Input-output Controllability Analysis

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Input-output Controllability Analysis Idea: Find out how well the process can be controlled - without having to design a specific controller kogestad, ``A procedure for SISO controllability analysis - with application to design of pH neutrali , 20, 373-386, 1996.

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Input-output Controllability Analysis. Idea: Find out how well the process can be controlled - without having to design a specific controller. Reference: S . Skogestad, ``A procedure for SISO controllability analysis - with application to design of pH neutralization processes'', - PowerPoint PPT Presentation

Transcript of Input-output Controllability Analysis

Page 1: Input-output Controllability Analysis

Input-output Controllability Analysis

Idea: Find out how well the process can be controlled - without having to

design a specific controller

Reference: S. Skogestad, ``A procedure for SISO controllability analysis - with application to design of pH neutralization processes'', Comp.Chem.Engng., 20, 373-386, 1996.

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• Two main rules.• Rule 1: speed of response– Fast response required to reject large disturbance– BUT: Response time is limited by effective time

delay• Rule 2: Input constraints– Also: Large disturbance rejection may give input

saturation

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Ideal controller inverts the plant• y = g(s)u + d

• Ideal controller inverts the plant g(s):– Think feedforward, u = cff(s) (ys-d)

– Perfect control: want y=ys ) cff = 1/g(s) = g-1

Limitations on perfect control: Inverse cannot always be realized:1. Input saturation , |u| > |umax|2. Time delay, g=e-µs .

• g-1= eµs = prediction (not possible)• Solution: Omit

3. Inverse response, g = -Ts+1. • g-1 = 1/(-Ts+1) = unstable (not possible as u will be unbounded)• Solution: Omit

4. More poles than zeros, g = 1/(¿ s+1), • g-1 = ¿ s + 1 = pure differentiation (not possible as u will be unbounded).• Solution: Replace by: (¿ s + 1)/(¿c+1) where ¿c< ¿ is a tuning parameter

– Example. g(s) = 5 (-0.5s+1) e-2s / (3s+1).• Realizable inverse (feedforward): 0.2 (3s+1)/(¿

c+1). E.g. choose ¿

c=0.5

• So we know what limits us from having perfect control– Same limitations apply to feedback control

• Controllability analysis: Want to find out what these limitations imply in terms of “acceptable control”, |y-ys|· ymax

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WANT TO QUANTIFY!

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Scaling

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Note: !B=!c=1/¿c

SCALED MODEL

!d

MAIN REASON FOR CONTROL: DISTURBANCES!

L=gcGd !c

Need control up to frequency !d where |Gd|=1 -> Need !c > !d(!c is frequency where |L|=1)Proof: y = Gd/(1+L)d, where L = GC (“llop”)Worst-case is |d|=1. Want |y| <1, so want

|Gd|<|1+L| ¼ |L| (approximation holds at low frequencies where |L| is large). Conclusion: Need |L|>|Gd| at frequencies where |Gd|>1

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SCALED MODEL

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Rules for speed of response (assuming control with integral action)

• Define !c=1/¿c = closed-loop bandwidth = where |gc| is 1• Rule 1a: Need !c >!d (¿c < 1/!d)

– Where !d is frequency where |g’d|=1.

– Rule 1 is for typical case where |g’d| is highest at low frequencies

• Rule 1b: Need !c < 1/µ (¿c > µ) – Where µ is effective time delay

• Rule 1c: Need !c > p (¿c < 1/p)– Where p is unstable pole, g(s) =k/(s-p)…

This order is OK according to rule 1:1/µ

!!dp

!c must be in this range

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10-3

10-2

10-1

100

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10-2

10-1

100

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s=tf('s')g = 500/((50*s+1)*(10*s+1))gd = 9/(10*s+1)w = logspace(-3,1,1000);[mag,phase]=bode(g,w);[magd,phased]=bode(gd,w);loglog(w,mag(:),'blue',w,magd(:),'red',w,1,'black'), grid on

SCALED MODELEXAMPLE

!d=0.9

|G|

|Gd|

PI-control: 1/µ = 1/5 = 0.2PID-control: 1/µ = 1/0 = 1

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CHECK CONTROLLABILITY ANALYSIS WITH SIMULATIONS

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PI control not acceptable*

0 50 100 150 200 250 300 350 400 450 5000

1

2

3

4

5

6

§ 1

y

s=tf('s')g = 500/((50*s+1)*(10*s+1))gd = 9/(10*s+1)% SIMC-PI with tauc=theta=5Kc=(1/500)*(55/(5+5)); taui=55; taud=0;

SCALED MODEL

*As expected since need !c > !d= 0.9, but can only achieve !c<1/µ = 1/5 = 0.2

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PID control acceptable: y and u are § 1

0 50 100 150 200 250 300 350 400 450 500-1

-0.5

0

0.5

1

1.5

2

y

u

g = 500/((50*s+1)*(10*s+1))gd = 9/(10*s+1)%SIMC-PID (cascade form) with tauc=wd=1:Kc=(1/500)*(50/(1+0)); taui=50; taud=10;

SCALED MODEL

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If process is not controllable: Need to change the design

• For example, dampen disturbance by adding buffer tank: Level control unimportant,

but need good mixing

Level control is NOT tight-> level varies

Integral action is not recommended for averaging level control

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Problem 1SCALED MODEL

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Problem 2SCALED MODEL

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Problem 3-

SCALED MODEL

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Problem 4SCALED MODEL

g = 200/((20*s+1)*(10*s+1)*(s+1))gd = 4/((3*s+1)*(s+1)^3)Kc=(1/200)*20/1,taui=20,taud=10.5

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Problem 5-

SCALED MODEL

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Problem 6pH-Neutralization.y = cH+ - cOH- (want=0§10-6 mol/l,pH=7§1)u = qbase (cOH-=10mol/l, pH=15)d = qacid (cH+ =10mol/l, pH=-1)

Using tanks in series, Acid and base in tank 1.

Scaled model: kd = 2.5e6Each tank: ¿ = 1000sControl: µ = 10s (meas. delay for pH)

Problem: How many tanks?

! [rad/s]

SCALED MODEL

n=2n=3

n=1

|Gd|

Reference for more applications of controllability analysis: Chapter 5 in book by Skogestad and Postlethwaite (2005)

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Control system• 3 tanks: Neutralization (base addition) only in tank 1 gives large effective

delay (>> 10s) because of tank dynamics in g(s)• Suggested solution is to add (a little) base also in the other tanks:

pH 5

pH 2