Model SIMC-tunings Tight control Smooth control Level control

PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway September 2008
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, , 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^ *Mo + 1.0 r = 2*A*abs(b/(1-b)) k = (1/Kc0) * abs(b/(1-b)) theta = tp*[ *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)

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 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 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

Quiz: SIMC PI-tunings SIMC-tunings 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

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, = = 0.148 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= =0.22, = = 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), (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: 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)

QUIZ: What are the benefits of adding a flow controller (inner cascade)?
qs Extra measurement y2 = q q z Counteracts nonlinearity in valve, f(z) With fast flow control we can assume q = qs Eliminates effect of disturbances in p1 and p2

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

Alternative form with detuning factor F

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”

Tunings from Step response “by eye” model
SIMC-tunings Kc=2.9, tauI=10 Kc=9.5, tauI=10 OUTPUT y INPUT u Tunings from Step response “by eye” model 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

Half-rule approach: Approximation of zeros depends on tauc!
QUIZ Half-rule approach: Approximation of zeros depends on tauc!

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) 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 Cascade control

Tuning of cascade controllers
• Want to control y (primary CV), but have “extra” measurement y 1 2 • Idea: Secondary variable (y ) may be tightly controlled and this 2 helps control of y . 1 • 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 ( G1’ = G1 G21 K2 (I+K2G22)-1 ) experimentally after closing each loop • One exception: Serial process, G21 = G22 2 – Inner (secondary - 2) loop may be modelled with gain=1 and effective delay=( t + q ) c 2 See next slide

Special case: Serial cascade
Cascade control Special case: Serial cascade y2 = T2 r2 + S2d2, T2 = G2K2(I+G2K2)-1 K2 is designed based on G2 (which has effective delay 2) then y2 = T2 r2 + S2 d2 where S2 ¼ 0 and T2 ¼1 · e-(2+c2)s T2: gain = 1 and effective delay = 2+c2 NOTE: If delay is in meas. of y2 (and not in G2) then T2 ¼ 1 ·e-c2s SIMC-rule: c2 ≥ 2 Time scale separation: c2 ≤ c1/5 (approximately) K1 is designed based on G1’ = G1T2 same as G1 but with an additional delay 2+c2

Example: Cascade control serial process
G1 u y1 K1 ys G2 K2 y2 y2s Use SIMC-rules! Without cascade With cascade

Tuning cascade control

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

Alternative closed-loop approach: Setpoint overshoot method
Procedure: Switch to P-only mode and make setpoint change Adjust controller gain to get overshoot about 0.30 (30%) Record “key parameters”: 1. Controller gain Kc0 2. Overshoot = (Δyp-Δy∞)/Δy∞ 3. Time to reach peak (overshoot), tp 4. Steady state change, b = Δy∞/Δys. Estimate of Δy∞ without waiting to settle: Δy∞ = 0.45(Δyp + Δyu) Advantages compared to Ziegler-Nichols: * Not at limit to instability * Works on a simple second-order process. Closed-loop step setpoint response with P-only control. M. Shamsuzzoha and S. Skogestad, ``The setpoint overshoot method: A simple and fast method for closed-loop PID tuning'', Journal of Process Control, 20, xxx-xxx (2010)

Proposed PI settings (including detuning factor F)
Setpoint overshoot method Summary setpoint overshoot method From P-control setpoint experiment record “key parameters”: 1. Controller gain Kc0 2. Overshoot = (Δyp-Δy∞)/Δy∞ 3. Time to reach peak (overshoot), tp 4. Steady state change, b = Δy∞/Δys Proposed PI settings (including detuning factor F) Choice of detuning factor F: F=1. Good tradeoff between “fast and robust” (SIMC with τc=θ) F>1: Smoother control with more robustness F<1 to speed up the closed-loop response.

Example: High-order process
Setpoint overshoot method Example: High-order process P-setpoint experiments Closed-loop PI response

Example: Unstable plant
Setpoint overshoot method Example: Unstable plant First-order unstable process No SIMC settings available Closed-loop PI response

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

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

Model SIMC-tunings Tight control Smooth control Level control

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