1. Practical plantwide process control: PID tuning
Sigurd Skogestad, NTNU

2. Part 4: PID tuning
PID controller tuning: It pays off to be systematic!
1. Obtaining first-order plus delay models
Open-loop step response
From detailed model (half rule)
From closed-loop setpoint response
2 . Derivation SIMC PID tuning rules
Controller gain, Integral time, derivative time
3. Special topics
Integrating processes (level control)
Other special processes and examples
When do we need derivative action?
Near-optimality of SIMC PID tuning rules
Non PID-control: Is there an advantage in using Smith Predictor? (No)
Feedforward control and Two degrees-of-freedom control 

3. 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)

4. PID controller Time domain (“ideal” PID)
Laplace domain (“ideal”/”parallel” form)
For our purposes. Simpler with cascade form
Usually τD=0. Then the two forms are identical.
Only two parameters left (Kc and τI)
How difficult can it be to tune???
Surprisingly difficult without systematic approach!

5. Disadvantages Ziegler-Nichols: Aggressive settings No tuning parameter
Trans. ASME, 64, (Nov. 1942).
Comment:
Similar to SIMC for integrating process with ¿c=0 (aggressive!):
Kc = 1/k’ 1/µ
¿I = 4 µ
Disadvantages Ziegler-Nichols:
Aggressive settings
No tuning parameter
Poor for processes with large time delay (µ)

6. Disadvantage IMC-PID (=Lambda tuning):
Many rules
Poor disturbance response for «slow» processes (with large ¿1/µ)

7. Motivation for developing SIMC PID tuning rules
The tuning rules should be well motivated, and preferably be model-based and analytically derived.
They should be simple and easy to memorize.
They should work well on a wide range of processes.

8. SIMC PI tuning rule
Approximate process as first-order with delay (e.g., use “half rule”)
k = process gain
¿1 = process time constant
µ = process delay
Derive SIMC tuning rule*:
Open-loop step response
c ¸ - : Desired closed-loop response time (tuning parameter)
Integral time rule combines well-known rules:
IMC (Lamda-tuning): Same as SIMC for small ¿1 (¿I = ¿1)
Ziegler-Nichols: Similar to SIMC for large ¿1 (if we choose ¿c= 0; aggressive!)
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”

9. 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
Recommend: Use second-order model only if ¿2>µ

10. 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)

11. First-order plus delay process
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 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

12. 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

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

14. MODEL, Approach 1B
Shams’ method: Closed-loop setpoint response with P-controller with about 20-40% overshoot
Kc0=1.5
Δys=1
Δy∞
OBTAIN DATA IN RED (first overshoot
and undershoot), and then:
tp=4.4, dyp=0.79; dyu=0.54, Kc0=1.5, dys=1
dyinf = 0.45*(dyp + dyu)
Mo =(dyp -dyinf)/dyinf % Mo=overshoot (about 0.3)
b=dyinf/dys
A = 1.152*Mo^ *Mo + 1.0
r = 2*A*abs(b/(1-b))
%2. OBTAIN FIRST-ORDER MODEL:
k = (1/Kc0) * abs(b/(1-b))
theta = tp*[ *exp(-0.61*r)]
tau = theta*r
3. CAN THEN USE SIMC PI-rule
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 PID-book 2012)

15. 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 

16. MODEL, Approach 2

17. MODEL, Approach 2
Example 1
Half rule

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

19. MODEL, Approach 2 2
half rule

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

21. Original 7th order, g0
1st-order, half rule, g1
2nd-order, half rule, g2
1st-order, closed loop, gcl

22. 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
Alternative and improved method for approximating zeros:
Simple Analytic PID Controller Tuning Rules Revisited
J Lee, W Cho, TF Edgar - Industrial & Engineering Chemistry Research 2014, 53 (13), pp 5038–5047

23. 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:

24. 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:

25. SIMC-tunings
NOTE: Setting the steady-state gain = 1 in T will result in integral action in the controller!

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

27. SIMC-tunings
Integral time
Found: Integral time = dominant time constant (I = 1) (IMC-rule)
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

28. 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

29. 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

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

31. 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

32. Cascade PID -> Ideal PID

33. SIMC-tunings

34. 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:

35. TIGHT CONTROL

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

37. 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

38. TIGHT CONTROL Approximate as first-order model:
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
Approximate as second-order model:
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

39. 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).

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

41. u Kc SMOOTH CONTROL Minimum controller gain for PI-and PID-control:
min |c(j)| = Kc

42. SMOOTH CONTROL
Rule: Min. controller gain for acceptable disturbance rejection: Kc ¸ |ud|/|ymax| often ~1 (in span-scaled variables)
|ymax| = allowed deviation for output (CV)
|ud| = required change in input (MV) for disturbance rejection (steady state)
= observed change (movement) in input from historical data

43. SMOOTH CONTROL
Rule: Kc ¸ |ud|/|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”

44. 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)

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

46. PID: More (Special topics)
Integrating processes (level control)
Other special processes and examples
When do we need derivative action?
Near-optimality of SIMC PID tuning rules
Non PID-control: Is there an advantage in using Smith Predictor? (No)
April 4-8, 2004
KFUPM-Distillation Control Course

47. 1. Application of smooth control
LEVEL CONTROL
1. 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.

48. 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.

49. Level control: Can have both fast and slow oscillations
Slow oscillations (Kc too low): P > 3¿I
Fast oscillations (Kc too high): P < 3¿I
Here: Consider the less common slow oscillations

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

51. 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

52. LEVEL CONTROL

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

54. 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.

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

56. Pure time delay process: Minor improvement by adding D-action
Optimal PI
= I-control
θ=1
Time delay process: Setpoint and disturbance responses same + input response same

57. Pure time delay process

58. ) Two alternative “Improved SIMC”-rules
Alt. 1. Improved PI-rule (iSIMC-PI): Add θ/3 to 1
Alt. 2. Improved PID-rule (iSIMC-PID): Add θ/3 to D
iSIMC-PI and iSIMC-PID are identical for pure delay process (¿1=0)
BUT in general: iSIMC-PID is better (1 > θ/3)

59. Example: Integrating process

60. 4. Optimality of SIMC rules
How good are the SIMC-rules compared to optimal PI/PID?
High controller gain (“tight control”)
Low controller gain (“smooth control”)
Multiobjective. Tradeoff between
Output performance
Robustness
Input usage
Noise sensitivity
Our choice:
Quantification
Output performance:
Rise time, overshoot, settling time
IAE or ISE for setpoint/disturbance
Robustness: Ms, Mt, GM, PM, Delay margin, …
Input usage: ||KSGd||, TV(u) for step response
Noise sensitivity: ||KS||, etc.
J = avg. IAE for
setpoint/disturbance
Ms = peak sensitivity

61. Performance (J):

62. Robustness (Ms):

63. JIAE vs. Ms for optimal PI/PID (-) and SIMC (¢¢) for 4 processes
CONCLUSION: SIMC almost «Pareto-optimal»
Note: “PID” weightings used for JIAE

64. 5. Better with IMC, Smith Predictor or MPC?
Suprisingly, the answer is: NO
IMC and SP are actually worse than well-tuned PID
Poor for integrating process
Poor time delay margin
SP = Smith Predictor

65. IMC controller (tuned for setpoints)
The Smith Predictor
Where K is a “normal” PID controller
IMC controller (tuned for setpoints)
= Special case of Smith Predictor where K is a PI controller with the parameters
¿1 > 0
¿1 = 0
Kc = ¿1/(k ¿c)
¿I = ¿1
Kc =0
Ki = Kc/¿I = 1/¿c
(I-controller)

66. Comparison of J vs. Ms for optimal and SIMC for 4 processes
IMC and SP:
Gets worse
as process
becomes
integrating
CONCLUSION: i-SIMC is generally better than IMC & SP!

67. Step responses with delay error
time
time
time
Smith Predictor and IMC: Sensitive to both positive and negative time delay error

68. Reason: SP & IMC can have multiple GM, PM, DM

69. CONCLUSION COMPARISON WITH SP & IMC
Well-tuned PI or PID (using i-SIMC rules) is better than Smith Predictor or IMC!!
Especially for integrating processes

70. 6. Feedforward control Model: y = g u + gd d
Measurement dm
cff
gdm
Model: y = g u + gd d
Measured disturbance: dm = gdm d
Feedforward controller: u = cFF dm
Get y = (g cFF gdm + gd) d
Ideal feedforward:
y = 0 ) cFF,ideal = - g-1 gd gdm-1 = - (gd / (gdm g)
In practice: cFF(s) must be realizable
Order pole polynomial ¸ order zero polynomial
No prediction allowed (µ can not be negative)
And must avoid that CFF has too high gain to avoid (to avoid aggressive input changes)
Common simplification: cFF = k (static gain)
General. Approximate cFF,ideal as :

71. When use feedforward? When feedback alone is not good
For example, if the measurement of y has a long delay
When the disturbance response (gd) is “slow” compared to the input response (g)
cFF,ideal = - gd / (gdm g) is then realizable, and the feedforward has “more time” to take the right action
For example, if gd has a larger delay than g (so that cFF,ideal has a delay) or if gd has a larger time constant than g

72. Example d2 d1 d1 d2

73. Example
«Chicken factor»

74. Feedforward control (measure d1 & d2)
d1: cff = -(20*s+1)/(2*s+1)
d2: cff2 = -1 (not shown) d1 d2

75. Simulation feedforward
100
200
300
400
500
600
0.5 1
1.5 2
2.5 3
3.5 4
FEEDBACK ONLY y1
FEEDFORWARD ADDED FOR d1 and d2
Setpoint
change d2 d1

76. Main problem feedforward: Need good models “If process gain increases by more than a factor 2, then ideal feedforward control is worse than no control”
Proof: y = g u + gd d where u=cFF d
Let cFF = - gd/g
Ideal: Get y = g cFF d + gd d = 0
so term “g cFF d” is equal to “-gd d”
Real: If g has increased by a factor x then
y = x(-gd d) + gd d = (-x+1) gd d
And we have that |-x+1|>1 (worse than no control) for x>2.
Example, x=2.1, gd d=1,
Ideal y = gd d = 1
Real: y = (-2.1+1)*1 = -1.1 (which is greater than 1 in magnitude, so y overshoots y by more than 1 on the other side)

77. 7. Two degrees-of-freedom control
Typically, the feedforward block is A = G--1Fr where G- is the invertible part of G.
A typical choice for the prefilter is Fr = 1/(τrs+1)
Example. G=(-3s+1)exp(-4s)/(7s+1)^2, Fv=1 (perfect measurement) and taur=2. Use “IMC-like” design and write G = G+ G- where G+=exp(-4s)(-3s+1)/(3s+1) and G-=(3s+1)/(7s+1)^2. Gives A = (7s+1)^2/(3s+1)(2s+1)
We want to choose B such that A and K can be designed independently!!
Solution (Lang and Ham ,1955): Choose B=FyGA so that transfer function from r to e is zero (with perfect model) !!
The feedback will then only take action if the feedforward is not working as expected (due to model error).
We must have B(0)=I so that we will have no offset (y=r at steady state) even with model error for G
Example. B(s) = FyGA = FyG+Fr = exp(-4s)(-3s+1)/(3s+1)(2s+1). Note that B(0)=1.
The feedback controller K can be designed for disturbance rejection and robustness, e.g., using SIMC rules.
Example. Approximate as first-order with delay process with theta=4+3+7/2=10.5 and tau1=7+7/2=10.5, and use SIMC! With tauc=theta get Kc=0.5 and taui=10.5.