not so well sited for highly oscillatory systems like this one. c. The least vale onnq, for which or problem is feasible, is 7. d. See Figre 2.5 for a plot of the cost fnction vale verss the order of the Q filters. WhenNQ reaches arond 2, the control will gain very little from increased complexity of the Q filter. We can then say that we have a good estimate of the limit of performance, i.e. lowest cost that linear controller can achieve given the problem setp. The maximm vale of the control signal will decrease as the order of the Q 7

8 Soltions 2. Sensitivity fnction - - Freqency [rad/s] Figre 2.2 The sensitivity fnction of the system..5 Nyqist Plot.5 Im Re Figre 2.3 sed. The open-loop Nyqist crve and the M s circle when the optimal controller is filters goes p. ForNQ= [7,, 5, 2], we get max = [6., 4.9, 3.8, 3.7]. This means that the more complex the controller becomes, the more freedom it will have to choose its control signal. As it is good to have a control signal that is low on energy (de to the cost fnction), it is also likely that it goes down in magnitde if it has the possibility. e. If we start ot by setting ρ = (i.e.weight_ =.), then we do not pnish the control signal energy at all, which means that we may get very aggressive and poorly damped control. The constraint on max will to some extent prevent this, bt if max is made arbitrarily large then we can get a step responses like the one in Figre 2.6. If instead γ = while ρ remains, then the soltion will remain fairly nchanged. The reason is that both the constraints on rise time and the cost of having a large e will force to be qite active 8

9 Soltions 2. 5 Bode Diagram 4 Magnitde (abs) Freqency (rad/s) Figre 2.4 Bode diagram of the controller Cost fnction vale N Q =n q =n q2 Figre 2.5 Cost fnction vale plotted against the complexity of the Q filters. still. If the step response constraints are made inactive, the soltion will be fairly close to the one of the nominal LQG controller. f. Looking in the files we find thatgcl corresponds to the closed loop transfer matrix. We can plot the magnitde crves with the command bodemag(gcl). We see that the transfer fnction in the middle, H o n, does not have high 9

10 Soltions 2. Position of the first mass, d.8 Amplitde Time (sec) Figre 2.6 p de to a reference step when ρ =. The control is very aggressive. freqency roll-off. High freqency measrement noise, n, may therefore lead to a very noise control signal. This shows that it is very important to take all signals in a system into consideration and that a soltion, even thogh optimal, might not be good. Remember, yo get what yo ask for. If we were to modify the problem, a good idea wold be to pt constraints on this closed-loop transfer fnction as well. g. First define the constraint for the step response d p, b_cl_diststep = min(, exp(-.2*(t-5))); and then add it to the optimization problem with -b_cl_diststep <= cl_stepresp(:,,3) <= b_cl_diststep; If we try to solve the problem, we will see that it is has become infeasible. By increasing the order of the Q-filters by pttingnq=3, yo will be able to find a soltion. See figres 2.7 and 2.8.

11 Soltions Time [s] Figre 2.7 Distrbance step when the controller has been designed with respect to the distrbance rejection constraint (ble) and the original controller (red). 5 Bode Diagram 4 Magnitde (abs) Freqency (rad/s) Figre 2.8 Controller gain when the controller has been designed with respect to a distrbance rejection constraint (ble) and original controller (red). Note that the low freqency gain is mch higher for the controller that was designed for step distrbance rejection, this can be seen as added integral action.