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PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway

PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway. Tuning of PID controllers. SIMC tuning rules (“Skogestad IMC”) (*) Main message: Can usually do much better by taking a systematic approach One tuning rule! Easily memorized References:

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PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway

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  1. PID Tuning and ControllabilitySigurd SkogestadNTNU, Trondheim, Norway

  2. Tuning of PID controllers • SIMC tuning rules (“Skogestad IMC”)(*) • Main message: Can usually do much better by taking a systematic approach • One tuning rule! Easily memorized References: (1) S. Skogestad, “Simple analytic rules for model reduction and PID controller design”, J.Proc.Control, Vol. 13, 291-309, 2003 (2) S. Skogestad, ``Tuning for smooth PID control with acceptable disturbance rejection'', Ind.Eng.Chem.Res, 45 (23), 7817-7822 (2006). (*) “Probably the best simple PID tuning rules in the world” • k’ = k/τ1 = initial slope step response • c ≥ 0: desired closed-loop response time (tuning parameter) • For robustness select (1): c¸ (gives Kc ≤ Kc,max) • For acceptable disturbance rejection select (2) : Kc ≥ Kc,min = ud0/ymax

  3. Model: Dynamic effect of change in input u (MV) on output y (CV) First-order + delay model for PI-control Second-order model for PID-control Need a model for tuning

  4. Step response experiment • Make step change in one u (MV) at a time • Record the output (s) y (CV)

  5. RESULTING OUTPUT y (CV) k’=k/1 STEP IN INPUT u (MV) Step response experiment First-order plus delay process : Delay - Time where output does not change 1: Time constant - Additional time to reach 63% of final change k : steady-state gain =  y(1)/ u k’ : slope after response “takes off” = k/1

  6. Start with complicated stable model on the form Want to get a simplified model on the form Most important parameter is usually the “effective” delay  Model reduction of more complicated model

  7. half rule

  8. Approximation of zeros

  9. PI-controller (based on first-order model) For second-order model add D-action. For our purposes it becomes simplest with the “series” (cascade) PID-form: Derivation of SIMC-PID tuning rules

  10. Basis: Direct synthesis (IMC) Closed-loop response to setpoint change Idea: Specify desired response T = (y/ys)desired and from this get the controller. Algebra:

  11. IMC Tuning = Direct Synthesis

  12. d u y c g Integral time • Found: Integral time = dominant time constant (I = 1) • Works well for setpoint changes • Needs to be modified (reduced) for integrating disturbances Example. “Almost-integrating process” with disturbance at input: G(s) = e-s/(30s+1) Original integral time I = 30 gives poor disturbance response Try reducing it!

  13. IMC rule: I = 1 =30 Reduce I to improve performance. This value just avoids slow oscillations: I = 4 (c+) = 8  Example: Integral time for “slow”/integrating process (see derivation next page)

  14. Derivation integral time: Avoiding slow oscillations for integrating process . • Integrating process: 1 large • Assume 1 large and neglect delay : • G(s) = k e- s /(1 s + 1) ¼ k/(1s) = k’/s • PI-control: C(s) = Kc (1 + 1/I s) • Poles (and oscillations) are given by roots of closed-loop polynomial • 1+GC = 1 + k’/s ¢ Kc ¢ (1+1/Is) = 0 • or I s2 + k’ KcI s + k’ Kc = 0 • Can be written on standard form (02 s2 + 2 0 s + 1) with • To avoid oscillations must require ||¸ 1: • Kc¢ k’ ¢I¸ 4 or I¸ 4 / (Kc k’) • With choice Kc = (1/k’) (1/(c+)) this gives I¸ 4 (c+) • Conclusion integrating process: Want I small to improve performance, but must be larger than 4 (c+) to avoid slow oscillations

  15. Summary: SIMC-PID Tuning Rules One tuning parameter: c

  16. Some special cases One tuning parameter: c

  17. Note: Derivative action is commonly used for temperature control loops. Select D equal to time constant of temperature sensor

  18. Selection of tuning parameter c Two cases • Tight control: Want “fastest possible control” subject to having good robustness • (1) S. Skogestad, “Simple analytic rules for model reduction and PID controller design”, J.Proc.Control, Vol. 13, 291-309, 2003 • Smooth control: Want “slowest possible control” subject to having acceptable disturbance rejection • (2) S. Skogestad, ``Tuning for smooth PID control with acceptable disturbance rejection'', Ind.Eng.Chem.Res, 45 (23), 7817-7822 (2006).

  19. (1) TIGHT CONTROL

  20. TIGHT CONTROL Example. Integrating process with delay=1. G(s) = e-s/s. Model: k’=1, =1, 1=1 SIMC-tunings with c with ==1: IMC has I=1 Ziegler-Nichols is usually a bit aggressive Setpoint change at t=0 Input disturbance at t=20

  21. TIGHT CONTROL • Approximate as first-order model with k=1, 1 = 1+0.1=1.1, =0.1+0.04+0.008 = 0.148 • Get SIMC PI-tunings (c=): Kc = 1 ¢ 1.1/(2¢ 0.148) = 3.71, I=min(1.1,8¢ 0.148) = 1.1 2. Approximate as second-order model with k=1, 1 = 1, 2=0.2+0.02=0.22, =0.02+0.008 = 0.028 Get SIMC PID-tunings (c=): Kc = 1 ¢ 1/(2¢ 0.028) = 17.9, I=min(1,8¢ 0.028) = 0.224, D=0.22

  22. (2) SMOOTH CONTROL (2) Will derive Kc,min. From this we can get c,max using SIMC tuning rule Tuning for smooth control • Tuning parameter: c = desired closed-loop response time • (1) Selecting c=(“tight control”) is reasonable for cases with a relatively large effective delay  • Other cases: Select c >  for • slower control • smoother input usage • less disturbing effect on rest of the plant • less sensitivity to measurement noise • better robustness • Question: Given that we require some disturbance rejection. • What is the largest possible value for c ? • Or equivalently: The smallest possible value for Kc?

  23. SMOOTH CONTROL Closed-loop disturbance rejection d0 -d0 ymax -ymax

  24. Kc u SMOOTH CONTROL Minimum controller gain for PI-and PID-control: Kc¸ Kc,min = |ud0|/|ymax| |ud0|: Input magnitude required for disturbance rejection |ymax|: Allowed output deviation

  25. SMOOTH CONTROL Minimum controller gain: Industrial practice: Variables (instrument ranges) often scaled such that Minimum controller gain is then (span) Minimum gain for smooth control) Common default factory setting Kc=1 is reasonable !

  26. SMOOTH CONTROL Example c is much larger than =0.25 Does not quite reach 1 because d is step disturbance (not not sinusoid) Response to disturbance = 1 at input

  27. SMOOTH CONTROL LEVEL CONTROL q LC V Application of smooth control • Averaging level control If you insist on integral action then this value avoids cycling Reason for having tank is to smoothen disturbances in concentration and flow. Tight level control is not desired: gives no “smoothening” of flow disturbances. Derivation of tunings: Minimum controller gain for acceptable disturbance rejection: Kc ¸≥ Kc,min = |ud0|/|ymax| Where in our case |ud0| = |q0| - expected flow change [m3/s] (input disturbance) |ymax| = |Vmax| - largest allowed variation in level [m3] Integral time: From the material balance (dV/dt = q – qout), the model is g(s)=k’/s with k’=1. Select Kc=Kc,min = |q0| |Vmax| . SIMC-Integral time for integrating process: I = 4 / (k’ Kc) = 4 |Vmax|/ |q0| = 4 ¢ residence time provided tank is nominally half full (so |Vmax| = V) and q0 is equal to the nominal flow.

  28. SMOOTH CONTROL Rule: Kc¸ |ud0|/|ymax| ( ~1 in scaled variables) • Exception to rule: Can have even smaller Kc (< 1) if disturbances are handled by the integral action. • Happens when c > I = 1 • Applies to: Process with short time constant (1 is small) • For example, flow control Kc: Assume variables are scaled with respect to their span

  29. SMOOTH CONTROL Summary: Tuning of easy loops • Easy loops: Small effective delay (¼ 0), so closed-loop response time c (>> ) is selected for “smooth control” • Flow control: Kc=0.5, I = 1 = time constant valve (typically, 10s to 30s) • Level control: Kc=2 (and no integral action) • Other easy loops (e.g. pressure control): Kc = 2, I = min(4c, 1) • Note: Often want a tight pressure control loop (so may have Kc=10 or larger) Kc: Assume variables are scaled with respect to their span

  30. LEVEL CONTROL More on level control • Level control often causes problems • Typical story: • Level loop starts oscillating • Operator detunes by decreasing controller gain • Level loop oscillates even more • ...... • ??? • Explanation: Level is by itself unstable and requires control.

  31. LEVEL CONTROL Integrating process: Level control

  32. LEVEL CONTROL How avoid oscillating levels? Actually, 0.1 = 1/π2

  33. LEVEL CONTROL Case study oscillating level • We were called upon to solve a problem with oscillations in a distillation column • Closer analysis: Problem was oscillating reboiler level in upstream column • Use of Sigurd’s rule solved the problem

  34. LEVEL CONTROL

  35. Identifying the model

  36. BOTTOM V: -0.5 63% xB 1=16 min / 19 min Model identification from step response • General: Need 3 parameters: • Delay  • 2 of the following: k, k’, 1(Note relationship k’ = k/1) • Two approaches for obtaining 1: • Initial response: ( + 1) is where initial slope crosses final steady-state • (+1)= 63% of final steady state • If disagree: Initial response is usually best, but to be safe use smallest value for 1 (“fast response”) as this gives more conservative PID-tunings • 1 = 16 min for this case k’ k

  37. Model identification from step response • Generally need 3 parameters • BUT For control: Dynamics at “control time scale” (c) are important • “Slow” (integrating) process (c< 1/5): May neglect 1and use only 2 parameters: k’ and  • g(s) = k’ e- s / s • “Delay-free” process (c > 5θ): Need only 2 parameters: k and 1 • g(s) = k / (1 s+1) c = desired response time 1¼ 200,   1 c time c < τ1/5 = 40: “Integrating” (neglect τ1) c > 5θ = 5: “Delay-free” (neglect θ)

  38. “Integrating” process (c < 1/5): Need only two parameters: k’ and  From step response: Example. Step change in u: u = 0.1 Initial value for y: y(0) = 2.19 Observed delay:  = 2.5 min At T=10 min: y(T)=2.62 Initial slope: Response on stage 70 to step in L y(t) 2.62-2.19 7.5 min =2.5 t [min]

  39. BOTTOM V: -0.5 63% xB 16 min “Delay-free” process (c > 5 ) • Example: • First-order model: • = 0 • k= (0.134-0.1)/(-0.5) = -0.068 (steady-state gain) • 1= 16 min (63% of change)

  40. Step response experiment: How long do we need to wait? • NORMALLY NO NEED TO RUN THE STEP EXPERIMENT FOR LONGER THAN ABOUT 10 TIMES THE EFFECTIVE DELAY ()

  41. Conclusion PID tuning

  42. Cascade control

  43. Tuning of cascade controllers

  44. ys y2 y2s u2 G2 G1 K1 K2 y1 Cascade control serial process(“exception”) d=6

  45. ys y2 y2s u G2 G1 K1 K2 y1 Without cascade With cascade Cascade control serial process d=6

  46. Tuning cascade control: serial process • Inner fast (secondary) loop: • P or PI-control • Local disturbance rejection • Much smaller effective delay (0.2 s) • Outer slower primary loop: • Reduced effective delay (2 s instead of 6 s) • Time scale separation • Inner loop can be modelled as gain=1 + 2*effective delay (0.4s) • Very effective for control of large-scale systems

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