PID Tuning using the SIMC rules

Sigurd SkogestadNTNU, Trondheim, Norway

September 2008

1. Model2. SIMC-tunings3. Tight control4. Smooth control5. Level control6. Discussion points

Operation: Decision and control layers

cs = y1s

MPC

PID

y2s

RTO

u (valves)

CV=y1; MV=y2s

CV=y2; MV=u

Min J (economics); MV=y1s

Stepwise procedure plantwide control

I. TOP-DOWNStep 1. DEGREES OF FREEDOMStep 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!

e

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 responseInitial 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 ≥

Let’s start with the CONCLUSION

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

MODEL

1. Step response experiment

Make step change in one u (MV) at a time Record the output (s) y (CV)

MODEL, Approach 1A

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 change1: Time constant - Additional time to reach 63% of final changek : steady-state gain = y(∞)/ u k’ : slope after response “takes off” = k/1

MODEL, Approach 1A

STEP IN INPUT u

RESULTING OUTPUT y

: Delay - Time where output does not change1: Time constant - Additional time to reach

63% of final changek = y(∞)/ u : Steady-state gain

Δy(∞)

Δu

MODEL, Approach 1A

Step response integrating process

Δy

Δt

MODEL, Approach 1A

Model from closed-loop response with P-controller

Kc0=1.5Δys=1

Δyu=0.54

Δyp=0.79

tp=4.4

dyinf = 0.45*(dyp + dyu)Mo =(dyp -dyinf)/dyinfb=dyinf/dys

A = 1.152*Mo^2 - 1.607*Mo + 1.0r = 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

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)

Δy∞

MODEL, Approach 1B

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

half rule

2

MODEL, Approach 2

original

1st-order+delay

2nd-order+delay

MODEL, Approach 2

Original 7th order, g0

1st-order, half rule, g1

2nd-order, half rule, g2

1st-order, closed loop, gcl

PID

PID

PIhalf-rule

PIhalf-rule

PIcl

PIcl

Approximation of zeros

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

MODEL, Approach 2

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)

cc

c

c c

c

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:

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

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 responseTry reducing it!

gc

dyu

SIMC-tunings

Integral Time

I = 1

Reduce I to this value:I = 4 (c+) = 8

SIMC-tunings

Setpoint change at t=0 Input disturbance at t=20

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.

SIMC-tunings

Details: See www.nt.ntnu.no/users/skoge/publications/2003/tuningPID Remark 13 II

Conclusion: SIMC-PID Tuning Rules

One tuning parameter: c

SIMC-tunings

Some insights from tuning rules

1. 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)

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 fast process (with small time constant τ1)

4. 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

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

SIMC-tunings

Selection of tuning parameter c

Two main cases1. TIGHT CONTROL: Want “fastest possible

control” subject to having good robustness• Want tight control of active constraints (“squeeze and shift”)

2. 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:

SIMC-tunings

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=∞SIMC-tunings with c with ==1:

IMC has I=∞

Ziegler-Nichols is usually a bit aggressive

Setpoint change at t=0c 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 = 0.148Get 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 = 0.028Get 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

Will derive Kc,min. From this we can get c,max using SIMC tuning rule

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?

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

Closed-loop disturbance rejectiond0

ymax

-d0

-ymax

SMOOTH CONTROL

10−2

10−1

100

101

102

100

101

102

103

Frequency

|d0|/|y

max|

|cPI

|

|cPID

|

Kcu

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)

SMOOTH CONTROL

|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

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 ( ≈ 0). For example, flow control

Then I = 1 is small so integral action is “large”

SMOOTH CONTROL

Summary: Tuning of easy loops

Easy loops: Small effective delay ( ≈ 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(4c, 1)

Note: Often want a tight pressure control loop (so may have Kc=10 or larger)

SMOOTH CONTROL

Application of smooth control Averaging level control

VqLC

Reason for having tank is to smoothen disturbances in concentration and flow. Tight level control is not desired: gives no “smoothening” of flow disturbances.

SMOOTH CONTROL LEVEL CONTROL

If you insist on integral actionthen this value avoids cycling

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 timeprovided 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.

LEVEL CONTROL

How avoid oscillating levels?LEVEL CONTROL

• 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 ≈ 1/2

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

LEVEL CONTROL

Conclusion PID tuningSIMC 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

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

Quiz: SIMC PI-tunings

(a) The Figure shows the response (y) from a test where we made a step change in theinput (Δu = 0.1) at t=0. Suggest PI-tunings for (1) τc=2,. (2) τc=10.

t [s]

y

0 10 20 30 40 50 600.1

0.105

0.11

0.115

0.12

0.125

0.13

Step response

y

Time t

SIMC-tunings

(b) Do the same, given that the actual plant is

QUIZ

Solution

Actual plant:

QUIZ

Approximation of step response

Approximation ”bye eye”

QUIZ