Control System Design - Interaction

Transcription

1 Department of Automatic Control LTH, Lund University

2 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

3 Introduction Centralized or distributed The process control experience A few important variables were controlled using the single loop paradigm: one sensor, one actuator and one controller Loops were added, feedforward, cascade, midrange, selectors Not obvious how to associate sensors and actuators - The pairing problem Early process control systems were distributed and decentralized, centralization came with computer control The state-feedback paradigm - centralized control Decentralized control popular for large systems What happens when loops are interacting? Interaction measures - The relative gain array RGA, Singular values Decoupling: static, dynamic (different time scales), different physical mechanisms, mass balance, energy balance

4 A Large Process

5 The Shell Standard Control Problem

6 Paper Machine Control Primary quality: Basis weight and moisture Head box control: Pressure and flow rate

7 Boiler Control Who is the boss? Boiler leading turbine or turbine leading boiler.

8 A Small Process Chiller

9 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

10 Interaction of Simple Loops u 1 y sp1 C 1 y 1 Process y sp2 C 2 u 2 y 2 Y 1 (s)=p 11 (s)u 1 (s)+p 12 U 2 (s) Y 2 (s)=p 21 (s)u 1 (s)+p 22 U 2 (s), What happens when the controllers are tuned individually?

11 An Example ControllerC 1 is a PI controller with gainsk 1 =1,k i =1, and thec 2 is a proportional controller with gainsk 2 =0, 0.8, and y u The second controller has a major impact on the first loop!

12 Analysis Y 1 (s)= Y 2 (s)= 1 2 (s+1) 2U 1(s)+ (s+1) 2U 2(s) 1 1 (s+1) 2U 1(s)+ (s+1) 2U 2(s). P-control of second loopu 2 (s)= k 2 Y 2 (s) gives Y 1 (s)= cl 11 (s)u 1(s)= s 2 +2s+1 k 2 (s+1) 2 (s 2 +2s+1+k 2 ) U 1(s). The gaink 2 in the second loop has a significant effect on the dynamics in the first loop. The static gain is cl 11 (0)=1 k 2 1+k 2. Notice that the gain decreases with increasingk 2 and becomes negative fork 2 >1.

13 Bristol s Relative Gain A simple way of measuring interaction based on static properties Edgar H. Bristol On a new measure of interaction for multivariable process control IEEE TAC 11(1967) Idea: What is effect of control of one loop on the steady state gain of another loop? Consider one loop when the other loop is under perfect control Y 1 (s)=p 11 (s)u 1 (s)+p 12 U 2 (s) 0=p 21 (s)u 1 (s)+p 22 U 2 (s).

14 Bristol s Relative Gain Consider the first loopu 1 y 1 when the second loop is in perfect control (y 2 =0) Y 1 (s)=p 11 (s)u 1 (s)+p 12 U 2 (s) 0=p 21 (s)u 1 (s)+p 22 U 2 (s). EliminatingU 2 (s) from the second equation gives Y 1 (s)= p 11(s)p 22 (s) p 12 (s)p 21 (s) U 1 (s). p 22 (s) The ratio of the static gains of loop 1 when the second loop is open and closed is λ= p 11 (0)p 22 (0) p 11 (0)p 22 (0) p 12 (0)p 21 (0). Parameter λ is called Bristol s interaction index

15 Many Loops Assumeninputs andnoutputs. Pick an input output pair and relabel so that the output isy 1, let the remaining outputs bey 2 =0. Let the input beu 2 and the remaining inputs beu 1. y 1 =p 11 u 1 +p 12 u 2 0=p 21 u 1 +p 22 u 2 Solving fory 1 gives (notice the Schur complement) y 1 =(p 12 p 11 p 1 21 p 22)u 2, r 12 = p 12 p 12 p 11 p 1 21 p 22 Compare P= p 11 p 12, P 1 =Q= p 21 p 22 p 12 p 11 p 1 21 p 22 The relative gain array isr=p P T (Hadamard or Schur product)

16 Bristol s Relative Gain Array Let P(s) be an n n matrix of transfer functions. The relative gain array is Λ=P(0) P T (0) The product is element by element product (Schur or Hadamard product). Properties (A B) T =A T B T P diagonal gives Λ=I Insight and use A measure of static interactions for square systems which tells how the gain in one loop is influenced by perfect feedback on all other loops Dimension free. Row and column sums are 1. Negative elements correspond to sign reversals due to feedback of other loops

17 Pairing When designing complex systems loop by loop we must decide what measurements should be used as inputs for each controller. This is called the pairing problem. The choice can be governed by physics but the relative gain can also be used Consider the previous example P(0)= 1 2, P 1 (0)= 1 1 Λ=P(0) P T (0)= , Negative sign indicates the sign reversal found previously Better to use reverse pairing, i.e. letu 2 controly 1

18 Pairing... Consider P(s)= 1 2 (s+1) 2 (s+1) (s+1) 2 (s+1) 2 Introducing the feedbacku 1 = k 2 y 2 gives Y 1 (s)= cl 12 (s)u 2(s)= 2s 2 +4s+2+k 2 (s+1) 2 (s 2 +2s+1+k 2 ) U 2(s), Zero frequency gain decreases from 2 to 1 whenk 2 ranges from 0 to. Discuss how dynamics changes withk 2! Use rootlocus!

19 Step Responses with Reverse Pairing 1 y u ( U 2 = 1+ 1 ) (Y sp1 Y 1 ) s u 1 = k 2 y 2 withk 2 =0, 0.8, and 1.6.

20 Summary for2 2 Systems λ=1 No interaction λ=0 Closed loop gainu 1 y 1 is zero. Pairu 1 andy 2 instead 0<λ<1 Closed loop gainu 1 y 1 is larger than open loop gain. Interaction strongest for λ=1 λ>1 Closed loop gainu 1 y 1 is smaller than open loop gain. Interaction increases with increasing λ. Very difficult to control both loops independently if λ is very large. λ<0 The closed loop gainu 1 y 1 has different sign than the open loop gain. Opening or closing the second loop has dramatic effects. The loops are counteracting each other. Such pairings should be avoided for decentralized control and the loops should be controlled jointly as a multivariable system.

21 Singular Values Let A be an k n matrix whose elements are complex variables. The singular value decompostion of the matrix is A=U ΣV where denotes transpose and complex conjugation,u andv are unitary matrices (UU =I andvv =I is. The matrix Σ is ak n matrix such that Σ ii = σ i and all other elements are zero. The elements σ i are called singular values. The largest σ=max i σ i and smallest σ=min i σ singular values are of particular interest. The number σ/σ is called the condition number. The singular values are the square roots of the eigenvalues ofa A. Example: A real2 2 matrix can be written as A= cosθ 1 sinθ 1 σ 1 0 cosθ 2 sinθ 2 sinθ 1 cosθ 1 0 σ 2 sinθ 2 cosθ 2

22 Singular DecompositionA=U ΣV The columnsu i ofu represent the output directions The columnsv i ofv represent the input directions We haveav=u Σ, orav i = σ i u i. An input in the directionv i thus gives the output σ i u i, i.e. in the directionu i Since the vectorsu i andv i are of unit length the gain ofa for the inputu i is σ i The largest gain is σ=max i σ i There are efficient numerical algorithms svd in Matlab Singular values can be applied to nonsquare matrices A natural way to define gain for matricesa and transfer function matrices G(s) gain=max v Av v = σ(a), gain=max σ(g(iω)) ω

23 Interaction Analysis Consider a system with the scaled zero frequency gain y u 1 y 2 = u y 3 Relative gain array Λ= Singular values: σ 1 =1.6183, σ 2 = and σ 3 = Condition number κ=166. Only two outputs can be controlled. What variables should be chosen? u 3

24 Interaction Analysis We havey=usv T. How to pick two input output pairs SV T = U= The matrixsv T shows thatu 1 andu 2 are obvious choices of inputs. There are two choices of outputs:y 1,y 3 ory 2,y 3, based on the angles between rows,y 1,y 2 is not a good choice because the corresponding rows ofus are almost parallel. Singular values: Selectiony 1,y 3 u 1,u 2 Condition number κ = 1.51 tion number κ = 1.45 Selectiony 2,y 3 u 1,u 2 Condi Λ= Λ=

25 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

26 Zeros of Multivariable Systems Roughly speaking transmission zeros are values ofswhere the transmission of the signale st is blocked Y(s)=P(s)ve st, 0=P(s)v There is a nontrivialv only if the matrixp(s) is singular. For a square system the zeross i are given by detp(s)=0 and the zero directions are the corresponding right eigenvectors of P(s i ).

27 Rosenbrock s Example There is a nice collection of linear multivariable systems with interesting properties. Here is one of them 1 2 P(s)= s+1 s s+1 s+1 Very benign subsystems. Relative gain array R= 1 2/3 3 3 = 3 2, The transmission zeros are given by detp(s)= 1 ( 1 s+1 s+1 2 ) = s+3 1 s (s+1) 2 (s+3) =0. Difficult to control the system with gain crossover frequencies larger than ω c =0.5.

28 Stability Region - P in Both Loops Discuss commissioning and windup!

29 Rosenbrock s Example y1 y u1 2 u PI controllers withk p =2 andk i =2 in both loops. Systems becomes unstable if gains are increased by a factor of 3.

30 Interactions Can be Beneficial P(s)= p s 1 s 11(s) p 12 (s) = (s+1)(s+2) (s+1)(s+2) p 21 (s) p 22 (s) 6 s 2. (s+1)(s+2) (s+1)(s+2) The relative gain array R= 1 0, 0 1 Transmission zeros detp(s)= (s 1)(s 2)+6s (s+1) 2 (s+2) 2 = s2 +4s+2 (s+1) 2 (s+2) 2 Difficult to control individual loops fast because of the zero ats=1. Since there are no multivariable zeros in the RHP the multivariable system can easily be controlled fast but the system is not robust to loop breaks.

31 Interactions Can be Beneficial... y1 y2 u u PI controllers with gainsk=100 andk i =2000 in both loops

32 Stability Region - P in Both Loops Discuss commissioning and windup!

33 The Quadruple Tank 1 γ 1 1 γ 2 Tank 3 (A1) Tank 4 (B1) y 3 y 4 γ 1 γ 2 Pump 1 (BP) Tank 1 (A2) Tank 2 y 1 (B2) y 2 u 1 u 2 Pump 2 (AP)

34 Transfer Function of Linearized Model Transfer function fromu 1,u 2 toy 1,y 2 γ 1 c 1 (1 γ 2 )c 1 1+sT 1 (1+sT 1 )(1+sT 3 ) P(s)= (1 γ 1 )c 2 γ 2 c 2 (1+sT 2 )(1+sT 4 ) 1+sT 2 Transmission zeros (1+sT 3 )(1+sT 4 ) (1 γ 1)(1 γ 2 ) γ detp(s)= 1 γ 2 (1+sT 1 )(1+sT 2 )(1+sT 3 )(1+sT 4 ) No interaction if γ 1 = γ 2 =1 Minimum phase if1 γ 1 + γ 2 2 Nonminimum phase if0< γ 1 + γ 2 1. Intuition?

35 The Quadruple Tank Zero frequency gain P(0)= γ 1c 1 (1 γ 2 )c 1 (1 γ 1 c 2 γ 2 c 2 Relative gain array R=P(0) P(0) 1 = λ 1 λ 1 λ λ where λ= γ 1 γ 2 γ 1 + γ 2 1 Recall RHP zero if γ 1 + γ 2 <1. Physical interpretation!

36 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

37 Decoupling Simple idea: Find a compensator so that the system appears to be without coupling. Many versions Input decoupling Q = PD or output decoupling Q = D P Conventional (Feedforward) Inverse (Feedback) Static Important to consider windup, manual control and mode switches. Keep the decentralized philosophy

38 Decoupling may not be useful Flight Control Think about what happens when the plane turns! An autopilot is typically split into two subsystems: lateral (pitch) longitudinal (yaw and roll)

39 Feedforward (Conventional) Decoupling D=P 1 1 (s)q(s)= p 22(s) p 12 (s) Q(s) det P(s) p 21 (s) p 11 (s) Q(s) is the desired transfer function of decoumpled process. This decoupler has difficulties to deal with anti-windup, manual control and mode changes (auto-tuning). ControllerC 1 has no information about v 2.

40 PID Controller with Tracking and Feedforward No tracking ifw=v!

41 Anti-windup for Controller with Tracking Mode y e KT d s K Σ v Actuator model u Actuator K/T i Σ 1/s + Σ 1/T t e s SP MV TR PID v Actuator model u Actuator Notice that there is no tracking effect ifu=v! The tracking input can be used in many other ways

42 Feedback (Inverted) Decoupling Simple decoupler, easy to deal with anti-windup, manual control and mode changes (auto-tuning) ifd 11 =d 22 =1. Why?

43 Inverted (Feedback) Decoupling u=v+du (I D)u=v u=(i D) 1 v Q=P(I D) 1 P=Q QD Blockdiagram vector case Easy to solve ford fb also for systems with many inputs and outputs. Example2 2, pickq=diag(p 11,p 22 ), why? p 11 p 12 = p 11 0 p d 12 p 21 p 22 0 p 22 0 p 22 d 21 0 Hence d 12 = p 12 p 11 d 21 = p 21 p 22

44 The Wood-Berry Distillation Column 12.8e s P(s)= 16.7s e 7s 10.9s+1 Choosed 11 =d 22 =1 18.9e 3s 21.0s e 3s. 14.4s+1 d 11 =1 d 12 = p 12 = s+1 p s+1 e 2s d 21 = p 21 p 22 = s s+1 e 4s d 22 =1 Easy to implement. What can go wrong?

45 Wood and His Column U Alberta

46 Properties of Inverted (Feedback) Decoupling Simple decoupler even for systems with many inputs and many outputs Easy to deal with anti-windup, manual control and other mode changes (auto-tuning) Decoupler may be noncausal (pure predictor). Try different pairing or add extra time delay ind ii. Since it is a feedback coupling there may be instabilities

47 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

48 Systems with Parallel Actuation Motor drives for papermachines and rolling mills Trains with several motors or several coupled trains Power systems Electric cars

49 A Prototype Example J dω dt +Dω=M 1+M 2 M L, Proportional control C 1 A 1 w sp Gearbox ω M 1 =M 10 +K 1 (ω sp ω) M 2 =M 20 +K 2 (ω sp ω) The proportional gains tell how the load is distributed J dω dt +(D+K 1+K 2 )ω=m 10 +M 20 M L +(K 1 +K 2 )ω sp. A first order system with time constantt=j/(d+k 1 +K 2 ) Discuss response speed, damping and steady state ω= ω 0 = K 1+K 2 D+K 1 +K 2 ω sp + M 10+M 20 M L D+K 1 +K 2. C 2 A 2

50 Integral Action? PI Motor 1 M 1 ω sp Σ M s 1 s ω PI Motor 2 M 2 Prototypes for lack of reachability and observability!

51 Integral Action ω sp ω M 1 M n 0 M L Notice thatm 2 is breaking fort>22

52 Better Integral Action M 10 ω sp PI α Motor 1 Σ Σ 1 s ω 1 α Motor 2 Σ M 20

53 Better Integral Action? ω sp ω M M n 0 M L

54 Power Systems - Massive Parallellism Edison s experience Two generators with governors having integral action Many generators supply power to the net. Frequency control Voltage control Isochronous governors (integral action) and governors with speed-drop (no integral action)

55 Turbine Governors

56 Load Sharing - Through Proportional Action In Sweden the final adjustment is made manually

57 Interaction 1 Introduction 2 Interaction of Simple Loops 3 Multivariable zeros 4 Decoupling 5 Parallel Systems 6 Summary Theme: When the wires get crossed

58 Summary All real systems are coupled Relative gain array and singular values give insight Never forget process redesign Multivariable zeros and zero directions Why decouple Simple system. SISO design, tuning and operation can be used What is lost? Multivariable design Dont forget windup and operational aspects (tuning, manual control,... ) Parallel systems One integrator only!

59 To Learn More F. G. Shinskey Controlling Multivariable Processes. ISA. Research Triangle Park P. Kundur Power System Stability and Control, 1994 Harold L. Wade Inverted decoupling: a neglected technique. ISA Transactions 36:1(1997)3-10. S. Skogestad and I Postlethwaite Multivariable Feedback Control - Analysis and Design. Wiley 1996, 2nd Edition T. J. McAvoy Interaction Analysis - Principles and Applications ISA Research Triangle Park 1983 Bristol, E. On a new measure of interaction for multivariable processes. IEEE TAC-11 (1966) M. Morari and E. Zafiriou Robust Process Control. Prentice Hall 1989.