Recent Advances in Positive Systems: The Servomechanism Problem

Transcription

1 Recent Advances in Positive Systems: The Servomechanism Problem 47 th IEEE Conference on Decision and Control December 28. Bartek Roszak and Edward J. Davison Systems Control Group, University of Toronto University of Toronto 1

2 Outline: Motivation - why positive systems? Introduction and background to positive LTI systems SISO results and examples MIMO results and examples Extensions and future development References University of Toronto 2

3 Motivation - Why Positive Systems: University of Toronto 3

4 Motivation - Intravenous Maintenance of Anesthesia: Intravenous Anesthetic u := continuous infusion transfer F 12 Compartment 1 transfer F 31 F 21 Compartment 2 Compartment 3 F 1 F 13 University of Toronto 4

5 Motivation - Water Tanks: u γ Water Tanks 1 γ 1 2 pump 1 φ 3 φ 6 y 4 5 University of Toronto 5

6 Introduction to positive systems: Motivation - why positive systems? Introduction and background to positive LTI systems Positive LTI System Definition "Almost" Positive LTI Systems System of Interest University of Toronto 6

7 Definition 1 A linear system Positive LTI System Definition: ẋ = Ax + Bu y = Cx + Du (1) where A R n n, B R n m, C R r n, and D R r m is considered to be a positive linear system if for every nonnegative initial state and for every nonnegative input the state of the system and the output remain nonnegative. University of Toronto 7

8 Positive LTI System Definition: It turns out that Definition 1 has a very nice interpretation in terms of the matrix quadruple (A,B,C,D). Theorem 1 A linear system (1) is positive if and only if the matrix A is a Metzler matrix, and B, C, and D are nonnegative matrices. A matrix A is Metzler if all the off-diagonal terms are nonnegative. University of Toronto 8

9 Almost Positive LTI Systems: An arbitrary linear system is considered to be an almost-state (output) positive linear system with respect to x if for any given δ = (δ 1,δ 2,...,δ n(r) ) R n(r) + \ {} there exists a u δ such that the state x (output y) of the system satisfies x i (t)(y i (t)) δ i, i = 1, 2,...,n(r), t [, ). x i x i t δ i University of Toronto 9

10 Problem statement: Motivation - why positive systems? Introduction and background to positive LTI systems SISO results and examples Servomechanism problem for SISO positive LTI systems University of Toronto 1

11 System of Interest:. SISO system ẋ = Ax + bu + e ω ω y = cx + du + fω (2) e := y y ref u R m is the input, x R n + is the state, y R + is the output to be regulated, ω R Ω are the disturbances, y ref R + is the tracking signal and e R r is the error in the system. Matrix A is stable Metzler, and matrices b, c, e ω ω, fω, and d are nonnegative. University of Toronto 11

12 Assumption 1 Given (2) assume that Key Assumptions:. rank(d ca 1 b) = 1 and for all tracking and disturbance signals in question, it s assumed that the steady-state of the system (2) given by x ss u ss = A b c d 1 and has the property that u ss R +. e ω f 1 ω y ref (3) University of Toronto 12

13 Servomechanism problem for SISO positive LTI systems:. Problem: Consider the plant (2), with initial condition x R n +, under Assumption 1. Find a nonnegative controller u that (a) guarantees closed loop stability; (b) ensures the plant (2) is nonnegative, i.e. the states x and the output y are nonnegative for all time; and (c) ensures tracking of the reference signals, i.e. e = y y ref, as t, y ref Y ref and ω Ω. In addition, (d) assume that a controller has been found so that conditions (a), (b), (c) are satisfied; then for all perturbations of the nominal plant modal which maintain properties (a) and (b), it is desired that the controller can still achieve asymptotic tracking and regulation, i.e. property (c) still holds. University of Toronto 13

14 Servomechanism problem for SISO positive LTI systems:. Problem: Consider the plant (2), with initial condition x R n +, under Assumption 1. Find a nonnegative controller u that (a) guarantees closed loop stability; (b) ensures the plant (2) is nonnegative, i.e. the states x and the output y are nonnegative for all time; and (c) ensures tracking of the reference signals, i.e. e = y y ref, as t, y ref Y ref and ω Ω. In addition, (d) the controller is robust. University of Toronto 14

15 Servomechanism problem for "almost" positive LTI systems: Remark 1 In the sequel when almost-state and almost-output positivity will be considered, then in the previous problem the words state and output should be replaced by almost-state and almost-output, respectively. Additionally, the constraint of nonnegativity on the input will be lifted, i.e. the input can be bidirectional. We call this problem the servomechanism problem for "almost" positive LTI systems. University of Toronto 15

16 Clamping Tuning Regulator:. η = ǫ(y ref y), η = u = kη, (4) where k = { η 1 η > and η = and ǫ (,ǫ ], ǫ R + \ {}. University of Toronto 16

17 Clamping Tuning Regulator:. Unknown System y ref + η η y ǫ Positive LTI 1 s System η = ǫ(y ref y), η = u = kη University of Toronto 17

18 Key Assumptions:. Finding rank(d ca 1 b) = 1. Algorithm 1 It is assumed that the output of the system is measurable and the input is excitable the disturbance set to zero, i.e. ω =. 1. Apply an input u = u to (2), with u having a nonzero steady-state value. 2. Measure the corresponding steady-state value of the output y = y. 3. If y, then the existence condition holds true. University of Toronto 18

19 Key Assumptions:. Remark on u ss. ẋ = = Ax ss + bu ss + e ω ω η = ǫ(y ref y) = = cx ss + du ss + fω y ref Isolating for x ss we get: x ss = A 1 bu ss A 1 e ω ω. By substituting x ss and isolating for u ss we obtain: u ss = ca 1 e ω ω fω + y ref d ca 1 b. University of Toronto 19

20 Key Assumptions:. Remark on u ss. ẋ = = Ax ss + bu ss + e ω ω η = ǫ(y ref y) = = cx ss + du ss + fω y ref Isolating for x ss we get: x ss = A 1 bu ss A 1 e ω ω. Therefore: ca 1 e ω ω fω + y ref. University of Toronto 2

21 Solution:. Theorem 2 Consider system (2) under the clamping tuning regulator. Further assume that rank(d ca 1 b) = 1 x R n + u ss >. Then there exists an ǫ such that for all ǫ (,ǫ ] the clamping tuning regulator solves the servomechanism problem. University of Toronto 21

22 Algorithm:. Algorithm 2 1. Check the existence condition rank(d ca 1 b) = 1 by Algorithm 1. (a) If Algorithm 1 returns y =, then there does not exist a solution to the servomechanism problem. (b) Otherwise, go to Step Apply the clamping regulator to the unknown plant. (a) If the clamping controller remains at zero for t [t +, ), where t +, and no tracking/regulation occurs, then the servomechanism problem is not solvable under any control law. (b) Otherwise, the clamping regulator solves the servomechanism problem. University of Toronto 22

23 Intravenous Anesthetic Intravenous Anesthetic u := continuous infusion transfer F 12 Compartment 1 transfer F 31 F 21 Compartment 2 Compartment 3 F 1 F 13 University of Toronto 23

24 Intravenous Anesthetic stable, rank(d ca 1 b) = 1 (f 1 + f 21 + f 31 ) f 12 f 31 f 21 f 12 f 31 f 13 ẋ = x + 1 u + e ω ω y = [1 ]x transfer F 12 Compartment 1 transfer F 31 F 21 Compartment 2 Compartment 3 F 1 F 13 University of Toronto 24

25 Intravenous Anesthetic Output y = x time (min) Time f 1 f 21 f 12 f 31 f 13 e ω ω [, 5) [5, 1) [1, 15] University of Toronto 25

26 Input Intravenous Anesthetic input u time (min) Time f 1 f 21 f 12 f 31 f 13 e ω ω [, 5) [5, 1) [1, 15] University of Toronto 26

27 Water Tanks:. u γ Water Tanks 1 γ 1 2 pump 1 φ 3 φ 6 y 4 5 University of Toronto 27

28 Water Tanks:. ẋ = x u ω Also, assume the output y is of the form y = [ 1 ] x * ǫ =.5. University of Toronto 28

29 Simulation:. 1.4 epsilon = output y and input u y u time (s) University of Toronto 29

30 Tuning Regulator: Almost What about almost positivity? positivity η = y y ref u = ǫη (5) where η = and ǫ (,ǫ ], ǫ R + \ {}. y ref + η η y ǫ Positive LTI 1 s Unknown System System University of Toronto 3

31 Servomechanism for Almost positivity The controller: η = y y ref u = ǫη (6) where η = and ǫ (,ǫ ], ǫ R + \ {}, under the assumption that rank(d ca 1 b) = 1, x ss R n + and x R n + solves Problem 1 under Remark 1, i.e. the servomechanism problem for almost positivity can be attained under (5). University of Toronto 31

32 Optimal approach: LQcR control.. Consider the same problem under the controller: η = y y ref, η = [ x u = max{[k x K η ] η ], } where K x R 1 n and K η R are found by solving the cheap control problem: ǫ 2 e T e + u T udτ (7) where ǫ > University of Toronto 32

33 Optimal approach: LQcR control.. Consider the same problem under the controller: η = y y ref, η = [ x u = max{[k x K η ] η ], } for the system: [ ẍ ė ] [ ] [ A ẋ = c e [ ] x e = [ 1] e ] + [ b d ] u University of Toronto 33

34 Optimal Approach: LQcR control 4.5 epsilon = 12 4 output y and input u u y time (s) University of Toronto 34

35 Optimal Approach: LQcR control We cannot blindly use the standard LTI approach! E.g. 3.5 epsilon = 1e output y time (s) University of Toronto 35

36 Experimental results: LQcR control A u ω 1 ω 2 y Tank 1 Tank 2 C B F pump D G Tank 3 Tank 4 E pump University of Toronto 36

37 Experimental results: LQcR control University of Toronto 37

38 Results and examples: Motivation - why positive systems? Introduction and background to positive LTI systems SISO results and examples MIMO results and examples University of Toronto 38

39 System of Interest: MIMO case ẋ = Ax + Bu + Eω y = Cx + Du + Fω e := y ref y (8) u R m is the input, x R n + is the state, y R + is the output to be regulated, ω R Ω + are the disturbances, y ref R + is the tracking signal and e R r is the error in the system. Matrix A is stable Metzler, and matrices B, C, D, E, F are nonnegative with m = r. University of Toronto 39

40 Problem of Interest: Find a controller u R m + for all reference tracking signals y ref R r + and for all disturbance signals ω R Ω + such that (a) closed loop stability is maintained; (b) nonnegativity of states x and outputs y occurs for all time; (c) tracking of reference signals occurs, i.e. e = y ref y, as t, y ref R r + and ω R Ω +. (d) assume that an LTI controller has been found so that conditions (a), (b), (c) are satisfied; then for all perturbations of the nominal plant modal which maintain properties (a) and (b), it is desired that the controller can still achieve asymptotic tracking and regulation, i.e. the controller is robust. University of Toronto 4

41 In General No Solution: Theorem: There does not exist a solution to the problem of interest for almost all positive systems (8). Reason: u ss = K r y ref + K d ω = (D CA 1 B) 1 y ref (D CA 1 B) 1 (F CA 1 E)ω University of Toronto 41

42 Key Assumption: Assumption 2 Given (8) assume that rank(d CA 1 B) = r and for all tracking and disturbance signals in question, it s assumed that the steady-state of the system (8) is given by x ss u ss = A B C D 1 E F I ω y ref and has the property that u ss = K r y ref + K d ω R m +. University of Toronto 42

43 Key Assumptions: Finding K r 1. Apply an input vector u = [... u i... ] T to (8), i = 1,...,m. 2. Measure the corresponding steady-state value of the output vectors y = y i R r, i = 1,...,m. 3. Solve the equation: K 1 u 1... u = y 1 1 y y 1 m y 2 1 y y 2 m u m y r 1 y r 2... y r m for K 1 = (D CA 1 B). Note K r = K 1 1. University of Toronto 43

44 Measurable Disturbances: Finding K d 1. Apply a disturbance vector ω = [... ω i... ] T to (8), i = 1,..., Ω. 2. Measure the corresponding steady-state value of the output vectors y = y i R r, i = 1,..., Ω. 3. Solve the equation: K 2 ω 1... ω = y 1 1 y y 1 Ω y 2 1 y y 2 Ω ω Ω y r 1 y r 2... y r Ω for K 2 = (F CA 1 E). Note: K d = K r K 2. University of Toronto 44

45 Tuning Regulator: Tuning Regulators and Feedforward control: η = ǫ(y ref y) u tr = (D CA 1 B) 1 η (9) where ǫ (,ǫ ], ǫ R + \ {}. Feedforward Control: u = (D CA 1 B) 1 y ref (D CA 1 B) 1 (F CA 1 E)ω University of Toronto 45

46 Tuning Regulators and Feedforward control: y ref + η ǫ s η (D CA 1 B) 1 + u tr + Unknown System Positive LTI System y ω + (D CA 1 B) 1 + (D CA 1 B) 1 (F CA 1 E) u ff University of Toronto 46

47 New Problem and Solution: New Problem: Obtain the largest subclass of tracking signals y ref Y ref R r + and disturbance signals ω Ω R Ω + such that the original Problem of Interest is satisfied. Theorem: The original problem is solvable if and only if (y ref, ω) Y ref Ω := {(y ref, ω) R r + R Ω + K r y ref > K d ω component-wise}. (1) Moreover, it suffices to use the feedforward compensator and the tuning regulator control as the control input u, i.e. u = u ff + u tr. University of Toronto 47

48 Water Tanks: u γ Water Tanks 1 γ 1 2 pump 1 φ 3 φ 6 y 4 5 University of Toronto 48

49 Water Tanks: ẋ = x u ω 7 5 Also, assume the output y is of the form y = x * ǫ =.1. University of Toronto 49

50 Simulation: 2 input u input time (s) University of Toronto 5

51 Simulation: 7 output y 6 5 output (L) time (s) University of Toronto 51

52 Extensions and future development: Motivation - why positive systems? Introduction and background to positive LTI systems SISO results and examples MIMO results and examples Extensions and future development University of Toronto 52

53 Extensions and future development: Assume model is known - want to solve the same type of problem using optimal control techniques Previous study, aside from almost positivity, has been for positive systems constraint by nonnegative control - want to solve the same type of problem for nonpositive and bidirectional control - presently ongoing! University of Toronto 53

54 Water Tanks Example with nonnegative input Water Tanks u γ 1 γ 1 2 pump 1 φ 3 φ 6 y 4 5 University of Toronto 54

55 Results for Optimal Control: Water Tanks Example (Figure 1) 1.4 Closed Loop MPC Servo Controller Output Input u * no constraints, y ref = 1, ω =, violation of u University of Toronto 55

56 Results for Optimal Control: Water Tanks Example (Figure 2) 2 Closed Loop MPC Servo Controller 1.5 Output Input u * same as Figure 1, but u is satisfied University of Toronto 56

57 Results for Optimal Control: Water Tanks Example (Figure 3) 1.4 Closed Loop MPC Servo Controller Output Input u * same as Figure 2, but an improved controller University of Toronto 57

58 Water Tanks Example with nonnegative input Water Tanks u γ 1 γ 1 2 pump 1 φ 3 φ 6 y 4 ω University of Toronto 58 5

59 Results for Optimal Control: Water Tanks Example (Figure 4) 1.4 Closed Loop MPC Servo Controller Output Input u * same as Figure 3, but e ω = [ 1 ] T University of Toronto 59

60 Water Tanks Example with nonnegative input Water Tanks u γ 1 γ 1 2 pump fω 1 φ 3 φ y 6 4 University of Toronto 6 5

61 Results for Optimal Control: Water Tanks Example (Figure 5) 1.2 Closed Loop MPC Servo Controller 1 Output Input u * no constraints are applied, fω =.5, y ref = 1, e ω = * violation occurs! University of Toronto 61

62 Results for Optimal Control: Water Tanks Example (Figure 6) 1.2 Closed Loop MPC Servo Controller 1 Output Input u * same as Figure 5, but input constraint is satisfied University of Toronto 62

63 Results for Optimal Control: Building Example u 1 y 1 u 2 ref = 1 yref 2 = x 1 = x 2 = y = [1 ]x y 4 ref = y 3 ref = x 4 = x 3 = u 4 u 3 ω outside (decrease of outside temperature) University of Toronto 63

64 Results for Optimal Control: Building Example 1.2 Tracking yref output y; state FB Tracking yref input u: state FB Time (sec) University of Toronto 64

65 References: Bartek Roszak Edward Davison The servomechanism problem for unknown MIMO LTI positive systems: feedforward and robust tuning regulators. American Control Conference, June 11-13, 28, Seattle, USA. The servomechanism problem for unknown SISO positive systems using clamping, 17th IFAC World Congress, July 6-11, 28, Seoul, Korea. Tuning regulators for tracking SISO positive linear systems, Proceedings of the European Control Conference pp , July 27 University of Toronto 65

66 Conclusion: Motivation - why positive systems? Introduction and background to positive LTI systems SISO results and examples MIMO results and examples Extensions and future development References University of Toronto 66