Hydraulic Servo Systems : Dynamic Properties and Control

Hydraulic Servo Systems Dynamic Properties and Control

by Closed loop stiffness for a Karlposition-Erik servo Rydberg with velocity feedback

s2 2 K h s 1 K s K K qi K 2 K vfv f sav A K FL vfv h vfv h p qi S Kvfv 1 K fv Ksav c X K V A p ce t p 2 1 s Ap 4eKce

s s2 2 Steady state loop gain [1/s] 1 h s 1 2 2 Ap Kvv Kvfv h Kvfv h Kqi 1 Sc Kvv Kvfv Kvv Ksav K f K V Ap Kvfv ce 1 t s K K (without velocity feedback) K-E Rydberg 4 K Feedbacks in Electro-Hydraulicvv v Servo Systems 7 e ce

For the same amplitude margin, Kv must have the same value in the system with and without velocity feedback.

Velocity feedback increases the steady state stiffness with the factor Kvfv.FL K æ V ö ce ç1+ t s÷ Karl-Erik Rydberg, Linköping University, Sweden 2 ç 21 ÷ Ap è 4be K ce ø Threshold Saturation - . u i i imax Kqi 1 xp xp c K r v 1 1 + + sav s + s2 2d ei Ap 1+ + h s + 1 s - - n wv 2 w wh h

Velocity feedback Kfv

Position feedback Closed loop stiffness for a position servoK fwith

velocityFigure 12: A feedbacklinear valve controlled position servo with velocity feedback Am = 6 dB If the bandwidth of the valve is relatively high and threshold and saturation is neglected With velocity feedback, K = 20 1/s, K = 9.0 the velocity feedbackvv will givevfv the effect on the hydraulic resonance frequency and damping as shown in Figure 13.

Without velocity feedback, Kv =20 1/s FL

æ ö Kqi Kce hv =V 387t rad/s ç1+ s÷ Kvfv = 1 + Kfv Ksav 2 ç ÷ Ap Ap è 4be K ce ø - . 1/ Kvfv x x uc iv Kqi p 1 p + Ksav s2 2d Ap + + h s + 1 s - 2 K w Kvfv wh vfv h

Position feedback Kf

Karl-ErikFigure Rydberg, 13: Linköping A linear University, position Sweden servo with velocity feedback22 included

From Figure 13 the new resonance frequency and damping (hv and hv) caused by the velocity feedback can be evaluated as

1 K qi K , , where the velocity loop gain is K 1 K K . hv h vfv hv h vfv fv sav A K vfv p +11 Designing the position control loop for the same amplitude margin as without velocity feedback gives the following relations:

K qi Steady state loop gain without velocity feedback: K v K sa K f Ap

K qi Steady state loop gain with velocity feedback: K vv K sav K f Ap K vfv

A certain amplitude margin means that K v h h . In this case h h hv hv , which

implies that K v K vv and thereby the servo amplifier gain K sav K sa K vfv . With

velocity feedback, the servo amplifier gain (Ksav) can be increased in proportion to the velocity loop gain Kvfv and the servo amplifier gain without velocity feedback, Ksa.

The open loop gain (Au(s)) for a position servo without (Kv = 20) and with velocity feefback (Kvfv = 10 and Kvv =20) is shown in Figure 14. K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Linköping November 6, 2014 Revised October 27, 2016

Karl-Erik Rydberg Professor, PhD

Department of Management and Engineering Linköping University SE-581 83 LINKÖPING

ISBN: 978-91-7685-620-8

2 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Hydraulic Servo Systems – Dynamic Properties and Control

Table of Contents Hydraulic Servo Systems - Dynamic Properties and Control ...... 1 1 Introduction ...... 5 1.1 What is a servo? ...... 5 1.2 Technology comparisons ...... 5 1.3 Capabilities of electro-hydraulic servos ...... 7 1.4 Different electro-hydraulic concepts ...... 7 1.5 Servo system efficiency ...... 9 1.5.1 Servo valve efficiency ...... 9 2 Configuration of electro-hydraulic servos ...... 12 2.1 Position servo ...... 13 2.2 Velocity and force servos ...... 13 3 Servo valves and their characteristics ...... 15 3.1 Number of lands and ports ...... 15 3.2 Types of valve center ...... 16 3.2.1 Valve sleeve ...... 17 3.3 Examples of electro-hydraulic servo valves ...... 18 3.3.1 Type of feedback ...... 18 3.3.2 Number of stages ...... 19 3.4 General steady state valve characteristics ...... 24 3.4.1 Valve Coefficients ...... 25 3.5 Critical center four-way valve ...... 26 3.5.1 Practical null coefficients for a critical center valve ...... 26 3.5.2 Leakage characteristics of a practical critical center four-way valve ...... 27 3.5.3 Blocked line pressure sensitivity curve ...... 27 3.5.4 Leakage flow curves ...... 28 3.5.5 Real flow gain characteristics ...... 29 3.6 Open center spool valve ...... 30 3.7 Three-way spool valve analysis ...... 32 3.8 Dynamic response of servo valves ...... 34 4 Position servos with valve-controlled cylinders ...... 36 4.1 Asymmetric cylinder ...... 36 4.1.1 Example ...... 38 Variation in resonance frequency for an asymmetric cylinder with line volumes ...... 38 Parameters ...... 38 4.2 Valve controlled symmetric cylinder ...... 39 4.2.1 Servo system stability and bandwidth ...... 42 4.4 Influence from flow forces on valve spools ...... 45 4.5 Position servo with mechanical springs at connectors ...... 47 4.5.1 Simulation of position servo with mechanical springs ...... 48 5.1 Four-way valve controlled motor with position feedback ...... 52 5.2 Valve controlled motor for an angular velocity servo ...... 54 5.3 Pump controlled motor ...... 56

3 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

5.4 Pump controlled motor with variable displacement ...... 58 5.5 Pump controlled symmetric cylinder ...... 59 6 Hydraulic systems with complex load dynamics ...... 61 6.1 Loads with one degree of freedom ...... 62 6.2 Loads with two degrees of freedom ...... 64 7 Feedbacks in Electro-Hydraulic Servo Systems ...... 75 7.1 Linear valve controlled position servo ...... 75 7.1.1 Influence of valve dynamics ...... 76 7.1.2 Closed loop stiffness ...... 77 7.2 Valve controlled position servo with load pressure feedback ...... 78 7.3 Valve controlled angular position servo with acc. feedback ...... 79 7.4 Velocity feedback in position control servos ...... 81 7.5 Valve controlled velocity servo ...... 83 7.6 Proportional valves with integrated position and pressure transducers .... 84 7.7 Electro-hydraulic servo actuators ...... 84 7.8 Design examples ...... 87 7.9 Summary of servo system design criterions ...... 90 7.9.1 Control loop dynamics – possible improvements ...... 90 8 Nonlinearities in Hydraulic Servo Systems ...... 92 8.1 How to handle nonlinear properties in linear models? ...... 92 8.2 Common Nonlinearities in Hydraulic Systems ...... 93 8.2.1 Saturation and its effect on system performance ...... 93 8.2.2 Dead-band ...... 96 8.2.3 Threshold and Hysteresis ...... 96 8.3.4 Nonlinear friction ...... 97 9 Controller Design for Hydraulic Servo Systems ...... 99 9.1 General structure of the controller ...... 99 9.2 Feed forward gain for reduction of velocity error in position servos ...... 100 9.3 PID Controller ...... 101 9.3.1 Proportional gain ...... 102 9.3.2 Integral gain ...... 102 9.3.3 Derivative gain ...... 103 9.3.4 Implementation and tuning of PID-controllers ...... 103 9.4 A commercial digital controller ...... 104 References ...... 106 Appendix 1 ...... 107 Design of a linear position servo ...... 107

4 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control

______K-E Rydberg Hydraulic Servo Systems 1 ______1 Introduction

WhenHydraulic closed-loop hydraulicServo Systems control systems 2 Theory first began and to appearApplications in industry, the applications were generally those in which very high performance was required. While hydraulic1. servo Introduction systems are still heavily used in high-performance applications such as the machine-tool industry, they are beginning to gain wide acceptance in a variety of industries.When closed-loop Examples arehydraulic material control handling, systems mobile first began equipment, to appear plastics, in industry, steel the plants, applications were generally those in which very high performance was required. While mining,hydraulic oil exploration servo systems and are automotive still heavily testing. used in high-performance applications such as

the machine-tool industry, they are beginning to gain wide acceptance in a variety of Closedindustries. loop servo Examples drive are technology material handling, is increasingly mobile equipment, becoming plastics, the norm steel in plants, machine automation,mining, oil where exploration the operators and automotive are demanding testing. greater precision, faster operation and simpler adjustment. There is also an expectation that the price of increasing the level of automationClosed loopshould servo be contained drive technology within isacceptable increasingly limits. becoming the norm in machine automation, where the operators are demanding greater precision, faster operation and simpler adjustment. There is also an expectation that the price of increasing the level of 1.1 automationWhat is should a servo? be contained within acceptable limits. In its simplest form a servo or a servomechanism is a control system, which measures What is a servo? its own output and forces the output to quickly and accurately follow a command In its simplest form a servo or a servomechanism is a control system which measures its signal,own se output Figure and 1forces-1. In the this output way, to thequickly effect and of accurately anomalies follow in the a command control signal,device se itself and Figurein the load1-1. Incan this be way, minimised the effect as of well anomalies as the in influence the control of deviceexternal itself disturbances. and in the A servomechanismload can be can minimised be designed as well to control as the almost influence any physical of external quantities, disturbances. e.g. motion, A force,servomechanism pressure, temperature, can be designed electrical to control voltage almost or current. any physical quantities, e.g. motion, force, pressure, temperature, electrical voltage or current.

Power source

Command signal Power Mechanical Motion Servo Actuator + S electronics modulator load -

Feedback transducer

FigureFigure 1 1-1:-1: BasicBasic servomechanism

1.2 TechnologyTechnology comparisons comparisons The potential for alternative technologies should be assessed in the light of the well- The knownpotential capabilities for alternative of electro-pneumatic technologies should and be electro-mechani assessed in thecal light servos. of the High well- knownperformance capabilities actuation of system electro is- pneumaticcharacterised andby wide electro bandwidth-mechanical frequency servos. response, High performancelow resolution actuation and high system stiffness. is characterised Additional requirements by wide bandwidth may include frequency demanding re dutysponse, low resolution and high stiffness. Additional requirements may include demanding duty cycles and minimisation of size and weight. The last mentioned requirements are of

5 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______K-E Rydberg Hydraulic Servo Systems 2 ______special interests in aerospace applications. The most important selection criteria can be summarised as follows: cycles and minimisation of size and weight. The last mentioned requirements are of special• Customer interests in performance aerospace applications. The most important selection criteria can be summarised• Cost as follows: • • Size Customer and weight performance • • Duty Cost cycle • • Environment Size and weight: vibration, shock, temperature, etc. The performance• Duty cycle available with electro-hydraulic servos encompasses every industrial and aerospace• Environment application.: vibration, As indicated shock, temperature, in Figure etc. 1-2 electro-hydraulic servos will coverThe performance applications available with higher with electro-hydraulic performance then servos electro encompasses-mechanical every and industrial electro- pneumaticand aerospace servos. applicati This on. is easily As indicated explained in Figure because 1-2 electro electro-hydraulic-hydraulic servo servos systems will cover applications with higher performance then electro-mechanical and electro- havepneumatic been designedservos. This and is easily developed explained to accomplishbecause electro-hydraulic essentially everyservo systems task that have has appeared.been designed and developed to accomplish essentially every task that has appeared. 100

Electro-hydraulic actuation limit

Electro-mechanical actuation limit 1 Electro-pneumatic actuation limit Actuation power [kW]

0.1 1 510 50 100 Actuated load dynamics [Hz]

Figure 1-2: Typical performance characteristics for different types of servo actuators Figure 1-2: Typical performance characteristics for different types of servo actuators The above figure indicates that applications in the lower range of power and dynamic response may also be satisfied with electro-pneumatic servos. However, the best choice Theis abovealways figuredetermined indicates by cons thatiderations, applications such asin thosethe lower selection range criteria of power discussed and above.dynamic responseIn most may applications also be satisfiedthe aspect with of cost electro is generally-pneumatic dominant. servos. However, the best choice is Experience always determined indicates that by considerations, electro-mechanical such or aselectro-pneumatic those selection actuators criteria tends discussed to above.have Inlower most cost applications than electro-hydraulic the aspect ofactua costtors is ingenerally the low dominant.performance range. This cost difference rapidly dissipates for applications that require high power and/or high Experiencedynamic response. indicates that electro-mechanical or electro-pneumatic actuator tends to haveIn comparinglower cost costs, than electro one must-hydraulic be careful actuators to consider in the the low total performance cost of entire range. servo- This costactuation difference system. rapidly The dissipateshigher cost for of applicationsan electro-hydraulic that require servo highoften power results and/orfrom the high dynamicpower response. conversion equipment needed to provide high pressure fluid with low contamination level. It is also clear that the relative cost of an alternative actuation In system comparing designed costs, for one a speci mustfic be application careful to will consider depend, the primarily, total cost on of the entire actuation servo- actuationpower level. system. The higher cost of an electro-hydraulic servo often results from the power conversion equipment needed to provide high-pressure fluid with low

6 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______contamination level. It is also clear that the relative cost of an alternative actuation system designed for a specific application will depend, primarily, on the actuation power level.

1.3 Capabilities of electro-hydraulic servos When rapid and precise control of sizeable loads is required an electro-hydraulic servo is often the best approach to the problem. Generally speaking, the hydraulic servo actuator provides fast response, high force and short stroke characteristics. The main advantages of hydraulic components are. • Easy and accurate control of work table position and velocity • Good stiffness characteristics • Zero backlash • Rapid response to change in speed or direction • Low rate of wear

There are several significant advantages of hydraulic servo drives over electric motor drives: ♦ Hydraulic drives have substantially higher power to weight ratios resulting in higher machine frame resonant frequencies for a given power level. ♦ Hydraulic actuators are stiffer than electric drives, resulting in higher loop gain capability, greater accuracy and better frequency response. ♦ Hydraulic servos give smoother performance at low speeds and have a wide speed range without special control circuits. ♦ Hydraulic systems are to a great extent self-cooling and can be operated in stall condition indefinitely without damage. ♦ Both hydraulic and electric drives are very reliable provided that maintenance is followed. ♦ Hydraulic servos are usually less expensive for system above several horsepower, especially if the hydraulic power supply is shared between several actuators.

1.4 Different electro-hydraulic concepts In electro-hydraulic applications different concepts will be used in order to meet the actual requirements. One example of a system where the weight is of great importance is an Electro Hydraulic Actuator (EHA) to be used in aircraft applications. A typical EHA-concept is shown in Figure 1-3. This EHA consists of an electric motor, a speed controlled pump with low displacement, a cylinder and an accumulator used as a tank. In a real application there is also a need for additional functions, such as by-pass damper and safety facilities, not shown in the figure. In order to save weight no cooler is applied and there will be a risk for too high fluid temperature with failure of the EHA

7 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______K-E Rydberg Hydraulic Servo Systems 4 ______asK-E a Rydbergconsequence. Therefore, the Hydrauliclosses and Servo thereby Systems the temperature of the fluid is 4of great______importance in this application. Uc - Motor cont. UU Xc - Motor cont. np U EM X Load np EM Load

FigureFigure 1-3:1-3: Electro Hydraulic ActuActuatorator withwith electricelectric motormotor controlcontrol Figure 1-3: Electro Hydraulic Actuator with electric motor control Other more conventional concepts are a pump or a valve controlled actuator, as shown Other more conventional concepts are a pump or a valve controlled actuator, as shown in Figure 1-4. The main difference between those systems is that the pump controlled insystemOther Figure moreis one1- 4conventional .separate The main unit differenceconcepts supplied are bybetween aan pump electric those or awire systemsvalve to thecontrolled is motor that the (theactuator, pump same controlled asas shownfor the systemin Figure is in one 1-4 Figure separate. The 1-3) main unit and difference supplied the valve bybetween an controlled electric those wire actuatorsystems to the is is motor that supplied the (the pump same by acontrolled as constant for the systempressuresystem is inhydraulic one Figure separate 1line.-3) unit andIn thesupplied the last valve case by an cthontrolled electrice EHA wireis actuator feed to theby is amotor central supplied (the hydraulic same by a as constant forsupply the pressureunit.system inhydraulic Figure line. 1-3) In and the the last valve case ctheontrolled, EHA is actuator feed by isa central supplied hydraulic by a constant supply pressure hydraulic line. In the last case the EHA is feed by a central hydraulic supply unit. Uc Uc unit. - - Pump U Constant U cont. X X Uc supply Uc - pressure np εp - EM Pump U Load Constant U Load cont. X supply X pressure np εp EM Load Load

Figure 1-4: Electro Hydraulic Actuators with pump and valve control respectively

Figure 1-4: Electro Hydraulic Actuators with pump and valve control respectively ComparingFigure the electric1-4: Electro motor Hydraulic controlled Actuators E HAwith pump in Figure and valve 1-3 control and the respectively pump controlled EHA in Figure 1-4, the overall efficiency will be better in the first case, where the pump shaftComparing speed (nthe) electricis controlled. motor Thecontrolled efficiency EHA curves in Figure for similar 1-3 and systems the pump are controlled shown in ComparingEHA in Figure thep 1-4,electric the overallmotor conefficiencytrolled willEHA be in better Figure in the1-3 first and case, the pumpwhere controlledthe pump Figure 1-5. Maximum pump flow is the same in both cases and pressure drop over the EHAshaft inspeed Figure (n 1) -is4, thecontrolled. overall efficiencyThe efficiency will becurves better for in similarthe first systemscase, where are shownthe pump in directional valvep is included as losses. shaftFigure speed 1-5 . (nMaximump) is controlled. pump flow The isefficiency the same curvesin both for cases similar and pressuresystems dropare shown over the in FigureLookingdirectional 1 -at5 . the valveMaximum variations is included pump in overallas flo losses.w is efficiency the same itin is both clear cases that andspeed pressure control drop has aover favour the over displacement control, especially in the power range up to 50% of maximum power. directionalLooking at valve the variations is included in asoverall losses. efficiency it is clear that speed control has a favour However, there are other problems to overcome in the pump speed control concept. For over displacement control, especially in the power range up to 50% of maximum power. Lookingexample, atthe the amplitude variations of in the overall flow efficiencypulsations itfrom is clear the pumpthat speed must control be very has low a favourat low However, there are other problems to overcome in the pump speed control concept. For shaft speeds in order to avoid problems with low frequency vibrations in the system. overexample, displacement the amplitude control, of the especially flow pulsati in ons the from power the range pump up must to be 50% very of low maximum at low This, require a special design of the pump. One suitable pump design is the inner gear power.shaft speeds However, in order there to are avoid other problems problems with to low overcome frequency in vibrations the pump in speed the system. control concept. In such a pump both kinematic and compressibility dependent flow pulsations concept.This, require For example, a special the design amplitude of the ofpump. the flow One pulsationssuitable pump from design the pump is the must inner be gearvery are extremely low compared to piston pumps. lowconcept. at low In shaft such speeds a pump in both order kinematic to avoid andproblems compressibility with low frequencydependent vibrationsflow pulsations in the system.are extremely This, requirelow compared a special to piston design pumps. of the pump. One suitable pump design is the inner gear concept. In such a pump both kinematic and compressibility dependent flow pulsations are extremely low compared to piston pumps.

8 KK-E-E Rydberg Hydraulic Servo Systems Hydraulic – DynamicServo Systems Properties and Control 5 ______

Pump speed controlF Pump displacement control F εp np

1.0 1.0

0.9 0.9

0.8 0.8

0.7 0.7 Pressure difference: 15 MPa Pressure difference: 15 MPa Pump disp. setting: 1.0 Pump shaft speed: 1500 rpm Overall efficiency [-] Overall efficiency Overall efficiency [-] Overall efficiency 0.6 0.6

0.5 0.5 0 1000 2000 3000 0 0.2 0.4 0.6 0.8 1 Pump speed [rpm] Pump displacement setting [-] FigureFigure 11-5:-5: Overall efficienciesefficiencies with pump speed control and displacement settingsetting controlcontrol

1.52. Servo Configuration system efficiency of an electro-hydraulic servo The overallbasic elements efficiency of anof aelectro-hydraulic servo system depends servo isupon shown the insystem Figure configuration, 2-1. The output which of hasthe servo been demonstratedis measured with in Chaptera transducer 1.4. However,device to convert the criteria it to foran selectionelectric signal. of system This feedback signal is compared with the command signal. The resulting error signal is then concept to a specific application belongs to the system power level, system bandwidth, amplified by the regulator and the electric power amplifier and then used as an input control accuracysignal to ,the number servo of valve. axis toThe control servo and valve the controls installation the fluiconditions.d flow to the actuator Inin proportiongeneral, it’s to a the well drive-known current fact from that pumpthe amp controllifier. isThe much actuator more then energy forces efficient the load than to move. Thus, a change in the command signal generates an error signal, which causes the valve control. A speed or displacement controlled pump has much lower response than load to move in an attempt to zero the error signal. If the amplifier gain is high, the aoutput servo will valve vary at rapidlythe same and nominal accurate power.ly following Especially the command at high powersignal. (> 50 kW) pump control will drastically reduce the bandwidth of the system compared to the dynamic External performance of a servo valve. Therefore, for servo systemsdisturbances with requirements on high response and accurate control, valve controlled actuators are the best choice. Also, if - severalCommand axes haveError toServo be controlled ampl. simultaneously, valve control is less costly instead of signal signal Servo Mechanical Output using one Spump for eachand axis. Actuator + S + regulator valve load -

1.5.1 Servo valve efficiency Feedback transducer The use of servo valves is not energy efficient, because they introduce quite heavy throttling losses Figure into 2-1: the Components system. in T hisan electro-hydraulic can be illustrated servomechanism by looking at valve specification.External disturbances The nominal (forces flow or torque) capacity can of cause a servo the valveload to is move specified without at a any total changes valve pressurein the command drop of signal.70 bar. InAssuming order to aoffset supply the pressure disturbance of 210 input bar an gives actuator that output only the is valveneeded losses in the represents opposite 33direction% of the (see input Figure power 2-1). at To nominal provide flow this conditions opposing and output only a 67%finite remains error signal for load is required. actuations. The magnitude of the required error signal is minimised if the amplifier gain is high. Ideally, the amplifier gain would be set high enough that the Theaccuracy most of simple the servo way to becomes supply dependent a servo valve only isupon to use the a accuracy fixed displacement of the transducer pump, whichitself. However,deliver a constantsince the flow control (qs) and loop the gain pump is proportional pressure (p s) to is theadjusted amplifier by a gain, pressure this relief valve, as shown in Figure 1-6. The figure also shows the servo valve efficiency.

9 Servo valve efficiency

K-E Rydberg HydraulicFixed Servo Systemsdisplacement – Dynamic Properties pumpand Control ______ServoLoad flow valve efficiencySupply flow (max valve flow) 1 1 qL = Cq " w " xv0 (ps ! pL ) qs = Cq " w" xv0 ps # !

0.385 ps " pL ! pL #sv =

p p Fixed s ! s q p pump [-] L ! L Max valve efficiency acc. to pL: sv " " = d " 2 sv ( sv ) 0 p p = ! L = s q ! p d(pL ) 3 s s

Karl-Erik Rydberg, IEI/Linköping University 9 Figure 1-6: Servo valve with supply unit and correspondingqL = load valve efficiency flow The servo valve efficiency is defined as load power over supply power, Variable pL = load pressure pump q ⋅ p η = L L (1-1) sv q = supply flow qs ⋅ ps s The flow equation for a four port symmetric and zero-lapped valve is expressed as, Ps = supply pressure 1 qL = Cq ⋅ w⋅ xv0 (ps − pPL ) = constant (1-2) δ s

The constant pump flow willKarl-Erik have theRydberg, same IEI/Linköping value as Universitymax valve flow, which gives, 8 1 q = C ⋅ w⋅ x p (1-3) s q v0 δ s The above equations make it possible to express the valve efficiency as a function of the load pressure pL and ps as a parameter,

ps − pL ⋅ pL ηsv = (1-4) ps ⋅ ps The load pressure that gives max efficiency is found from the efficiency/pressure derivative as,

d(ηsv ) 2 = 0 ⇒ pL = ps (1-5) d(pL ) 3 From Figure 1-6 it can be noticed that max valve efficiency is 38,5%, which can’t be acceptable for a modern servo system. In today’s servo system the supply unit is commonly a variable displacement pump equipped with a constant pressure controller, see Figure 1-7. The pump line is also supplied by an hydraulic accumulator. In a servo system it is important to keep the supply pressure constant, because variations in ps has an impact system response. The accumulator helps the pump to control its pressure when the servo valve are opened or closed very fast.

10 Servo valve efficiency

Fixed q p pump L ! L "sv = K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control qs ! ps ______

qL = load flow

Variable pL = load pressure pump qs = supply flow

Ps = supply pressure

Ps = constant

Figure 1-6: Servo valve supplied by a variable pressureKarl-Erik controlled Rydberg, pump IEI/Linköping and accumulator University 8

The use of a variable displacement pump has the advantage that the pump flow always is adjusted to fit the load flow, qs = qL. According to equation (1-1), the servo valve efficiency can be expressed as,

pL ηsv = (1-6) ps

Nominal load flow gives a valve efficiency of ηsv = 0,67. In practice, the valve efficiency will be lower than the above figures, because of the fact that the valve leakage flow has been neglected. For single stage servo valve (zero- lapped) the leakage flow loss is about 2 % of nominal flow and for a two or three stage valve up to 5 %. Under-lapped valves can have much higher losses.

11 K-E Rydberg Hydraulic Servo Systems 5 ______

Pump speed controlF Pump displacement control F εp np

1.0 1.0

0.9 0.9

0.8 0.8 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control 0.7 0.7 Pressure difference: 15 MPa Pressure difference: 15 MPa ______Pump disp. setting: 1.0 Pump shaft speed: 1500 rpm Overall efficiency [-] Overall efficiency Overall efficiency [-] Overall efficiency 0.6 0.6

0.5 0.5 0 1000 2000 3000 0 0.2 0.4 0.6 0.8 1 2 ConfigurationPump speed of[rpm] electro- hydraulic Pump servo displacements setting [-] Figure 1-5: Overall efficiencies with pump speed control and displacement setting control The basic elements of an electro-hydraulic servo are shown in Figure 2-1. The output of the2. servo Configuration is measured with a oftransducer an electro-hydraulic device to convert it to servo an electric signal. This feedbackThe basic signal elements is compared of an electro-hydraulic with the command servo signal. is shown The in resultingFigure 2-1 error. The signal output is of then ampliftheied servo by isthe measured regulator with and a transducerthe electric de vicepower to convertamplifier it to and an electricthen used signal. as anThis input feedback signal is compared with the command signal. The resulting error signal is then controlamplified signal by to thethe regulator servo valve. and theThe electric servo powervalve controlsamplifier theand fluid then flowused toas thean inputactuator in proportioncontrol signal to theto the drive servo current valve. fromThe servo the amplifier.valve controls The the actuator fluid flow then to forcesthe actuator the load to move.in proportion Thus, a to change the drive in currentthe command from the signalamplifier. generates The actuator an error then signal, forces thewhich load causes to the loadmove. to Thus, move a changein an attempt in the command to zero thesignal error generates signal. an If error the amplifiersignal, which gain causes is high, the the load to move in an attempt to zero the error signal. If the amplifier gain is high, the outputoutput will will vary vary rapidly rapidly and and accurately accurately following the the command command signal. signal.

External disturbances

- Command Error Servo ampl. signal signal Servo Mechanical Output S and Actuator + S + regulator valve load -

Feedback transducer

FigureFigure 2- 12-1:: Components Components inin anan electro-hydraulicelectro-hydraulic servomechanism servomechanism External disturbances (forces or torque) can cause the load to move without any changes Externalin the disturbances command signal. (forces In order or torque)to offset canthe disturbance cause the input load an to actuator move without output is any changesneeded in in the the command opposite direction signal. In (see order Figure to 2-1). offset To the provide disturbance this opposing input outputan actuator a finite error signal is required. The magnitude of the required error signal is minimised if outputthe isamplifier needed gain in theis high. opposite Ideally, direction the amplif (seeier gainFigure would 2-1). be Toset highprovide enough this that opposing the outputaccuracy a finite of errorthe servo signal becomes is required. dependent The only magnitude upon the of accuracy the required of the error transducer signal is minimiseditself. However, if the amplifier since the gain control is high. loop gainIdeally, is proportional the amplifier to the gain amplifier would gain,be set this high enough that the accuracy of the servo becomes dependent only upon the accuracy of the transducer itself. However, since the control loop gain is proportional to the amplifier gain, this gain is limited by stability considerations. In some applications, stability may be critical enough that the desired performance is not possible to reach. The three common types of electro-hydraulic servos are: • Position servo (linear or angular) • Velocity or speed servo (linear or angular) • Force or torque servo

12 K-E Rydberg Hydraulic Servo Systems 6 ______gain is limited by stability considerations. In some applications, stability may be critical enough that the desired performance is not possible to reach. The three common types of electro-hydraulic servos are: K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control • Position servo (linear or angular) ______• Velocity or speed servo (linear or angular) • Force or torque servo 2.1 Position servo Position servo ProbablyProbably the most basic closed-loopclosed-loop controcontroll system is is a a position position servo. A schematic diagram of a complete pospositionition servo is shown in Figure 2-22-2.. Fix reference xp Ap Ap Mt FL V1 V2

P1 P2 Position transducer Servo amplifier uc + i Kf - Ksa uf Ps = const.

Figure 2-2: Symbol circuit of a position servo Figure 2-2: Symbol circuit of a position servo In Figure 2-2 the actuator or load position is measured by a position transducer, which In Figure 2-2 the actuator or load position is measured by a position transducer, which gives an electric signal (uf) in voltage as an output. The servo amplifier compares the gives an electric signal (uf) in voltage as an output. The servo amplifier compares the command signal (uc) in voltage with the feedback signal (uf). Then, the resulting error signalcommand are gainedsignal (uthec) within voltage the factor with K thesa. The feedback output signal current (u signalf). Then, (i) fromthe resul the tingamplifier error willsignal control is gained the servowith thevalve. factor Ksa. The output current signal (i) from the amplifier will control the servo valve. Velocity and force servos Another2.2 Velocity common andtypes force of closed servos loop contro l systems are velocity (speed) and force (torque) servos. The configuration of these systems are identical to the position servo depictedAnother commonin Figure types2-2, expect of closed that loopthe tran controlsducer systems measures are velocityvelocity or(speed) force insteadand force of position(torque) servos. and that The the configuration controller ma ofy havethese differentsystems characteristics.are identical to Figurethe position 2-3 showsservo bothdepicted a speed in Figure and a 2 force-2, expect servo. that It isthe notable transducer that measuresthe same velocitytype of servoor force valve instead can beof usedposition in all and of these that the applications. controller As may indicated have different in Figure characteristics. 2-3, velocity orFigure speed 2servos-3 shows are moreboth acommonly speed and used a force to controlservo. Itthe is shaftnotable speed that of the an same hydraulic type ofmotor servo than valv toe controlcan be linearused in velocity. all of these applications. As indicated in Figure 2-3, velocity or speed servos are Inmore the commonlyvelocity servo used the to servo control amplifier the shaft is of speed integrating of a hydraulic type, as motorshown than in Figure to control 2-3. Comparedlinear velocity. to a position servo the velocity servo has no integration between servo valve displacement and the output velocity. Therefore, the integration in a velocity servo is generallyIn the velocity provided servo electronicallythe servo amplifier in the is amplifier.of integra ting The type, integration as shown is in desirable Figure 2 - to3. minimiseCompared static to a positionerrors and servo to maintain the velocity stability. servo has no integration between servo valve displacement and the output velocity. Therefore, the integration in a velocity servo is generally provided electronically in the amplifier. The integration is desirable to minimise static errors and to maintain stability.

13 KK-E-E Rydberg Hydraulic Servo Systems Hydraulic – DynamicServo Systems Properties and Control 7 ______

. TL vp qm Ap Ap Dm Jt Mt FL Bm Speed Force transducer transducer

Kf Kf Integrating Servo servo ampl. amplifier uc + i uc + i K ____sa P = konst. Ksa u - s s Ps = const. f uf - Servo system with speed feedback Servo system with force feedback FigureFigure 2-3:2-3: Symbol circuitcircuit ofof aa speedspeed andand aa forceforce servoservo In a real force servo the transducer measures the output force and this signal is fed back Into athe real amplifier. force servo A morethe transducer simpler waymeasures to implement the output a forceforce and servo this is signal to use is fed the back load topressure the amplifier. in the actuator A mor eas simplea feedback way signal to implement. This is aquite force close servo to isa true to use force the servo load pressureexcept from in thethe actuatorfriction forceas a feedbackin the actuator. signal. This is quite close to a true force servo except from the friction force in the actuator. 3. Servo valves and their characteristics The heart of the hydraulic servo system is the servo valve and it is essential that its characteristics be thoroughly understood. A servo valve is a component which work as an interface between an electrical (or mechanical) input signal and the hydraulic power represented by the product of flow and pressure. Depending of the application there are different types of servo valves to use.

3.1 Number of lands and ports The most widely used valve is the sliding valve employing spool type construction. Typical spool valve configurations are shown in Figure 3-1. As explained in the figure, spool valves can be classified by the numbers of ways the flow can enter and leave the valve and the number of lands. Because all valves require a supply, a return and at least one line to the load, valves are either of three-port or four-port type.

AB A B

P T P T Two-land, four-port Three-land, four-port

AB A

P T T P Four-land, four-port Two-land, three-port Figure 3-1: Typical confuguration of the main stage of hydraulic servo valves

14 K-E Rydberg Hydraulic Servo Systems 7 ______

. TL vp qm Ap Ap Dm Jt Mt FL Bm Speed Force transducer transducer

Kf Kf Integrating Servo servo ampl. amplifier uc + i uc + i K ____sa P = konst. Ksa u - s s Ps = const. f uf - Servo system with speed feedback Servo system with force feedback

K-E Rydberg HydraulicFigure 2-3: Servo Symbol Systems circuit – Dynamic of a speed Properties and a force and servo Control In______a real force servo the transducer measures the output force and this signal is fed back to the amplifier. A more simpler way to implement a force servo is to use the load pressure in the actuator as a feedback signal. This is quite close to a true force servo except3 Servo from the frictionvalves force and in the their actuator. characteristics 3.The heart Servo of the valves hydraulic and servo systemtheir ischaracteristics the servo valve and it is essential that its Thecharacteristics heart of the be hydraulicthoroughly servo understood. system isA theservo servo valve valve is a andcomponent it is essential, which thatacts itsas characteristicsan interface between be thoroughly an electrical understood. (or mechanical) A servo valveinput signalis a component and the hydraulic which work power as anrepresen interfaceted bybetween the product an electri of flowcal (or and mechan pressure.ical) Depending input signal of andthe applicationthe hydraulic there power are representeddifferent types by ofthe servo product valves of flow to use. and pressure. Depending of the application there are different types of servo valves to use. 3.1 Number of lands and ports 3.1 Number of lands and ports The most widely used valve is the sliding valve employing spool type construction. The most widely used valve is the sliding valve employing spool type construction. Typical spool valve configurations are shown in Figure 3-1. As explained in the figure, Typical spool valve configurations are shown in Figure 3-1. As explained in the figure, spool valves can be classclassifiedified by , thethe numbersnumbers ofof waysways thethe flowflow cancan enterenter andand leaveleave the valve and the number of lands. BBecauseecause all valves require a ssupply,upply, a return and at least one line to the load, valves are eitheitherer of three-portthree-port oror four-portfour-port type.type.

AB A B

P T P T Two-land, four-port Three-land, four-port

AB A

P T T P Four-land, four-port Two-land, three-port Figure 3-1:3-1: TypicalTypical confugurationconfiguration of the main stage of hydraulichydraulic servoservo valvesvalves

15 K-E Rydberg Hydraulic Servo Systems 8 ______

The four-port valves in Figure 3-1 have two, three and four lands. For mechanical positioning of the main spool, two or three lands can be used. With four (or at least three) lands on the spool it is possible to use hydraulic pressure for positioning of a four-port valve. This is the most common concept for high response servo valves. Special valves may have more than four lands. AK- E three-port Rydberg valve,Hydraulic shown Servo down Systems to the – righDynamict in Properties Figure 3-1, and Control requires a bias pressure acting______on one side of an unsymmetrical cylinder for direction reversal. Usually the head- side piston area is twice the rod-side area and supply pressure acts on the smaller area to provide the bias force for reversal. 3.2 Types of valve center 3.2The type Types of valve of center valve is centerdefined by the width of the land compared to the width of Thethe port type in of the valve valve center sleeve are when defined the byspool the is width in neutral of the position. land compared If the width to the of widththe land of theis smaller port in thanthe valve the port, sleeve the when valve the is saidspool to is have in neutral an open position.-center Ifor the to bewidth under of -thelapped land, isas smaller shown than in Figure the port, 3 -the2. Avalve critical is said-center to have or anzero open-center-lapped valve or to hasbe under-lapped a land width, asidentical shown to in the Figure port width. 3-2. A A critical-center valve with a landor zero-lapped width greater valve than has the a port land w idthwidth is identical to the port width. A valve with a land width greater than the port width is called closed-center or over-lapped. called closed-center or over-lapped.

xv xv xv

Under-lap Zero-lap Over-lap FigureFigure 3-2:3-2: DifferentDifferent valvevalve lappinglapping whenwhen thethe spoolspool isis inin neutralneutral positionposition

The flow characteristics of the valve may be directly related to the type of valve center. Corresponding to Figure 3-2 there are three important flow gain characteristics, with the Corresponding to Figure 3-2 there are three important flow gain characteristics, with the shape shown in Figure 3-3. In fact, it is better to define the type of valve center from theshape shape shown of the in flowFigure gain 3- near3. In neutralfact, it posis betterition thanto define from geometricalthe type of valveconsiderations. center from A criticalthe shape center of the valve flow may gain be near defined neutral as position the geometrical than from fit geometrical required to considerations. achieve a linear A flowcritical gain center in the valve vicinity may of be neutral defined positi as theon, geowhichmetrical usually fit necessrequireditates to achievea slight aoverlap linear toflow offset gain the in effectthe vicinity of radial of clearance.neutral position, which usually necessitates a slight overlap Ato majorityoffset the of effect four-way of radial servo clearance. valves are manufactured with a critical center because of the emphasis on the linear flow gain. Closed center valves are not desirable because of A majority of four-way servo valves are manufactured with a critical center because of the dead-band characteristics in the flow gain. With a proportional amplifier the dead- bandthe emphasis results inon steady the linear state flow error gain. and Closed can cause center backlash valves whichare not maydesirable lead tobecause stability of problemsthe dead- bandin the characteristics servo loop. It isin possiblethe flow to gain. compensate With a proportionalfor dead-band amplifier electronically the dead but- itband will results at least in influence steady state the re errorsponse and time can of cause the servo backlash valve., which may lead to stability problems in the servo loop. It is possible to compensate for dead-band electronically but it will at least influence the response time of the servo valve.

16 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control K-E Rydberg Hydraulic Servo Systems 9 ______

QL Open Critical center center

Underlap Closed region center

xv Overlap region

Figure 3-3: Flow gain (load flow, QL versus spool stroke, xv) of different center types Figure 3-3: Flow gain (load flow, QL versus spool stroke, xv) of different center types

Open center valves are used in applications which require a continuous flow to maintain Openan acceptable center valves fluid temperature are used in and/or applications an increase, which of the require hydraulics a damping. continuous However, flow to maintainthe large an power acceptable loss in fluid neutral temperature position, the and/or decrease an increase in flow gainof the outside hydraulic the under-lap damping. However,region and the thelarge decreased power losspressure in neutral sensitivity position, of open the centerdecrease valves in flow restrict gain their outside use to the underspecial-lap applications. region and the decreased pressure sensitivity of open center valves restrict their use to special applications. Valve sleeve 3.2.1Since Valve the flow sleeve characteristics of a servo valve is of great importance in a servo system the valve must be manufactured with high precision. That close and matching tolerances Sincefor thespool flow lands, characteristics ports and radial of a clearancesservo valve mu isst of also great be importance held as constant in a servo as possible system theduring valve operating must be conditions. manufactured In order with to compen high sate precision. for the influence That close from andpressure matching and temperature the spool is working in a sleeve (bushing), which is surrounded by the tolerances for spool lands, ports and radial clearances must also be held as constant as valve housing. The sleeve or bushing, shown in Figure 3-4, is pressurised both inside possibleand outside during and operating therefore, conditions.the pressure Inwill order not influence to compensate the radial for clearances. the influence from pressure and temperature the spool is working in a sleeve (bushing), which is surrounded by the valve housing. The sleeve or bushing, shown in Figure 3-4, is pressurised both inside and outside and therefore, the pressure will not influence the radial clearances.

Figure 3-4: Main spool and bushing of a servo valve

Figure 3-4: Main spool and bushing of a servo valve, (MOOG)

17 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

In Figure 3-4 it can also be seen that the port in the bushing is formed as a slot (flow metering slot). Each port will include at least two or up to four slots arranged symmetricallyK-E Rydberg around the bushing. Hydraulic This means Servo Systemsthat the flow metering area, for a spool 10 with______the diameter, ds and the spool displacement, xv will be calculated as,

A(xv ) = f sc ⋅π ⋅ d s ⋅ xv In Figure 3-4 it can also be seen that the port in the bushing is formed as a slot (flow where f is the fraction of spool circumference that is opened to the slots. Normally, meteringsc slot). Each port will include at least two or up to four slots arranged thissymmetrically parameter willaround be inthe the bushing. following This interval: means 0.25that the≤ fsc flow ≤ 0.5. metering area, for a spool with the diameter, ds and the spool displacement, xv will be calculated as, 3.3 Examples of electro-hydraulic servo valves A()xv = f sc ⋅π ⋅ d s ⋅ xv The electro-hydraulic servo valve connects the electronic and hydro-mechanical where fsc is the fraction of spool circumference that is opened to the slots. Normally, portionsthis parameter of a hydraulic will be in system. the following Such ainterval: valve has 0.25 electric ≤ fsc ≤ current0.5. as input signal. This electric signal is, then transformed proportionally by different types of feedback loops, to3.3 a mechanical Examples or a hydraulic of electro-hydraulic signal. servo valves The electro-hydraulic servo valve connects the electronic and hydro-mechanical 3.3.1portions Type of a ofhydraulic feedback system. Such a valve has electric current as input signal. This electric signal is then transformed proportionally, by different types of feedback loops, Theto a shapemechanical of the or steady a hydraulic state signal.flow-pressure curves of a servo valve is given from the typeType of of feedback feedback used in the valve. Three types of feedback can be identified, which are spoolThe shape position of the, load steady pressure state flow-pressureand load flow curves feedback of a. servo In an valve ordinary are given flow from direction the controlledtype of feedback servo valve,used in which the valve. is commonly Three types used of in feedback position, can velocity be identified, and force which servos, are thespool main position spool, positionload pressure is proportional and load to flow the input feedback signal.. In The an ordinaryposition flowfeedback direction used incontrolled this case servocan be valve, realised which in differentis commonly ways, used such in as position, direct mechanical velocity and feedback, force servos, force feedbackthe main orspool electrical position position is proportional feedback. to In th forcee input feedback signal. servoThe position valves thefeedback main spoolused in this case can be realised in different ways, such as direct mechanical feedback, force positionfeedback is or converted electrical toposition a force feedback. by a spring In force and feedback this force servo is balancedvalves the at main the torquespool motorposition armature is converted against to thea force torque by adue spring to the and input this current.force is balancedWith pressure at the compensationtorque motor orarmature pressure against feedback the torque the load due flow to the or i loadnput current.pressure With can be pressure maintained compensation to varying or proportionallypressure feedback to the theinput load signal. flow However, or load the pressurese types canof valves be maintained will be best to suited varying for specialproportionally applications to the such input as signal. constant However, flow or th constantis types ofpressure valves control.will be bestThe suitedpressure for- special applications such as constant flow or constant pressure control. The pressure- flow characteristics of the mentioned valves are illustrated in Figure 3-5. flow characteristics of the mentioned valves is illustrated in Figure 3-5.

QL Position feedbackQL Load pressure feedbackQL Load flow feedback

i i i / imax = 1,0 i / imax = 1,0 i / imax = 1,0

pL/ps 1,0 pL/ps 1,0 pL/ps 1,0

FigureFigure 33-5:-5: PressurePressure-flow-flow curves of a servo valve using position,position, loadload pressurepressure andand loadload flow flow feedback. feedback. There are also servo valves where different characteristics are combined. One of these There are also servo valves where different characteristics are combined. One of these valves is called the dynamic pressure feedback servo valve. This valve has a valvescharacteristic is called of a the position dynamic feedback pressure valve feedback at low frequencies servo valve and. characteristics This valve has of a a load pressure feedback valve at higher frequencies. This behaviour is useful in some applications to increase the damping of valve-actuator combinations. 18 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______characteristic of a position feedback valve at low frequencies and characteristics of a load pressure feedback valve at higher frequencies. This behaviour is useful in some applications to increase the damping of valve-actuator combinations.

3.3.2 Number of stages

Servo valves may also be broadly classified as either single-stage, two-stage or three- stage. Single-stage servo valves consist of a torque motor or a linear force motor directly attached for positioning of the spool. Because torque or force motors have limited power capability, this in turn limits the hydraulic power capacity of single-stage servo valves. In some applications the single-stage concept may also lead to stability problems. This is the case if the flow forces acting on the spool are close to the force produced by the electro-magnetic motor. Flow forces are proportional to the flow and the square root of the valve pressure drop, which gives a limitation in hydraulic power.

Single-stage valves A single-stage servo valve with a linear force motor is shown in Figure 3-6. The valve illustrated in the figure is a valve, which employs just one linear force motor (proportional magnet) to move the spool either side of the central position. The electric signal from a position transducer is then used for closed loop control of the spool position. In addition, the spool has a “power off” position whereby when no power is applied to the magnet, the bias spring pushes the spool fully over to the right side. I “power off” position all ports (A, B, P and T) are closed. In normal operation the spool will operate either side the null position but in the event of a power failure or machine shut-down, the spool will move to the “power off” position. The maximum pressure for this type of valve is, normally 350 bars and the maximum flow is less then 80 litre/min.

Figure 3-6: Single-stage servo valve with electrical position feedback, Vickers

Figure 3-7 shows another single stage servo valve with a linear force motor which can actively stroke the spool from its spring centred position in both directions. This is an

19 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______advantage compared with proportional solenoids with one force direction only, as in figure 3-6. The closed loop spool position electronics and pulse width modulated (PWM) drive electronics are integrated into the valve. This permits control directly from, for example, a machine control without the use of additional interface electronics. The valve in Figure 3-7 has a quite strong force motor. High spring stiffness and resulting centring force plus external forces (flow forces and friction forces) must be overcome during out-stroking. During backstroking to centre position the spring force adds to the motor force and provides additional spool driving force which makes the drive very less contamination sensitive. The relatively high force from the force motor also means that the influence from flow forces on the spool position control is very small. This is important to avoid reduction of the hydraulic damping caused by dominant flow forces.

Figure 3-7: Single-stage servo valve with electrical position feedback, MOOG Direct Drive (DDV)

Two-stage valves One of the most common types of servo valve is the two-stages. The servo valve, shown in Figure 3-8 uses an electrical torque motor, a double-nozzle pilot stage and a sliding spool second stage. Electrical current in the torque motor gives proportional displacement of the second stage spool. The flapper in the pilot stage attaches to the centre of the armature and extends down, inside the flexure tube. A nozzle is located on each side of the flapper so that flapper motion varies the nozzle openings. Differential pressures caused by flapper movement between the nozzles are applied to the ends of the valve main spool.

20 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 3-8: Two-stage servo valve (MOOG) Operation: Electrical current in the torque motor coils causes either clockwise or counter-clockwise torque, as shown in Figure 3-9, on the armature. This torque displaces the flapper between the two nozzles. The differential nozzle flow moves the spool to either the right or left. The spool continues to move until the feedback torque counteracts the electromagnetic torque. At this point the armature/flapper is returned to centre, so the spool stops and remains displaced until the electric input changes to a new level.

Figure 3-9: Valve responding to change in electrical input for a two-stage servo valve, MOOG

Instead of flapper nozzle pilot stage a jet pipe stage can be used as illustrated in Figure 3-10. The “servo-jet” consists mainly of a torque motor, jet pipe and receiver. A current through the coil displaces the jet pipe from neutral position.

21 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 3-10: Two-stage servo valve with jet-pipe pilot stage, MOOG

This displacement combined with the special shape of the nozzle direct focussed fluid jet more into one receiver than the other. The jet now produces a pressure difference in the control ports. This pressure difference results in a pilot flow, which in turn causes a spool displacement. One advantage for the jet pipe pilot valve is the less sensitivity for contamination than for example a flapper nozzle valve. The position control loop for the main stage spool is closed by the integrated electronics. An electrical command signal is applied to the integrated position controller, which drives the valve coil. A position transducer measures the position of the main spool. This signal is then fed back to the controller where it is compared with the command signal. The controller drives the pilot valve until the error between the command and feedback signal is zero. Thus the position of the main spool is proportional to the electrical command signal.

Three-stage valves For high flow capacity the required power to drive the main spool will be high. In such applications (flow capacity over 150 litre/min and maximum pressure about 350 bar) a three-stage valve will be used. Three-stage means that a two-stage servo valve, is used as a pilot valve for the main stage, just as shown in Figure 3-11. This valve has an electrical feedback for the main spool position control. In the controller this signal is compared with the command signal. The controller drives the pilot valve, in this case a two-stage servo valve, until the error between the command and feedback signal is zero. In the same way as for the foregoing valves presented, the position of the main spool will be proportional to the electrical command signal.

22 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 3-11: Three-stage servo valve with electrical feedback for main spool position control, MOOG It is also notable that the valves in the Figures 3-10 and 3-11 have no sleeve for the main spool. Therefore, these valves are often called proportional valves instead of servo valves. Today, the main difference between, proportional and servo valves, is just the sleeve. A proportional valve is not manufactured by such precision as a servo valve and therefore also less expensive and mainly used for open loop control. However, the valves shown above have quite linear characteristics through zero position of the main spool and are well suited for closed loop control as ordinary servo valves.

P/Q-valve In some applications there is a need for both pressure and flow control. Velocity control can be required for a part of the working cycle and pressure control for another part of the cycle. Other requirements can be pressure limiting during velocity control. A valve capable of handling these requirements is shown in Figure 3-12.

Figure 3-12: MOOG PQ-proportional control valve for pressure and flow control

23 K-E Rydberg Hydraulic Servo Systems 15 ______

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______Figure 3-12: MOOG PQ-proportional control valve for pressure and flow control

3.43.4 GeneralGeneral steady steady statstatee valve characteristicscharacteristics TheThe design design of of a aservo servo valve valve depends depends upon upon the the requirements onon hydraulichydraulic power power capacity,capacity, accuracy accuracy and and dynamic dynamic response. response. The The valve valve characteristics characteristics will will also also depends depends on the onport the lapping port lapping and the and type the oftype internal of internal feedback feedback control control loops loops in inthe the valve. valve. As mentionedAs mentioned before, before, the the most most common common type type ofof servoservo valve isis aa four four-port-port critical critical-center-center valvevalve with with spool spool position position control. control. The The main main ststageage of this typetype ofof valve valve is is schematically schematically shownshown in Figurein Figure 3-13 3-13. This. This valve valve is is a adevices device thatthat useuses mechanical motion motion to to control control the the hydraulic power from a source to an actuator. hydraulic power from a source to an actuator.

ps qL qL ps pL

FigureFigure 3-13: 3-13 Four-port: Four-port critical critical center center spool spool valvevalve with threethree land land

The load flow through the valve is expressed by the equation

1 ⎛ x ⎞ q = C wx ⎜ p − v p ⎟ (3-1) L q v ⎜ s x L ⎟ ρ ⎝ v ⎠

where, Cq = flow coefficient, [-] pL = p1 – p2 = load pressure dif., [Pa]

ps = supply pressure, [Pa] xv = spool displacement, [m] w = area gradient, [m] ρ = fluid density, [kg/m3]

Assume constant supply pressure (ps) and let the load pressure vary in the range –ps ≤ pL ≤ ps and the spool displacement in the range –xvmax ≤ xv ≤ xvmax. With these assumptions the load flow characteristics will be as illustrated in Figure 3-14.

24 K-E Rydberg Hydraulic Servo Systems 16 ______

The load flow through the valve is expressed by the equation æ ö 1 ç xv ÷ qL = Cq wxv ç ps − pL ÷ (3-1) ρ è xv ø where, Cq = flow coefficient, [-] pL = p1 J p2 = load pressure dif., [Pa] ps = supply pressure, [Pa] xv = spool displacement, [m] w = area gradient, [m] ρ = fluid density, [kg/m3] K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control Assume constant supply pressure (ps) and let the load pressure vary in the range Jps ≤ pL ______≤ ps and the spool displacement in the range Jxvmax ≤ xv ≤ xvmax. With this assumptions the load flow characteristics will be as illustrated in Figure 3-14.

qLmax

xv /xvmax = 1,0

xv /xvmax = 0,5 -1,0 pL /ps 1,0

xv /xvmax = -0,5

xv /xvmax = -1,0

-q Lmax FigureFigure 33-14:-14: Load flow characteristics for a 44-port-port critical center valve

If the valve has an electrical input signal (iv) and the spool displacement xv is Ifproportional the valve to has iv (x anv = electricalKiiv), the valve input flow signal equation (iv) and can thebe written spool as displacement xv is proportional to iv (xv = Kiiv), the valve flow equation can be written as æ ö 1 ç xv ÷ qL = Ki Cq wiv ps − pL (3-2) 1 ⎛ç xv ⎞÷ q = K C wi ρ ⎜è p − xv p ⎟ø (3-2) L i q v ⎜ s x L ⎟ ρ ⎝ v ⎠ Valve Coefficients 3.4.1The valveValve coefficients Coefficients for an ideal critical center valve can be obtained by differentiation of the equation for the pressure-flow curves (equation 3-1) or graphically Thefrom valvea plot of coefficients the curves (Fig. for 3.14). an ideal These critical partial centerderivatives valve defi canne the be most obtained important by differentiationparameters of theof thevalve. equation Assuming for the constant pressure supply-flow pressurecurves (equation (ps) they 3 are:-1) or graphically from a plot of the curves (Fig. 3.14). These partial derivatives define the most important parameters of the valve. Assuming constant supply pressure (ps) they are: ⎡∂q ⎤ Flow gain L (3-3) Kq = ⎢ ⎥ ∂xv ⎣ ⎦ pL =const ⎡ ∂q ⎤ Flow-pressure coefficient L (3-4) Kc = ⎢− ⎥ ∂pL ⎣ ⎦ xv =const

⎡∂p ⎤ Kq Pressure sensitivity L (3-5) K p = ⎢ ⎥ = ∂xv Kc ⎣ ⎦ qL =const For an electro-hydraulic servo valve (electrical input) the valve coefficients will be defined as

⎡∂q ⎤ ⎡ ∂q ⎤ ⎡∂p ⎤ K qi L , L , L (3-6) K qi = ⎢ ⎥ Kci = ⎢− ⎥ K pi = ⎢ ⎥ = ∂iv ∂pL ∂iv Kc ⎣ ⎦ pL =const ⎣ ⎦ iv =const ⎣ ⎦ qL =const

Now, the linearized equation of the valve pressure-flow curve becomes as,

25 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

ΔQL = K q ΔX v − K c ΔPL (3-7)

or ΔQL = K qi Δiv − K ci ΔPL (3-8)

3.5 Critical center four-way valve Assuming a symmetrical valve (symmetrical orifices around zero, which means that

A1(xv) = A2(-xv)), only positive spool displacement (xv) can be studied. The valve flow equation can now be simplified as 1 1 qL = Cq wxv (ps − pL ) or qL = K iCq wiv (ps − pL ) (3-9) ρ ρ The valve coefficients for an ideal critical center valve can be obtained by differentiation of (3-9).

1 1 Flow gain K q = Cq w (ps − pL ) , K qi = K iCq w (ps − pL ) (3-10) ρ ρ

Cq wxv KiCq wiv Flow-pressure coeff Kc = , Kci = (3-11) 2 ρ(ps − pL ) 2 ρ(ps − pL )

2(ps − pL ) 2(ps − pL ) Pressure sensitivity K p = , K pi = (3-12) xv iv

As system design parameters, the null operating point is the most important. Evaluation of these coefficients at the point qL = pL = xv = 0 gives the null coefficients for the ideal critical center valve as,

ps Theoretical null coeff. K = K C w , K = 0 , K pi0 = ∞ (3-13) qi0 i q ρ ci 0

These null coefficients are just theoretical, but the computed null flow gain has been amply verified by tests of practical critical center valves and may be used with confidence. However, the theoretical values for Kci0 and Kpi0 are far from that obtained in tests. It is possible to compute more realistic values for these null coefficients once the leakage characteristics around zero operation for such valve has been investigated.

3.5.1 Practical null coefficients for a critical center valve It has been shown that the valve coefficients vary a lot with the operating point. The most important operating point is the origin of the flow-pressure curves (pL=qL=iv=0) because system operation usually occurs near this region. The null coefficients are of special interest, since they are used as design parameters for the servo system where the

26 K-E Rydberg Hydraulic Servo Systems 18 ______

p Theoretical null coeff. K = K C w s , K = 0 , K = ∞ (3-13) qi0 i q ρ ci0 pi0 These null coefficients are just theoretical, but the computed null flow gain has been amply verified by tests of practical critical center valves and may be used with confidence. However, the theoretical values for Kci0 and Kpi0 are far from that obtained in tests. It is possible to compute more realistic values for these null coefficients once theK-E leakageRydberg characteristicsHydraulic around Servo Systems zero operati – Dynamicon for Properties such valve and Control has been investigated. ______Practical null coefficients for a critical center valve

Itservo has valvebeen shownis used. that The the servo valve loop coefficients gain is proportional vary a lot withto the the valve operating flow gainpoint. (K Theqi0). mostThis nullimportant coefficient operating represents point isthe the highest origin flow of the gain, flow-pressure which gives curves the m aximum(pL=qL=i vloop=0) because system operation usually occurs near this region. The null coefficients are of gain to obtain stability. Most of the hydraulic damping comes from the valve flow- special interest, since they are used as design parameters for the servo system where the pressure coefficient (K ). The null-coefficient gives the lowest K -value and limits the servo valve is used. Theci0 servo loop gain is proportional to the valvec flow gain (Kqi0). Thisloop null gain coefficient for stability. represents The pressure the highest sensitivity flow gain, (Kpi0 )which determines gives the maximumstiffness of loop the gainservo to system. obtain Or stability. in other Most words, of thehow hydraulic well load dampingdisturbances comes can from de controlled the valve out flow- by pressurethe closed coefficient loop system. (Kci0 ). The null-coefficient gives the lowest Kc-value and limits the loop gain for stability. The pressure sensitivity (Kpi0) determines the stiffness of the servo system. Or in other words, how well load disturbances can de controlled out by the3.5.2 closed Leakage loop system. characteristics of a practical critical center four-way valve It is the leakage characteristics, which actually differentiate a practical valve from an Leakage characteristics of a practical critical center four-way spool valve ideal once. An ideal valve has perfect geometry so that leakage flow are zero. The real Itvalve is the has leakage radial clearance characteristics, and perhaps which minute actually under differentiate- or overlap. a practical In order valveto compensate from an idealfor the once. radial An clearance ideal valve a critical has perfect valve geometry normally sohas that a small leakage overlap flow ofare about zero. 5The-25 µrealm. valve has radial clearance and perhaps minute under- or overlap. In order to compensate The leakage performance of such valves dominates their behaviour and the associated for the radial clearance a critical valve normally has a small overlap of about 5-25 µm. Theflow leakage-pressure performance curves at of small such valve valves openings, dominates let their say behaviour |xv| ≤ 0.02 and⋅x vmaxthe associated(or |iv| ≤ 0.02⋅ivmax). Therefore, the null coefficients, Kci0 and Kpi0 are strongly dependent of the flow-pressure curves at small valve openings, let say |xv| ≤ 0.02⋅xvmax (or |iv| ≤ 0.02⋅ivmax). Therefore,valve leakage the performanc null coefficients,e. On the K otherci0 and hand Kpi0 for are larger strongly valve openingsdependent equation of the (3valve-11) leakageand (3-12) performance. fits quite well. On the other hand, for larger valve openings equations (3-11) and (3-12) fits quite well. Consider a four-way, assumed to have matched and symmetrical orifices, as shown in ConsiderFig 3-15 . aWith four-way, blocked assumed load ports to have it can matched be seen a ndtha symmetricalt the two gaps orifices, for each as landshown from in Fig 3-15. With blocked load ports it can be seen that the two gaps for each land from supply port (P) to tank (T) will be equal when the spool is in neutral position, xv = 0. supply port (P) to tank (T) will be equal when the spool is in neutral position, xv = 0. For For an out-stroked spool, xv > 0, one gap for each land will be opened up and the an out-stroked spool, xv > 0, one gap for each land will be opened up and the leakage flowleakage (qle flow) is restricted (qle) is restricted by the other by the gap. other gap. Blocked load ports Blocked load ports AB AB

qle qle

P T P T x = 0 x > 0 v v Figure 3-15:3-15: LeakageLeakage flowflow inin aa valvevalve withwith blocked blocked load load ports ports

3.5.3 Blocked line pressure sensitivity curve

If its true that all the four gaps from P to T are equal at xv = 0, the leakage flow over each land will be equal and also the load pressures pA and pB will be equal. In order to introduce a load pressure difference, pL > 0 (pL = pA - pB), the spool must be moved from neutral position. By stroking the valve and recording load pressure, pL versus input current (iv) the pressure sensitivity can be measured as demonstrated in Fig 3-16.

27 K-E Rydberg Hydraulic Servo Systems 19 ______K-E Rydberg Hydraulic Servo Systems 19 ______

Blocked line pressure sensitivity curve

IfBlocked its true thatline allpressure the four sensitivitygaps from P curveto T are equal at xv = 0, the leakage flow over each land will be equal and also the load pressures pA and pB will be equal. In order to IfK- Eits Rydberg true that all theHydraulic four gaps Servo from Systems P to– Dynamic T are equal Properties at x vand = Control0, the leakage flow over introduce a load pressure difference, pL > 0 (pL = pA - pB), the spool must be moved each______land will be equal and also the load pressures pA and pB will be equal. In order to from neutral position. By stroking the valve and recording load pressure, pL versus input introduce a load pressure difference, pL > 0 (pL = pA - pB), the spool must be moved current (iv) the pressure sensitivity can be measured. from neutral position. By stroking the valve and recording load pressure, pL versus input current (i ) the pressure sensitivity can be measured. v pL p pLs

iv ps

iv Kpi0 ps 1 Kpi0 ps pL i 1 v pL iv

-ps

Figure 3-16:3-16: BlockedBlocked lineline pressurepressure sensitivitysensitivity -p forfors aa criticalcritical centercenter valvevalve

The load pressureFigure difference, 3-16: Blocked pL line quickly pressure increases sensitivity to for full a critical supply center pressure valve after a very The load pressure difference, pL quickly increases to full supply pressure after a very smallThe load increasing pressure of difference, input current. p quickly For a goodincreases servo to valve full supply a typical pressure value afterof pressure a very small increasing of input current.L For a good servo valve a typical value of pressure sensitivitysmall increasing is Kpi0 of= 0.8input⋅ps /(0.01current.⋅ivmax For). As a good an example servo valve ps = 350 a typical bar and value ivmax of = pressure 50 mA, sensitivity is K =10 0.8 p /(0.01 i ). As an example p = 350 bar and i = 50 mA, gives Kpi0 =5.6⋅pi010 Pa/A.⋅ s Pressure⋅ vmax sensitivity (gain) is usuallys specified asvmax the average sensitivity is Kpi0 =10 0.8⋅ps/(0.01⋅ivmax). As an example ps = 350 bar and ivmax = 50 mA, gives Kpi0 =5.6⋅1010 Pa/A. Pressure sensitivity (gain) is usually specified as the average slopegives Kofpi0 the =5.6 of ⋅load10 Pa/A.pressure Pressure drop vers sensitivityus input (gain) current is inusually the region specified between as the ± average40% of maximumslope of the load of pressureload pressure drop. drop versusversus ininputput current in the region between ±40% of maximum load pressure drop. Leakage flow curves

ByLeakage3.5.4 stroking Leakage flow the valve curves flow (i vcurves) and measuring the supply flow (qs) for a given supply pressure and with blocked load ports the leakage flow curve can be plotted, as shown in Figure By stroking stroking the the valve valve (i v) (i andv) and measuring measuring the supply the supply flow (q flows) for (q as )given for asupply given pressure supply 3-17. andpressure with andblocked with loadblocked ports load the ports leakage the leakageflow curve flow can curve be plotted, can be plotted,as shown as in shown Figure in 3-17. i Figure 3-17. v Leakage flow iv qLeakagec flow ps qs q ) ( c ps qs pL

) ( pL Pilot valve flow

Pilot valve flowiv Figure 3-17: Leakage flow characteristic for a two-stage four-way servo valveiv

The leakageFigureFigure flow 3-17:3is-17 maximum: Leakage flow (qc )characteristic at valve neutral for a two-stagetwo (i-vstage = 0) four-wayfour and-way decreases servoservo valvevalve rapidly with valveThe leakage stroke becauseflow is maximumthe spool land (q ) overlapat valve the neutral return (i valve = 0) orifice,and decreases see Fig. rapidly3-15. It with can The leakage flow is max (q ) at valvec neutral (i = 0) vand decreases rapidly with valve bevalve observed stroke thatbecause the flow the spool consumptionc land overlap of the the pilot returnv stage valve of the orifice, valve see is constant Fig. 3-15. because It can stroke because the spool land, overlap the return valve orifice, see Fig. 3-15. It can be ofbe theobserved constant that supply the flow pressure. consumption of the pilot stage of the valve is constant because observed that the flow consumption of the pilot stage of the valve is constant because of of the constant supply pressure. the constant supply pressure.

Looking at the supply flow (qs), it can be shown theoretically that the following expression is valid:

∂qs ∂qL = − = Kc (3-14) ∂ps ∂pL

28 K-E Rydberg Hydraulic Servo Systems 20 ______

Looking at the supply flow (qs), it can be shown theoretically that the following Kexpression-E Rydberg is valid:Hydraulic Servo Systems – Dynamic Properties and Control ______∂qs ∂qL = − = K c (3-14) ∂ps ∂pL From equation (3-14) it is obvious that by measuring the center leakage flow (qc) versus From equation (3-14) it is obvious that by measuring the center leakage flow (qc) versus the supply pressure (ps) it is possible to calculate the null coefficient Kci0. the supply pressure (ps) it is possible to calculate the null coefficient Kci0.

Kci Practical Kc-curve

Kci0 Theoretical Kc-curve

0 iv

FigureFigure 3-18:3-18: Flow-pressureFlow-pressure coefficient versusversus inputinput currentcurrent

The null coefficient, Kci0 is determined by the leakage characteristics only. However, for The null coefficient, Kci0 is determined by the leakage characteristics only. However, higher stroking of the valve Kci will follow equation (3-11). for higher stroking of the valve Kci will follow equation (3-11). About the null coefficient it is important to note that the Kc-value varies as the valve Aboutwears. theFor nulla new coefficient the center it leakageis important flow tocan note be nearlythat the laminar. Kc-value For varies a worn as thevalve valve the wears.center flowFor awill new be the more center and leakage more turbulent.flow can beThe nearly center laminar. leakage For can a beworn expressed valve the in centerprincipal flow as, will be more and more turbulent. The center leakage can be expressed in principal as, qc = cons tant ⋅ µ ⋅ ps (laminar flow) (3-15)

qc = cons tant ⋅ µ ⋅ ps (laminar flow) (3-15) qc = cons tant ⋅ ps (turbulent flow) (3-16) q = cons tant ⋅ p (turbulent flow) (3-16) where µ is the dynamic viscosityc of the fluid.s whereThis fact µ is also the givesdynamic implication viscosity onof thethe fluid.null pressure sensitivity, Kpi0. An approximate expression for this null coefficient of a real critical center valve can be obtained by This fact also gives implication on the null pressure sensitivity, K . An approximate dividing equation (3-10) by (3-15) and (3-16) respectively, pi0 expression for this null coefficient of a real critical center valve can be obtained by dividingLaminar equationcenter flow: (3-10) by (3-15)K pi 0and= cons(3-16)tan respectively,t ⋅ µ ⋅ ps (3-17)

LaminarTurbulent center center flow: flow: K piK0 pi=0 cons= constantant ⋅ µt ⋅⋅ps ps (3(3-18)-17)

TurbulentIt is perhaps center worth flow: mentioning thatK equatipi0 = conson (3-17)tant ⋅ pands (3-17) represents two extremes(3-18) and for a practical servo valve, measurement of Kpi0 related to the supply pressure (ps) Itwill is beperhaps something worth between mentioning the equationthat equation (3-17) (3 and-17) (3-18). and (3-17) represents two extremes and for a practical servo valve, measurement of Kpi0 related to the supply pressure (ps) willFlow be gain something between the equation (3-17) and (3-18).

The flow gain of a servo valve with electrical input represents the slope of the control 3.5.5flow (q LReal) versus flow input gain current characteristics (iv) curve at any specific operation region. For a practical critical center valve the flow gain varies depending on non-linearities and hyteresis, as The flow gain of a servo valve with electrical input represents the slope of the control illustrated in Figure 3-19. flow (qL) versus input current (iv) curve at any specific operation region. For a practical critical center valve the flow gain varies depending on nonlinearities and hysteresis, as illustrated in Figure 3-19.

29 K-EK-E Rydberg Hydraulic Servo Systems Hydraulic – DynamicServo Systems Properties and Control 21 ______K-E Rydberg Hydraulic Servo Systems 21 ______

qL q L y iv ps = constant x i pL = 0 y v p = constant x p s Hysteresis s pL = 0 ps qL ) ( Hysteresis Null bias iv qL ) ( LinearityNull bias y / x = Flow gain iv Linearity y / x = Flow gain

FigureFigure 3-19:3-19: FlowFlow gaingain characteristicscharacteristics forfor aa four-wayfour-way criticalcritical centercenter valvevalve Figure 3-19: Flow gain characteristics for a four-way critical center valve The straight dotted line in Figure 3-19 drawn from the zero flow point, throughout the The straight dotted line in Figure 3-19 drawn from the zero flow point, throughout the range of rated current is called Normal Flow Gain. The operating conditions for the Therange straight of rated dotted current line is in called Figure Nomin 3-19 aldraw Flown from Gain the. The zero operating flow point, conditions throughout for the flowrange curve of rated in the current figure isabove called is Normalconstant Flowsupply Gain pressure. The and operating no load conditionspressure (p forL = the0). Flowflow curvegain as in demonstrated the figure above in the is constantfigure is supplythe normal pressu flowre andgain no with load zero-load, pressure which(pL = 0). is flow curve in the figure above is constant supply pressure and no load pressure (pL = 0). calledFlow gain gainNo-Load asas demonstrateddemonstrated Flow Gain in.in The the the figureno-load figure is is flowthe the normal gain,normal K flowqi0 flow will gain gain vary with with with zero-load, zero supply-load, pressurewhich which is as described by equation (3-13). calledis called No-Load No-Load Flow Flow Gain Gain. The. no-load The no -flowload gain, flow K gain,qi0 will K qi0vary will with vary supply with pressure supply aspressure described as described by equation by equation(3-13). (3-13). 3.6 Open center spool valve Consider3.6 Open the four-way centercente spoolr spoolspool valve valvevalve shown in Figure 3-20. When the valve is centred, the underlap of the supply and the return ports are identical with the value U. Consider the four-wayfour-way spool valve shown in Figure 3-20.3-20. When the valve is centred, the underlapunder-lap of of the the supply supply and and the the return return por portsts are are identical identical with with the the value value U. U.

xv 1 4 3 2

1 4 3 2 U U U U ps ps U qL U U qL U ps pL ps qL qL pL

Figure 3-20: Four-way open center spool valve with underlap U

If it is assumed FigurethatFigure the 33-20:- 20valve: FourFour-way -isway matc openhed center and spool symmetrical valve with underlapunder and- lap if U U only the underlap region is studied, x ≤ U , the orifice areas are (observe the direction of xv in Fig. 3-20), If it is assassumedumed thatv the valvevalve isis matchedmatched and and symmetrical symmetrical and and if if only only the the under underlap-lap (observe the direction of x in Fig. 3-20) region is studied, xv ≤ U , the orificeorifice areasareas areare (observe the direction of xv in Fig. 3-20),, xv ≤ U A1 = w()U + xv = A3 v (3-19)

A1 = w()U + xv = A3 (3-19) AA21 = w()(U −+ xxvv )== AA43 (3-20)(3-19)

A2 = w()U − xv = A4 (3-20) A2 = w(U − xv ) = A4 (3-20)

30 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

TheK-E subsc Rydbergripts of the area refer to Hydraulicthe number Servo at Systems the ports in Figure 3-20 and arrows 22 at the______ports indicate flow directions. The general flow equation for positive and negative spool stroke ( xv ≤ U ) and load pressure pL can be written as,

The subscripts of the area refer to1 the number at the ports in Figure1 3-20 and arrows at the ports indicate flow directions. The general flow equation for positive and negative qL = Cq w(U + xv ) (ps − pL ) − Cq w(U − xv ) (ps + pL ) (3-21) spool stroke ( xv ≤ U ) and load pressureρ pL can be written as, ρ

This is the equation for the pressure1 -flow curves of an open1 center four-way spool qL = Cq w()U + xv ()ps − pL − Cq w()U − xv ()ps + pL (3-21) valve for operation within the underρ -lap region. In normalisedρ manner equation (3-21) becomesThis is the equation for the pressure-flow curves of an open center four-way spool valve for operation within the underlap region. In normalised manner equation (3-21) becomes q ⎛ x ⎞ p ⎛ x ⎞ p L = ⎜1+ v ⎟ 1− L − ⎜1− v ⎟ 1+ L (3-22) æ ö æ ö qL p ⎝ Uxv ⎠ pLs ⎝ xUv ⎠ ppL s C wU s = ç1+ ÷ 1− − ç1− ÷ 1+ (3-22) q p è U ø p è U ø p C wU ρ s s s q ρ Equation (3-21) is plotted in Figure 3-21. These curves are quite linear compared to Equation (3-21) is plotted in Figure 3-21. These curves are quite linear compared to thosethose for for a acritical critical center center valve valve inin thethe nullnull displacementdisplacement region.region. However,However, outside outside the the underunderlap-lap region region thethe valvevalve actsacts asas aa criticalcritical centercenter valve,valve, bbecauseecause only only two two orifices orifices are are actingacting at ata time.a time.

xv /U = 1,0

xv /U = 0 -1,0 pL/ps 1,0

xv /U = -1,0

FigureFigure 3- 213-21:: Pressure Pressure-flow-flow curves curves for for under underlap-lap regionregion of of an an open open center center four-way four-way spool spool valve valve TheThe valve valve coefficient coefficient can can be be obtained obtained by by differentiating (3-21). (3-21). Evaluating Evaluating the the derivatives at qL = pL = xv = 0 will gives the null coefficients. The coefficients are, derivatives at qL = pL = xv = 0 will gives the null coefficients. The coefficients are, 1 ps Flow gain K q = Cq w []ps − pL + ps + pL , K q0 = 2Cq w (3-23) 1ρ ρps Flow gain K q = Cq w [ ps − pL + ps + pL ] , K q0 = 2Cq w (3-23) ρ é ù ρ Cq wxv U + xv U − xv Cq wU Flow-pr. coeff. K c = ê + ú , K c0 = (3-24) 2 ρ ëê ps − pL ps + pL ûú ρ ⋅ ps Cq wxv ⎡ U + x U − x ⎤ Cq wU Flow-pr. coeff. K v v , K (3-24) c = ⎢ K + 2 p⎥ c0 = 2 ρ p − p q p + p s p Pressure sensitivity ⎣⎢ Ks p = L , K ps0 = L ⎦⎥ ρ ⋅ s (3-25) K c U

K q 2 ps Pressure sensitivity K p = , K p0 = (3-25) K c U

31 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control

K-E______Rydberg Hydraulic Servo Systems 23 ______As discussed before the null coefficients are the most important ones. It can be noted that the flow gain in the under-lap region is twice of that for a critical center valve. As discussed before the null coefficients are the most important ones. It can be noted thatFurther, the flowKc0 depends gain in on the the underlap area gradient region w is and twice the under of that-lap for U aand critical Kp0 is center independent valve. Further,of w. Kc0 depends on the area gradient w and the underlap U and Kp0 is independent ofLeakage w. flow curves similar to those defined for the critical center valve can be made Leakagefor the openflow centercurves valve.similar Theto those total defined center flowfor the through critical the center valve valve is useful can be since made it for the open center valve. The total center flow through the valve is useful since it describe the power loss at null operation. At this point, pL = xv = 0 the orifice areas are describe the power loss at null operation. At this point, pL = xv = 0 the orifice areas are A1 = A2 = wU, which gives the total center flow as A1 = A2 = wU, which gives the total center flow as

ps qc = 2Cq wU ps (3-26) qc = 2Cq wU ρ (3-26) ρ

3.7 Three-way spool valve analysis 3.7 Three-way spool valve analysis AA three-way three-way spool spool valves valves must must be be used used to togethergether with with a a differential differential area area piston, piston, to to provideprovide directiondirection reversal.reversal. TheThe areaarea ratioratio isis normallynormally two, two, as as shown shown in in Figure Figure 3-22. 3-22 .

qL pc Ah xv Ah/2

pT=0 ps

Figure 3-22: Three-way spool valve with differential piston Figure 3-22: Three-way spool valve with differential piston The rod and the head side areas of the piston is such that the steady state control The rod and the head side areas of the piston is such that the steady state control pressure acting on the head side is about pressure acting on the head side is about ps pc0 ≈ p (3-27) p ≈ 2 s (3-27) c0 2 This design relation allows the control pressure pc to rise and fall and to provide equal accelerationThis design andrelation deceleration allows the capability. control pressure With no ploadsc to rise on andthe pistonfall and this to isprovide satisfied equal by makingacceleration the headand decele area twiceration thecapability. rod area With (Ah no= 2Aloadr). on This the rule piston is generallythis is satisfied used , forby pistonmaking sizing the headeven areawith load twice forces. the rod However, area (A hthis = 2Atyper). of This valve rule and is piston generally combination used for isp istonusually sizing preferred even within hydro-mechanical load forces. However, servos this where type the of external valve and load piston is quite combination small. Foris usually a critical preferred center invalve hydro the- mechanicalpressure-flow servos curves where can thebe expressedexternal load as is quite small. For a critical center valve the pressure2-flow curves can be expressed as qL = Cq wxv ()ps − pc for xv ≥ 0 ρ2 qL = Cq wxv (ps − pc ) for xv ≥ 0 ρ 2 qL = Cq wxv pc for xv < 0 (3-28) ρ2 qL = Cq wxv pc for xv < 0 (3-28) The null operating point for a three-way valveρ is defined by qL = xv = 0 and pL = ps/2. Evaluating the derivatives at this point gives the null coefficients for a critical center three-way valve as 32 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

The null operating point for a three-way valve is defined by qL = xv = 0 and pL = ps/2. Evaluating the derivatives at this point gives the null coefficients for a critical center three-way valve as,

p Flow gain K = C w s (3-29) q0 q ρ

Cq wxv Flow-pressure coefficient Kc0 = = 0 (3-30) ρ ⋅ ps

ps Pressure sensitivity K p0 = = ∞ (3-31) xv Comparing the null coefficients, it can be noted that the flow gain is the same but the pressure sensitivity is half of that for a four-way critical center valve. Therefore load forces will cause twice the static error compared with a four-way valve. For a practical valve the null coefficients becomes,

ps K q0 Practical null coeff. K q0 = Cq w , Kc0 > 0 , K p0 = (3-32) ρ K c0

Three-way valves can also be of open center type. For operation within the underlap region, the pressure-flow curves is given by

2 2 qL = Cq w(U + xv ) (ps − pc ) − Cq w(U − xv ) pc (3-33) ρ ρ

Evaluating the derivatives at qL = xv = 0 and pL = ps/2 will gives the null coefficients. The null coefficients of a three-way open center valve are,

p Flow gain K = 2C w s (3-34) q0 q ρ

2Cq wU Flow-pressure coefficient K c0 = (3-35) ρ ⋅ ps p Pressure sensitivity K = s (3-36) p0 U

33 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

3.8 Dynamic response of servo valves The dynamic characteristic of servo valves is of great importance for a proper design of a closed loop servo system. The valve must be fast enough to transmit its command signal into the load actuator system with a frequency higher than the highest frequency in the actuator/load system. In other words, the bandwidth of the valve must be 2-5 times higher than the bandwidth of the rest of the servo system. The bandwidth of a servo valve can be determined by measuring the frequency at which the phase lag between the inputServoServo current and valvevalve output flow dynamicsdynamics reaches -90 ̊. The measurements are normally presented in a bode-diagram as illustrated in Figure 3-23.

Valve transfer function

(of max Slow valve ( < 50 Hz) input sv ampl.)

] QL Ksv o G (s) sv i s v 1 sv

Fast valve (sv > 50 Hz) Gain[dB] Phaseshift [ QL Ksv Gsv (s) 2 iv 2 sv s 1 2 sv sv

Valve bandwidth = sv Frequency [Hz] Figure 3-23: Frequency response of a servo valve at different input signal amplitudes For 2- or 3-stage valves the bandwidth depends on the supply pressure to the pilot stage: sv ps As can be seen from Figure 3-23 the frequency response varies with input signal amplitude. Because of saturation in the valve spool actuator system the valve bandwidth will be reduced at increased signal amplitude. The requirements on valve bandwidth, depends upon the application. However, for valve selection a typical peak-to-peak signal amplitude is 80% of the valve rated input signal. For all valves with electro-hydraulic pilot stages the valve bandwidth is also affected by the supply pressure ps. Since the loop gain of the pilot stage is proportion to its flow gain the valve frequency behave as,

ωsv ∝ ps (3-37) ServoServo valvevalve efficiencyefficiency Also the fluid temperature will influence the valve dynamics. Low temperature gives high fluid viscosity and thereby increased friction, which reduces the bandwidth.

Fixed pump xv qL pL ps qs 34 Load qL pL ps sv qs ps pT = 0

qL = load flow Variable pump xv qL p = load pressure pL L ps qs Load

qs = supply flow

Ps = supply pressure

pT = 0 Ps = constant

4 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

For modelling of valve dynamics the following transfer functions can be used:

Valve bandwidth lower than 50 Hz:

QL Ksv Valve bandwidth higher than 50 Hz: Gsv (s) = = 2 iv 2δ sv s 1+ + 2 ωsv ωsv More details about servo valve definitions and specifications can be found in referense [2].

35 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control K-E Rydberg Hydraulic servo systems 25 ______

4 Position servos with valve-controlled cylinders 4 Position servos with valve-controlled cylinders 4.1 Asymmetric Asymmetric cylinder cylinder Consider a valve-controlledvalve-controlled piston wwithith position feedback as shown in Figure 4-14-1. Fix reference xp be A1 A2 be Bp Mt FL P1 V1 V2 P2

qL1 qL2 Position xv transducer Servo amplifier uc + i Kf - Ksa uf Ps = const.

Figure 4-1: Asymmetrical cylinder controlled by four-way critical center spool valve Figure 4-1: Asymmetrical cylinder controlled by four-way critical center spool valve If it is assumed that the servo valve is matched and symmetrical and the supply pressure If it is assumed that the servo valve is matched and symmetrical and the supply pressure (ps) is constant and the tank pressure is approximately zero. (ps) is constant and the tank pressure is approximately zero. The general flow equations for positive spool stroke (xv) can be written as, The general flow equations for positive spool stroke (xv) can be written as, 2 2 qL1 Cq wxv 2 ps p1 and qL2 Cq wxv 2 p2 (4-1) qL1 = Cq wxv (ps − p1 ) and qL2 = Cq wxv p2 (4-1) ρ ρ The valve coefficient can be obtained by differentiating (4-1). For an operating point 0 theThe coefficients valve coefficient are, can be obtained by differentiating (4-1). For an operating point 0 the coefficients are, 2 2 Flow gain K C w p p , K C w p (4-2) q1 q 2 s 10 q2 q 2 20 Flow gain K q1 = Cq w (ps − p10 ) , K q2 = Cq w p20 (4-2) ρ ρ Cq wxv0 Cq wxv0 Flow-pressure coeff. K c1 , K c2 (4-3) Cq wxv0 Cq wxv0 2 ps p10 2 p20 Flow-pressure coeff. Kc1 = , K c2 = (4-3) 2ρ ⋅ (ps − p10 ) 2ρ ⋅ p20 The two linearised and laplace-transformed equations describing the valve flow becomesThe two linearized and laplace-transformed equations describing the valve flow becomes QL1 K q1X v K c1P1 (4-4) ΔQL1 = K q1ΔX v − K c1ΔP1 (4-4) QL2 K q2 X v K c2 P2 (4-5) ΔQL2 = K q2 ΔX v + K c2 ΔP2 (4-5) Considering no leakage flow in the cylinder gives the linearised and laplace- transformedConsidering continuity no leakage equations flow for in the volumes cylinder V 1 gives and V the2 as linearized and laplace- transformed continuity equations for the volumes V1 and V2 as V1 QL1 A1sX p sP1 (4-6) e

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

K-E Rydberg Hydraulic servo systemsV 26 ΔQ = A sΔX + 1 sΔP (4-6) ______L1 1 p 1 β e V − ΔQ = −A sΔX + 2 sΔP (4-7) L2 2 p Vβ 2 Q A sX 2e sP (4-7) L2 2 p 2 The final equation for the actuator arises from the foe rces of the piston. The linearized andThe laplace final equation-transformed for the force actuator equation arises will fro bem written the forces as of the piston. The linearised and laplace-transformed force equation will be written as 2 A1ΔP1 − A2ΔP2 = M t s ΔX p + Bp sΔX p + ΔFL (4-8) 2 A1P1 A2 P2 M t s X p B p sX p FL (4-8) Combining equation (4-4) to (4-7) gives Combining equation (4-4) to (4-7) gives ⎛ V1 ⎞ K q1ΔX v = A1sΔX p + ⎜ Kc1 + s⎟ΔP1 (4-9) ⎜ V1 ⎟ β e (4-9) K q1X v A1sX p ⎝K c1 s⎠P1 e ⎛ V2 ⎞ − K q2ΔX v = −A2 sΔX p + ⎜ Kc2 + s⎟ΔP2 (4-10) ⎜ V2 ⎟ β e (4-10) K q2 X v A2 sX p ⎝K c2 s⎠P2 e

Introducing the position feedback gain Kf, the servo amplifier gain Ksa and the transfer Introducing the position feedback gain Kf, the servo amplifier gain Ksa and the transfer function of the servo valve Gv(s) the spool displacement of the valve (xv) becomes function of the servo valve Gv(s) the spool displacement of the valve (xv) becomes

ΔX v = (U c − K f ΔX p )K saGv (s) (4-11) X v U c K f X p K saGv (s) (4-11)

ByBy usingusing thethe equationsequations (4(4-8)-8) –Q (4(4-11)-11) a blockblock diagram of the closed loop system willwill bebe asas shownshown inin FigureFigure 44-2-2..

A1 - 1 P1 Kq1 + V1 A1 Kc1 + s be + . u i x xp 1 xp c K G (s) v F 1 _ + sa v L - s - - Mt s + Bp 1 K A q2 - V2 2 Kc2 + s P2 + be A2

FigureFigure4-2: 4-2: Block diagram of a positionposition servoservo withwith aa valvevalve controlledcontrolled asymmetric asymmetric cylinder cylinder In the above equations the leakage in the cylinder has been neglected. If leakage is taken In the above equations the leakage in the cylinder has been neglected. If leakage is into account an extra term will be added to the Kc-values. This can be modelled as taken into account an extra term will be added to the Kc-values. This can be modelled as K ce1 K c1 C p1 and K ce2 K c2 C p2 . K ce1 = K c1 + C p1 and K ce 2 = K c2 + C p2 . A problem with the asymmetric cylinder is that the behaviour of the system will not be the same for both stroking directions. When the direction of output velocity is changed it will also be a pressure jump because of the fact that different piston areas are used to produce the actuator force acting on the load. 37

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

A problem with the asymmetric cylinder is that the behaviour of the system will not be K-E theRydberg same for both stroking directions. Hydraulic When servo the systems direction of output velocity is changed 27 ______it will also be a pressure jump because of the fact that different piston areas are used to produce the actuator force acting on the load.

Example:4.1.1 Example Variation in resonance frequency for an asymmetric cylinder with line volumes Variation in resonance frequency for an asymmetric cylinder with line volumes

A massA mass loaded loaded asymmetric asymmetric cylinder cylinder with with lineline volumesvolumes isis shown shown in in Figure Figure 4- 34-3. .

A1 A2 Mt V1 V2

VL1 VL2

Figure 4-3: A double acting hydraulic cylinder with line volumes Figure 4-3: A double acting hydraulic cylinder with line volumes

ParametersParameters Piston displacement: 0 ≤ xp ≤ xpmax, xpmax = 1 m Piston displacement: 0 xp xpmax, xpmax = 1 m 2 Piston area: A1 = 0.02 m2 , A2 = 0.75A1 Piston area: A1 = 0.02 m , A2 = 0.75A1 Line volumes: VL1 = VL2 = k⋅A1⋅ xpmax, 0 < k < 1 Line volumes: VL1 = VL2 = kA1 xpmax, 0 < k < 1 Bulk modulus: βe = 1000 MPa Bulk modulus: e = 1000 MPa K ⎡ A 2 A 2 ⎤ Resonance frequency: ω = h where 1 2 h K h = βe ⎢ 2 + 2 ⎥ KMh V AA x V A x A x Resonance frequency: t where ⎣ L 1 + 11 p L 2 + 2( p max2 − p )⎦ h K h e M t V L 1 A 1x p V L 2 A 2x p max x p

Figure 4-4 shows the relative frequency ωh/ωhmin versus piston displacement (xp) for

Figurek = 4-4 0.05 shows (the highestthe relative curve) frequency and k = 0.10h/ hmin (the versus lowest piston curve). displacement It is notable (x thatp) for the k =cylinder 0.05 (the will highestbe much curve) stiffer when and kthe = piston 0.10 is (the close lowest to end curve). positions. It is notable that the cylinder will be much stiffer when the piston is close to end positions.

2.5

2 Wh/Whmin 1.5 38

1 0 0.2 0.4 0.6 0.8 1 Piston displacement, Xp [m]

Figure 4-4: Relative resonance frequency versus piston displacement for two different line volumes

!"#$%&'()*+$$ ,&'*-./01$2)*34$2&25)62$ 78$ 999999999999999999999999999999999999999999999999999999999999999999999999999999999999$

Example: Variation in resonance frequency for an asymmetric cylinder with line volumes

:$6-22$/4-')'$-2&66)5*01$1&/0;')*$<05=$/0;)$34/.6)2$02$2=4<;$0;$Figure 4-3>$$

:A :7 D5 ?A ?7

?@A ?@7 $

E0+.*)$F"GH$:$'4.(/)$-150;+$=&'*-./01$1&/0;')*$<05=$/0;)$34/.6)2$ $ Parameters

I0254;$'02C/-1)6);5H$$J$$BC$$BC6-BK$BC6-B$L$A$6$ 7 I0254;$-*)-H$$:A$L$J>J7$6 K$$$:7$L$J>8M:A$

@0;)$34/.6)2H$$?@A$L$?@7$L$N:A$BC6-BK$$J$O$N$O$A$

P./N$64'./.2H$$)$L$AJJJ$DI-$ $ K A 7 A 7 %)24;-;1)$Q*)R.);1&H$ h $<=)*)$ A 7 $ h K h e M t V L A A Ax p V L 7 A 7x p 6-B x p $

FigureK-E Rydberg 4-4$2=4<2$5=)$*)/-503)$Q*)R.);1&$ Hydraulic Servo Systems – Dynamic=S=60; Properties$3)*2.2$C0254;$'02C/-1)6);5$TB and Control CU$Q4*$ N$______L$ J>JM$ T5=)$ =0+=)25$ 1.*3)U$ -;'$ N$ L$ J>AJ$ T5=)$/4<)25$1.*3)U>$V5$02$;45-(/)$5=-5$5=)$ 1&/0;')*$<0//$()$6.1=$250QQ)*$<=);$5=)$C0254;$02$1/42)$54$);'$C420504;2>$

2.5

hmin k = 0.05 2 ! / h Wh/Whmin ! 1.5

k = 0.10 1 0 0.2 0.4 0.6 0.8 1 K-E Rydberg Piston HydraulicPisto ndisplacement, di spservolace systemsment, X p X [m ][m] $ 28 p ______E0+.*)$F"FH$%)/-503)$*)24;-;1)$Q*)R.);1&$3)*2.2$C0254;$'02C/-1)6);5$Q4*$5<4$'0QQ)*);5$/0;)$34/.6)2 $ Figure 4-4: Relative resonance frequency versus piston displacement for two different line volumes $ 4.2 Valve controlled symmetric cylinder A position servo with a symmetricsymmetric cylinder is shown in Figure 4-54-5.. $ Fix reference xp be Ap Ap be Bp Mt FL P1 V1 V2 P2

qL1 qL2 x Position v transducer Servo amplifier uc + i Kf - Ksa uf Ps = const.

FigureFigure 4-5:4-5: Position servo with a valve controlled symmetric cylindercylinder The general flow equations for the four-wayfour-way critical centercenter valvevalve cancan bebe writtenwritten as,as, 1 q C wx p p where pL = p1 – p2 (4-12) qL = Cq wxv (ps − pL ) where pL = p1 – p2 (4-12) ρ From equation (4-12) the valve coefficient can be obtained as From equation (4-12) the valve coefficient can be obtained as 1 Flow gain K C w p p (4-13) q q 1 s L0 Flow gain K q = Cq w (ps − pL0 ) (4-13) ρ Cq wxv0 Flow-pressure coeff. K c (4-14) Cq wxv0 2 ps pL0 Flow-pressure coeff. K c = (4-14) 2 ρ ⋅ (ps − pL0 ) The linearised and laplace-transformed equations describing the valve flow becomes

QL K q X v K c PL (4-15) Considering no external leakage flow in the39 cylinder (just across the piston) gives the linearised and laplace-transformed continuity equations for the volumes V1 and V2 as

V1 QL1 C p P1 P2 Ap sX p sP1 (4-16) e

V2 QL2 C p P1 P2 Ap sX p sP2 (4-17) e Because of the symmetric cylinder it is possible to calculate the load flow as Q Q Q L1 L2 (4-18) L 2

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

The linearized and laplace-transformed equations describing the valve flow becomes

ΔQL = K q ΔX v − K c ΔPL (4-15)

Considering no external leakage flow in the cylinder (just across the piston) gives the linearized and laplace-transformed continuity equations for the volumes V1 and V2 as

V1 ΔQL1 − C p (ΔP1 − ΔP2 ) = Ap sΔX p + sΔP1 (4-16) β e

V2 − ΔQL2 + C p (ΔP1 − ΔP2 ) = −Ap sΔX p + sΔP2 (4-17) β e Because of the symmetric cylinder it is possible to calculate the load flow as ΔQ + ΔQ ΔQ = L1 L2 (4-18) L 2

Assume that the piston is in centred position (xp=0) and the volumes are, V1 = V2 = Vt/2, where Vt is the total pressurised volume in the cylinder. Combining equation (4-

16) to (4-18) using the definition of the load pressure difference ΔPL = ΔP1 − ΔP2 gives

Vt ΔQL = Ap sΔX p + C p ΔPL + sΔPL (4-19) 4β e Combining equation (4-15) and (4-19) and with electrical input to the servo valve gives

⎛ V ⎞ K X A s X ⎜ K t s⎟ P (4-20) qi Δ v = p Δ p + ⎜ ce + ⎟Δ L ⎝ 4β e ⎠ where the total flow-pressure coefficient is K ce = K c + C p .

The final equation for the actuator arises from the forces of the piston. The linearized and laplace-transformed force equation will be written as

2 Ap (ΔP1 − ΔP2 ) = Ap ΔPL = M t s ΔX p + Bp sΔX p + ΔFL (4-21)

Introducing the position feedback gain Kf, the servo amplifier gain Ksa and the transfer function of the servo valve Gv(s) the spool displacement of the valve (xv) becomes

ΔX v = (U c − K f ΔX p )K saGv (s) (4-22)

By using the equations (4-20) – (4-22) a block diagram of the closed loop system will be as shown in Figure 4-6.

40 K-E Rydberg Hydraulic servo systems 29 ______

Assume that the piston is in centred position (xp=0) and the volumes are V1 = V2 = Vt/2, where Vt is the total pressurised volume in the cylinder. Combining equation (4-16) to

(4-18) using the definition of the load pressure difference PL P1 P2 gives

Vt QL Ap sX p C p PL sPL (4-19) 4 e Combining equation (4-15) and (4-19) and with ilectrical input to the servo valve gives

V t K qi X v Ap sX p K ce sPL (4-20) 4 e where the total flow-pressure coefficient is K ce K c C p .

The final equation for the actuator arises from the forces of the piston. The linearised and laplace-transformed force equation will be written as

2 Ap P1 P2 Ap PL M t s X p B p sX p FL (4-21)

Introducing the position feedback gain Kf, the servo amplifier gain Ksa and the transfer function of the servo valve Gv(s) the spool displacement of the valve (xv) becomes

K -E Rydberg Hydraulic ServoX v Systems U c – KDynamicf X p KPropertiessaGv (s) and Control (4-22) ______By using the equations (4-20) S (4-22) a block diagram of the closed loop system will be as shown in Figure 4-6. FL - . u 1 xp 1 xp c i G K 1 A _ + Ksa v qi + p + M s + B - Vt PL t p s - Kce + s 4be Ap

FigureFigure 44-6:-6: BlockBlock diagradiagramm of a position servo with aa valvevalve controlledcontrolled symmetricsymmetric cylinder cylinder

ConsiderConsider aa firstfirst orderorder transfertransfer function for the servo valve,valve, writtenwritten asas 1 ( ) 1 (4-23) Gv (s) = (4-23) v ss 1+ v ω v The block-diagram in Figure 4-4 can now be reduced to the following form, shown in The block-diagram in Figure 4-4 can now be reduced to the following form, shown in Figure 4-7. Figure 4-7.

Figure 4-7: Block diagram of a position servo with a valve controlled symmetric cylinder

2 If the term Bp Kce / Ap is smaller than unit the hydraulic resonance frequency ωh and the hydraulic damping δh shown in the block-diagram (Figure 4-5) will be expressed as

2 2 β e Ap ⎛ 1 1 ⎞ 4β e Ap ⎜ ⎟ V1 = V2 = Vt/2 gives ω h = ⎜ + ⎟ ω h = M t ⎝V1 V2 ⎠ M tVt

Kce β e M t Bp Vt and δ h = + Ap Vt 4Ap β e M t

2 If the term Bp Kce / Ap is included the resonance frequency and damping becomes

' B p K ce ' δ h ω h = ω h 1+ 2 and δ h = Ap Bp Kce 1+ 2 Ap

41 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

4.2.1 Servo system stability and bandwidth

In order to study the stability of the servo system the open loop gain Au(s) must be analysed. Figure 4-5 yields

K sa K qi K f / Ap K v Au (s) = = (4-24) ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎝ ωv ⎠ ⎝ω h ωh ⎠ ⎝ ωv ⎠ ⎝ω h ω h ⎠

Kv expresses the steady state loop gain and the value of this parameter must be set to a certain level to make sure that the control system will be stable.

The critical parameter in this servo system is the amplitude margin Am, which is expressed as

10 Kv Am = −20 log [dB] (4-25) − 2δ hω h In other words, the control system will be stable if the amplitude margin is positive, which gives the stability criteria as

Kv max < 2δ h minωh min (4-26 a) or for a specified amplitude margin,

A − m 20 Kv max = 10 2δ h minωh min (4-26 b)

TheBode-diagram open loop gain of the position of the servo open with Kv = loop δhωh (Am gain = 6 dB)of isthe shown in Figure 4-8. position servo

Figure 4-8: Bode-diagram of the open loop gain of the position servo depicted in Figure 4-5 Karl-Erik Rydberg, Linköping University, Sweden 10

42 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

The bode-diagram in Figure 4-8 shows the two stability margins, amplitude margin o (Am) and phase margin (φm). Since, φm is close to 90 , this is not a critical margin in the actual system. The critical stability margin is A . From the figure it can also be Closed loop transferm function observed that the marked crossover frequency ωc (when the amplitude curve cross the zero-line, 0 dB) is equal to Kv, the amplitude at the frequency ω =1 rad/s

The closed loop transfer can be derived as illustrated in Figure 4-9a. X X c A (s) p + u X p Au (s) 1 - Gc (s) = = = Xc 1+ Au (s) 1/ Au (s)+1 1

Figure 4-9a: Block diagram and corresponding closed loop transfer function

Au(s) as expressed in eq. (4-24), but without valve dynamics, gives the closed loop transfer function as, 1 1 G (s) = ≈ (4-27) c " 2 % " % " 2 % s s 2δh s s 2δh ⋅$ 2 + s +1'+1 $ +1'⋅$ 2 + s +1' Kv #ωh ωh & # Kv & #ωh ωh &

The first order low-pass filter, 1/(s/Kv + 1) gives the bandwidth of the system. The Bode-diagramKarl-Erik of Rydberg, the Linköping closed University, Sweden loop gain of 11 bandwidth is ωb = Kv. the position servo The bode-diagram of the closed loop system is shown in Figure 4-9b.

Figure 4-9b: Bode-diagram of the closed loop gain of the position servo depicted in Figure 4-5

Karl-Erik Rydberg, Linköping University, Sweden 12 The closed system bandwidth ωb is shown in Figure 4-9b. Because of the -1 slope of the open loop gain curve (Fig 4-8) up to just above the crossover frequency, there will be a well-defined relation between these frequencies and Kv, which is,

ωb = ωc = Kv

43 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______K-E Rydberg Hydraulic servo systems 32 K-E Rydberg Hydraulic servo systems 32 ______4.3 Three-way valve controlled cylinder with force feedback 4.3 Three-way valve controlled cylinder with force feedback A4.3 three Three-way-way spool valves valve must controlled be used together cylinder with with a differential force feedback area piston, to provideA three-way directi on spool reversal. valves The must area be ratio used is tonormallygether withtwo, aas differentialshown in Figure area piston, 4-10. In to A three-way spool valves must be used together with a differential area piston, to theprovide figure direction the position reversal. feedback The areais implemented ratio is normally by using two, a as force shown feedback in Figure (spring) 4-10 .on In provide direction reversal. The area ratio is normally two, as shown in Figure 4-10. In thethe spoolfigure ofthe the position control feedback valve. is The implemented control valve by using is equipped a force withfeedback a proportional (spring) on thethe figure spool the of position the control feedback valve. is The implemented control valve by using is equippe a forced feedback with a proportional(spring) on magnet,the spool which of the gives control a force valve. (Fm ) Theproportional control valveto the isinput equippe currentd with (i). Then a proportional the force magnet, which gives a force (Fm) proportional to the input current (i). Then the force balancemagnet, onwhich the valve gives spool a force means (Fm )that proportional the piston toposition the input (x ) cuwillrrent be (i).proportional Then the toforce the balance on the valve spool means that the piston position (xpp) will be proportional to the inputbalanceinput current. current. on the valve spool means that the piston position (xp) will be proportional to the input current. xp xp be Ah Ar = Ah/2 be Ah Ar = Ah/2 FL Mt FL Mt pc Vh pc Vh xv Kf xv Kf K1 i Fm K1 i Fm ps p s FigureFigure 4 4-10:-10: Linear Linear actuator actuator with with aa threethree-way-way spoolspool valve,valve, differentialdifferential pistpistonon andand forceforce feedbackfeedback Figure 4-10: Linear actuator with a three-way spool valve, differential piston and force feedback Neglecting flow forces the forces acting on the valve spool is given by the equation: NeglectingNeglecting flowflow forcesforces the forces acting on the valve spool is given by the equation: Fm K f X p ΔFm − K f ΔX p Fm K f X p X v K1X v 0 X v Fm K f X p (4-28) ΔFm − K f (ΔX p − ΔX v )− K1ΔX v = 0 ⇒ ΔX v = (4-28) Fm K f X p X v K1X v 0 X v K f K1 (4-28) K f + K1 K f K1 AssumingAssuming thatthat the magnet forceforce isis a a linear linear function function of of input input current, current, as asF m K m i the Assuming that the magnet force is a linear function of input current, as FΔFmK= Kim theΔi block diagram of the system will become as shown in Figure 4-11. m m theblock block diagram diagram of the of thesystem system will will become become as shown as shown in Figure in Figure 4-11 4.- 11. FL FL K æ V ö ce æç + h ö÷ Kce2 ç1 Vh s÷ A ç + be K ce ÷ h2 çè1 s÷ø Ah è be K ce ø - 1 . i F x Kq .xp 1 xp K m 1 v - s2 21 i m Fm+ 1 xv Kq + + dh + xp 1 xp K Kf + K1 Ah s22 2d s 1 s m + - + w + wh s + 1 s Kf + K1 Ah 2h w - wh h Position feedback Kf K Position feedback f Figure 4-11: Block diagram of an actuator with a three-way spool valve, differential piston Figure 4-11: Block diagram of an actuator with a three-way spool valve, differential piston Figure 4-11: Block diagram of an actuatorand force with feedback a three-way spool valve, differential piston and force feedback and force feedback In the system above there is only one cylinder volume acting as a spring, namely Vh. If In the system above there is only one cylinder volume acting2 as a spring, namely Vh. If Inalso the the system viscous above friction there ofis onlythe piston one cylinder is low volume( B p K ce /actingA2h 1as) thea spring, hydraulic namely resonance Vh. If also the viscous friction of the piston is low ( B p K ce / A2h 1) the hydraulic resonance alsofrequency the viscous h and friction the hydraulic of the piston damping is low h shown( Bp K cein/ Atheh

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

frequency ωh and the hydraulic damping δh shown in the block-diagram (Figure 4-11) will be expressed as

2 β e Ah Hydraulic resonance frequency: ωh = M tVh

Kce β e M t Bp Vh Hydraulic damping: δ h = + 2Ah Vh 2Ah β e M t

2 If the term Bp Kce / Ah is included the resonance frequency and damping becomes

Bp Kce δ ' and ' h ωh = ωh 1+ 2 δ h = Ah Bp K ce 1+ 2 Ah

In order to study the stability of the servo system the open loop gain Au(s) must be ⎛ 1 ⎞ analysed. Including valve dynamics ⎜G ⎟ Figure 4-11 yields ⎜ v = ⎟ ⎝ 1+ s /ωv ⎠

K f K q

(K f + K1 )Ap K v Au (s) = = (4-29) ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎝ ω v ⎠ ⎝ω h ω h ⎠ ⎝ ω v ⎠ ⎝ω h ω h ⎠

Kv expresses the steady state loop gain and the value of this parameter must be set to a certain level, according to equation (4-27), to make sure that the control system will be stable. This type of actuators, are commonly used for spool control in big valves and for displacement control of variable pumps and motors. Because of the fact that the cylinder pressures are only controlled in one chamber, the stiffness of the actuator will be reduced compared to an actuator with two-chamber control. That means less accuracy in system with heavy external forces.

4.4 Influence from flow forces on valve spools In an actuator system with a direct controlled servo valve the flow forces acting on the valve spool will influence the spool position. The flow forces act as a positive load pressure feedback, which means reduced hydraulic damping in the system. To reduce this influence the valve spool positioning system must be stiff enough so that the deflexion of the spool, caused by the flow forces, is small. By introducing a spring between the electric armature (magnet) and the valve spool, the influence from flow forces can be studied. This is illustrated in Figure 4-12.

45 KK-E-E Rydberg Hydraulic Servo Systems Hydraulic – Dynamic servo systems Properties and Control 34 ______K-E Rydberg Hydraulic servo systems 34 ______Fix reference xp be Ap Ap be Fix reference Mt xpFL Pb1e VA1p AVp 2 bPe2 Mt FL xa xv P1 V1 V2 P2 Position transducer Servo amplifier xa Kxva uc + Kf Position i transducer ServoK - sa K uuf amplifier a Ps = const. c + i Kf - Ksa uFigurefFigure 4-12:4-12: Position servo with a weakPs = const. spool control of thethe servoservo valvevalve

Referring to FigureFigure 4-12: 44-12-12 Position the armature servo with positiona weak spool (xa )control can of be the expressed servo valve as xa = i ⋅ K m , where Km is thethe forceforce coefficientcoefficient ofof thethe magnet.magnet. InIn thethe fourfour-port-port valve the flow forces Referring to Figure 4-12 the armature position (x ) can be expressed as x i K , remains from two orifices, which gives the total steadya state flow forceforce as, a m where Km is the force coefficient of the magnet. In the four-port valve the flow forces remainsF ft 2 fromCq w twoxv ( porifices,s p1 ) cos(which) gives 2Cq thew xtotalv p steady2 cos( state) 2flowCq w force xv ( pas,s pL )cos() Fft = 2Cq w⋅ xv ( ps − p1 )cos(ϕ) + 2Cq w⋅ xv ⋅ p2 ⋅ cos(ϕ) = 2Cq w⋅ xv ( ps − pL )cos(ϕ) The force equation for the valve spool can now be expressed as, TheF ftforce 2C equationq w xv ( p fors thep1 ) valvecos( spool) 2C canq w nowxv p be2 cos(expressed) 2 Cas,q w xv ( ps pL )cos()

i K m xv K a 2Cq w xv ( ps pL )cos() i K m xv K a K ft xv ( ps pL ) 0 The(i ⋅ forceK m − equationxv )K a − 2forCq thew⋅ xvalvev ( ps spool− pL ) cancos( nowϕ) = be(i ⋅ expressedK m − xv )K as,a − K ft ⋅ xv ( ps − pL ) = 0 A linearised expression of the force equation, according to the variable xv and pL gives, i K m xv K a 2Cq w xv ( ps pL )cos() i K m xv K a K ft xv ( ps pL ) 0 A linearized expression of the force equation, according to the variable xv and pL gives, i K X K K ( p p 0 )X K x 0 P 0 A linearised expressionm of thev forcea equation,ft s accordingL v to theft variablev L xv and pL gives, (Δi ⋅ K m − ΔX v )K a − K ft ( ps − pL0 )ΔX v + K ft xv0 ΔPL = 0 In this equation the valve dynamics is ignored. This can be done if the bandwidth of the i K m X v K a K ft ( ps pL0 )X v K ft xv0 PL 0 Invalve this isequation higher the than val ve the dynamics dominant is resignored.onance This frequency can be done of theif the system. bandwidth The of spool the valvedisplacementIn this isequation higher will the than become valve the dynaas, dominant mics is resonanceignored. This frequency can be done of theif the system. bandwidth The of spool the displacementvalve is higher will thanbecome the as, dominant resonance frequency of the system. The spool displacement will become ias, K m K a K ft xv0 PL i K m K a K fp PL X (4-30) v Δi ⋅ K K + K x ΔP Δi ⋅ K K + K ΔP K a m Kaft ( ps ft pv0L0 ) L Km a a K fx fp L ΔX v = i K m K a K ft xv0 PL = i K m K a K fp PL (4-30) X v K a + K ft ( ps − pL0 ) K a + K fx (4-30) Implementation of equationK a (4-30) K ft ( p ins a p blocL0 ) k diagramK ofa theK fx servo system gives the Implementationresult presented in of Figure equation 4-13 (4.- 30) in a block diagram of the servo system gives the Implementation of equation (4-30) in a block diagram of the servo system gives the result presented in Figure 4-13. Kfp result presented in Figure 4-13. FL - - . u i x + 1 x K 1 xp 1 xp c K K a K v K fp 1 A _ + sa m a + K +K q + p + FL s a fx - Vt PL Mt s . - - Kce + s - u i x + 1 x 4b 1 xp 1 xp c K K a K v K 1 e A _ + sa m a + q + p + Ka+Kfx - Vt PL A Mt s s - Kce + s p 4be AKpf Figure 4-13: Block diagram of a position servo with flow forces acting on the Kf spool control of the servo valve FigureFigure 44-13:-13: Block diagram of a position servo withwith flowflow forcesforces acting acting on on the the spool control of the servo valve Figure 4-13 shows that the flowspool force control act ofas thea positive servo valve load pressure (pL) feedback. It is well known that the flow/pressure-coefficient of the valve and cylinder (Kce) act as a Figure 4-13 shows that the flow force act as a positive load pressure (pL) feedback. It is well known that the flow/pressure-coefficient of the valve and cylinder (Kce) act as a 46

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______K-E Rydberg Hydraulic servo systems 35 ______Figure 4-13 shows that the flow force acts as a positive load pressure (pL) feedback. It is well known that the flow/pressure-coefficient of the valve and cylinder (Kce) act as a K-E Rydberg Hydraulic servo systems 35 negative load pressure feedback. Therefore, it’s interesting to combine these effects and negative______load pressure feedback. Therefore, itBs interesting to combine these effects and see what happens with the resulting Kce-value. This can be done by reduction of the see what happens with the resulting Kce-value. This can be done by reduction of the block diagram as shown in Figure 44-14.-14. negative load pressure feedback. Therefore, itBs interesting to combine these effects and K see what happens with the resulting Kce-value.q This can be done by reduction of the Kfp block diagram as shown in Figure 4-14.Ka +Kfx FL

Kq - . 1 x + K xp v K K +K 1 fp A 1 + K +K q + a fx p +F a fx - Vt PL L Mt s Kce + s - . 1 x + 4be xp v K 1 A 1 + q + p + Ka+Kfx - Vt PL Ap Mt s Kce + s 4be Figure 4-14: Block diagram showing the load pressure feedback from flow forces Figure 4-14: Block diagram showing the load pressure feedbackAp from flow forces

If the positive load pressure feedback is co-operated into the main transfer function If the positiveFigure load 4-14: pressure Block diagram feedback showing is the co load-operated pressure feedback into the from main flow forces transfer function from flow to load pressure, this will give the resulting Kce-value as, fromIf theflow positive to load loadpressure, pressure this feedbackwill give isthe co-operated resulting K intoce-value the mainas, transfer function from flow to load pressure, this will give the resulting Kce-value as, * K q K fp * K q K fp K ce K ce (4-31) K ce = K ce − (4-31) K a K fx * Ka q+K fp fx K ce K ce (4-31) * * K a K fx Equation (4-31) shows that K ce K ce and since the hydraulic damping is proportional Equation (4-31) shows that K ce < K ce and since the hydraulic damping is proportional to the Kce-value it is true that* flow forces acting on a valve spool will reduce the to Equation the K - value(4-31) itshows is true that that K ce flow K ce forces and since acting the hydraulic on a valve damping spool is will proportional reduce the hydraulicce damping. For a good valve itBs therefore important to design the armature so to the K -value it is true that flow forces acting on a valve spool will reduce the hydraulicthat stiffness damping.ce (in this For model a good K )valve is much it’s highertherefore than important the spring to designcoefficient the armatureof the flow so hydraulic damping. For a gooda valve itBs therefore important to design the armature so thatforces, stiffness Kfx. With (in thissuch model a design Ka )the is influencemuch higher from than flow the forces spring will coefficient be small. of the flow that stiffness (in this model Ka) is much higher than the spring coefficient of the flow forces, Kfx. With such a design the influence from flow forces will be small. forces, Kfx. With such a design the influence from flow forces will be small.

4.5 Position servo with mechanical springs at connectors 4.5 Position servo with mechanical springs at connectors 4.5 Position servo with mechanical springs at connectors Figure 4-14 shows a linear hydraulic position servo. For the cylinder the weakness of Figure 4-14 shows a linear hydraulic position servo. For the cylinder the weakness of theFigure mechanical 4-14 shows part is a relatedlinear hydraulicto spring positisystemson servo.in the Forrear the mounting cylinder end the (Kweakness1) and in of the the mechanical part is related to spring systems in the mounting end (K1) and in the pistonthe mechanicalrod (K ) as part shown is related in the to figure. spring systems in the rear mounting end (K1) and in the piston rod (KL) as shown in the figure. piston rod (KL L) as shown in the figure. Kf x1 Kf xp x2 xL x1 xp x2 xL K1 V1 Ap V2 KL K1 V1 Ap V2 KL Mt Mt p1 p2 p1 p2 Servo Servo amplifieramplifier x ufu - xvv f - ii KKsasa ucuc++

FigureFigure 4-14: 4-14: Valve Valve controlled controlled piston withwith mechanical mechanical springs springs Figure 4-14: Valve controlled piston with mechanical springs AccordingAccording to toFigure Figure 4-14 4-14 the the block block diagradiagram for thethe system,system, without without position position feedback, feedback, willwill be bedeveloped developed as as in in Figure Figure 4-15. 4-15. 47

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

K-E Rydberg Hydraulic servo systems 36 AccordingK-E Rydberg to Figure 4-14 the block Hydraulic diagram servo for systemsthe system, without position feedback, 36 ______will______be developed as in Figure 4-15.

Kce Kce - Kce . X 1 P ⎡ ⎛ ⎤ Xp X v - _____ L 1 2 1 1 ⎞ . . _1 p Kq - 1 Ap ⎢M t s ⎜ + ⎟ + 1⎥ X X Xv + _____Vt PL 1 2 1 1 Xp _1s p Xv ______1 PL M 1t s 2⎝K11 K1L ⎠ p 1 Xp Kq s Ap ⎣M t s 1⎦ _ Kq + - Vt Ap M t s 1 s + ___4βVe ts M t s K 1 K L s - ___ s M t s K 1 K L - 4e 4e QL Ap QQL L AAp Figure 4-15: Block diagram for a valvep controlled cylinder with elastic mountings Figure 4-15: Block diagram for a valve controlled cylinder with elastic mountings Figure 4-15: Block diagram for a valve controlled cylinder with elastic mountings Reduction of the block diagram in Figure 4-15 and completing with the transfer Reduction of the block diagram in Figure 4-15 and completing with the transfer functionReduction from of X p the to X blockL and diagramthe position in feedback Figure 4-15 loop and gives completing the following with diagram. the transfer function from Xp to XL and the position feedback loop gives the following diagram. function from Xp to XL and the position feedback loop gives the following diagram.

1 1 M s2 1+ 1 +1 . t 2 . 1 uc i X K M t s K +K +1 Xp 1 Xp 1 X u + K v q 1 L X _ L c + sa i Gv Xv K q K1 K L p _1 Xp 2 1 1 XL Ksa Gv A s2 d ' s M s 1+ 1 + 1 - Ap s2 + 2 hds' +1 s Mt s2K +K + 1 - p '2 2 ' h s 1 t 1 L wh '2 + wh ' + K1 K L wh wh

Kf Kf Figure 4-16: Complete block diagram for a valve controlled cylinder servo with elastic mountings FigureFigure 4 -4-16:16: Complete Complete block block diagram diagram for for a avalve valve controlled controlled cylinder cylinder servo servo with with elastic elastic mountings mountings ' ' K e 1 V t 1 1 ' K ceM t h ' where ' K ; 1 V 1 1 and ' K M wwherehere h M e ; K 4 At 2 K K and h Ace2 t 2 h h Mt Ke 4e Ap 2 K1 KL h Ap 2 2 t e e p 1 L p If can be noted that the effective spring gradient Ke is derived from the series ItIf can can be be noted noted that that the the effective effective spring spring gradient gradient K Ke e 2isis derivedderived fromfrom thethe seriesseries 4eA p 2 2 connection between the hydraulic spring gradient 4eA p and the two mechanical 4βe Ap connectionconnection between between the the hydraulic hydraulic spring spring gradient gradient Vt and and thethe twotwo mechanicalmechanical Vt springs K1, KL. Vt springssprings K K,1 ,K K.L . 1 L

Simulation of position servo with mechanical springs Simulation of position servo with mechanical springs 4.5.1The following Simulation parameter of po valuessition is servoused in with the simulation mechanical model: springs TheThe following following parameter parameter-3 2 values values is is used used in in the the simulation simulation model: 9model: Ap = 2,5·10 -3 m 2 e = 1,0·10 9Pa Ap = 2,5·10-3 m2 e = 1,0·109 Pa Ap = 2,5·10 m βe = 1,0 ·10 Pa Bp = 0 Kf = 25 V/m Bp = 0 Kf = 25 V/m Bp = 0 -11 5 Kf = 25 V/m 3 Kce = 1,0·10 -11 m /Ns5 Kqi = 0,02 m /As3 KKce = = 1,0 1,0·10·10-11 m m5/Ns/Ns K Kqi = = 0,02 0,02 m m3/As/As Kcesa = 0,05 A/V Mqit = 1500 kg Ksa = 0,05 A/V Mt = 1500 kg Ksa = 0,05 A/V-3 3 Mt = 1500 kg Vt = 1,0·10 -3 m 3 v = 0,005 s Vt = 1,0·10-3 m3 7 v = 0,005 s Vt = 1,0·10 m τv = 0,005 s K1 = KL = 5,0·10 7N/m KK 1= = K K L= = 5,0 5,0·10·107 N/m N/m 1 L

The transfer function from valve flow to piston position (Xp) has the form: The transfer function from valve flow to piston position (X ) has the form: The transfer function2 from valve flow to piston position (X )p has the form: s p s 2 2 L s 1 2 2 L 1 1 L 2 L s and from Xp to XL we have: G hxp 2 G hxL 2 1 s L L and from Xp to XL we have: s G hxp 2 h G hxL 2 L 2 2 s 1 2 2 s 1 s h 48 s L h 2 2 h s 1 L 2 2 L s 1 h h L L

K-E Rydberg Hydraulic servo systems 36 ______

Kce - . 1 X X Xv _____ PL 1 2 1 1 p _1 p Kq Ap M s 1 + ___Vt M s t K K s - s t 1 L 4e Q L A p Figure 4-15: Block diagram for a valve controlled cylinder with elastic mountings

Reduction of the block diagram in Figure 4-15 and completing with the transfer function from Xp to XL and the position feedback loop gives the following diagram.

1 1 M s2 + +1 . 1 u t X c + i Xv K q K1 K L p _1 Xp XL Ksa Gv 1 1 A s2 d ' s M s2 + + 1 - p 2 h s 1 t '2 + ' + K1 K L wh wh

Kf Figure 4-16: Complete block diagram for a valve controlled cylinder servo with elastic mountings

' ' K e 1 V t 1 1 ' K ceM t h where h ; 2 and h 2 M t Ke 4e A p K 1 K L A p 2

If can be noted that the effective spring gradient Ke is derived from the series 4 A 2 connection between the hydraulic spring gradient e p and the two mechanical Vt springs K1, KL.

Simulation of position servo with mechanical springs The following parameter values is used in the simulation model: -3 2 9 Ap = 2,5·10 m e = 1,0·10 Pa

Bp = 0 Kf = 25 V/m -11 5 3 Kce = 1,0·10 m /Ns Kqi = 0,02 m /As

Ksa = 0,05 A/V Mt = 1500 kg -3 3 Vt = 1,0·10 m v = 0,005 s 7 K -E RydbergK1 = KL = 5,0·10Hydraulic N/m Servo Systems – Dynamic Properties and Control ______

The transfer function from valve flow to piston position (Xp) has the form: s 2 2 L s 1 2 1 G L L and from Xp to XL we have: G hxp s 2 hxL s2 K-E Rydberg 2 h s 1 Hydraulic servo systems 2 L s 1 37 2 2 ______h h L L where ωh = 91 rad/s, δh = 0.11 and ωL = 123 rad/s, δL = 0.01 (δL belongs to the viscous friction on the load side). where h = 91 rad/s, h = 0.11 and L = 123 rad/s, L = 0.01 (L belongs to the viscous friction on the load side). 1 The valve dynamics is modelled as G = v s1 1 The valve dynamics is modelled as G τ v + v s 1 v K v K sa K qi Form the parameter list it can be seen that the steady state loop is K v = = 10 , K v K sa K qi Ap Form the parameter list it can be seen that the steady state loop is K v 10 , which gives an amplitude margin of about 6 dB. Ap which gives an amplitude margin of about 6 dB.

Simulation of the system with piston position feedback, Xp Simulation of the system with piston position feedback, Xp

A DYMOLA-model of the servo system with position feedback (Xp or XL) is shown in A DYMOLA-model of the servo system with position feedback (Xp or XL) is shown in FigureFigure 4 -4-1717. By. By setting setting the the feedbackfeedback gaingain to 11 oror 00 thethe feedbackfeedback loop loop can can be be activated activated or oreliminated. eliminated.

FigureFigure 4- 4-17:17: DYMOLA DYMOLA-model-model of of thethe positionposition servo withwith position position feedback feedback of of piston piston or orload load (X p(X orp XorL )X L)

Simulation results with piston position feedback (X ) are shown in Figure 4-18 and 4-19 Simulation results with piston position feedback (Xp p, versus time in [s]) are shown in respectively. It can be noted that the amplitude of oscillation is higher for X then for Figure 4-18 and 4-19 respectively. It can be noted that the amplitude of oscillationL is Xp, because of the dynamics between these positions. higher for XL then for Xp, because of the dynamics between these positions. Integ1.y[1] 0.006Integ1.y[1] 0.006

0.004 0.004 Xp Xp 0.002 0.002

0 0

0 0.25 0.5 0 0.25 0.5 Figure 4-18: Step response (0 - 5 mm) of piston position Time (X [s]p) with piston position feedback, Xp

Figure 4-18: Step response (0 - 5 mm) of piston position (Xp) with piston position feedback, Xp

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

GhxL.y 0.006

0.004 XL

0.002

0 0.25 0.5 Time [s]

Figure 4-19: Step response (0 - 5 mm) of load position (XL) with piston position feedback, Xp

Simulation of the system with load position feedback, XL

Simulation results with load position feedback (XL) are shown in Figure 4-20 and 4-21 respectively. From the simulation model (Figure 4-17) it is obvious that the introduced load damping (δL) will not affect the amplitude of XL, but the piston position (Xp) will be better damped if the load damping is increased.

Integ1.y[1] 0.006

0.004

0.002

0 0.25 0.5 Time [s]

Figure 4-20: Step response (0 - 5 mm) of piston position (Xp) with load position feedback, XL

50 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

GhxL.y 0.006

0.004

0.002

0 0.25 0.5 Time [s]

Figure 4-21: Step response (0 - 5 mm) of load position (XL) with load position feedback, XL

The results shows that feedback of Xp gives less oscillations amplitude than if XL is fed back. The reason is that with feedback of Xp acts ωL as an anti-resonance and reduces the resonance-top from ωh. A request for the reduction is that ωL is quite close to ωh, in this case is ωL = 1.4 ωh (observe that ωL always is greater than ωh in the given system).

51 K-E Rydberg Hydraulic servo systems 39 ______

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control The results shows that feedback of Xp gives less oscillations amplitude than if XL is fed ______back. The reason is that with feedback of Xp acts L as an anti-resonance and reduces the resonance-top from h. A request for the reduction is that L is quite close to h, in this case is L = 1.4 h (observe that L always is greater than h in the given system). 5 Servo systems with valve or pump controlled 5 motors Servo systems with valve or pump controlled 5.1 motorsFour-way valve controlled motor with position feedback The5.1 hydraulic Four-way actuator valve composed controlled of a valve motor controlled with rotary position motor feedbackis a widely used combination.The hydraulic All actuator the non composedlinearities ofexisting a valve in controlled a valve controlled rotary motor cylinder is a widelysystem usedwill excombination.ist even here. All One the special non-linearities problem existingis that the in displacement a valve controlled volume cylinder in the motor system is willnot constantexist even but here. varies One in special a discontinuous problem is fashion that the with displacement the shaft rotation. volume inBy the using motor a motor is not constant but varies in a discontinuous fashion with the shaft rotation. By using a motor withwith a a high high number number of of "pumping" "pumping" elements elements (pistons)(pistons) thethe amplitudeamplitude ofof thethe kinematikinematicc displacementdisplacement variationvariation willwill bebe reducedreduced andand cancan bebe ignoredignored inin dynamicdynamic calculations.calculations. AA simplifiedsimplified angularangular positionposition servoservo isis shownshown inin FigureFigure 55-1-1::

Kf - K Uc sa qm + V1 ps pL Dm Jt TL

Figure 5-1: Valve controlled angular position servo Figure 5-1: Valve controlled angular position servo The linearised and laplace-transformed equations describing the valve flow becomes The linearized and laplace-transformed equations describing the valve flow becomes

QL K qi X v K c PL (5-1) ΔQL = K qi ΔX v − K c ΔPL (5-1) Assume that the volumes between valve and motor are equal, V1 = V2. The total Assume that the volumes between valve and motor are equal, V = V . The total pressurised volume is Vt = V1 + V2. From the cylinder case, equation1 (4-16)2 to (4-18) pressurised volume is Vt = V1 + V2. From the cylinder case, equation (4-16) to (4-18) and the definition of the load pressure difference PL P1 P2 gives and the definition of the load pressure difference ΔPL = ΔP1 − ΔP2 gives Vt QL Dm s m Cm PL sPL (5-2) 4Vt ΔQ = D sΔθ + C ΔP + e sΔP (5-2) L m m m L 4β L Combining equation (5-1) and (5-2) gives e Combining equation (5-1) and (5-2) gives

Vt K qi X v Dm s m K ce sPL (5-3) ⎛ Vt ⎞ K X D s ⎜K 4 e s⎟ P (5-3) qi Δ v = m Δθ m + ⎜ ce + ⎟Δ L ⎝ 4β e ⎠ where the total flow-pressure coefficient is K ce K c Cm . where the total flow-pressure coefficient is Kce = Kc + Cm . The final equation for the actuator arises from the torque of the motor. The linearised Theand laplace-transformedfinal equation for the force actuator equation arises will from be writtenthe torque as of the motor. The linearized and laplace-transformed force equation will be written as

2 Dm (ΔP1 − ΔP2 ) = Dm ΔPL = J t s Δθ m + Bm sΔθ m + ΔTL (5-4)

52 K-E Rydberg Hydraulic servo systems 40 ______K-E Rydberg Hydraulic servo systems 40 K______-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______2 Dm P1 P2 Dm PL J t s m Bm s m TL (5-4) 2 Dm P1 P2 Dm PL J t s m Bm s m TL (5-4) IntroducingIntroducing thethe positionposition feedbackfeedback gaingain Kff,, thethe servoservo amplifier gain Ksa and the transfertransfer functionfunction ofof thethe servoservo valvvalvee GGv(s)(s) thethe spool displacement of the valvevalve ((xxv)) becomesbecomes Introducing the position feedbackv gain Kf, the servo amplifier gain Ksav and the transfer function of the servo valve Gv(s)X the U spool K displacementX K G (ofs) the valve (xv) becomes(5-5) ΔX vv = (U c − K f ΔX p )Ksa Gv (s) (5-5)

X v U c K f X p K saGv (s) (5-5) ByBy usingusing thethe equationsequations (5(5-3)-3) –I (5(5-5)-5) a block diagram of the closed loop system willwill bebe asasBy shownshown using ininthe FigureFigure equations 55-2-2.. (5-3) I (5-5) a block diagram of the closed loop system will be as shown in Figure 5-2. TL - . uc 1 PL TL 1 qm 1_ qm Ksa Gv Kqi Dm . + + Kce + Vt s / 4be + - Jt s + Bm s u - - 1 PL 1 q q c K G K D m 1_ m + sa v qi K + V s / 4b m - + - ce t e + Jt s + Bm s Dm D Kfm

K FigureFigure 55-2:-2: Block diagram of a valvef controlled angularangular positionposition servoservo The block-diagramFigure in 5-2: Figure Block 5-2 diagram can beof areduced valve controlled to the following angular position form, servo shown in Figure The block-diagram in Figure 5-2 can be reduced to the following form, shown in 5-3. Gv is a typical low pass filter describing the servo valve dynamics (eq. 4-23). FigureThe block-diagram 5-3. Gv is a intypical Figure low 5-2 pass can filterbe reduced describing to the the following servo valve form, dynamics shown in (eq.Figure 4- 23).5-3 . Gv is a typical low pass filter describingT theL servo valve dynamics (eq. 4-23). K æ V ö ce ç + TLt ÷ 2ç1 s÷ D 4be K ce Km èæ V øö ce ç + t ÷ . 2ç1 - s÷ uc iv Kqi Dm è 4be K ce ø 1 qm 1 qm + Ksa Gv s2 2d . Dm + - + h s + 1 s uc - iv Kqi 2 w1 qm 1 qm wh h + Ksa Gv Au(s) s2 2d Dm + + h s + 1 s - 2 w wh h K Au(s) f Figure 5-3: Reduced block diagram of a positionKf servo with a valve controlled motor

2 If the termFigureFigure Bm 5 5-3:K-3ce: Reduced Reduced/ Dm is block blocksmaller diagram diagram than ofof unit aa positionposition the hydraulic servoservo withwith aresonancea valvevalve controlledcontrolled frequency motormotor h and the hydraulic damping 2h shown in the block-diagram (Figure 5-3) will be expressed as If the term B K / D2 is smaller than unit the hydraulic resonance frequency h and If the term BmmK cece/ Dmm is smaller than unit the hydraulic resonance frequency ωh and the hydraulic damping h2 shown in the block-diagram (Figure 5-3) will be2 expressed as the hydraulic damping eδDh mshown 1 in1 the block-diagram (Figure 5-3) will4 e Dbem expressed as h V1 = V2 = Vt/2 gives h 2 2 J t 2 V1 V2 J tVt 2 e Dm 1 1 4 e Dm β e Dm ⎛ 1 1 ⎞ 4β e Dm h V1 = V2 = Vt/2 gives h ω h = ⎜ + ⎟ V1 = V2 = Vt/2 gives ω h = JJ t ⎜VV1 VV2 ⎟K J B V JJVtVt t and⎝ 1 2 ⎠ ce e t m t t t h D V 4D J Km tJ Bm eV t and Kcece β eeJ tt Bmm Vtt 2and δ hh = + If the term B K / D is includedD them resonanceVt 4 Dfrequencym e J t and damping becomes m ce m Dm Vt 4Dm β e J t

22 If the term B K /' D is includedB K the resonance frequency and damping becomes If the term BmmK cece/ Dmm is includedm thece resonance 'frequencyh and damping becomes h h 1 2 and h Dm Bm K ce ' Bm K ce ' h ' Bm Kce ' 1 δ h 2 h h 1 2 and and h ω h = ω h 1+ 2 δ h = Dm Dm B K Dm BmmK cece 1 2 1+ 2 Dm Dm

53 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control K-E Rydberg Hydraulic servo systems 41 K-E______Rydberg Hydraulic servo systems 41 ______

In order to study the stability of the servo system the open loop gain Au(s) must be analysed.In order toFigure study 5 -the3 yields stability of the servo system the open loop gain Au(s) must be Inanalysed. order to Figure study 5-3 the yields stability of the servo system the open loop gain Au(s) must be analysed. Figure 5-3 yields K sa K qi K f / Dm K A (s) = K K K / D = v (5-6) u sa qi 2 f m K v2 Au (s) ⎛ s K⎞ sa K⎛qi Ks f / D2mδ ⎞ ⎛ s ⎞ ⎛Ks 2δ ⎞ (5-6) A (s) ⎜1+ ⎟⋅ s ⋅⎜ 2 + 2 h s +1⎟ ⎜1+ ⎟ ⋅ s ⋅⎜ v2 + 2 h s +1⎟ (5-6) u ⎜ s ⎟ ⎜ s22 h ⎟ ⎜ s ⎟ ⎜ s22 h ⎟ 1 ωsv s ωs h 2ωh s 1 1 ωsv s ωs h 2ωh s 1 ⎝1 ⎠ s ⎝ 2 h s 1⎠ ⎝1 ⎠ s ⎝ 2 h s 1⎠ v 2h h v 2h h v h h v h h Kv expresses the steady state loop gain and the value of this parameter must be set to a certainKv expresses level tothe make steady sure state that loop the control gain and system the value will beof stable.this parameter must be set to a Kcertainv expresses level tothe make steady sure state that loop the controlgain and system the value will beof stable.this parameter must be set to a certain level to make sure that the control system will be stable.

5.2 Valve controlled motor for an angular velocity servo 5.2 Valve controlled motor for an angular velocity servo 5.2A valve Valve controlled controlled, motor motor is often for used an for angular velocity velocity (shaft speed) servo control. If an A valve controlled motor is often used for velocity (shaft speed) control. If an Aintegrating valve controlled amplifier is motor used isin a often velocity used servo for the velocity loop gain (shaft A u(s) speed) will be control. in principle If an integrating amplifier is used in a velocity servo the loop gain Au(s) will be in principle integratingthe same same as as amplifier for for a a position position is used servo servoin a velocity with with prop proportional servoortional the loop control. control. gain A Such Suchu(s) awill a velocity velocity be in principle servo is theshown same in Figure as for a55-4 -4 position. servo with proportional control. Such a velocity servo is shown in Figure 5-4. Kf K - f - Ksa . Uc Ksa q. U + s V m c s 1 qm + V1 ps ps pL Jt TL pL Jt TL

V2 V2 Figure 5-4: Angular velocity servo FigureFigure 5-4:5-4: AngularAngular velocityvelocity servoservo A block diagram of the velocity servo is shown in Figure 5-5. An integrating amplifier Ameans block that diagram the control of the error velocity will servobe integrated is shown and in theFigure steady 5-5.5- 5.state An control integrating error amplifierbecomes meanszero. that that the the control control error error will will be integrated be integrated and the and steady the state steady control state errorcontrol becomes error zero.becomes zero. TL TL Kce K Gl Amplifier with ce2 Dm2 Gl Amplifierintegral gain with Threshold Saturation Dm integral gain Threshold Saturation - . i uc 1 ir i max Kqi - q.m imax G uc+ Ksa 1 ir i Kqi v + Gh qm K s Dm G G + - sa s ein D v + h - ein m Au(s) Au(s) Kf K f Figure 5-5: Angular velocity servo FigureFigure 5-5:5-5: AngularAngular velocityvelocity servoservo The transfer functions in the above block-diagram are: The transfer functions in the above block-diagramblock-diagram are:

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control

K-E______Rydberg Hydraulic servo systems 42 ______

1 Vt 1 Gv (s) = , G1 (s) = 1+ s , Gh (s) = 2 s 4β K s 2δ h 1+1 Vet ce 1 G (s) , G (s) 1 s , G (s) 2 + s +1 v ω 1 h 2 sv 4 K ωs h 2ωh 1 e ce h s 1 2 In order to study thev stability of the servo system the open loop h gainh Au(s) must be

Inanalysed. order to Figure study 5 the-5 yields stability the ofsame the openservo loop system gain theas foropen the loopposition gain servo, Au(s) shown must bein analysed.equation (5Figure-6). Therefore,5-5 yields the same design open criteria loop forgain stability as for the will position be the servo, same shown as for in a equationposition servo (5-6). with Therefore, proportional the designcontrol. criteria for stability will be the same as for a position servo with proportional control. By using an integrating servo amplifier the steady state stiffness will be considerably By using an integrating servo amplifier the steady state stiffness will be considerably increased. The integration of the velocity error will cancel out the disturbance. The increased. The integration of the velocity error will cancel out the disturbance.. The . stiffness of the closed loop system describes the controlled signal deflection Δθ m due stiffness of the closed loop system describes the controlled signal deflection m due to variations in the disturbance torque ΔTL. By setting Uc = 0 in the block-diagram in to variations in the disturbance torque TL. By setting Uc = 0 in the block-diagram in Figure 5-5 5-5 and and if threshold if threshold and andsaturation saturation are neglected are neglected the new the block-diagram new block -becomesdiagram asbecomes in Figure as in 5-6 Figure. 5-6. . æ ö 1 DTL Kce Vt Dqm ç1+ s÷ s2 2 2ç 4b K ÷ - + dh s + 1 Dm è e ce ø - 2 w wh h

K qi 1 1 s K K Dm 1+ sa s f wv

FigureFigure 5-6:5-6: Block diagram for stiffness calculation ofof aa velocityvelocity servoservo withwith integratingintegrating controllercontroller

TL − ΔTL TheThe stiffness ofof thethe closedclosed looploop servoservo isis defined defined as as Sc . .. NeglectingNeglecting the valve Sc = . m Δθ m dynamics (G = 1) the block diagram in Figure 5-6 gives the closed loop stiffness as dynamics (Gv = 1) the block diagram in Figure 5-6 gives the closed loop stiffness as

3 2 3 2 2 s 2 h 2 s s s 2 h s 2δ h 2 s ⎛ s 1⎞⎛ s δ h s 1⎞ 2 2 s 1 2 ⎜ 1⎟ ⎜ 2 s 1⎟ 2 2 + s + +1 2 + ⋅ 2 + + Dm K v h K v h K v Dm ⎜ K v ⎟ ⎜ h h ⎟ Dm K vω h K vω h K v Dm K v ω h ω h (5-7) Sc K v K v ⎝ ⎠ ⎝ ⎠ (5-7) Sc = K v ⋅ ≈ K v ⋅ K ce Vt K ce s K ce ⎛ Vt ⎞ K ce ⎛ s ⎞ s ⋅⎜1+ s⎟ s ⋅⎜1+ ⎟ ⎜ 4 e K ce ⎟ ⎜ 2 h h ⎟ ⎝ 4β e K ce ⎠ ⎝ 2δ hω h ⎠ where the steady state loop gain K = K K K /D . where the steady state loop gain Kvv = KsasaKqiKff/Dm. T The steady state stiffness is defined as S − ΔTLL . Equation (5-7) gives The steady state stiffness is defined as Sc s0 . . Equation (5-7) gives c s→0 = . m s0 Δθ m s→0 2 2 Dm 1 S K Dm 1 (5-8) Sc s0 = Kv ⋅ = ∞ (5-8) c s→0 v K s Kce s s0 ce s→0 From equation (5-8) it can be noted that the steady state stiffness goes to infinity because of the integrating controller. It has to be observed that this is true only at very low frequencies (steady state conditions). In practical applications the steady state stiffness also will be limited by the resolution of the velocity transducer. 55

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

From equation (5-8) it can be noted that the steady state stiffness goes to infinity because of the integrating controller. It has to be observed that this is true only at very low frequencies (steady state conditions). In practical applications the steady state K-Estiffness Rydberg also will be limited by the Hydraulic resolution servo ofsystems the velocity transducer. 43 ______

5.3 Pump controlled motor 5.2Pump Pump controlled controlled motors are motorthe preferred power element in applications which require Pumpconsiderable controlled horsepower motors are forthe preferred control purposes.power element This in type applications of closed which hydrostatic require considerabletransmission horsepower gives much for higher control efficiency purposes. compared This to type a valve of closed controlled hydrostatic actuator, transmissionsince there are gives no muchflow orif higherices in efficiency the main compared circuit. However, to a valve the controlled comparatively actuator, slow since there are no flow orifices in the main circuit. However, the comparatively slow response of the pump displacement controller, limit their use in high performance response of the pump displacement controller, limit their use in high performance systems.systems. In In Figure Figure 5-7 5- 7a apump pump controlled controlled motor motor used used as as an an angular angular position position servo servo is is shown.shown.

p1 ep V1 wp Dm Jt TL Dp

qm uc + p V i 2 2 Angular - Ksa transducer uf pr Kf

FigureFigure 5-7: 5-7 Pump: Pump controlled controlled motor motor used used as as an an angular angular position position servo servo

As shown in the figure the pump displacement setting (p) is controlled by a position As shown in the figure the pump displacement setting (εp) is controlled by a position servo. During normal operation the pressure in one line between pump and motor will servo. During normal operation the pressure in one line between pump and motor will be at replenishing pressure (pr) and the other pressure will modulate to match the load. Thebe attwo replenishing lines will switch pressure functions (pr) and if the the other load dictatespressure a will pressure modulate reversal. to match It is possiblethe load. forThe both two line lines pressures will switch to vary functions simultaneously if the load if dic transientstates a pressure are rapid reversal. and load It isreversals possible occur.for both However, line pressures for system to vary modelling simultaneously it is assumed if transients that only are one rapid pressure and loadvaries reversals at the sameoccur. time However, and that for both system sides modellingare identical. it is assumed that only one pressure varies at the

Ifsame it is timeassumed and thatthat bothp1 in sides Figure are 5-7 identical. denotes the high pressure the continuity equation for the volume V1 (constant volume) is If it is assumed that p1 in Figure 5-7 denotes the high pressure the continuity equation

for the volume V1 (constant volume) is V1 p D p p Ctp P1 Ctm P1 Dm s m sP1 (5-9) e V1 Δε p Dpω p − CtpΔP1 − CtmΔP1 − Dm sΔθ m = sΔP1 (5-9) Introducing the total leakage coefficient for pump and motor, Cβt e Ctp Ctm gives Introducing the total leakage coefficient for pump and motor, C = C + C gives V1 t tp tm p D p p Ct P1 Dm s m sP1 (5-10) e V1 Δε p Dpω p − Ct ΔP1 − Dm sΔθ m = sΔP1 (5-10) Assuming lumped constants to describe the load, Newton'sβ esecond law is used to obtain the torque balance equation for the motor shaft. If, the friction torque is described only by a viscous friction coefficient, Bm the torque equation is given as

2 Dm P1 Pr Pr cons tan t Dm P1 J t s m Bm s m TL (5-11) 56 The transfer function for the pump displacement controller from input current (iv) to displacement setting (p) is

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Assuming lumped parmeters to describe the load, Newton's second law is used to obtain the torque balance equation for the motor shaft. If, the friction torque is described only by a viscous friction coefficient, Bm the torque equation is given as

2 Dm (ΔP1 − ΔPr ) = Pr = cons tant = Dm ΔP1 = J t s Δθ m + Bm sΔθ m + ΔTL (5-11) K-E Rydberg [ Hydraulic] servo systems 44

The______transfer function for the pump displacement controller from input current (iv) to displacement setting (εp) is

Δε p 1 p 1 = K psG ps = K ps (5-12) K psG ps K ps s (5-12) Δiips s ps 11+ ω ps ps wherewhere KKpsps isis thethe proportionalproportional gaingain forfor thethe pumppump servoservo andand GGpsps isis aa lowlow-pass-pass filterfilter describingdescribing thethe dynamicsdynamics ofof thethe pumppump servo.servo. CombiningCombining equationequation (5(5-10)-10) toto (5(5-12)-12) givesgives thethe blockblock diagramdiagram shownshown inin FigureFigure 55-8-8..

TL C æ V ö t ç + 1 ÷ 2 ç1 s÷ Dm è be C t - . uc ips ep Dp wp 1 qm 1 qm + Ksa KpsGps s2 2d Dm + + h s + 1 s - 2 w wh h Au(s)

FigureFigure 5 5-8:-8: Block Block diagram diagram of of a a pump pump controlled controlled motor motor used used as as an an angular angular positionposition servoservo 2 2 If the term BmCt / Dm is smaller than unit the hydraulic resonance frequency h and the If the term BmCt / Dm is smaller than unit the hydraulic resonance frequency ωh and the hydraulic damping h shown in the block-diagram (Figure 5-8) will be expressed as hydraulic damping δh shown in the block-diagram (Figure 5-8) will be expressed as 2 2 D2 1 D2 eD m ⎛ 1 ⎞ eD m h β e m . If V1 = V2 = V0 , h β e m ω h = ⎜ ⎟. If V1 = V2 = V0 , ω = J t ⎜V1⎟ h J tV0 J t ⎝V1 ⎠ J tV0 C J B V C t eJ t B m V 0 and h t β e t m 0 and δ h = + 2Dm V0 2Dm e J t 2Dm V0 2Dm β e J t 2 If the term BmCt / D2m is included the resonance frequency and damping becomes If the term BmCt / Dm is included the resonance frequency and damping becomes

' BmCt ' h B C ' δ 'h h 1 m 2 t and h h ω h = ω h 1+ and δ h = D2m BmCt Dm 1BmCt 1+ 2 D2m Dm

In order to study the stability of the servo system the open loop gain Au(s) must be Inanalysed. order to Figure study 5-8 the yields stability of the servo system the open loop gain Au(s) must be analysed. Figure 5-8 yields / K sa K ps D p p K f Dm K v Au (s) (5-13) s s 2 2 s s 2 2 1 s h s 1 1 s h s 1 2 2 ps h h 57 ps h h

Kv expresses the steady state loop gain and the value of this parameter must be set to a certain level to make sure that the control system will be stable. In this system it will be noticed that both the hydraulic resonance frequency and its damping are reduced because of the fact that the pressure is varying in only one volume between pump and motor.

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

K sa K ps Dpω p K f / Dm Kv Au (s) = = (5-13) ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎛ s ⎞ ⎛ s 2 2δ ⎞ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜1+ ⎟ ⋅ s ⋅⎜ + h s +1⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎝ ω ps ⎠ ⎝ωh ωh ⎠ ⎝ ω ps ⎠ ⎝ωh ωh ⎠

5.4 Pump controlled motor with variable displacement If the motor in Figure 5-7 is replaced with a unit that has variable displacement, the output speed range can be gained without oversizing of the pump. Therefore, the system concept shown in Figure 5-9 is commonly used as hydrostatic drives in heavy vehicle applications. In the figure, the closed hydrostatic transmission is equipped with a closed loop speed control unit.

Figure 5-9: Pump controlled variable motor used as an angular velocity servo

Changing the fixed displacement motor to a variable unit means that max motor . displacement Dm has to be replaced with the factor εm Dm, where εm is the motor displacement setting. The motor setting can varies in the range, 0 < εm <= 1,0. . Implementation of the partial displacement (εm Dm) of the variable motor in equation (5- 9) to (5-11) gives the block diagram illustrated in Figure 5-10.

58 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 5-10: Block diagram of a pump controlled variable motor used as an angular velocity servo

2 2 If the term BmCt / (εmDm ) is smaller than unit the hydraulic resonance frequency ωh and the hydraulic damping δh shown in the block-diagram (Figure 5-10) is expressed as

2 2 ! $ 2 2 βeεmDm 1 βeεmDm ωh = # & . If V1 = V2 = V0 , ωh = Jt "V1 % JtV0

Ct βeJt Bm V0 and δh = + 2εmDm V0 2εmDm βeJt

According to Figure 5-10 and equation (5-12) yields the open loop gain Au(s) for the angular velocity servo as,

KsaK psDpω pK f / (εmDm ) Kv Au (s) = = (5-14) ! s $ ! s2 2δ $ ! s $ ! s2 2δ $ #1 & s h s 1 #1 & s h s 1 # + &⋅ ⋅# 2 + + & # + &⋅ ⋅# 2 + + & " ω ps % "ωh ωh % " ω ps % "ωh ωh %

It is interesting to note that the motor displacement setting (εm) has a great impact on both the loop gain and system dynamics. In order to keep the amplitude margin of the control loop constant when the motor setting is changed the servo amplifier gain Ksa has to be multiplied with the actual motor setting, εm.

5.5 Pump controlled symmetric cylinder It is possible to replace the motor in Figure 5-7 with a symmetric cylinder, without any changes in the rest of the system. In that case we will have a closed hydrostatic transmission with a linear actuator, which is shown in Figure 5-11.

59 K-E Rydberg Hydraulic servo systems 45 ______

5.3 Pump controlled symmetric cylinder K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______It is possible to replace the motor in Figure 5-8 with a symmetric cylinder, without any changes in the rest of the system. In that case we will have a closed hydrostatic transmission with a linear actuator, which is shown in Figure 5-9.

xc xp xL

Kc Ka FL Mt

Position transducer

Kf Pm

Dp xpr uc + - Ksa uf

FigureFigure 55-9:-11 :Pump Pump controlled controlled cylinder cylinder used used as as a a linear linear position position servo servo Figure 5-9 also illustrates the valves needed to control maximum level of the high Figurepressure 5 -(pressure11 also illustratesrelief valves) the valvesand to maintain needed to constant control low maximum pressure level (cooling of the valve). high pressureThe dynamics (pressure of this relief system valves) will and be similarto maintain to that constant described low in pressure Figure 5-8. (cooling valve). The dynamics of this system will be similar to that described in Figure 5-8.

60 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

6 Hydraulic systems with complex load dynamics The design of hydraulic systems is often based on a simple load model, represented by single lumped parameters. This type of load model can be used if the connection between the hydraulic system and the mechanical load is stiff. However, in many applications, the mechanical system, which the hydraulic power elements are connected to becomes weak compared to the stiffness of the hydraulic system. Such a weak mechanical structures cause resonances, which can be lower than the hydraulic natural frequency. If the structural resonances dominate the frequency response of the servo system, it is extremely important to take this fact into account in system design. The main reasons are that the stability of the systems and the bandwidths are limited by the lowest natural frequency in the control loop.

A simple valve-controlled hydraulic cylinder with a load represented by a mass Mt and an arbitrary load force FL is shown in Figure 6-1. The four-way valve is assumed to be a servo-valve with constant flow gain Kq.

Figure 6-1: Valve-controlled hydraulic cylinder with a mass load

For the special case of centred piston, the oil volumes between the piston and the valve are V1 = V2 = Vt/2. If then the load pressure is defined as pL = p1 - p2 the following linearized and laplace transformed equations can be derived:

⎛ Vt ⎞ KqXv = ApsXp + ⎜Kce + s⎟PL (6-1) ⎝ 4βe ⎠

61 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Linköpings universitet 2 2008-10-13 2 Mts Xp = ApPL − FL (6-2) LinköpingsIEI / FluMeS, universitet K-E Rydberg 2008-10-13 2 IEI / FluMeS, K-E Rydberg

TheseThese equations equations resulted resulted in inth ethe block block diagram diagram in inFigure Figure 2. 6-2. These equations resulted in the block diagram in Figure 2. Hydraulic system Mechanical system - Load Hydraulic system Mechanical system - Load K ce FL K - ce -FL . X Xv - _____ 1 PL - ____ 1 . p _1 Xp Kq A X 1 P p Xp 1 Xp v + ______Vt L + ____M 1t s _s Kq - s Ap + ___4Vt + Mt s s - βe s 4βe QL Q Ap L A p Figure 2: Block diagram of a valve-controlled hydraulic cylinder with a mass load FigureFigure 6 -2:2: Block Block diagram diagram of of a a valve-controlled valve-controlled hydraulic hydraulic cylinder cylinder with with a a mass mass load load The question is now to define a load model, which is simple to connect to the hydraulic The question is now to define a load model, which is simple to connect to the hydraulic system.The question As can is be now seen to fromdefine Figure a load 2 model, the tran whichsfer functionis simple of to theconnect mechanical to the hydraulic load can be system. As can be seen from Figure 2 the transfer function of the mechanical load can be expressedsystem. asAs can be seen from Figure 6-2 the transfer function of the mechanical load can expressed as be expressed as, QL G s ; Q A sX (3) m Q L p QLp PL Gms ; GQLm (s )A =p sXp ; QL = ApsXp (3)(6- 3) PL PL

1. Loads with one degree of freedom 1. 6.1 LoadsLoads with with one one degree degree of freedom of freedom

AnA hydraulic hydraulic cylinder cylinder acting, acting on on a simplea simple mass mass load load is illustratedis illustrated in inFigure Figure 3. This 6-3. isThis a load is a An hydraulic cylinder acting on a simple mass load is illustrated in Figure 3. This is a load withload one with degree one ofdegree freedom. of freedom. with one degree of freedom.

Ap Ap A A p p Mt FL M p1 C1 C2 p2 t FL p p 1 xCp 1 C2 2

FigureFigure 6 -3:3: SymmetricSymmetric hydraulic hydraulic cylinder cylinder with with a a mass mass load load Figure 3: Symmetric hydraulic cylinder with a mass load

For a double acting symmetric cylinder loaded by a mass Mt and an external force FL For a double acting symmetric cylinder loaded by a mass Mt and an external force FL For a double acting symmetric cylinder loaded by a mass Mt and an external force FL according to Figure 3, the piston position Xp is given by equation (2) as according to Figure 6-3, the piston position Xp is given by equation (6-2) as according to Figure 3, the piston position Xp is given by equation (2) as 2 2 A P FApPL − FL A s A1p s 1 p XL = L ; F = 0 ⇒ G 2p(s) = = (4)(6-4) Xp A P p F ; 2 FL 0 L Gm s Am s 1 2 M p L 2 L M s p 2 Mts t Xp Mts ; t FL 0 Gms Mts t s s (4) M s2 M s2 M2t 2 t t A s Ap 2p Ap

62 Linköpings universitet 2008-10-13 3 IEI / FluMeS, K-E Rydberg K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______The load dynamics, from force to piston velocity, is in this case represented by a pure integrator. Introducing the total capacitance of the cylinder CL the transfer function Gm(s) can be rewrittenThe load asdynamics from force to piston velocity is in this case, represented by a simple integrator. Introducing the total capacitance of the cylinder CL the transfer function QL CL CL Gm(s) can be rewritten as Gms (5) PL CLMt s s Q C C 2 2 L L L A Gm(s) h= = = (6-5) p PL CLMt s s 2 2 Ap ωh The capacitance of the cylinder CL is definded as The capacitance of the cylinder CL is defined as 1 1 1 V1 V2 ; C och C (6) C C C 1 1 1 1 2V1 V2 L 1 2 = +e ; C1 = aned C2 = (6 -6) CL C1 C2 βe βe

Centered cylinder piston gives the Capacitance CL and the hydraulic natural frequency h as Centred cylinder piston gives the Capacitance CL and the hydraulic natural frequency

ωh as 2 Vt Vt 4eAp V V C 2 (7) 1 2 2 V L 4 V h V M 4β A t e t t t e p V1 = V2 = ⇒ CL = ⇒ ωh = (6-7) 2 4βe VtMt Figure 4 shows a more general load situation where the mass load is completed with a spring Figure 6-4 shows a more general load situation where the mass load is completed with gradient KL and a viscous damping coefficient BL. The cylinder is also defective with viscous a spring gradient KL and a viscous damping coefficient BL. The cylinder is also friction, the coefficient B . defective with viscousp friction, the coefficient B . p KL Ap Ap Mt FL p1 C1 C2 p2 xp Bp BL

FigureFigure 6 4:-4 Symmetric: Symmetric hydraulic hydraulic cylinder cylinder with with a mass,a mass, spring spring and and damping damping load load

The load transfer function will be expressed in the same way as in equation (6-5), The load transfer function will be expressed in the same way as in equation (5) which gives which gives 2 2 Ap Ap s s Q Q C sC s K K L L L L L L Gm Gsm (s ) = = = (8)(6-8) PL P CLMCt M CLBCe B CLKCL K Mt M Be B L L s2t 2 L s e L L s2t 2 s e 1 2 s + 2 s + 2 K s K + s + 1 A 2 A 2 A 2 L KL L KL p Ap p Ap p Ap

where the total viscous friction coefficient is Be = BL + Bp. where the total viscous friction coefficient is Be = BL + Bp.

63 Linköpings universitet 2008-10-13 5 IEI / FluMeS, K-E Rydberg

Kce - . 1 X X Xv _____ PL 1 2 1 1 p _1 p Kq Ap M s 1 + ___Vt M s t K K s - s t 1 L 4e Q L A p Figure 7: Block diagram for a valve controlled cylinder with elastic mountings

Reduction of the block diagram in Figure 7 and completing with the transfer function from Xp to XL gives the following diagram.

1 1 M s2 1 . 1 t X Xv K q K1 K L p _1 Xp XL 2 ' 1 1 A s s M s2 1 p 2 h 1 t K K '2 ' 1 L h h Figure 8: Complete block diagram for a valve controlled cylinder with elastic mountings

' ' K e 1 V t 1 1 ' K ceM t h where h ; 2 and h 2 M t Ke 4e A p K 1 K L A p 2

If can be noted that the effective spring gradient Ke is derived from the series connection K-E Rydberg Hydraulic Servo Systems –2 Dynamic Properties and Control 4eA p between______the hydraulic spring gradient and the two mechanical springs K1, KL. Vt

2. 6.2 LoadsLoads with with two two degrees degrees of freedom of freedom a) Mechanical flexibility between two masses on a piston rod a) Mechanical flexibility between two masses on a piston rod

AssumeAssume that that the the load load consists consists of of two two masses masses (M (M1 and1 and M 2 M), 2 of), of which which the the first first one one is fixed is mountedfixed mounted to the piston to the and piston. the secondThe second mass masses, is connected is connect by aed spring to the (K firstL) and mass a by viscous a damperspring (B (KL)L to) and the afirst viscous mass. damper (BL).

KL Ap Ap M1 M2 FL p1 C1 C2 p2 BL xp x L

FigureFigure 6 9:-5 :Hydraulic Hydraulic cylinder cylinder with with a a load load with with two two degrees degrees of of freedom freedom

The force balance equations for the masses M1 and M2 are for FL = 0 2 2 M1s Xp + M2s XL = ApPL (6-9) 2 M2s XL = −BLs (XL − Xp) − KL(XL − Xp) (6-10)

The piston position Xp can be expressed as

⎛ M2 2 BL ⎞ ⎜ s + s + 1⎟ ApPL ⎝ KL KL ⎠ X = (6-11) p M M B ( ) 2 ⎢⎡ 1 2 2 L ⎥⎤ M1 + M2 s ⎢ s + s + 1⎥ ⎣ KL(M1 + M2) KL ⎦

With the load flow QL = ApsXp we obtain the transfer function

⎛ M2 2 BL ⎞ CL ⎜ s + s + 1⎟ QL ⎝ KL KL ⎠ Gm(s) = = (6-12) PL CL ⎡ M1M2 2 BL ⎤ (M1 + M2) s ⎢ s + s + 1⎥ 2 ⎢ K (M + M ) KL ⎥ Ap ⎣ L 1 2 ⎦

Equation (6-12) can also be written in the following form

64 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

2 ⎛ s 2δa ⎞ ⎜ + s + 1⎟ 2 2 ωa Ap ⎝ ωa ⎠ G (s) = (6-13) m ( ) 2 M1 + M2 s ⎛ s 2δ1 ⎞ ⎜ + s + 1⎟ 2 ω1 ⎝ ω1 ⎠

KL KL(M1 + M2) M2 ωa = , ω1 = = ωa 1 + M2 M1M2 M1 where BL 1 BL M1 + M2 M2 δa = , δ1 = = δa 1 + 2 KLM2 2 KLM1M2 M1

From equation (6-13) it can be seen that the two, coupled masses cause one natural frequency of the fundamental mode ω1 and one natural frequency of nodes of vibration

ωa. For these frequencies, it is always evident that ωa < ω1 and the damping ratio δa < δ1. A Bode diagram for equation (6-13) is shown in Figure 6-6.

101

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

Phase -50

-100 100 101 102 Frequency [rad/s]

Figure 6-6: Frequency response according to equation (6-13). Solid line is valid for M1 = M2 and dashed line for M1 = M2/2

Figure 6-6 explains how the amplitude and the phase curve change when the mass M1 is reduced in proportion to M2. If M1 instead increases in proportion to M2 the frequency response will be influenced in the way as expressed in Figure 6-7.

65 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

Phase -50

-100 100 101 102 Linköpings universitet 2008-10-13 8 Frequency [rad/s] IEI / FluMeS, K-E Rydberg Figure 6-7: Frequency response according to equation (16). Solid line is valid for M1 = M2 and dashed line for M1 = 2M2 It isIt convenient is convenient to combineto combine equati equationon (16) (6 with-13) thewith hydraulic the hydraulic part of part the of system. the system. If the Ifexternal the forceexternal FL = 0,force the FvalveL = 0,cylinder the valve combination cylinder combination (compare with (compare Figure with1) and Figure the load 1) and with the two massesload givewith thetwo block masses diagra givem the illustrated block diagram in Figure illustrated 12. in Figure 6-8.

Ap . X Kq ______P Xp v ___ L A ______1 Vt p GLX(s) Ap + K + ___ s (M + M ) s - ce 1 2 4βe

22 ss 22 δaa + ss + 11 22 ωaa ωaa G (ss) = where LXLX 22 ss 22 δ11 + ss + 11 22 ω11 ω11

Figure 6-8: Block diagram for a valve-controlled cylinder with a load of two masses (FL = 0) Figure 12: Block diagram for a valve-controlled cylinder with a load of two masses (FL = 0)

From Figure 12 the transfer function for the hole system, with valve opening Xv as input From Figure 6-8 the transfer function for the hole system, with valve opening Xv as signal and piston velocity dX /dt as output signal, can be derived as input signal and piston velocityp dX /dt as output signal, can be derived as p sXp Kq GLXs G s HL X A 2 (17) v p s 2 h s GLXs 2 h h

2 66 4eAp Kce eM1 M2 where ; (18) h V M M h A V t 1 2 p t

The frequency response for equation (17), when the hydraulic natural frequency h is lower than the structural resonance[s (a and 1) is shown in Figure 13 (see next page). The dashed lines in the diagrams illustrated the response when the mass M2 is increased 5 times compared 2 to the situation of the solid lines. The hydraulic spring rate Kh (=4eAp /Vt) has the same value for both cases.

In this system, the frequency response will be dominated by the hydraulic system. In the low frequency range is GLX(s) 1 and the system can be treated as a one-mass system with the load mass Mt = M1 + M2.

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

sXp Kq GLX(s) G (s) = = HL X A 2 (6-14) v p s 2 δh + s + GLX(s) 2 ωh ωh 2 4βeAp Kce βe(M1 + M2) where ω = ; δ = (6-15) h ( ) h A V Vt M1 + M2 p t

The frequency response for equation (6-14), when the hydraulic natural frequency ωh is

lower than the structural resonance’s (ωa and ω1) is shown in Figure 6-9. The dashed

lines in the diagrams illustrated the response when the mass M2 is increased 5 times 2 compared to the situation of the solid lines. The hydraulic spring rate Kh (=4βeAp /Vt) has the same value for both cases. In this system, the frequency response will be dominated by the hydraulic system. In Linköpings universitet 2008-10-13 9 IEI /the FluMeS, low K-E frequency Rydberg range is GLX(s) ≈ 1 and the system can be treated as a one-mass system with the load mass Mt = M1 + M2. 101 100

10-1

Amplitude 10-2 ωh ωa ω1 10-3 100 101 102 Frequency [rad/s]

-50 -100 Phase -150 -200 100 101 102 Frequency [rad/s]

FigureFigure 6 13:-9: Frequency response for GHL(s)(s) accordingaccording toto equatioequationn (17).(6-14 Solid). Solid line: line: M 1M =1 M= M2 =2 1.= 1. . DashedDashed line:line: MM11 == 11 andand MM22 =5.=5. KKhh == 0,10,1.KKLL forfor bothboth curvescurves

If weIf we now now consider conside thatr that the the structuralstructural resonanceTsresonances are are lower lower than than the the hydraulic hydraulic natural natural frequency,frequency, the thesystem system response response will willchange change drastically, drastically which, which can becan seen be seenin Figure in Figure 14. 6-

10. 101

100 67

Amplitude ωa ω1 ω'h 10-1 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

Figure 14: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = M0. Dashed line: M1 = M0 and M2 =5M0. Kh = 10.KL for both curves Linköpings universitet 2008-10-13 9 IEI / FluMeS, K-E Rydberg

101 100

10-1

Amplitude 10-2 ωh ωa ω1 10-3 100 101 102 Frequency [rad/s]

-50 -100 Phase -150 -200 100 101 102 Frequency [rad/s]

Figure 13: Frequency response for GHL(s) according to equation (17). Solid line: M1 = M2 = 1. Dashed line: M1 = 1 and M2 =5. Kh = 0,1.KL for both curves K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control If we______now consider that the structural resonanceTs are lower than the hydraulic natural frequency, the system response will change drastically, which can be seen in Figure 14.

101

100

Amplitude ωa ω1 ω'h 10-1 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

FigureFigure 6 -14:10: FrequencyFrequency responseresponse forfor GGHLHL(s)(s) accordingaccording toto equatioequationn (17).(6-14 Solid). Solid line: line: M 1M =1 M= 2M =2 M= 0M. 0. . Dashed line:line: M11 = M00 andand MM22 =5M0.. KKh = 10.KL for both curves

In Figure 6-10 the hydraulic spring rate Kh is 10 times greater than the mechanical

spring rate KL, which means that ωh will be the highest frequency. If we look at the

amplitude of the transfer function GLX (see Figure 6-8) with frequencies greater than

ω1, this equation will reach a constant value

2 ⎛ ω1 ⎞ M2 ⏐GLX(s)⏐ω>ω = ⎜ ⎟ = 1 + (6-16) 1 ⎝ ωa ⎠ M1

Since the hydraulic resonance, occur at a frequency higher than the structural

resonances, the value of ωh have to be influenced by the function GLX written as

equation (6-16). For this case, the hydraulic frequency and damping, here named ω´h

and δ´h, may be changed from the original expression (equation 6-16) to

2 , M2 Kh 4βeAp ωh = ωh 1 + = = M1 M1 VtM1 (6-17)

, M2 Kce βeM1 δh = δh/ 1 + = M1 Ap Vt

68 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

This equation shows that the dynamics of the hydraulic system only is dependent on the mass M1. Increasing of the mass M2, of course, will lower the mechanical resonances but the hydraulic frequency ω´h is not influenced by change of this mass. The reason is that the movement of the mass M2 is approximately zero at such high frequencies as

ω´h. The hydraulic natural frequency is solely determined by the mass M1, which is stretched between the hydraulic spring in the cylinder Kh and the load spring KL.

However, the low value of KL compared to Kh means that KL can not be seen in equation (6-17).

If the system is changed so that only the hydraulic spring constant Kh (see eq 6-17) increases, the variation in amplitude and phase shift according to the mechanical structure will be reduced. This situation is shown in Figure 6-11. From equation (6-14) it can also be seen that, if the structural resonances (GLX(s)) are dominant, the transfer function from valve opening to piston speed approaches

GHL(s) = sXp/Xv ≈ Kq/Ap (6-18)

101

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

Figure 6-11: Frequency response for GHL(s) according to equation (6-14). Solid line: Kh = Kh0. Dashed line: Kh = 5.Kh0. M1 = M2/2 for both curves

Another situation where ωh increases and makes the hydraulic system stiffer will arise if the mass M1 is reduced. Figure 6-12 shows how the frequency response develops when M1 is reduced from M1 = M10 to M1 = 0,2.M10. Since M1 also influences the mechanical resonances this rise of ωh will not cause a reduction of the structural resonances as in the case of increased hydraulic stiffness.

69 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

101

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

Figure 6-12: Frequency response for GHL(s) according to equation (6-14). Solid line: M1 = M10. Dashed line: M1 = 0.2.M10. Kh and KL are the same for both curves Linköpings universitet 2008-10-13 12 IEI / FluMeS, K-E Rydberg The conclusion to be drawn from these examples is that the structural resonance’s are reduced by a stiff hydraulic system if ωh is always higher than the mechanical Thefrequencies. conclusion Thisto be stiffnessdrawn from can thesebe achieved examples by islow that hydraulic the structural capacitance resonanceKs for the are valve reduced by cylinder a stiff hydrauliccombination system and/or if feedbackh is always control. higher than the mechanical frequencies. This stiffness can be achieved by low hydraulic capacitance for the valve cylinder combination and/orb) Two feedback-mass control.system with mechanical flexibility in the rear mounting end b) ATwo-mass further example system with of a mechanical hydraulic cylinder flexibility loaded in the by rear a two mounting-mass system end is shown in Figure 6-13. A further example of a hydraulic cylinder loaded by a two-mass system is shown in Figure 17.

K1 M1 Ap Ap M2 FL p1 C1 C2 p2 B1 x xp x 1 2

FigureFigure 6 17:-13 :Hydraulic Hydraulic cylinder cylinder with with flexible flexible rear rear flange flange and and a a load load with with two two degrees degrees of of freedom freedom

The transfer function for this mechanical system is The transfer function for this mechanical system is 2 s 2a s 1 2 2 a QL Ap a G s m P M s 2 (22) L 2 s 21 s70 1 2 1 1

K1 K1 M2 where a , 1 a 1 M1 M2 M1 M1

B1 1 B1 1 M2 , 1 a 2 K M M 1 2 K M a M 1 1 2 1 1 1

According to equation (17) a valve controlled cylinder yields

sXp Kq GLXs G s (23) HL X A 2 v p s 2 h s GLXs 2 h h

2 s 2 a s 1 2 a 2 a 4eAp Kce eM2 where G s ; ; LX 2 h V M h A V s 2 1 t 2 p t s 1 2 1 1 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

2 ⎛ s 2δa ⎞ ⎜ + s + 1⎟ 2 2 ωa QL Ap ⎝ ωa ⎠ G (s) = = m P M s 2 (6-19) L 2 ⎛ s 2δ1 ⎞ ⎜ + s + 1⎟ 2 ω1 ⎝ ω1 ⎠

K1 K1 M2 where ωa = , ω1 = = ωa 1 + M1 + M2 M1 M1

B1 1 B1 1 M2 δ = , δ = = δ 1 + a 2 K (M M ) 1 2 K M a M 1 1 + 2 1 1 1

According to equation (6-14) a valve controlled cylinder yields

sXp Kq GLX(s) G (s) = = (6-20) HL X A 2 v p s 2 δh + s + GLX(s) 2 ωh ωh

2 s 2 δa + s + 1 2 ωa 2 ωa 4βeAp Kce βeM2 where G (s) = ; ω = ; δ = LX 2 h V M h A V s 2 δ1 t 2 p t + s + 1 2 ω1 ω1 The Bode diagram for equation (6-20) is shown in Figure 6-14. In this figure the hydraulic resonance ωh is the dominant frequency, lower than ωa. If the hydraulic spring rate Kh is changed, the value of ωh will also be changed which is illustrated in the figure.

71 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

101 Amplitude

10-3 100 101 102 Frequency [rad/s]

-100 Phase

-200 100 101 102 Frequency [rad/s]

Figure 6-14: Frequency response for GHL(s) according to equation (6-20). Solid line: Kh = 0,04 KL. Dashed line: Kh = 0,1 KL. M1 = 1, M2 = 2 and KL = 1000 (same for both curves)

If the hydraulic cylinder is stiffer than the mechanical structure (Kh > KL), the system behaviour will be similar to that described in Figure 6-11. An increase of Kh is shown in Figure 6-15 and it can be seen that the structural resonances are reduced by the large value of ωh.

101

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

Figure 6-15: Frequency response for GHL(s) according to equation (6-20). Solid line: Kh = 4 KL. Dashed line: Kh = 10 KL. M1 = 1, M2 = 2 and KL = 100 (same for both curves)

72 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

With this stiff hydraulic system, the real hydraulic frequency and damping, ω´h and δ´h , are changed from the original expression (compare with eq. 6-17) to,

2 , M2 4βeAp(M1 + M2) ωh = ωh 1 + = M1 VtM1M2 (6-21)

M2 Kce βeM1M2 δ, = δ / 1 + = h h M A ( ) 1 p Vt M1 + M2

Figure 6-16 shows the frequency response when the mass M2 is increased five times.

Note that the reduction of ωh increases the variation in amplitude and phase shift for the mechanical structure, which can also be seen from the expression of ωa and δa (eq. 6- 19).

101

100

10-1 Amplitude

10-2 100 101 102 Frequency [rad/s]

100

Phase -100

-200 100 101 102 Frequency [rad/s]

Figure 6-16: Frequency response for GHL(s) according to equation (6-20). Solid line: M1 = M2 = 1. Dashed line: M1 = 1, M2 = 5. Kh = 4 KL (= 100) for both curves.

73 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control Linköpings universitet 2008-10-13 15 IEILinköpings ______/ FluMeS, universitet K-E Rydberg 2008-10-13 15 IEI / FluMeS, K-E Rydberg c) c)Hydraulic Hydraulic motor motor with with mechanical mechanical flexibility flexibility between between two two inertia inertia loads loads c) Hydraulic motor with mechanical flexibility between two inertia loads

B Bm m KL KL Dm J1 J2 TL Dm J1 J2 TL B θm BL θL θm L θL

Figure 21: Hydraulic motor with a load with two degrees of freedom Figure 6-17: Hydraulic motor with a load with two degrees of freedom Figure 21: Hydraulic motor with a load with two degrees of freedom

ThisThis hydraulic hydraulic motor motor system system with withtwo inertia two inertia loads is loads identical is iden withtical the with cylinder the application cylinder This hydraulic motor system with two inertia loads is identical with the cylinder application illustratedapplica tionin Figure illustrated 9. in Figure 6-5. illustrated in Figure 9.

ConsiderConsid theer the system system shown shown in in Figure Figure 21 6 and-17 andlet thelet motorthe motor be controbe controlledlled by aby 4-port a 4-port servo Consider the system shown in Figure 21 and let the motor be controlled by a 4-port servo servo valve with the coefficients Kq and Kc. The block diagram of the overall system valve with the coefficients Kq and Kc. The block diagram of the overall system will be valve with the coefficients Kq and Kc. The block diagram of the overall system will be developedwill be developedas in Figure as 22. in Figure 6-18. developed as in Figure 22. TL TL GLT(s) GLT(s) Dm . ______- X Kq Dm P - θ.m v ___Kq ______L ______1 ) θ Xv ______Vt PL Dm ______1 GLθ(s m Dm + K + Vt s Dm + (J + J ) s GLθ(s) D + Kce + ___ s + ( 1 2 m - ce 4βe J1 + J2) s - 4βe

FigureFigure 6 22:-18 Block: Block diagram diagram for for a avalve-controlled valve-controlled motor motor with with two two inertia inertia loads loads and and a aload load torque torque (T (TL)L ) Figure 22: Block diagram for a valve-controlled motor with two inertia loads and a load torque (TL)

From Figure 6-18 it can be seen that the mechanical structure has influence on the From Figure 22 it can be seen that the mechanical structure has influence on the torque Fromtorque Figure disturbance 22 it can (TbeL seen) by thatthe transferthe mech functionanical structure GLT(s). hasThe in transferfluence function on the torque GLθ(s) is disturbance (TL) by the transfer function GLT(s). The transfer function GL(s) is similar to disturbancesimilar to (T GLLX) by(s) thein Figure transfer 6- function5. These GtransferLT(s). The functions transfer can function be expressed GL(s) as is similar to GLX(s) in Figure 9. These transfer functions can be expressed as2 GLX(s) in Figure 9. These transfer functions can be expressed as 2δ s s a + s + 1 1 + 2 2 ωa 2δ1ω1 s 2 2aωa G (s) = s ; Gs ( s) =2 a s 1 (25) LT1 s2 Lθ2 2 s 1 1 2s 2δ 2 as 2δ 21 1 a a a 1 (25) GLTs 1 1+ ; s +G 1Ls a + s + 1 G s 2 2 ω a ; G s 2 2 ω1 (25) LT s 2 2ωa a L s 2 21ω1 s 2a s 1 s 21 s 1 2 a s 1 2 1 s 1 a2 a 12 1 a 1 KL KL(J1 + J2) J2 where ω = , ω = = ω 1 + a K J 1 K J JJ J a J J KL 2 KL J1 J12 2 J2 1 where a L , 1 L 1 2 a 1 2 where a J2 , 1 J1J2 a 1 J1 BJL2 1 BL J1J2 J1+ J2 J1 J2 δ = , δ = = δ 1 + a 2 K J 1 2 K J J a J BL 1 L 2 BL J1 J2L 1 2 J2 1 BL 1 , BL J1 J2 1 J2 a 2 K J , 1 2 K J J a 1 J a 2 KL J2 1 2 KL J1 J2 a J1 L 2 L 1 2 1

74 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 1 K-E______Rydberg Feedbacks in Electro-Hydraulic Servo Systems 1

7FEEDBACKS Feedbacks IN ELECTRO- in ElectroHYDRAULIC-Hydraulic SERVO Servo SYSTEMSSystems FEEDBACKS IN ELECTRO-HYDRAULIC SERVO SYSTEMS 7.11. LinearLinear valve valve controlled controlled position position servo servo 1. Linear valve controlled position servo A linearlinear valvevalve controlledcontrolled positionposition servo servo is is shown shown in in Figure Figure 7 -11.. LeakageLeakage flow over the A linear valve controlled position servo is shown in Figure 1. Leakage flow over the piston with the the flow flow-pressure-pressure coefficient coefficient C Cpp andand aa viscousviscous frictionfriction coefficient Bp are pistonincluded with in the the model. flow-pressure The servo coefficient amplifier C(controller)p and a viscous is proportional friction withcoefficient the gain B Kp are. included in the model. The servo amplifier (controller) is proportional with the gain Ksasa. included in the model. The servo amplifier (controller) is proportional with the gain Ksa. Fix reference Fix reference Cp Cp xp be Ap Ap be Bp xp be A A be Bp p p Mt Mt FL P1 V1 V2 P2 FL P1 V1 V2 P2 qL1 qL2 qL1 qL2 x Position xv transducerPosition v transducer Servo amplifierServo uc + amplifier i Kf uc + Kf - Ksa i uf - Ksa Ps = const. uf Ps = const.

Figure 1: Valve controlled position servo FigureFigure 7 1:-1 Valve: Valve controlled controlled position position servo servo The transfer functions (in the frequency domain) of the components in the position The transfer functionsfunctions (in the frequency domain) of the components in the position servo are are illustrated illustrated in in FigureFigure 7 2-2. . Threshold Threshold and and saturation saturation in in the the servo valve are are servoincluded. are illustrated in Figure 2. Threshold and saturation in the servo valve are included. FL FL K æ V ö ce æç + t ö÷ Kce2 ç1 Vt s÷ A ç + 4be K ce ÷ p2 çè1 s÷ø Ap 4be K ce Threshold Saturation è ø Threshold Saturation - . u i i imax Kqi 1 x. p xp c r v i 1 - 1 uc+ Ksa ir iv max Kqi 1 s2 21d xp 1 xp Ap s + h s + - Ksa ein 1+ s22 + 2d s + 1 Ap wsv + w whh s ein 1+ h2 + s + 1 - wv w wh Au(s) h Au(s) Kf K f Figure 2: Block-diagram of a linear position servo including valve dynamics and non-linearityLs FigureFigure 7 2:-2 :Block-diagram Block-diagram of of a alinear linear position position servo servo including including valve valve dynamics dynamics and and non-linearityLs non-linearitie s 1 The transfer function of the valve is G (s) 1 . The hydraulic resonance frequency The transfer function of the valve is v 1 s . TheThe hydraulichydraulic resonance frequency G v (s) = v 1 s 1+ v v ω v 4 A2 K M B V e 2p2 ce e t p t and damping is expressed as: h 4 e Ap and h K M B p V . 4β e Ap Kce βe Mtt Bp Vtt and damping is expressed as: h M V and h A V 4A M . and damping is expressed as: ω h = t t and δ h = p t + p e t . M tVVt Ap Vt 4Ap e M t The parameter values of the system are ast follows:t Ap Vt 4Ap β e M t The parameter .values-3 2 of the system are as follows: 9 The parameterAp = 2,5. 10values-3 m2 of the system are as follows:e = 1,0·109 Pa Ap = 2,5.10-3 m2 e = 1,0·10 9 Pa ABp = 2,50 10 m β Kef == 1,025 ·V/m10 Pa Bp = 0 -11 5 Kf = 25 V/m 3 BKpce = = 0 1,0·10 -11 m5/Ns Kfqi = = 25 0,02V/m m 3/As Kce = 1,0·10 -11 m 5/Ns Kqi = 0,02 m3/As Kcesa = 1,00,1· A/V10 m /Ns K Mqit == 15000,02 mkg/As Ksa = 0,1 A/V-3 3 Mt = 1500 kg Vt = 1,0·10-3 m3 v = 1/v = 0,005 s Vt = 1,0·10 m v = 1/v = 0,005 s 75 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 2 Ksa = 0,1 A/V Mt = 1500 kg -3 3 Vt = 1,0·10 m τv = 1/ωv = 0,005 s

These parameter values gives ωh = 129 rad/s and δhh == 0.1550.155..

The open loop gain (Au(s)) of the positionposition servoservo withwith KKvv == δhhωhh == 2020 1/s1/s (A(Amm == 6 6 dB) dB) is is shownshown in Figure 3.7 -Observe3. Observe that that the thebandwidth bandwidth of theof thevalve valve v =ω 1v =v 1=/τ 200v = 200rad/s rad/s is is higher than the hydraulic resonance frequency ωhh..

Kv = dh = 0,155 wh

Am Amplitude [dB] Phase shift [degrees]

Frequency [w/wh] Figure 3: Bode-diagram of the open loop gain of the position servo depicted in Figure 2 Figure 7-3: Bode-diagram of the open loop gain of the position servo depicted in Figure 7-2 when the servo valve is assumed to be very fast when the servo valve is assumed to be very fast

Influence of valve dynamics 7.1.1 Influence of valve dynamics To really make use of the actuator capability of controlling the load it is very important Tothat really the servo make valve use of is thefast actuator enough. capability Normally of the controlling selected valvethe load will it haveis very a bandwidthimportant that the servo valve is fast enough. Normally the selected valve will have a bandwidth (v) of at least twice as high as the hydraulic resonance frequency (h). Figure 4 shows (theωv )open of at loop least gain twice of asthe high position as the servo hydraulic depict ed resonance in Figure frequency 2, with an (ω hordinary). Figure valve 7-4 shows the open loop gain of the position servo depicted in Figure 7-2, with an ordinary (v=200 rad/s) and a valve with slow response (v = 20 rad/s). valve (ωv=200 rad/s) and a valve with slow response (ωv = 20 rad/s). 2 2 10 10

0 0 10 10 Amplitude Amplitude

−2 −2 10 10

0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s]

−50 −50

−100 −100 −150 −150 −200 −200

Phase Phase −250 −250 −300

−300 −350

−350 −400 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s] a) Normal valve bandwidth, v = 200 rad/s b) Valve with low bandwidth, v = 20 rad/s Figurea) Normal 4: Bode-diagram valve bandwidth, of the openωv = loop200 rad/sgain of a position b) Valve servo with with low a) bandwidth,fast valve and ωv b)= 20slow rad/s valve FromFigure Figure 7-4: Bode 4 -itdiagram can be of recognisedthe open loop thatgain ofthe a positionopen loop servo gain with anda) fast thereby valve and the b) slowamplitude valve margin will be change because of the valve dynamics. For a slow valve (v < h) the open loop gain can be approximated as

K v Au , which gives Kvmax = v for a reasonable stability margin. 1 s /v s 76 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

From Figure 7-4 it can be recognised that the open loop gain and thereby the amplitude margin will be change because of the valve dynamics. For a slow valve (ωv < ωh) the open loop gain can be approximated as

Kv K-E RydbergAu ≈ Feedbacks, which ingives Electro-Hydraulic Kvmax = ωv Servofor a Systemsreasonable stability margin.3 (1+ s /ωv )s

Closed7.1.2 Closed loop stiffnessloop stiffness TheThe most important characteristic of the serservovo system is the closed loop stiffness. The stiffness of the closed loop system dedescribesscribes the controlled signal deflection ΔXpp due to variations in the disturbance force ΔFL.. ByBy settingsetting UUcc = 0 in the block-diagramblock-diagram inin Figure 27 -the2 the new new block-diagram block-diagram becomes becomes as asin inFigure Figure 5. 7-5.

. æ ö 1 DFL Kce Vt xp 1 DXp ç1+ s÷ s2 2 2 ç 4b K ÷ - + dh s + 1 s Ap è e ce ø - 2 w wh h

Saturation Threshold Kqi 1 imax A s Ksa Kf p 1+ w ei v n

FigureFigure 7 5:-5 Block-diagram: Block-diagram describing describing the the s tiffnessstiffness of of a aclosed closed loop loop position position servo servo

−ΔFFLL The stiffnessstiffness ofof thethe closedclosed looploop servoservo isis defined defined as as SScc = .. IfIf the valve dynamics ΔXX pp and the thresholdthreshold are neglected thethe stiffnessstiffness becomesbecomes

s 33 2 s ⎛ ss ⎞ ⎛ ss22 22δ ⎞ s δhh 22 s hh 2 + ss + +11 2 ⎜ +11⎟⋅⎜ + ss+11⎟ A 2 K 22 K K AA2 ⎜ KK ⎟ ⎜ 22 ⎟ pp Kvvωhh Kvvωhh Kvv pp ⎝ vv ⎠ ⎝ωhh ωhh ⎠ Scc = K vv ⋅ ≈ KKvv ⋅ K Vt KK cece 1 t s cece ⎛ ss ⎞ 1+ s ⎜11+ ⎟ 4βe K ce ⎜ 2 ⎟ e ce ⎝ 2δhhωhh ⎠ where the steady statestate looploop gaingain KKvv == KKsasaKqiqiKff/A/Ap.. The closed loop stiffness includingincluding valve dynamics is shown shown in in FigureFigure 7 6-6. . The The amplitude amplitude curve curve is is normalised normalised as as A22 Scc App ,, wherewhere KKss = KKvv ⋅ K ss KKcece

2 2 10 10

1 1 10 10

0 0 10 10 Amplitude, (Sc/Ks) Amplitude, (Sc/Ks)

−1 −1 10 10 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s]

200 200

150 150

100 100

Phase Phase 50

50 0

0 −50 0 1 2 3 0 1 2 3 10 10 10 10 77 10 10 10 10 Frequency [rad/s] Frequency [rad/s] a) Normal valve bandwidth, v = 200 rad/s b) Valve with low bandwidth, v = 20 rad/s Figure 6: Bode-diagram of the closed loop stiffness with a) fast valve and b) slow valve In Figure 6b) it can be seen that the valve dynamics reduce the stiffness just at frequencies around the bandwidth of the valve (v = 20 rad/s). K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 3

Closed loop stiffness The most important characteristic of the servo system is the closed loop stiffness. The stiffness of the closed loop system describes the controlled signal deflection Xp due to variations in the disturbance force FL. By setting Uc = 0 in the block-diagram in Figure 2 the new block-diagram becomes as in Figure 5.

. æ ö 1 DFL Kce Vt xp 1 DXp ç1+ s÷ s2 2 2 ç 4b K ÷ - + dh s + 1 s Ap è e ce ø - 2 w wh h

Saturation Threshold Kqi 1 imax A s Ksa Kf p 1+ w ei v n Figure 5: Block-diagram describing the stiffness of a closed loop position servo

FL The stiffness of the closed loop servo is defined as Sc . If the valve dynamics X p and the threshold are neglected the stiffness becomes

s 3 2 s s s 2 2 h s 2 1 1 h s 1 A2 2 A2 2 p K v h K v h K v p K v h h Sc K v K v K Vt K s ce 1 s ce 1 4 K e ce 2 h h where the steady state loop gain Kv = KsaKqiKf/Ap. The closed loop stiffness including Kvalve-E Rydberg dynamics isHydraulic shown Servo in SystemsFigure – 6Dynamic. The Properties amplitude and curveControl is normalised as 2 ______Sc Ap , where K s K v K s K ce

2 2 10 10

1 1 10 10

0 0 10 10 Amplitude, (Sc/Ks) Amplitude, (Sc/Ks)

−1 −1 10 10 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s]

200 200

150 150

100 100

Phase Phase 50

50 0

0 −50 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s] a) Normal valve bandwidth, = 200 rad/s b) Valve with low bandwidth, = 20 rad/s a) Normal valve bandwidth, ωv = 200 rad/s b) Valve with low bandwidth, ωvv = 20 rad/s FigureFigure 7 6:-6 :Bode-diagram Bode-diagram of of the the closed closed loop loop stiffness stiffness with with a) a) fast fast valve valve and and b) b) slow slow valve valve In Figure Figure 7 6b)-6b) it it can can be be seen seen that that the the valve valve dynamics dynamics reduce reduce the the stiffness just at frequencies around the bandwidth of the valve ( = 20 rad/s). frequenciesK-E Rydberg around the bandwidthFeedbacks inof Electro-Hydraulic the valve (ωv = Servo 20 rad/s). Systems 4

The threshold of the servo valve will also cause a position error ΔXpε. If the threshold is

ε ⋅in ε⋅in the position error is ΔX pε = , where in is nominal valve input current. The threshold of the servo valve willK sa K alsof cause a position error Xp. If the threshold is i i the position error is X n , where i is nominal valve input current. n p n K sa K f 7.2 Valve controlled position servo with load pressure 2. Valvefeedback controlled position servo with load pressure feedback The load pressure feedback is used to increase the hydraulic damping in the system. A The load pressure feedback is used to increase the hydraulic damping in the system. A negative load pressure signal acts in the same way as a Kc-value (flow-pressure negative load pressure signal acts in the same way as a Kc-value (flow-pressure coefficient) ooff the servo valve. Load pressurepressure feedback can be of proportional or dynamic type. Proportional pressure feedback is shown in Figure 7-7. dynamic type. Proportional pressure feedback is shown in Figure 7.

Proportional pressure feedback

Kpf FL - - . uc + i + 1 + 1 xp _1 xp Ksa Gv Kqi Ap Kce + Vt s / 4be s - - PL Mt s

Figure 7: Block-diagram of a linear position servo with proportional pressure feedback (Bp = 0) Figure 7-7: Block-diagram of a linear position servo with proportional pressure feedback (Bp = 0) Load pressure feedfeedbackback will mainly increase the hydraulic damping. It works just as a Kc-value. In the above block diagram the proportional pressure feedback will increase Kc-value. In the above block diagram the proportional pressure feedback will increase ' ' the effective Kc-value as follows, K ce K ce K pf K saGv K qi . The resulting bode the effective Kc-value as follows, Kce = Kce + K pf KsaGv Kqi . The resulting bode diagram of the open loop gain (A (s)) and the closed loop stiffness (S (s)) for a diagram of the open loop gain (Au(s)) and the closed loop stiffness (Sc(s)) for a hydraulic damping of h = 0,46 is shown in Figure 8. One negative effect of proportional pressure feedback is that the steady state stiffness will be reduced.

2 2 10 78 10

1 0 10 10

Amplitude 0 10

−2 Amplitude, (Sc/Ks) 10

−1 10 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s]

−50 200 −100

−150 150

−200 100 Phase −250 Phase

−300 50 −350

−400 0 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s] Figure 8: Open loop gain (to the left) and closed loop stiffness of a position servo with load pressure feedback

Dynamic pressure feedback is shown in Figure 9. The idea of using dynamic pressure feedback is that the feedback signal shall reach its maximum value at a frequency, which has to be damped (the hydraulic frequency h). Therefore, the pressure signal K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 4

The threshold of the servo valve will also cause a position error Xp. If the threshold is

in in the position error is X p , where in is nominal valve input current. K sa K f

2. Valve controlled position servo with load pressure feedback The load pressure feedback is used to increase the hydraulic damping in the system. A negative load pressure signal acts in the same way as a Kc-value (flow-pressure coefficient) of the servo valve. Load pressure feedback can be of proportional or dynamic type. Proportional pressure feedback is shown in Figure 7.

Proportional pressure feedback

Kpf FL - - . uc + i + 1 + 1 xp _1 xp Ksa Gv Kqi Ap Kce + Vt s / 4be s - - PL Mt s

Figure 7: Block-diagram of a linear position servo with proportional pressure feedback (Bp = 0) Load pressure feedback will mainly increase the hydraulic damping. It works just as a K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control Kc-value. In the above block diagram the proportional pressure feedback will increase ______' the effective Kc-value as follows, K ce K ce K pf K saGv K qi . The resulting bode diagram of the open loop gain (Au(s)) and the closed loop stiffness (Sc(s)) for a hydraulic damping of of δhh = = 0,46 0,46 is is shown shown in in FigureFigure 7 - 88.. One One negative negative effect effect of proportional pressure feedback is that the steady statestate stiffnessstiffness willwill bebe reduced.reduced.

2 2 10 10

1 0 10 10

Amplitude 0 10

−2 Amplitude, (Sc/Ks) 10

−1 10 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s]

−50 200 −100

−150 150

−200 100 Phase −250 Phase

−300 50 −350

−400 0 0 1 2 3 0 1 2 3 10 10 10 10 10 10 10 10 Frequency [rad/s] Frequency [rad/s] FigureFigure 7 8:-8 :Open Open loop loop gain gain (to (to the the left) left) an andd closed closed loop loop stiffness stiffness of of a aposition position servo servo with load pressure feedbackfeedback

DynamicK-E Rydberg pressure pressure feedback feedbackFeedbacks is isshown in shown Electro-Hydraulic in Figure in Figure 9 .Servo The 7-9 Systemsidea. The of ideausing of dynamic using dynamicpressure5 feedbackpressure feedback is that the is feedback that the signalfeedback shall signal reach shall its maximum reach its maximumvalue at a value frequency, at a whichfrequency, has to which be damped has to (the be damped hydraulic (the frequency hydraulic h). frequency Therefore, ω theh). pressure Therefore, signal the willpressure be high-pass signal wi filtered.ll be high At- passlow frequencies filtered. At the low pressure frequencies feedback the pressuresignal is feedbacklow and thesignal reduction is low and of the reduction steady state of the stiffness steady will state be stiffness very low will compared be very low to compared proportional to pressureproportional feedback. pressure feedback.

Dynamic pressure feedback

s/wf Kpf s/wf + 1 FL . - - x x uc i 1 1 p _1 p + Ksa Gv Kqi + Ap + - Vt PL Mt s s - Kce + s 4be Ap

Figure 9: Block-diagram of a linear position servo with dynamic pressure feedback (Bp = 0) Figure 7-9: Block-diagram of a linear position servo with dynamic pressure feedback (Bp = 0)

3. Valve controlled angular position servo with acc. feedback 7.3 Valve controlled angular position servo with acc. feedback Acceleration feedback works in principal as dynamic pressure feedback. When the load startsAcceleration oscillate feedback there will works be ain feedback principal signal,as dynamic which pressure increase feedback. the hydraulic When dampingthe load juststarts at oscillate the resonance there willfrequency. be a feedback The good signal, thing which with accelerationincrease the feedback hydraulic is damping that the steadyjust at thestate resonance stiffness frequency.will not be Theaffected. good Anthing angular with accelerationposition servo feedback with acceleration is that the feedbacksteady state is shownstiffness in willFigure not 10be affected.and the corresponding An angular position block- diagramservo with is expressedacceleration in Figurefeedback 11 is. shown in Figure 7-10 and the corresponding block-diagram is expressed in Figure 7-11. Kf K 79ac - .. - G reg qm qm Uc + V1 + ps pL Jt TL

Figure 10: An angular position servo with acceleration feedback (Bm = 0) From Figure 11 the effect of the acceleration feedback can be expressed as a change in 1 the second order transfer function of the hydraulic system, Gh (s) 2 . s 2 h 2 s 1 h h This transfer function will now change to 1 . Gh (s) s 2 2 K h K K qi G (s)s 1 2 ac sa v h h Dm K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 5

will be high-pass filtered. At low frequencies the pressure feedback signal is low and the reduction of the steady state stiffness will be very low compared to proportional pressure feedback.

Dynamic pressure feedback

s/wf Kpf s/wf + 1 FL . - - x x uc i 1 1 p _1 p + Ksa Gv Kqi + Ap + - Vt PL Mt s s - Kce + s 4be Ap

K f

Figure 9: Block-diagram of a linear position servo with dynamic pressure feedback (Bp = 0)

3. Valve controlled angular position servo with acc. feedback Acceleration feedback works in principal as dynamic pressure feedback. When the load starts oscillate there will be a feedback signal, which increase the hydraulic damping just at the resonance frequency. The good thing with acceleration feedback is that the K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control steady state stiffness will not be affected. An angular position servo with acceleration ______feedback is shown in Figure 10 and the corresponding block-diagram is expressed in Figure 11.

Kac - .. - G reg qm qm Uc + V1 + ps pL Jt TL

Figure 10: An angular position servo with acceleration feedback (Bm = 0) Figure 7-10: An angular position servo with acceleration feedback (Bm = 0) From Figure 11 the effect of the acceleration feedback can be expressed as a change in From Figure 7-11 the effect of the acceleration feedback can be expressed as a1 change the second order transfer function of the hydraulic system, Gh (s) 1 . s 2 2 in the second order transfer function of the hydraulic system, Gh (s) = 2 h . s 2 2 s 1 δ h 2h + h s +1 1 ω h ω h This transfer function will now change to G (s) . This transfer function will now change to h 2 1 K . K-E Rydberg Feedbacks in Electro-HydraulicGh (s) = sServo Systems2 h qi 6 2 K K G (s)s 1 s 2 ⎛2δ h ac sa K qi v ⎞ h + ⎜ h + K K Dm G (s)⎟s +1 2 ⎜ ac sa v ⎟ ω h ⎝ ω h Dm ⎠

TL æ V ö Kce ç t ÷ 2 ç1+ s÷ Dm è 4be K ce ø - . uc iv Kqi 1 1 qm 1 qm + + Ksa s s2 2d Dm 1+ + + h s + 1 s - - wv 2 w wh h

Acceleration feedback Kac.s Position feedback Kf

Figure 11a: Block-diagram of an angular position servo with acceleration feedback (Bm = 0) Figure 7-11a: Block-diagram of an angular position servo with acceleration feedback (Bm = 0) 2 4e Dm With 2 , Gv(s) = 1,0 and Bm = 0 the effective hydraulic damping (including h 4βe Dm With ω = J tVt , Gv(s) = 1,0 and Bm = 0 the effective hydraulic damping h J V t t K J (includingacceleration feedback) acceleration will follow feedback) the equation: will* ce followe t K K theK equation:e . h D V ac sa qi V J K J m t t t * ce βe t βe . δ h = + Kac Ksa Kqi * ConstantDm accelerationVt feedbackVt J t gain (Kac) means that the total damping ( h ) varies according to variations in the inertia load Jt, as shown in Figure 11b. * Constant acceleration feedback gain (Kac) means that the total damping (δ h ) varies according to variations in the inertia load Jt, as shown in Figure 11b.

Figure 11b: Damping in an angular position servo with acceleration feedback (Bm = 0)

4. Velocity feedback in position control servos Pressure and acceleration feedback is used to increase the hydraulic damping and this makes it possible to increase the steady state loop gain Kv and the closed loop stiffness will increase. Another way to increase the stiffness of a position servo is to introduce a velocity feedback. A block-diagram of a linear position servo with velocity feedback is shown in Figure 12. K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 7-11b: Damping in an angular position servo with acceleration feedback (Bm = 0)

7.4 Velocity feedback in position control servos Pressure and acceleration feedback is used to increase the hydraulic damping and this makes it possible to increase the steady state loop gain Kv and the closed loop stiffness willK-E Rydbergincrease. Another wayFeedbacks to increase in Electro-Hydraulic the stiffness Servoof a positionSystems servo is to introduce7 a velocity feedback. A block-diagram of a linear position servo with velocity feedback is shown in Figure 7-12. FL K æ V ö ce ç + t ÷ 2 ç1 s÷ Ap è 4be K ce ø Threshold Saturation - . u i i imax Kqi 1 xp xp c K r v 1 1 + + sav s + s2 2d ei Ap 1+ + h s + 1 s - - n wv 2 w wh h

Velocity feedback Kfv

Position feedback Kf

FigureFigure 7 12:-12 :A A linear linear valve valve controlled controlled position position servo servo with with velocity velocity feedback feedback If the bandwidth of the valve isis relativelyrelatively highhigh and threshold and saturation is neglected the velocity feedback will give the effect effect on on the the hydraulic hydraulic reso resonancenance frequency and damping as shown in Figure 137-13. . FL æ ö K Kce Vt qi ç1+ s÷ Kvfv = 1 + Kfv Ksav 2 ç ÷ Ap Ap è 4be K ce ø - . 1/ Kvfv x x uc iv Kqi p 1 p + Ksav s2 2d Ap + + h s + 1 s - 2 K w Kvfv wh vfv h

Position feedback Kf 81 Figure 13: A linear position servo with velocity feedback included

From Figure 13 the new resonance frequency and damping (hv and hv) caused by the velocity feedback can be evaluated as

1 K qi K , , where the velocity loop gain is K 1 K K . hv h vfv hv h vfv fv sav A K vfv p Designing the position control loop for the same amplitude margin as without velocity feedback gives the following relations:

K qi Steady state loop gain without velocity feedback: K v K sa K f Ap

K qi Steady state loop gain with velocity feedback: K vv K sav K f Ap K vfv

A certain amplitude margin means that K v h h . In this case h h hv hv , which implies that K v K vv and thereby the servo amplifier gain K sav K sa K vfv . With velocity feedback, the servo amplifier gain (Ksav) can be increased in proportion to the velocity loop gain Kvfv and the servo amplifier gain without velocity feedback, Ksa.

The open loop gain (Au(s)) for a position servo without (Kv = 20) and with velocity feefback (Kvfv = 10 and Kvv =20) is shown in Figure 14. K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 7

FL K æ V ö ce ç + t ÷ 2 ç1 s÷ Ap è 4be K ce ø Threshold Saturation - . u i i imax Kqi 1 xp xp c K r v 1 1 + + sav s + s2 2d ei Ap 1+ + h s + 1 s - - n wv 2 w wh h

Velocity feedback Kfv

Position feedback Kf Figure 12: A linear valve controlled position servo with velocity feedback K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control If the bandwidth of the valve is relatively high and threshold and saturation is neglected ______the velocity feedback will give the effect on the hydraulic resonance frequency and damping as shown in Figure 13.

FL æ ö K Kce Vt qi ç1+ s÷ Kvfv = 1 + Kfv Ksav 2 ç ÷ Ap Ap è 4be K ce ø - . 1/ Kvfv x x uc iv Kqi p 1 p + Ksav s2 2d Ap + + h s + 1 s - 2 K w Kvfv wh vfv h

Position feedback Kf

FigureFigure 7 13:-13 :A A linear linear position position servo servo with with velocity velocity feedback feedback included included

From Figure 13 the new resonance frequency and damping (hv and hv) caused by the From Figure 13 the new resonance frequency and damping (ωhv and δhv) caused by the velocity feedback can be evaluated as

K qi 1 , where the velocity loop gain is K 1 K K qi . hv h K vfv , hv h , where the velocity loop gain is K vfv = 1+ K fv K sav . ω hv = ω h K vfv , δ hv = δ h vfv fv sav A K vfv Ap K vfv p Designing the position control loop forfor thethe samesame amplitudeamplitude marginmargin asas withoutwithout velocityvelocity feedbackfeedback givesgives the follfollowingowing relations: K Kqi Steady state loop gain without velocity feedback: K v K sa K f v = sa A f Ap K Kqiqi Steady state loop gain with velocity feedback: K vv K sav K f vv = sav A K f Ap Kvfvvfv A certain amplitude margin means that K . In this case , which A certain amplitude margin means that Kvv ∝ ωhδ h . In this case ω hδ h = ωhvδ hv, which implies that K K and thereby the servo amplifier gain K K K . With implies that Kvv = Kvvvv and thereby the servo-amplifier gain is, Ksavsav = Ksasa Kvfvvfv . With velocity feedback, the servo amplifier gain (Ksav) can be increased in proportion to the velocity feedback, the servo amplifier gain (Ksav) can be increased in proportion to the velocity loop gain Kvfv and the servo amplifier gain without velocity feedback, Ksa. velocity loop gain Kvfv and the servo amplifier gain without velocity feedback, Ksa. K-EThe Rydberg open loop gain (AuFeedbacks(s)) for ain positionElectro-Hydraulic servo withoutServo Systems (Kv = 20) and with velocity8 The open loop gain (Au(s)) for a position servo without (Kv = 20) and with velocity feefback (Kvfv = 10 and Kvv =20) is shown in Figure 14. feedback (Kvfv = 10 and Kvv =20) is shown in Figure 7-14.

2 10

0 10 Amplitude

−2 10

0 1 2 3 10 10 10 10 Frequency [rad/s]

−50

−100

−150

Phase −200

−250

−300 0 1 2 3 10 10 10 10 Frequency [rad/s] Figure 14: Open loop gain for a position servo without and with velocity feefback (Kv = Kvv) Figure 7-14: Open loop gain for a position servo without and with velocity feedback (Kv = Kvv)

5. Valve controlled velocity servo 82 If an integrating amplifier is used in a velocity servo the loop gain Au(s) will be in principle the same as for a position servo with proportional control. Such a velocity servo is shown in Figure 15.

vp Ap Ap Mt FL

Velocity transducer Integrating K servo ampl. f uc + i ____Ksa s Ps = const. uf - Figure 15: A linear valve controlled velocity servo

A block diagram of the velocity servo is shown in Figure 16. FL

Kce 2 Gl Integrating Ap amplifier Threshold Saturation . i - uc 1 ir i max Kqi xp K G G + sa s A v + h - ein p

Au(s) K f

Figure 16: Block-diagram of a linear valve controlled velocity servo K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 8 K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 8

2 10

0 10

0 10 Amplitude

−2

Amplitude 10

−2 10 0 1 2 3 10 10 10 10 Frequency [rad/s] 0 1 2 3 10 10 10 10 Frequency [rad/s] −50

−10050

−−100150

K-E Rydberg −Phase −150200 Hydraulic Servo Systems – Dynamic Properties and Control

Phase −−200250 ______−−250300 0 1 2 3 10 10 10 10 −300 Frequency [rad/s] 0 1 2 3 10 10 10 10 Figure 14: Open loop gain for a position servoFrequency without [rad/s] and with velocity feefback (K = K ) v vv Figure 14: Open loop gain for a position servo without and with velocity feefback (Kv = Kvv) 7.5 Valve controlled velocity servo 5. Valve controlled velocity servo 5.If an Valve integrating controlled amplifier velocity is used in servo a velocity servo the loop gain A (s) will be in If an integrating amplifier is used in a velocity servo the loop gain Au (s) will be in principle the same as for a position servo with proportional control. Suchu a velocity Ifprinciple an integrating the same amplifier as for a is position used in servo a velocity with servo proportional the loop control. gain A Suchu(s) willa velocity be in principleservo is shown the same in Figure as for 7 - a15 position. servo with proportional control. Such a velocity servo is shown in Figure 15. servo is shown in Figure 15.

vp Ap Ap vp Ap Ap Mt F Mt L FL

Velocity transducerVelocity Integrating transducer Kf Integratingservo ampl. uc + Kf servoK ampl. i uc + ____sa s i Ps = const. uf ____Ksa - s Ps = const. uf - FigureFigure 7 -15:15: AA linearlinear valvevalve controlled controlled velocity velocity servo servo Figure 15: A linear valve controlled velocity servo AA blockblock diagramdiagram ofof thethe velocityvelocity servoservo isis shownshown inin FigureFigure 716-16. . A block diagram of the velocity servo is shown in Figure 16. FL FL Kce Gl Integrating KAce2 p Gl Integratingamplifier Threshold Saturation 2 Ap . amplifier Threshold Saturationi - uc 1 ir i max Kqi .xp Ksa i Gv - G uc+ 1s ir i max K Aqi + h xp - Ksa ein p Gv G + s A + h - ein p Au(s) Au(s) Kf K f Figure 16: Block-diagram of a linear valve controlled velocity servo FigureFigure 7 16:-16 :Block-diagram Block-diagram of of a alinear linear valve valve controlled controlled velocity velocity servo servo The transfer functions in the above block-diagram are:

1 V 1 G (s) = , G (s) = 1+ t s , G (s) = v s 1 4 K h s 2 2δ +1 β e ce h 2 + s +1 ω v ω h ω h

An integrating amplifier means that the control error will be integrated and the steady state control error becomes zero.

83 K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 9

The transfer functions in the above block-diagram are: 1 V 1 G (s) , G (s) 1 t s , G (s) v s 1 4 K h s 2 2 1 e ce h s 1 2 K-E Rydberg Hydraulicv Servo Systems – Dynamic Properties and Control h h ______An integrating amplifier means that the control error will be integrated and the steady state control error becomes zero.

7.66. ProportionalProportional valves valves with with integrated integrated position andand pressure transducers transducers In all fluid power applications a load has to be controlled by an actuator in respect of In all fluid power applications a load has to be controlled by an actuator in respect of speeds and forces. A new dimension of the ways to look upon these control aspects is to speeds and forces. A new dimension of the ways to look upon these control aspects is to useuse a acontrol control valve valve (proportional (proportional oror servoservo valve),valve), whichwhich isis capablecapable ofof controlling controlling both both flowflow and and pressure pressure in in the the actuator actuator portsports (two(two portsports for a doubledouble cylindercylinder or or motor). motor). Such Such a proportionala proportional valve valve has has been been developed developed byby Ultronics. TheThe prinprincipleciple design design of of the the valve valve is isshown shown in in Figure Figure 7 17-17. .

U X

Uc Load

- + Pressure and position signals X X Valve + U U controller Output signals

- Supply pressure FigureFigure 7- 17:17: ApplicationApplication with with Ultronics Ultronics proportional proportional valve valve

FromFrom Figure Figure 7 -1717 itit cancan bebe seenseen thethe valvevalve hashas twotwo spools,spools, whichwhich make make it it possible possible to to controlcontrol meter meter-in-in and and meter meter-out-out flowflow ofof anyany actuatoractuator independently. ThisThis facility facility gives gives thethe opportunity opportunity of of smooth smooth acceleration acceleration and and ddecelerationeceleration control control of of the the load load by by individualindividual pressure pressure control control in in each each cylindercylinder chamber.chamber. The pressurepressure transducers transducers can can also also be used for load pressure feedback to increase the hydraulic damping. By measurement be used for load pressure feedback to increase the hydraulic damping. By measurement of the pressure drop (p) over a spool the load flow (qL) can be controlled by of the pressure drop (Δp) over a spool the load flow (qL) can be controlled by calculation of the spool displacement (x ) from the flow equation of the valve, which calculation of the spool displacement (x v) from the flow equation of the valve, which gives v gives qL xv qL2 xv = C w p q 2 C w Δp q ρ

7.7 Electro-hydraulic servo actuators Today electro-hydraulic actuators are normally manufactured as integrated units. The servo valve is connected to the actuator (cylinder or motor) and all the transducers needed for close loop control are integrated in the valve and actuator. An industrial actuator for linear position control is depicted in Figure 7-18. The control card for this actuator includes connectors for all feedback signals and the controller is implemented

84 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______in a microprocessor. The input signals to the control card are electric power supply and a set point signal and than the card deliver a current signal (i) to the servo valve. The hydraulic part of the actuator system has two connectors, one hydraulic supply line and one return flow line. In many industrial applications there is a need for multiple degrees of freedom control of the load. One application, which requires advanced control, is motion simulator platforms. This type of platform is often used for dynamic simulation of aircrafts and cars. A common way to design a platform, which can be moved in a 3D-space, is to use 6 electro-hydraulic linear actuators as shown in Figure 7-19.

Figure 7-18: Industrial electro-hydraulic linear position control actuator, MOOG

Figure 7-19: Electro-hydraulic motion platform with 6 degrees of freedom, Rexroth

85 K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 11

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

For low powerFigure applications 19: Electro-hydraulic (low loadmotion weights) platform with the 6 platformdegrees of freedom, shown Rexroth in Figure 7-19 is oftenFor realisedlow power by applicationsusing electro (low-mechanical load weights) actuators the platform (electric shown motor in and Figure a ball 19 screw). is often realised by using electro-mechanical actuators (electric motor and a ball screw). A similar control strategy as for the 6 DOF plat-form can be used for crane (or industrialA similar robot) control tip strategy control as .for El theectro 6 -DOFhydraulic platform control can be of used a lorry for crane crane (or is industrial shown in Figurerobot) 7tip-20 control. . Electro-hydraulic control of a lorry crane is shown in Figure 20.

Mechanical System

Z3 Optronic Sensor

Computer Based Hydraulic Measurement and System Control System X3 K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 12 FigureFigure 7- 2020:: Crane tip control withwith optronic optronic sensor sensor for for vertical vertical position position measurement measurement

The strategy for 2 DOF crane tip control is shown in Figure 7-21. A range camera The strategy for 2 DOF crane tip control is shown in Figure 21. A range camera (optronic sensor) is used to measure the vertical distance (h) between the camera and (optronic sensor) is used to measure the vertical distance (h) between the camera and the object. Z3 is the vertical co-ordinate from the base line of the crane to the crane tip. the object. Z3 is the vertical co-ordinate from the base line of the crane to the crane tip. The reference value for the vertical crane tip position is calculated as Z3ref = Z3+href–h. The reference value for the vertical crane tip position is calculated as Z3ref = Z3+hrefMh. TheThe ki kinematicsnematics of of the the crane crane structurestructure isis calculated byby usingusing thethe signals signals from from position position transducerstransducers in in the the hydraulic hydraulic cylinders cylinders andand a geometric descriptiondescription of of the the crane crane structure. structure. However,However, this this will will not not give give the the truetrue tiptip positionposition of the cranecrane tip tip because because of of the the weakness weakness inin the the mechanical mechanical structure. structure. ByBy usingusing a range cameracamera itit isis possiblepossible to to compensate compensate the the verticalvertical position position control control according according toto the mechanical weakness.weakness. Lead- - filter X3ref Xp1r PI- xv1r X3 Inverse S Gcyl1 + contr Xp1 Kine- Kine- Z X matics href 3ref matics Xp2r PI- xv2r p2 Z3 S S Gcyl2 + - + contr + - q Lead- 3 filter h Range Camera

FigureFigure 7 -21:21: Control Control strategy strategy for for crane crane tip tip positioning positioning

8. Design examples 86 Hydraulically operated boom with lumped masses The figure shows a valve controlled cylinder used for operation of a mechanical arm. The total mass of the moving arm is ML. The distance from the gravity centre of the mass to the joint (0) is L. The lever length for the hydraulic cylinder is e, which will vary according to xp. The piston area is Ap and its pressurised volume is VL and this volume varies according to the piston position. The effective bulk modulus is e. The pressure on the piston rod side is assumed as constant, pR = constant. The mass of the cylinder housing is M0 and the mechanical spring coefficient for the connection is KL. L 0 ML q e pR = constant

xp Ap VL pL M0 KL

Figure 22: Application with variable mechanical gearing between cylinder and load

K-E Rydberg Feedbacks in Electro-Hydraulic Servo Systems 12

The strategy for 2 DOF crane tip control is shown in Figure 21. A range camera (optronic sensor) is used to measure the vertical distance (h) between the camera and the object. Z3 is the vertical co-ordinate from the base line of the crane to the crane tip.

The reference value for the vertical crane tip position is calculated as Z3ref = Z3+hrefMh. The kinematics of the crane structure is calculated by using the signals from position transducers in the hydraulic cylinders and a geometric description of the crane structure. However, this will not give the true tip position of the crane tip because of the weakness in the mechanical structure. By using a range camera it is possible to compensate the vertical position control according to the mechanical weakness. Lead- - filter X3ref Xp1r PI- xv1r X3 Inverse S Gcyl1 + contr Xp1 Kine- Kine- Z X matics href 3ref matics Xp2r PI- xv2r p2 Z3 S S Gcyl2 + - + contr + - q Lead- 3 filter h Range Camera K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______Figure 21: Control strategy for crane tip positioning

8.7.8 Design Design examples examples

Hydraulically operated operated boom boom with withlumped lumped masses masses The figure shows a valve controlled cylinder used for operation of a mechanical arm. The total mass of the moving arm is ML. The distance from the gravity centre of the mass to the joint (0) is L. The lever length for the hydrauhydrauliclic cylinder is e, which will vary according to xp. The piston area is Ap and its pressurised volume is VL and this volume varies according to the piston position. The effective bulk modulus is βe. The pressure on the piston rod side is assumed as constant, pR = constant. The mass of the cylinder housing is M0 and the mechanical spring coefficient forfor thethe connectionconnection isis KKLL.. L 0 ML q e pR = constant

xp Ap VL pL M0 KL

FigureFigure 7 22:-22 :Application Application with with variable variable mech mechanicalanical gearing gearing between between cylinder cylinder and and load load

Equivalent cylinder mass The equivalent mass loading the piston rod is found from the torque equation for the joint (0).

.. 2 2 Inertia : J = M L .. .. t L X p ⎛ L ⎞ .. With θ = ⇒ pL Ap = M L ⎜ ⎟ X p . 2 e ⎝ e ⎠ Torque : Tθ = M L L θ = pL Ap e L Introducing the mechanical gear U = , the equivalent cylinder mass can be expressed e 2 as, M t = M LU

Hydraulic resonance frequency and damping

Assuming ML as the dominant mass the resonance frequency can be calculated as,

K 1 K 2 e e 1 VL 1 K L β e Ap M0 << ML gives ωh = = where = 2 + ⇒ Ke = 2 M t U M L Ke β e Ap K L K LVL + β e Ap

Kce βe M L Low mechanical friction gives the hydraulic damping: δh = U 2Ap VL

87 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

According to the gear U it can be observed that an increase of the gear gives a reduced

ωh but an increased δh. Studying the product δh·ωh the expression is,

2 Kce βe M L 1 βe Ap K ce β e δ hωh ≈ U ⇒ δ hω h ≈ , for KL >> Kh 2Ap VL U VL M L 2 VL0 + Ap x p

Cylinder design according to max pressure level This example is aimed to demonstrate how the cylinder design will influence the hydraulic frequency and damping. Figure 7-23 shows a system with a stiff mechanical structure and the cylinder is loaded by one mass (ML).

Figure 7-23: Cylinder controlled mass with mechanical gear

M L g As in Fig. 22 the mechanical gear U = L/e. The piston area is selected as, Ap =U . pL

X L The cylinder volume depends of the load displacement (XL) as, V = A . For the 0 p U β A2 hydraulic resonance frequency the basic equation is, e p . If the cylinder is ωh = 2 V0 M LU designed for some maximum load pressure (pLmax), with Ap and V0 as described above, the hydraulic frequency will follow the expression:

βe ⋅ g ωh = . X L ⋅ pLmax

2 K M U Kce ⋅U βe ⋅ pLmax The hydraulic damping is described as, ce βe L or , δ h = δh = 2Ap V0 2Ap X L ⋅ g where the flow/pressure coefficient (Kce) is assumed to be constant.

Kce βe Kce βe ⋅ pLmax The product δh·ωh is expressed as, δ hωh = = . 2 V0 2 M t ⋅ g ⋅ X L

88 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Figure 7-24 shows how the frequency, damping and the product varies according to the design parameter max load pressure, pLmax.

Figure 7-24: Hydraulic resonance frequency and damping versus max load pressure From the equations it can be noticed that the hydraulic damping will be proportional to 3/ 2 pL max and the product δ hωh ∝ pLmax . This indicates that the cylinder-load response will show less oscillations when the max load pressure is increased. The system response for different pLmax is illustrated in Figure 7-25.

Figure 7-25: Response of the cylinder-load dynamics with cylinder design for max load pressure of 100, 200 and 300 bar respectively

89 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

7.9 Summary of servo system design criterions The most important design criterions for servo systems are stability, accuracy, repeatability, reliability and robustness. Stability and control accuracy has to be analysed using standard control theory and critical nonlinearities must be included (see Chapter 8). The robustness and reliability criterions belong to each component as well as the supply unit. The simplest way to improve robustness reliability of the hydraulic hardware is to use a supply unit with high-quality filtering and cooling systems. Finally, the use of a high-quality hydraulic fluid is of great importance to reduce wear and increase service life of the system.

7.9.1 Control loop dynamics – possible improvements

Figure 7-26 illustrates a block diagram for an electro-hydraulic position servo with a symmetric servo valve and cylinder, loaded by the mass Mt. The controller has three different gains, one proportional gain KP of the position error, the gain KD for the negative velocity feedback and KAC for the negative acceleration feedback. Servo valve bandwidth is much higher than ωh.

Figure 7-26: Response of the cylinder-load dynamics

The velocity and acceleration feedbacks shown in Figure 7-26 will influence the second order transfer function (resonance frequency and damping) as follows:

1/ Kvfv 1/ Kvfv = , where Kvfv = 1+ Kfv·KD·Kqi/Ap 2 2 ' ⎛ s ⎛ 2 K ⎞ ⎞ ⎛ s 2δ h ⎞ ⎜ ⎜ δ h fAC ⎟ ⎟ ⎜ + s +1⎟ 2 + + s +1 ⎜ '2 ' ⎟ ⎜ K ⎜ K K ⎟ ⎟ ωh ωh ⎝ ωh vfv ⎝ ωh vfv vfv ⎠ ⎠ ⎝ ⎠ ' ' ' ωh ⎛ 2δh K fAC ⎞ and KfAC = Kfv·KAC·Kqi/Ap. → ω = ω K and δ = ⎜ + ⎟ . h h vfv h 2 ⎜ K K ⎟ ⎝ωh vfv vfv ⎠ -3 With the following figures, Kvfv = 6, KfAC = 8,0·10 s, ωh = 100 rad/s and δh = 0,15, the new frequency and damping became as:

' ω ⎛ 2δ K fAC ⎞ ω' = ω K = 245 rad/s and δ ' = h ⎜ h + ⎟= 0,225. h h vfv h 2 ⎜ K K ⎟ ⎝ωh vfv vfv ⎠

90 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

The velocity and acceleration feedback gives the steady state loop gain as:

' ' Kqi K f Kv = KP ⋅ ⋅ . Ap Kvfv

Kqi With position feedback only, the steady state loop gain is, Kv = K P ⋅ ⋅ K f . Ap

Designing for a specific amplitude margin (Am), for example, Kv = δhωh ⇒ Am = 6dB ' ' ' and Kv = δ hωh ⇒ Am = 6dB, means that the use of velocity and acceleration feedbacks results in a much higher Kv-value. The difference is Kv = δhωh =15 1/s and ' ' ' Kv = δ hωh = 55 1/s, which gives an increasing factor of 3,67 and accordingly the system bandwidth (ωb) will increase with the same factor.

When it comes to steady state stiffness of the closed loop system (|Sc|s->0) the difference is much higher.

2 − ΔFL ' Ap Velocity and acceleration feedbacks gives: Sc = = Kv Kvfv , resulting in ΔX p Kce s→0 ' Sc,va Kv Kvfv 55 ⋅ 6 the ratio: = = = 22. The steady state stiffness increases 22 times. S K 15 c s→0 v The full state feedbacks, shown in Figure 7-26 gives the possibility to improve system dynamics to meet requirements on control accuracy, bandwidth and stiffness. However, a successful control performance requires accurate feedback signals and a high response servo-valve with low hysteresis and threshold. Instead of acceleration feedback a dynamic load pressure feedback can be used in order to improve the hydraulic damping. However, the load pressure signal has to be properly filtered and the feedback gain has to be adjusted according to the actual resonance frequency, otherwise the damping and stiffness will be affected.

91 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

8 Nonlinearities in Hydraulic Servo Systems

All real systems contain nonlinear elements to some extent. Mechanical mechanisms have nonlinear friction in all contacts with relative motion between surfaces. Saturation is another nonlinear phenomena, which set a limit for max output velocity/speed in all kind of systems. Hydraulic systems have saturation in flow and pressure and electrical systems in current and voltage. Analysis of nonlinear system is a quite complex task. The major problem is to determine the effect on performance of the nonlinear elements. By using digital computer simulation most kind of nonlinearities can be handled sufficiently. Commercial simulation program package usually contains tool-boxes with functions which can handle the most common nonlinearities. However, nonlinear systems are very difficult to analyse, because they have no unique equilibrium point and it is difficult to quantify the stability criteria for these systems. In opposite to nonlinear systems a linear system has only one unique equilibrium point and the stability criteria are well defined. FromNonlinearities control engineering point of in view, hydraulic there is a great advantage systems of using linear system descriptions. Linear differential equations have unique solutions and all aspects of system performance, such as response and stability are predictable. By using a method to linearize the existing nonlinearities the full system can be treated as a linear one. That makes the control system design much more understandable and easier to perform.

8.1 How to handle nonlinear properties in linear models? As mentioned above, hydraulic servo system includes a number of nonlinear functions. M The servo valve has nonlinear flow gain, threshold and flow saturation, 0see Figure 8-1. The mechanical part of the system contains nonlinear friction and maybe, also backlash.

Figure 8-1: Block-diagram of a linear position servo including valve nonlinearities

Describing Functions There are different ways to implement nonlinearities in a control loop with linear transfer functions. One common method is to use Describing Functions. A describing function is derived as a quasi-linearization of a nonlinear function. Introducing a describing function in a control loop means that a nonlinear element is replaced by a linear function except for a dependence on the amplitude of the input waveform. A control loop including nonlinearities as a describing function (Gdf(α,β)) can be depicted as illustrated in Figure 8-2.

92 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______Describing function – Gdf(#) Control loop containing nonlinearities Control error Nonlinearities

U E M sin(!t) G ( , ) f ("t +!) Y + G1(s) df ! " G2 (s) -

Feedback signal

Figure 8-2: Block-diagram of a control loopY includingG ( sa )nonlinear!G (" function)!G (Gsdf)(α,β) Closed loop transfer function: = 1 df 2 The transfer function (output over input signal)U 1 of+ Gthe1( closeds)!Gdf loop(") servo!G2 ( ssystem,) with unit feedback, is derived as, Characteristic equation: Gch (i!) =1+ G1(i!)#Gdf (")#G2 (i!) Y G (s)⋅G (α, β)⋅G (s) = 1 df 2 (81 -1) Roots of char. equation:U 1 +GGch1((si)"⋅G)d=f (α0,%β)⋅GG12((is") )$G2 (i") = # Gdf (!) Linear differential equations can be solved from the characteristic equation of the system. The requirement on linearity also means that the nonlinearities must be linearized, which is the case when describing functions are used. Since the describing function only is defined for sine-wave signals, the Laplace-operator s has to be substituted by iω. The characteristic equation of the closed loop system, shown in Figure 8-2, can directly be derived from the transfer function as the denominator in (8- 1),

Gce (iω) =1+G1(iω)⋅Gdf (α, β)⋅G2 (iω) (8-2) The roots of the characteristic equation, addresses the stability of the system. The characteristic equation equated to zero (Gce (iω) = 0 ) gives the roots and the relation between the describing function and the other transfer functions as, 1 G1(iω)⋅G2 (iω) = − (8-3) Gdf (α, β) The solutions to equation (8-3) are easily found by plotting the left and right side in a Nyquist diagram (amplitude-phase plane). If intersections, of the two loci exists a limit cycle will occur, with a frequency given by the left side of eq. (8-3) and amplitude given by the right side.

8.2 Common Nonlinearities in Hydraulic Systems

8.2.1 Saturation and its effect on system performance

Saturation is the most common nonlinearity in servo system, since it exists in all types of systems. The saturation level (S) and its impact on a sine-wave signal (M.sin ωt) are demonstrated in Figure 8-3. It can be noted that the saturation element just will cut the top of the sine-wave signal when S < M. In practice, saturation in a control system means a reduction of the steady state loop gain (Kv) and the saturation level gives a limit for the actuator velocity/speed.

93 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______Saturation gives gain reduction Reference: Merritt H E, ”Hydraulic Control Systems”

Gdf

FigureFourier 8-3: Saturation analysis of a sine-wave gives signal, [1]

1/ 2 Fourier analysis of the output' signal from the saturation element2 $ givesS/M the following= 1,0 => G = 1,0 => -1/G = -1.0 2 ! 41 S S - 3 S 0 * ! df df describing function,Gdf ( 6) = &sin + +141 . ( # 5 ! M M ,+ 2 M / )( ! S/M = 0,4 => G = 0,5 => -1/G = -2.0 % ( 2 , " df df 2 * −1 S S " S % * Gdf = )sin + 1−$ ' - (8-4) M M M π +* # & .*

Numerical values of S and M in eq. (8-3) gives Gdf = 1.0 for S = M. Reduction of the saturation level increases Gdf, which gives the range 0 < Gdf <= 1. From equation (8-3) it can now be stated that the highest value of the right side is -1 and reduced saturation level gives lower values (only real numbers), which gives the criteria,

1 − ≤ −1 (8-5) Gdf (α)

In the above equation the describing function is expressed as Gdf(α) because, the saturation element just includes one variable (with real number), the amplitude ratio S/M. Saturation and its impact on system dynamics, such as oscillation frequency and amplitude, depend upon the system loop gain characteristics. Looking at a hydraulic linear position servo with proportional control and a servo valve with quite high response theControl control system loopcan be described for asposition in Figure 8-4. servo Control error Saturation function 1 U E Y K G ( ) ' s2 2) $ + v df ! % h s 1" s - % 2 + + "! &(h (h # Feedback signal

Figure 8-4: Block diagram of a hydraulic position servo with proportional control and saturation K = 40 1/s Loop gain included v Kv Kv = 20 1/s Au (s) = ' s2 2!0.2 $ % s 1" s % 2 + + "! &100h 100 # 94

Gdf (!) =1.0 K K A (i ) v v u # #=100 = = 2!#h !" h 40 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______Nyquist diagram for position servo with Plotting the left andsaturation right side of equation in the (8-3) controlfor this system loop in a Nyqui st diagram gives the result as shown in Figure 8-5.

Intersection Loop gain -1 Kv Au (s) = ' s2 2!0.2 $ 1 % + s +1"! s " % 2 " G (!) ! =$ &100 100 # df ! =100 rad/s phase shift Saturation char. ! = 50 !1 ! " !1.0 Gdf (#) Kv = 40 1/s Kv = 20 1/s

Saturation in position servo Figure 8-5: Nyquist diagram of a hydraulic position servo with proportional control and saturation characteristics

From the open loop gain plotted in Figure 8-5 it can be noted that Kv = 20 1/s is equal . to the product, ωh δh = 20 1/s, which gives a stability-margin/amplitude-margin of Am = 6 dB for the control loop. Kv = 40 1/s gives Am = 0 dB, which is the border between stability and instability. For normal setting of Kv (20 1/s) the real number of the loop gain in bigger than -1 and no intersection with the saturation characteristics can occur. In other words, saturation can’t provide limit cycle oscillations in this system. Kv = $h "#h = 20 ! Am = 6 dB

Kv = 20 (no sat.)

Step

Kv = 40 (no sat.) Position [m] Kv = 40, sat. = 0.5

Time [s] Figure 8-6: Step response of a hydraulic position servo with proportional control and saturation

From Figure 8-6 it can be observed, that the highest steady state gain, Kv = 40 1/s and no saturation results in on-going oscillations after that the final position has been reached. When a saturation element is included in the control loop the oscillation amplitude will be drastically reduced. Compared to the amplitude without saturation,

95 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

the saturation ratio is S/M ≈ 0.33 which gives this high suppression of the oscillation amplitude. However, the saturation element will not change the stability criteria, so the oscillations will still continue. Compensation of deadband q 8.2.2 Dead-band Dead-band: In a servo valve dead-band is caused by friction and overlap of valve ports. Valve overlap causes a dead-bandEx: Valvearound neutral characteristic spool position (position at zero input D- D+ signal). The dead-band impact on the valve flow gain and how dead-band can i be Compensation ofcompensated deadband for is demonstrated in Figure 8-7. q q i

+ Ex: Valve characteristic D - + % D D i D- iref iref

Figure 8q- 7: Step response of a hydraulic position servo with+ proportional control and saturation - i iref > 0 " i = iref + D iref < 0 " i = iref + D A dead-band in a servo valve means that the valve flow gain is zero inside the dead- D+ band region. By adjusting the valve command signal it is simple to compensate for % dead-band. Figure 8-7 shows that a negative reference signal, iref < 0 gives the D- iref iref - - command signal as i= iref + D (where D has negative sign) and a positive reference + signal, iref > 0 gives the command signal as i = iref + D . In practical applications it is very important to implement a small threshold around zero, otherwise the dead-band compensator can start to oscillate because of noise on the reference signal. + - iref > 0 " i = iref + D Deadiref <-band 0 " compensation i = iref + D is easy to implement in a digital controller. However, it has to be remembered that a dead-band will not remains constant. The range of a valve dead- band varies according to supply pressure and temperature. Also the valve response has to be quite fast, otherwise the dead-band compensation will cause a significant time delay in the control loop.

8.2.3 Threshold and Hysteresis

Threshold: The increment of input current required to produce a change in valve output is named as valve threshold. Valve threshold is usually expressed in percent of rated current. In opposite to dead-band the threshold level must be passed in the full operation range to change the valve output. To overcome a threshold value in a control loop the absolute value of the control signal must be bigger than zero. Therefore, the threshold will cause a control error in a servo system with only proportional control, see page 72.

The threshold effect on the valve spool displacement (xv) versus input current (i) is illustrated in Figure 8-8.

96 Hysteresis

K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

!iTH xv

Figure 8-8: Threshold effect valve spool displacement

The figure demonstrates that the threshold characteristic is similar to backlash and as well as backlash it creates a hysteresis profile. This kind of nonlinearity is multivalued, since it has impact on both amplitude and phase. However, in a servo valve a typical threshold value is ΔiTH = 1 % of rated current iN, which give a phase shift less than 2 degrees. Karl-Erik Rydberg, IEI/Linköping University 14

8.3.4 Nonlinear friction

In practice mechanical friction has always a nonlinear velocity dependency. At zero velocity the friction force belongs to dry friction effects and at higher velocities the friction is developed to full viscous nature. Real characteristics of friction forces are often defined as a “StribeckNon-linear curve”, which is illustrated friction in Figure 8-8.

Ff [N]

Bf < 0 Bf > 0

v [m/s]

Bf > 0 Bf < 0

Figure 8-8: Friction force versus velocity, Stribeck characteristics Hydraulic damping in position servo – Kce !eM t B f Vt Valve controlled symmetric cylinder: "h = + In the figure, Bf in [Ns/m] represents the viscousA p frictionVt coefficient4Ap !,e M whicht is the gradient of the force-velocity profile. Using Bf to define the slope of the force curve it can be seen that the numerical value of Bf will be negative at low velocities. This effect has to be taken into account at design of servo systems with relatively high friction forces in actuator and load. Friction can have a great impact on the damping of oscillations in servo systems. This can be demonstrated by looking at the expression of the relative hydraulic damping for a system with a symmetric cylinder controlled by a four-port symmetric valve, see equation (8-7).

97 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Kce βeMt Bf Vt δh = + (8-7) Ap Vt 4Ap βeMt

Fully developed viscous friction, Bf > 0, will just increase the effective damping, but in the low velocity range, Bf < 0, friction will cause lower damping. Analysing the two terms on the right side of eq. (8-7) it can be observed that increased effective bulk modulus (βe) and reduced volume (Vt) will increase the first term and reduce the second.

Improvements of the hydraulic stiffness (low hydraulic capacitance, Ch = Vt/βe) will reduce the control problem caused by nonlinear friction in actuators and loads. Therefore, a safe way to handle nonlinear friction is to make sure that the first term in eq. (8-7) is dominant for all output velocities/speeds.

98 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

9 Controller Design for Hydraulic Servo Systems

9.1 General structure of the controller The most general controller of conventional type is the PID-controller. However, even with this controller there can still be a need of more dynamic compensations in the control loop. In a hydraulic system the relative damping is often quite low. A stabilisation feedback (load pressure or acceleration feedback) can be used to increase the damping. Depending of the variation of the command signal there will be a delay between the derivative of the command signal and the output signal. This delay can be reduced to a minimum by use of a feed forward gain. The action of the PID-controller means that the derivative gain increases proportionally to the frequency. In spite of this behaviour it is important to reduce the gain of the D- action at high frequencies. Otherwise, the high frequency disturbances on the signals will be amplified to a level, which can mainly influence the function of the system. A forward loop filter is used to reduce the derivative gain at high frequencies. From the above discussion the general structure of the controller will be as shown in Figure 9-1.

Figure 9-1: Structure of a PID controller with feed forward gain and stabilisation feedback.

99 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

9.2 Feed forward gain for reduction of velocity error in position servos Assume a linear position servo with valve-controlled piston. In this case a plain proportional controller is suitable to use and easy to adjust for stability. However, if the command signal is changed there will be a phase lag from input to output signal in the servo. In the position servo the phase lag cause a position error proportional to time derivative of the command signal (velocity). If the feed forward gain introduces a derivative of the command signal it will be possible to more or less eliminate the phase lag. This feed forward gain helps the servo control loop (servo valve) to react quickly to a change in the command signal. Implementation of a feed forward gain in a position servo is shown in the “simulink- model” in Figure 9-2. The feed forward gain is represented by the transfer function

Gff(s) = s/Kv, where Kv is the steady state gain in the control loop from feed forward -1 input to system output signal. In this case Kv = 20 sec and 1/Kv = 0.05 sec. The feed forward gain also includes a low-pass filter with a break frequency of 1000 rad/s (compare with the forward loop filter in Figure 9-1).

Figure 9-2: Simulink-model of a valve controlled cylinder with position feedback and feed forward gain.

The command signal in Figure 9-2 is a sine wave. The simulation results in Figure 9-3 shows that the output signal can follow the command signal with a very small phase lag. The oscillations at start, depends on the relatively low hydraulic damping (δh = 0.155) in the system.

Figure 9-3: Command and output signal with feed forward gain.

100 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

The effect of the feed forward gain can preferable be studied by plotting the output signal (Y) versus command signal (X), as illustrated in Figure 9-4.

Figure 9-4: Output versus command signal without (to the left) and with feed forward gain in a position servo with proportional control. A notable behaviour of the feed forward gain is that its action is like a pre-filter, which not affect the control loop gain and the stability margins.

9.3 PID Controller The Proportional–Integral–Derivative controller (PID controller) is a control loop feedback mechanism widely used in industrial control system. A PID controller attempts to correct the error between a measured system variable and a desired command signal by calculating and then outputting a corrective action that can adjust the process accordingly. A PID controller and its control algorithm are shown in Figure 9-5.

P_action

k={1} Output_Y Input_U I_action Sum Saturation +1 +1 + I +1 startTime={0.2} k={3} uMax={2}

D_action

DT1

k={0} 1 t dU (t) Y(t) = K ⋅U(t) + U (τ )dτ +T P T ∫ D dt I t0 Figure 9-5: PID Controller.

101 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

9K-E.3.1 Rydberg Proportional gain Controller design 4 K-E Rydberg Controller design 4 Proportional______gain is used for all tuning situations. It introduces a control signal that is ______proportional to the error signal. As proportional gain increases, the error decreases and the feedback signal tracks the command signal more closely, see Figure 9-6. Proportionalsystem response gain increases by boosting system the response effect of by the boosting error the signal. effect However, of the error too signal. much system response by boosting the effect of the error signal. However, too much However,proportional too gain much can proportional cause the system gain can to becomecause the unstable. system to become unstable. proportional gain can cause the system to become unstable. Command signal Output signal Command signal Output signal

Too low gain Optimal gain Too high gain Too low gain Optimal gain Too high gain Figure 6: Effects of proportional gain. FigureFigure 96:-6 :Effects Effects of of proportional proportional gain. gain. Integral gain Integral9.3.2 Integral gain gain With an integral control mode the error signal will be integrated over time, which With an integral control mode the error signal will be integrated over time, which Withimproves an integralmean level control response mode during the errordynamic signal operation. will be Integr integratedal gain over increases time, system which improves mean level response during dynamic operation. Integral gain increases system improvesresponse duringmean levelsteady resp stateonse or duringlow-frequency dynamic operation operation. and The maintain integral the gain means, increases value at response during steady state or low-frequency operation and maintain the mean value at systemhigh-frequency response operation. during steady The I-gainstate or adjustment low-frequency determines operation how and mu maintach timein itthe takes mean to high-frequency operation. The I-gain adjustment determines how much time it takes to valueimprove at high the -frequency mean level operation. accuracy. The Higher I-gain adjustmentintegral gain determines settings how increase much systemtime it improvetakes to improve the mean the mean level level accuracy. accuracy. Higher Higher integral integral gain gain settings settings increase increase systemsystem response, but too much gain can cause slow oscillations, as shown in Figure 7. response,response, butbut tootoo muchmuch gaingain cancan causecause slowslow oscillations,oscillations, asas shownshown inin FigureFigure 79.- 7. Command signal Output signal Command signal Output signal

OptimalOptimal gaingainHighHigh gaingain Too Too highhigh gaingain

Figure 7: Effects of integral gain. FigureFigure 97:-7 :Effects Effects of of integral integral gain. gain. The integrator output signal depends upon the I-gain and the input signal level, see The integrator output signal depends upon the I-gain and the input signal level, see TheFigure integrator 8. It is very output important signal, dependsto set a limit upon for the the I- gainoutput and signal, the input as shown signal in level,Figure see 8, FigureFigure 89.- 8It. isIt isvery very important important to toset set a limita limit for for the the output output signal, signal, as asshown shown in Figurein Figure 8, to prevent the integrator for “windup”. to9- 8,prevent to prevent the integrator the integrato for r“windup”. for “windup”.

Figure 9-8: Integrator action with different input signals. Figure 8: Integrator action with different input signals. Figure 8: Integrator action with different input signals.

An An “Anti-windup”“Anti-windup” implementationimplementation foforr aa PIDPID102 controllercontroller isis shownshown inin FigureFigure 99.. K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

An “Anti-windup” implementation for a PID controller is shown in Figure 9-9.

K-E Rydberg Controller design 5 ______

Figure 9-9: “Anti-windup” implementation for I-action in a PID controller.

The anti-windup implementation shown in Figure 8-9 is using the saturation level for the system to adjust the integrator gain so that the control signal stays inside the saturation margins. Figure 9: “Anti-windup” implementation for I-action in a PID controller.

9.3.3 Derivative gain Derivative gain With a derivative control mode the feedback signal means it anticipates the rate of changeWith a ofderivative the feedback control and mode slows the feedback the system signal response means at it high anticipates rates of the change. rate of Derivativechange of gain the provides feedback stability and slows and reduces the system noise responseat higher proportional at high rates gain of settings. change. TheDerivative D-gain gain tends provides to amplify stability noise and from reduces sensors noise and at to higher decrease proportional system response gain settings. when setThe is D-gain too high tends, see to Figure amplify 9- 10noise. Too from much sens derivativeors and to gain decrease can create system instability response at whenhigh frequencies.set is too high. Too much derivative gain can create instability at high frequencies. Overshoot

Ringing

Low rate Optimum rate Too much rate FigureFigure 9 10:-10 : Effects Effects of of derivative derivative gain gain

9 .3.4 Implementation and tuning of PID-controllers I literature there is a number of “thumb-rules” which can be used for tuning of PID- controllers. However, most of them are based on conventional control theory with the assumption that the system dynamics is linear. In all practical hydraulic applications, the system dynamics includes a number of non-linear properties such as, saturation, hysteresis, non-linear flow gain and non-linear friction characteristics. Before tuning of a PID-controller the implementation of the controller must be checked. One common controller implementation, described in the frequency domain is,

103 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

1 1 TDs Y (s) = KP ⋅ e(s)+ ⋅ ⋅ e(s)⋅ f (aw)+ ⋅ufb (s) T s TD I 1+ s N The above equation shows that the integrator gain is multiplied with the control error e(s) and an anti-windup function f(aw). Derivative action is only used on the feedback signal ufb(s) and then a low-pass filter is added where the time constant TD/N has to be adjusted to suitable limit of the derivative gain at high frequencies. As pointed out before, the tuning of a PID-controller for a hydraulic system is not a simple task. A first attempt for tuning a digital can be to use a method developed by Takahashi et al. [3]. Start to run the system with a simple P-controller and increase the gain until self- oscillation occurs. At this point the proportional gain is KC and the periodic time of oscillation TC. Now the parameters in the PID-controller can be tuned as follows,

TD = 0.075⋅ KC ⋅TC 1 K =1.2 ⋅ C TI TC

Ts KP = 0.60 ⋅ KC − 0.50 ⋅ TI where Ts = sampling time in [s].

9.4 A commercial digital controller M3000 – A Motion-Control Toolbox for Hydraulic Axes There are several advancedBy digitalDr.-Ing. C. controllers Boes and Dipl. on-Ing. theJ. We market.iblen One digital controller especially developed for hydraulicMoog multi GmbH,- axesBöblingen, control Germany is Moog M3000, described in ref. [4]. When compared to analog control systems, the implementation of nonlinear control In [4],structures the M3000 is relatively system easyis summarized in digital systems.as a system Figure that 3 iscontrasts optimally the configured frequency for automationresponses tasks of control in hydraulic valves with an danalog electrical and digital drive electronics. engineering. The M3000 system is designed to handle a number of transducer signals from each axis. The connection of one servo3.1 Hardwarevalve controlled & software axis structureto the M3000 -system is illustrated in Figure 9-11.

CAN / SSI / Encoder M12 x 5 D S

4 U M8 x 4 P 4 11+PE M8 x 4 U Signale P 11+PE + Supply

4 A B U S 2 5 P T CANopen M12 x 5 DDV 4 U M8 x 4 P

Elektronik Figure 4: Axis-Figurecontrol 9valve-11: M3000 connected to a valve-controlled axis, [4]

The axis-control valve's interfaces were configured in such a way that all major axis- control functions familiar from servohydraulics104 can be realized. As shown in Figure 4, the hydraulic axis can read-in the position and pressure signals of the cylinder chambers as well as the valve's supply pressure.

The depiction of the axis-control valve in Figure 5 demonstrates that additional interfaces would lead to cabling requirements that are no longer manageable under practical conditions. The axis's position signal can be read-in over a second CAN interface, an SSI interface, or an encoder interface. With this arrangement, the following control circuits are easy to establish:

- Position - Speed - Pressure - Differential pressure / force - Parallel

Prepared for the 4th International Fluidpower Conference Dresden, Germany on 25-26 March 2004. 4

M3000 – A Motion-Control Toolbox for Hydraulic Axes K-E Rydberg HydraulicBy Servo Dr.-Ing. Systems C. Boes – Dynamicand Dipl.- PropertiesIng. J. Weiblen and Control ______Moog GmbH, Böblingen, Germany "acceleration". The integrated feedback is activated only under certain conditions that the user must parameterize. It is possible to parameterize yielding and a delaying A controller design is based on block diagram form. The implementation of a position controllerfeedback forof thea valve controller-controlled fault for axis high is- frequencyshown in Figuredrives. 9-12.

Kra

Acceleration

Krv

Velocity

Kp, Tp

Kd, Td

X1 X0 Ki Uimax, Uimin

Kx Kpos, Kneg Umax, Umin Position Trajectory Reference Generator Position Valve Reference

1, Tfilt Tdiff Kv

Position Feedback

Tdiff Ka

Figure 6: StructureFigure 9- 1of2 :the Structure implemented of position position controller controller for a valve -controlled axis, [4]

ThisA major figure component illustrate that of theM3000 position has all controller the facilities is the needed trajectory to implement generator. a This controller block structureconditions like the the incoming module positionshown in set Figure points 9 in-1. such a way that a smooth path with limited speed and acceleration values will be specified.

3.2.2 Example "Pressure-control closed loop"

As shown in Figure 7, the pressure controller exhibits the structure /1//2/, as known from common sources. The controller consists of a PI core with a nonlinear anti-wind-up function and yielding feedback of the actual pressure value.

Although the controller structure shown here may appear to be very simple, in practical

situations there are often difficulties achieving the optimal setting of the Ki, Kd, and Td controller parameters.

Prepared for the 4th International Fluidpower Conference Dresden, Germany on 25-26 March 2004. 6

105 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

References [1] Herbert E. Merritt: Hydraulic Control Systems. ISBN: 978-0-471-59617-2. 368 pages. Wiley. January 1991. [2] MOOG: Electrohydraulic Valves – A Technical Look. MOOG Industrial Controls Division. East Aurora, NY, US [3] Y. Takahashi, C. S. Chan, D. M Auslander: Parametereinstellung bie Linearen DDC-Algorithmen. Regelungstchnik und Process-Daten- verarbeitung, Vol. 19, No. 6, 1971, pp. 237-284. [4] C. Boes, J. Weiblen: M3000 – A Motion-Control Toolbox for Hydraulic Axes. Moog GmbH, Böblingen, Germany. Technical paper, March 2004.

106 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

Appendix 1 DesignDesign of a linear positionof hydraulic servo position servo

Parameter values:

FLmax = 28 kN, Mt = 900 kg, vmax = 0,75 m/s, ps = 21 MPa, "e = 1000 MPa,

Cylinder stroke = 1,0 m, Position accuracy !Xp = 2,0 mm / 10 kN, Amplitud e margin: Am = 6,0 dB, Servo bandwidth: #b = 12 rad/s 5 Karl-Erik Rydberg, IEI/Linköping University a) Piston area and servo-valve size? Piston area? Assume steady state load conditions.

FL max Ap pL max = FL max " Ap = pLmax = ? pL max 3 28 !10 2 Select pLmax = 2/3 ps = 14 MPa " A = = 0,002 m p 14 !106 Valve size – flow capacity?

The servo valve nominal flow qN is given at !pv = (ps - pL) = 70 bar

1 Ap ! vmax qL = Cqwxv (ps ! pL ) qL = qN = KqiNiN = Ap ! vmax K = " qiN iN

0, 002 ! 0, 75 3 Nominal valve input current, iN = 50 mA " KqiN = = 0,03 m /As 0, 050 . 3 Nominal valve flow qN = 0,002 0,75 = 0,0015 m /s = 90 litre/min p 210 Max flow gain? p = 0: K = K " s = 0, 03 = 0,052 m3/As L qi0 qiN 70 70

6 Karl-Erik Rydberg, IEI/Linköping University

107 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

b1) Control loop stability and bandwidth

Required amplitude margin, Am = 6 dB. This requirement has to be compared with the req. on bandwidth, #b = 12 rad/s.

Valve bandwidth,

#v >> #h

K A 2010 log v Definition of Am: m = # # 2"h!h

Kqi0 A / 20 Am gives the req. on $ m Kv max = Ksa # K f = 10 #2#"h min #!h min steady state loop gain: Ap

7 Karl-Erik Rydberg, IEI/Linköping University b2) Control loop stability and bandwidth

Am = 6 dB $ Kv max = "h min #!h min

!bmax = Kv max ! !bmax = "h min #!h min where

2 4!e Ap Kce min !eMt Bp min Vt "h min = "h min = + M tVt Ap Vt 4Ap !eMt

!e Bp = 0 and Kcemin = Kc0 gives: "bmax = 2 # Kc0 # Vt V Required K -value: !bmax ! t c0 Kc0 = 2"e 12 0, 002 m5 Gives a leakage flow of ! "11 3 Kc0 = 6 =1, 2 !10 0,00025 m /s at 21 MPa, 2 !1000 !10 Ns a quite high value.

8 Karl-Erik Rydberg, IEI/Linköping University

108 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

b3) Control loop stability and bandwidth

Bp = 0 and Kcemin = Kc0 gives min resonance frequency and damping as:

4!1, 0 !109 ! 0, 0022 rad 1, 2 !10"11 1, 0 !109 ! 900 ! = = 94,3 !hmin = = 0,127 hmin 900 ! 0, 002 s 0, 002 0, 002 1 and ! !" = 94,3! 0,127 =12 (as stated) hmin hmin s

The control loop is stable with these values, but the damping, $hmin = 0,127 is very low, which gives a quite oscillative step response. Calculation of controller gain:

Ksa ! Kqi0 Kvmax ! Ap Kvmax = K f " Ksa = Ap Kqi0 ! K f

2 Ap = 0,002 m , Kvmax = 12 1/s, 12 " 0, 002 ! Ksa = = 0,046 A/V K = 0,052 m3/As, K = 10 V/m 0, 052 "10 qi0 f Karl-Erik Rydberg, IEI/Linköping University 9 c1) Steady state stiffness

Closed loop stiffness (including position feedback)!

Requirement: Max pos. error !Xp = 2 mm / !FL = 10 kN

2 2 " !F Ap # "F Ksa ! Kqi0 Ap S L K L = K ! = K ! K ! K ! A c s#0 = = v " f sa p0 f p !X p Kce "X p Ap Kce s#0 s$0

!pL Kq Where the valve pressure gain is: K p0 = = !xv Kc qL =0

9 Karl-Erik Rydberg, IEI/Linköping University

109 K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

c2) Steady state stiffness

Ksa ! Kqi0 Kv max ! Ap Kv max = K f " Ksa = and Ap Kqi0 ! K f

Steady state stiffness for Kce = Kc0 # "FL F A2 0, 0022 "X p !" L p 6 s$0 K 12 4 10 N / m K = v = !11 = $ p0 = [Pa/m] "X K 1, 2 $10 Ksa ! K f ! Ap p s#0 c0

. 6 Required stiffness is at least, 10000/0,002 = 5 10 N/m Since the required stiffness is higher than the calculated value the system design has to be improved to fulfilled the requirement on position accuracy. The problem can be solved by introducing a dynamic load pressure!"#$%&'()*+ feedback$ ,))'(-./0$12$#3).4*5"6&'*-731.$8)*95$8&04):0signal into the controller,$; see figure$ on next slide. <133$()$=1+=">-00$?134)*)'@$A4$35<$?*)B7)2.1)0$4=)$>*)007*)$?))'(-./$01+2-3$10$35<$-2'$ 4=)$ *)'7.4152$ 5?$ 4=)$ 04)-'&$ 04-4)$ 041??2)00$ <133$ ()$ 9)*&$ 35<$ .5:>-*)'$ 45$ >*5>5*4152-3$ 10 c3) Steady>*)007*)$?))'(-./@$ stateKarl-Erik stiffnessRydberg, IEI/Linköping University Dynamic pressure feedback

s/wf Kpf s/wf + 1 FL . - - x x uc i 1 1 p _1 p + Ksa Gv Kqi + Ap + - Vt PL Mt s s - Kce + s 4be Ap

K f $ $ The pressure,1+7*)$CD$E35./"'1-+*-:$5?$-$312)-*$>5014152$0)*95$<14=$'&2-:1.$>*)007*)$?))'(-./$FE feedback provides a frequency dependent>$G$HI Kc-value. At !"#$%&'()*+$ $ ,))'(-./0$12$#3).4*5"6&'*-731.$8)*95$8&04):0$;$ low frequencies Kce = Kc0 and for high # Kce > Kc0. !"# !"#$%&'()*+(##%,&")-.#"+&/(01*1()&0%+$(&21*3&"''4&5%%,6"'7& . -11 5 By usingA..)3)*-4152$?))'(-./$<5*/0$12$>*12.1>-3$-0$ a valve with Kc0 = 0,6 10 '&2-:1.$>*)007*)$?))'(-./@$J=)2$4=)$35-'$ m /Ns the steady state stiffness 04-*40$ 50.133-4)$ 4=)*)$. 6 <133$ ()$ -$ ?))'(-./$ 01+2-3K$ <=1.=$ 12.*)-0)$ 4=)$ =&'*-731.$ '-:>12+$ will increase to 8 10 N/m, which also giveTL space to handle threshold. L704$-4$4=)$*)052-2.)$?*)B7)2.&@$M=)$+55'$4=12+$<14=$-..)3)*-4152$?))'(-./$10$4=-4$4=)$ 04)-'&$04-4)$041??2)00$<133$254$()$-??).4)'@$A2$-2+73-*$K æ V >5014152$0)*95$<14=$-..)3)*-4152$ö The stiffness requirement can alsoce ç + be tsatisfied÷ by using ?))'(-./$10$0=5<2$12$!"#$%&'()$-2'$4=)$.5**)0>52'12+$(35./"2 ç1 s÷ '1-+*-:$10$)N>*)00)'$12$ Dm è 4be K ce ø acceleration!"#$%&'((@$ feedback: .• Kf - K 1 Xq p X p uc iv Kqiqi 1 m 1 qm K 2 + + sa sKac s 2d Dm + + + h s + s - - - A 1 w 2 1 p v w .. wh - G h reg qm qm Uc + V1 + . Acceleration feedback ps Kac s pL J T Position feedback t L K f $ V2 $ ,1+7*)$<<-=$>35./"'1-+*-:$5?$-2$-2+73-*$@5014152$0)*95$A14B$-..)3)*-4152$?))'(-./$C> $D$EF$ : 12 ,1+7*)$OHD$A2$-2+73-*$>5014152$0)*95$<14=$-..)3)*-4152$?))'(-./$FEKarl-Erik Rydberg, IEI/Linköping University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

,1+7*)$<<(=$O-:@12+$12$-2$-2+73-*$@5014152$0)*95$A14B$-..)3)*-4152$?))'(-./$C>:$D$EF$ $ !"# !"#$%&'()*""+,-%.)&/)0$1&'&$/)%$/'2$#)1"23$1) P*)007*)$-2'$-..)3)*-4152$?))'(-./$10$70)'$45$12.*)-0)$4B)$B&'*-731.$'-:@12+$-2'$4B10$

:-/)0$14$@5001(3)$45$12.*)-0)$4B)$04)-'&$04-4)$355@$+-12$!9$-2'$4B)$.350)'$355@$041??2)00$ A133$12.*)-0)M$Q254B)*$A-&$45$12.*)-0)$4B)$041??2)00$5?$-$@5014152$0)*95$10$45$124*5'7.)$-$ 9)35.14&$?))'(-./M$Q$(35./"'1-+*-:$5?$-$312)-*$@5014152$0)*95$A14B$9)35.14&$?))'(-./$10$ 0B5A2$12$!"#$%&'(*M$ K-E Rydberg Hydraulic Servo Systems – Dynamic Properties and Control ______

c4) Steady state stiffness and threshold

Position error (!Xp) according to valve threshold: Threshold: i = % of i to change # TH N !i = K K !X the main spool position TH f sa p

Accept a position error caused by threshold to !Xp = 0,0005 m:

! "iTH =10 # 0, 046 # 0, 0005 = 0, 00023 A

. !iTH = 0,00023 A is 0,0046 iN or 0,5 % of iN.

!iTH = 0,5 % of iN is a typical value for an advanced servo valve, so it is acceptable.

Total position error (!Xp,tot) is the sum of disturbance force (F ) error and threshold error: X X X L ! p,tot = ! p,FL + ! p,TH

. -11 5 Kc0 = 0,6 10 m /Ns gives: !X p,tot = 0, 00125+ 0, 0005 = 0, 00175 m = 1,75 mm

Karl-Erik Rydberg, IEI/Linköping University 13

111