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PID Tuning 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 Key: Look at initial part of step response Initial slope: k’ = k/1 One tuning rule! Easily memorized c ¸ 0: desired closed-loop response time (tuning parameter) For robustness select: c ¸ Reference: S. Skogestad, “Simple analytic rules for model reduction and PID controller design”, J.Proc.Control, Vol. 13, 291-309, 2003 (*) “Probably the best simple PID tuning rules in the world” Need a model for tuning 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 Step response experiment Make step change in one u (MV) at a time Record the output (s) y (CV) First-order plus delay process RESULTING OUTPUT y (CV) k’=k/1 STEP IN INPUT u (MV) Step response experiment : 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 Model reduction of more complicated model 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 Example Half rule half rule original 1st-order+delay 2nd-order+delay Approximation of zeros Derivation of SIMC-PID tuning rules 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: Basis: Direct synthesis (IMC) Closed-loop response to setpoint change Idea: Specify desired response: and from this get the controller. Algebra: IMC Tuning = Direct Synthesis Integral time Found: Integral time = dominant time constant (I = 1) Works well for setpoint changes Needs to be modified (reduced) for integrating disturbances d u y c g 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! Integral Time I = 1 Reduce I to this value: I = 4 (c+) = 8 Integral time Want to reduce the integral time for “integrating” processes, but to avoid “slow oscillations” we must require: Derivation: Conclusion: SIMC-PID Tuning Rules One tuning parameter: c Some insights from tuning rules 1. The effective delay θ (which limits the achievable closed-loop time constant τ2/2 ) is independent of the dominant process time constant τ1 It depends on τ2/2 (PI) or τ3/2 (PID) 2. Use (close to) P-control for integrating process Beware of large I-action (small τI) for level control 3. Use (close to) I-control for time delay process Some special cases One tuning parameter: c Another special case: IPZ process IPZ-process may represent response from steam flow to pressure Rule T2: SIMC-tunings These tunings turn out to be almost identical to the tunings given on page 104-106 in the Ph.D. thesis by O. Slatteke, Lund Univ., 2006 and K. Forsman, "Reglerteknik for processindustrien", Studentlitteratur, 2005. Note: Derivative action is commonly used for temperature control loops. Select D equal to 2 = time constant of temperature sensor Selection of tuning parameter c Two main cases TIGHT CONTROL: 1. TIGHT CONTROL: Want “fastest possible control” subject to having good robustness • Want tight control of active constraints (“squeeze and shift”) 2. SMOOTH CONTROL: SMOOTH CONTROL: Want “slowest possible control” subject to acceptable disturbance rejection • Want smooth control if fast setpoint tracking is not required, for example, levels and unconstrained (“self-optimizing”) variables • THERE ARE ALSO OTHER ISSUES: Input saturation etc. TIGHT CONTROL TIGHT CONTROL Typical closed-loop SIMC responses with the choice c= 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 TIGHT CONTROL 1. 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 TIGHT CONTROL SMOOTH CONTROL Tuning for smooth control Tuning parameter: c = desired closed-loop response time 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? Will derive Kc,min. From this we can get c,max using SIMC tuning rule SMOOTH CONTROL Closed-loop disturbance rejection d0 -d0 ymax -ymax SMOOTH CONTROL Kc u Minimum controller gain for PI-and PID-control: Kc ¸ Kc,min = |u0|/|ymax| |u0|: Input magnitude required for disturbance rejection |ymax|: Allowed output deviation SMOOTH CONTROL Minimum controller gain: Industrial practice: Variables (instrument ranges) often scaled such that (span) Minimum controller gain is then Minimum gain for smooth control ) Common default factory setting Kc=1 is reasonable ! 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 step disturbance = 1 at input SMOOTH CONTROL LEVEL CONTROL Application of smooth control Averaging level control q V If you insist on integral action LC 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. Let |u0| = | q0| – expected flow change [m3/s] (input disturbance) |ymax| = |Vmax| - largest allowed variation in level [m3] Minimum controller gain for acceptable disturbance rejection: Kc ¸ Kc,min = |u0|/|ymax| From the material balance (dV/dt = q – qout), the model is g(s)=k’/s with k’=1. Select Kc=Kc,min. SIMC-Integral time for integrating process: I = 4 / (k’ Kc) = 4 |Vmax| / | q0| = 4 ¢ residence time provided tank is nominally half full and q0 is equal to the nominal flow. 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. LEVEL CONTROL Integrating process: Level control q V LC • Level control problem has – y = V, u = qout, d=q • Model of level: V(s) = (q - qout)/s – G(s)=k’/s, Gd(s)= -k’/s (with k’=1) • Apply PI-control: u = c(s) (ys-y); c(s) = Kc(1+1/Is) • Closed-loop response to input disturbance: y/d = gd / (1+gc) = I s / (I/k’ s2 + Kc I s + Kc) • The denominator can be rewritten on standard form This is the basis – (02 s2 + 2 0 s + 1) with 02 = I/k’¢Kc and 2 0 = I for the SIMC-rule – Algebra gives: for the minimum integral time – To avoid oscillations we must require 1, or Kc¢k’ ¢I > 4 • The controller gain Kc must be large to avoid oscillations! LEVEL CONTROL How avoid oscillating levels? • Simplest: Use P-control only (no integral action) • If you insist on integral action, then make sure the controller gain is sufficiently large • If you have a level loop that is oscillating then use Sigurds rule (can be derived): To avoid oscillations, increase Kc ¢I by factor f=0.1¢(P0/I0)2 where P0 = period of oscillations [s] I0 = original integral time [s] 0.1 ¼ 1/2 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 LEVEL CONTROL SMOOTH CONTROL Rule: Kc ¸ |u0|/|ymax| =1 (in scaled variables) Exception to rule: Can have Kc < 1 if disturbances are handled by the integral action. Disturbances must occur at a frequency lower than 1/I Applies to: Process with short time constant (1 is small) and no delay ( ¼ 0). Then I = 1 is small so integral action is “large” For example, flow control Kc: Assume variables are scaled with respect to their span 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” ASSUME VARIABLES HAVE BEEN SCALED WITH RESPECT TO THEIR SPAN SO THAT |u0/ymax| = 1 (approx.). 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(4c, 1) Note: Often want a tight pressure control loop (so may have Kc=10 or larger) Selection of c: Other issues Input saturation. Problem. Input may “overshoot” if we “speedup” the response too much (here “speedup” = /c). Solution: To avoid input saturation, we must obey max “speedup”: A little more on obtaining the model from step response experiments 1 ¼ 200 (may be neglected for c < 40) “Factor 5 rule”: Only dynamics within a factor 5 from “control 0.9954 time scale” (c) are important 0.9953 0.9953 0.9952 Integrating process (1 = 1) 0.9952 Time constant 1 is not important if it is much larger than the desired response time c. More 0.9951 precisely, may use 0.9951 0.995 1 =1 for 1 > 5 c 0.995 0.9949 0 10 20 30 40 50 60 time Delay-free process (=0) ¼1 Delay is not important if it is much smaller than (may be neglected for c > 5) the desired response time c. More precisely, may use c = desired response time ¼ 0 for < c/5 “Integrating process” (c < 0.2 1): Need only two parameters: k’ and From step response: Example. Response on stage 70 to step in L 2.8 Step change in u: u = 0.1 2.7 Initial value for y: y(0) = 2.19 2.6 Observed delay: = 2.5 min At T=10 min: y(T)=2.62 2.5 Initial slope: y(t) 2.4 2.62-2.19 2.3 7.5 min 2.2 =2.5 t [min] 2.1 0 2 4 6 8 10 First-order with delay process (c > 0.2 1) Need 3 parameters: Delay 2 of the following: k, k’, 1 (Note relationship k’ = k/1) Example: BOTTOM V: -0.5 0.14 0.135 First-order model: 0.13 = 0 k= (0.134-0.1)/(-0.5) = -0.068 (steady-state gain) 63% 0.125 xB 1= 16 min (63% of change) 0.12 0.115 0.11 Gives k’ = 0.068/16 = 0.00425 (could alternatively find k’ by observing the initial 0.105 response) 0.1 0.095 16 min 0 50 100 150 200 250 300 Step response experiment: How long do we need to wait? RULE: May stop at about 10 times of effective delay FAST TUNING DESIRED (“tight control”, c = ): NORMALLY NO NEED TO RUN THE STEP EXPERIMENT FOR LONGER THAN ABOUT 10 TIMES THE EFFECTIVE DELAY () EXCEPTION: LET IT RUN A LITTLE LONGER IF YOU SEE THAT IT IS ALMOST SETTLING (TO GET 1 RIGHT) SIMC RULE: I = min (1, 4(c+)) with c = for tight control SLOW TUNING DESIRED (“smooth control”, c > ): HERE YOU MAY WANT TO WAIT LONGER TO GET 1 RIGHT BECAUSE IT MAY AFFECT THE INTEGRAL TIME BUT THEN ON THE OTHER HAND, GETTING THE RIGHT INTEGRAL TIME IS NOT ESSENTIAL FOR SLOW TUNING SO ALSO HERE YOU MAY STOP AT 10 TIMES THE EFFECTIVE DELAY () Conclusion PID tuning SIMC tuning rules 1. Tight control: Select c= corresponding to 2. Smooth control. Select Kc ¸ Note: Having selected Kc (or c), the integral time I should be selected as given above Cascade control Tuning: 1. First tune TC (based on response from V to T) 2. Close TC and tune CC (based on response from Ts to xB) Ts TC Primary controller (CC) sets setpoint to secondary controller (TC). xB CC CC: Primary controller (“slow”): y1 = xB (“original” CV), u1 = y2s (MV) TC: Secondary controller (“fast”): y2 = T (CV), u2 = V (“original” MV) Tuning of cascade controllers • Want to control y 1 (primary CV), but have “extra” measurement y 2 • Idea: Secondary variable (y 2) may be tightly controlled and this helps control of y1. • Implemented using cascade control: Input (MV) of “primary” controller (1) is setpoint (SP) for “secondary” controller (2) • Tuning simple: Start with inner secondary loops (fast) and move upwards • Must usually identify new model experimentally after closing each loop. • One exception: Serial process with “original” input (u) and outputs (y1) at opposite ends of the process, and y 2 in the middle. – Inner (secondary-2) loop may be modelled with gain=1 and effective delay=(c+2 Cascade control serial process d=6 ys y2s u2 y2 K1 K2 G2 G1 y1 Cascade control serial process d=6 ys y2s u y2 K1 K2 G2 G1 y1 Without cascade With cascade 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 CONTROLLABILITY Controllability (Input-Output) “Controllability” is the ability to achieve acceptable control performance (with any controller) “Controllability” is a property of the process itself Analyze controllability by looking at model G(s) What limits controllability? CONTROLLABILITY Controllability Recall SIMC tuning rules 1. Tight control: Select c= corresponding to 2. Smooth control. Select Kc ¸ Must require Kc,max > Kc.min for controllability ) max. output deviation initial effect of “input” disturbance y reaches k’ ¢ |d0|¢ t after time t y reaches ymax after t= |ymax|/ k’ ¢ |d0| CONTROLLABILITY Controllability • More general disturbances. Requirement becomes (for c=): Following step disturbance d 0: Time it takes for output y to reach max. deviation • Conclusion: The main factors limiting controllability are – large effective delay from u to y ( large) – large disturbances (k’d |d0| / ymax large) • Can generalize using “frequency domain”: |Gd(j¢0.5/max)| ¢|d0| = |ymax| CONTROLLABILITY Example: Distillation column Response to 20% increase in feed rate (disturbance) with no control -3 x 10 16 14 Data for “column A” 12 Product purities: xD = 0.99 § 0.002, xB =0.01 § 0.005 10 xB(t) (mole fraction light component) 8 Small reboiler holdup, MB/F = 0.5 min 6 ymax=0.005 4 2 xD(t) 0 -2 Max. delay in feedback loop, max = 3/2 = 1.5 min 0 1 2 3 4 5 6 7 8 9 10 time [min] time to exceed bound = ymax/k’d |d| = 3 min Controllability: Must close a loop with time constant (c) faster than 1.5 min to avoid that bottom composition xB exceeds max. deviation If this is not possible: May add tank (feed tank?, larger reboiler volume?) to smooth disturbances CONTROLLABILITY Example: Distillation column Increase reboiler holdup to MB/F = 10 min x 10 -3 Original holdup x 10 -3 Larger holdup 16 16 14 14 xB(t) 12 12 10 10 8 8 xB(t) 6 6 4 4 2 2 0 0 -2 -2 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 3 min 5.8 min time [min] With increased holdup: Max. delay in feedback loop: = 2.9 min

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posted: | 12/4/2011 |

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