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: (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

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

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 

half rule

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 T = (y/ys)desired 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 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

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

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

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

Some special cases One tuning parameter: c

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

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).

(1) TIGHT CONTROL

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

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? (2) 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 = |ud0|/|ymax| |ud0|: 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 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 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. 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.

Rule: Kc ¸ |ud0|/|ymax| ( ~1 in scaled variables) 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

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” 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

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? Actually, 0.1 = 1/π2

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

Identifying the model

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 BOTTOM V: -0.5 xB 1=16 min / 19 min 63% k’ k

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 1 and 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 1 ¼ 200,   1 c time c < τ1/5 = 40: “Integrating” (neglect τ1) c > 5θ = 5: “Delay-free” (neglect θ)

“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]

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

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 ()

Conclusion PID tuning

Cascade control

Tuning of cascade controllers

Cascade control serial process (“exception”) 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

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 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

Distillation example: Frequency domain analysis CONTROLLABILITY Distillation example: Frequency domain analysis

Other factors limiting controllability Input limitations (saturation or “slow valve”): Can sometimes limit achievable speed of response Unstable process (including levels): Need feedback control for stabilization ) Makes sure inputs do not saturate in stabilizing loops