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PID Tuning

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Tuning of PID controllers

SIMC tuning rules (Skogestad IMC)(*)

Main message: Can usually do much better by taking asystematic 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 0: desired closed-loop response time (tuning parameter)

For robustness select: c

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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

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Step response experiment

Make step change in one u (MV) at a time

Record the output (s) y (CV)

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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

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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

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Example

Half rule

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half rule

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original

1st-order+delay

2nd-order+delay

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Approximation of zeros

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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:

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Basis: Direct synthesis (IMC)

Closed-loop response to setpoint change

Idea: Specify desired response:

and from this get the controller. Algebra:

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IMC Tuning = Direct Synthesis

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Integral time

Found: Integral time = dominant time constant (I = 1) Works well for setpoint changes

Needs to be modified (reduced) for integratingdisturbances

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!

gc

dyu

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Integral Time

I =

1

Reduce I to this value:

I = 4 (c+) = 8

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Integral time

Want to reduce the integral time for integrating

processes, but to avoid slow oscillations we must

require:

Derivation:

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Conclusion: SIMC-PID Tuning Rules

One tuning parameter: c

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Some insights from tuning rules

1. The effective delay (which limits the achievable

closed-loop time constant 2/2 ) is independent of

the dominant process time constant 1 It depends on 2/2 (PI) or3/2 (PID)

2. Use (close to) P-control for integrating process Beware of large I-action (small I) for level control

3. Use (close to) I-control for time delay process

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Some special cases

One tuning parameter: c

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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.

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Note: Derivative action is commonly used for temperature control loops.

Select D equal to 2 = time constant of temperature sensor

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Selection of tuning parameter cTwo main cases

1. TIGHT CONTROL: Want fastest possiblecontrol subject to having good robustness

Want tight control of active constraints (squeeze and shift)

2. SMOOTH CONTROL: Want slowest possiblecontrol 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: Inputsaturation etc.

TIGHT CONTROL:

SMOOTH CONTROL:

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TIGHT CONTROL

TIGHT CONTROL

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

TIGHT CONTROL

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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

TIGHT CONTROL

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1. 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

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TIGHT CONTROL

SMOOTH CONTROL

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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 largeeffective 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 forc ?

Or equivalently: The smallest possible value for Kc?

SMOOTH CONTROL

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Closed-loop disturbance rejectiond0

ymax

-d0

-ymax

SMOOTH CONTROL

SMOOTH CONTROL

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Minimum controller gain:

I ndustri al practice:Variables (instrument ranges) often scaled such that

Minimum controller gain is then

Minimum gain for smooth control)

Common default factory setting Kc=1 is reasonable !

SMOOTH CONTROL

(span)

SMOOTH CONTROL

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Example

Does not quite reach 1 because d is

step disturbance (not not sinusoid)

Response to step disturbance = 1 at input

c is much larger than =0.25

SMOOTH CONTROL LEVEL CONTROL

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Application of smooth control

Averaging level controlVqLC

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.

If you insist on integral action

then this value avoids cycling

LEVEL CONTROL

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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 andrequires control.

LEVEL CONTROL

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Integrating process: Level control

Level control problem has y = V, u = qout, d=q

Model of level: V(s) = (q - qout)/s G(s)=k/s, Gd(s)= -k/s (with k=1)

Apply PI-control: u = c(s) (ys-y); c(s) = Kc(1+1/Is) Closed-loop response to input disturbance:y/d = gd / (1+gc) = I s / (I/k s

2 + Kc I s + Kc)

The denominator can be rewritten on standard form (0

2 s2 + 2 0 s + 1) with 02 = I/kKc and 2 0 = I

Algebra gives:

To avoid oscillations we must require 1, or Kck I > 4

The controller gain Kc must be large to avoid oscillations!

VqLC

VqLC

qLC

This is the basisfor the SIMC-rulefor the minimum

integral time

LEVEL CONTROL

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How avoid oscillating levels?

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

LEVEL CONTROL

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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 Sigurds rule solved the problem

LEVEL CONTROL

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SMOOTH CONTROL

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Rule: Kc |u0|/|ymax| =1 (in scaledvariables)

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 ( 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

SMOOTH CONTROL

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Summary: Tuning of easy loops

Easy loops: Small effective delay ( 0), so closed-loop response time c (>> ) is selected for smoothcontrol

ASSUME VARIABLES HAVE BEEN SCALEDWITH 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(4c, 1) Note: Often want a tight pressure control loop (so may have

Kc=10 or larger)

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Selection ofc: 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 maxspeedup:

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A little more on obtaining the model

from step response experiments

Factor 5 rule: Only dynamicswithin a factor 5 from controltime 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. Moreprecisely, 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

0 for < c/5

0 10 20 30 40 50 600.9949

0.995

0.995

0.9951

0.9951

0.9952

0.9952

0.9953

0.9953

0.9954

1

(may be neglected forc > 5)

1 200(may be neglected forc < 40)

time

c = desired response time

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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 ()

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Integrating process (c < 0.2 1):

Need only two parameters: k and

From step response:

0 2 4 6 8 102.1

2.2

2.3

2.4

2.5

2.6

2.7

2.8

Response on stage 70 to step in L

7.5 min

2.62-2.19

=2.5

Example.

Step change in u: u = 0.1Initial value for y: y(0) = 2.19

Observed delay: = 2.5 min

At T=10 min: y(T)=2.62

Initial slope: y(t)

t [min]

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Conclusion PID tuning

SIMC tuning rules

1. Tight control: Select c= corresponding to

2. Smooth control. Select Kc

Note: Having selected Kc (orc), the integral time I should beselected as given above

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Cascade control

CC

TCTs

xB

CC: Primary controller (slow): y1 = xB (original CV), u1 = y2s (MV)

TC: Secondary controller (fast): y2 = T (CV), u2 = V (original MV)

Tuning:

1. First tune TC(based on response from V to T)

2. Close TC and tune CC

(based on response from Ts to xB)

Primary controller (CC)

sets setpoint to secondary

controller (TC).

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Tuning of cascade controllers

Want to control y1 (primary CV), but have extra measurement y2 Idea: Secondary variable (y2) may be tightly controlled and this

helps control of y1.

Implemented using cascade control: Input (MV) of primarycontroller (1) is setpoint (SP) for secondary controller (2)

Tuning simple: Start with inner secondary loops (fast) and moveupwards

Must usually identify new model experimentally after closing eachloop.

One exception: Serial process with original input (u) and outputs(y1) at opposite ends of the process, and y2 in the middle. Inner (secondary-2) loop may be modelled with gain=1 and effective

delay=(c+2

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Cascade control serial process

d=6

G1u2

y1

K1ys

G2K2y2y2s

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Cascade control serial processd=6

Without cascade

With cascade

G1u

y1

K1ys

G2K2y2y2s

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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

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Controllability

(Input-Output) Controllability is the ability toachieve 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

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Recall SIMC tuning rules

1. Tight control: Select c= corresponding to

2. Smooth control. Select Kc

Must require Kc,max > Kc.min for controllability

)

Controllability

initial effect of input disturbance

max. output deviation

y reaches k |d0| t after time t

y reaches ymax after t= |ymax|/ k |d0|

CONTROLLABILITY

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Controllability

More general disturbances. Requirementbecomes (forc=):

Conclusion: The main factors limitingcontrollability are

large effective delay from u to y ( large)

large disturbances (kd |d0| / ymax large)

Can generalize using frequency domain:|Gd(j0.5/max)| |d0| = |ymax|

Following step

disturbance d0:Time it takes for output y

to reach max. deviation

CONTROLLABILITY

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Example: Distillation column

0 1 2 3 4 5 6 7 8 9 10-2

0

2

4

6

8

10

12

14

16x 10

-3

time [min]

Response to 20% increase in feed rate (disturbance) with no control

xB(t)

xD(t)

ymax=0.005

time to exceed bound = ymax/kd |d| = 3 min

Controllability: Must close a loop with time constant (c) faster than 1.5 min toavoid that bottom composition xB exceeds max. deviation

If this is not possible: May add tank (feed tank?, larger reboiler volume?)

to smooth disturbances

Data for column A

Product purities:

xD = 0.99 0.002, xB=0.01 0.005

(mole fraction light component)

Small reboiler holdup, MB/F = 0.5 min

Max. delay in feedback

loop, max = 3/2 = 1.5 min

CONTROLLABILITY

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Example: Distillation column

0 1 2 3 4 5 6 7 8 9 10-2

0

2

4

6

8

10

12

14

16x 10

-3

0 1 2 3 4 5 6 7 8 9 10-2

0

2

4

6

8

10

12

14

16x 10

-3

time [min]

Increase reboiler holdup to MB/F = 10 min

xB(t)

3 min 5.8 min

xB(t)

Original holdup Larger holdup

With increased holdup: Max. delay in feedback loop: = 2.9 min

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Conclusion controllability

If the plant is not controllable then improved

tuning will not help

Alternatives

1. 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.

2. Give up some of your performance requirements

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