research article quadratic optimal regulator...

Hindawi Publishing CorporationModelling and Simulation in EngineeringVolume 2013 Article ID 862190 8 pageshttpdxdoiorg1011552013862190

Research ArticleQuadratic Optimal Regulator Design of a PneumaticControl Valve

Mohammad Heidari1 and Hadi Homaei2

1 Mechanical Engineering Group Islamic Azad University Aligudarz Branch PO Box 159 Aligudarz Iran2 Faculty of Engineering Shahrekord University PO Box 115 Shahrekord Iran

Correspondence should be addressed to Mohammad Heidari moh104337yahoocom

Received 30 May 2013 Revised 20 September 2013 Accepted 4 October 2013

Academic Editor Abdelali El Aroudi

Copyright copy 2013 M Heidari and H Homaei This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Pneumatic control valves are still the most used devices in the process industries due to their low cost and simplicity This paperpresents a regulator for pneumatic control valves using pole-placement method optimal control full-order state observer andminimum-order state observer and their responses will be compared with each other Bondgraph method has been used to modelthe control valve Simulation results have been made for four models of regulator The results show that minimum overshoot andsettling time are achieved using optimal regulator of pneumatic valve

1 Introduction

Process plants consist of hundreds or even thousands ofcontrol loops all networked together to produce a productto be offered for sale Each of these control loops is designedto keep some important process variables such as pressureflow level and temperature within a required operatingrange to ensure the quality of the end product Each ofthese loops receives and internally creates disturbances thatdetrimentally affect the process variable and interaction fromother loops in the network provides disturbances that influ-ence the process variable To reduce the effect of these loaddisturbances sensors and transmitters collect informationabout the process variable and its relationship to some desiredset point A controller then processes this information anddecides what must be done to get the process variable back towhere it should be after a load disturbance occurs When allthe measuring comparing and calculating are done sometype of final control element must implement the strategyselected by the controller The most common final controlelement in the process control industries is the control valveThe control valve manipulates a flowing fluid such as gassteam water or chemical compounds to compensate for theload disturbance and keep the regulated process variable asclose as possible to the desired set point Control valves adjust

the temperature pressure flow rate and so forth by changingthe flow rate Figure 1 shows a sliding-stempneumatic controlvalve Pneumatic control valves are still the most used valvesin the process industries due to their low cost and simplicityPneumatic valves are used extensively in various industriestoday Industry standard has been established that detailsthe vibration humidity thermal salt spray and temperatureextremes that these valves must operate within This makesthe design of valve control systems a very challenging taskControl valves have two major components valve bodyhousing and the actuation unit One factor in the qualityof the final end product is the improvement of the controlloop performance A critical component in the loop is thefinal control element the control valve package Optimizedactuator parameters play a vital role in the dynamic per-formance of the pneumatic control valve Champagne andBoyle [1] reviewed the pneumatic actuator and positionerparameters that affect the control package performance Thisis done through the use of a control valve package computermodel to assess the dynamic performance The attributes ofspring return versus double acting actuators are illustratedThe effects of supply pressure step size load margin flowactuator volume anddesign style are investigated through theuse of mathematical simulations of pneumatic control valvedynamic performance Hagglund presented a procedure that

2 Modelling and Simulation in Engineering

Figure 1 Pneumatic control valve with diaphragm

compensates for static friction (stiction) in pneumatic controlvalves [2] The compensation is obtained by adding pulsesto the control signal The characteristics of the pulses aredetermined from the control action

The compensator is implemented in industrial controllersand control systems and the industrial experiences showthat the procedure reduces the control error during stick-slip motion significantly compared to standard control with-out stiction compensation The oscillations caused by staticfriction (stiction) in pneumatic control valves cause lossesin quality and expense of raw materials The input-outputbehavior of a pneumatic control valve affected by stiction invalve De Souza et al [3] presented a well-known stictioncompensationmethod that reduces variability both at processvariable and pneumatic valve stemmovementThe two-movemethod is revisited in this paper and it was shown thatassumptions on the knowledge of steady-state stem positionof control valve that assures equality of set point and thecontrolled variable were not easily achievable Bondgraph isa graphical representation of a physical dynamics systemIt is similar to the better known block diagram and signalflow with the major difference that the arcs in bondgraphsrepresent bidirectional exchange of physical energy whilethose in block diagrams and signal-flow graphs representunidirectional flow of information Also bondgraphs aremultidomain and domain neutral This means that a bond-graph can incorporate multiple domains simultaneouslyBondgraphs were devised by Paynter [4] at MIT in April 1959and subsequently developed into a methodology togetherwith Karnopp et al [5] Early prominent promoters ofbondgraph modeling techniques among others were Thoma[6] Dixhoorn and Dransfield Athanasatos and Costopoulos[7] used the bondgraphmethod for finding the proactive faultin 43 way direction control valve of a high pressure hydraulicsystemThe accuracy of the bondgraphmodel was verified bycomparing its response to the response of an actual hydraulicsystem Dıaz-Zuccarini et al [8] utilized the bondgraph asboundary condition for a detailed model of an idealized

I

1 121 3

4

5

6

R

SE TF

C

Figure 2 Bondgraph control valve actuator

mitral valve A specific application in cardiovascular model-ing was demonstrated by focusing on a specific example a3D model of the mitral valve coupled to a lumped parametermodel of the left ventricle

In this study a pneumatic control valve is modeled bybondgraph method A regulator has been designed usingpole-placement method optimum regulator full-order stateobserver and minimum-order state observer for this systemIn continuing their responses have been compared with eachother

This research is organized as follows Section 2 recalls thebondgraph model of valve and proposes equations of motionof valve Section 3 develops quadratic optimal regulatorof pneumatic valve Sections 4 and 5 present designing aregulator by pole-placement technique full and minimumorder observers respectively Section 6 shows the results offull and minimum order regulator observers and quadraticoptimal regulator of valve and finally some conclusions aregiven in Section 7

2 Bondgraph Model of Valve andTransfer Function

The bondgraph model of the valve is shown in Figure 2In this model SE is the inlet pressure of the system Thepressure changes to force by multiplying in effect area of thediaphragm In bondgraph this transformer is modeled byTF Element 119877 is the friction of the system Element 119868 is themovable mass of valve and diaphragm

Element 119862 represents the spring of the valve actuatorAlso 1-junction is a common flow junction 1-junctions haveequality of flows and the efforts sum up to zero with thesame power orientation In fact junctions can connect twoor more bonds The direction of the half arrows () denotesthe direction of power flow given by the product of the effortand flow variables associatedwith the power bondThe bondsin a bondgraphmay be numbered sequentially using integersstarting with 1 The two 1-junctions in the bondgraph showncan be uniquely identified as (S 1 2) and (S 4 5 6) similarlysymbols like 119878119864

1and 119877

6can be used to identify a particular

element This system has two state variables 1198754and 119902

5sdot 1199025is

the displacement of valve stem and the variation of the springlength Also V

4= 11987541198684is the velocity of the valve stem The

Modelling and Simulation in Engineering 3

equations of motion are derived using bondgraph method asfollows

4= 119860 times 119878119864

1minus 11987031199025minus

1198776

1198684

1198754 (1)

1199025=

1198754

1198684

(2)

Now if the velocity and position of stem are zero in the initialcondition119883(0) then we have

119883 (0) = [1198754(0)

1199025(0)

] = [0

0] (3)

By derivation of relation (2) with respect to time we have

1199025=

4

1198684

(4)

By substitution of 4from (1) into (4) we have

1199025=

119860 times 1198781198641minus 11987031199025minus (11987761198684) 1198754

1198684

(5)


1199025=

1

1198684

(119860 times 1198781198641minus 11987031199025minus 1198776

1199025) (6)

Using laplace transformation of (6) we have

1199025(119904)

1198781198641(119904)

=(1198601198684)

1199042 + (11987761198684) 119904 + (119870

31198684) (7)

Equation (7) is the transfer function of the valve The resultsof bondgargh model of valve show that the response of thesystem is identical with the result in [9 10]

3 Quadratic Optimal Regulatorof Pneumatic Valve

Let us consider the system that is defined as form of a state-space representation as follows

= 119860119909 + 119861119906 (8)

119910 = 119862119909 + 119863119906 (9)

where 119909 is called the state vector the derivative of the statevector with respect to time 119910 the output vector and 119906(119905) theinput or control vector Also 119860 is the system matrix 119861 theinput matrix 119862 the output matrix and119863 direct transmissionmatrix We will now consider the optimal control problemthat given the system (8) thematrix119870 should be determinedso that

119906 (119905) = minus119870119909 (119905) (10)

It means that the control signal is determined by the state ofthatmomentThedimension of the state feedback gainmatrix

is 1 times 119899 and 119899 is the number of the states This minimizes theperformance index (119869) as

119869 = int

infin

0

(119909119879119876119909 + 119906

119879119877119906) 119889119905 (11)

where 119876 is a positive-definite (or positive semidefinite)Hermitian or real symmetric matrix and 119877 is a positive-definite Hermitian or real symmetric matrix [11]

Note that the second term on the right side of (11)accounts for the expenditure of the energy of the signalcontrol In this problem we assume that the control vector119906(119905) is unconstrained The linear control law given by (11) isthe optimal control law Therefore if the unknown elementsof the matrix 119870 are determined so as to minimize theperformance index then 119906(119905) = minus119870119909(119905) is optimal for anyinitial state 119909(0) Hence

119870 = 119877minus1119861119879119875 (12)

Equation (12) gives the optimal matrix Thus the optimalcontrol law to the quadratic optimal control problem wherethe performance index is given by (11) is linear and given by

119906 (119905) = minus119870119909 (119905) = minus119877minus1119861119879119875119909 (119905) (13)

The matrix 119875 in (12) must satisfy the following reducedequation

119860119879119875 + 119875119860 minus 119875119861119877

minus1119861119879119875 + 119876 = 0 (14)

Equation (14) is called the Riccati reduced-matrix equation

4 Designing a Regulator byPole-Placement Technique

Suppose that the system is defined by (8) and the controlsignal is given by (10) The feed back gain matrix K thatforces the eigenvalues of 119860 minus 119861119870 to be 120583

1and 120583

2(desired

values) can be determined by the following steps Firstly thecontrollability of the system is checked The controllabilitymatrix is defined as follows

119872 = [119860 119860119861] (15)

For this system the rank of the controllability matrix is twoIn the next step the characteristic polynomial of matrix 119860

|119904119868 minus 119860| = 1199042+ 1198861119904 + 1198862 (16)

determines the values of 1198861and 1198862 Now the desired closed-

loop poles 1205831and 120583

2are determined based on the transient

response or frequency response requirements such as speeddamping ratio and steady state requirements The desiredcharacteristic equation becomes

(119904 minus 1205831) (119904 minus 120583

2) = 1199042+ 1205721119904 + 1205722 (17)

The values of 1205721and 120572

2are determined from (17) The

required state feedback gain matrix 119870 can be determinedfrom (18)

119870 = [1205722 minus 1198862

1205721minus 1198861] 119879minus1 (18)


where

119879 = 119872119882 (19)

Matrix119882 is given by

119882 = [1198861

1

1 0] (20)

5 Full- and Minimum-Order State Observer

If not all state variables are available for feedback unavailablestate variables should be estimated by observer Supposethat observer observes all state variables of the systemregardless of whether some state variables are available fordirect measurement For this situation wemust use full-orderstate observer and observe all state variables State observerscan be designed if and only if the observability condition issatisfied Necessary and sufficient condition for observabilitycondition is that the dual of the original system must becompletely state controllable [9]

The complete state observability for this system is that therank of

119873 = [119862119879

119860119879119862119879] (21)

is 2 because the system is a two-degree-of-freedom systemCharacteristic equation can be obtained from (16)We shoulddetermine two groups of desired poles as each one has twopoles The first group relates to system state variables andthe other one relates to observer variables Therefore stateobserver gain matrix can be obtained for both situations Wechoose two dominant poles based on the transient responserequirements and remained poles assignment so enough farfrom two dominant poles State observer gain matrix can bedetermined from (19)ndash(23) Also matrix 119882 can be obtainedby using (16)ndash(20) after specifying observer desired polesSubsequently we have

119870119890= (119882119873

119879)minus1

[1205722 minus 1198862

1205721minus 1198861]119879

(22)

The transfer function of the full-order controller observer is

119880 (119904)

minus119884 (119904)= 119870(119904119868 minus 119860 + 119870

119890119862 + 119861119870)

minus1

119870119890 (23)

For designing of minimum-order state observer we shouldtraverse the following stages By selecting eigenvalues forobserver matrix the characteristic equation is

1003816100381610038161003816119904 minus 119860119887119887

+ 119870119890119860119886119887

1003816100381610038161003816 = 119904 + 1 (24)

where 1is the desired eigenvalue for the minimum-order

observer The minimum observer gain119870119890can be determined

by choosing the desired eigenvalue for the minimum-orderobserver and then the procedure developed for the full-orderobserver with appropriate modifications We have

119860 = [119860119886119886

119860119886119887

119860119887119886

119860119887119887

] 119861 = [119861119886

119861119887

] (25)

The minimum observer gain is defined as follows

119870119890= (

119879)minus1

(1minus 1198861) (26)

where

= 119860119886119887 = 119886

1 (27)

Note that 1198861is the coefficient in the characteristic equation

for the state equation1003816100381610038161003816119904 minus 119860

119887119887

1003816100381610038161003816 = 119904 + 1198861 (28)

The transfer function for minimum-order state observer isgiven by the following formula

119880 (119904)

minus119884 (119904)= minus [119862(119904 minus 119860)

minus1

119861 + 119863] (29)

where

119860 = 119860 minus 119865119870119887

119861 = 119861 minus 119865 (119870119886+ 119870119887119870119890)

119862 = minus119870119887

119863 = minus (119870119886+ 119870119887119870119890)

(30)

where

119860 = 119860119887119887

minus 119870119890119860119886119887

119861 = 119860119870119890+ 119860119887119886

minus 119870119890119860119886119886

119865 = 119861119887minus 119870119890119861119886

(31)

6 Results and Discussion

Table 1 shows the parameters of a sliding-stem pneumaticcontrol valve

By substitution of Table 1 into (7) we have

1199025(119904)

1198781198641(119904)

=653

1199042 + 3333119904 + 2263333 (32)

Matrices of the state space equations of the valve are asfollows

119860 = [0 1

minus22633333 minus3333]

119861 = [0

9142] 119862 = [1 0] 119863 = 0

(33)

In this section we consider the design of regulator systemwith full- and minimum-order observers and then quadraticoptimal regulator of pneumatic valve The detail of designinga full and minimum order observer is given in the previoussection Let us design a regulator which achieves a 10maximum overshoot and settle time less than 05 second foroutput when the initial condition is 119909(0) = [0 01]

119879


Table 1 Valve parameters [9]

Name of variable Parameter ValueEffective area of diaphragm 119860 0196 ft2

Spring constant 119870 6790Movable mass 119868(119872) 003 slugResistance and frictioncoefficient 119877 1 lbsdotSft

Air pressure SE 140 lbft2

A 10 overshoot and a settling time of 05 second yield120585 = 0591 and 120596

119899= 1353 thus the characteristic equation

for dominant poles is 1199042+16119904+1831 = 0 where the dominantpoles are located at minus8 plusmn 1198951091 [12] Hence choose thedesired closed-loop poles at 119904 = 120583

119894(119894 = 1 2) where

1205831= minus8 + 1198951091 120583

2= minus8 minus 1198951091 (34)

In the pole placement method we must first check the rankof the controllability matrix from (15)

119872 = [0 9142

9142 minus30470] (35)

Since the rank of matrix 119872 is 2 arbitrary pole placement ispossible In this case for determining the state feedback gainmatrix 119870 = [119896

11198962] the characteristic polynomial for the

desired system is

|119904119868 minus 119860 + 119861119870|

=

10038161003816100381610038161003816100381610038161003816

[119904 0

0 119904] minus [

0 1

minus22633333 minus3333] + [

0

9142] [1198961 119896

2]

10038161003816100381610038161003816100381610038161003816

= 1199042+ (9142119896

2+ 3333) 119904 + 9142119896

1+ 226331198905

(36)

The characteristic polynomial must be equal to

(119904 + 1205831) (119904 + 120583

2) = (119904 + 8 + 1198951091) (119904 + 8 minus 1198951091)

= 1199042+ 16119904 + 1831

(37)

By equating the coefficients of the terms of the like powers of119904 we obtain

1198961= minus24734 119896

2= minus0019 (38)

or

119870 = [minus24734 minus0019] (39)

Response to initial condition with pole placement method isshown in Figure 3 Note that themaximumovershoot of valveis about 04 for output and settling time is about 05 secondThe design is thus acceptable

The input 119906(119905) = minus119896119909(119905) can be calculated from thepreviously calculated matrices 119896 and 119909 The control input119906(119905) which is a force is applied to the valve stem The largerthe control input magnitude the bigger the energy spent by

0 01 02 03 04 05 06 07 08 09 1minus2

0

2

4 Response to initial conditiontimes10minus3

Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0005

01015 Response to initial condition

Stat

e var

iabl

etimes2

t (s)

(b)

Figure 3 Response to initial condition

the actuator in generating the control input and the higherthe cost of control As shown in Figure 6 themaximum valueof input control signal is about 09 Observability matrix canbe obtained from (21) as

119873 = [1 0

0 1] (40)

Since the rank of observability matrix is 2 the given systemis completely state observable Hence full-order observer isapplicable Assume that the desired eigenvalues for full-orderobserver are 120582

1= minus116 and 120582

2= minus116 The state feedback

gain matrix119870 for this case is obtained as follows

119870 = [minus2473751 minus0019] (41)

The full observer gain matrix119870119890is calculated as follows

119870119890= [200 minus219500]

119879

(42)

The transfer function of the full-order observer controller(119866119904119891) is obtained as follows

119866119904119891(119904) =

minus331800s minus 52710000

1199042 + 1343119904 + 213300 (43)

Figure 4 shows results of the initial response of stem valvewith full-order observer controller Note that the maximumovershoot of valve is about 015 for output and settling timeis 05 second The design is thus acceptable In this case themaximum value of input control signal is about 04

The minimum order observer is of first order Assumethat the desired eigenvalue for minimum order observer is1205821= minus40The state feedback gain matrix 119870 is obtained as

119870 = [minus2473751 minus0019] (44)


Table 2 Settling time overshoot and maximum control input of the response to initial condition of the valve

Method for regulator design Settling time (sec) Overshoot () Maximum control inputPole placement 05 04 09Full-order observer 05 015 04Minimum-order observer 05 025 06Optimal regulator 01 0018 0005

0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01

015Response to initial condition

Stat

e var

iabl

etimes2

(b)

Figure 4 Response to initial condition with full-order observer

0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)

Figure 5 Response of system to initial condition with minimum-order observer

And the minimum observer gain 119870119890can be obtained as

119870119890= 667 (45)

The transfer function of the minimum-order observer con-troller (119866

119904119898) is obtained as follows

119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)

Figure 5 shows results of the initial response of valve stemwith minimum-order observer controller The initial condi-tion is assumed as 119909(0) = [0 01]

119879 Note from Figure 5that the maximum overshoot of valve is about 025 foroutput and settling time is about 05 second The design isthus acceptable In this case control input 119906(119905) is shown inFigure 6 and the maximum value is about 06

Figure 6 shows the comparison of control input signalof the valve with pole placement method (pole) full-orderobserver (full) and minimum-order observer (mini)

We will now consider the optimal control problem ofvalve For this purpose we choose by trial and error 119877 = 1

and 119876 = [001 0

0 001] Solving (14) for 119875 we have

119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)

Substituting 119875 into (12) the optimum gain matrix will be asfollows

119870 = [0 007] (48)

The initial condition is assumed as 119909(0) = [0 01]119879 The

results of the overshoot and settling time to step input forclosed loop system are shown in Figure 7 As shown inFigure 7 the overshoot is very small and less than 0018Also the settling time is 01 sec

The input 119906(119905) = minus119896119909(119905) can be calculated from thepreviously calculated matrices 119896 and 119909 The control inputfor the value of the optimal regulator gain matrix is shown inFigure 8 Note that the maximum input value is about 0005

Transient response parameters are given in Table 2 forcomparing four methods of regulator design with each otherAll designing of regulators of the valve are under the sameinitial condition

7 Conclusion

The aim of this study was the development of designsome regulators to meet transient response specificationsof a pneumatic control valve The mathematical model of


minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time

Comparison of control inputs with three methods

PoleMiniFull

u(t)

Figure 6 The control input 119906(119905) of the system with three methods

0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2

Response to initial condition

(b)

Figure 7 Closed-loop initial response of valve for optimal regulator designed with 119876 and 119877 = 1

0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)

Figure 8 The control input 119906(119905) of the system (with quadraticoptimal)

system and subsequently state equations were derived usingbondgraph method Regulator has been designed usingpole-placement method optimal control full-order state

observer and minimum-order state observer for this systemAlso their responses and then important transient responseparameters were derived and a comparison between thefour designed regulators was made It was observed thatresponses of the pole-placement technique and minimum-order observer are near to each otherThe full order observerregulator has the most high overshoot percentage amongthe other regulators The results show that the minimumovershoot and settling time are achieved in quadratic optimalregulator of pneumatic valve

References

[1] R P Champagne and S J Boyle ldquoOptimizing valve actuatorparameters to enhance control valve performancerdquo ISA Trans-actions vol 35 no 3 pp 217ndash223 1996

[2] T Hagglund ldquoA friction compensator for pneumatic controlvalvesrdquo Journal of Process Control vol 12 no 8 pp 897ndash9042002

[3] M A de Souza L Cuadros C J Munaro and S MunaretoldquoImproved stiction compensation in pneumatic control valvesrdquoComputers and Chemical Engineering vol 38 pp 106ndash114 2012

[4] H Paynter Analysis and Design of Engineering Systems MITPress 1959


[5] D C Karnopp R C Rosenberg and D L Margolis SystemDynamics Modeling Simulation and Control of MechatronicSystems John Wiley amp Sons 5 edition 2012

[6] J U Thoma Simulation By Bondgraphs Introduction To AGraphical Method Springer 2012

[7] P Athanasatos and T Costopoulos ldquoProactive fault finding ina 43-way direction control valve of a high pressure hydraulicsystem using the bond graph method with digital simulationrdquoMechanism and Machine Theory vol 50 pp 64ndash89 2012

[8] V Dıaz-Zuccarini D Rafirou J LeFevre D R Hose and P VLawford ldquoSystemic modelling and computational physiologythe application of Bond Graph boundary conditions for 3D car-diovascular modelsrdquo Simulation Modelling Practice andTheoryvol 17 no 1 pp 125ndash136 2009

[9] J C Mackanic Design Construction and Evaluation of ASimulated Geothermal Flow System University of California1980

[10] M Heidari and H Homaei ldquoStem control of a sliding-stempneumatic control valve using a recurrent neural networkrdquoAdvances in Artificial Neural Systems vol 2013 Article ID410870 7 pages 2013

[11] K OgataModern Control Engineering Prentice Hall 5 edition2010

[12] N S Nise Control System Engineering John Wiley amp Sons 6edition 2010

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Figure 1 Pneumatic control valve with diaphragm

compensates for static friction (stiction) in pneumatic controlvalves [2] The compensation is obtained by adding pulsesto the control signal The characteristics of the pulses aredetermined from the control action

The compensator is implemented in industrial controllersand control systems and the industrial experiences showthat the procedure reduces the control error during stick-slip motion significantly compared to standard control with-out stiction compensation The oscillations caused by staticfriction (stiction) in pneumatic control valves cause lossesin quality and expense of raw materials The input-outputbehavior of a pneumatic control valve affected by stiction invalve De Souza et al [3] presented a well-known stictioncompensationmethod that reduces variability both at processvariable and pneumatic valve stemmovementThe two-movemethod is revisited in this paper and it was shown thatassumptions on the knowledge of steady-state stem positionof control valve that assures equality of set point and thecontrolled variable were not easily achievable Bondgraph isa graphical representation of a physical dynamics systemIt is similar to the better known block diagram and signalflow with the major difference that the arcs in bondgraphsrepresent bidirectional exchange of physical energy whilethose in block diagrams and signal-flow graphs representunidirectional flow of information Also bondgraphs aremultidomain and domain neutral This means that a bond-graph can incorporate multiple domains simultaneouslyBondgraphs were devised by Paynter [4] at MIT in April 1959and subsequently developed into a methodology togetherwith Karnopp et al [5] Early prominent promoters ofbondgraph modeling techniques among others were Thoma[6] Dixhoorn and Dransfield Athanasatos and Costopoulos[7] used the bondgraphmethod for finding the proactive faultin 43 way direction control valve of a high pressure hydraulicsystemThe accuracy of the bondgraphmodel was verified bycomparing its response to the response of an actual hydraulicsystem Dıaz-Zuccarini et al [8] utilized the bondgraph asboundary condition for a detailed model of an idealized

I

1 121 3

4

5

6

R

SE TF

C

Figure 2 Bondgraph control valve actuator

mitral valve A specific application in cardiovascular model-ing was demonstrated by focusing on a specific example a3D model of the mitral valve coupled to a lumped parametermodel of the left ventricle

In this study a pneumatic control valve is modeled bybondgraph method A regulator has been designed usingpole-placement method optimum regulator full-order stateobserver and minimum-order state observer for this systemIn continuing their responses have been compared with eachother

This research is organized as follows Section 2 recalls thebondgraph model of valve and proposes equations of motionof valve Section 3 develops quadratic optimal regulatorof pneumatic valve Sections 4 and 5 present designing aregulator by pole-placement technique full and minimumorder observers respectively Section 6 shows the results offull and minimum order regulator observers and quadraticoptimal regulator of valve and finally some conclusions aregiven in Section 7

2 Bondgraph Model of Valve andTransfer Function

The bondgraph model of the valve is shown in Figure 2In this model SE is the inlet pressure of the system Thepressure changes to force by multiplying in effect area of thediaphragm In bondgraph this transformer is modeled byTF Element 119877 is the friction of the system Element 119868 is themovable mass of valve and diaphragm

Element 119862 represents the spring of the valve actuatorAlso 1-junction is a common flow junction 1-junctions haveequality of flows and the efforts sum up to zero with thesame power orientation In fact junctions can connect twoor more bonds The direction of the half arrows () denotesthe direction of power flow given by the product of the effortand flow variables associatedwith the power bondThe bondsin a bondgraphmay be numbered sequentially using integersstarting with 1 The two 1-junctions in the bondgraph showncan be uniquely identified as (S 1 2) and (S 4 5 6) similarlysymbols like 119878119864

1and 119877

6can be used to identify a particular

element This system has two state variables 1198754and 119902

5sdot 1199025is

the displacement of valve stem and the variation of the springlength Also V

4= 11987541198684is the velocity of the valve stem The



4= 119860 times 119878119864

1minus 11987031199025minus

1198776

1198684

1198754 (1)

1199025=

1198754

1198684

(2)


119883 (0) = [1198754(0)

1199025(0)

] = [0

0] (3)


1199025=

4

1198684

(4)


1199025=


1198684

(5)


1199025=

1

1198684


1199025) (6)


1199025(119904)

1198781198641(119904)

=(1198601198684)

1199042 + (11987761198684) 119904 + (119870

31198684) (7)




= 119860119909 + 119861119906 (8)

119910 = 119862119909 + 119863119906 (9)


119906 (119905) = minus119870119909 (119905) (10)



119869 = int

infin

0

(119909119879119876119909 + 119906

119879119877119906) 119889119905 (11)



119870 = 119877minus1119861119879119875 (12)




119860119879119875 + 119875119860 minus 119875119861119877

minus1119861119879119875 + 119876 = 0 (14)




1and 120583

2(desired


119872 = [119860 119860119861] (15)


|119904119868 minus 119860| = 1199042+ 1198861119904 + 1198862 (16)





(119904 minus 1205831) (119904 minus 120583

2) = 1199042+ 1205721119904 + 1205722 (17)




119870 = [1205722 minus 1198862

1205721minus 1198861] 119879minus1 (18)


where

119879 = 119872119882 (19)


119882 = [1198861

1

1 0] (20)




119873 = [119862119879

119860119879119862119879] (21)


119870119890= (119882119873

119879)minus1

[1205722 minus 1198862

1205721minus 1198861]119879

(22)


119880 (119904)

minus119884 (119904)= 119870(119904119868 minus 119860 + 119870

119890119862 + 119861119870)

minus1

119870119890 (23)


1003816100381610038161003816119904 minus 119860119887119887

+ 119870119890119860119886119887

1003816100381610038161003816 = 119904 + 1 (24)




119860 = [119860119886119886

119860119886119887

119860119887119886

119860119887119887

] 119861 = [119861119886

119861119887

] (25)


119870119890= (

119879)minus1

(1minus 1198861) (26)

where

= 119860119886119887 = 119886

1 (27)



119887119887

1003816100381610038161003816 = 119904 + 1198861 (28)


119880 (119904)


minus1

119861 + 119863] (29)

where

119860 = 119860 minus 119865119870119887

119861 = 119861 minus 119865 (119870119886+ 119870119887119870119890)

119862 = minus119870119887

119863 = minus (119870119886+ 119870119887119870119890)

(30)

where

119860 = 119860119887119887

minus 119870119890119860119886119887

119861 = 119860119870119890+ 119860119887119886

minus 119870119890119860119886119886

119865 = 119861119887minus 119870119890119861119886

(31)




1199025(119904)

1198781198641(119904)

=653

1199042 + 3333119904 + 2263333 (32)


119860 = [0 1


119861 = [0

9142] 119862 = [1 0] 119863 = 0

(33)


119879









119894(119894 = 1 2) where

1205831= minus8 + 1198951091 120583

2= minus8 minus 1198951091 (34)


119872 = [0 9142

9142 minus30470] (35)



desired system is

|119904119868 minus 119860 + 119861119870|

=

10038161003816100381610038161003816100381610038161003816

[119904 0

0 119904] minus [

0 1

minus22633333 minus3333] + [

0

9142] [1198961 119896

2]

10038161003816100381610038161003816100381610038161003816

= 1199042+ (9142119896

2+ 3333) 119904 + 9142119896

1+ 226331198905

(36)


(119904 + 1205831) (119904 + 120583

2) = (119904 + 8 + 1198951091) (119904 + 8 minus 1198951091)

= 1199042+ 16119904 + 1831

(37)


1198961= minus24734 119896

2= minus0019 (38)

or

119870 = [minus24734 minus0019] (39)



0 01 02 03 04 05 06 07 08 09 1minus2

0

2


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0005


Stat

e var

iabl

etimes2

t (s)

(b)



119873 = [1 0

0 1] (40)


1= minus116 and 120582



119870 = [minus2473751 minus0019] (41)


119870119890= [200 minus219500]

119879

(42)


119866119904119891(119904) =


1199042 + 1343119904 + 213300 (43)



119870 = [minus2473751 minus0019] (44)




0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01


Stat

e var

iabl

etimes2

(b)


0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)



119870119890= 667 (45)



119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)





and 119876 = [001 0


119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)


119870 = [0 007] (48)





7 Conclusion



minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation











4= 119860 times 119878119864

1minus 11987031199025minus

1198776

1198684

1198754 (1)

1199025=

1198754

1198684

(2)


119883 (0) = [1198754(0)

1199025(0)

] = [0

0] (3)


1199025=

4

1198684

(4)


1199025=


1198684

(5)


1199025=

1

1198684


1199025) (6)


1199025(119904)

1198781198641(119904)

=(1198601198684)

1199042 + (11987761198684) 119904 + (119870

31198684) (7)




= 119860119909 + 119861119906 (8)

119910 = 119862119909 + 119863119906 (9)


119906 (119905) = minus119870119909 (119905) (10)



119869 = int

infin

0

(119909119879119876119909 + 119906

119879119877119906) 119889119905 (11)



119870 = 119877minus1119861119879119875 (12)




119860119879119875 + 119875119860 minus 119875119861119877

minus1119861119879119875 + 119876 = 0 (14)




1and 120583

2(desired


119872 = [119860 119860119861] (15)


|119904119868 minus 119860| = 1199042+ 1198861119904 + 1198862 (16)





(119904 minus 1205831) (119904 minus 120583

2) = 1199042+ 1205721119904 + 1205722 (17)




119870 = [1205722 minus 1198862

1205721minus 1198861] 119879minus1 (18)


where

119879 = 119872119882 (19)


119882 = [1198861

1

1 0] (20)




119873 = [119862119879

119860119879119862119879] (21)


119870119890= (119882119873

119879)minus1

[1205722 minus 1198862

1205721minus 1198861]119879

(22)


119880 (119904)

minus119884 (119904)= 119870(119904119868 minus 119860 + 119870

119890119862 + 119861119870)

minus1

119870119890 (23)


1003816100381610038161003816119904 minus 119860119887119887

+ 119870119890119860119886119887

1003816100381610038161003816 = 119904 + 1 (24)




119860 = [119860119886119886

119860119886119887

119860119887119886

119860119887119887

] 119861 = [119861119886

119861119887

] (25)


119870119890= (

119879)minus1

(1minus 1198861) (26)

where

= 119860119886119887 = 119886

1 (27)



119887119887

1003816100381610038161003816 = 119904 + 1198861 (28)


119880 (119904)


minus1

119861 + 119863] (29)

where

119860 = 119860 minus 119865119870119887

119861 = 119861 minus 119865 (119870119886+ 119870119887119870119890)

119862 = minus119870119887

119863 = minus (119870119886+ 119870119887119870119890)

(30)

where

119860 = 119860119887119887

minus 119870119890119860119886119887

119861 = 119860119870119890+ 119860119887119886

minus 119870119890119860119886119886

119865 = 119861119887minus 119870119890119861119886

(31)




1199025(119904)

1198781198641(119904)

=653

1199042 + 3333119904 + 2263333 (32)


119860 = [0 1


119861 = [0

9142] 119862 = [1 0] 119863 = 0

(33)


119879









119894(119894 = 1 2) where

1205831= minus8 + 1198951091 120583

2= minus8 minus 1198951091 (34)


119872 = [0 9142

9142 minus30470] (35)



desired system is

|119904119868 minus 119860 + 119861119870|

=

10038161003816100381610038161003816100381610038161003816

[119904 0

0 119904] minus [

0 1

minus22633333 minus3333] + [

0

9142] [1198961 119896

2]

10038161003816100381610038161003816100381610038161003816

= 1199042+ (9142119896

2+ 3333) 119904 + 9142119896

1+ 226331198905

(36)


(119904 + 1205831) (119904 + 120583

2) = (119904 + 8 + 1198951091) (119904 + 8 minus 1198951091)

= 1199042+ 16119904 + 1831

(37)


1198961= minus24734 119896

2= minus0019 (38)

or

119870 = [minus24734 minus0019] (39)



0 01 02 03 04 05 06 07 08 09 1minus2

0

2


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0005


Stat

e var

iabl

etimes2

t (s)

(b)



119873 = [1 0

0 1] (40)


1= minus116 and 120582



119870 = [minus2473751 minus0019] (41)


119870119890= [200 minus219500]

119879

(42)


119866119904119891(119904) =


1199042 + 1343119904 + 213300 (43)



119870 = [minus2473751 minus0019] (44)




0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01


Stat

e var

iabl

etimes2

(b)


0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)



119870119890= 667 (45)



119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)





and 119876 = [001 0


119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)


119870 = [0 007] (48)





7 Conclusion



minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










where

119879 = 119872119882 (19)


119882 = [1198861

1

1 0] (20)




119873 = [119862119879

119860119879119862119879] (21)


119870119890= (119882119873

119879)minus1

[1205722 minus 1198862

1205721minus 1198861]119879

(22)


119880 (119904)

minus119884 (119904)= 119870(119904119868 minus 119860 + 119870

119890119862 + 119861119870)

minus1

119870119890 (23)


1003816100381610038161003816119904 minus 119860119887119887

+ 119870119890119860119886119887

1003816100381610038161003816 = 119904 + 1 (24)




119860 = [119860119886119886

119860119886119887

119860119887119886

119860119887119887

] 119861 = [119861119886

119861119887

] (25)


119870119890= (

119879)minus1

(1minus 1198861) (26)

where

= 119860119886119887 = 119886

1 (27)



119887119887

1003816100381610038161003816 = 119904 + 1198861 (28)


119880 (119904)


minus1

119861 + 119863] (29)

where

119860 = 119860 minus 119865119870119887

119861 = 119861 minus 119865 (119870119886+ 119870119887119870119890)

119862 = minus119870119887

119863 = minus (119870119886+ 119870119887119870119890)

(30)

where

119860 = 119860119887119887

minus 119870119890119860119886119887

119861 = 119860119870119890+ 119860119887119886

minus 119870119890119860119886119886

119865 = 119861119887minus 119870119890119861119886

(31)




1199025(119904)

1198781198641(119904)

=653

1199042 + 3333119904 + 2263333 (32)


119860 = [0 1


119861 = [0

9142] 119862 = [1 0] 119863 = 0

(33)


119879









119894(119894 = 1 2) where

1205831= minus8 + 1198951091 120583

2= minus8 minus 1198951091 (34)


119872 = [0 9142

9142 minus30470] (35)



desired system is

|119904119868 minus 119860 + 119861119870|

=

10038161003816100381610038161003816100381610038161003816

[119904 0

0 119904] minus [

0 1

minus22633333 minus3333] + [

0

9142] [1198961 119896

2]

10038161003816100381610038161003816100381610038161003816

= 1199042+ (9142119896

2+ 3333) 119904 + 9142119896

1+ 226331198905

(36)


(119904 + 1205831) (119904 + 120583

2) = (119904 + 8 + 1198951091) (119904 + 8 minus 1198951091)

= 1199042+ 16119904 + 1831

(37)


1198961= minus24734 119896

2= minus0019 (38)

or

119870 = [minus24734 minus0019] (39)



0 01 02 03 04 05 06 07 08 09 1minus2

0

2


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0005


Stat

e var

iabl

etimes2

t (s)

(b)



119873 = [1 0

0 1] (40)


1= minus116 and 120582



119870 = [minus2473751 minus0019] (41)


119870119890= [200 minus219500]

119879

(42)


119866119904119891(119904) =


1199042 + 1343119904 + 213300 (43)



119870 = [minus2473751 minus0019] (44)




0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01


Stat

e var

iabl

etimes2

(b)


0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)



119870119890= 667 (45)



119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)





and 119876 = [001 0


119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)


119870 = [0 007] (48)





7 Conclusion



minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation

















119894(119894 = 1 2) where

1205831= minus8 + 1198951091 120583

2= minus8 minus 1198951091 (34)


119872 = [0 9142

9142 minus30470] (35)



desired system is

|119904119868 minus 119860 + 119861119870|

=

10038161003816100381610038161003816100381610038161003816

[119904 0

0 119904] minus [

0 1

minus22633333 minus3333] + [

0

9142] [1198961 119896

2]

10038161003816100381610038161003816100381610038161003816

= 1199042+ (9142119896

2+ 3333) 119904 + 9142119896

1+ 226331198905

(36)


(119904 + 1205831) (119904 + 120583

2) = (119904 + 8 + 1198951091) (119904 + 8 minus 1198951091)

= 1199042+ 16119904 + 1831

(37)


1198961= minus24734 119896

2= minus0019 (38)

or

119870 = [minus24734 minus0019] (39)



0 01 02 03 04 05 06 07 08 09 1minus2

0

2


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0005


Stat

e var

iabl

etimes2

t (s)

(b)



119873 = [1 0

0 1] (40)


1= minus116 and 120582



119870 = [minus2473751 minus0019] (41)


119870119890= [200 minus219500]

119879

(42)


119866119904119891(119904) =


1199042 + 1343119904 + 213300 (43)



119870 = [minus2473751 minus0019] (44)




0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01


Stat

e var

iabl

etimes2

(b)


0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)



119870119890= 667 (45)



119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)





and 119876 = [001 0


119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)


119870 = [0 007] (48)





7 Conclusion



minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation












0 01 02 03 04 05 06 07 08 09 1minus1

0

1


Stat

e var

iabl

etimes1

t (s)

(a)

0 01 02 03 04 05t (s)

06 07 08 09 1minus005

0

005

01


Stat

e var

iabl

etimes2

(b)


0 01 02 03 04 05 06 07 08 09 1minus1

0

1

2

3 times10minus3

t (s)

x1

(a)

0 01 02 03 04 05 06 07 08 09 1minus005

0

005

01

015

t (s)

x2

(b)



119870119890= 667 (45)



119866119904119898

(119904) =minus2483119904 minus 15500

119904 + 6267 (46)





and 119876 = [001 0


119875 = [3945 21119890 minus 6

21119890 minus 6 15119890 minus 5] (47)


119870 = [0 007] (48)





7 Conclusion



minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation










minus02

0

02

04

06

08

1

0 02 04 06 08 1 12Time


PoleMiniFull

u(t)


0 005 01 015 02 025 03 035 04 045 05minus2

minus1

0

1


t (s)

x1

(a)

0 005 01 015 02 025 03 035 04 045 05minus01

minus005

0

005

01

t (s)

x2


(b)


0 005 01 015 02 025 03 035 04 045 05minus8

minus6

minus4

minus2

0

2

4

6

u(t)

times10minus3

t (s)




References
















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation




















RoboticsJournal of





Journal of



RotatingMachinery





VLSI Design



Shock and Vibration







Journal of



Volume 2014


SensorsJournal of





Propagation









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