EECE 460 Decentralized Control of MIMO Systems Guy A. Dumont Department of Electrical and Computer Engineering University of British Columbia January 2011 Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 1 / 32
Contents university-logo 1 Introduction 2 Decentralized Control Example 3 Pairing of Inputs and Outputs Robustness Issues Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 2 / 32
Introduction Introduction In the case of SISO control, we found that one could use a wide variety of synthesis methods. Some of these carry may over directly to the MIMO case. However, there are several complexities that arise in MIMO situations. For this reason, it is often desirable to use synthesis procedures that are in some sense automated. Full MIMO control design is the topic of courses such as EECE 568. In this course we investigate when, if ever, SISO techniques can be applied to MIMO problems directly. We will study Decentralized control as a mechanism for directly exploiting SISO methods in a MIMO setting Robustness issues associated with decentralized control. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 3 / 32
Decentralized Control Completely Decentralized Control university-logo Before proceeding to a fully interacting multivariable design, it is often useful to check on whether a completely decentralized design can achieve the desired performance objectives. When applicable, the advantage of a completely decentralized controller, compared to a full MIMO controller, is that it is often simpler to understand is often easier to maintain can be enhanced in a straightforward fashion in case of a plant upgrade. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 4 / 32
Decentralized Control Decentralized Control university-logo Exposure to practical control reveals that a substantial proportion of real-world systems utilize decentralized architectures. Thus, one is led to ask the question, is there ever a situation in which decentralized control will not yield a satisfactory solution? In Chapter 22 of the textbook several real-world examples that require MIMO thinking to get a satisfactory solution are presented. As a textbook example of where decentralized control can break down, consider the following MIMO example. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 5 / 32
Example Decentralized Control Example Consider a two-input, two-output plant having the transfer function [ G 0 G 0 (s) = 11 (s) G 0 12 (s) G 0 21 (s) G0 22 (s) ] G 0 11(s) = G 0 21(s) = 2 s 2 + 3s + 2 k 21 s 2 + 2s + 1 G 0 12(s) = k 12 s + 1 G 0 6 22(s) = s 2 + 5s + 6 Assume that k 12 and k 21 depend on the operating point (a common situation, in practice). Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 6 / 32
Example Decentralized Control Example Operating point 1 (k 12 = k 21 = 0) Clearly, there is no interaction at this operating point. Thus, we can safely design two SISO controllers. To be specific, say we aim for the following complementary sensitivities: T 01 (s) = T 02 (s) = 9 s 2 + 4s + 9 The corresponding controller transfer functions are C 1 (s) and C 2 (s), where C 1 (s) = 4.5( s 2 + 3s + 2 ) ; C 2 (s) = 1.5( s 2 + 5s + 6 ) s(s + 4) s(s + 4) The two independent loops perform as predicted by the choice of complementary sensitivities. university-logo Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 7 / 32
Example Decentralized Control Example Operating point 2 (k 12 = k 21 = 0.1) We leave the controller as previously designed for operating point 1. We apply a unit step in the reference for output 1 at t = 1 and a unit step in the reference for output 2 at t = 10. The closed-loop response is shown on the next slide. The results would probably be considered very acceptable, even though the effects of coupling are now evident in the response. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 8 / 32
Example Decentralized Control Example Figure: Effects of weak interaction in control loops with SISO design. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 9 / 32
Example Decentralized Control Example Operating point 3 (k 12 = 1,k 21 = 0.5) With the same controllers and for the same test as used at operating point 2, we obtain the results on the next slide. We see that a change in the reference in one loop now affects the output in the other loop significantly. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 10 / 32
Example Decentralized Control Example Figure: Effects of strong interaction in control loops with SISO design. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 11 / 32
Example Decentralized Control Example Operating point 4 (k 12 = 2,k 21 = 1) Now a simulation with the same reference signals indicates that the whole system becomes unstable. We see that the original SISO design has become unacceptable at this final operating point. The strong interactions need to be accounted for in the control design Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 12 / 32
Pairing of Inputs and Outputs Pairing of Inputs and Outputs university-logo If one is to use a decentralized architecture, then one needs to pair the inputs and outputs in the best possible way. In the case of an m m plant transfer function, there are m! possible pairings. However, physical insight can often be used to suggest sensible pairings. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 13 / 32
Pairing of Inputs and Outputs RGA: A Simple Example One method that can be used to suggest pairings is a quantity known as the (RGA). Consider a simple 2 2 system y 1 = K 11 u 1 + K 12 u 2 y 2 = K 21 u 1 + K 22 u 2 Assume we want to control y 1 with u 1. When the other loop is open, i.e. u 2 = 0, we have y 1 = K 11 u 1 Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 14 / 32
Pairing of Inputs and Outputs Now consider the situation when the other loop provides perfect control, i.e. y 2 = 0. This happens when u 2 = K 21u 1 K 22 We then have [ y 1 = K 11 K ] 21 K 12 u 1 K 22 This means that closing the other loop has changed the gain between y 1 and u 1. The gain change is characterized by the ratio λ 11 = 1 1 K 12K 21 K 11 K 22 Intuitively, for decentralized control, we prefer to pair variables u j and y i so that λ ij is close to 1, because this means that the gain from u j to y i is unaffected by closing the other loops. university-logo Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 15 / 32
Pairing of Inputs and Outputs For a system with matrix transfer function G 0 (s), the RGA is defined as a matrix Λ with the ij th element λ ij = [G 0 (0)] ij [ G 1 0 (0)] ji where [G 0 (0)] ij and [ G 1 0 (0)] ji denote the ijth element of the plant d.c. gain and the ji th element of the inverse of the d.c. gain matrix respectively. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 16 / 32
RGA Pairing of Inputs and Outputs Note that [G 0 (0)] ij corresponds to the d.c. gain from the i th input, u i, to the j th output, y j, while the rest of the inputs, u 1 for 1 {1, 2,..., i 1, i + 1,..., m} are kept constant. Also [ G 1 ] 0 ij is the reciprocal of the d.c. gain from the ith input, u i, to the j th output, y j, while the rest of the outputs, y 1 for 1 {1, 2,..., j 1, j + 1,..., m} are kept constant. Thus, the parameter λ ij provides an indication of how sensible it is to pair the i th input with the j th output. One usually aims to pick pairings such that the diagonal entries of Λ are large. One also tries to avoid pairings that result in negative diagonal entries in Λ. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 17 / 32
RGA Example Pairing of Inputs and Outputs Consider again the system G 0 (s) = 2 s 2 +3s+2 k 21 s 2 +2s+1 k 12 s+1 6 s 2 +5s+6 The RGA is then Λ = 1 1 k 12 k 21 k 12 k 21 1 k 12 k 21 k 12 k 21 1 k 12 k 21 1 1 k 12 k 21 Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 18 / 32
RGA Example Pairing of Inputs and Outputs For 1 > k 12 > 0, 1 > k 21 > 0, the RGA suggests the pairing (u 1, y 1 ), (u 2, y 2 ). We recall from our earlier study of this example that this pairing worked very well for k 12 = k 21 = 0.1 and quite acceptably for k 12 = 1, k 21 = 0.5. In the latter case, the RGA is Λ = 1 3 [ 2 1 1 2 ] Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 19 / 32
RGA Example Pairing of Inputs and Outputs However, for k 12 = 2, k 21 = 1 we found that the centralized controller based on the pairing (u 1, y 1 ), (u 2, y 2 ) was actually unstable. The corresponding RGA in this case is [ 1 2 Λ = 2 1 which indicates that we probably should have changed to the pairing (u 1, y 2 ), (u 2, y 1 ). ] Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 20 / 32
Pairing of Inputs and Outputs Frequency-Dependent RGA The frequency-dependent RGA Λ(s) is defined as λ ij (s) = [G 0 (s)] ij [ G 1 0 (s)] ji Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 21 / 32
Pairing of Inputs and Outputs Frequency-Dependent RGA Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 22 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus Consider the quadruple-tank apparatus shown below. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 23 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus Recall that this system has an approximate transfer function, G(s) = 3.71γ 1 62s+1 4.7(1 γ 1 ) (30s+1)(90s+1) 3.7(1 γ 2 ) (23s+1)(62s+1) 4.7γ 2 90s+1 The RGA for this system is [ Λ = λ 1 λ 1 λ λ ] where λ = γ 1γ 2 γ 1 γ 2 1 Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 24 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus For 1 < γ 1 + γ 2 < 2, we recall from a previous lecture that the system is minimum phase. If we take, for example, γ 1 = 0.7 and γ 2 = 0.6, then the RGA is [ 1.4 0.4 Λ = 0.4 1.4 This suggests that we can pair (u 1, y 1 ) and (u 2, y 2 ). Because the system is of minimum phase, the design of a decentralized controller is relatively easy in this case. For example, the following decentralized controller gives the results shown on the next slide. ] ( C 1 (s) = 3 1 + 1 ) ( ; C 2 (s) = 2.7 1 + 1 ) 10s 20s university-logo Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 25 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus Figure: Decentralized control of a minimum-phase four-tank system. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 26 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus For 0 < γ 1 + γ 2 < 1, we recall from a previous lecture that the system is nonminimum phase. If we take, for example γ 1 = 0.43 and γ 2 = 0.34, then the system has a NMP zero at s = 0.0229, and the relative gain array becomes [ ] 0.64 1.64 Λ = 1.64 0.64 This suggests that (y 1, y 2 ) should be commuted for the purposes of decentralized control. This is physically reasonable, given the flow patterns produced in this case. This leads to a new RGA of [ 1.64 0.64 Λ = 0.64 1.64 ] Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 27 / 32
Pairing of Inputs and Outputs Quadruple-tank Apparatus Note, however, that control will still be much harder than in the minimum-phase case. For example, the following decentralized controllers give the results below. ( C 1 (s) = 0.5 1 + 1 ) ( ; C 2 (s) = 0.3 1 + 1 ) 30s 50s Figure: Decentralized control of a nonminimum-phase four-tank system. university-logo Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 28 / 32
Pairing of Inputs and Outputs Robustness Issues Robustness Issues in Decentralized Control One way to carry out a decentralized control design is to use a diagonal nominal model. The off-diagonal terms then represent under-modelling. Thus, we say we have a model G 0 (s), then the nominal model for decentralized control could be chosen as G d 0 (s) = diag{ g 0 11,..., g 0 mm(s) } and the additive model error would be G (s) = G 0 (s) G d 0 (s); G I(s) = G (s) [ G d 0 (s)] 1 A sufficient condition for robust stability is σ (G 1 (jω)t 0 (jω)) < 1 ω R where σ (G 1 (jω)t 0 (jω)) is the maximum singular value of G 1 (jω)t 0 (jω). university-logo Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 29 / 32
Example Pairing of Inputs and Outputs Robustness Issues Consider a MIMO system with G(s) = 1 s+1 0.25 10s+1 (s+1)(s+2) 0.25 10s+1 (s+1)(s+2) 2 s+2 ; G 0 (s) = 1 s+1 0 0 2 s+2 We first observe that the RGA for the nominal model G 0 (s) is given by [ 1.0159 0.0159 Λ 0.0159 1.0159 ] Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 30 / 32
Pairing of Inputs and Outputs Robustness Issues university-logo This value of the RGA might lead to the hypothesis that a correct pairing of inputs and outputs has been made and that the interaction is weak. We thus proceed to do a decentralized design leading to a diagonal controller C(s) to achieve a complementary sensitivity T 0 (s), where T 0 (s) = 9 [ 1 0 s 2 + 4s + 9 0 1 ] [ 9(s+1) ; C(s) = 0 s(s+4) 0 9(s+2) 2s(s+4) ] Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 31 / 32
Pairing of Inputs and Outputs Robustness Issues university-logo However, this controller, when applied to control the full plant G(s), leads to closed-loop poles located at 6.00, 2.49 ± j4.69, 0.23 ± j1.36, and 0.50 an unstable closed loop! The lack of robustness in this example can be traced to the fact that the required closed-loop bandwidth includes a frequency range where the off-diagonal frequency response is significant. More sophisticated techniques are required to adequately control this system. Guy A. Dumont (UBC EECE) EECE 460 - Decentralized Control January 2011 32 / 32