1. PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway

2. PID controller Time domain Laplace domain
Only two parameters (Kc and τI)…
but surprisingly difficult to tune without systematic approach

3. 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
(*) “Probably the best simple PID tuning rules in the world”
c ¸ - : desired closed-loop response time (tuning parameter)
For robustness select: c ¸ 

4. 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

5. Step response experiment
Make step change in one u (MV) at a time
Record the output (s) y (CV)

6. 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

7. Step response integrating processΔy Δt

8. 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 

10. Example 1
Half rule

11. half rule

12. 2
half rule

13. original
1st-order+delay
2nd-order+delay

14. Approximation of zeros
Correction: More generally replace θ by τc in the above rules

15. 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:

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

18. IMC Tuning = Direct Synthesis
Algebra:

19. 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

20. Integral Time
I = 1
Reduce I to this value:
I = 4 (c+) = 8 

21. Integral time
Want to reduce the integral time for “integrating” processes, but to avoid “slow oscillations” we must require:
Derivation:

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

23. 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)

24. Some special cases
One tuning parameter: c

25. 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.

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

28. Quiz: SIMC PI-tunings y y Step response t [s] Time t
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

29. Solution
Actual plant: g = 0.35*(50*s+1)/((4*s+1)*(75*s+1)*(3*s+1)*(0.3*s+1))*exp(-0.2*s);

30. Tunings from Step response
Kc=2.9, tauI=10
Kc=9.5, tauI=10
OUTPUT y
INPUT u
Tunings from Step response
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

31. 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:

32. TIGHT CONTROL

33. Typical closed-loop SIMC responses with the choice c=
TIGHT CONTROL
Typical closed-loop SIMC responses with the choice c=

34. 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

35. 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

36. TIGHT CONTROL

37. 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

38. Closed-loop disturbance rejection
SMOOTH CONTROL
Closed-loop disturbance rejection d0
-d0
ymax
-ymax

39. u Kc SMOOTH CONTROL Minimum controller gain for PI-and PID-control:
Kc ¸ Kc,min = |u0|/|ymax|
|u0|: Input magnitude required for disturbance rejection
|ymax|: Allowed output deviation

40. 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 !

41. Example SMOOTH CONTROL Response to step 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 step disturbance = 1 at input

42. 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.
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.

43. 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.

44. Integrating process: Level control

45. How avoid oscillating levels?
LEVEL CONTROL
How avoid oscillating levels?
0.1 ¼ 1/2

46. 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

47. LEVEL CONTROL

48. Rule: Kc ¸ |u0|/|ymax| =1 (in scaled variables)
SMOOTH CONTROL
Rule: Kc ¸ |u0|/|ymax| =1 (in scaled variables)
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 ( ¼ 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

49. 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)
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)

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

51. 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

52. 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”:

53. 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

54. 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 ()

55. “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]

56. 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

57. Cascade control

58. Tuning of cascade controllers

59. Cascade control serial processG1 u2 y1 K1 ys G2 K2 y2
y2s

60. Cascade control serial processG1 u y1 K1 ys G2 K2 y2
y2s
Without cascade
With cascade

61. 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

62. 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?

63. 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|

64. CONTROLLABILITY
Controllability

65. Example: Distillation column
CONTROLLABILITY
Example: Distillation column

66. Example: Distillation column
CONTROLLABILITY
Example: Distillation column

67. 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