CHAPTER 10: STABILITY &TUNING


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1 When I complete this chapter, I want to be able to do the following. Determine the stability of a process without control Determine the stability of a closedloop feedback control system Use these approaches to learn how dead time affects stability.
2 Outline of the lesson. Define stability Review determining the roots of the characteristic equation Introduce the Bode stability method Apply to determine some general trends in feedback systems
3 MV ( t) = 1 d CV Kc E( t) E( t') dt' Td I TI 0 dt 20 v1 TC v2 No! Yes! or We influence stability when we implement control. How do we achieve the influence we want?
4 bounded CHAPTER 10: STABILITY &TUNIN First, let s define stability: A system is stable if all bounded inputs to the system result in bounded outputs. Sample Sample Inputs Process Outputs Feed T1 F1 T2 T4 T5 T3 T6 P1 L1 Vapor product F2 Process fluid F3 Steam A1 L. Key Liquid product unbounded unbounded bounded
5 (s) = Y(s)/X(s) Y(s) = [N(s)/D(s)] X(s) Let s review how we determine the stability of a model. With α i the solution to the denominator of the transfer function being zero, D(s) = 0 giving s = α 1, α 2, α i.... Y ( t ) = A 0 A e 1 α t 1 A e 2 α t 2... ( B 0 B t 1 B 2 t 2..) e α t p... [ C 1 cos( ω t ) C 2 sin( ω t )] e α t q... Real, distinct α i Complex α i If all α i are???, Y(t) is stable If any one α i is???, Y(t) is unstable Real, repeated α i Class exercise
6 With α i the solutions to D(s) = 0, which is a polynomial. Y( t) = A 0 A e 1 α t 1 A e 2 α t 2... ( B 0 B t 1 B 2 t 2..) e α t p... [ C 1 cos( ωt) C2 sin( ωt)] e α t q If all real [α i ] are < 0, Y(t) is stable If any one real [α i ] is 0, Y(t) is unstable 2. If all α i are real, Y(t) is overdamped (does not oscillate) If one pair of α i are complex, Y(t) is underdamped
7 Quick review of model for closedloop feedback system. D(s) d (s) SP(s) E(s) MV(s) C (s) v (s) P (s)  CV m (s) S (s) CV(s) Transfer functions C (s) = controller v (s) = valve P (s) = feedback process S (s) = sensor d (s) = disturbance process Variables CV(s) = controlled variable CV m (s) = measured value of CV(s) D(s) = disturbance E(s) = error MV(s) = manipulated variable SP(s) = set point
8 D(s) d (s) SP(s)  E(s) C (s) MV(s) CV m (s) v (s) P (s) CV(s) S (s) Set point response Disturbance Response CV SP( s) = 1 p p v v c c S CV D( s) = 1 p v d c S The denominator determines the stability of the closedloop feedback system! We call it the characteristic equation.
9 Direction Solution for the Roots to determine the stability Controller is a Ponly controller. Is the system stable? Let s evaluate the roots of the characteristic equation s 3 ( 1 τs) ( 1 5s) 3 p K C K P v 75s = 1 2 c K 15s C S ( ) = K c = 0 F S solvent F A pure A AC
10 0.5 CHAPTER 10: STABILITY &TUNIN Plot of real and imaginary parts of the roots of the characteristic equation  three roots for cubic Imaginary Kc As the controller gain, K C, is increased, some roots approach, then cross the boundary (s=0) between stable and Unstable. Shaded is the unstable region Real Stable Unstable
11 The denominator determines the stability of the closedloop feedback system! CV SP( s) Set point response = 1 p p v v c c S 3 2 c For the mixer : 125s 75s 15s K = 0 Bode Stability Method Calculating the roots is easy with standard software. However, if the equation has a dead time, the term e θs appears. Therefore, we need another method. Th method we will use next is the Bode Stability Method.
12 Bode Stability: To understand, let s do a thought experiment F S solvent Loop open F A pure A AC SP
13 Bode Stability: To understand, let s do a thought experiment SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) open S (s) CV(s) F S solvent Loop open F A pure A AC SP
14 Bode Stability: To understand, let s do a thought experiment No forcing!! F S solvent Loop closed F A pure A AC SP
15 Bode Stability: To understand, let s do a thought experiment No forcing!! SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) No forcing!! F S solvent Loop closed F A pure A AC SP
16 Bode Stability: To understand, let s do a thought experiment No forcing!! SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) No forcing!! F S solvent Loop closed F A pure A AC SP
17 Bode Stability: To understand, let s do a thought experiment SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) Under what conditions is the system stable (unstable)? Hint: think about the sine wave as it travels around the loop once.
18 Bode Stability: To understand, let s do a thought experiment SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) Under what conditions is the system stable (unstable)? If the sine is larger in amplitude after one cycle; then it will increase each time around the loop. The system will be unstable. Now: at what frequency does the sine most reinforce itself?
19 Bode Stability: To understand, let s do a thought experiment SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) Now: at what frequency does the sine most reinforce itself? When the sine has a lag of 180 due to element dynamics, the feedback will reinforce the oscillation (remember the  sign). This is the critical frequency.
20 Bode Stability: To understand, let s do a thought experiment SP(s) E(s) MV(s) C (s) v (s) P (s)  Loop CV m (s) closed S (s) CV(s) Let s put the results together. OL (s) includes all elements in the closed loop. At the critical frequency: OL (ω c j) = 180 The amplitude ratio: OL (ω c j) < 1 for stability OL (ω c j) > 1 for stability See textbook for limitations
21 Bode Stability: Let s do an example: threetank mixer with 5 minutes dead time added OL (ω c j) = 180 OL (ω c j) < 1 for stability OL = c v p s = K P e θ s ( 1 τs) 3 K c 1 1 T s I K P Process = % A/% open τ = 5 min (each tank) θ = 5 min Controller tuning w/o dead time K T I c = 30% open/% A = 11min From Ciancone correlations
22 Bode Stability: OL (ω c j) = OL (ω c j) < 1 for stability Amplitude Ratio OL (ω c j) = 0.75 Conclusion? Frequency, w (rad/time) 50 Phase Angle (degrees) Critical frequency Frequency, w (rad/time)
23 Bode Stability: OL (ω c j) = OL (ω c j) < 1 for stability Amplitude Ratio OL (ω c j) = 0.75 < 1 Conclusion: stable!! Phase Angle (degrees) Frequency, w (rad/time) The sine will decrease in amplitude each time around the loop. Critical frequency Frequency, w (rad/time)
24 2 SLOOP plots deviation variables (IAE = ) Controlled Variable Stable, but performance poor, why? Time 60 Manipulated Variable Time
25 2 SLOOP plots deviation variables (IAE = ) Controlled Variable Stable, but performance poor, why? Manipulated Variable Time Key 60 lesson: Stability is required, but more is 40 required for good performance. 20 PI tuning was for the process without dead time. The process with dead time is more difficult to control. Must make controller less agressive! Time
26 Bode calculations can be done by hand, easier with S_LOOP ************************************************************* * S_LOOP: SINLE LOOP CONTROL SYSTEM ANALYSIS * * BODE PLOT OF OL(s) = p(s)c(s) * * * * Characteristic Equation = 1 OL(s) * ************************************************************* SELECT THE APPROPRIATE MENU ITEM MODIFY... PRESENT VALUES 1) Lowest Frequency ) Highest Frequency ) Create Bode plot and calculate the results at critical frequency 4) Return to main menu Enter the desired selection: ************************************************* Critical frequency and amplitude ratio from Bode plot of OL ************************************************* Caution: 1) cross check with plot because of possible MATLAB error in calculating the phase angle 2) the program finds the first crossing of 180 The critical frequency is between and The amplitude ratio at the critical frequency is Or, write your own program in MATLAB. Amplitude Ratio Phase Angle (degrees) Frequency, w (rad/time) Frequency, w (rad/time)
27 Let s review what we have accomplished so far. We can evaluate the stability of a process without control by evaluating the roots of char. equation We can evaluate the stability of a process under feedback by either  evaluating the roots of char. equation  Bode method (required for process with dead time) These are local tests, caution about nonlinearity Stability does not guarantee good performance!!!! Unstable system performance always poor!!!
28 1. What else can we do with this neat technology? Tune controllers F S solvent MV ( t) = 1 d CV Kc E( t) E( t') dt' Td I TI 0 dt F A pure A AC ZieglerNichols Tuning We can tune controllers. The basic idea is to keep a reasonable margin from instability limit. This reasonable margin might give good performance.
29 1. What else can we do with this neat technology? Tune controllers Controller Kc TI Td Ponly Ku/ PI Ku/2.2 Pu/ PID Ku/1.7 Pu/2.0 Pu/8 ain margin is approximately 2 Integral mode is required for zero ss offset Derivative has stabilizing effect
30 F S Controlled Variable Manipulated Variable solvent F A pure A SLOOP plots deviation variables (IAE = ) Time Time AC ZieglerNichols tuning enerally, Ziegler Nichols tuning is not the best initial tuning method. However, these two guys were real pioneers in the field! Its taken 50 years to surpass their guidelines.
31 2. What else can we do with this neat technology? Understand why detuning is required for tough processes. 10 KcKp 1 As dead time increases, we must detune the controller In this plot, (θτ) is constant and θ/ (θτ) is changed. ZieglerNichols fraction dead time Ciancone
32 3. What else can we do with this neat technology? Understand need for robustness. After we tune the controller, we change the flow of solvent. What happens? F S solvent F S = 3.0 to 6.9 m3/min F A pure A AC
33 10 0 CHAPTER 10: STABILITY &TUNIN 3. What else can we do with this neat technology? Understand need for robustness. amplitude ratio Must consider the model error when selecting controller 105 tuning phase angle frequency (rad/time) Range of critical frequencies. Smallest is most conservative frequency (rad/time)
34 3. What else can we do with this neat technology? Understand need for robustness. Tune for the process response that is slowest, has highest fraction dead time, and largest process gain. This will give least aggressive controller. F S solvent F S = 3.0 to 6.9 m3/min F A pure A AC
35 Match your select of tuning method to tuning goals!
36 CHAPTER 10: TUNIN & STABILITY WORKSHOP 1 The data below is a process reaction curve for a process, plotted in deviation variables. Determine the tuning for a PID controller using the ZieglerNichols method. Controlled Variable Time v1 TC Manipulated Variable v Time
37 CHAPTER 10: TUNIN & STABILITY WORKSHOP 2 Answer true or false to each of the following questions and explain each answer. A. A closedloop system is stable only if the process and the controller are both stable. B. The Bode stability method proves that the closedloop system is stable for only sine inputs. C. OL (s) is the process model, P (s), and sensor, final element, and signal transmission dynamics D. A process would be stable if it had three poles with the following values: 1, .2, and 0.
38 CHAPTER 10: TUNIN & STABILITY WORKSHOP 3 The PID controller has been tuned for a threetank mixer. Later, we decide to include another mixing tank in the process. If we do not retune the controller, will the control system be stable with the fourtank mixer? F S solvent F A pure A AC K c = 30 T I = 11 T d = 0.8
39 When I complete this chapter, I want to be able to do the following. Determine the stability of a process without control Determine the stability of a closedloop feedback control system Use these approaches to learn how dead time affects stability. Lot s of improvement, but we need some more study! Read the textbook Review the notes, especially learning goals and workshop Try out the selfstudy suggestions Naturally, we ll have an assignment!
40 CHAPTER 10: LEARNIN RESOURCES SITE PCEDUCATION WEB  Instrumentation Notes  Interactive Learning Module (Chapter 10)  Tutorials (Chapter 10) S_LOOP  You can perform the stability and frequency response calculations uses menudriven features. Then, you can simulate in the time domain to confirm your conclusions.
41 CHAPTER 10: SUESTIONS FOR SELFSTUDY 1. Determine the stability for the example in textbook Table 9.2 (recommended tuning). Use the nominal process parameters. How much would K C have to be increased until the system became unstable? 2. Determine the ZieglerNichols tuning for the threetank mixer process. Simulate the dynamic response using S_LOOP. 3. Discuss applying the Bode stability method to a process without control.
42 CHAPTER 10: SUESTIONS FOR SELFSTUDY 4. We do not want to operate a closedloop system too close to the stability limit. Discuss measures of the closeness to the limit and how they could be used in calculating tuning constant values.
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