Program for North American Mobility In Higher Education

Module 5 – Controllability Analysis 11

PIECENAMPProgram for North American Mobility In Higher EducationProgram for North American Mobility In Higher Education

NAMPNAMP

Introducing Process integration for Environmental Control in Engineering CurriculaIntroducing Process integration for Environmental Control in Engineering Curricula

Controllability Controllability AnalysisAnalysis

Module 5Module 5

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Program for North American Mobility in Higher EducationProgram for North American Mobility in Higher Education NAMPNAMP

Process integration for Environmental Control in Engineering CurriculaProcess integration for Environmental Control in Engineering Curricula

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University of University of OttawaOttawa

École École Polytechnique Polytechnique de Montréalde Montréal

Instituto Instituto Mexicano del Mexicano del

PetrPetróóleoleo

North Carolina North Carolina State State

UniversityUniversity

PapricanPaprican

Universidad Universidad AutAutóónoma de noma de

San Luis PotosSan Luis Potosíí

Texas A&M Texas A&M UniversityUniversity

Universidad de Universidad de GuanajuatoGuanajuato


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Module 5Module 5

This module was This module was created by:created by:Stacey WoodruffStacey Woodruff

Carlos Carlos CarreónCarreón

Host Host UniversityUniversity

FromFrom






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ObjectivesObjectives Create web-based modules to assist universities to Create web-based modules to assist universities to address the introduction to Process Integration into address the introduction to Process Integration into Engineering curriculaEngineering curriculaMake these modules widely available in each of the Make these modules widely available in each of the participating countriesparticipating countries

Participating institutionsParticipating institutions Six universities in three countries (Canada, Mexico and Six universities in three countries (Canada, Mexico and the USA)the USA)Two research institutes in different industry sectors: Two research institutes in different industry sectors: petroleum (Mexico) and pulp and paper (Canada)petroleum (Mexico) and pulp and paper (Canada)Each of the six universities has sponsored 7 exchange Each of the six universities has sponsored 7 exchange students during the period of the grant subsidised in students during the period of the grant subsidised in part by each of the three countries’ governmentspart by each of the three countries’ governments

Project SummaryProject Summary


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What is the structure of this module?What is the structure of this module?

All modules are divided into 3 tiers, each with a All modules are divided into 3 tiers, each with a specific goal:specific goal:

Tier I: Background InformationTier I: Background InformationTier II: Case Study ApplicationsTier II: Case Study ApplicationsTier III: Open-Ended Design ProblemTier III: Open-Ended Design Problem

These tiers are intended to be completed in that These tiers are intended to be completed in that particular order. In the first tier, students are quizzed particular order. In the first tier, students are quizzed at various points to measure their degree of at various points to measure their degree of understanding, before proceeding to the next two understanding, before proceeding to the next two tiers.tiers.

Structure of Module 5Structure of Module 5


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What is the purpose of this module?What is the purpose of this module?

It is the objective of this module to cover the It is the objective of this module to cover the basic aspects of basic aspects of Controllability AnalysisControllability Analysis. It . It is targeted to be an integral part of a is targeted to be an integral part of a fundamental/and or advanced fundamental/and or advanced ControlControl course. course.This module is intended for students with This module is intended for students with some basic understanding of the fundamental some basic understanding of the fundamental concepts of control. concepts of control.

Purpose of Module 5Purpose of Module 5


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Tier IBackground Information


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• Statement of IntentStatement of Intent

– Define StabilityDefine Stability– Demonstrate simple methods for stability Demonstrate simple methods for stability

analysis, mostly for Single-Input Single-Output analysis, mostly for Single-Input Single-Output (SISO) systems(SISO) systems

– Understand interaction between control loops in Understand interaction between control loops in Multiple-Input Multiple-Output (MIMO) systems Multiple-Input Multiple-Output (MIMO) systems

– Demonstrate the Relative Gain ArrayDemonstrate the Relative Gain Array– Investigate controllability analysis for Investigate controllability analysis for

continuous and discrete systems continuous and discrete systems – Comprehend singular value decomposition (SVD)Comprehend singular value decomposition (SVD)


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StabilityStability

A dynamic system is stable if the system output response is bounded for all bounded inputs. A stable system will tend to return to its equilibrium point following a disturbance. Conversely, an unstable system will have the tendency to move away from its equilibrium point following a disturbance.


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• Why is the stability of a system important??Why is the stability of a system important?? When a system becomes unstable it can be When a system becomes unstable it can be

A DISASTER!!!!!A DISASTER!!!!!


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• ExampleExampleThe concept of stability is illustrated in the following figure. The sphere in (a) is stable as it will return to its original equilibrium after a small disturbance whereas the sphere in (b) is unstable as it moves away from its equilibrium point and never comes back. The sphere in (c) is said to be marginally stable.

(a) (b) (c)


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Quiz #1Quiz #1

• Why is it important that a system Why is it important that a system is stable?is stable?

• List two examples of systems that List two examples of systems that have become unstable.have become unstable.


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There are many ways of determining if a There are many ways of determining if a system is stable such as :system is stable such as :

Roots of Characteristic EquationRoots of Characteristic Equation Bode DiagramsBode Diagrams Nyquist PlotsNyquist Plots SimulationSimulation


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• Roots of Characteristic EquationRoots of Characteristic Equation

One can determine if a system is stable based on the One can determine if a system is stable based on the nature of the roots of its characteristic equations. nature of the roots of its characteristic equations. Consider the following system:Consider the following system:

3G (s)D(s)

M(s) Y(s)+

+

Ym(s)

Y*(s) (s)+-

U(s)1G (s) 2G (s)

4G (s)

CG (s)


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From the previous diagram, we can see that the From the previous diagram, we can see that the output Y is influenced in the following manner. output Y is influenced in the following manner.

GGOLOL is the open loop transfer function.is the open loop transfer function.

c 1 2 3

OL OL

OL c 1 2 4

G G G GY(s) = Y*(s) + D(s)

1 + G 1 + G

Where

G = G G G G


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For the moment, let’s consider that there is only a change in set For the moment, let’s consider that there is only a change in set point, therefore, the previous equation reduces to the closed loop point, therefore, the previous equation reduces to the closed loop transfer function,transfer function,

The roots rThe roots r11, r, r22, r, r33… r… rnn are those of the characteristic equation are those of the characteristic equation

1+G1+GccGG11GG22GG44 =0 =0

and and (s) is a function that arises from the rearrangement. The roots (s) is a function that arises from the rearrangement. The roots of the characteristic equation (denominator) are the poles of the of the characteristic equation (denominator) are the poles of the transfer function whereas the roots of the numerator are the zeros.transfer function whereas the roots of the numerator are the zeros.

c 1 2 c 1 2 c 1 2

c 1 2 4 c 1 2 4 1 2 3 n

G G G G G G G G G (s)1Y(s) = Y*(s) = =

1 + G G G G s 1 + G G G G s(s - r )(s - r )(s - r )...(s - r )


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• The nature of the roots of the characteristic equation can dictate if a The nature of the roots of the characteristic equation can dictate if a system is stable or not due to the fact that if there is one (or more) root system is stable or not due to the fact that if there is one (or more) root on the right half of the complex plane, the response will contain a term on the right half of the complex plane, the response will contain a term that grows exponentially, leading to an unstable system.that grows exponentially, leading to an unstable system.

StableStableRegionRegion

StableStableRegionRegion

ReaReal l

ParPartt

ImaginarImaginary Party Part

UnstableUnstableRegionRegion

Real Real PartPart

timetime

φφ



Real Real PartPart

Negative real root


Real Real PartPart

Complex Roots (Negative real parts)

φφ

timetime


Real Real PartPart

Positive real root

φφ

timetime

Complex Roots (Positive real parts)

φφtimetime


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• Routh TestRouth TestThe Routh test (Routh stability criterion) is a very useful The Routh test (Routh stability criterion) is a very useful tool in determining whether or not a closed-loop system is tool in determining whether or not a closed-loop system is stable provided the characteristic equation is available. stable provided the characteristic equation is available. The Routh stability criterion is based on a characteristic The Routh stability criterion is based on a characteristic equation that is in the formequation that is in the form

A necessary (but not sufficient) condition of stability is that A necessary (but not sufficient) condition of stability is that all of the coefficients (aall of the coefficients (a00, a, a11, a, a22, …etc.) must be positive., …etc.) must be positive.

n n-1n n-1 1 0a s + a s + ... + a s + a = 0


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Routh Array

When all coefficients are positive, a Routh Array must When all coefficients are positive, a Routh Array must be constructed as follows:be constructed as follows:

The system is stable if ALL the elements in the first The system is stable if ALL the elements in the first column are positive!column are positive!

n n -2 n -4

n -1 n -3 n -5

1 2 3

1 2 3

Row

1 a a a ...

2 a a a ...

3 b b b ...

4 c c c ...

n + 1

The first two rows are filled in using the coefficients of the characteristic equation. Subsequent rows are calculated as shown in the next page.

}


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After the coefficients of the characteristic equation are After the coefficients of the characteristic equation are input in the array, the coefficients, binput in the array, the coefficients, b11, b, b2 2 … b… bnn and and subsequently csubsequently c11…c…cnn should be calculated as follows and should be calculated as follows and input into the array.input into the array.

Routh Array

n -1 n -2 n n -31

n -1

a a - a ab =

an n -2 n -4

n -1 n -3 n -5

1 2 3

1 2 3

Row

1 a a a ...

2 a a a ...

3 b b b ...

4 c c c ...

n + 1

n -1 n -4 n n -52

n -1

a a - a ab = ...

a

1 n -3 n -1 21

1

b a - a bc =

b

1 n -5 n -1 32

1

b a - a bc = ...

b

Pivot to calculate all bi


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Routh Test TheoremsRouth Test TheoremsTheorem 1Theorem 1-- The necessary and sufficient condition for stability (i.e. All The necessary and sufficient condition for stability (i.e. All

roots with negative real parts) is that all elements of the first roots with negative real parts) is that all elements of the first column of the Routh Array must be positive and non zero.column of the Routh Array must be positive and non zero.

Routh Test Example 1- Routh Test Example 1- Consider the following characteristic equation:Consider the following characteristic equation:

All of the elements in the first All of the elements in the first column column of this Routh Array of this Routh Array are positive,are positive,

therefore the system is therefore the system is stable.stable.

3 2s + 4.583s + 6.38s + 15.625 = 0Row

1(s3) 1 6.38

2(s2) 4.583 15.625

3(s1) 2.97 0

4(s0) 15.625 0


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Routh Test Example 2- Routh Test Example 2- It is possible to determine for which values It is possible to determine for which values of Kof Kcc the system remains stable the system remains stable

3 2 c1+Ks + 4.583s + 6.38s + = 0

0.384

29.24-(1-Kc)/0.384>0 → Kc

<10.23

1+Kc >0 → Kc>-1 (Kc is positive)

Row

1(s3) 1 6.38

2(s2) 4.583 (1+Kc)/0.384

3(s1) 0

4(s0) (1+Kc)/0.384 0

c29.24 - (1+K )/0.384

4.583


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Theorem 2- If some of the elements of the first column are Theorem 2- If some of the elements of the first column are negative, the number of roots on the right hand side of the negative, the number of roots on the right hand side of the imaginary axis is equal to the number of sign changes in the first imaginary axis is equal to the number of sign changes in the first column.column.

Routh Test Example 3 – If the characteristic equation of a system Routh Test Example 3 – If the characteristic equation of a system is given by the following equation, is the system stable?is given by the following equation, is the system stable?

There are 2 sign changes. Therefore, the system has two roots in the right-hand plane, and the system is unstable.

4 3 2s + 6s + 11s + 36s + 120 = 0Row

1(s4) 1 11 120

2(s3) 6 36 0

3(s2) 5 120

4(s1) -108 0

5(s0) 120


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Theorem 3Theorem 3- If one pair of roots is on the imaginary - If one pair of roots is on the imaginary axis, equidistant from the origin, and all the other axis, equidistant from the origin, and all the other roots are in the left-hand plane, all the elements of the roots are in the left-hand plane, all the elements of the nnthth row will vanish. The location of the pair of row will vanish. The location of the pair of imaginary roots can be found by solving the auxiliary imaginary roots can be found by solving the auxiliary equation: equation:

where the coefficients C and D are the elements of the where the coefficients C and D are the elements of the array in the (n-1)array in the (n-1)thth row. These roots are also the roots row. These roots are also the roots of the characteristic equation.of the characteristic equation.

Cs2+D=0


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Routh Test Example 4 – Routh Test Example 4 – Determine the stability of the system Determine the stability of the system having the following characteristic equation:having the following characteristic equation:

4 3 2s + 3s + 6s + 12s + 8 = 0

There are no sign changes in the first column, indicating that there are no roots located on the right-hand side of the plane.

The derivative taken indicates that a 4 should be placed in the s row (Row 4). The procedure is carried out.

ssds

d4)82( 2

Row

1(s4) 1 6 8

2(s3) 3 12

3(s2) 2 8

4(s1) 0

4(s1) 4

5(s0) 8


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Quiz #2Quiz #2• In what cases can the Routh test be used to In what cases can the Routh test be used to

determine stability?determine stability?• Is the system having the following characteristic Is the system having the following characteristic

equation stable?equation stable?

• If a system has two negative real roots, is the system If a system has two negative real roots, is the system stable?stable?

• If a system has one negative real root and one If a system has one negative real root and one positive real root is the system stable?positive real root is the system stable?

4 3 2s + 7s + 6s + 1 = 0


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Frequency ResponseFrequency Response• One very useful method of determining system stability, One very useful method of determining system stability,

even when transportation lags exist, is Frequency even when transportation lags exist, is Frequency Response.Response.

• Frequency response is a method concerning the response Frequency response is a method concerning the response of a process or system to a sustained sinusoidal plot.of a process or system to a sustained sinusoidal plot.

• Frequency Response Stability CriteriaTwo principal criteria:

1. Bode Stability Criterion2. Nyquist Stability Criterion


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Bode stability criterionA closed-loop system is unstable if the Frequency Response of the open-loop Transfer Function, GOL

=GCG1G2G4, has an amplitude ratio greater than one at the critical frequency, ωωcc. Otherwise the

closed-loop system is stable.

Note: ωc is the value of ω where the open-loop phase angle is -1800.

Thus,

The Bode Stability criterion provides information on the closed-loop stability from open-loop frequency response information.


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Bode Stability Criterion- Example 1Bode Stability Criterion- Example 1A process has the following transfer function:A process has the following transfer function:

With a value of With a value of G1=0.1 and G4=10. If proportional control is used, determine closed-loop stability for 3 values of Kc: 1, 4, and 20. GOL

=GCG1G2G4

Solution:

2 3

2G (s) =

(0.5s + 1)

cOL c 1 2 4 c 3 3

2K2G = G G G G = (K )(0.1) (10) =

(0.5s+1) (0.5s+1)

Kc AROL for Kc Stable?Stable?

11 0.250.25 YesYes

44 11 MarginallyMarginally

2020 55 NoNo

You will find the Bode plots on the next slide


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Bode plots for GOL = 2Kc/(0.5s + 1)3


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• Nyquist Stability CriterionNyquist Stability CriterionThe Nyquist stability criterion is the most powerful stability test The Nyquist stability criterion is the most powerful stability test that is available for linear systems described by transfer that is available for linear systems described by transfer function models.function models.

Consider an open-loop transfer function, GOL(s) that is proper and has no unstable pole-zero cancellations. Let N be the number of times that the Nyquist plot of GOL(s) encircles the (-1, 0) point in a clockwise direction. Also, let P denote the number of poles of GOL(s) that lie to the right of the imaginary axis. Then, Z=N+P, where Z is the number of roots (or zeros) of the characteristic

equation that lie to the right of the imaginary axis.

The closed-loop system is stable, if and only if Z=0.


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Example 9.2 – Find the amplitude ratio and the phase lag of the following process for = 0.1 and 0.4.

1

5s + 1-0.3se 3 2

1.2

s + 2.3s + 1.7s + 0.4

U(s) X(s) Y(s)Z(s)

-1 -1

2 2 2 2 2

:

1 1 1AR = = = ; = tan (- ) = tan (-5 )

ω + 1 (5) ( ) + 1 25( ) +

First s

ystem

1

180AR = 1 ; = -

Second s

=

ystem:

-0.3

3 2 2 3

3-1

22 22 3

1.2 1.2G(j ) = =

(j ) + 2.3(j ) + 1.7(j ) + 0.4 0.4 - 2.3 + 1.7 - j

1.2 - (1.7 - )AR = ; =

Third system:

tan 0.4 - 2.30.4 - 2.3 + 1.7 -


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Example 9.2 – Find AR and (from known equations)

1 2 n

2 2 22 3

G(jω) = G (jω) G (jω) ... G (jω)

1 1.2G(jω) = 1

25( ) + 1 0.4 - 2.3 + 1.7 -

1 2 n

3-1 -1

2

G(jω) = G (jω) + G (jω) + ... + G (jω)

- (1.7 - )G(jω) = tan (-5 ) - 0.3 + tan

0.4 - 2.3

-1 o = 0.1 AR = 2.60 ; = - 0.915 s or - 52.4 -1 o = 0.4 AR = 0.87 ; = - 2.75 s or - 157.3

If (0.4 – 2.33) < 0 then – or – 180o


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Example 9.2 – Find AR and … Nyquist plot

0

90

180

270

0 1 2 3

Im

Re

=0.4

=0.1


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Quiz #3Quiz #3

• Name two methods of determining stability Name two methods of determining stability using frequency response.using frequency response.

• What does an amplitude ratio (AR) of 1 signify? What does an amplitude ratio (AR) of 1 signify? An amplitude ratio of less than 1?An amplitude ratio of less than 1?

• What does a value of Z=0 signify?What does a value of Z=0 signify?


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• Multiple Input Multiple Output (MIMO) Multiple Input Multiple Output (MIMO) SystemsSystems

FEED PUMPS

CRUDE OIL FEED STORAGE TANKS

Air FuelGas

FURNACE

PIPESTILL FRACTIONATOR

High boiling Residue

Heavy gas oil

Light gas oil

Naptha

Cooling unit Reflux Receiver


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When dealing with Multiple Input Multiple When dealing with Multiple Input Multiple Output systems, we have to ask ourselves Output systems, we have to ask ourselves two main questions.two main questions.

1. How to pair the input and output 1. How to pair the input and output variablesvariables

2. How to design the individual single-loop 2. How to design the individual single-loop controllerscontrollers


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Let’s consider the following system:Let’s consider the following system:

Gc1

+

-

Loop 2

m2

m1 y1

y2

Loop 1

y1(s) = G11(s)m1(s) + G12(s)m2(s)

y2(s) = G21(s)m1(s) + G22(s)m2(s)

G11

G12

G22

G21

+

-

++

+Gc2

+


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We will perform 2 small “experiments” to demonstrate MIMO We will perform 2 small “experiments” to demonstrate MIMO system interactions.system interactions.

Let´s consider mLet´s consider m11 as a candidate to pair with y as a candidate to pair with y11..

Experiment #1Experiment #1When a unit step change is made to the input variable mWhen a unit step change is made to the input variable m11, with , with all loops open, the output yall loops open, the output y11 will change, and so will y will change, and so will y22, but for , but for now, we are primarily concerned with the effect on ynow, we are primarily concerned with the effect on y11. After . After steady-state is reached, let’s consider the change in ysteady-state is reached, let’s consider the change in y11 as a as a result of the change in mresult of the change in m11, , yy1m1m ; this will represent the main ; this will represent the main effect of meffect of m11 on y on y11. .

ΔΔyy1m1m = K = K1111

Keep in mind that no other input variables have been changed, Keep in mind that no other input variables have been changed, and that all loops are open, so no feedback control is required.and that all loops are open, so no feedback control is required.


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Experiment #2-Experiment #2-Unit step change in mUnit step change in m11 with Loop 2 closed. with Loop 2 closed.

These things will happen as a result of the unit step change in These things will happen as a result of the unit step change in mm11..

1- y1- y11 changes because of G changes because of G1111, but because of interactions via , but because of interactions via the element Gthe element G2121, y, y22 changes as well. changes as well.2- Under feedback control, Loop 2 wards off this interaction 2- Under feedback control, Loop 2 wards off this interaction effect on yeffect on y22 by manipulating m by manipulating m22 until y until y22 is returned to its initial is returned to its initial state before the disturbance.state before the disturbance.3-The changes in m3-The changes in m22 will now affect y will now affect y11 via the G via the G1212 transfer transfer element.element.

The changes in yThe changes in y11 are from two different sources. are from two different sources.(1) the DIRECT INFLUENCE of m(1) the DIRECT INFLUENCE of m11 on y on y11 ( (ΔΔyy1m1m))(2) the (2) the Indirect InfluenceIndirect Influence, from the retaliatory action from , from the retaliatory action from Loop 2 in warding off the interaction effect of mLoop 2 in warding off the interaction effect of m11 on y on y22 ( (ΔΔyy1r1r))


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After dynamic transients die away and steady-state is After dynamic transients die away and steady-state is reached, the net change observed in yreached, the net change observed in y11 is given by: is given by:

ΔΔyy11*= *= ΔΔyy1m1m++ ΔΔyy1r1r

This net change is the sum of the main effect of mThis net change is the sum of the main effect of m11 on y on y11 and the interactive effect provoked by mand the interactive effect provoked by m11 interacting with interacting with the other loop.the other loop.

A good measure of how well a system can be controlled A good measure of how well a system can be controlled (λ) if m(λ) if m11 is used to control y is used to control y11 is: is:

*1* 112211

211211 K

KK

KKKy

rm

mm

yy

y

y

y

11

1111 *


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Loop Pairing on the Basis of Interaction AnalysisLoop Pairing on the Basis of Interaction Analysis

Case 1 : λCase 1 : λ1111=1=1This case is only possible if This case is only possible if yy1r 1r is equal to zerois equal to zero. In physical . In physical

terms, this means that the main effect of mterms, this means that the main effect of m1 1 on yon y11, when all , when all the loops are opened, and the total effect, measured when the loops are opened, and the total effect, measured when the other loop is closed, are identical.the other loop is closed, are identical.

This will be the case if:This will be the case if:• mm11 does not affect y does not affect y22, and thus, there is no retaliatory control , and thus, there is no retaliatory control

action from maction from m22, or, or• mm11 does affect y does affect y22, but the retaliatory control action from m, but the retaliatory control action from m2 2

does not cause any change in ydoes not cause any change in y11 because m because m22 does not affect does not affect yy11..Under these circumstances, mUnder these circumstances, m11 is the perfect input is the perfect input

variable to control yvariable to control y11 because there will be NO because there will be NO interaction problems.interaction problems.


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Case 2 : λCase 2 : λ1111=0=0

This condition indicates that This condition indicates that mm11 has no effect on y has no effect on y11, , therefore therefore y y1m 1m willwill be zero in response to a change be zero in response to a change in min m11. Note that under these circumstances, m. Note that under these circumstances, m22 is is the perfect input variable for controlling ythe perfect input variable for controlling y22, , NOTNOT y y11. . Since mSince m11 does not affect y does not affect y11, y, y11 can be controlled can be controlled with mwith m22 without any interaction with y without any interaction with y11..


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Case 3 : 0 < λCase 3 : 0 < λ1111< 1< 1This condition indicates that the direction of the This condition indicates that the direction of the

interaction effect is in the interaction effect is in the same same direction as that direction as that of the main effect. In this case the total effect is of the main effect. In this case the total effect is greater than the main effect. For greater than the main effect. For λλ1111>0.5>0.5, the , the main effect contributes MOREmain effect contributes MORE to the total to the total effect than the interaction effect, and as the effect than the interaction effect, and as the contribution of the main effect increases, the contribution of the main effect increases, the closer to a value of closer to a value of 11 λ λ1111 becomes. For becomes. For λλ1111<0.5<0.5, , the contribution from the the contribution from the interaction effect interaction effect dominatesdominates, as this contribution increases, λ, as this contribution increases, λ1111 moves closer to zero. For moves closer to zero. For λλ1111=0.5=0.5, the , the contributions of the main effect and the contributions of the main effect and the interaction effect are interaction effect are equalequal..


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Case 4 : λCase 4 : λ1111>1>1This is the condition where This is the condition where yy1r1r is the opposite sign of is the opposite sign of yy1m1m, but it is , but it is smaller in absolute value. In this case smaller in absolute value. In this case yy11* (* (yy1r1r + +yy1m1m) is less than ) is less than the main effect the main effect yy1m1m, and therefore a larger controller action m, and therefore a larger controller action m11 is is needed to achieve a given change in yneeded to achieve a given change in y11 in the closed loop than in in the closed loop than in the open loop. For a very large and positive λthe open loop. For a very large and positive λ1111 the interaction the interaction effect almost cancels out the main effect and closed-loop control of effect almost cancels out the main effect and closed-loop control of yy11 using m using m11 will be very difficult to achieve. will be very difficult to achieve.

Case 5 : λCase 5 : λ1111< 0< 0This is the case when This is the case when y y1r 1r is not only opposite in sign, but also is not only opposite in sign, but also larger in absolute value to larger in absolute value to y y1m1m.. The pairing of m The pairing of m11 with y with y1 1 in this in this case is case is notnot very desirable because the direction of the effect of m very desirable because the direction of the effect of m1 1

on yon y11 in the open loop is opposite to the direction in the closed in the open loop is opposite to the direction in the closed loop. The consequences of using such a pairing could be loop. The consequences of using such a pairing could be catastrophiccatastrophic..


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Quiz#4Quiz#4

• What is a MIMO system?What is a MIMO system?

• What does λWhat does λ1111=1 signify? If this is the case, is =1 signify? If this is the case, is mm11 a good input variable to control y a good input variable to control y11??

• If λIf λ1111 is very large and positive, is m is very large and positive, is m1 1 a good a good input variable to control yinput variable to control y11??


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Relative Gain Array (RGA)Relative Gain Array (RGA)The quantity λThe quantity λ1111 is defined as the is defined as the Relative GainRelative Gain between between

input minput m11 and output y and output y11..

λλijij is defined as the relative gain between output y is defined as the relative gain between output y ii and and input minput mjj, as the ratio of two steady-state gains:, as the ratio of two steady-state gains:

j

all loopsopen

all loops closedexceptforthem loop

i

j

ij

i

j

ym

ym

open-loopgainclosed-loopgainij

j

for loop i under

the control of m


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When the relative gain is calculated for all of the When the relative gain is calculated for all of the input/output combinations of a multivariable input/output combinations of a multivariable system, the results are placed into a matrix as system, the results are placed into a matrix as follows and this array produces follows and this array produces

THE RELATIVE GAIN ARRAYTHE RELATIVE GAIN ARRAY

nnnn

n

n

21

22221

11211


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• Properties of the Relative Gain ArrayProperties of the Relative Gain Array

1. The elements of the RGA across any row, or 1. The elements of the RGA across any row, or down any column sum up to 1. i.e.:down any column sum up to 1. i.e.:

2. λ2. λij ij is dimensionless; therefore, neither the units, is dimensionless; therefore, neither the units, nor the absolute value actually taken by the nor the absolute value actually taken by the variables mvariables mjj, or y, or yi i affect it.affect it.

111

n

jij

n

iij

PROPERTIES OF THE RELATIVE GAIN ARRAY


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3. 3. The value λThe value λijij is a measure of the is a measure of the steady-state interaction expected in the steady-state interaction expected in the iithth loop of the multivariable system if its loop of the multivariable system if its output (youtput (yii) is paired with input (m) is paired with input (mjj); in ); in particular, λparticular, λij ij =1 indicates that m=1 indicates that mj j affects affects yyi i without interacting with the other without interacting with the other loops. Conversely, if λloops. Conversely, if λijij=0 this indicates =0 this indicates that mthat mjj has no effect on y has no effect on yii..



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4. 4. Let KLet Kijij* represent the loop i steady-state gain * represent the loop i steady-state gain when all loops (other than loop i) are closed, when all loops (other than loop i) are closed, whereas, Kwhereas, Kij ij represents the normal open loop represents the normal open loop gain.gain.

This equation has the very important implication: This equation has the very important implication: that 1/λthat 1/λijij tells us by what factor the open loop tells us by what factor the open loop gain between output ygain between output yii and input m and input mjj will be will be changed when the loop are closed.changed when the loop are closed.

ijij

ij KK1

*



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5. 5. When λWhen λijij is negative, it indicates a is negative, it indicates a situation in which loop i, with all loops situation in which loop i, with all loops open, will produce a change in yopen, will produce a change in yii in in response to a change in mresponse to a change in mjj in totally the in totally the opposite direction to that when all the opposite direction to that when all the other loops are closed. Such other loops are closed. Such input/output pairings are potentially input/output pairings are potentially unstable and should be avoided.unstable and should be avoided.



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• Calculating the Relative Gain ArrayCalculating the Relative Gain Array

There are two ways of calculating the Relative There are two ways of calculating the Relative Gain ArrayGain Array

1.1. The “First Principles” MethodThe “First Principles” Method

2.2. The Matrix MethodThe Matrix Method

COMPUTING THE RELATIVE GAIN ARRAY


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•First Principles MethodFirst Principles Method

Let’s consider a 2x2 system as we encountered before. Let’s consider a 2x2 system as we encountered before. First, we must observe that the Relative Gain Array deals First, we must observe that the Relative Gain Array deals with steady-state systems, and therefore , must only be with steady-state systems, and therefore , must only be concerned with the steady state form of this model which concerned with the steady state form of this model which is:is:

In order to calculate the λIn order to calculate the λ1111 we defined earlier, we need to we defined earlier, we need to evaluate the partial derivatives as was explained on slide evaluate the partial derivatives as was explained on slide 47.47.

Recall:Recall:


1 11 1 12 2

2 21 1 22 2

y =K m +K m

y =K m +K m

j

all loopsopen

all loops closedexceptforthe m loop

i

j

ij

i

j

ym

ym

(Eq. (Eq. 1a1a))(Eq. (Eq. 1b1b))


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Due to the fact that the equations found on the previous slide Due to the fact that the equations found on the previous slide represent steady-state, open-loop conditions, the represent steady-state, open-loop conditions, the differentiation for the numerator portion of the relative gain differentiation for the numerator portion of the relative gain is:is:

The second partial derivative (the denominator) requires Loop 2 The second partial derivative (the denominator) requires Loop 2 to be closed, so that in response to changes in mto be closed, so that in response to changes in m11 , the , the second control variable msecond control variable m22 can be used to restore y can be used to restore y22 to its to its initial value of 0. To obtain the second partial derivative, we initial value of 0. To obtain the second partial derivative, we first find from Eq. 1b the value of the mfirst find from Eq. 1b the value of the m22 must be to maintain must be to maintain yy22=0 in the face of changes in m=0 in the face of changes in m11, what effect this will have , what effect this will have on yon y11 is deduced by substituting this value of m is deduced by substituting this value of m22 into Equation into Equation 1a.1a.


111

1 Km

y

open loops all


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The computation of the denominator of λThe computation of the denominator of λ1111

Set ySet y22=0 and solve m=0 and solve m22 in Eq. 1b. in Eq. 1b.

Substituting this value of mSubstituting this value of m22 into Eq. 1a. gives: into Eq. 1a. gives:

Having eliminated mHaving eliminated m22 from the equation, we now may from the equation, we now may differentiate with respect to mdifferentiate with respect to m11..

212 1

22

Km m

K

122

21121111 m

K

KKmKy

2211

211211

21

1 1KK

KKK

m

y

closedloop



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We then substitute the numerator and denominator into We then substitute the numerator and denominator into the definition of λthe definition of λ1111 which yields: which yields:

This equation simplifies to the form:This equation simplifies to the form:

wherewhere


1111

12 2111

11 22

K

K KK 1-

K K

1

111 12 21

11 22

K K=

K K


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This exercise should be repeated for all This exercise should be repeated for all λλijij’s so that the RGA can be constructed.’s so that the RGA can be constructed.

For Practice, repeat this exercise and For Practice, repeat this exercise and verify the following.verify the following.

andand

12112

1

11122



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• Thus the RGA for this 2x2 system is given by:Thus the RGA for this 2x2 system is given by:

Note, that if we defineNote, that if we define

The RGA can be rewritten as followsThe RGA can be rewritten as follows

1

1

1

11

1

1

111

1

1



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• The Matrix Method for Calculating RGAThe Matrix Method for Calculating RGA

Let K be the matrix of steady-state gains of the Let K be the matrix of steady-state gains of the transfer function matrix G(s) i.e.:transfer function matrix G(s) i.e.:

Whose elements are KWhose elements are Kijij, further, let R be the , further, let R be the transpose of the inverse of this steady state transpose of the inverse of this steady state matrix (K)matrix (K)


KsGs

)(lim0

TKR 1


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With elements rWith elements rijij it is possible to show that the it is possible to show that the elements or the RGA can be obtained from the elements or the RGA can be obtained from the elements of these two matrices as:elements of these two matrices as:

It is important to note that the equation above It is important to note that the equation above indicates an element-by-element multiplication indicates an element-by-element multiplication of the corresponding elements of the two of the corresponding elements of the two matrices, K and R, DO NOT TAKE THE PRODUCT matrices, K and R, DO NOT TAKE THE PRODUCT OF THESE MATRICES!OF THESE MATRICES!


ijijij rK


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•Example- Matrix Method of Calculating RGA.Example- Matrix Method of Calculating RGA.Find the RGA for the 2x2 system represented by Find the RGA for the 2x2 system represented by Equations 1a and 1b and compare them with the results Equations 1a and 1b and compare them with the results obtained using the First Principles Method.obtained using the First Principles Method.

Solution:Solution:

For this system, the steady-state gain matrix (K) is the For this system, the steady-state gain matrix (K) is the following.following.


2221

1211

KK

KKK


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From the definition of the inverse matrix we know thatFrom the definition of the inverse matrix we know that

Where the determinant of K, |K| is:Where the determinant of K, |K| is:

Therefore, by taking the transpose of the KTherefore, by taking the transpose of the K-1-1 matrix, we matrix, we obtain the R matrixobtain the R matrix


1121

12221 1

KK

KK

KK

21122211 KKKKK

1112

21221 1

KK

KK

KKR

T


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Since we now have the R and K matrices, we can perform Since we now have the R and K matrices, we can perform an element by element multiplication to obtain the an element by element multiplication to obtain the elements (λelements (λijij) of the RGA (Λ)) of the RGA (Λ)

OR OR

here is the first element of the matrix. Try on your own to here is the first element of the matrix. Try on your own to compute the other 3 elements of the RGA.compute the other 3 elements of the RGA.


11 2211=

K KK

11 2211

11 22 12 21

=-

K KK K K K

11 22 12 21

21 12 22 11

-

-

K K K KK K

K K K KK K


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• Example of RGA for the Wood and Berry Example of RGA for the Wood and Berry Distillation, using the Matrix MethodDistillation, using the Matrix Method

Find the RGA for Wood and Berry Distillation column Find the RGA for Wood and Berry Distillation column whose transfer function matrix is whose transfer function matrix is

SolutionSolution:: For this system, the steady-state gain matrix is For this system, the steady-state gain matrix is easily extracted from the transfer function matrix by easily extracted from the transfer function matrix by setting s=0.setting s=0.

4.196.6

9.188.12)0(GK

14.14

4.19

19.10

6.610.21

9.18

17.16

8.12

)( 37

3

s

e

s

es

e

s

e

sG ss

ss


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The next step is to determine the inverse of the matrix K:The next step is to determine the inverse of the matrix K:

Once the inverse is calculated, the transpose of this matrix Once the inverse is calculated, the transpose of this matrix must be calculated to yield the matrix R.must be calculated to yield the matrix R.

After these two matrices are computed, it is time to calculate After these two matrices are computed, it is time to calculate the RGA by multiplying the matrices element by element.the RGA by multiplying the matrices element by element.

104.0053.0

153.0157.01K

104.0153.0

053.0157.0)( 1 TKR

21

12Note that all of the Note that all of the rows and columns rows and columns sum to one.sum to one.


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• Loop Pairing using the RGALoop Pairing using the RGA

Now that we know how to compute the RGA, we will now Now that we know how to compute the RGA, we will now consider how it can be used to guide the pairing of consider how it can be used to guide the pairing of input and output variables in order to obtain the control input and output variables in order to obtain the control configuration with minimal loop interaction.configuration with minimal loop interaction.

On the following slides, we will investigate how to On the following slides, we will investigate how to interpret the elements of the RGA (λinterpret the elements of the RGA (λijij). We will use the ). We will use the five scenarios presented early to interpret the five scenarios presented early to interpret the implications of the values of λimplications of the values of λijij

LOOP PAIRING USING THE RELATIVE GAIN ARRAY


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Case 1: λCase 1: λijij=1, the open loop gain is the =1, the open loop gain is the equal to the closed loop gain.equal to the closed loop gain.

Loop interactions implications :Loop interactions implications : This situation This situation indicates that loop i will not be subject to retaliatory indicates that loop i will not be subject to retaliatory effects from other loops when they are closed, effects from other loops when they are closed, therefore mtherefore mjj can control y can control yii without interference from without interference from other control loops. If any of the other elements in the other control loops. If any of the other elements in the transfer function matrix are nonzero, the itransfer function matrix are nonzero, the ithth loop will loop will experience some disturbances from other control loops, experience some disturbances from other control loops, but these are NOT provoked from actions in the ibut these are NOT provoked from actions in the ithth loop. loop.

Recommendation for pairing:Recommendation for pairing: In this case, the pairing In this case, the pairing if mif mjj with y with yii would be ideal. would be ideal.



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Case 2: λCase 2: λijij=0, the open loop gain between =0, the open loop gain between mmjj and y and yii is zero. is zero.

Loop interactions implications :Loop interactions implications : m mjj has no direct has no direct influence on yinfluence on yii (keep in mind that m (keep in mind that mjj may still have an may still have an effect on other control loops)effect on other control loops)

Recommendation for pairing: Recommendation for pairing: Do Do NOT NOT pair ypair yii with m with mjj, , it would be more advantageous to pair mit would be more advantageous to pair mjj with another with another output variable, since we are led to believe that youtput variable, since we are led to believe that yi i will will not be influenced by the loop containing mnot be influenced by the loop containing mjj..



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Case 3: 0<λCase 3: 0<λijij<1, the open loop gain <1, the open loop gain between ybetween yii and m and mjj is is smallersmaller than the than the closed loop gain.closed loop gain.

Loop interactions implications :Loop interactions implications : The closed loop The closed loop gain is the sum of the open loop gain gain is the sum of the open loop gain andand the the retaliatory effect, from the other loops,retaliatory effect, from the other loops,

a)a) The loops are interacting, but The loops are interacting, but

b)b) They interact in such a way that the retaliatory effect They interact in such a way that the retaliatory effect from the other loops is in the same direction as the from the other loops is in the same direction as the main effect of mmain effect of mjj on y on yii. .



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Loop interactions implications :Loop interactions implications :

The loop interactions “assist” mThe loop interactions “assist” mjj on controlling y on controlling yii, The extent , The extent of this assistance is dependent on how close λof this assistance is dependent on how close λij ij is to 0.5is to 0.5

When:When:

λλij ij =0.5: the main effect of m=0.5: the main effect of mjj on y on yii is exactly the same as the is exactly the same as the retaliatory effect.retaliatory effect.

0.5<λ0.5<λij ij <1, the retaliatory effects are less than the main effect<1, the retaliatory effects are less than the main effect

0<λ0<λijij<< 0.5, the retaliatory effect is larger than the main effect.0.5, the retaliatory effect is larger than the main effect.

Recommendation for pairing:Recommendation for pairing: If possible, avoid pairing yIf possible, avoid pairing yii with mwith mjj if λ if λijij<<0.50.5


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Case 4: λCase 4: λijij>1, the open loop gain between y>1, the open loop gain between yii and m and mjj is is largerlarger than the closed loop gain. than the closed loop gain.

Loop interactions implications :Loop interactions implications : The loops interact, and the The loops interact, and the retaliatory effect from the other loops acts in retaliatory effect from the other loops acts in opposition opposition to to the main effect of mthe main effect of mjj on y on yii, (which means that the loop gain , (which means that the loop gain will be reduced when the other loops are closed), but the will be reduced when the other loops are closed), but the main effect is still dominant, otherwise λmain effect is still dominant, otherwise λijij would be would be negative. For large values of λnegative. For large values of λijij, the controller gain for loop i , the controller gain for loop i will have to be chosen much larger than when all loops are will have to be chosen much larger than when all loops are open. This would cause loop i to be stable when the other open. This would cause loop i to be stable when the other loops are open.loops are open.

Recommendation for pairing: Recommendation for pairing: The higher the value of λThe higher the value of λijij , , the greater the opposition mthe greater the opposition mjj experiences from the other experiences from the other loops in trying to control yloops in trying to control yii. Therefore try . Therefore try notnot to pair y to pair yi i with with mmjj with if the value of λ with if the value of λijij isis large. large.



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Case 5: λCase 5: λijij<0, the open loop and closed loop <0, the open loop and closed loop gains between ygains between yjj and m and mi i have opposite have opposite signs.signs.

Loop interactions implications :Loop interactions implications : The loops interact, The loops interact, and the retaliatory effect from the other loops is not only in and the retaliatory effect from the other loops is not only in opposition, but it is greater in absolute value to the main opposition, but it is greater in absolute value to the main effect of meffect of mjj on y on yii. This is potentially dangerous because if . This is potentially dangerous because if the other loops are opened, loop i could become very the other loops are opened, loop i could become very unstable.unstable.

Recommendation for pairing:Recommendation for pairing: Avoid pairing m Avoid pairing mj j with ywith yii

because of the retaliatory effect that mbecause of the retaliatory effect that mjj provokes from the provokes from the other loops acts in opposition to, and dominates the main other loops acts in opposition to, and dominates the main effect on yeffect on yii..



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Quiz#5Quiz#5

• What advantages does the Matrix Method have What advantages does the Matrix Method have over the First Principles Method?over the First Principles Method?

• What does λ with a value of 1 signify, and What does λ with a value of 1 signify, and should mshould mjj and y and yii be paired together? be paired together?

• What does λ with a value less than zero of What does λ with a value less than zero of signify, and should msignify, and should mjj and y and yii be paired be paired together?together?


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• Basic Loop Pairing RulesBasic Loop Pairing RulesFrom what we have learned about loop pairing, it is natural From what we have learned about loop pairing, it is natural

that the ideal RGA would take the formthat the ideal RGA would take the form

This is known as the identity matrix, in which each row and This is known as the identity matrix, in which each row and column only contains one non-zero element whose value is column only contains one non-zero element whose value is unity (1). This ideal RGA is produced when the transfer unity (1). This ideal RGA is produced when the transfer matrix G(s) has one of two forms, only a diagonal element, matrix G(s) has one of two forms, only a diagonal element, or is in lower triangular from. The first situation indicates or is in lower triangular from. The first situation indicates that there is no interaction between the loops. The second that there is no interaction between the loops. The second case indicates that there is a case indicates that there is a one-wayone-way interaction (which is interaction (which is explained on the next slide).explained on the next slide).

1000

0

0100

0010

0001


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If the G(s) indicates that there is a If the G(s) indicates that there is a one-wayone-way interaction( the interaction( the transfer function matrix is in triangular form), it will yield an RGA transfer function matrix is in triangular form), it will yield an RGA of the identity matrix, but it can not be treated as if there are no of the identity matrix, but it can not be treated as if there are no interactions or influences. interactions or influences. Please consider the following example.Please consider the following example.

yields an RGAyields an RGA

Note that since the element gNote that since the element g1212(s) is zero, the input m(s) is zero, the input m22 does not does not have an effect on the output yhave an effect on the output y11, however, the input m, however, the input m11 does does influence the output yinfluence the output y22 as can be seen due to the fact that the g as can be seen due to the fact that the g2121 element is nonzero. Upsets in Loop 1 requiring action by melement is nonzero. Upsets in Loop 1 requiring action by m11 would would have to also be handled by the controller of Loop 2. So, even have to also be handled by the controller of Loop 2. So, even though the RGA is ideal, Loop 2 would be at a disadvantage. though the RGA is ideal, Loop 2 would be at a disadvantage. Thus, in deciding on loop pairing, one should distinguish between Thus, in deciding on loop pairing, one should distinguish between ideal RGAs produced from diagonal or triangular transfer function ideal RGAs produced from diagonal or triangular transfer function matrices.matrices.

14

4

13

3

01

1

)(

ss

ssG

10

01


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• RULE RULE #1#1

Pair input and output variables that have positive Pair input and output variables that have positive RGA elements closest to 1.0.RGA elements closest to 1.0.

Consider the following examples to demonstrate Consider the following examples to demonstrate this rule.this rule.

For a 2x2 system with output variables yFor a 2x2 system with output variables y11 and y and y22, to be , to be paired with mpaired with m11 and m and m22

If the RGA is…If the RGA is…

Then it is recommended to pair mThen it is recommended to pair m11 with y with y11 and m and m22 with y with y22, , which is quite often referred to a the 1-1/2-2 pairing.which is quite often referred to a the 1-1/2-2 pairing.

8.02.0

2.08.0


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Now, consider the 2x2 system whose transfer matrix is:Now, consider the 2x2 system whose transfer matrix is:

In this case, a 1-1/2-2 pairing is preferred as to avoid In this case, a 1-1/2-2 pairing is preferred as to avoid pairing on a negative RGA element. Usually, we will try to pairing on a negative RGA element. Usually, we will try to avoid pairing on RGA elements greater than 1, but pairing avoid pairing on RGA elements greater than 1, but pairing on negative RGA elements is worse.on negative RGA elements is worse.

Recall the Wood and Berry distillation column example we Recall the Wood and Berry distillation column example we saw on Slide 65, it’s RGA issaw on Slide 65, it’s RGA is::

5.15.0

5.05.1

21

12 In this case, it In this case, it is desirable for is desirable for a 1-1/2-2 a 1-1/2-2 pairingpairing


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On the other hand, for the 2x2 systems whose RGA isOn the other hand, for the 2x2 systems whose RGA is

yy1 1 should be paired with mshould be paired with m22 and y and y22 should be paired with should be paired with mm11, this is referred to as 1-2/2-1 pairing. (as the , this is referred to as 1-2/2-1 pairing. (as the elements 1-2,2-1 are closer to a value of 1 and all elements 1-2,2-1 are closer to a value of 1 and all elements in the RGA are positive.)elements in the RGA are positive.)

3.07.0

7.03.0


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Let’s consider the following 3x3 matrix:Let’s consider the following 3x3 matrix:

The same general guidelines, we applied to the 2x2 The same general guidelines, we applied to the 2x2 systems can also be applied here. It can be seen that systems can also be applied here. It can be seen that although the diagonal elements are all greater than 1, although the diagonal elements are all greater than 1, the other elements are all negative, suggesting that a the other elements are all negative, suggesting that a 1-1/2-2/3-3 pairing would be preferable.1-1/2-2/3-3 pairing would be preferable.

52.123.029.0

22.088.166.0

3.065.095.1


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Niederlinski IndexNiederlinski Index

Pairing Rule #1 is usually sufficient in Pairing Rule #1 is usually sufficient in most cases, it is often necessary to use most cases, it is often necessary to use this rule in conjunction with the this rule in conjunction with the theorem found on the next slide theorem found on the next slide developed by Niederlinski and later developed by Niederlinski and later modified by Grosdidier et al. This modified by Grosdidier et al. This theorem is especially useful if the theorem is especially useful if the system is 3x3 or larger.system is 3x3 or larger.

NIEDERLINSKI INDEX


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Consider the Consider the n x nn x n multivariable system whose input- multivariable system whose input-output variables have been paired youtput variables have been paired y11-u-u11, y, y22-u-u22…..y…..ynn-u-unn, , resulting in a transfer function model of the form:resulting in a transfer function model of the form:

..

y(s)=G(s) u(s)y(s)=G(s) u(s)

Let each element of G(s), gLet each element of G(s), gijij(s) be,(s) be,

1.1.Rational, andRational, and

2.2.Open-loop stableOpen-loop stable

NIEDERLINSKI INDEX


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Let Let nn individual feedback controllers (which have integral individual feedback controllers (which have integral action) be designed for each loop so that each one of action) be designed for each loop so that each one of the resulting the resulting nn feedback loops is stable when all of the feedback loops is stable when all of the other n-1 loops are open.other n-1 loops are open.

Under closed-loop conditions in all n loops, the Under closed-loop conditions in all n loops, the multivariable will be unstable for all possible values of multivariable will be unstable for all possible values of controller parameters if the Niederlinski Index N defined controller parameters if the Niederlinski Index N defined below is negative.below is negative.

0)0(

)0(

1

ii

n

ig

GN (Eq. N)

On the following slides there are important points to help us use this result properly.


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Important Points for us to consider:Important Points for us to consider:1.The result is both necessary and sufficient for 2x2 1.The result is both necessary and sufficient for 2x2

systems; for higher dimensional systems, it only systems; for higher dimensional systems, it only provides sufficient conditions (provides sufficient conditions (ifif Equation N holds it is Equation N holds it is definitely unstable, but definitely unstable, but ifif Eq. N does Eq. N does notnot hold, the hold, the system may or may not be unstable: the stability will be system may or may not be unstable: the stability will be dictated by the values taken by the controller dictated by the values taken by the controller parameters).parameters).

2.For 2x2 systems the Niederlinski index becomes2.For 2x2 systems the Niederlinski index becomes

where where ζ ζ defined as follows as defined as follows as

seen on Slide 57seen on Slide 57

NIEDERLINSKI INDEX

1N 12 21

11 22

K KK K


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2. For a 2x2 system with a negative relative gain, 2. For a 2x2 system with a negative relative gain, ζ ζ >1, the >1, the Niederlinski index is always negative; hence Niederlinski index is always negative; hence 2x2 systems 2x2 systems paired with negative relative gains are ALWAYS paired with negative relative gains are ALWAYS structurally unstable.structurally unstable.

3. This theorem is designed for systems with rational transfer 3. This theorem is designed for systems with rational transfer function elements, therefore, this technically excludes function elements, therefore, this technically excludes systems containing time-delays. However, since Eq.N systems containing time-delays. However, since Eq.N depends on Steady State gains (s=0, therefore, the gains depends on Steady State gains (s=0, therefore, the gains are independent of time-delays). Due to this fact, the results are independent of time-delays). Due to this fact, the results of this theorem also provide important information about of this theorem also provide important information about time-delay systems as well, but is not very rigorous. time-delay systems as well, but is not very rigorous. USE USE CAUTION WHEN APPLYING Eq.N TO SYSTEMS WITH CAUTION WHEN APPLYING Eq.N TO SYSTEMS WITH TIME DELAYS.TIME DELAYS.

NIEDERLINSKI INDEX


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•RULE #2RULE #2

Any loop pairing is Any loop pairing is unacceptable if it leads to a unacceptable if it leads to a control system configuration control system configuration for which the Niederlinski for which the Niederlinski Index is negative.Index is negative.


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Summary of using RGA for Loop PairingSummary of using RGA for Loop Pairing

1.1. Given the transfer matrix G(s), obtain the steady-state Given the transfer matrix G(s), obtain the steady-state gain matrix K=G(0), and from this obtain the RGA, Λ, gain matrix K=G(0), and from this obtain the RGA, Λ, also calculate the determinant of the K and the also calculate the determinant of the K and the product of the elements on the main diagonalproduct of the elements on the main diagonal

2.2. Use Rule #1 to obtain tentative loop pairing Use Rule #1 to obtain tentative loop pairing suggestions from the RGA by pairing the suggestions from the RGA by pairing the positivepositive elements which are closest to one.elements which are closest to one.

3.3. Use the Niederlinski condition (Eq. N) to verify the Use the Niederlinski condition (Eq. N) to verify the stability status of the of the control configuration stability status of the of the control configuration obtained using Step 2, if the selected pairing is obtained using Step 2, if the selected pairing is unacceptable, choose anotherunacceptable, choose another..


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•Applying Loop Pairing RulesApplying Loop Pairing RulesLoop Pairing Example 1:Loop Pairing Example 1: Calculate the RGA for the system Calculate the RGA for the system whose steady-state gain matrix is given below and investigate whose steady-state gain matrix is given below and investigate the loop pairing suggested upon applying Rule #1.the loop pairing suggested upon applying Rule #1.

51 1

31

1 13

11 1

3

K = G(0) =K = G(0) =


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First, we need to take the inverse of this matrix, then First, we need to take the inverse of this matrix, then take the transpose of this matrix to obtain R, being:take the transpose of this matrix to obtain R, being:

The next step is to determine the RGA by multiplying the The next step is to determine the RGA by multiplying the elements of the K and R matrices.elements of the K and R matrices.

15.45.4

5.415.4

5.45.410

35.45.4

5.435.4

5.45.46

R


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Rule Rule #1 would suggest a 1-1,2-2,3-3 pairing#1 would suggest a 1-1,2-2,3-3 pairingTo calculate the Niederlinski Index we need to find :To calculate the Niederlinski Index we need to find :• The determinant of the K matrix which is :|K|=-0.148The determinant of the K matrix which is :|K|=-0.148• The product of the main diagonal which is :The product of the main diagonal which is :

It is clear that when the determinant is divided by the It is clear that when the determinant is divided by the product of the elements of the main diagonal it will product of the elements of the main diagonal it will yield a negative number which leads to a… yield a negative number which leads to a…

NEGATIVE NEGATIVE NIEDERLINSKI INDEX which NIEDERLINSKI INDEX which violatesviolates Rule #2.Rule #2.

27

5

3

1

3

1

3

5

1

n

iiiK


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This example provides a situation where the pairing This example provides a situation where the pairing suggested by Rule #1 is disqualified by Rule #2. Due suggested by Rule #1 is disqualified by Rule #2. Due to this fact, we need to investigate another loop to this fact, we need to investigate another loop pairing. Let’s try the possible pairing of 1-1,2-3,3-2, pairing. Let’s try the possible pairing of 1-1,2-3,3-2, which would give a RGA of:which would give a RGA of:

5.415.4

15.45.4

5.45.410


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The new K is:The new K is:

It is clear that the element in 2-2 has been It is clear that the element in 2-2 has been interchanged with the element 2-3 and the interchanged with the element 2-3 and the element 3-3 has been interchanged with the old element 3-3 has been interchanged with the old element 2-2.element 2-2.

13

11

3

111

113

5

)0(GK


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We need to calculate the determinant and product of the We need to calculate the determinant and product of the elements of the main diagonal of the new matrix K:elements of the main diagonal of the new matrix K:

|K|=0.1481 while the product of the elements is equal to |K|=0.1481 while the product of the elements is equal to 5/3.5/3.

Therefore, the Niederlinski Index isTherefore, the Niederlinski Index is

Clearly, this Niederlinski Index is positive, so we Clearly, this Niederlinski Index is positive, so we come to the conclusion that this system is no come to the conclusion that this system is no longer structurally unstable.longer structurally unstable.

03/5

148.0

1

n

iiiK

KN


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Loop Pairing Example 2:Loop Pairing Example 2: Consider the system with the Consider the system with the steady state gain matrix as seen below steady state gain matrix as seen below

• The determinant of this matrix is 0.53.The determinant of this matrix is 0.53.

The RGA is :The RGA is :

132

121.0

1.011

)0(GK

59.361.502.3

89.102.313.0

7.059.389.1


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From the RGA seen, there is only one feasible pairing, From the RGA seen, there is only one feasible pairing, because all of the other pairings violate Rule 2. The because all of the other pairings violate Rule 2. The only feasible pairing is a 1-1,2-2,3-3 pairing, but you only feasible pairing is a 1-1,2-2,3-3 pairing, but you will notice that this pairing violates Rule 1, as the RGA will notice that this pairing violates Rule 1, as the RGA element 1-1 is negative, but according to the element 1-1 is negative, but according to the Niederlinski Theorem this system would NOT be Niederlinski Theorem this system would NOT be structurally unstable.structurally unstable.

If the first loop is opened (the mIf the first loop is opened (the m11, y, y11 elements dropped elements dropped from the process model) the new steady-state gain from the process model) the new steady-state gain matrix relating the 2 remaining input variables with the matrix relating the 2 remaining input variables with the 2 remaining output variables is:2 remaining output variables is:

13

12~

K


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It is easy to see that if the first loop is open, the It is easy to see that if the first loop is open, the Niederlinski Index of the remaining two loops would be Niederlinski Index of the remaining two loops would be negative, indicating that the system would be negative, indicating that the system would be structurally unstable. As a consequence, this system structurally unstable. As a consequence, this system will only be stable if all loops are CLOSED, such a will only be stable if all loops are CLOSED, such a system is said to have system is said to have a low degree of integritya low degree of integrity..

There are some examples of higher order There are some examples of higher order systems where it is possible to pair on negative systems where it is possible to pair on negative RGA values and still have a structurally stable RGA values and still have a structurally stable system (this is NOT possible for 2x2 systems).system (this is NOT possible for 2x2 systems).


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• Summary of Loop Pairing using RGASummary of Loop Pairing using RGAAlways pair on positive RGA elements that are Always pair on positive RGA elements that are

the closest to 1 in value. Thereafter, use the the closest to 1 in value. Thereafter, use the Niederlinski Index to check if the resulting Niederlinski Index to check if the resulting configuration is structurally stable. Whenever configuration is structurally stable. Whenever possible, try to avoid pairing on negative RGA possible, try to avoid pairing on negative RGA elements; for 2x2 systems such pairings elements; for 2x2 systems such pairings always always lead to unstable configurations, while lead to unstable configurations, while for systems of higher dimension, they can for systems of higher dimension, they can lead to a condition which, at best has a low lead to a condition which, at best has a low degree of integrity.degree of integrity.


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Quiz Quiz #6#6

• What does a positive Niederlinski Index What does a positive Niederlinski Index indicate?indicate?

• According to Rule 1, should elements be According to Rule 1, should elements be paired on positive or negative elements?paired on positive or negative elements?

• In what case should a favourable pairing In what case should a favourable pairing from Rule 1 be discarded?from Rule 1 be discarded?


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•Loop Pairing for Non-linear systems.Loop Pairing for Non-linear systems.Example 1- RGA and Loop pairing of non-linear systems.Example 1- RGA and Loop pairing of non-linear systems. The process shown is a blending process, the objective is The process shown is a blending process, the objective is to control both the total product flow rate (F) and the to control both the total product flow rate (F) and the product composition (x) as calculated in terms of the mole product composition (x) as calculated in terms of the mole fraction of A in the blend. Obtain the RGA for the system fraction of A in the blend. Obtain the RGA for the system and suggest which input variable to pair with each output.and suggest which input variable to pair with each output.

LOOP PAIRING FOR NON-LINEAR SYSTEMS

BlendingBlending

FFCC

GGCC

AnalyzAnalyzererFFAA

FFBB

FF

xx


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Total Mass BalanceTotal Mass Balance::

Mass Balance on Component AMass Balance on Component A

Solution:Solution: Notice that for this system, the two output variables Notice that for this system, the two output variables are F and x, and the input variable are Fare F and x, and the input variable are FAA and F and FBB, from now , from now on, we will refer to the input variables as mon, we will refer to the input variables as m11 and m and m22 for the for the input feeds of A and B respectively.input feeds of A and B respectively.

Therefore, our Overall Mass Balance becomesTherefore, our Overall Mass Balance becomes

(Eq 1) (which is (Eq 1) (which is linear)linear)

And the Component A Mass Balance becomesAnd the Component A Mass Balance becomes

(Eq 2) (which is NON-(Eq 2) (which is NON-linear)linear)

21 mmF

21

1

mm

mx

FFF BA

BA

A

FF

Fx


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Since this is a 2x2 system, we only need to obtain the Since this is a 2x2 system, we only need to obtain the (1,1) element of the RGA given by:(1,1) element of the RGA given by:

Recall:Recall:

To calculate the numerator, take the derivative of the first To calculate the numerator, take the derivative of the first equation with both loops open with respect to mequation with both loops open with respect to m11 , , yieldingyielding

closed loop second1

open loopsboth 1

m

F

m

F

1 open loopsboth 1

m

F


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In order to calculate the denominator, loop 2 must be closed, In order to calculate the denominator, loop 2 must be closed, and we will have to determine the value of mand we will have to determine the value of m22 so that when a so that when a change occurs in mchange occurs in m11, x will return to its steady state value (x*)., x will return to its steady state value (x*).

To determine the value of mTo determine the value of m22 in this case, we must set x=x* in in this case, we must set x=x* in Equation 2 and solve for mEquation 2 and solve for m22 in terms of m in terms of m11 and x*, the result is: and x*, the result is:

When loop 2 is closed, the mole fraction of the the component When loop 2 is closed, the mole fraction of the the component A in the output at x*, mA in the output at x*, m22 will respond to changes in m will respond to changes in m11, to , to determine the relationship, we have to substitute the value of determine the relationship, we have to substitute the value of mm22 above into the Overall Mass Balance (Equation 1) yielding: above into the Overall Mass Balance (Equation 1) yielding:

12 1

mm = -m

x*

1 11 1

m mF=m + -m or F=

x* x*


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The next step is to differentiate the expression of F obtained in The next step is to differentiate the expression of F obtained in the last step with respect to mthe last step with respect to m11 yielding: yielding:

If the numerator and denominator are substituted into the If the numerator and denominator are substituted into the statement for the relative gain (λ), we get:statement for the relative gain (λ), we get:

For a 2x2 matrix recall that the RGA is given by…For a 2x2 matrix recall that the RGA is given by…

Therefore the RGA of this system is:Therefore the RGA of this system is:

second loop1 closed

1*

Fm x

**/1

1x

x

1

1

**1

*1*

xx

xx Where x* is Where x* is the desired the desired mole fraction mole fraction of A in the of A in the product.product.


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Some things to consider about these results:Some things to consider about these results:

1.1. The RGA is dependent on the steady-state value of x* The RGA is dependent on the steady-state value of x* desired for the composition of the blend; it is NOT desired for the composition of the blend; it is NOT constant as it was in the linear systems we dealt with constant as it was in the linear systems we dealt with before.before.

2.2. It is implied that the recommended loop-pairing will It is implied that the recommended loop-pairing will depend on the steady-state operating point.depend on the steady-state operating point.

3.3. Due to the fact that x* is a mole fraction, it is bounded Due to the fact that x* is a mole fraction, it is bounded between 0 and 1 (0 between 0 and 1 (0 << x* x*<< 1) and therefore, none of the 1) and therefore, none of the elements in the RGA will be negative. The implication of elements in the RGA will be negative. The implication of this fact, is that in the worst possible scenario is that this fact, is that in the worst possible scenario is that there will be large interactions between the input there will be large interactions between the input variables if the input and output variables are paired variables if the input and output variables are paired improperly, but the system will not become unstableimproperly, but the system will not become unstable ..


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A loop pairing strategy for this system is as follows:A loop pairing strategy for this system is as follows:

1. If x* is close to 1, the first implication is that m1. If x* is close to 1, the first implication is that m11 is larger than is larger than mm2 .2 . If we look at the RGA, the following pairing would be If we look at the RGA, the following pairing would be recommended, F-mrecommended, F-m11, x-m, x-m22.(ie. The larger flow rate is used to .(ie. The larger flow rate is used to control the total flow rate out and the smaller flow rate is used control the total flow rate out and the smaller flow rate is used to control the composition.)to control the composition.)

2. This is the most reasonable pairing because: when the product 2. This is the most reasonable pairing because: when the product composition is close to one (x* close to 1), we have almost pure composition is close to one (x* close to 1), we have almost pure A coming out of the system, and so we can modify the flow rate A coming out of the system, and so we can modify the flow rate out quite easily by changing the flow rate of A into the blending out quite easily by changing the flow rate of A into the blending without changing the composition of the blend significantly. without changing the composition of the blend significantly. Similarly if we alter the composition, the additional small Similarly if we alter the composition, the additional small amounts of material B will not have a significant impact on the amounts of material B will not have a significant impact on the flow rate of the blend out of the system. Thus, the flow flow rate of the blend out of the system. Thus, the flow controller will not interact strongly with the composition controller will not interact strongly with the composition controller if the pairing : F-mcontroller if the pairing : F-m11 and x-m and x-m22 is used, but if the is used, but if the opposite pairing was used, the interaction would be severe.opposite pairing was used, the interaction would be severe.


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3.3. When the steady-state product composition is closer to When the steady-state product composition is closer to 0, the RGA suggests that the loop pairing stated in 0, the RGA suggests that the loop pairing stated in point 2 should be switched, i.e. mpoint 2 should be switched, i.e. m22 (F (FBB) should be paired ) should be paired with the outgoing flow rate (F-mwith the outgoing flow rate (F-m22) and m) and m11(F(FAA) should be ) should be paired with the composition (x-mpaired with the composition (x-m11). If you analyze the ). If you analyze the effects that each variable has as done in point 2, you effects that each variable has as done in point 2, you will see that the physics of this system dictates such a will see that the physics of this system dictates such a pairing.pairing.

4. An interesting situation arises when the composition 4. An interesting situation arises when the composition (x*) is equal to 0.5 (x*=0.5). In this case it does not (x*) is equal to 0.5 (x*=0.5). In this case it does not matter which input variable is used to control which matter which input variable is used to control which output variable. The observed interactions will be equal output variable. The observed interactions will be equal and significant in either case.and significant in either case.


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Loop Pairing for Systems with Pure Loop Pairing for Systems with Pure Integrator Modes:Integrator Modes:

Since we have seen that interaction analysis using the Since we have seen that interaction analysis using the RGA is carried out using steady-state information, an RGA is carried out using steady-state information, an interesting situation occurs when dealing with systems interesting situation occurs when dealing with systems that contain pure integrator elements (i.e. if s was set that contain pure integrator elements (i.e. if s was set to zero, an element would become undefined), since to zero, an element would become undefined), since pure integrator elements show no steady-state. Several pure integrator elements show no steady-state. Several suggestions are available to deal with this problem, but suggestions are available to deal with this problem, but we will use the industrial application of the a de-we will use the industrial application of the a de-ethanizer to demonstrate one method to recommend a ethanizer to demonstrate one method to recommend a loop pairing strategy. loop pairing strategy.

LOOP PAIRING FOR PURE INTEGRATOR MODES


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Pure Integrator System Example 1 - Pure Integrator System Example 1 - The transfer function The transfer function for a 2x2 subsystem extracted from a larger system for an for a 2x2 subsystem extracted from a larger system for an industrial de-ethanizer is given below. Obtain the RGA industrial de-ethanizer is given below. Obtain the RGA and use it to recommend loop pairings.and use it to recommend loop pairings.

Solution-Solution- Our usual course of action to determine the RGA Our usual course of action to determine the RGA is to normally calculate the K matrix which is G(s) when is to normally calculate the K matrix which is G(s) when s=0. Unfortunately, we can see that elements (1,2) and s=0. Unfortunately, we can see that elements (1,2) and (2,2) contain pure integrator elements represented by 1/s, (2,2) contain pure integrator elements represented by 1/s, which if we set s=0 would yield an undefined number.which if we set s=0 would yield an undefined number.

ssss

s

s

e

s

e

sG

ss

36.0

)15.6)(110)(127(

)1182(0385.0

3120

318.1

)(

45.2


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Let’s make the substitution,Let’s make the substitution,

If I is substituted into G(s), K becomes:If I is substituted into G(s), K becomes:

The relative gain parameter (λ)The relative gain parameter (λ)

1I

s

0

I1.318

lim lim 30.038 0.36I

s IK

1lim

0.038 x 0.3331

1.138 x 0.36

I II


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We can see that in the λ term the We can see that in the λ term the IIs cancel out, so we s cancel out, so we obtainobtain

λ=0.97λ=0.97Therefore the resulting RGA isTherefore the resulting RGA is

It is quite obvious that it is desirable to pain in a 1-It is quite obvious that it is desirable to pain in a 1-1,2-2 fashion.1,2-2 fashion.

If you encounter a system in which there the If you encounter a system in which there the IIs do not s do not cancel out, you will have to consult another reference.cancel out, you will have to consult another reference.

97.003.0

03.097.0


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• Loop Pairing for Non-Square SystemsLoop Pairing for Non-Square Systems

In the previous slides, we have discussed how obtain In the previous slides, we have discussed how obtain RGAs and how to use them for input/output pairings RGAs and how to use them for input/output pairings when the process has an when the process has an equal equal number of input and number of input and output variables (output variables (square systemssquare systems).).

There are some cases, where multivariable systems do There are some cases, where multivariable systems do not have the same number of input and output not have the same number of input and output variables, these are referred to as variables, these are referred to as non-square non-square systems.systems.

The most obvious problem with non-square systems is The most obvious problem with non-square systems is that after the input/output pairing, there will always be that after the input/output pairing, there will always be either an input or an output that is not paired (a either an input or an output that is not paired (a residual ).residual ).

LOOP PAIRING FOR NON-SQUARE SYSTEMS


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With non-square systems, we are faced With non-square systems, we are faced with two questions.with two questions.

1) Which input/output variables should be 1) Which input/output variables should be paired together?paired together?

2) Which variables are redundant and 2) Which variables are redundant and which take an active role in control?which take an active role in control?


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Classifying Non-Square SystemsClassifying Non-Square SystemsWe have 2 types of non-square systems, We have 2 types of non-square systems,

1)1) UnderdefinedUnderdefined-- there are fewer input variables than there are fewer input variables than output variables.output variables.

2)2) OverdefinedOverdefined-- there are more input variables than there are more input variables than output variables.output variables.

Thus, a multivariable system withThus, a multivariable system with nn output andoutput and mm input variables, whose transfer function matrix input variables, whose transfer function matrix willwill therefore betherefore be n x mn x m in dimension is:in dimension is:

UNDERDEFINED if m<nUNDERDEFINED if m<n and and OVERDEFINED if OVERDEFINED if m>nm>n


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B 4

Underdefined SystemsUnderdefined Systems

As seen in the system above, there are less inputs As seen in the system above, there are less inputs mm than than there are outputs there are outputs nn, thus is defined as an , thus is defined as an underdefined system.underdefined system.

m=the number of inputs = 2m=the number of inputs = 2

n=the number of outputs = 4n=the number of outputs = 4

m inputsm inputs n outputsn outputs

m<nm<n


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Underdefined SystemsUnderdefined Systems

The main issue with underdefined systems is that The main issue with underdefined systems is that not all outputs can be controlled, since we do not all outputs can be controlled, since we do not have enough input variables.not have enough input variables.

The loop pairing is easier if we make the following The loop pairing is easier if we make the following considerationconsideration

By economic considerations, or other such By economic considerations, or other such means, decide which m of the n output means, decide which m of the n output variables are the most important, these m variables are the most important, these m output variables should be paired with the m output variables should be paired with the m input variables; the less important (n-m) output input variables; the less important (n-m) output variables will not be under any control.variables will not be under any control.


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B 4

Overdefined SystemsOverdefined Systems

m inputsm inputs n n outpuoutputsts

As seen in the system above, there are less inputs As seen in the system above, there are less inputs mm than than there are outputs there are outputs nn, thus is defined as an , thus is defined as an underdefined system.underdefined system.

m=the number of inputs = 3m=the number of inputs = 3

n=the number of outputs = 2n=the number of outputs = 2

m>nm>n


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Overdefined SystemsOverdefined SystemsDeciding the loop pairing of overdefined systems presents Deciding the loop pairing of overdefined systems presents a real challenge. In this case, there is an excess of input a real challenge. In this case, there is an excess of input variables, therefore we can achieve arbitrary control of variables, therefore we can achieve arbitrary control of the fewer output variables in the fewer output variables in more than one way.more than one way.

The situation we are faced with is as follows: since there The situation we are faced with is as follows: since there are are mm input variables to control input variables to control nn output variable output variable (m>n)(m>n), , there are many more input variables to choose from in there are many more input variables to choose from in pairing the inputs and the outputs, and therefore, there pairing the inputs and the outputs, and therefore, there will be several different square subsystems from which will be several different square subsystems from which the pairing is possible. There are possible square the pairing is possible. There are possible square subsystems.subsystems.

m

n

m m!= (m-n)!

n!n Recall Recall

that:that:


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The Variable Pairing Strategy for Overdefined The Variable Pairing Strategy for Overdefined Systems is:Systems is:

1. Determine all of the subsystems from a given 1. Determine all of the subsystems from a given model.model.

2.Obtain the RGAs for each of the square subsystems.2.Obtain the RGAs for each of the square subsystems.

3.Examine the RGAs and chose the best subsystem on the 3.Examine the RGAs and chose the best subsystem on the basis of the overall character of its RGA (in terms of how basis of the overall character of its RGA (in terms of how close it is to the ideal RGA).close it is to the ideal RGA).

4. After determining the best subsystem, use its RGA to 4. After determining the best subsystem, use its RGA to decide which input variable within its subsystem to pair decide which input variable within its subsystem to pair with each output variable.with each output variable.

n

m


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Loop Pairing in the Absence of Process ModelsLoop Pairing in the Absence of Process Models

Sometimes, situations arise where a process model is not Sometimes, situations arise where a process model is not available, but it is still possible to determine their RGAs available, but it is still possible to determine their RGAs from experimental data. There are 2 approaches as from experimental data. There are 2 approaches as follows:follows:

Approach 1-Approach 1- Experimentally determine the steady-state Experimentally determine the steady-state gain matrix K, by implementing a step change in the gain matrix K, by implementing a step change in the process input variables, one at a time, and observing the process input variables, one at a time, and observing the ultimate change in each output variable.ultimate change in each output variable.Let Let yy1j1j be the observed change in the value of the output be the observed change in the value of the output variable 1 in response to a change of variable 1 in response to a change of m mjj in the j in the jth th input input variable mvariable mj j ; then , by definition of the steady-state gain:; then , by definition of the steady-state gain:

j

jj m

yk

1

1

LOOP PAIRING IN THE ABSENCE OF PROCESS MODELS


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In general, the steady-state gain between In general, the steady-state gain between the ithe ithth variable and the j variable and the jthth variable will be variable will be given bygiven by

Thus, the elements of the K matrix can be Thus, the elements of the K matrix can be

calculated, and once the K matrix is known, calculated, and once the K matrix is known, it is easy to calculate the RGA.it is easy to calculate the RGA.

j

ijij m

yk


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Approach 2-Approach 2- It is possible to determine each element of It is possible to determine each element of the RGA directly from experimentation.the RGA directly from experimentation.As you may recall, each RGA element (λAs you may recall, each RGA element (λijij) can be obtained by ) can be obtained by performing two experiments. The first experiment determines performing two experiments. The first experiment determines the open-loop steady-state gain by measuring the response of the open-loop steady-state gain by measuring the response of yyii to input m to input mjj , when all other loops are open. In the second , when all other loops are open. In the second experiment, all other loops are closed – using PI controllers to experiment, all other loops are closed – using PI controllers to ensure that there will be no steady-state offsets – and the ensure that there will be no steady-state offsets – and the response of yresponse of yii to input m to input mj j is redetermined. By definition, the is redetermined. By definition, the ratio of these two gains is the desired relative gain element ratio of these two gains is the desired relative gain element ( λ( λij ij ).).

The second approach is more time consuming, and involves too The second approach is more time consuming, and involves too many upsets to the process; for these reasons it is not many upsets to the process; for these reasons it is not desirable in practice. Therefore, the first approach is preferred.desirable in practice. Therefore, the first approach is preferred.


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Final Comments on the RGAFinal Comments on the RGA

1.The RGA requires only steady-state process information, 1.The RGA requires only steady-state process information, it is therefore easy to calculate and easy to use.it is therefore easy to calculate and easy to use.

2. The main criticism of the RGA is that the RGA only 2. The main criticism of the RGA is that the RGA only provides information about the steady-state interactions provides information about the steady-state interactions within a process systems, and therefore, dynamic within a process systems, and therefore, dynamic factors are not taken into account by the RGA analysis. factors are not taken into account by the RGA analysis.

3. The RGA only suggests input/output pairing such that the 3. The RGA only suggests input/output pairing such that the interaction effects are minimized; it provides no interaction effects are minimized; it provides no guidance about other factors which may influence the guidance about other factors which may influence the pairing.pairing.


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Other Factors Influencing the Choice of Loop PairingOther Factors Influencing the Choice of Loop Pairing

1.1.Constraints on the input variable:Constraints on the input variable: It is possible that It is possible that the best pairing obtained from the RGA will result in a the best pairing obtained from the RGA will result in a choice of input variable for ychoice of input variable for y ii that is severely limited by that is severely limited by some constraint (ex. maximum feed concentration) in a some constraint (ex. maximum feed concentration) in a way that it can not carry out the assigned control task.way that it can not carry out the assigned control task.

2.2.The presence of a time-delay, inverse-response, or The presence of a time-delay, inverse-response, or other slow dynamics in the best RGA pairing:other slow dynamics in the best RGA pairing: Since Since the RGA is based on steady-state information, the RGA is based on steady-state information, sometimes, the best RGA pairing results can result in sometimes, the best RGA pairing results can result in very slow closed-loop response if there are long time very slow closed-loop response if there are long time delays, significant inverse response or large time delays, significant inverse response or large time constants. If this is the case, it would be more suitable to constants. If this is the case, it would be more suitable to pair on more unfavourable RGA elements if the slow pair on more unfavourable RGA elements if the slow elements could be omitted to improve system elements could be omitted to improve system performance.performance.


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Other Factors Influencing the Choice of Loop PairingOther Factors Influencing the Choice of Loop Pairing

33. Timescale Decoupling of Loop Dynamics. Timescale Decoupling of Loop Dynamics: Often : Often timescale issues arisetimescale issues arise that can influence the choice of loop that can influence the choice of loop pairing. For example, in a 2x2 system, it may be that for a pairing. For example, in a 2x2 system, it may be that for a given pairing, the RGA indicates a serious loop interaction. given pairing, the RGA indicates a serious loop interaction. However, if at the same time, one of the loops responds a However, if at the same time, one of the loops responds a great deal faster than the other, there can be a timescale great deal faster than the other, there can be a timescale decoupling of the loops. This can occur if the fast loop decoupling of the loops. This can occur if the fast loop responds so fast that the effect on the slow loop seems to be responds so fast that the effect on the slow loop seems to be a constant disturbance, in opposition, the slow loop does not a constant disturbance, in opposition, the slow loop does not respond at all to the high-frequency disturbances coming from respond at all to the high-frequency disturbances coming from the fast loop. This indicates that loops with large differences the fast loop. This indicates that loops with large differences in closed-loop response times can be paired even when the in closed-loop response times can be paired even when the RGA indicates that the pairing is unfavourable.RGA indicates that the pairing is unfavourable.


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Quiz#7Quiz#7

• What system information is needed to construct What system information is needed to construct the RGA?the RGA?

• What is the difference between a underdefined What is the difference between a underdefined and overdefined system?and overdefined system?

• What is a difficulty in overdefined systemsWhat is a difficulty in overdefined systems??


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Controller Design Procedure-Multiloop Controller Controller Design Procedure-Multiloop Controller DesignDesign

There are 2 stages in the design of multiple single-loop There are 2 stages in the design of multiple single-loop controllers for multivariable systems:controllers for multivariable systems:

•Judicious choice of loop pairingJudicious choice of loop pairing•Controller tuning for each individual loopController tuning for each individual loop

We have discussed this first point a great deal in the past We have discussed this first point a great deal in the past slides, this should signify importance of the choice of loop slides, this should signify importance of the choice of loop pairing in controller design.pairing in controller design.Now, we must address the issue of tuning the individual Now, we must address the issue of tuning the individual controllers.controllers.


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It should be obvious that when the RGA for a It should be obvious that when the RGA for a process is close to ideal (ie. λprocess is close to ideal (ie. λijij is very close to 1) is very close to 1) that the multiloop controllers are very likely to that the multiloop controllers are very likely to function very well if they are designed properly.function very well if they are designed properly.

However, when the RGA indicates strong However, when the RGA indicates strong interactions for the chosen loop pairing (ie. λinteractions for the chosen loop pairing (ie. λijij is is very large or negative) the controller is not likely very large or negative) the controller is not likely to perform well even if it is tuned well.to perform well even if it is tuned well.


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•Controller Tuning for Multiloop SystemsController Tuning for Multiloop Systems

The main challenge in controller tuning is the interactions The main challenge in controller tuning is the interactions between the different control loops of a multi-loop system. Due between the different control loops of a multi-loop system. Due to this fact, it can be risky to adopt the obvious strategy of to this fact, it can be risky to adopt the obvious strategy of tuning each controller individually without considering the tuning each controller individually without considering the other controllers and hoping that when all the loops are closed other controllers and hoping that when all the loops are closed that the overall system performance will be adequate.that the overall system performance will be adequate.

The procedure that is normally followed in practice is the The procedure that is normally followed in practice is the following:following:

1.With the other loops on manual control, tune each control 1.With the other loops on manual control, tune each control loop independently until satisfactory closed-loop performance loop independently until satisfactory closed-loop performance is obtained.is obtained.2.Restore all the controllers to joint operation under automatic 2.Restore all the controllers to joint operation under automatic control and readjust the tuning parameters until the overall control and readjust the tuning parameters until the overall closed-loop performance is satisfactory in all the loops.closed-loop performance is satisfactory in all the loops.


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When the interactions between the control loops are not When the interactions between the control loops are not too significant, the procedure mentioned before can be too significant, the procedure mentioned before can be quite useful. However, for systems with significant quite useful. However, for systems with significant interactions, the readjustment of the tuning in Step 2 can interactions, the readjustment of the tuning in Step 2 can be difficult and tedious. One can cut down on the amount be difficult and tedious. One can cut down on the amount of guesswork that goes into such a procedure by noting of guesswork that goes into such a procedure by noting that in almost all cases, the controllers will need to be that in almost all cases, the controllers will need to be made more conservative (ie. the controller gains will have made more conservative (ie. the controller gains will have to be reduced and the integral times increased) when all to be reduced and the integral times increased) when all the loops are closed in comparison to when all of the the loops are closed in comparison to when all of the individual controllers are operating individually, with all of individual controllers are operating individually, with all of the other loops open. The process of this changing of the the other loops open. The process of this changing of the control parameters is referred to as “detuning”.control parameters is referred to as “detuning”.


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One method of “detuning” for a 2x2 system is as follows:One method of “detuning” for a 2x2 system is as follows:

1.Use any of the single-loop tuning rules (Ziegler-Nichols, 1.Use any of the single-loop tuning rules (Ziegler-Nichols, Cohen and Coon, etc) to obtain starting values for the Cohen and Coon, etc) to obtain starting values for the individual controllers; let the controller gains be Kindividual controllers; let the controller gains be Kcici*.*.

2. These gains should be reduced using the following 2. These gains should be reduced using the following expressions that depend on the relative gain parameter λ:expressions that depend on the relative gain parameter λ:

It may still be necessary to “retune” these controllers after It may still be necessary to “retune” these controllers after they have been put in operation; however, this will not they have been put in operation; however, this will not require as much effort as if one were starting from require as much effort as if one were starting from scratch.scratch.

2

2

( ) * 1.0

* 1.0

ci

ci

ci

KK

K


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Design of Multivariable ControllersDesign of Multivariable Controllers

In the next section, we will discuss the design of true In the next section, we will discuss the design of true multivariable controllers that utilize multivariable controllers that utilize allall of the available of the available process output information jointly to determine what the process output information jointly to determine what the complete input vector complete input vector uu should be. Thus each control should be. Thus each control command from the multivariable controller will be based command from the multivariable controller will be based onon allall of the output variables, not just based on of the output variables, not just based on oneone.. In In principle, it will be possible to eliminate all of the principle, it will be possible to eliminate all of the interactions between the process variables. The objective interactions between the process variables. The objective of the next section is to present some of the principles of the next section is to present some of the principles and techniques used for designing multivariable and techniques used for designing multivariable controllers, as designing multivariable controllers is one of controllers, as designing multivariable controllers is one of the more challenging problems faced in industrial process the more challenging problems faced in industrial process control. We will start by addressing control. We will start by addressing loop decouplingloop decoupling, the , the most widely used multivariable controller technique. We most widely used multivariable controller technique. We will then address Singular Value Decomposition (SVD) will then address Singular Value Decomposition (SVD) which is a means of determining when it is structurally which is a means of determining when it is structurally unstable to apply decoupling to a system.unstable to apply decoupling to a system.

DESIGN OF MULTIVARIABLE CONTROLLERS-Introduction


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ggc1c1

ggc1c1

ggI2I2

ggI1I1

gg1111

gg1212

gg2121

gg2222

Please consider the following system:Please consider the following system:

yydd

11

yydd

22

++ --

++

--

++ ++ ++++

++

++

++

++

εε11

εε22

vv11

vv22

uu11

uu22

yy11

yy22

Figure 1-DFigure 1-D


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Let’s assume that the input/output variable pairing has Let’s assume that the input/output variable pairing has been determined to be: ybeen determined to be: y11-u-u11, y, y22-u-u2 2 … y… ynn-u-un n pairings.pairings.

Under the multiple, independent, single-loop control Under the multiple, independent, single-loop control strategy, each controller gstrategy, each controller gci ci operates according to: operates according to:

uuii=g=gcici(y(ydidi-y-yii))

OROR

uuii=g=gciciεεii

The controller The controller transfertransferfunction multiplied by function multiplied by the difference in the the difference in the set point of yset point of yii(y(ydidi) and ) and the actual ythe actual yii output output The output errorThe output error

The The difference difference between the between the desired ydesired yii and and the actual ythe actual yii output.output.


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However, a true multivariable controller must decide on However, a true multivariable controller must decide on uuii, not using only ε, not using only εii, but using the entire set of ε, but using the entire set of ε11, ε, ε22 … ε … εnn. .

Thus, the controller actions are obtained by:Thus, the controller actions are obtained by:

uu11=f=f11 ( (εε11, , εε2 2 , … , … εεnn))

uu22=f=f22 ( (εε11, , εε22 , … , … εεnn))

uu33=f=f33 ( (εε11, , εε2 2 , … , … εεnn))

……

uunn=f=fnn((εε11, , εε2 2 , … , … εεnn))

The design problem is to find the The design problem is to find the ff11(.),f(.),f22(.)…f(.)…fnn(.) (.) so that so that each of the output variable errors is driven to zero.each of the output variable errors is driven to zero.


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Decoupling:Decoupling:In Decoupling, as seen in the Figure on Slide 132, In Decoupling, as seen in the Figure on Slide 132, additional transfer function blocks are introduced additional transfer function blocks are introduced between the single-loop controllers and the process, between the single-loop controllers and the process, functioning as links between the otherwise independent functioning as links between the otherwise independent controllers. The actual control action experienced by the controllers. The actual control action experienced by the process will now contain information from all of the process will now contain information from all of the controllers. For example, a 2x2 system, whose individual controllers. For example, a 2x2 system, whose individual controller outputs are gcontroller outputs are gc1c1εε11 and g and gc2c2εε22 if the decoupling if the decoupling blocks for each loop have transfer functions of gblocks for each loop have transfer functions of gI1I1 and g and gI2I2 respectively, then the control equations will be given by:respectively, then the control equations will be given by:

uu11=g=gc1c1εε11+g+gI1 I1 (g(gc2c2εε22))

uu22=g=gc2c2εε22+g+gI2 I2 (g(gc1c1εε11))

DECOUPLING INTRODUCTION


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Decoupling IntroductionDecoupling IntroductionWe know from our discussion of input/output pairing that We know from our discussion of input/output pairing that the pairing of ythe pairing of y11-u-u11, y, y22-u-u22,…y,…ynn-u-unn couplings are desirable; it couplings are desirable; it is however the yis however the yii-u-ujj cross-couplings, by which y cross-couplings, by which yii is is influenced by uinfluenced by ujj (for all i and all j with i≠j), that are (for all i and all j with i≠j), that are undesirable: they are responsible for the control loop undesirable: they are responsible for the control loop interactions.interactions.

It is clear that any technique that eliminates the It is clear that any technique that eliminates the undesired cross-coupling will improve the performance of undesired cross-coupling will improve the performance of control systems. It is however NOT possible to ELIMINATE control systems. It is however NOT possible to ELIMINATE the cross-couplings; that is a physical impossibility since it the cross-couplings; that is a physical impossibility since it will require altering the physical nature of the system. will require altering the physical nature of the system. Consider an example of this on the following slide.Consider an example of this on the following slide.


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ItIt is not possible to stop is not possible to stop the the hot streamhot stream from from affecting the affecting the temperature of the temperature of the stirred tank, even though stirred tank, even though the main objective of this the main objective of this stream is to maintain the stream is to maintain the tank level. It is also true tank level. It is also true that we can notthat we can not prevent prevent the the cold streamcold stream from from affecting the tank level affecting the tank level even though controlling even though controlling the temperature is its the temperature is its main responsibility.main responsibility.

Cold flow Cold flow raterate

Hot flow rateHot flow rate


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GGcc GGII GGεεyydd ++ vv uu yy

--

The main objective in decoupling is to compensate for the The main objective in decoupling is to compensate for the effect of interactions as a result of cross-coupling of the effect of interactions as a result of cross-coupling of the process variables. As shown in the figure above, this can be process variables. As shown in the figure above, this can be achieved by introducing an additional transfer function “block”( achieved by introducing an additional transfer function “block”( the Interaction Compensatorthe Interaction Compensator) between the ) between the Single Loop Single Loop ControllersControllers and the process. This and the process. This Interaction Compensator, Interaction Compensator, together with the Single Loop Controllers now form the together with the Single Loop Controllers now form the multivariable decoupling controller. In the ideal case, the multivariable decoupling controller. In the ideal case, the decoupler causes the control loops to act as if they are totally decoupler causes the control loops to act as if they are totally independent of each other, reducing the tuning task so that it independent of each other, reducing the tuning task so that it will be possible to use SISO design techniques.will be possible to use SISO design techniques.

Single Single Loop Loop ControllerController

Interaction Interaction CompensatiCompensationon


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The design problem is to find the element GThe design problem is to find the element GII (the (the compensator) to satisfy one of the following compensator) to satisfy one of the following objectives.objectives.

•Dynamic Decoupling-Dynamic Decoupling- To eliminate interactions from To eliminate interactions from all all control loops, at every instant in timecontrol loops, at every instant in time•Steady-State Decoupling- Steady-State Decoupling- To only eliminate steady-To only eliminate steady-state interactions from all loops; in this case dynamic state interactions from all loops; in this case dynamic interactions are tolerated. Although this type of decoupling interactions are tolerated. Although this type of decoupling is less rigorous than this dynamic decoupling, it leads to is less rigorous than this dynamic decoupling, it leads to much simpler decoupler designs.much simpler decoupler designs.•Partial Decoupling- Partial Decoupling- To eliminate dynamic or steady-To eliminate dynamic or steady-state interactions in a subset of the control loops. This state interactions in a subset of the control loops. This focuses only on the critical loops with the strongest focuses only on the critical loops with the strongest interactions, leaving those with weak interactions to act interactions, leaving those with weak interactions to act without decoupling.without decoupling.


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Design of Ideal Decouplers - Design of Ideal Decouplers - Simplified DecouplingSimplified Decoupling

First we will consider some important aspects of the block First we will consider some important aspects of the block diagram in Figure 1-D (found on slide 132)diagram in Figure 1-D (found on slide 132)

1.1. There are two compensator blocks gThere are two compensator blocks gII11 g gII22 , one for each , one for each loop.loop.

2. There is a new notation: the controller outputs are 2. There is a new notation: the controller outputs are now vnow v11 and v and v22, while the actual control action , while the actual control action implemented on the process remains as uimplemented on the process remains as u11 and u and u22. This . This distinction is necessary because the output of the distinction is necessary because the output of the controllers and the control action to be implemented controllers and the control action to be implemented on the process no longer have to be the same.on the process no longer have to be the same.

SIMPLIFIED DECOUPLING


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3. Without the compensator, u3. Without the compensator, u11=v=v11 and u and u22=v=v2 2 and the and the process model remains process model remains yy11=g=g1111uu11+g+g1212uu22

yy22=g=g1212uu11+g+g2222uu22

The interactions persist, as uThe interactions persist, as u22 is still cross-coupled with and is still cross-coupled with and affecting yaffecting y11 through the g through the g1212 element, and u element, and u11 affects y affects y22 by by cross-coupling through gcross-coupling through g2121..

4.4. With the interaction compensator, Loop 2 is “informed” of With the interaction compensator, Loop 2 is “informed” of changes in vchanges in v11 through g through gII22, so that u, so that u22 “what the process “what the process actually feels” actually feels” is adjusted accordingly. The same process is adjusted accordingly. The same process is preformed by Loop 1 by gis preformed by Loop 1 by gII11 which adjusts uwhich adjusts u11 from from information about vinformation about v22..

SIMPLIFIED DECOUPLING


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The design question is now posed:The design question is now posed:What should gWhat should gI1 I1 and gand gI2 I2 be if the effects of loop be if the effects of loop

interactions are to be completely neutralized?interactions are to be completely neutralized?

To answer this:To answer this:

Let’s consider Loop 1 in Figure 1-D where the process Let’s consider Loop 1 in Figure 1-D where the process model is :model is :

yy11=g=g1111uu11+g+g1212uu22

yy22=g=g1212uu11+g+g2222uu22

Because of the compensators, the equations governing Because of the compensators, the equations governing the control action are:the control action are:

uu11=v=v11+g+gII11vv22

uu22=v=v22+g+gII11vv11


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If we substitute the expressions for uIf we substitute the expressions for u11 and u and u22 into the into the expressions for yexpressions for y11 and y and y22 seen on the previous slide the seen on the previous slide the system is defined as:system is defined as:

yy11= g= g1111(v(v11+g+gII11vv22) ) + g+ g1212(v(v22+g+gII11vv1 1 ))

yy22=g=g1212(v(v11+g+gII11vv22) ) +g+g2222(v(v22+g+gII11vv1 1 ))

yy11=(g=(g1111+g+g1212ggII22)v)v11+(g+(g1111ggII11+g+g1212)v)v2 2 (Eq.1-D)(Eq.1-D)

yy22=(g=(g2121+g+g2222ggII22)v)v11+(g+(g2222+g+g1212ggII11)v)v2 2 (Eq.2-D)(Eq.2-D)

Which YieldsWhich Yields


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In order to only have vIn order to only have v11 affect y affect y11 and to eliminate the and to eliminate the effect of veffect of v22 on y on y11, we must choose a value of g, we must choose a value of gI1I1 so that so that the coefficient of vthe coefficient of v22 in in Eq.1-DEq.1-D will disappear i.e.: will disappear i.e.:

gg1111ggI1I1+g+g1212=0=0

Then solving for gThen solving for gI1I1

A similar procedure can be done for Loop 2, which A similar procedure can be done for Loop 2, which eliminates any influences of veliminates any influences of v1 1 on yon y22, with the , with the manipulation of Eq 2-D we obtain a value of:manipulation of Eq 2-D we obtain a value of:

1

12I

11

gg =-

g

2

21I

22

gg =-

g


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The transfer functions seen on the previous slide are the The transfer functions seen on the previous slide are the decouplers needed to exactly compensate for the effect decouplers needed to exactly compensate for the effect of loop interactions in the 2x2 system shown in Figure 1-of loop interactions in the 2x2 system shown in Figure 1-D.D.

If we now substitute our expressions for gIf we now substitute our expressions for gI1 I1 and gand gI2 I2 into into Equations 1-D and 2-D respectively we will yield:Equations 1-D and 2-D respectively we will yield:

12 211 11 1

22

g gy = g - v

g

12 212 22 2

11

g gy = g - v

g


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Now the system is completely decoupled with only vNow the system is completely decoupled with only v11

affecting yaffecting y11, and v, and v22 affecting y affecting y22. . We can see in the figure below the equivalent block We can see in the figure below the equivalent block diagram where the loops appear to act independently and diagram where the loops appear to act independently and therefore, can be individually tuned.therefore, can be individually tuned.

ggc1c1

ggcc

22

yyd1d1

yyd2d2

yy11

yy22

12 2111

22

g gg -

g

12 2122

11

g gg -

g

vv11

vv22

++

++

--

--


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Let’s consider that the closed loop system is under steady Let’s consider that the closed loop system is under steady state. If the steady state gain for an element gstate. If the steady state gain for an element g ij ij =K=Kijij, , observe how the system is expressed at steady-state. observe how the system is expressed at steady-state.

12 211 11 1

22

y = - vK K

KK

222 2y = v

K

Recall the definition of λ for a 2x2 system:Recall the definition of λ for a 2x2 system:

Then the system simplifies to:Then the system simplifies to:

111 1y = v

K

12 212 22 2

11

y = - vK K

KK

12 21

11 22

1

1K KK K


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When we examine the simplified decoupling , the effective When we examine the simplified decoupling , the effective closed-loop steady-state gain in each loop is the ratio of the closed-loop steady-state gain in each loop is the ratio of the open-loop gain and the relative gain parameter (λ). open-loop gain and the relative gain parameter (λ). Note that when λ is very large, the effective closed-loop Note that when λ is very large, the effective closed-loop gains become very small, and control system performance gains become very small, and control system performance may be jeopardized.may be jeopardized.It is important to note that when dealing with systems with It is important to note that when dealing with systems with dimensions larger than 2x2, the simplified decoupling dimensions larger than 2x2, the simplified decoupling method can become very tedious. For an N x N system there method can become very tedious. For an N x N system there are (Nare (N22-1) compensators. The same principles as used for a -1) compensators. The same principles as used for a 2 x 2 system are applicable, but the work becomes very 2 x 2 system are applicable, but the work becomes very cumbersome.cumbersome.On the next slide we will see an example of a 3 x 3 system, On the next slide we will see an example of a 3 x 3 system, which has 6 compensator blocks, it is clear that using which has 6 compensator blocks, it is clear that using simplified decoupling in this situation would be simplified decoupling in this situation would be very very tedioustedious..


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ggc1c1

ggc3c3

ggc2c2

yyd1d1

yyd2d2

yyd3d3

ggII1212

ggII2323

ggII1313

ggII2121

ggII3131

ggII3232

gg1111

gg3333

gg1212

gg2222

gg2323

gg3131

gg3232

gg2121

gg1313

++

++

++

++

++

++

++++

++

++++++

++

++

++

++++

++

++

++ ++

yy11

yy22

yy33

uu11

uu22

uu33

vv11

vv22

vv33

uu33 uu22 uu11

--

--


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Generalized DecouplingGeneralized DecouplingPlease refer to Figure 1-D which we will use this figure to Please refer to Figure 1-D which we will use this figure to outline a more generalized procedure for decoupler outline a more generalized procedure for decoupler design.design.

GENERALIZED DECOUPLING

ggc1c1

ggc1c1

ggI2I2

ggI1I1

gg1111

gg1212

gg2121

gg2222

yydd

11

yydd

22

++ --

++

--

++ ++ ++++

++

++

++

++

εε11

εε22

vv11

vv22

uu11

uu22 yy22

Figure 1-DFigure 1-D


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1.1. We can observe from Figure 1-D that:We can observe from Figure 1-D that:

So that:So that:

y=Guy=Gu

u=Gu=GIIvv

y=GGy=GGIIvv



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2. In order to eliminate all interactions, 2. In order to eliminate all interactions, yy must be related must be related to to vv through a diagonal matrix, let us call it G through a diagonal matrix, let us call it GRR(s), now we (s), now we must chose Gmust chose GII such that such that

GGGGII=G=GRR(s)(s)

And the compensated input/output relation becomes:And the compensated input/output relation becomes:

y=Gy=GRR(s)v(s)v

Where GWhere GRR represents the equivalent diagonal process that represents the equivalent diagonal process that the diagonal controllers Gthe diagonal controllers GCC are required to control. are required to control.



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3. Therefore, the compensator (G3. Therefore, the compensator (GII) must be given by:) must be given by:

GGII=G=G-1-1 G GRR

4. The compensator obtained depends on what G4. The compensator obtained depends on what GRR is is selected. The elements of Gselected. The elements of GRR should be chosen to should be chosen to provide the desired decoupled behaviour with the provide the desired decoupled behaviour with the simplest possible decoupler. A common choice for Gsimplest possible decoupler. A common choice for GRR is: is:

GGRR=Diag[G(s)]=Diag[G(s)]

Ie. The diagonal elements of G(s) are retained as the Ie. The diagonal elements of G(s) are retained as the elements of the diagonal matrix Gelements of the diagonal matrix GRR, however, other , however, other choices have been used.choices have been used.


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The Relationship between Generalized and The Relationship between Generalized and Simplified DecouplingSimplified Decoupling““Generalized” decoupling may be related to simplified Generalized” decoupling may be related to simplified decoupling, by noting that for simplified decoupling applied to decoupling, by noting that for simplified decoupling applied to a 2x2 system, the compensator transfer function matrix is a 2x2 system, the compensator transfer function matrix is given by:given by:

While for a 3x3 system, the compensatory matrix GWhile for a 3x3 system, the compensatory matrix GII takes the takes the form:form:

1

2

I

II

1 gG =

g 1

12 13

21 23

31 32

I I

I I I

I I

1 g g

G = g 1 g

g g 1


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Quiz Quiz #8#8

• What is the main objective of decoupling?What is the main objective of decoupling?

• What is a downfall of simple decoupling?What is a downfall of simple decoupling?

• Is it often easy to achieve perfect decoupling?Is it often easy to achieve perfect decoupling?


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Some Limitations of the Application of DecouplingSome Limitations of the Application of Decoupling

There are some limitations to the application of There are some limitations to the application of decoupling, and we must keep these in mind in order to decoupling, and we must keep these in mind in order to maintain a proper perspective when designing maintain a proper perspective when designing decouplers.decouplers.

Perfect decoupling is only possible if the process model is Perfect decoupling is only possible if the process model is perfect, which is hardly ever the case, so perfect perfect, which is hardly ever the case, so perfect decoupling in practice is impossible.decoupling in practice is impossible.

Perfect dynamic decouplers are based on model inverses. Perfect dynamic decouplers are based on model inverses. As such, they can only be implemented if such inverses As such, they can only be implemented if such inverses are both are both causalcausal and and stablestable..

LIMITATIONS OF DECOUPLING


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To illustrate the idea of stable and casual, please consider To illustrate the idea of stable and casual, please consider the 2x2 compensators we saw in Figure 1-D whose transfer the 2x2 compensators we saw in Figure 1-D whose transfer functions are Gfunctions are GI1I1 and G and GI2I2 must be casual must be casual (no e(no e++ααss terms) terms) and and stable. stable.

To satisfy To satisfy causalitycausality for the 2x2 system, any time delays in for the 2x2 system, any time delays in gg1111 must be smaller than the time delays in g must be smaller than the time delays in g1212 and a similar and a similar condition must hold for gcondition must hold for g2222 and g and g2121..

To satisfy To satisfy stabilitystability, a second condition that g, a second condition that g1111 and g and g2222 must must not have any right hand plane zeros and also gnot have any right hand plane zeros and also g1212 and g and g21 21

must not have any right hand plane poles. This leads to the must not have any right hand plane poles. This leads to the following general conditions that must be satisfied in order following general conditions that must be satisfied in order to implement simplified dynamic decoupling for N x N to implement simplified dynamic decoupling for N x N systems.systems.



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1.1.Causality:Causality: In order to ensure causality in the compensator In order to ensure causality in the compensator transfer functions the time-delay structure in G(s) must be such transfer functions the time-delay structure in G(s) must be such that the smallest time-delay in each row occurs on the diagonal. that the smallest time-delay in each row occurs on the diagonal. For simplified decoupling, this is an absolute requirement, but it For simplified decoupling, this is an absolute requirement, but it is possible to add delays to the inputs uis possible to add delays to the inputs u11,u,u22…u…unn, to satisfy the , to satisfy the requirement if the original process G does not comply. This is requirement if the original process G does not comply. This is equivalent to defining a modified process as Gequivalent to defining a modified process as Gmm::


Where D is a Where D is a diagonal matrix diagonal matrix of time delaysof time delays

11

22

nn

d s

d s

d s

e 0

eD(s)=

0 e

GGmm=GD=GD


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The simplified decoupler is then designed by using the The simplified decoupler is then designed by using the elements of Gelements of Gmm rather than G, and the matrix D must be rather than G, and the matrix D must be inserted into the control loop as shown below:inserted into the control loop as shown below:


GGcc GGII DD GG

Single Single Loop Loop ControllerControllerss

DecouplerDecoupler DelayDelays s

ProcesProcesss

εε vv uu

modified process Gmodified process Gmm

yyyydd

In the case of generalized decoupling, one may use the In the case of generalized decoupling, one may use the modified process Gmodified process Gmm as above, or alternatively, the time as above, or alternatively, the time delays in the diagonal matrix Gdelays in the diagonal matrix GRR can be adjusted, in order can be adjusted, in order that the elements of Gthat the elements of GII=(GD)=(GD)-1-1GGRR are casual. This is are casual. This is equivalent to requiring that Gequivalent to requiring that GRR

-1-1GD have the smallest GD have the smallest delay in each row on the diagonal.delay in each row on the diagonal.

++

--


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2. 2. Stability-Stability- In order to ensure the stability of the In order to ensure the stability of the compensator transfer functions, the causality compensator transfer functions, the causality condition must be satisfied and there are no condition must be satisfied and there are no Right Hand Plane zeros of the process G(s). This is Right Hand Plane zeros of the process G(s). This is an absolute requirement for simplified decoupling an absolute requirement for simplified decoupling and reduces to the condition that there are no and reduces to the condition that there are no Right Hand Plane zeros in the diagonal elements Right Hand Plane zeros in the diagonal elements of G and that the off-diagonal elements of G are of G and that the off-diagonal elements of G are stable. For generalized decoupling, this may be stable. For generalized decoupling, this may be performed by adjusting the dynamics of Gperformed by adjusting the dynamics of GRR in in order that the elements of Gorder that the elements of G II=G=G-1-1GGRR be stable. be stable.



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Partial DecouplingPartial DecouplingIf some loop interactions are weak or if some of the loops If some loop interactions are weak or if some of the loops do not need to achieve high performance, the partial do not need to achieve high performance, the partial decoupling is a method one should consider. If this is the decoupling is a method one should consider. If this is the case, only a subset of the control loops where the case, only a subset of the control loops where the interactions are important and high performance is interactions are important and high performance is important are focused on.important are focused on.

Typically partial decoupling is considered for 3x3 or higher Typically partial decoupling is considered for 3x3 or higher dimension systems. The main advantage is the reduction of dimension systems. The main advantage is the reduction of dimensionality. Partial decoupling is also applicable to 2x2 dimensionality. Partial decoupling is also applicable to 2x2 systems, in this case, one of the compensator blocks is set systems, in this case, one of the compensator blocks is set to zero for the loop that is to be excluded from decoupling.to zero for the loop that is to be excluded from decoupling.

PARTIAL DECOUPLING


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Steady-State DecouplingSteady-State Decoupling

The difference between dynamic decoupling and steady-The difference between dynamic decoupling and steady-state decoupling is that dynamic decoupling uses the state decoupling is that dynamic decoupling uses the complete, dynamic version of each transfer function complete, dynamic version of each transfer function element to obtain the decoupler, and steady-state element to obtain the decoupler, and steady-state decoupling only uses the steady-state gain portion of each decoupling only uses the steady-state gain portion of each of the transfer elements.of the transfer elements.

Therefore, if each transfer function element gTherefore, if each transfer function element g ijij(s), has a (s), has a steady-state gain term Ksteady-state gain term Kijij, and if the gain matrix is defined , and if the gain matrix is defined as K, the steady-state decoupling results in the same way as K, the steady-state decoupling results in the same way as it did for a 2x2 system that we discussed earlier.as it did for a 2x2 system that we discussed earlier.

STEADY-STATE DECOUPLING


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Simplified steady-state decoupling for a 2x2 Simplified steady-state decoupling for a 2x2 systemsystem

Here: andHere: and

These expressions to describe the transfer function of These expressions to describe the transfer function of the compensator block are simple, constant, the compensator block are simple, constant, numerical values so they will always be realizable and numerical values so they will always be realizable and can be implemented.can be implemented.

1

12I

11

Kg =-

K 2

21I

22

Kg = -

K

STEADY-STATE DECOUPLING FOR A 2X2 SYSTEM


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Simplified steady-state decoupling for a 2x2 systemSimplified steady-state decoupling for a 2x2 systemIn this case, the decoupler matrix is given by:In this case, the decoupler matrix is given by:

Where KWhere KRR is the steady-state version of G is the steady-state version of GRR(s). The (s). The inversion indicated is a matrix of numbers, and inversion indicated is a matrix of numbers, and therefore, the inversion will always be realizable and therefore, the inversion will always be realizable and easily implemented.easily implemented.

The main advantages of steady-state decoupling are The main advantages of steady-state decoupling are that the design involves simple numerical that the design involves simple numerical computations and that the resulting decouplers are computations and that the resulting decouplers are always realizable.always realizable.

-1I RG =K K

STEADY-STATE DECOUPLING FOR A 2X2 SYSTEM


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Quiz Quiz #9#9

• What 2 conditions must a system satisfy to What 2 conditions must a system satisfy to achieve perfect dynamic decoupling?achieve perfect dynamic decoupling?

• What is the main advantage of partial-What is the main advantage of partial-decoupling?decoupling?

• Why is steady-state decoupling a favorable Why is steady-state decoupling a favorable method if applicable?method if applicable?


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Singular Value DecompostionSingular Value DecompostionAny real Any real n x mn x m matrix K, it is possible to find orthogonal matrix K, it is possible to find orthogonal (unitary) matrices W and V such that(unitary) matrices W and V such that

WWTTAV=∑AV=∑Here ∑ is the m x n matrix described below:Here ∑ is the m x n matrix described below:

wherewhere

Where, for p=min(m,n), the diagonal elements of S: Where, for p=min(m,n), the diagonal elements of S:

σσ11>> σ σ22>> … … >> σ σrr>> 0,(r 0,(r >> p), together with σ p), together with σr+1r+1=0, σ=0, σpp=0 are =0 are called the singular values of A; these are the positive square called the singular values of A; these are the positive square roots of the eigenvalues of Aroots of the eigenvalues of ATTA; r is the rank of A .A; r is the rank of A .

s 0

0 0

1

2

0 0 0

0 0 0s

0 0 0 r

SINGULAR VALUE DECOMPOSITION


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W is the W is the m x mm x m matrix matrix

Whose columns wWhose columns wii, i=1,2,…,m are called the , i=1,2,…,m are called the left singular left singular vectorsvectors of A; these are normalized (orthonormal) of A; these are normalized (orthonormal) eigenvectors of AAeigenvectors of AATT..

V is the V is the n x n n x n matrix:matrix:

Whose n columns vWhose n columns vii, i=1,2,…,n are called the , i=1,2,…,n are called the right right singular vectorssingular vectors of A; these are normalized (orthonormal) of A; these are normalized (orthonormal) eigenvectors of Aeigenvectors of ATTA.A.

1 2 mW= w w w

1 2 nV= v v v



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Because they are composed of orthonormal vectors, the Because they are composed of orthonormal vectors, the matrices W and V are orthogonal (or unitary) matrices i.e.matrices W and V are orthogonal (or unitary) matrices i.e.

WWTTW=I=WWW=I=WWTT

So thatSo that WW-1-1=W=WTT

Also Also VVTTV=I=V VV=I=V VTT

So thatSo that VV-1-1=V=VTT



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By applying these properties of unitary matrices, we can By applying these properties of unitary matrices, we can obtain the relationship:obtain the relationship:

A=W ∑ VA=W ∑ VTT

Analogously to the eigenvalue/eigenvector expression for Analogously to the eigenvalue/eigenvector expression for square matrices, we have the more general pair of square matrices, we have the more general pair of expressionsexpressions

AvAvii= σ= σiiwwii

AATTiiwwii= σ= σiivvii



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The ratio of the largest to the smallest singular The ratio of the largest to the smallest singular value is designated the value is designated the condition member of A:condition member of A:ie.ie.

ThisThis gives the most reliable indication of how gives the most reliable indication of how close A is to being singular. Note that for a close A is to being singular. Note that for a singular matrix, κ(A)=∞, thus nearness to singular matrix, κ(A)=∞, thus nearness to singularity is indicated by excessively large (but singularity is indicated by excessively large (but finite) values for κ(A)finite) values for κ(A)

1( )p

A



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Example - Singular Value Decomposition of a 3x2 Example - Singular Value Decomposition of a 3x2 matrixmatrix

Therefore,Therefore,

1 2

A 2 1

2 1

SINGULAR VALUE DECOMPOSITION EXAMPLE

T 9 2A A=

2 6


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The eigenvalues are obtained as The eigenvalues are obtained as 1010 and and 5 5 , thus , thus the singular values of A arethe singular values of A are : :

σσ11,, σσ22=√10 and √5=√10 and √5

Ordered so that Ordered so that σσ11>σ>σ2 2 as required for SVD as required for SVD analysis, the next step is to determine the 3x2 analysis, the next step is to determine the 3x2 matrix ∑.matrix ∑.


10 0

0 5

0 0


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Right Singular ValuesRight Singular Values

The first eigenvector or AThe first eigenvector or ATTA corresponding to λA corresponding to λ11 is is obtained from adj(Aobtained from adj(ATTA- λA- λ11 I) I)

A possible choice for the eigenvector is the second A possible choice for the eigenvector is the second column. Normalizing this with column. Normalizing this with √2√222+1+12= 2= √5√5, the norm of the , the norm of the vector, we obtain the first right singular vector vvector, we obtain the first right singular vector v11

corresponding to corresponding to σσ11= √10= √10


T1

-4 2adj(A A- I) =

2 1

1

2

5v =

1

5


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In the same way, the second normalized eigenvalue In the same way, the second normalized eigenvalue corresponding tocorresponding to λλ22=5 is:=5 is:

Therefore:Therefore:


2

1

5v =

2

5

2 1

5 5v =

1 2

5 5


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From V we can determine VFrom V we can determine VT T to be:to be:

You can verify that V is a unitary matrix by You can verify that V is a unitary matrix by evaluating Vevaluating VTTV and confirming that the product V and confirming that the product is I.is I.


T

2 1

5 5V =

1 2

5 5


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Left Singular ValuesLeft Singular Values

For the given matrix:For the given matrix:

The Eigenvalues of this 3x3 matrix are obtained from the The Eigenvalues of this 3x3 matrix are obtained from the characteristic equation which in this case is:characteristic equation which in this case is:

(5-(5-λλ)) [(5- [(5- λλ))2-2--25-25]=0]=0

Ie. Ie. λλ11,,λλ22, , λλ33= 10,5,0= 10,5,0


T

5 0 0

AA 0 5 5

0 5 5


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Note that the non-zero eigenvalues of AANote that the non-zero eigenvalues of AATT are identical to are identical to the eigenvalues of Athe eigenvalues of ATTA.A.

For For λλ11=10=10

To find the adjoint of this matrix, we first find the To find the adjoint of this matrix, we first find the cofactors and take the transpose of the matrix of cofactors and take the transpose of the matrix of cofactors. In this case,cofactors. In this case,


T1

0 0 0

adj(AA - I) = 0 25 25

0 25 25

T1

5 0 0

(AA - I) = 0 5 5

0 5 5


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And by normalizing any of the non-zero columns, we And by normalizing any of the non-zero columns, we obtain the first left singular value of A, and by a similar obtain the first left singular value of A, and by a similar procedure the second and third eigenvectors can be procedure the second and third eigenvectors can be determined using values of determined using values of λλ22, =5 , =5 and and λλ33=0=0


1

0

1w =

21

2

2

1

w = 0

0

3

0

1w =

21

2


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When the 3 eigenvalues are combinedWhen the 3 eigenvalues are combined::

Now we have all of the elements desired to decompose Now we have all of the elements desired to decompose the matrix. You can verify that the matrix. You can verify that A=W ∑ VA=W ∑ VT T by multiplying by multiplying the elements we have determinedthe elements we have determined..


0 1 0

1 1W = 0

2 21 1

02 2


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Steady-State Decoupling by Singular Value Steady-State Decoupling by Singular Value DecompositionDecomposition

The Singular Value Decomposition (SVD) of the steady-state The Singular Value Decomposition (SVD) of the steady-state gain matrix of a process is another approach to steady-gain matrix of a process is another approach to steady-state decoupling.state decoupling.

The SVD of a process gain matrix K can be written as:The SVD of a process gain matrix K can be written as:

K=W ∑ VK=W ∑ VT T

then applying the SVD of K, the steady state model then applying the SVD of K, the steady state model becomes:becomes:

y=y= W ∑ VW ∑ VT T uu

STEADY-STATE DECOUPLING BY SINGULAR VALUE DECOMPOSITION


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We will multiply both sides by WWe will multiply both sides by WTT (recall the orthogonality (recall the orthogonality properties of W), our expression becomes:properties of W), our expression becomes:

WWTTy= y= ∑ V ∑ VT T uu

Recall that when the matrix K is a square matrix ∑ is a Recall that when the matrix K is a square matrix ∑ is a diagonal matrix of singular values. This allows us to diagonal matrix of singular values. This allows us to define a new define a new output variables ηoutput variables η and and new input new input variables μ variables μ where:where:

η= η= WWTTyyAndAnd

μμ =∑ V =∑ VT T uu



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Now, the process model becomesNow, the process model becomes

ηη = ∑ = ∑ μ μ

Because ∑ is diagonal, this indicates that the system is Because ∑ is diagonal, this indicates that the system is completely decoupled at steady state.completely decoupled at steady state.


WWTT GGcc∑∑ GG

yydd++ uuμμ yy

--

WWTT

VVηηdd

ηη


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The implication of this is the following: instead of The implication of this is the following: instead of controlling y with u, the transformed variables will controlling y with u, the transformed variables will convert the original system (with cross-coupling convert the original system (with cross-coupling among the process variables) to a system that has among the process variables) to a system that has no cross-coupling. The open-loop gain of each loop no cross-coupling. The open-loop gain of each loop of the transformed system is indicated clearly by of the transformed system is indicated clearly by the singular values and conditioning is the singular values and conditioning is automatically accessed from the condition number.automatically accessed from the condition number.

A controller can now be designed for the equivalent A controller can now be designed for the equivalent (steady-state) system which controls (steady-state) system which controls ηη by using by using μμ. . If this controller is designated If this controller is designated GGcc

∑∑ then the scheme then the scheme would be implemented as seen in the previous would be implemented as seen in the previous slide.slide.



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References:References:

• Ogunnaike,B.,Ray,W. Process Dynamics, Modeling, and Control. Oxford University Press, New York (1994)

• Seborg, D., et al. Process Dynamics and Control. John Wiley & Sons, Inc, United States of America

(2004)

• Thibault, Jules. Courses Notes, CHG 3335: Process Control. University of Ottawa, Ottawa (July 2004)

REFERENCES

Documents

Program for North American Mobility In Higher Education