PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway Model SIMC-tunings Tight control Smooth control Level control Discussion points

Operation: Decision and control layers RTO Min J (economics); MV=y1s cs = y1s CV=y1; MV=y2s MPC y2s PID CV=y2; MV=u u (valves)

Stepwise procedure plantwide control I. TOP-DOWN Step 1. DEGREES OF FREEDOM Step 2. OPERATIONAL OBJECTIVES Step 3. WHAT TO CONTROL? (primary CV’s c=y1) Step 4. PRODUCTION RATE II. BOTTOM-UP (structure control system): Step 5. REGULATORY CONTROL LAYER (PID) “Stabilization” What more to control? (secondary CV’s y2) Step 6. SUPERVISORY CONTROL LAYER (MPC) Decentralization Step 7. OPTIMIZATION LAYER (RTO) Can we do without it?

PID controller Time domain (“ideal” PID) Laplace domain (“ideal”/”parallel” form) Usually τD=0. Only two parameters left (Kc and τI)… How difficult can it be??? Surprisingly difficult without systematic approach!

Tuning of PID controllers Let’s start with the CONCLUSION 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 (Also reprinted in MIC) (*) “Probably the best simple PID tuning rules in the world” c ¸ - : desired closed-loop response time (tuning parameter) For robustness select: c ¸ 

MODEL 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

1. Step response experiment MODEL, Approach 1A 1. Step response experiment Make step change in one u (MV) at a time Record the output (s) y (CV)

MODEL, Approach 1A Δy(∞) RESULTING OUTPUT y STEP IN INPUT u Δu : Delay - Time where output does not change 1: Time constant - Additional time to reach 63% of final change k =  y(∞)/ u : Steady-state gain

Step response integrating process MODEL, Approach 1A Step response integrating process Δy Δt

Model from closed-loop response with P-controller MODEL, Approach 1B Model from closed-loop response with P-controller Kc0=1.5 Δys=1 Δy∞ dyinf = 0.45*(dyp + dyu) Mo =(dyp -dyinf)/dyinf b=dyinf/dys A = 1.152*Mo^2 - 1.607*Mo + 1.0 r = 2*A*abs(b/(1-b)) k = (1/Kc0) * abs(b/(1-b)) theta = tp*[0.309 + 0.209*exp(-0.61*r)] tau = theta*r Δyp=0.79 Δyu=0.54 tp=4.4 Example 2: Get k=0.99, theta =1.68, tau=3.03 Ref: Shamssuzzoha and Skogestad (JPC, 2010) + modification by C. Grimholt (Project, NTNU, 2010; see also new from PID-book 2011)

2. Model reduction of more complicated model MODEL, Approach 2 2. 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 

MODEL, Approach 2

MODEL, Approach 2 Example 1 Half rule

MODEL, Approach 2 original 1st-order+delay

MODEL, Approach 2 2 half rule

MODEL, Approach 2 original 1st-order+delay 2nd-order+delay

Approximation of zeros MODEL, Approach 2 Approximation of zeros To make these rules more general (and not only applicable to the choice c=): Replace  (time delay) by c (desired closed-loop response time). (6 places) c c c c c c Correction: More generally replace θ by τc in the above rules

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

Basis: Direct synthesis (IMC) SIMC-tunings Basis: Direct synthesis (IMC) Closed-loop response to setpoint change Idea: Specify desired response: and from this get the controller. ……. Algebra:

SIMC-tunings

IMC Tuning = Direct Synthesis SIMC-tunings IMC Tuning = Direct Synthesis Algebra:

SIMC-tunings 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 SIMC-tunings I = 1 Reduce I to this value: I = 4 (c+) = 8  Setpoint change at t=0 Input disturbance at t=20

SIMC-tunings Integral time Want to reduce the integral time for “integrating” processes, but to avoid “slow oscillations” we must require: Derivation: Setpoint response: Improve (get rid of overshoot) by “pre-filtering”, y’s = f(s) ys. Details: See www.nt.ntnu.no/users/skoge/publications/2003/tuningPID Remark 13 II

Conclusion: SIMC-PID Tuning Rules SIMC-tunings Conclusion: SIMC-PID Tuning Rules One tuning parameter: c

Some insights from tuning rules SIMC-tunings 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) Parameter variations: For robustness tune at operating point with maximum value of k’ θ = (k/τ1)θ

SIMC-tunings Some special cases One tuning parameter: c

Another special case: IPZ process SIMC-tunings 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.

SIMC-tunings Note: Derivative action is commonly used for temperature control loops. Select D equal to 2 = time constant of temperature sensor

SIMC-tunings

SIMC-tunings

Selection of tuning parameter c SIMC-tunings 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=0c 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

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 S. Skogestad, ``Tuning for smooth PID control with acceptable disturbance rejection'', Ind.Eng.Chem.Res, 45 (23), 7817-7822 (2006).

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: min |c(j)| = Kc

SMOOTH CONTROL Rule: Min. controller gain for acceptable disturbance rejection: Kc ¸ |u0|/|ymax| often ~1 (in span-scaled variables) |ymax| = allowed deviation for output (CV) |u0| = required change in input (MV) for disturbance rejection (steady state) = observed change (movement) in input from historical data

SMOOTH CONTROL Rule: Kc ¸ |u0|/|ymax| Exception to rule: Can have lower Kc 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). For example, flow control Then I = 1 is small so integral action is “large”

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; close to pure integrating!) Level control: Kc=2 (and no integral action) Other easy loops (e.g. pressure): Kc = 2, I = min(4c, 1) Note: Often want a tight pressure control loop (so may have Kc=10 or larger)

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. Proof: 1. 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| = |q0| / |Vmax| 2. 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.

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

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

Quiz: SIMC PI-tunings SIMC-tunings QUIZ y y Step response t [s] Time t (a) 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. (b) Do the same, given that the actual plant is

QUIZ Solution Actual plant:

Approximation of step response QUIZ Approximation of step response Approximation ”bye eye”