1. PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway
Model
SIMC-tunings
Tight control
Smooth control
Level control
Discussion points

2. 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)

3. 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?

4. 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!

5. 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 ¸ 

6. 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

7. 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)

8. 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

9. Step response integrating process
MODEL, Approach 1A
Step response integrating process Δy Δt

10. 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; see also new from PID-book 2011)

11. 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 

12. MODEL, Approach 2

13. MODEL, Approach 2
Example 1
Half rule

14. MODEL, Approach 2
original
1st-order+delay

15. MODEL, Approach 2 2
half rule

16. MODEL, Approach 2
original
1st-order+delay
2nd-order+delay

17. 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

18. 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:

19. 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:

20. SIMC-tunings

21. IMC Tuning = Direct Synthesis
SIMC-tunings
IMC Tuning = Direct Synthesis
Algebra:

22. 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

23. 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

24. 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

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

26. 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

27. SIMC-tunings
Some special cases
One tuning parameter: c

28. 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.

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

30. SIMC-tunings

31. SIMC-tunings

32. 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:

33. TIGHT CONTROL

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

35. 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

36. 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

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

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:
min |c(j)| = Kc

40. SMOOTH CONTROL
Rule: Min. controller gain for acceptable disturbance rejection: Kc ¸ |u0|/|ymax| often ~1 (in span-scaled variables)
|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

41. 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”

42. 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)

43. 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.

44. 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.

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

49. 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

50. QUIZ
Solution
Actual plant:

51. Approximation of step response
QUIZ
Approximation of step response
Approximation ”bye eye”