RELATIVE GAIN MEASURE OF INTERACTION

RELATIVE GAIN MEASURE OF INTERACTION

We have seen that interaction is important. It affects whether feedback control is possible, and if possible, its performance.

Do we have a quantitative measure of interaction?

The answer is yes, we have several! Here, we will learn about the RELATIVE GAIN ARRAY.

Our main challenge is to understand the correct interpretations of the RGA.

RELATIVE GAIN

We are here, and making progress all the time!

• Defining control objectives

• Controllability & Observability

• Interaction & Operating window

• The Relative Gain

• Multiloop Tuning

• Performance and the RDG

• SVD and Process directionality

• Robustness

• Integrity

• Control for profit

• Optimization-based design methods

• Process design- Series and self-regulation- Zeros (good/bad/ugly)- Recycle systems- Staged systems


Let’s start here to build understanding

OUTLINE OF THE PRESENTATION1. DEFINITION OF THE RGA

2. EVALUATION OF THE RGA

3. INTERPRETATION OF THE RGA

4. EXTENSIONS OF RGA

5. PRELIMINARY CONTROL DESIGN IMPLICATIONS OF RGA


+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)The relative gain between MVj and CVi is ij . It is defined in the following equation.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k What have we assumed about the other controllers?

Explain in words.


Now, how do we determine

the value?







j

i

CVj

i

MVj

i

CVMVCVKMV

MVCVMVKCV

ij

1

ij

kI

k

The relative gain array is the element-by-element product of K with K-1. ( = product of ij elements, not normal matrix multiplication)

jiijij kIkKK T

1

1. The RGA can be calculated from open-loop gains (only).

Open-loop

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

Closed-loop


1. The RGA can be calculated from open-loop values.

The relative gain array for a 2x2 system is given in the following equation.

2211

211211

1

1

KK

KK

What is true for the RGA to have 1’s on diagonal?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


2. The RGA elements are scale independent.

2

1

2

1

2

1

2

1

109

101

109

1010

MV

MV

CV

CV

MV

MV

CV

CV *

. *

Original units Modified units

What is the effect of changing the units of the CV, expressing CV as % of instrument range, or changing the capacity of the final element on ij ?

10910

91010

22

1

21

1

/or

or

CVCV

CV

MVMV

MVCan we prove that thisis general?


3. The rows and columns of the RGA sum to 1.0.

109

910

2

1

21

CV

CV

MVMV

For a 2x2 system, how many elements are independent?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k



closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

Class exercise: prove this statement.

Hint: a matrix and its inverse commute, i.e.,

K K-1 = K-1 K = I



closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

K K-1 = I = K-1 K

n

1i

n

1m1

n

1j

n

1m1

jrow of sum ji if 1 and ji if 0

i col of sum ji if 1 and ji if 0

ijmjjm

n

kkjik

ijmiim

n

kkjik

kIkkkI

kIkkIk

From the left hand equation, the elements of I are equal to

From the right hand equation, the elements of I are equal to


4. In some cases, the RGA is very sensitive to small errors in the gains, Kij.

2211

211211

1

1

KK

KK

When is this equation very sensitive to errors in the individual gains?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k



Change in FD used infinite difference for

derivative

11 for a positivechange in FD

11 for a negativechange in FD

Average 11 forpositive and negative

changes in FD

2% .796 .301 .5480.5% .673 .508 .5900.2% .629 .562 .5960.05% .605 .588 .597

We must perform a thorough study to ensure that numerical derivatives are sufficiently accurate!

The x must be sufficiently small (be careful about roundoff).

constant

constant

k

k

CVj

i

MVj

i

ij

MVCV

MVCV

From McAvpy, 1983



We must perform a through study to ensure that numerical derivatives are sufficiently accurate!

The convergence tolerance must be sufficiently small.

Convergencetolerance of

equations (some ofall errors squared)

11 for a positivechange in FD

11 for a negativechange in FD

Average 11 forpositive and negative

changes in FD

10-4 -4.605 8.080 -.88710-6 -.096 1.068 .50310-8 .556 .615 .58610-10 .622 .568 .59510-16 .629 .562 .596

constant

constant

k

k

CVj

i

MVj

i

ij

MVCV

MVCV

Average gains from +/-

From McAvpy, 1983


5. The relative gain elements are independent of the control design for the “ij” inputs and outputs being considered.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

AC

TC

Solvent

Reactant

FS >> FR


6. A permutation in the gain matrix (changing CVs and MVs) results in the same permutation in the RG Array.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

109

1010

2

1

2

1

MV

MV

CV

CV109

910

2

1

21

CV

CV

MVMV

Process gain RGA

1010

109

2

1

1

2

MV

MV

CV

CV ??


How do weuse values to

evaluate behavior?







MVj CVi

ij < 0 In this case, the steady-state gains have different signs depending on the status (auto/manual) of the other loops.

A

A

CA0

CA

CSTR with A B

Solvent

A

Discuss interaction and RGA in this system.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


MVj CVi

ij < 0 In this case, the steady-state gains have different signs depending on the status (auto/manual) of other loops

We can achieve stable multiloop feedback by using the sign of the controller gain that stabilizes the multiloop system.

Discuss what happens when the other interacting loop is placed in manual!

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


MVj CVi

ij < 0 the steady-state gains have different signs

For ij < 0 , one of three BAD situations occurs

1. Multiloop is unstable with all in automatic.

2. Single-loop ij is unstable when others are in manual.

3. Multiloop is unstable when loop ij is manual and other loops are in automatic

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

0 100 200 300 4000.975

0.98

0.985

0.99

0.995IAE = 0.3338 ISE = 0.0012881

XD

, D

istil

late

Lt K

ey

0 100 200 300 4000.005

0.01

0.015

0.02

0.025

0.03IAE = 0.58326 ISE = 0.0041497

XB

, B

otto

ms

Lt

Ke

y

0 100 200 300 400 50013.3

13.4

13.5

13.6

13.7

13.8

Time

Reb

oile

d V

ap

or

0 100 200 300 400 5008.5

8.6

8.7

8.8

8.9

9

Time

Ref

lux

Flo

w

FR XB

FV XD

Example of pairing on a negative RGA (-5.09). XB controller has a Kc with opposite sign from single-loop control! The system goes unstable when a constraint is encountered. But, we can achieve stable control with pairing on negative RGA!

FR

FV

XB

XD


MVj CVi

ij < 0 the steady-state gains have different signs

For ij < 0 , one of three situations occurs

1. The process gij(s) has a RHP zero

2. The overall plant has a RHP zero

3. The system with gij(s) removed has a RHP zero

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

See Skogestad and Postlethwaite, 1996


MVj CVi

ij = 0 In this case, the steady-state gain is zero when all other loops are open, in manual.

T

L

Could this control system work?

What would happen if one controller were in manual?

Heating tank without boiling

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


MVj CVi

0<ij<1 In this case, the multiloop (ML) steady-state gain is larger than the single-loop (SL) gain.

What would be the effect on tuning of opening/closing the other loop?

Discuss the case of a 2x2 system paired on ij = 0.1

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


MVj CVi

ij= 1 In this case, the steady-state gains are identical in both the ML and the SL conditions.

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)

What is generally true when ij= 1 ?

Does ij= 1 indicate no interaction?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k



AC

TC

CSTR with zero heat of reaction

Solvent

Reactant

FS >> FR

Determine the relative gain.

Discuss interaction in this system.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MV

CV

MV

CV

MV

CV

MV

CV

k

k



..

..

..

k

k

K22

11

........k

....

....

kk

k

K

n1

2221

11

0

0

0

IRGA

1

1

1

1

10

0Lower diagonal gainmatrix

Diagonal gainmatrixDiagonal gainmatrix

Both give an RGA that is diagonal!

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


MVj CVi

1<ij In this case, the steady-state multiloop (ML) gain is smaller than the single-loop (SL) gain.

What would be the effect on tuning of opening/closing the other loop?

Discuss a the case of a 2x2 system paired on ij = 10.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

FR XD

FRB XB

FD XD

FRB XB


XD, XB FeedComp.

RGA RGA

.998,.02 .25 46.4 .07

.998,.02 .50 45.4 .113

.998,.02 .75 66.5 .233

.98, .02 .25 36.5 .344

.98, .02 .50 30.8 .5

.98, .02 .75 37.8 .65

.98, .002 .25 66.1 .787

.98, .002 .50 46 .887

.98, .002 .75 48.8 .939

1. Do level loops affect the composition RGA’s?2. Does the process operation affect RGA’s?

Rel. vol = 1.2, R = 1.2 Rmin

From McAvpy, 1983


MVj CVi

ij= In this case, the gain in the ML situation is zero. We conclude that ML control is not possible.

How can we improve the situation?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

Have we seen this result before?







Let’s extend theconcept


The relative gain between MVj and CVi is ij .

constant

constant

k

k

CVj

i

MVj

i

ij

MVCV

MVCV

The basic definition involves steady-state gain information.

• Some plants are unstable

• Control performance is influenced by dynamics

• Many plants have an unequal number of MVs and CVs

• Control design involves structures other than single-loop

• Disturbances are not considered!


We can evaluate the RGA of a system with integrating processes, such as levels.

Redefine the output as the derivative of the level; then, calculate as normal. (Note that L is unstable, but dL/dt is stable.)

slurry of density

21

1

21

mm

mD

FmmAdt

dLA out

m2m1m

L

A

= density

D = density

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


We can evaluate the RGA of a system with integrating processes, such as levels.

Redefine the output as the derivative of the level; then, calculate as normal.

/)/(

)/(/

DDD

DD

mm

1

121

m2m1m

L

A

= density

D = density

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

FR XD

FRB XB


A frequency-dependent RGA can be calculated using the transfer functions in place of the steady-state gains.

+ +

+ +

G11(s)

G21(s)

G12(s)

G22(s)

Gd2(s)

Gd1(s)

D(s)

CV1(s)

CV2(s)MV2(s)

MV1(s)

We can evaluate the RGA of dynamics processes

10-4

10-2

100

102

10-5

100

frequency, radians/min

ampl

itude

, X

D(jw

)/F

R(jw

)

10-4

10-2

100

102

10-5

100

ampl

itude

, X

D(jw

)/F

V(jw

)

10-4

10-2

100

102

10-4

10-2

100

ampl

itude

, X

B(jw

)/F

R(jw

)

10-4

10-2

100

102

10-4

10-2

100

ampl

itude

, X

B(jw

)/F

V(jw

)

Bode plots of the individual transfer functions for a distillation tower

10

-310

-210

-110

010

110

210

-1

100

101

frequency, rad/time

ampl

itude

rat

io

frequency dependent RGA for distillation tower

Bode plot of the RGA 11 element. What frequency range is most important for feedback control?


The basic definition involves steady-state gain information.

• Some plants are unstable

• Control performance is influenced by dynamics

• Many plants have an unequal number of MVs and CVs

• Control design involves structures other than single-loop

• Disturbances are not considered!

Apparently, there is a lot more to learn.

We better plan to address these issues in the remainder of the course


Let’s evaluatesome design

guidelines basedon RGA







Proposed Guideline #1

Select pairings that do not have any ij<0

• Review the interpretation, i.e., the effect on behavior.

• What would be the effect if the rule were violated?

• Do you agree with the Proposed Guideline?

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k



Select pairings that do not have any ij=0




closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


• We conclude that the RGA provides excellent insight into the INTEGRITY of a multiloop control system.

• INTEGRITY: A multiloop control system has good integrity when after one loop is turned off, the remainder of the control system remains stable.

• “Turning off” can occur when (1) a loop is placed in manual, (2) a valve saturates, or (3) a lower level cascade controller no lower changes the valve (in manual or reached set point limit).

• Pairings with negative or zero RGA’s have poor integrity

RGA and INTEGRITY



Select a pairing that has RGA elements as close as possible to ij=1




closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k

FR XD

FRB XB

FD XD

FRB XB

0 50 100 150 2000.98

0.982

0.984

0.986

0.988IAE = 0.26687 ISE = 0.00052456

XD

, lig

ht k

ey

0 50 100 150 2000.02

0.021

0.022

0.023

0.024IAE = 0.25454 ISE = 0.0004554

XB

, lig

ht k

ey

0 50 100 150 2008.5

8.6

8.7

8.8

8.9

9SAM = 0.31512 SSM = 0.011905

Time

Ref

lux

flow

0 50 100 150 20013.5

13.6

13.7

13.8

13.9

14SAM = 0.28826 SSM = 0.00064734

Time

Reb

oile

d va

por

0 50 100 150 2000.98

0.982

0.984

0.986

0.988IAE = 0.059056 ISE = 0.00017124

XD

, lig

ht k

ey

0 50 100 150 2000.019

0.02

0.021

0.022

0.023IAE = 0.045707 ISE = 8.4564e-005

XB

, lig

ht k

ey

0 50 100 150 2008.46

8.48

8.5

8.52

8.54SAM = 0.10303 SSM = 0.0093095

Time

Ref

lux

flow

0 50 100 150 20013.5

13.6

13.7

13.8

13.9

14SAM = 0.55128 SSM = 0.017408

Time

Reb

oile

d va

por

RGA = 6.09 RGA = 0.39

For set point response, RGA closer to 1.0 is better

FR XD

FRB XBFD XD

FRB XB

0 50 100 150 200

0.975

0.98

IAE = 0.14463 ISE = 0.00051677

XD

, lig

ht k

ey

0 50 100 150 2000

0.005

0.01

0.015

0.02

0.025IAE = 0.32334 ISE = 0.0038309

XB

, lig

ht k

ey

0 50 100 150 2008.5

8.55

8.6

8.65

8.7SAM = 0.21116 SSM = 0.0020517

Time

Ref

lux

flow

0 50 100 150 20013.1

13.2

13.3

13.4

13.5

13.6SAM = 0.38988 SSM = 0.0085339

Time

Reb

oile

d va

por

RGA = 6.09 RGA = 0.39

0 50 100 150 2000.95

0.96

0.97

0.98

0.99IAE = 0.45265 ISE = 0.0070806

XD

, lig

ht k

ey

0 50 100 150 2000

0.005

0.01

0.015

0.02

0.025

0.03IAE = 0.31352 ISE = 0.0027774

XB

, lig

ht k

ey

0 50 100 150 2008

8.1

8.2

8.3

8.4

8.5

8.6SAM = 0.51504 SSM = 0.011985

Time

Ref

lux

flow

0 50 100 150 20011

11.5

12

12.5

13

13.5

14SAM = 4.0285 SSM = 0.6871

Time

Reb

oile

d va

por

For feed composition disturbance response, RGA farther from 1.0 is better


Using guidelines #1 and #2, the control possibilities for this example process were reduced from 36 to 4.


• Tells us about the integrity of multiloop systems and something about the differences in tuning as well.

• Uses only gains from feedback process!

• Does not use following information- Control objectives- Dynamics- Disturbances

• Lower diagonal gain matrix can have strong interaction but gives RGAs = 1

Can we design controls without this information?

The RGA gives useful conclusions from S-S information, but not enough to design process control

Powerful results from limited information!

“Interaction?”

Workshop on Relative Gain Array

INTERACTION IN FEEDBACK SYSTEMS

Workshop on Relative Gain Array: Problem 1

The RGA has been evaluated, but the regulatory control system (below the loops being analyzed using RGA) has been modified. Instead of adjusting a valve directly, one of the loops being evaluated will adjust a flow controller set point (which adjusts the same valve). How would you evaluate the new RGA?


You have decided to pair on a loop that has a negative RGA element. Discuss the tuning that is appropriate for this loop.

A

A

CA0

CA

CSTR with A B

Solvent

A


Discuss what information can be obtained from the RGA and some information that cannot.

closed loopsother

open loopsother

constant

constant

j

i

j

i

CVj

i

MVj

i

ij

MVCV

MVCV

MVCV

MVCV

k

k


You would like to evaluate the steady-state RGA for a process, but the feedback controllers must remain in automatic status. How can you obtain the needed data?

• Explain a procedure that might yield the needed information.

• Discuss the practicality of the approach.

1

2

3

13

14

15LC-1

LC-3

dP-1

dP-2

To flare

T5

T6

TC-7

AC-1

LAHLAL

PAH

PC-1

P3

F4

F7

F8

F9

PV-3

TAL

T10

AC-2


TC

LC

FC

VC

SP = 95% open

To the SP of the feed flow

CV to the VC controller

MV from the VC controller

MV from the TC controller

Determine the relative gain array for the TC and VC controllers. Discuss the behavior of this design and generalize the conclusions.

RGA WORKSHOP PROBLEM 6Three CSTR's with the configuration in the figure and with the design parameters below areconsidered in this example; the common data is given below, and the case-specific data andsteady-states are given in the table.

F=1 M3 , V=1 M3 , CA0=2.0 kg-moles/M3, Cp=1 cal/(g C), =106 g/M3 , ko = 1.0x1010

min-1 , E/R = 8330.1 K-1 , (Fc)s=15 M3/min , Cpc=1 cal/(g K) , c=106 g/M3 , b=0.5Case I (Example 3.10) II III

-?Hrxn 106 cal/(kg-mole) 130 13 -30

a (cal/min)/? K 1.678x106 1.678x106 0.7746x106

T0 ? K 323 370 370

Tcin ? K 365 365 420 (heating)

Ts ? K 394 368.3 392.7

CAs kg-mole/M3 0.265 0.80 0.28

Gain matrices for the three Cases, about the appropriate steady-state

Inputvariable

Case I Case II Case III

CA T CA T CA T

CA0 - 0.161 23.8 0.3615 1.309 0.2214 -6.267

Fc 0.0158 -1.28 .0034 -0.1144 -0.0085 0.657

T0 -.0026 0.211 -.0049 0.1678 -.0032 0.243

F -.0948 26.33 0.4251 1.864 .4527 -16.30

RG

A W

OR

KS

HO

P. P

RO

BL

EM

6

(co

nti

nu

ed)

A. Calculate the relative gain arrays for the three cases shown in the table below.

CASE I CASE II CASE III

CA0 FC CA0 FC CA0 FC

CA

T

Notes:1. CA0 is controlled by adjusting the reactant valve2. FC is achieved by adjusting the valve on the pipe to the heat exchanger coils.3. For Cases I and II, the coils provide coling, and for Case III, the coils provide heating.4. The feed total flow and temperature controllers are in operation for Cases I-III

B. Determine conclusions for control design for each case. Explain each result on physicalgrounds.

C. Discuss relationships among the cases.

Date post:	11-Jan-2016
Category:	Documents
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RELATIVE GAIN MEASURE OF INTERACTION

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