Alternative form with detuning factor F

Quiz: SIMC PI-tunings

(a) 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

SIMC-tunings

(b) Do the same, given that the actual plant is

QUIZ

Solution

Actual plant:

QUIZ

Approximation of step response

Approximation ”bye eye”

QUIZ

OUTPUT y INPUT uTunings from Step response “by eye” modelSetpoint 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

SIMC-tunings

Half-rule approach: Approximation of zeros depends on tauc!

QUIZ

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).

Cascade control

Tuning of cascade controllers

See next slide

Cascade control

• 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 ( G1’ = G1 G21 K2 (I+K2G22)-1 )

experimentally after closing each loop

• One exception: Serial process, G21 = G22 2– Inner (secondary-2) loop may be modelled with gain=1 and effective

delay=(c+

Special case: Serial cascade

K2 is designed based on G2 (which has effective delay 2)

then y2 = T2 r2 + S2 d2 where S2 ¼ 0 and T2 ¼1 · e-(2+c2)s

T2: gain = 1 and effective delay = 2+c2

NOTE: If delay is in meas. of y2 (and not in G2) then T2 ¼ 1 ·e-c2s

SIMC-rule: c2 ≥ 2

Time scale separation: c2 ≤ c1/5 (approximately)

K1 is designed based on G1’ = G1T2

same as G1 but with an additional delay 2+c2

y2 = T2 r2 + S2d2, T2 = G2K2(I+G2K2)-1

Cascade control

Example: Cascade control serial process d=6

Without cascade

With cascade

G1u

y1

K1

ys G2K2

y2y2s

Use SIMC-rules!

Cascade control

Tuning cascade controlCascade control

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

Cascade control

py

pt

y sy

t0t

uy

sy

Procedure:• Switch to P-only mode and make

setpoint change• Adjust controller gain to get overshoot

about 0.30 (30%)

Record “key parameters”:1. Controller gain Kc0

2. Overshoot = (Δyp-Δy∞)/Δy∞

3. Time to reach peak (overshoot), tp

4. Steady state change, b = Δy∞/Δys. Estimate of Δy∞ without waiting to settle:

Δy∞ = 0.45(Δyp + Δyu) Advantages compared to Ziegler-Nichols:* Not at limit to instability * Works on a simple second-order process.

Alternative closed-loop approach:Setpoint overshoot method

Closed-loop step setpoint response with P-only control.

Setpoint overshoot method

M. Shamsuzzoha and S. Skogestad, ``The setpoint overshoot method: A simple and fast method for closed-loop PID tuning'',Journal of Process Control, 20, xxx-xxx (2010)

Choice of detuning factor F: F=1. Good tradeoff between “fast and robust” (SIMC with τc=θ) F>1: Smoother control with more robustness F<1 to speed up the closed-loop response.

pI pτ =min 0.86A ,b

t t1-b

2.44

F

From P-control setpoint experiment record “key parameters”:1. Controller gain Kc0

2. Overshoot = (Δyp-Δy∞)/Δy∞

3. Time to reach peak (overshoot), tp

4. Steady state change, b = Δy∞/Δys

Summary setpoint overshoot method

0c cK = K A F2A= 1.152( ) - 1.6overshoot overs07( ) + 1.0hoot

Proposed PI settings (including detuning factor F)

Setpoint overshoot method

1

1 0.2 1 0.04 1 0.008 1g

s s s s

0 5 10 15 200

0.25

0.5

0.75

1

1.25

time

OU

TP

UT

y

Proposed method with F=1 (overshoot=0.104)Proposed method with F=1 (overshoot=0.292)Proposed method with F=1 (overshoot=0.598)SIMC (

c=

effective=0.148)

Example: High-order process

P-setpoint experiments

Closed-loop PI response

Setpoint overshoot method

5 1

seg

s

First-order unstable process

0 20 40 60 800

0.5

1

1.5

2

time

OU

TP

UT

y

Proposed method with F=1 (overshoot=0.10)Proposed method with F=1 (overshoot=0.30)Proposed method with F=1 (overshoot=0.607)

Example: Unstable plant

• No SIMC settings available

Closed-loop PI response

Setpoint overshoot method

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

CONTROLLABILITY