Controllability of Processes with Large Gains

Sigurd SkogestadAntonio C. B. de Araújo

NTNU, Trondheim, NorwayAugust, 2004

12th Nordic Process Control Workshop

Introduction

• McAvoy and Braatz (2003)(*) based on a study case with valve stiction:For control purposes the magnitude of steady-state process gain (maximum singular value) should not exceed about 50.

• If correct, it has important implications for design of many processes.

• However, seems intuitively it must be wrong:Condsider control of liquid level: Has infinite steady-state gain due to integrator, but is easily controllable

• The objective of our work is to study this in more detail

(*) T. A. McAvoy and R. D. Braatz, 2003, ”Controllability of processes with large singular values”, Ind. Eng. Chem. Res., 42, 6155-6165.

Claims by McAvoy and Braatz:

1. Systems with very high gain are sensitive. Impossible in practice to get the fine manipulation of the control valves that is required for control because the valves would be limited to move in a very small region

2. Upper limit for σ1(G) should be imposed. Suggest that a reasonable limit is 50 because essentially all control systems are eventually implemented with analogue devices which typically have an accuracy on the order of 0.5%.

3. Valve stiction: May get rid of oscillations by detuning the controller.

Study Case (McAvoy and Braatz, 2003)

• Plant is given by:

σ1(G) = 101.2 and σn(G) = 1.196

• ... and the controller is given by:

11100

011

110s

eG

s

sTK

sTK

K

RC

RC

22

11

110

01

1This controller is first tuned such that KCi=0.154 and TRi=7.7s.

• The block diagram of the system was built using Simulink as follows:

Study Case (McAvoy and Braatz, 2003)

y2sp

y2

y2

y1sp

y1

y1

u2quant

u2quant

u2

u2

u1quant

u1quant

u1

u1

t

To Workspace

Stiction 2

Stiction 1In1

In2

Out1

Out2

Plant

In1

In2

Out1

Out2

Controller

Clock

• Valve stiction induces a ”disturbance” at the input.• Here: Reproduce using Simulink Quantizer which

discretizes the input:

• Smooth signal into a stair-step output: uq = q * round(u/q),

where q is the Quantization interval parameter

• q=0.01: Reproduces results by McAvoy and Braatz (2003).

Study Case (McAvoy and Braatz, 2003)

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8 10-1.5

-1

-0.5

0

0.5

1

1.5

Quantizer

u uq

• First we reproduce the same results as in the paper for the case where a step change of 0.23 is introduced in the setpoint of y1 and KC1 = KC2 = 0.154.

0 10 20 30 40 50 60 70 80-0.2

0

0.2

0.4

0.6

0.8

1

1.2

- With perfect valve

- With valve inaccuracy

y1

y2

Study Case (McAvoy and Braatz, 2003)

• Eliminate oscillations by detuning?

• Simulation with KC1 reduced by a factor 3 to 0.0513:

It looks very nice with no oscilation on y2.

0 10 20 30 40 50 60 70 80-0.2

0

0.2

0.4

0.6

0.8

1

1.2- Original tuning

- Controller for y1 detuned

y1

y2

That is the orignal final time McAvoy and Braatz (2003) choose for this simulation.

Study Case (McAvoy and Braatz, 2003)

• Let’s make the time interval longer......

The oscilations in y2 start after about 95s.

0 20 40 60 80 100 120 140 160 180 200-0.2

0

0.2

0.4

0.6

0.8

1

1.2

- Original tuning

- Controller for y1 detuned

y1

y2

Study Case (McAvoy and Braatz, 2003)

Actually, with integral action oscillations will always appear - it may just takes longer time if the controller gain is reduced

• This is what input 1 is doing:

Study Case (McAvoy and Braatz, 2003)

0 20 40 60 80 100 120 140 160 180 2000.01

0.012

0.014

0.016

0.018

0.02

0.022

0.024

0.026

0.028

0.03

u1

y1

u1

Desired steady-state output: yss = 0.23 Required average steady-state input: uss = yss/11 = 0.02091.So u1 must cycle between 0.02 and 0.03 (9.1% of the time)

• Consider simpler SISO example

where T=0.05 or smaller. T = ”effective delay” • K(s): PI-controller that cancels dominant time

constant at 1.• The Quantizer step is 1, representing an on/off

valve (the ”worst-case valve”).

Study Case (Skogestad and Araujo, 2004)

;)1(

1

1

1.4)(

2

TsssG ,

1111)(

s

s

ssK

• Setpoint response with T=0.05• Cycling because of integral action in controller• Average input: uss = yss/k = 1/4.1 = 0.24 (24% at 1; 76% at 0)• Magnitude of oscillations in y: a = 0.063

Study Case (Skogestad and Araujo, 2004)

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1 2a = 0.126

Pu = 0.4

y

• The magnitude is clearly related to T. Why?• We need a light!!!

T a = oscillations in y (±)

0.05 0.063

0.005 0.0063

0.0005 0.00063

Study Case (Skogestad and Araujo, 2004)

• Again: Magnitude and frequency of oscillations independent of controller tuning

• But depend on plant dynamics; simulating for various T (effective delay):

• Oscillations in the outputs can be generated by relay feedback (on/off-controller).

• Cycles at natural frequency Pu = 1/w180.• Furthermore, from relay formula (Åstrøm, 1988), the

corresponding ultimate controller gain is

where d is the relay amplitude (input) and a is the amplitude (output) of the oscillations.

In our case, d = 0.5 (half of the Quantizer step)

Relay Feedback MethodRelay

ControllerProcess y(t)u(t)

r(t)

a

dKu

4

• Relay formula

• Can also find Ku from frequency domain analysis:

Study Case (Skogestad and Araujo, 2004)

a

dKu

4

T a = 0.13*T

0.05 0.065 (observed 0.063)

0.005 0.0065 (observed 0.0063)

0.0005 0.00065 (observed 0.00063)

Inaccurate valve with quantization d:

Assume that we require a < amax (max output variation).Gives controllability requirement:

Gives upper limit on plant gain at frequency where L = -

Note: amax=1 and d = 0.016 (1.6% valve error) gives |G(jwL180)|<50

Controllability with inaccurate valve

Conclusions

1. Systems with very high gain are sensitive. Impossible in practice to get the fine manipulation of the control valves that is required for control because the valves would be limited to move in a very small region

2. Upper limit for σ1(G) should be imposed. Suggest that a reasonable limit is 50 because essentially all control systems are eventually implemented with analogue devices which typically have an accuracy on the order of 0.5%.

3. My get rid of oscillations by detuning the controller.

OK

Only true for feedforward without pulsing. No problem with feedback (must accept some cycling)

Only at bandwidth frequency – no limit at steady-state

No. Always oscillations if controller has integral action

Paper: Large process gain• Introduction, Previous work. Briefly mention MB-paper (1 page)• Input disturbances is main potential problem (1 page):

– A. Load disturbance. |G(jwb)| < bound (previous work; easy to derive)– B. Valve inaccuracy, |G(jwL180)|< bound (have derived)– Nothing at steady-.state . Counterexample: Liquid level

• A. Input load disturbance (2 pages)– Derive bound

• Haig gain -> Require high bandwidth. (wb=closed-loop bandwidth) because |G(jw)| drops with w.• If not possible with high bandwidth: Must redesign• Example: pH-neutralization (ssgain = 1e6). High bw not possible. Must redesign: Add more tanks (ssgain same, but drops at high freq)

• B. Valve inaccuracy (“stiction”) (6 pages)– Use new example and results from this paper– Start with example. Use example with a high gain– Theory

• Always get oscillations if integral action in controller• Gain at wL180 (NOT same as wB; wB is where |L|=1; well-designed control system: very close)• Note wL180 approx wG180 because phase of controller (PI or PID) is close to zero at wG180)• Magnitude of oscillations: a = (4/pi) * |G(jwL180)| * d• NOTATION? A -> ya, E ?, d-> dq , uq ???• This is the maximum (worst-case), likely to happen because for some operations valve is likely to be in mid-range.

– Challenge: Prove that this is the MAX we get (with “fully developed” sinusoids – midrange input)• Important difference from A: Bandwidth wb (Controller gain) has no effect• So how can you avoid oscillations from inaccurate valve / stiction?

– From formula: Only thing you can do is change valve (smaller dq) – or take away integral action, but must then accept offset (PROBABLY of magnitude a)– Or redesign process (put an extra tank)

– Stress that controller tuning normally does not matter• Gain: no effect• Integral time: No effect if reasonably tuned• P-controller: May get rid of oscillations

• Discussion (1 page)– MB-paper. Too short simulation time– “Not fully developed sinusoids”– MIMO

• Conclusion

• System with inaccurate valve (quantifier).1. Controller tuning has no effect on output variations due to

quantizer. Thus, detuning does NOT help, because the integral action will in any case force the system into cycling.

2. Magnitude of the disturbance or setpoint may have some effect, especially on Pu, due to its influence on the steady-state:a. If steady-state input is reasonably in the "middle" between to

quantification values and we get "fully" developed sinusoidsb. If steady-state is close to one of the quantification values (e.g. u =

0.21 with uq1=0.2 and uq2=0.3, then it is only f = 0.1 of the time at 0.3) then sinusoid is probably not fully developed. May try step response analysis

Conclusions

Singular Value Decomposition (SVD)

• Antonio: ”In order to make life beautiful God created man, man has created lots of problems, and man had the brilliant idea of creating SVD to solve some of them”.

• Any matrix can be decomposed into the SVD:G=UΣVH,

U and V: unitary matrices Σ: diagonal matrix of of singular values σi

• SVD gives useful information about input and output directions.G=UΣVH GV=UΣ Gvi= σiui, for column i.

• Furthemore for any input direction v:

• v1 and u1 and σ1: strong direction • vn and un and σn:: weake direction.• Condition number:

vGv

GvGn ),()( 1

2

2

).(/)()( 1 GGG n

σ1(G) and σn(G)

• Skogestad and Postlethwaite (1996): Need σn(G) ≥ 1 to avoid input saturation (assuming unitary scaling).

• Skogestad and Postlethwaite (1996): ”A large condition number may be caused by a small value of σn, which is generally undesirable. On the other hand, a large value of σ1 is not necessarily a problem.”

• McAvoy and Braatz (2003): Also a large value of σ1 should be avoided. Their claim is based on an example (”study case”) with valve stiction

• Okay! • But what about McAvoy and Braatz- example?• Loop 1 gives us:

• Observed oscillations in the output y1:a = (0.266-0.236)/2 = 0.005 with period P = 13.

• Theory (relay formula): a=0.009 with period P = 1/w180 = 4

• Does not quite agree...

Study Case (McAvoy and Braatz - revisited)

11 110

11ue

sy s

• Thee sinusoids are not fully developed• Give us another light, please!

Study Case (McAvoy and Braat, 2003 - revisited)

0 10 20 30 40 50 60 70 800

0.05

0.1

0.15

0.2

0.25

y1

Pu = 13 2a = 0.01

• Try ”step response” analysis which should be better for not fully developed sinusoids.

Study Case (McAvoy and Braat, 2003 - revisited)

20 40 60 800.1

0.15

0.2

0.25

0 50 1000.02

0.025

0.03

0.035

0.04

Process

For short response times: Approximate G as integrating process: G = 11/10s.Response: y1(t) = (11/10)*d*t.How long (T) does pulse last?

We need to find the fraction of time, f, T is at the maximum value (0.03).The steady-state input is uss = yss/11 = 0.23/11 = 0.02091.So pulse lasts 9.1% of the period of 13s.

Thus, T = 0.091*13 = 1.18s, so y(T) = (11/10)*0.005*1.18 = 0.0065 (OK!)