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PID Tuning using the SIMC rules
Sigurd SkogestadNTNU, Trondheim, Norway
September 2008
1. Model2. SIMC-tunings3. Tight control4. Smooth control5. Level control6. Discussion points
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Operation: Decision and control layers
cs = y1s
MPC
PID
y2s
RTO
u (valves)
CV=y1; MV=y2s
CV=y2; MV=u
Min J (economics); MV=y1s
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Stepwise procedure plantwide control
I. TOP-DOWNStep 1. DEGREES OF FREEDOMStep 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?
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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!
e
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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 (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 ≥
Let’s start with the CONCLUSION
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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
MODEL
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1. Step response experiment
Make step change in one u (MV) at a time Record the output (s) y (CV)
MODEL, Approach 1A
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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(∞)/ u k’ : slope after response “takes off” = k/1
MODEL, Approach 1A
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STEP IN INPUT u
RESULTING OUTPUT y
: Delay - Time where output does not change1: Time constant - Additional time to reach
63% of final changek = y(∞)/ u : Steady-state gain
Δy(∞)
Δu
MODEL, Approach 1A
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Step response integrating process
Δy
Δt
MODEL, Approach 1A
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Model from closed-loop response with P-controller
Kc0=1.5Δys=1
Δyu=0.54
Δyp=0.79
tp=4.4
dyinf = 0.45*(dyp + dyu)Mo =(dyp -dyinf)/dyinfb=dyinf/dys
A = 1.152*Mo^2 - 1.607*Mo + 1.0r = 2*A*abs(b/(1-b))k = (1/Kc0) * abs(b/(1-b))theta = tp*[0.309 + 0.209*exp(-0.61*r)]tau = theta*r
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)
Δy∞
MODEL, Approach 1B
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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
MODEL, Approach 2
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MODEL, Approach 2
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Example 1
Half rule
MODEL, Approach 2
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original
1st-order+delay
MODEL, Approach 2
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half rule
2
MODEL, Approach 2
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original
1st-order+delay
2nd-order+delay
MODEL, Approach 2
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Original 7th order, g0
1st-order, half rule, g1
2nd-order, half rule, g2
1st-order, closed loop, gcl
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PID
PID
PIhalf-rule
PIhalf-rule
PIcl
PIcl
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Approximation of zeros
Correction: More generally replace θ by τc in the above rules
MODEL, Approach 2
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)
cc
c
c c
c
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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:
SIMC-tunings
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Basis: Direct synthesis (IMC)
Closed-loop response to setpoint change
Idea: Specify desired response:
and from this get the controller. ……. Algebra:
SIMC-tunings
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SIMC-tunings
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IMC Tuning = Direct Synthesis Algebra:
SIMC-tunings
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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
dyu
SIMC-tunings
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Integral Time
I = 1
Reduce I to this value:I = 4 (c+) = 8
SIMC-tunings
Setpoint change at t=0 Input disturbance at t=20
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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.
SIMC-tunings
Details: See www.nt.ntnu.no/users/skoge/publications/2003/tuningPID Remark 13 II
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Conclusion: SIMC-PID Tuning Rules
One tuning parameter: c
SIMC-tunings
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Some insights from tuning rules
1. 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)
4. Parameter variations: For robustness tune at operating point with maximum value of k’ θ = (k/τ1)θ
SIMC-tunings
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Some special cases
One tuning parameter: c
SIMC-tunings
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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.
SIMC-tunings
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Note: Derivative action is commonly used for temperature control loops. Select D equal to 2 = time constant of temperature sensor
SIMC-tunings
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SIMC-tunings
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SIMC-tunings
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Selection of tuning parameter c
Two 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:
SIMC-tunings
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TIGHT CONTROL
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Typical closed-loop SIMC responses with the choice c=
TIGHT CONTROL
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Example. Integrating process with delay=1. G(s) = e-s/s.Model: k’=1, =1, 1=∞SIMC-tunings with c with ==1:
IMC has I=∞
Ziegler-Nichols is usually a bit aggressive
Setpoint change at t=0c Input disturbance at t=20
TIGHT CONTROL
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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
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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?
S. Skogestad, ``Tuning for smooth PID control with acceptable disturbance rejection'', Ind.Eng.Chem.Res, 45 (23), 7817-7822 (2006).
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Closed-loop disturbance rejectiond0
ymax
-d0
-ymax
SMOOTH CONTROL
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10−2
10−1
100
101
102
100
101
102
103
Frequency
|d0|/|y
max|
|cPI
|
|cPID
|
Kcu
Minimum controller gain for PI-and PID-control:min |c(j)| = Kc
SMOOTH CONTROL
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Rule: Min. controller gain for acceptable disturbance rejection:
Kc ≥ |u0|/|ymax|often ~1 (in span-scaled variables)
SMOOTH CONTROL
|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
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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 ( ≈ 0). For example, flow control
Then I = 1 is small so integral action is “large”
SMOOTH CONTROL
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Summary: Tuning of easy loops
Easy loops: Small effective delay ( ≈ 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(4c, 1)
Note: Often want a tight pressure control loop (so may have Kc=10 or larger)
SMOOTH CONTROL
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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.
SMOOTH CONTROL LEVEL CONTROL
If you insist on integral actionthen this value avoids cycling
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 timeprovided tank is nominally half full and q0 is equal to the nominal flow.
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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
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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 ≈ 1/2
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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
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LEVEL CONTROL
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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
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Quiz: SIMC PI-tunings
(a) The Figure shows the response (y) from a test where we made a step change in theinput (Δu = 0.1) at t=0. Suggest PI-tunings for (1) τc=2,. (2) τc=10.
t [s]
y
0 10 20 30 40 50 600.1
0.105
0.11
0.115
0.12
0.125
0.13
Step response
y
Time t
SIMC-tunings
(b) Do the same, given that the actual plant is
QUIZ
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Solution
Actual plant:
QUIZ
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Approximation of step response
Approximation ”bye eye”
QUIZ