Control of Hydraulic Servo Robot

Transcription

1 Control of Hydraulic Servo Robot -Develop control strategies for a hydraulic actuated robot arm - Project Report INTRO MCE764 Aalborg University Department of Energy Technology Pontoppidanstræde 101 DK-9220 Aalborg

2 Copyright c Aalborg University 2012 This document has been created using LaTex. For simulation SIMULINK have been used, together with MATLab. The figures have been made using data logged with LabView, and plotted in MATLab.

3 Title: Control of Hydraulic Servo Robot Semester: 1st Semester theme: Control of hydraulic actuated mechanical systems Project period: to ECTS: 10 Supervisor: Hamidreza Shaker Project group: INTRO-MCE764 SYNOPSIS: Aline Saint Svein Erik Svendsen Triantafyllos Dernektsis Different controllers has been developed and tested for a one degree of freedom hydraulic actuated servo robot. The system, hydraulic and mechanic, will be modeled. These models were nonlinear, and implemented in SIMULINK. The models were linearised for control design purposes. Different controllers has been developed using the linear models, and tested on the nonlinear models in SIMULINK. The best controllers will be implemented on the real physical robot. Copies: 2 Pages, total: 32 Appendix: 51 Supplements: CD w/ models, references,.m-files and LabView program By signing this document, each member of the group confirms that all participated in the project work and thereby all members are collectively liable for the content of the report.

4

5 Contents Preface v 1 Introduction 1 2 Modeling Hydraulic Modeling Linearization of non-linear hydraulic mathematical model Mechanical Modeling Controllers 13 4 Implementation 15 5 Conclusion 19 Bibliography 23 A Modeling 25 A.1 Model of hydraulic system A.2 Non-linear Mathematical Model A.3 Orifice equations of the proportional/servo directional valve A.4 Continuity equations of the valve-cylinder combination A.5 Momentum equation A.6 Symbols of hydraulic nonlinear mathematical model A.7 Simulink implementation of hydraulic model A.8 Linearization of non-linear hydraulic mathematical model A.9 Linearization of orifice equations A.10 Linearization of continuity equations A.11 Transfer function of valve-cylinder combination A.12 Addition of high pass filter into the hydraulic system A.13 Validation of linear model A.14 Validation of nonlinear model A.15 Adjustable parameters in the nonlinear model A.16 Validation with the usage of a sinusoidal input iii

6 iv Contents A.17 Mechanical Modeling A.18 General mechanics/newton s second law A.19 Determination of the torque A.19.1 Application of Al-Kashi s law (law of cosines) A.20 Denavit Hartberg Parameters B Controllers 55 B.1 Control B.2 Signal flow in the hydraulic system B.3 Linear Controllers B.3.1 Relative Stability Specifications B.4 Proportional Controller B.4.1 Open loop frequency response without high pass filter B.4.2 Open loop frequency response with high pass filter B.4.3 Proportional Controller tested in the nonlinear model B.5 Proportional Integral Controller B.5.1 Integral Windup B.5.2 Open loop frequency response without high pass filter B.5.3 Open loop frequency response with high pass filter B.5.4 Proportional-Integral Controller tested in the nonlinear model 63 B.6 Proportional Lag Controller B.6.1 Open loop frequency response without high pass filter B.6.2 Open loop frequency response with high pass filter B.6.3 Proportional-Lag Controller tested in the nonlinear model.. 65 C Implementation 67 C.1 Physical setup C.2 P controller C.3 PI controller C.4 P-lag controller D Conclusion 75 D.1 Conclusion

7 Preface For our 7th semester, M.S.c. students are supposed to work on a 4 months project. We picked a project around Mechatronics. The project is Control of a hydraulic servo robot two D.O.F. To succeed in the control we had to model the hydraulic and the mechanics. Then we made a model of different controllers. The laboratory work consisted in the validation of our models and then in the implementation of the controllers. This report contain all the informations we used to deliver the project Aalborg University, December 14, 2012 Aline Saint <asaint12@student.aau.dk> Svein Erik Svendsen <ssvend12@student.aau.dk> Triantafyllos Dernektsis <tderne11@student.aau.dk> v

8 vi Preface

9 Chapter 1 Introduction The project will deal with one hydraulic actuator, which is mounted on a two link robot. This will give one degree of freedom for the system. Controlling a hydraulic actuator is not straight forward. Hydraulic systems have several parameters like pressure, flow and volume. A small change in any of these parameters can cause severe problems for the controlling of the system. To create a successful control it will require good modeling of the system, a careful approach with linearizing, and a stable and functional controller. This system is interesting to study. It will give experience of modeling a physical system, and dealing with physical systems. It will also give valuable knowledge of the workings of hydraulics. The project s objective is to successfully control the hydraulic actuated robot. To achieve this several steps are needed. A good model needs to be created for the system. The model will be linearized for control purposes. For control development and implementation, a classical PID-controller will be used. Information regarding the PID controller will be collected from courses, previous studies done on the system, together with the group s research to create an adequate controller. This will ensure a good controller, with the time given for the project. 1

10 2 Chapter 1. Introduction

11 Chapter 2 Modeling This chapter will explain the modeling of the hydraulic servo robot. The hydraulic and the mechanical part will be discussed Hydraulic Modeling Hydraulic servo systems are used as drives in many industrial applications. That is because of the power ratios and their ability to develop large torques and forces. For instance they can be used as robot drives, a topic which will be studied in the following chapter. In general the dynamics of a hydraulic servo system are very nonlinear. Also these systems characterized by low damping ratio. The hydraulic servo system which is used as a drive for the hydraulic servo robot is presented at Figure 2.1. In order to model the hydraulic system, some assumptions will be taken into account. These assumptions will result in a less complicated mathematical model. The hydraulic components which will be neglected are Safety valves Accumulators Pressure relief valves Additionally, the bulk modulus of the oil is considered constant The two hydraulic cylinders are identical, hence only the first one will be used for the mathematical model. The valve-cylinder-load schematic is presented at Figure

12 4 Chapter 2. Modeling Figure 2.1: Valve Cylinder Arrangment [2] The basic equations describing the dynamics of the valve-cylinder-load arrangement are the orifice equations of the proportional/servo directional valve, the continuity equations of system s volumes and the momentum equation deriving from Newton s 2nd law of motion. The servo valve driving the hydraulic cylinders is MOOG D761 series valve. The input signal to the valve is directly proportional with the position of the spool. The applied input signal has the maximum amplitude of 10 V. At this amplitude the maximum valve opening is achieved. Accordingly at -10 V there is the maximum opening in the opposite direction. is The pressure drop at nominal flow of 19 liters minute or m s 1000psi or P a Using these data the valve coefficient K v will be determined. The equation which describes the relationship between the flow through the valve and the pressure drop is given by the orifice equation: 2 Q N = C d A pn Q N = K v pn (2.1) ρ

13 5 Hence, inserting the nominal flow and the nominal pressure difference into the equation above, the valve coefficient is easily determined K v = 1.2 m s P a. Furthermore, the dynamics of the valve are described by a second order transfer function in the form of: x s v = ωn 2 s 2 + 2ζω n s + ωn 2 (2.2) In such a way the input signal v is related to the travel of the spool. The value of the natural frequency ω n = 3146 rad/s and the damping ζ = 1 are given by the data sheet of the valve.[1] At this stage some assumptions must be made: The geometry of the valve is assumed ideal and the orifices are matched and symmetrical The discharge coefficients for the orifices are equal The return pressure is neglected because it is smaller than the other pressures of the system The neutral position of the spool is chosen to be the position where all the flows are equal to zero. This position is defined at x s =0 The compressibility of the fluid is considered zero The orifice areas vary linear with the travel of the spool Taking into account these assumptions, the equations which describe the flows from the valve to the cylinder and reversely are } Q 1 = K v x s ps p 1 for x s >0 (2.3) Q 2 = K v x s p2 p t and } Q 1 = K v x s p1 p t Q 2 = K v x s ps p 2 for x s <0 (2.4)

14 6 Chapter 2. Modeling The continuity equation from the valve to the 1st cylinder chamber is defined as Q 1 0 = A p ẋ p + C l (p 1 p 2 ) + V 1 p 1 β F t (2.5) 0 Q 2 = A p ẋ p C l (p 1 p 2 ) + V 2 p 2 β F t (2.6) However, the volumes V 1 and V 2 are not constant as they vary with the piston position. For this reason they can be modeled using the equations V 1 = V initial1 + A p x p (2.7) V 2 = V initial2 A p x p (2.8) Finally, applying the Newton s second law of motion at the cylinder and the load, the resulting momentum equation defined as: Mẍ p = p 1 A p p 2 A p Bẋ p F L (2.9) All the symbols which are used for the hydraulic nonlinear mathematical modeling are presented at Table Linearization of non-linear hydraulic mathematical model At this chapter the non-linear hydraulic model will be linearized using the Taylor series. Initially some assumption must be considered. These assumptions are: The gravitational, centrifugal and coriolis effects are neglected The volumes in the servo valve and hoses are neglected. The volumes under consideration are the volumes at the cylinder chambers. The response of servo valve is considered much faster in contrast with the rest of the system. For this reason it will be neglected during the linearisation. The flow entering into the cylinder is equal to the flow getting out from the cylinder (steady state assumption) Coloumb friction is neglected The aim of the linearization is the derivation of valve-cylinder transfer function. Knowing the transfer function of the linear system, the frequency response can be determined and the controller can be designed. The linearisation of orifice equations can be achieved making a steady state assumption regarding the flows from the valve to the load and reversely. The steady state assumption states that the flow entering the cylinder is equal to the flow getting out from the cylinder, for positive travel of the spool x s > 0.

15 7 Symbol Q N C d A ρ p N K v Q 1 Q 2 p s p t p 1 p 2 x s A p x p ẋ p ẍ p V 1 V 2 C l V initial1 V initial2 M F L B Description nominal flow through the valve discharge coefficient of the valve orifice area density of the fluid nominal pressure difference valve coefficient flow from the valve to cylinder and reverse flow from the cylinder to valve and reverse supply pressure tank pressure pressure in the first chamber pressure in the second chamber normalized displacement of the spool piston effective area position of the piston velocity of the piston acceleration of the piston volume of the 1st chamber, connecting line, valve volume of the 2nd chamber, connecting line, valve total leakage coefficient initial volume at the 1st cylinder chamber initial volume at the 2nd cylinder chamber mass of the load load force viscous friction coefficient of the piston Table 2.1: Symbols of hydraulic nonlinear mathematcal model Linearization result: p s sign( x s0 )p L0 Q L = K v x s + K v x s0 sign( x s0 ) p L (2.10) 2 4 ps 2 p L0sign( x s0 ) 2 Q L = k q x s k qp p L (2.11) The partials in the equation above define two important parameters. The first partial is defined as the flow gain k q k q = Q L x s = K v p s sign( x s0 )p L0 2 (2.12)

16 8 Chapter 2. Modeling and the second partial is defined as the flow-pressure coefficient k qp k qp = Q L p L = K v x s0 sign( x s0 ) 4 ps 2 p L0sign( x s0 ) 2 (2.13) The two parameters above are called valve coefficient and are very important in the dynamic analysis of the valve-cylinder combination. The flow gain affects the open loop gain of the system and for this reason influence the stability of the system. The flow-pressure coefficient affects the damping in the valve-cylinder combination. The values of the linearisation load pressure p L0 and the linearisation nominal spool travel x s0 are chosen as: p L0 = 40e5 Pa and x s0 = Linearization of continuity equations V t = V 1 + V 2 = V 1initial + A p x p + V 2initial A p x p = V 1initial + V 2initial = 2V 0 0 (2.14) Laplace transforming will give the end result: Mechanical Modeling Q L (s) = A p ẋ p (s) + C l p L (s) + V t 4β F sp L (s) (2.15) This is the process for the mechanical modeling. The mechanical aspects of the robot will be examined and a complete model describing the mechanics will be derived. To simplify the modeling an angle of 90 degrees was set for the second joint. By locking the second of the two possible joints, the system will now only have 1 moving joint, meaning it only have 1 degree of freedom. For modeling purposes, the robot is now regarded as a pendulum, which will greatly simplify the modeling of the system. For the mechanical modeling an expression of the angle between the robot arm and the actuator is needed. The angle will be expressed in an "universal" reference frame. This experssion will include the piston position of the actuator. It is then possible to combine the mechanical and the hydraulic models to make a complete model of the system. With this combination of models it will be possible to design the controller. The controller can only be designed after the model equations have been linearised. In the mechanical modeling is it essential to also know which forces are applied on the robot arm to design a hydraulic system strong enough to make it move and stay and in the wanted position. The forces applied on the robot arm, described here as a free body diagram All this information will be used in MATLAB to build a model. With this model it will be possible to anticipate the reaction of the robot. The Newton s second law for rigid body will be applied at the point O which is the rotation center. J tot θ = τ Mtot g OG (2.16)

17 9 Figure 2.2: expression of the angle between the robot arm and the actuator [1] Figure 2.3: forces applied on the robot [1] with where is the total inertia, J tot = J O1 + m 2 L J B2 J tot J O1

18 10 Chapter 2. Modeling is the inertia of link 1 at the point O and J B2 is the inertia of link 2 at the point B In order to express this analysis, two models will be created as in the Figure 2.4 Figure 2.4: Modelisation of the position of the center mass This model is reduced as shown in Figure 2.5 where Figure 2.5: Torque application M tot = m 1 + m 2 The model need to be linearised in order to design a controller for the robot. The method chosen for this system, is to look at energy conservation. 1 2 M eqv 2 eq = 1 2 MV 2 (2.17)

19 11 The real system can be represented as: Figure 2.6: Nonlinear mechanical system Linearisation will result in a new representation of the system: Figure 2.7: Linear mechanical system The velocities V eq and V need to be expressed with known variables and parameters. From Figure 2.6 V eq can be expressed as: From figure 2.6 V can be expressed as: V eq = θ 1 a (2.18) V = ( θ 1 (a + b)) (2.19)

20 12 Chapter 2. Modeling To derive the linear model, the worst possible angle is chosen. This angle will be when the robot arm is closest to the ground, thus creating maximum force seen from the actuator. The angle is changed up and down with small values to measure the force in the linearisation point. The equivalent mass is also calculated (a + b)2 M eq = a 2 M (2.20) All the symbols which are used for the mechanical mathematical modeling are presented at Table 2.2 Symbol M eq V eq M V a b Description mass in the linear model velocity in the linear model real mass of the system real velocity of the system length from fixed rotation point to applied force length from applied force to center of load mass Table 2.2: Symbols of mechanical mathematical model

21 Chapter 3 Controllers The aim of design of linear controller is to force the robot to follow a specific position reference inside the the working area. The controllers will be tuned with and without a high pass filter. Relative stability specifications for the system is one of the most important parts, and must be designed with great care. For an adequate transient response with reduced overshoot, the relative stability margins must be increased. Also another reason for large stability margin is the inaccuracies in the mathematical model. The most common values of gain and phase margin for reasonable control system are GM > 6 8 db and P M > degrees. The proportional controller is a pure gain and the transfer function is given by: G c (s) = K P (3.1) Adding the gain compensator into the system, the bode plot magnitude will be shifted up or down (with no change in the magnitude shape) according with the value of gain compensator. The phase shape will be unchanged. As much as concerning the response of the system, the addition of the P controller will improve the transient response of the system. In order to determine the gain, the Matlab software will be used and specifically the Control Systems Toolbox. For the specified gain and phase margin, the compensator proportional gain is found to be K P = 10 The open loop frequency response of the system combined with the proportional controller is presented at Figure 3.1 The gain and the phase margin are GM=9.21 db, PM=89.3 degrees. The controller s gain for the system combined with the high pass filter has been determined as: K P = 40 13

22 14 Chapter 3. Controllers Figure 3.1: Open loop frequency response with P-controller The open loop frequency response combined with the proportional controller and filter is presented at Figure 3.2 Figure 3.2: Open loop frequency response with P-controller and filter The gain and the phase margin are GM=20.7 db, PM=47.6 degrees.

23 Chapter 4 Implementation The P controller is tested in the laboratory. A step signal is applied on the robot and the response of the system is presented at Figure 4.1. Figure 4.1: Tested step response with P controller 15

24 16 Chapter 4. Implementation The system respond fast with a very small steady state error which is considered insignificant. In order to investigate the difference (error) between the reference signal and the nonlinear/physical setup signals, two plots will be used. The first presents the difference between the reference and the tested signal at Figure 4.2 and the second plot presents the signal difference between the reference and the simulated nonlinear model at Figure 4.3. The error tolerance is considered 1 cm. Figure 4.2: Difference between reference and tested signal for P controller It is obvious form the figure above, that the error between the reference and the tested signal is 2.5 cm. It is interesting to note that as the time is increasing the error is decreasing, something encouraging for further testing the system with P controllers and sinusoidal tracking.

25 17 Figure 4.3: Difference between reference and simulated signal for P controller The error between the reference and the simulated is relative small, less than 0.5 cm.

26 18 Chapter 4. Implementation

27 Chapter 5 Conclusion The implementation of a P-controller on the robot have been discussed. By looking at the test results it can be concluded that the p-controller will follow the referemce best. This is the same result as expected, based on the hydraulic control theory. The P-controller is fairly easy to implement, but it may not fulfill the design requirements for the system. In order to reach the required preformance, a more complicated controller may be needed. To further increase the preformance of the robot, an adaptive controller or a sliding mode controller can be implemented. 19

28 20 Chapter 5. Conclusion

29 Acknowledgment The authors would like to thank Henrik C. Pedersen, Torben Ole Andersen, Michael M. Bech and Hamidreza Shaker at Aalborg University for theory on hydraulic, mechanic and controlling. 21

30 22 Chapter 5. Conclusion

31 Bibliography [1] Servo Robot data sheet [2] Michael H. Sørensen, Morten H. Sørensen, Anders R. Pedersen and Mikael Højen, Control Strategies for a Hydraulic Servo Robot Aalborg University, Denmark, [3] Charles L. Phillips, John M. Parr, Feedback Control Systems 5th edition Prentice Hall, 1 Lake Street, Upper Saddle River, NJ 07458, 2011 [4] Michael Rygaard Hansen, Torben Ole Andersen, Fluid Power Systems, 2nd edition Aalborg University, Denmark, March 2003 [5] Kristian Holm Nielsen, Lasse Schmidt, Comparison of Advanced Control Schemes Implemented on Hydraulic Actuated Robot Manipulator Aalborg University, Denmark, 2008 [6] Dennis Bertelsen, Nikolaj Kristensen, Thomas Hammerholt Madsen and Claus Vad, Trajectory Control of a Hydraulic Planar Elbow Servo Robot Aalborg University, Denmark,

32 24 Bibliography

33 Appendix A Modeling A.1 Model of hydraulic system Hydraulic servo systems are used as drives in many industrial applications. That is because of the power ratios and their ability to develop large torques and forces. For instance they can be used as robot drives, a topic which will be studied in the following chapter. In general the dynamics of a hydraulic servo system are very nonlinear. Also these systems characterized by low damping ratio. The hydraulic servo system (circuit) which is used as a drive for the hydraulic servo robot is presented at Figure A.1. Figure A.1: Hydraulic Circuit of Servo Robot [1] 25

34 26 Appendix A. Modeling A.2 Non-linear Mathematical Model In order to model the hydraulic system, some assumptions will be taken into account. These assumptions will result in a less complicated mathematical model. The hydraulic components which will be neglected are Safety valves Accumulators Pressure relief valves Additionally, the bulk modulus of the oil is considered constant The valve-cylinder-load schematic is presented at Figure A.2. Figure A.2: Valve Cylinder Load Arrangment

35 A.3. Orifice equations of the proportional/servo directional valve 27 The basic equations describing the dynamics of the valve-cylinder-load arrangement are the orifice equations of the proportional/servo directional valve, the continuity equations of system s volumes and the momentum equation deriving from Newton s 2nd law of motion. A.3 Orifice equations of the proportional/servo directional valve The servo valve driving the hydraulic cylinders is MOOG D761 series valve. The input signal to the valve is directly proportional with the position of the spool. The applied input signal has the maximum amplitude of 10 V. At this amplitude the maximum valve opening is achieved. Accordingly at -10 V there is the maximum opening in the opposite direction. The valve flow for maximum opening of the valve as a function of the pressure drop is presented at Figure A.3 Figure A.3: Valve flow for maximum opening of servo valve versus pressure drop [6]

36 28 Appendix A. Modeling is According to the figure A.3, the pressure drop at nominal flow of 19 liters minute or m s 1000psi or P a Using these data the valve coefficient K v will be determined. The equation which describes the relationship between the flow through the valve and the pressure drop is given by the well-known orifice equation: 2 Q N = C d A pn Q N = K v pn ρ (A.1) Hence, inserting the nominal flow and the nominal pressure difference into the equation above, the valve coefficient is easily determined K v = 1.2 m s P a. Furthermore, the dynamics of the valve are described by a second order transfer function in the form of: x s v = ωn 2 s 2 + 2ζω n s + ωn 2 (A.2) In such a way the input signal v is related to the travel of the spool. The value of the natural frequency ω n = 3146 rad/s and the damping are given by the data sheet of the valve. ζ = 1 At this stage some assumptions must be made: The geometry of the valve is assumed ideal and the orifices are matched and symmetrical The discharge coefficients for the orifices are equal The return pressure is neglected because it is smaller than the other pressures of the system The neutral position of the spool is chosen to be the position where all the flows are equal to zero. This position is defined at x s =0 The compressibility of the fluid is considered zero

37 A.4. Continuity equations of the valve-cylinder combination 29 The orifice areas vary linear with the travel of the spool Taking into account these assumptions, the equations which describe the flows from the valve to the cylinder and reversely are } Q 1 = K v x s ps p 1 for x s >0 (A.3) Q 2 = K v x s p2 p t and } Q 1 = K v x s p1 p t Q 2 = K v x s ps p 2 for x s <0 (A.4) A.4 Continuity equations of the valve-cylinder combination The continuity equation from the valve to the 1st cylinder chamber is defined as Q 1 0 = A p ẋ p + C l (p 1 p 2 ) + V 1 p 1 β F t (A.5) 0 Q 2 = A p ẋ p C l (p 1 p 2 ) + V 2 p 2 β F t (A.6) However, the volumes V 1 and V 2 are not constant as they vary with the piston position. For this reason they can be modeled using the equations V 1 = V initial1 + A p x p (A.7) V 2 = V initial2 A p x p (A.8) A.5 Momentum equation Finally, applying the Newton s second law of motion at the cylinder and the load, the resulting momentum equation defined as: M L ẍ p = p 1 A p p 2 A p Bẋ p F L (A.9) A.6 Symbols of hydraulic nonlinear mathematical model All the symbols which are used for the hydraulic nonlinear mathematical modeling are presented at Table A.1

38 30 Appendix A. Modeling Symbol Q N C d A β f ρ p N K v Q 1 Q 2 p s p t p 1 p 2 x s A p x p ẋ p ẍ p V 1 V 2 C l V initial1 V initial2 M L F L B Description nominal flow through the valve discharge coefficient of the valve orifice area bulk modulus density of the fluid nominal pressure difference valve coefficient flow from the valve to cylinder and reverse flow from the cylinder to valve and reverse supply pressure tank pressure pressure in the first chamber pressure in the second chamber normalized displacement of the spool piston effective area position of the piston velocity of the piston acceleration of the piston volume of the 1st chamber, connecting line, valve volume of the 2nd chamber, connecting line, valve total leakage coefficient initial volume at the 1st cylinder chamber initial volume at the 2nd cylinder chamber mass of the load load force viscous friction coefficient of the piston and load Table A.1: Symbols of hydraulic nonlinear mathematical model A.7 Simulink implementation of hydraulic model The nonlinear hydraulic model has been implemented in Matlab simulink and is presented at Figure A.4 A.8 Linearization of non-linear hydraulic mathematical model At this chapter the non-linear hydraulic model will be linearized using the Taylor series. Initially some assumption must be considered. These assumptions are: The gravitational, centrifugal and coriolis effects are neglected

39 A.8. Linearization of non-linear hydraulic mathematical model 31 Figure A.4: Nonlinear hydraulic model The volumes in the servo valve and hoses are neglected. The volumes under consideration are the volumes at the cylinder chambers. The response of servo valve is considered much faster in contrast with the rest of the system. For this reason it will be neglected during the linearisation. The flow entering into the cylinder is equal to the flow getting out from the cylinder (steady state assumption) Coloumb friction is neglected The aim of the linearization is the derivation of valve-cylinder transfer function. Knowing the transfer function of the linear system, the frequency response can be determined and the controller can be designed.

40 32 Appendix A. Modeling A.9 Linearization of orifice equations The linearisation of orifice equations can be achieved making a steady state assumption regarding the flows from the valve to the load and reversely. The steady state assumption states that the flow entering the cylinder is equal to the flow getting out from the cylinder, for positive travel of the spool x s > 0. Hence: Q 1 = Q 2 K v x s ps p 1 = K v x s p2 p t p s p 1 = p 2 (A.10) Also, under the assumption that the load pressure p L is equal to the pressure difference of p 1 and p 2, the pressure p 1 can be arranged as p L = p 1 p 2 p L = p 1 (p s p 1 ) = 2p 1 p s p 1 = p L + p s 2 Substituting the value of p 1 into the orifice equations of Q 1 gives: ps p L Q 1 = K v x s for x s > 0 2 ps + p L Q 1 = K v x s for x s < 0 2 The above equations can be rearranged into one equation in the form of: Q 1 = K v x s p s sign( x s )p L 2 The same method can be used for negative travel of the spool x s < 0. Q 1 = Q 2 K v x s p1 p t = K v x s ps p 2 p 1 = p s p 2 (A.11) (A.12) (A.13) (A.14) (A.15) and using p L = p 1 p 2 p L = p s p 2 p 2 = p s 2p 2 p 2 = p s p L 2 Substituting the value of p 2 into the orifice equations of Q 2 gives: ps p L Q 2 = K v x s for x s > 0 2 (A.16) (A.17) ps + p L Q 2 = K v x s for x s < 0 (A.18) 2 The above equations can be rearranged into one equation in the form of: Q 2 = K v x s p s sign( x s )p L 2 (A.19)

41 A.9. Linearization of orifice equations 33 gives Substituting the derived flow equations Q 1 and Q 2 into the load flow equation Q L = Q 1 + Q 2 2 Q L = K v x s p s sign( x s )p L 2 (A.20) (A.21) The load flow Q L is a function of spool position x s and load pressure p L. If the spool position and the load pressure change by a small amount about a operating point (x 0, p L0, Q L0 ), the load flow equation can be linearized using the Taylor series. Hence: Q L = Q L0 + Q L x s This yields xs= x s0 p L =p L0 ( x s x s0 ) + Q L p L xs= x s0 p L =p L0 (p L p L0 ) (A.22) p s sign( x s0 )p L0 Q L = K v x s + K v x s0 sign( x s0 ) p L 2 4 ps 2 p L0sign( x s0 ) 2 (A.23) Q L = k q x s k qp p L (A.24) The partials in the equation above define two important parameters. The first partial is defined as the flow gain k q k q = Q L x s = K v p s sign( x s0 )p L0 2 and the second partial is defined as the flow-pressure coefficient k qp (A.25) k qp = Q L p L = K v x s0 sign( x s0 ) 4 ps 2 p L0sign( x s0 ) 2 (A.26) The two parameters above are called valve coefficient and are very important in the dynamic analysis of the valve-cylinder combination. The flow gain affects the open loop gain of the system and for this reason influence the stability of the system. The flow-pressure coefficient affects the damping in the valve-cylinder combination. The values of the linearisation load pressure p L0 and the linearisation nominal spool travel x s0 are chosen as: p L0 = 40e5 Pa and x s0 = 0.02.

42 34 Appendix A. Modeling A.10 Linearization of continuity equations The first step in the linearization of continuity equations is the definition of the average contained volume of each cylinder chamber V 0. Also is defined the total volume of both cylinder chambers V t. Hence V t = V 1 + V 2 = V 1initial + A p x p + V 2initial A p x p = V 1initial + V 2initial = 2V 0 (A.27) because the cylinder is symmetrical. Rewriting the continuity equations using only the volumes in each cylinder chamber gives: Q 1 0 = A p ẋ p + C l (p 1 p 2 ) + V 0 p 1 β F t (A.28) 0 Q 2 = A p ẋ p C l (p 1 p 2 ) + V 0 p 2 β F t Substituting the two equations above into the equation of load flow Q L gives: (A.29) or and Laplace transforming Q L = A p ẋ p + C l p L + V 0 (p 1 p 2 ) 2β F t Q L = A p ẋ p + C l p L + V t p L 4β F t Q L (s) = A p sx p (s) + C l p L (s) + V t 4β F sp L (s) (A.30) (A.31) (A.32) A.11 Transfer function of valve-cylinder combination Before deriving the valve-cylinder combination transfer function, the Newton s second law equation should be Laplace transformed. Ms 2 x p (s) = p L A p Bsx p (s) (A.33) At this point, combining the linearized valve flow equation, the Laplace transformed continuity equation and the Laplace transformed Newton s second law equation, the valve-cylinder transfer function can be derived. G(s) cyl = x p x s = 1 s k qa p MV t 4β F s 2 + s Mkqp+MC l+ BV t 4β F MV t + Bkqp+BC l +A2 p MV t 4β F 4β F (A.34)

43 A.11. Transfer function of valve-cylinder combination 35 Hence, the natural frequency of the system can be expressed as: ω n = Bk qp + BC l + A 2 p MV t 4β F (A.35) the damping is given by: ζ = 4β F (Mk qp + MC l + BVt 4β F ) 2ω n MV t (A.36) and the gain k q A p K = Bk qp + BC l + A 2 p (A.37) Inserting the system parameters into the valve-cylinder transfer function, the frequency response of the valve-cylinder combination is plotted at Figure A.5.

44 36 Appendix A. Modeling Figure A.5: Frequency response of valve-cylinder arrangement The block diagram of the valve-cylinder linear model is presented at Figure A.6.

45 A.11. Transfer function of valve-cylinder combination 37 Figure A.6: Block diagram of valve-cylinder linear model

46 38 Appendix A. Modeling A.12 Addition of high pass filter into the hydraulic system It is obvious from Figure A.5 that the frequency response of the system presents resonance(without the high pass filter). The peak on the bode plot (resonance) is related to the amount of overshoot in the step response and as a result in the transient response. Generally, the higher resonance presents, the higher the overshoot will be. For this reason, the frequency response of system should not present significant resonance. One common way (in hydraulics) of removing the resonance is to apply a pressure feedback using a high pass filter. The transfer function of the high pass filter is given by: G f = K f s τ f s + 1 Adding the G f into the system, the load flow equation becomes: (A.38) Q L = k q x s k qp p L G f p L (A.39) and the valve-cylinder transfer function: G(s) cyl = x p x s = 1 s k qa p MV t 4β F s 2 + s Mkqp+MC l+mg f + BV t 4β F MV t + Bkqp+BC l +BG f +A2 p MV t 4β F 4β F (A.40) In the frequency domain, the time constant of the high pass filter is being chosen to be less than the natural frequency of the system and the gain is determined using an iterative process for a desired response. As much as concerning the time domain, the gain and the time constant will be determined taking into account the oscillations in the load pressure p L. Testing the linear system(without the filter) with a step as an input, the oscillations of load pressure are presented at Figure A.7

47 A.12. Addition of high pass filter into the hydraulic system 39 Figure A.7: Load pressure without the high pass filter An appropriate value for the time constant could be τ = 0.25s. The gain should be a very small value K f = and is determined after an iterative process. Hence: G f = s (A.41) 0.25s + 1 The frequency response of the filter is given at Figure A.8 Figure A.8: Frequency response of high pass filter

48 40 Appendix A. Modeling Adding the high pass filter into the hydraulic system, and testing the linear system with the same step input as before, the load pressure of the hydraulic system is presented at Figure A.9 Figure A.9: Load pressure with the high pass filter

49 A.12. Addition of high pass filter into the hydraulic system 41 Furthermore, the frequency response of the system together with the high pass filter is presented at Figure A.10 Figure A.10: Frequency response of valve-cylinder arrangement together with high pass filter

50 42 Appendix A. Modeling Hence with the addition of the high pass filter, the frequency response of the system does not present resonance. As a result the system will not present extended overshoot and the process of controller s design will be more feasible. The block diagram of the valve-cylinder linear model together with the high pass filter is presented at Figure A.11. A.13 Validation of linear model At this stage, the validation of the linear will be tested. This can be achieved by applying the same input signal to the linear and nonlinear model and studying the piston movement and load pressure.more specifically, applying the same step input signal (0 to 0.2 at 1 second) into the models, the load pressure can be seen at Figure A.12 Furthermore, applying the same sinusoidal input signal (amplitude=1, angular frequency=6.3 rad/s) into the models, the load pressure can be seen at Figure A.13

51 A.13. Validation of linear model 43 Figure A.11: Block diagram of valve-cylinder linear model with the high pass filter

52 44 Appendix A. Modeling Figure A.12: Load pressure of linear and non-linear hydraulic model for step input Figure A.13: Load pressure of linear and non-linear hydraulic model for sin input From the figure above, the load pressure of the linear model is oscillating at the beginning of the simulation. These oscillations can be explained by the linearization process. As much as concerning the nonlinear load pressure, oscillates the same way in every cycle. This can be explained from the differential equations describing the hydraulic system, which is characterized as highly dynamic.

53 A.14. Validation of nonlinear model 45 The piston displacement of the linear and nonlinear model is presented at Figure A.14 Figure A.14: Piston displacement of linear and non-linear hydraulic model The amplitude of piston displacement waveform in both models has almost the same value. One important thing which should be considered is the fact that there is no change in the amplitude of the displacement waveform in both models. This happens because we have neglected the gravitational, centrifugal and coriolis effects, but also coulomb friction. Finally, the comparison between the nonlinear and linear model is considered sufficient enough for designing the linear controllers. A.14 Validation of nonlinear model The validation of the nonlinear model will be accomplished by comparing the nonlinear model with the real system at the laboratory. At this stage, it is highly important that the response of the nonlinear model due to arbitrary inputs to have the same characteristics with the response of the system at the laboratory. That is because the tuned controllers (tuned using the linear model) will be tested in the nonlinear model and in the laboratory. As much as concerning the response of the simulated and tested system, it is expected to have some differences due to the simplifications which had been made in the modeling process and due to disturbances in the physical system. A.15 Adjustable parameters in the nonlinear model The simulation of the nonlinear system has been made considering some crucial parameters as constants. These parameters highly affect the dynamic response of the system. These parameters are:

54 46 Appendix A. Modeling 1. Viscous friction coefficient B = 6000 kg/s 2. Leakage coefficient C l = m 3 /(s P a) 3. Valve coefficient K v = m 3 /(s P a) 4. Bulk modulus β f = N/m 2 5. Fluid density ρ = 860 kg/m 3 Hence, during the validation of the nonlinear model, some of the above parameters have been adjusted in order to get the desired response. A.16 Validation with the usage of a sinusoidal input The same sinusoidal function(amplitude=1, angular frequency=8 rad/s) is applied as an input in both nonlinear model and physical setup. The response of the system is presented at figure A.15 Figure A.15: Piston position of tested and simulated system As it was expected, the robot is moving sinusoidally. Additionally, it is interesting to note that the robot is directed toward a lower position. The initial position of the piston was at 0.20 meters, and is decreasing. This downward movement can be explained from the gravitational forces and the moment of the robot s second link, that drag the robot to the lower position. Furthermore, the validation of pressures will be performed using the load pressure. A step input (final value 0.1) is applied and the response of the simulated and tested system is presented at Figure A.16

55 A.17. Mechanical Modeling 47 Figure A.16: Load pressure of tested and simulated system It is obvious that there is not a steady state error of the load pressure response. However, a phase shift is presented. The phase shift is present due to measurement inaccuracies of the transducers and due to the noisy nature of the measured signal. Based on the results above, the nonlinear model is considered to be sufficiently accurate for designing the controllers. A.17 Mechanical Modeling The process for the mechanical modeling. To simplify the modeling we decide to put an angle of 90 degrees for the elbow. We took for hypothesis that the shoulder got only one degree of freedom. For the mechanical part we need to know the correlation of the angle between the robot arm and the actuator. For that we will express the angle in an "universal" referential. Then we will use this expression of the angle in the hydraulic modeling. With this combination of modelings we will be able to design the controller. The mechanical modeling is essential also to know which forces are applied on the robot arm to design a hydraulic system strong enough to make it move and stay and in the wanted position. For that we will express the robot as an assembly of links Then we will look at the forces applied on the robot arm We will use all these information in MATLAB to build a model. With this model we will be able to anticipate the reaction of the robot.

56 48 Appendix A. Modeling Figure A.17: expression of the angle between the robot arm and the actuator [1] Figure A.18: link model of the robot A.18 General mechanics/newton s second law First of all it is necessary to describe Newton s second law for mass to model the mechanical system. Then the Newton s second law for rigid body will be applied. At the point O which is the rotation center J tot θ = τ Mtot g OG (A.42) with J tot = J O1 + m 2 L 2 + J B2

57 A.18. General mechanics/newton s second law 49 Figure A.19: forces applied on the robot where is the total inertia, is the inertia of link 1 at the point O and J tot J O1 is the inertia of link 2 at the point B In order to express this analysis, two models will be created as in figure A.20 J B2 τ = F D(x) (A.43) with D the distance, it depend of the position of the piston and the angle of the piston and the "universal" referential

58 50 Appendix A. Modeling Figure A.20: Modelisation of the position of the center mass This model is reduce as in figure A.21 where Figure A.21: Torque application M tot = m 1 + m 2 First of all it is necessary to find the position of the global central mass center hypothesis : (from data sheet) L OR = 0.483m L OB = 1.206m L BCx = 0.353m L BCy = 0.130m

59 A.18. General mechanics/newton s second law 51 Figure A.22: Global center mass position of center mass link 1 : point R : position of center mass link 2 : point C : Barycentrum : point G x R = 0 y R = L OR x C = L BCx y C = L OB + L BCy x G = (x R + x C )/2 = L BCx /2 = m y G = (y R + y C )/2 = (L OR + L OB + L BCy )/2 = m

60 52 Appendix A. Modeling This is the expression of the center mass in the referential of the robot arm, it moves with the robot arm. It is necessary to express it in a "universal" referential the link between the two referential is like describe in figure A.22 The equation below is deduce x G = cos(θ)x 0 (A.44) y G = sin(θ)y 0 (A.45)

61 A.19. Determination of the torque 53 A.19 Determination of the torque A.19.1 Application of Al-Kashi s law (law of cosines) Figure A.23: Al-Kashi figure Then cos(α H ) = m2 2 + x2 m 2 1 2m 1 x (A.46) F τ = cos(α F )F with α F = π/2 α H Then equation (A.46) is injected in equation (A.47). (A.47) So F τ = cos( π 2 cos 1 ( m2 2 + x2 m 2 1 ))F 2m 2 x τ = cos( π 2 cos 1 ( m2 2 + x2 m 2 1 ))F m 2 2m 2 x (A.48) A.20 Denavit Hartberg Parameters To determine the DH parameters it is necessary to derive the DH matrix from the figure. The DH parameters are given from this matrix cos(θ n ) sin(θ n ) cos(α n ) sin(θ n ) sin(α n ) a n cos(θ n ) sin(θ n ) cos(θ n ) cos(α n ) cos(θ n ) sin(α n ) a n sin(θ n ) 0 sin(α n ) cos(α n ) d n

62 54 Appendix A. Modeling The transformation matrix from first joint to second joint: 0 1A = cos(θ 1 ) sin(θ 1 ) 0 a 1 cos(θ 1 ) sin(θ 1 ) cos(θ 1 ) 0 a 1 sin(θ 1 ) The transformation matrix from second joint to TCP 1 2A = a Final transformation matrix is found by multiplying the two transformation matrices. This is done to get the transformation from the TCP to the base cordinate system. 0 2T = 0 1 A 12 A = sin(θ 1 ) cos(θ 1 ) 0 a 2 sin(θ 1 ) + a 1 cos(θ 1 ) cos(θ 1 ) sin(θ 1 ) 0 a 2 cos(θ 1 ) + a 1 sin(θ 1 )

63 Appendix B Controllers B.1 Control The control of the hydraulic servo robot will be implemented by controlling the proportonal directional valves. These valves are actuated by an electrical control signal which is proportional to the position of the spool. The control signal adjusts the flow rate and the flow direction into the hydraulic system. In such a way, the movement of piston of the hydraulic cylinder can be controlled according to the attitude of the load. Furthermore, a second control signal is applied to the proportional pressure relief valve for continually controlling system s pressure. B.2 Signal flow in the hydraulic system The signal flow in the hydraulic system is presented at Figure B.1. 55

64 56 Appendix B. Controllers Figure B.1: Signal flow in the hydraulic system According to the figure above 1. A voltage (mostly from -10 V to 10 V) is applied on the electrical amplifier 2. The amplifier converts the voltage into current 3. The current affects the proportional solenoid 4. The proportional solenoid actuates the proportional valve 5. The proportional valve control the fluid flow and fluid direction of the hydraulic cylinder according the load demand 6. The cylinder converts the fluid power into kinetic energy

65 B.3. Linear Controllers 57 B.3 Linear Controllers The aim of linear controller s design is to force the robot to follow a specific position reference inside the the working area. The controllers will be tuned with and without the high pass filter. B.3.1 Relative Stability Specifications For an adequate transient response with reduced overshoot, the relative stability margins must be increased. Also another reason for large stability margin is the inaccuracies in the mathematical model. The most common values of gain and phase margin for reasonable control system are GM > 6 8 db and P M > degrees. B.4 Proportional Controller The proportional controller is a pure gain and the transfer function is given by: G c (s) = K P (B.1) Adding the gain compensator into the system, the bode plot magnitude will be shifted up or down (with no change in the magnitude shape) according with the value of gain compensator. The phase shape will be unchanged. As much as concerning the response of the system, the addition of the P controller will improve the transient response of the system. B.4.1 Open loop frequency response without high pass filter In order to determine the gain, the Matlab software will be used and specifically the Control Systems Toolbox. For the specified gain and phase margin, the compensator proportional gain is found to be K P = 10 The open loop frequency response of the system combined with the proportional controller is presented at Figure B.2

66 58 Appendix B. Controllers Figure B.2: Open loop frequency response with P-controller B.4.2 The gain and the phase margin are GM=9.21 db, PM=89.3 degrees. Open loop frequency response with high pass filter The controller s gain for the system combined with the high pass filter has been determined as: K P = 40 The open loop frequency response combined with the proportional controller and filter is presented at Figure B.3 Figure B.3: Open loop frequency response with P-controller and filter

67 B.4. Proportional Controller 59 B.4.3 The gain and the phase margin are GM=20.7 db, PM=47.6 degrees. Proportional Controller tested in the nonlinear model The proportional controller has been tuned and will be tested in the nonlinear model. A step input is applied at the system and the step response is presented at Figure B.4 Figure B.4: Step response of non-linear model with P controller The piston of the cylinder reaches the reference signal without overshoot and in an acceptable rise time. Furthermore, a sinusoidal input signal is applied at the system in order to increase the accuracy of the determined proportional gain. The sinusoidal response is presented at Figure B.5 Figure B.5: Sinusoidal response of non-linear model with P controller Figure B.5 reviles a steady state error as the piston does not track the reference appropriately.