PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway

PID controller Time domain Laplace domain Only two parameters (Kc and τI)… but surprisingly difficult to tune without systematic approach

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 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” c ¸ - : desired closed-loop response time (tuning parameter) For robustness select: c ¸ 

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 Step response experiment k’=k/1 STEP IN INPUT u (MV) RESULTING OUTPUT y (CV) : 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

Step response integrating process Δy Δt

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 the “effective” delay 

Example 1 Half rule

half rule

2 half rule

original 1st-order+delay 2nd-order+delay

Approximation of zeros Correction: More generally replace θ by τc in the above rules

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 Algebra:

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! g c d y u

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 The effective delay θ (which limits the achievable closed-loop time constant τc) is independent of the dominant process time constant τ1 It depends on τ2/2 (PI) or τ3/2 (PID) Use (close to) P-control for integrating process Beware of large I-action (small τI) for level control Use (close to) I-control for fast process (with small time constant τ1)

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

Quiz: SIMC PI-tunings y y Step response t [s] Time t The Figure shows the response (y) from a test where we made a step change in the input (Δu = 0.1) at t=0. Suggest PI-tunings for (1) τc=2,. (2) τc=10

Solution Actual plant: g = 0.35*(50*s+1)/((4*s+1)*(75*s+1)*(3*s+1)*(0.3*s+1))*exp(-0.2*s);

Tunings from Step response Kc=2.9, tauI=10 Kc=9.5, tauI=10 OUTPUT y INPUT u Tunings from Step response Setpoint change at t=0, input disturbance = 0.1 at t=50 Tunings from Half rule (Somewhat better) Kc=2, tauI=5.5 Kc=6, tauI=5.5

Selection of tuning parameter c Two main cases TIGHT CONTROL: Want “fastest possible control” subject to having good robustness Want tight control of active constraints (“squeeze and shift”) 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: SMOOTH CONTROL:

TIGHT CONTROL

Typical closed-loop SIMC responses with the choice c= TIGHT CONTROL Typical closed-loop SIMC responses with the choice c=

Example. Integrating process with delay=1. G(s) = e-s/s. 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 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

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

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

u Kc SMOOTH CONTROL Minimum controller gain for PI-and PID-control: Kc ¸ Kc,min = |u0|/|ymax| |u0|: Input magnitude required for disturbance rejection |ymax|: Allowed output deviation

Common default factory setting Kc=1 is reasonable ! 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 !

Example SMOOTH CONTROL Response to step disturbance = 1 at input 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

Application of smooth control LEVEL CONTROL Application of smooth control Averaging level control V q LC 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. 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.

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.

Integrating process: Level control

How avoid oscillating levels? LEVEL CONTROL How avoid oscillating levels? 0.1 ¼ 1/2

Case study oscillating level 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

Rule: Kc ¸ |u0|/|ymax| =1 (in scaled variables) 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

Summary: Tuning of easy loops 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=0.2, I = 1 = time constant valve (typically, 2 to 10s) 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)

Conclusion PID tuning 3. Derivative time: Only for dominant second-order processes

Some discussion points Selection of τc: some other issues Obtaining the model from step responses: How long should we run the experiment? Cascade control: Tuning Controllability implications of tuning rules

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 time scale” (c) are important Integrating process (1 = 1) Time constant 1 is not important if it is much larger than the desired response time c. More precisely, may use 1 =1 for 1 > 5 c Delay-free process (=0) Delay  is not important if it is much smaller than the desired response time c. More precisely, may use  ¼ 0 for  < c/5 time  ¼ 1 (may be neglected for c > 5) c = desired response time

Step response experiment: How long do we need to wait? RULE: May stop at about 10 times 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 ()

“Integrating process” (c < 0.2 1): 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]

Example (from quiz) OUTPUT y INPUT y tauc=10 tauc=2 Step response Δu=0.1 Assume integrating process, theta=1.5; k’ = 0.03/(0.1*11.5)=0.026 SIMC-tunings tauc=2: Kc=11, tauI=14 (OK) SIMC-tunings tauc=10: Kc=3.3, tauI = 46 (too long because process is not actually integrating on this time scale!) OUTPUT y INPUT y tauc=10 tauc=2

Cascade control

Tuning of cascade controllers

Cascade control serial process G1 u2 y1 K1 ys G2 K2 y2 y2s

Cascade control serial process G1 u y1 K1 ys G2 K2 y2 y2s 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

A comment on 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 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

Example: Distillation column CONTROLLABILITY Example: Distillation column

Example: Distillation column CONTROLLABILITY Example: Distillation column

Conclusion controllability If the plant is not controllable then improved tuning will not help Alternatives Change the process design to make it more controllable Better “self-regulation” with respect to disturbances, e.g. insulate your house to make y=Tin less sensitive to d=Tout. Give up some of your performance requirements