1. PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway

2. 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, , 2003
(2) S. Skogestad, ``Tuning for smooth PID control with acceptable disturbance rejection'', Ind.Eng.Chem.Res, 45 (23), (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

3. 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

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

5. 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

6. 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 

9. half rule

10. Approximation of zeros

11. 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:

12. 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:

13. IMC Tuning = Direct Synthesis

14. 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

15. 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)

16. 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

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

18. Some special cases
One tuning parameter: c

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

21. 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, , 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), (2006).

22. (1) TIGHT CONTROL

23. 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

24. 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

25. 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

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

27. 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

28. 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 !

29. 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

30. 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.

31. 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

32. 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

33. 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.

34. Integrating process: Level control

35. How avoid oscillating levels?
LEVEL CONTROL
How avoid oscillating levels?
Actually, 0.1 = 1/π2

36. 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

37. LEVEL CONTROL

38. Identifying the model

39. 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

40. 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)

41. “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)
7.5 min
=2.5
t [min]

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

43. 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 ()

44. Conclusion PID tuning

46. Cascade control

47. Tuning of cascade controllers

48. Cascade control serial process (“exception”)G1 u2 y1 K1 ys G2 K2 y2
y2s

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

50. 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

51. 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?

52. 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|

53. CONTROLLABILITY
Controllability

54. Example: Distillation column
CONTROLLABILITY
Example: Distillation column

55. Example: Distillation column
CONTROLLABILITY
Example: Distillation column

56. Distillation example: Frequency domain analysis
CONTROLLABILITY
Distillation example: Frequency domain analysis

57. 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