PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway.
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Transcript of PID Tuning using the SIMC rules Sigurd Skogestad NTNU, Trondheim, Norway.
PID Tuning using the SIMC rules
Sigurd SkogestadNTNU, Trondheim, Norway
PID controller
Time domain
Laplace domain
Only two parameters (Kc and τI)… but surprisingly difficult to tune without systematic
approach
e
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 responseInitial 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 ¸ - : desired closed-loop response time (tuning parameter)•For robustness select: c ¸
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
Step response experiment Make step change in one u (MV) at a time Record the output (s) y (CV)
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 change1: Time constant - Additional time to reach 63% of final changek : steady-state gain = y(1)/ u k’ : slope after response “takes off” = k/1
Step response integrating process
Δy
Δt
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
Example 1
Half rule
half rule
half rule
2
original
1st-order+delay
2nd-order+delay
Approximation of zeros
Correction: More generally replace θ by τc in the above rules
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:
IMC Tuning = Direct Synthesis Algebra:
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 responseTry reducing it!
gc
d
yu
Integral TimeI = 1
Reduce I to this value:I = 4 (c+) = 8
Integral time
Want to reduce the integral time for “integrating” processes, but to avoid “slow oscillations” we must require:
Derivation:
Conclusion: SIMC-PID Tuning Rules
One tuning parameter: c
Some insights from tuning rules1. 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)
2. Use (close to) P-control for integrating process Beware of large I-action (small τI) for level control
3. Use (close to) I-control for fast process (with small time
constant τ1)
Some special cases
One tuning parameter: c
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.
Note: Derivative action is commonly used for temperature control loops. Select D equal to 2 = time constant of temperature sensor
Quiz: SIMC PI-tunings
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
t [s]
y
Step response
y
Time t
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);
OUTPUT y INPUT uTunings from Step responseSetpoint change at t=0, input disturbance = 0.1 at t=50
Tunings from Half rule (Somewhat better)
Kc=9.5, tauI=10
Kc=2.9, tauI=10
Kc=2, tauI=5.5
Kc=6, tauI=5.5
Selection of tuning parameter cTwo main cases1. TIGHT CONTROL: Want “fastest possible
control” subject to having good robustness• Want tight control of active constraints (“squeeze and shift”)
2. 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:
TIGHT CONTROL
Typical closed-loop SIMC responses with the choice c=
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
TIGHT CONTROL
1. Approximate as first-order model with k=1, 1 = 1+0.1=1.1, =0.1+0.04+0.008 = 0.148Get 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 = 0.028Get 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
TIGHT CONTROL
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 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?
Closed-loop disturbance rejection d0
ymax
-d0
-ymax
SMOOTH CONTROL
Kc
u
Minimum controller gain for PI-and PID-control:Kc ¸ Kc,min = |u0|/|ymax|
|u0|: Input magnitude required for disturbance rejection|ymax|: Allowed output deviation
SMOOTH CONTROL
Minimum controller gain:
Industrial 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)
Example
Does not quite reach 1 because d isstep disturbance (not not sinusoid)
Response to step disturbance = 1 at input
c is much larger than =0.25
SMOOTH CONTROL
Application of smooth control Averaging level control
VqLC
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 timeprovided tank is nominally half full and q0 is equal to the nominal flow.
SMOOTH CONTROL LEVEL CONTROL
If you insist on integral actionthen this value avoids cycling
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.
LEVEL CONTROL
Integrating process: Level control
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’ s2 + Kc I s + Kc)• The denominator can be rewritten on standard form
– (02 s2 + 2 0 s + 1) with 0
2 = I/k’¢Kc and 2 0 = I
– Algebra gives:
– To avoid oscillations we must require 1, or Kc¢k’ ¢I > 4• The controller gain Kc must be large to avoid oscillations!
VqLC
VqLC
qLC
This is the basis for the SIMC-rule for the minimumintegral time
How avoid oscillating levels?
LEVEL CONTROL
• 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 ¼ 1/2
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
LEVEL CONTROL
LEVEL 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
SMOOTH CONTROL
Kc: Assume variables are scaled with respect to their span
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(4c, 1) Note: Often want a tight pressure control loop (so may have
Kc=10 or larger)
SMOOTH CONTROL
Conclusion PID tuningSIMC tuning rules
1. Tight control: Select c= corresponding to
2. Smooth control. Select Kc ¸
Note: Having selected Kc (or c), the integral time I should be selected as given above
3. Derivative time: Only for dominant second-order processes
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
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”:
A little more on obtaining the model from step response experiments “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
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
¼ 1(may be neglected for c > 5)
1 ¼ 200(may be neglected for c < 40)
time
c = desired response time
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: = min (, 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 ()
“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
2.8Response on stage 70 to step in L
7.5 min
2.62-2.19
Example.
Step change in u: u = 0.1Initial value for y: y(0) = 2.19Observed delay: = 2.5 minAt T=10 min: y(T)=2.62Initial slope:
y(t)
t [min]
Example (from quiz)
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!)
tauc=2
tauc=10
OUTPUT y INPUT y
Step responseΔu=0.1
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).
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 “primary” controller (1) is setpoint (SP) for “secondary” controller (2)
• Tuning simple: Start with inner secondary loops (fast) and move upwards
• Must usually identify new model experimentally after closing each loop.
• 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+
Cascade control serial process
d=6
G1u2
y1
K1
ys G2K2
y2y2s
Cascade control serial process d=6
Without cascade
With cascade
G1u
y1
K1
ys G2K2
y2y2s
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
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?
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
)
Controllability
initial effect of “input” disturbance
max. output deviation
y reaches k’ ¢ |d0|¢ t after time ty reaches ymax after t= |ymax|/ k’ ¢ |d0|
CONTROLLABILITY
ControllabilityCONTROLLABILITY
• More general disturbances. Requirement becomes (for c=):
• Conclusion: The main factors limiting controllability are – large effective delay from u to y ( large)– large disturbances (k’d |d0| / ymax large)
• Can generalize using “frequency domain”: |Gd(j¢0.5/max)| ¢|d0| = |ymax|
Following step disturbance d0:Time it takes for output yto reach max. deviation
Example: Distillation column
CONTROLLABILITY
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/k’d |d| = 3 min
Controllability: Must close a loop with time constant (c) faster than 1.5 min toavoid that bottom composition xB exceeds max. deviationIf 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 feedbackloop, max = 3/2 = 1.5 min
Example: Distillation column
CONTROLLABILITY
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
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