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PID Tuning and Controllability Sigurd Skogestad NTNU, Trondheim, Norway

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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 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).``Tuning for smooth PID control with acceptable disturbance rejection'', (*) Probably the best simple PID tuning rules in the world k = k/ τ 1 = initial slope step response c 0: desired closed-loop response time (tuning parameter) For robustness select (1) : c ¸ (gives K c K c,max ) For acceptable disturbance rejection select (2) : K c K c,min = u d0 /y max

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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

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

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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 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

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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

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half rule

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Approximation of zeros

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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 it becomes simplest with the series (cascade) PID-form:

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Basis: Direct synthesis (IMC) Closed-loop response to setpoint change Idea: Specify desired response T = (y/y s ) desired and from this get the controller. Algebra:

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IMC Tuning = Direct Synthesis

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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 response Try reducing it! gc d y u

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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)

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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/( 1 s) = k/s PI-control: C(s) = K c (1 + 1/ I s) Poles (and oscillations) are given by roots of closed-loop polynomial 1+GC = 1 + k/s ¢ K c ¢ (1+1/ I s) = 0 or I s 2 + k K c I s + k K c = 0 Can be written on standard form ( 0 2 s s + 1) with To avoid oscillations must require | | ¸ 1: K c ¢ k ¢ I ¸ 4 or I ¸ 4 / (K c k) With choice K c = (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

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Summary: SIMC-PID Tuning Rules One tuning parameter: c

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Some special cases One tuning parameter: c

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

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Selection of tuning parameter c Two cases 1. 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, , 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).``Tuning for smooth PID control with acceptable disturbance rejection'',

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(1) TIGHT CONTROL

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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=0Input 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, = = Get SIMC PI-tunings ( c = ): K c = 1 ¢ 1.1/(2¢ 0.148) = 3.71, I =min(1.1,8¢ 0.148) = Approximate as second-order model with k=1, 1 = 1, 2 = =0.22, = = Get SIMC PID-tunings ( c = ): K c = 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 (2) Will derive K c,min. From this we can get c,max using SIMC tuning rule (2) 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 K c ?

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Closed-loop disturbance rejection d0d0 y max -d 0 -y max SMOOTH CONTROL

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KcKc u Minimum controller gain for PI-and PID-control: K c ¸ K c,min = |u d0 |/|y max | |u d0 |: Input magnitude required for disturbance rejection |y max |: Allowed output deviation SMOOTH CONTROL

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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 K c =1 is reasonable ! SMOOTH CONTROL (span)

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Exampl e Does not quite reach 1 because d is step disturbance (not not sinusoid) Response to disturbance = 1 at input c is much larger than =0.25 SMOOTH CONTROL

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Application of smooth control Averaging level control V q LC 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 ¸ K c,min = |u d0 |/|y max | Where in our case |u d0 | = | q 0 | - expected flow change [m3/s] (input disturbance) |y max | = | V max | - largest allowed variation in level [m3] Integral time: From the material balance (dV/dt = q – q out ), the model is g(s)=k/s with k=1. Select K c =K c,min = | q 0 | | V max |. SIMC-Integral time for integrating process: I = 4 / (k K c ) = 4 | V max | / | q 0 | = 4 ¢ residence time provided tank is nominally half full (so | V max | = V) and q 0 is equal to the nominal flow. SMOOTH CONTROLLEVEL CONTROL If you insist on integral action then this value avoids cycling

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Rule: K c ¸ |u d0 |/|y max | ( ~1 in scaled variables) Exception to rule: Can have even smaller K c (< 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 SMOOTH CONTROL K c : Assume variables are scaled with respect to their span

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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: K c =0.5, I = 1 = time constant valve (typically, 10s to 30s) Level control: K c =2 (and no integral action) Other easy loops (e.g. pressure control): K c = 2, I = min(4 c, 1 ) Note: Often want a tight pressure control loop (so may have K c =10 or larger) SMOOTH CONTROL K c : Assume variables are scaled with respect to their span

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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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Integrating process: Level control LEVEL CONTROL

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How avoid oscillating levels? LEVEL CONTROL Actually, 0.1 = 1/π 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 Sigurds rule solved the problem LEVEL CONTROL

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Identifying the model

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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: 1. Initial response: ( + 1 ) is where initial slope crosses final steady- state 2. ( + 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 xBxB 1 =16 min / 19 min 63% k k

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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 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 > 5θ = 5: Delay-free (neglect θ) 1 ¼ 200, 1 time c = desired response time c c < τ 1/ 5 = 40: Integrating (neglect τ 1)

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Integrating process ( c < 1 /5): Need only two parameters: k and From step response: Response on stage 70 to step in L 7.5 min 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: y(t) t [min]

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Delay-free process ( c > 5 ) Example: First-order model: = 0 k= ( )/(-0.5) = (steady-state gain) 1 = 16 min (63% of change) BOTTOM V: -0.5 xBxB 16 min 63%

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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 ( )

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Conclusion PID tuning

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Cascade control

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Tuning of cascade controllers

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Cascade control serial process (exception) d=6 G1G1 u2u2 y1y1 K1K1 ysys G2G2 K2K2 y2y2 y 2s

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Cascade control serial process d=6 Without cascade With cascade G1G1 u y1y1 K1K1 ysys G2G2 K2K2 y2y2 y 2s

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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

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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

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Recall SIMC tuning rules 1. Tight control: Select c = corresponding to 2. Smooth control. Select K c ¸ Must require K c,max > K c.min for controllability ) Controllability initial effect of input disturbance max. output deviation y reaches k ¢ |d 0 |¢ t after time t y reaches y max after t= |y max |/ k ¢ |d 0 | CONTROLLABILITY

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Controllability CONTROLLABILITY

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Example: Distillation column CONTROLLABILITY

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Example: Distillation column CONTROLLABILITY

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Distillation example: Frequency domain analysis CONTROLLABILITY

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Other factors limiting controllability 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

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