Development and Control of an Electro-Hydraulic Actuator System for an Exoskeleton Robot

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Article Development and Control of an Electro-Hydraulic Actuator System for an Exoskeleton Robot

Dongyoung Lee, Buchun Song, Sang Yong Park and Yoon Su Baek * Department of Mechanical Engineering, Yonsei University, Seoul 03722, Korea; [email protected] (D.L.); [email protected] (B.S.); [email protected] (S.Y.P.) * Correspondence: [email protected]; Tel.: +82-2-2123-2827

Received: 25 July 2019; Accepted: 4 October 2019; Published: 12 October 2019

Abstract: Exoskeleton robots have been developed in various fields and are divided into electric and hydraulic exoskeletons according to the actuator type. In the case of hydraulic robots, because a unidirectional pump is applied, there are limitations to the wearer’s walking. In addition, robot systems are complicated, because a directional control valve is required to change the direction of the actuator. To solve these problems, we designed the electro-hydraulic actuator (EHA) system which has both the hydraulic and electric advantages. The EHA system consists of a hydraulic bidirectional pump, a motor, a hydraulic cylinder, and various valves. For the development of the piston pump, we analyzed the gait cycle and considered the flow rate and pulsation rate. In order to reduce the size and weight of the EHA system, the valves were made from one manifold, and the hydraulic circuit was simplified. We verified that the developed EHA system is applicable to robots through position and force control experiments. Because the hydraulic system is nonlinear, we designed a sliding mode control (SMC) and compared it with the proportional integral derivative (PID) controller.

Keywords: electro-hydraulic actuator; exoskeleton robot; sliding mode control

1. Introduction Exoskeleton robot has been developed in various fields of industrial and rehabilitation training applications. These robot systems are largely divided into muscle strengthening and muscle assisting systems, according to the purpose of use. Muscle-assisted exoskeleton robots are applied to rehabilitation training and welfare fields for paralyzed patients and elderly people to provide muscular strength. In other words, these exoskeleton robots assist people to walk without help. So, the actuator in the exoskeleton robot needs only power to support the weight of a wearer. Therefore, it can be operated and powered by most electric actuators [1–4]. Muscle-strengthening exoskeleton robots are used in industrial applications to carry heavy loads that are difficult to perform using human strength. They are classified into electric and hydraulic type considering their use in the robot. In the case of the electric type, it is a structure in which a motor and a reducer are combined, and it operates the robot by the power of the motor. A large reducer with a large gear ratio is installed additionally for high power, and each joint has a complicated structure [5–7]. However, this increases the noise, volume, and weight of the robot as well as its development costs. Also, the range of motion of the joint is limited and the wearer feels uncomfortable. Because the power is small compared to the hydraulic type, it is not applicable to the muscle strengthening exoskeleton robots. On the other hand, hydraulic types can produce higher power than the electric types and are inexpensive. Hydraulic systems consist of several hydraulic parts, such as servo valves, which control the unidirectional pump and flow rate [8,9]. However, hydraulic systems are complicated and thus difficult to maintain. Because the hydraulic pump is unidirectional, it must continue to operate, regardless of whether or not the hydraulic actuator operates. Because of this problem,

Appl. Sci. 2019, 9, 4295; doi:10.3390/app9204295 www.mdpi.com/journal/applsci Appl. Sci. 2019, 9, 4295 2 of 17 noise and vibration occur. Also, it consumes energy constantly, which lowers the energy efficiency. However, this has been studied in the applications of muscle-strengthening robots. This is because hydraulic pump robots are more powerful than electric type robots [9–11]. Most muscle strengthening exoskeleton robots use conventional hydraulic systems because of the high power ratio of torque and mass, and these systems use a unidirectional pump and a control valve. Because the pump cannot change direction, when the robot strengthens a muscle during a Sit to Stand in stand phase, an actuator is driven in active mode, but in other cases of a stand to sit in the swing phase, it is operated in passive mode. That is, for the wearer to bend the knee, a controller opens solenoid valves and the wearer has to bend the knee joint of the robot. This makes the wearer feel tired and has the disadvantage of the slow response [12–14]. Such conventional hydraulic systems are complicated because of solenoid valves and have problems such as leakages. In addition, many gear pumps apply to exoskeleton robots. The gear pumps have the limitation of high-speed rotation and have the disadvantage of large noise and vibration instead of a simple structure. It is also difficult to produce high pressures because of the many internal leakages [15–17]. Recently, in order to solve the problems of the existing hydraulic systems and electric system, an electro-hydraulic actuator (EHA) which combines electric and hydraulic systems was replaced [18]. This system has more power and higher energy-efficiency than the electric type. Moreover, the system is equipped with a bidirectional pump, and the speed and position of the hydraulic actuator can be controlled through a motor attached to the pump. This EHA provides a high level of power, has the advantages of a conventional hydraulic system, and solves the disadvantages of the hydraulic and electric types. Compared with the hydraulic system, the EHA system does not need a directional control valve. Because the hydraulic circuit is simple, it is easy to manage leakage, and manufacturing costs are reduced. Therefore, it has been applied to various fields [10,19,20]. Therefore, the EHA system applied to the exoskeleton robot can solve the problems of the electric exoskeleton robot and the hydraulic exoskeleton robot. Because the robot’s knee joint is equipped with a cylinder, not a motor, it does not need an additional reducer. Also, the hydraulic circuit is simple. As a result, the size and weight can be reduced, and the maintenance is easy. Therefore, in order to solve the problems of the conventional hydraulic exoskeleton robots, it is necessary to study an EHA that is applicable to robots with a bidirectional piston pump. In this paper, we developed an EHA system for an exoskeleton robot considering volume, size, and weight. We designed the controller and verified its performance. First, a bidirectional piston pump was designed to overcome the limitation of the gear pump. The piston pump is advantageous for high-speed rotation and can obtain high pressure compared with the gear pump. Because the wearer lifts heavy weight on the back and moves, the hydraulic system needs to be minimized in weight and volume. For this development, we analyzed the gait cycles of ordinary people using data relative to gait analysis. Based on the results, we determined the stroke length of a hydraulic cylinder and the discharge flow rate of the pump. Also, a hydraulic pump was designed considering the pulsation rate, which affects the performance of the hydraulic system. In addition, we designed the hydraulic circuit with a simple structure and made one manifold in many valves to make them smaller and lighter. Therefore, the EHA system was developed by integrating the motor, pump, and hydraulic valves into one system. Because this hydraulic system has many uncertainties, robust control is required to minimize the effect of nonlinearity. As a result, we proposed sliding mode control for this and designed the control based on the mathematical modeling of the EHA system. So, the flow of the paper is shown in Figure1 and its composition is as follows. The developed axial piston pump is presented in Section2, and the hydraulic circuit is presented in Section3. Additionally, we modeled the system mathematically, and this is presented in Section4. Finally, we designed a controller and verified the performance of the proposed EHA system via experiments, which is presented in Section5. Appl. Sci. 2019, 9, 4295 3 of 17 Appl. Sci. 20192019,, 99,, xx FORFOR PEERPEER REVIEWREVIEW 33 ofof 1717

FigureFigure 1. 1. TheThe composition composition and and flow flow of of a a paper. paper. 2. Design of Axial Piston Pump 2. Design of Axial Piston Pump 2.1. Analysis of Gait and Calculation of the Required Flow Rate 2.1. Analysis of Gait and Calculation of the Required Flow Rate Analyzing a wearer’s gait cycle is very important when developing an exoskeleton robot. This is Analyzing a wearer’s gait cycle is very important when developing an exoskeleton robot. This becauseAnalyzing the specifications a wearer’s ofgait an cycle actuator is very in the important robot can when be determined, developing and an aexoskeleton control algorithm robot. This and is because the specifications of an actuator in the robot can be determined, and a control algorithm isthe because intention the of specifications the wearer can of an be actuator grasped. in This the studyrobot aimscan be to determined, develop an EHAand a system control that algorithm can be and the intention of the wearer can be grasped. This study aims to develop an EHA system that can andapplied the tointention an exoskeleton of the wearer robot thatcan canbe grasped. move with Th ais 40 study kg mass. aims Forto develop this purpose, an EHA specifications system that of can the be applied to an exoskeleton robot that can move with a 40 kg mass. For this purpose, specifications bepump applied were to determined an exoskeleton by calculating robot that thecan flow move rate wi accordingth a 40 kg mass. to the For analysis. this purpose, As shown specifications in Figure2, of the pump were determined by calculating the flow rate according to the analysis. As shown in ofthe the gait pump cycle were is divided determined into the by stance calculating phase and the swingflow rate phase. according The start to ofthe the analysis. stance phase As shown is a heel in Figure 2, the gait cycle is divided into the stance phase and swing phase. The start of the stance phase Figurestrike, and2, the during gait cycle this is phase, divided the into exoskeleton the stance robot phase supports and swing the weight.phase. The This start stage of the occupies stance 60% phase of is a heel strike, and during this phase, the exoskeleton robot supports the weight. This stage occupies isthe a entireheel strike, walking and cycle. during After this that,phase, the the swing exoske phaseleton starts robot from supports the moment the weight. of toe-o Thisff stage[21]. occupies 60% of the entire walking cycle. After that, the swinging phasephase startsstarts fromfrom thethe momentmoment ofof toe-offtoe-off [21].[21].

FigureFigure 2. 2. TheThe gait gait cycle cycle is is divided divided into into th thee stance stance phase phase and and swing swing phase. phase.

AtAt swingswing phase, phase, the the exoskeleton exoskeleton robot robot moves moves with thewith wearer the wearer according according to the wearer’s to the intentions.wearer’s intentions.intentions.This phase ThisThis is the phasephase section isis thethe in sectionsection which theinin whichwhich robot thethe isunder robotrobot is forceis underunder control forceforceby controlcontrol interaction byby interactioninteraction with the withwith wearer. thethe wearer.The angular The velocityangular ofvelocity the knee of jointthe knee is the joint fastest is th ine this fastest phase. in this Therefore, phase. theTherefore, hydraulic the actuator hydraulic on actuatorthe knee on joint the should knee workjoint should accordingly. work Figureaccordin3 showsgly. Figure that the 3 shows angle and that angular the angle velocity and angular of each velocityjoint plotted of each on thejoint sagittal plotted plane on the based sagittal on theplan datae based of “gender on the data diﬀerences of “gender in three-dimensional differences in three- gait dimensionalanalysis data gait from analysis 98 healthy data Korean from 98 adults healthy [22].” Korean This gait adults study [22].” included This gait people study 175 included cm tall, and people one 175cycle cm in tall, the stageand one is 1.5 cycle s, expressed in the stage as 0%is 1.5 to 100%.s, expressed After 60%as 0% of to the 100%. walking After cycle, 60% the of swingthe walking phase cycle,starts (toe-othe swingﬀ). The phase angular starts velocity (toe-off). is aboutThe angular 210 deg velocity/s. When is theabout stance 210 phasedeg/s. starts When (at the the stance heel strike),phase starts (at the heel strike), the angular velocity is about −225 deg/s. The change in the angle can be seen Appl. Sci. 2019, 9, 4295 4 of 17

Appl.the angularSci. 2019,, 9 velocity,, xx FORFOR PEERPEER is about REVIEWREVIEW225 deg/s. The change in the angle can be seen from 10 to 65 degrees.4 of 17 − In order to be implemented for general people, the angular velocity of the robot knee should be about from 10 to 65 degrees. In order to be implemented for general people, the angular velocity of the robot from230 deg 10 /tos. 65 degrees. In order to be implemented for general people, the angular velocity of the robot knee should be about 230 deg/s.

(a) (b)

FigureFigure 3. 3. TheThe angle angle and and angular angular velocity velocity of of each each joint joint in in the the sagittal sagittal plane plane according according to to the the gait gait cycle: cycle: ((aa)) is is the the angle, angle, and and ( (bb)) is is the the angular angular velocity. velocity.

BeforeBefore developing developing an an EHA EHA system system for for an an exos exoskeletonkeleton robot, robot, we we assumed assumed that that the the hydraulic actuatoractuator applied applied only only to to the the robot’ robot’ss knee knee and and that the robot wearer’s height height is is 175 cm tall. The The knee knee ofof the the exoskeleton exoskeleton robot robot is is bent bent and and stretched stretched by by the the forward forward and and backward backward motion motion of of the the hydraulic hydraulic cylinder.cylinder. TheThe kneeknee joint, joint, cylinder cylinder body body connecting connecting portion, portion, and cylinderand cylinder rod connecting rod connecting portion portion formed formedtwo triangles, two triangles, as shown as by shown the simple by the drawing simple in drawing Figure4 a.in TheFigure lengths 4a. The of the lengths linksare of thel1 = links360 mmare , l2 ==36088 mm,,l 3l==81.288 mmmm,, andl = 81.2l4 = mm56.4, andmm respectively.l = 56.4 mm The respectively. angle of the The knee angle joint of the is φknee, and joint we l1 360 mm , l2 88 mm , l3 81.2 mm , and l4 56.4 mm respectively. The angle of the knee joint canφ see that the angle varies by 65 degrees (Figure3a). The resulting change in the cylinder rod can is φ , and we can see thatq the angle varies by 65 degrees (Figure 3a). The resulting change in the 2 2 2 2 be represented by dist = l + l + l + l +2222bc cos(α φ). We were able to obtain the change of the 1 2 3 =++++4 22221 ()α −φ cylinder rod can be represented by distllllbc1234− cos() 1 . We were able to obtain the actual length in the cylinder, and the maximum1234 rod length of the cylinder 1 was about 115 mm. Based on change of the actual length in the cylinder, and the maximum rod length of the cylinder was about changethis, the of cylinder the actual with length the 120 in mm the strokecylinder, length and was the appliedmaximum (Figure rod 4lengthb). As of a result, the cylinder it could was be driven about 115 mm. Based on this, the cylinder with the 120 mm stroke length was applied (Figure 4b). As a 115to the mm. angle Based of the on kneethis, jointthe cylinder for the wearer. with the 120 mm stroke length was applied (Figure 4b). As a result, it could be driven to the angle of the knee joint for the wearer.

(a) (b)

FigureFigure 4. 4. ((aa)) This This is is the the knee knee structure structure of of the the exoskeleton exoskeleton robot robot and and its its simplification. simpliﬁcation. ( (bb)) This This is is the the changechange in in length length of of th thee cylinder cylinder and and the the speed. speed.

The displacement volume of an axial piston pump differs depending on the stroke length of the piston and the number of pistons. As shown in Figure 5, the radius of the dotted line in the valve plate is R , the cam angle is α , and then a piston stroke length is L as shown below [23]: = α LR= 2tanα . (1) Appl. Sci. 2019, 9, x FOR PEER REVIEW 5 of 17

The cross-section area of the piston is A , the number of the piston is z , and the displacement volume is Vp expressed as follows: == α VzALzARp 2tan. (2)

When the piston rotates by θ from the bottom dead center (BDC), the piston moving is expressed as: LL xR=−cosαθ =() 1 − cos . (3) 22 Lω The velocity of the piston v becomes sinθ in Equation (3). ω is the angular velocity of 2 the cylinder block. At this time, the instantaneous discharge amount Qt of the pump is expressed as follows [23]:

zz 00ωωzzπ −1  π ==ALθθπ = AL 00 + QAvtisin i sin sin  / sin . (4) ii==1022zz  2 = If the number of pistons z is an even number, zz0 /2 and the discharge amount Qt is ωππ =−AL θ Qt cos / sin . (5) 2 zz =± If it is an odd number, zz0 (1)/2 and the discharge amount Qt is ωππ =−AL θ Qt cos / sin . (6) 42zz

If z is an even number, the cylinder block pulsates z times when it rotates once, but if it is ε ()− anAppl. odd Sci. number,2019, 9, 4295 it pulsates 2z times. The pulsation rate is expressed as QQQmax nim/ mean 5, and of 17 when z is an even number, it is as follows:

The displacement volume of an axialQQ piston− pumpππ diﬀers depending on the stroke length of the ε ==ttmax min tan . (7) piston and the number of pistons. As shownQzz intmean Figure5, the radius2 of the dotted line in the valve plate is R, the cam angle is α, and then a piston stroke length is L as shown below [23]: When it is an odd number, it is as follows:

QQL −= 2R tan αππ. (1) ε ==ttmax min tan . (8) Qzztmean 24

Figure 5.5. StructureStructure of of an an axial axial piston piston pump. pump. The The pump pump consists consists of a driveof a shaft,drive ashaft, valve a plate, valve a cylinderplate, a cylinderblock, pistons, block, pistonpistons, shoes, piston and shoes, a swash and plate.a swash plate.

The cross-section area of the piston is A, the number of the piston is z, and the displacement volume is Vp expressed as follows:

Vp = zAL = 2zAR tan α. (2)

When the piston rotates by θ from the bottom dead center (BDC), the piston moving is expressed as: L L x = R cos α = (1 cos θ). (3) 2 − 2 − Lω The velocity of the piston v becomes 2 sin θ in Equation (3). ω is the angular velocity of the cylinder block. At this time, the instantaneous discharge amount Qt of the pump is expressed as follows [23]:

z z X0 ALωX0 ALω z π z 1 π Q = Av = sin θ = sin 0 sin θ + 0 − π / sin . (4) t i 2 i 2 z z 2 i=1 i=0

If the number of pistons z is an even number, z0 = z/2 and the discharge amount Qt is

ALω π π Qt = cos θ / sin . (5) 2 − z z

If it is an odd number, z = (z 1)/2 and the discharge amount Qt is 0 ± ALω π π Qt = cos θ / sin . (6) 4 − 2z z Appl. Sci. 2019, 9, 4295 6 of 17

If z is an even number, the cylinder block pulsates z times when it rotates once, but if it is an odd number, it pulsates 2z times. The pulsation rate ε is expressed as (Qmax Q )/Qmean, and when z is − min an even number, it is as follows:

Qtmax Q π π ε = − tmin = tan . (7) Qtmean z 2z

When it is an odd number, it is as follows:

Qtmax Q π π ε = − tmin = tan . (8) Qtmean 2z 4z

The pulsation rate of the discharge amount according to the number of pistons is calculated as shown in Table1.

Table 1. Pulsation rate according to the number of pistons.

z 5 6 7 8 9 10 11 12 ε [%] 4.98 14.0 2.53 7.81 1.53 4.97 1.02 3.45

For the calculated pulsation rate ε, as the number of pistons increases, the pulsation rate becomes small. However, if it is an odd number, it becomes much smaller. Therefore, it is common to choose an odd number of pistons, in terms of the pulsation rate. In this study, nine pistons were selected considering the size of the piston pump. The inner diameter of each piston is 6 mm and the cam angle is 12 degrees. At this time, according to Equation (2) and the number of pistons, the displacement volume discharged is about 1.5 cc in one rotation. Considering the maximum speed of the cylinder rod, the required discharge ﬂow rate is about 113 cc/s. At this time, because the discharge amount of the pump in one revolution is about 1.5 cc, the pump speed had to be over 75.3 rev/s. That is, the speed of the motor needed to be 4520 rpm or more.

2.2. Design of the Piston Pump The EHA system for an exoskeleton robot is composed of a motor, a hydraulic pump, a hydraulic cylinder, an oil tank, and a hydraulic circuit (hydraulic lines, hydraulic valves, etc.). Depending on the performance of the hydraulic pump, the operation of the robot is affected seriously. In this study, the hydraulic pump is a bidirectional swash plate piston pump, and it is designed based on the required flow rate according to the wearer’s gait cycle. Piston pump has excellent characteristics in the high-pressure region and is suitable for high-speed rotation. This is widely used in hydraulic systems that can vary the output capacity and produce high power [23]. Piston pumps are also divided into swash plate type axial pumps and bent axis type axial pumps, according to the field of use. Therefore, it is applied in various fields and researched and developed [24,25]. Such a piston pump can overcome the limitations of the gear pump. However, piston pump is complex and has many components. First of all, we developed a piston pump to compensate for the disadvantage of the existing piston pump and to apply it to the exoskeleton robot and the EHA system. The pump was compacted by integrating the pump housing and valve plate, and was simplified by integrating the cylinder block, spherical washer, and spline connection. The structure of the piston pump is shown in Figure5, and the diameter and length are 56 mm and 81 mm, respectively. The hydraulic pump mainly consists of a drive shaft, a valve plate, a cylinder block, pistons, piston shoes, and a swash plate. The swash plate is at a 12 degree fixed angle. The nine holes in the cylinder block are formed at angles of 40 degrees. Through Equations (1) and (2), the stroke distance of the piston was designed to be 6.38 mm, and the displacement volume was about 1.5 cc. The suction and discharge of the hydraulic pump change according to the rotation direction of the motor. First, when the motor rotates the drive shaft of the pump, the cylinder block connected Appl. Sci. 2019, 9, 4295 7 of 17 to the drive shaft rotates. At this time, the piston shoe connected to the piston and ball socket joint moves along the surface of the cam, and the piston moves up and down in the cylinder block. As a result, the hydraulic pressure is produced, and then the pump starts sucking and discharging the hydraulic fluids. Because the valve plate and the cylinder block rotate in surface contact, they are made of different materials in consideration of friction. The cylinder block is made of brass and the valve plate is made of iron. Furthermore, when the piston pushes the hydraulic fluids, the oil film is formed between the cylinder block and the valve plate, so that the cylinder block gently rotates. For the same reason, the piston was made of brass, and the hole was made in the piston and shoe to produce an oil film. The valve plate and the cylinder block contact each other, and the pistons with shoes and the cam are shown in Figure6b (the inside of the proposed pump). Appl. Sci. 2019, 9, x FOR PEER REVIEW 7 of 17

(a) (b)

Figure 6. An oil film ﬁlm is is formed formed on on the the su surfacerface where where the the respective respective parts parts are are in in contact contact with with each each other. other. ((aa)) ThisThis isis aa sideside viewview ofof thethe pump,pump, andand ((bb)) thisthis isis thethe insideinside viewview ofof thethe proposedproposed pump.pump.

When thethe exoskeleton exoskeleton robot robot supports supports a mass a mass of 40 of kg, 40 the kg, internal the internal pressure pressure of the cylinder of the becomescylinder 2 12.73becomes kgf /12.73cm . Atkgf/cm this time,2 . At thethis force time, F the on oneforce piston F on one can piston be obtained can be as obtained follows: as follows:

FFPA==+=+PA tantanααμ+ µPA PA= ()( μµ + tantan αα) PAPA. (9) (9)

The piston area A A is is 0.2830.283 cm22,, andand thethe cam angle αα is is 12 12 degrees. degrees. Assuming Assuming thatthat thethe frictionfriction coecoefficientﬃcient betweenbetween brassbrass andand ironiron isis 0.06,0.06, thethe forceforce isis

=+() ×22 × = = F = (0.06F +0.060.213 0.213) 12.73 12.73 kgf/ kgf/cmcm2 0.283 0.283 cm 2cm= 0.984 0.984 kgf kgf= 9.65 9.65 N. N . × × In order to produce the above pressure, the piston must push the hydraulic fluids by the force In order to produce the above pressure, the piston must push the hydraulic ﬂuids by the force F. F. The rotation radius of the piston r is 15 mm, and 9 pistons are operated, but half of the pistons are The rotation radius of the piston r is 15 mm, and 9 pistons are operated, but half of the pistons are sucked and half are discharged simultaneously. Therefore, the required torque is 608 mNm. For this sucked and half are discharged simultaneously. Therefore, the required torque is 608 mNm. For this purpose, the Maxon 400 Watt, EC60 BLDC motor was selected for the pump motor. The specifications purpose, the Maxon 400 Watt, EC60 BLDC motor was selected for the pump motor. The speciﬁcations of the desired pump and the motor are shown in Table 2. of the desired pump and the motor are shown in Table2.

Table 2. The specifications of the desired pump and the motor.

Pump Components Unit Value Motor Parameters Unit Value Diameter of Pump mm 56 Nominal Voltage V 48 Length of Pump mm 81 No Load Speed rpm 5370 Number of Pistons EA 9 Nominal Speed rpm 4960 Diameter of Piston mm 6 Nominal Torque mNm 768 Stroke Length mm 6.37 Nominal Current A 9.56 Volumetric Displacement cc/rev 1.51 Torque Constant mNm/A 84.9 Cam angle degree 12 Speed Constant rpm/V 113

Table 2. The speciﬁcations of the desired pump and the motor.

3. Design of the Hydraulic Circuit A hydraulic system considering a hydraulic exoskeleton robot mostly uses a unidirectional pump. This system needs a relief valve, a check valve, a pilot check valve, and a hydraulic cylinder. Because the pump is unidirectional, it needs an additional servo valve for the cylinder to change direction. So, the operation of the exoskeleton robots using the hydraulic system is divided into the active mode and the passive mode. In other words, because it is a semi-active hydraulic system, the pump rotates unidirectionally and the cylinder advances to lift the weight. However, in order for the cylinder to move backward, the flow control valve is opened to move the cylinder backward by the force of the wearer [12]. In other words, when the robot performs the sit-to-stand operation, the robot increases the strength of the wearer via the actuator. Conversely, for the stand-to-sit operation in swing phase, the actuator moves manually. The cylinder moves backward through the directional control valve while the motor continues to operate [13]. In this case, the system becomes complicated, the efficiency becomes low, and it has problems with noise, vibration, and leakage. In order to overcome these drawbacks, we developed a bidirectional pump that is applicable to the hydraulic exoskeleton robot, and we also designed a hydraulic circuit. Figure7a shows the designed hydraulic circuit. The circuit consists of the bidirectional pump, two pilot-operated check valves, four check valves, two relief valves, and the oil tank. Q1 and Q2 are the actual flow rates flowing in and out of the cylinder. QR1 and QR2 are drained through the relief valve. For system safety, the relief valve is set so that the internal pressure does not exceed 25 bar. When a motor rotates, the pump produces pressure P1 and the cylinder moves forward. QT and QP are sucked through the pump from the tank, and unlike the double-rod cylinder, the cross-sectional areas A1 and A2 in the single-rod cylinder are different. So, the flow rates Q1 and Q2 are different. The pilot-operated check valve compensates for this difference in flow, allowing the cylinder to move gently [26]. The P1 pressure operates the pilot-operated check valve on the Q2 line, which results in Q2 being equal to QPCV2 and QPCV2 flowing into the tank. There are also check valves between the cylinder and the pump, so that the flow from the cylinder does not directly enter the pump. By designing the hydraulic circuit in this way, the hydraulic fluids circulate in the system but can be viewed as an open circuit. So, the mathematical modeling is simplified. Also, if the motor is not driven, the pilot-operated check valve is not opened, and the cylinder can maintain a constant force because of the check valve. Figure7b shows the actual hydraulic circuit; the circuit was modularized with a compact size of 130 100 70 mm3. Because of this, the hydraulic hose between the valves was × × eliminated. So, the circuit has been simplified, and maintenance of leaks has become easier. The oil tank was made of acrylic, and its size is 70 70 110 mm3. The cylinder’s maximum pressure was × × 3.5 MPa, the inner diameter was 20 mm, and the stroke length was 120 mm. Appl. Sci. 2019, 9, x FOR PEER REVIEW 8 of 17 control valve while the motor continues to operate [13]. In this case, the system becomes complicated, the efficiency becomes low, and it has problems with noise, vibration, and leakage. In order to overcome these drawbacks, we developed a bidirectional pump that is applicable to the hydraulic exoskeleton robot, and we also designed a hydraulic circuit. Figure 7a shows the designed hydraulic circuit. The circuit consists of the bidirectional pump, two pilot-operated check valves, four check valves, two relief valves, and the oil tank. Q1 and Q2 are the actual flow rates flowing in and out of the cylinder. QR1 and QR2 are drained through the relief valve. For system safety, the relief valve is set so that the internal pressure does not exceed 25 bar. When a motor rotates, the pump produces pressure P1 and the cylinder moves forward. QT and QP are sucked through the pump from the tank, and unlike the double-rod cylinder, the cross- sectional areas A1 and A2 in the single-rod cylinder are different. So, the flow rates Q1 and Q2 are different. The pilot-operated check valve compensates for this difference in flow, allowing the cylinder to move gently [26]. The P1 pressure operates the pilot-operated check valve on the Q2 line, which results in Q2 being equal to QPCV 2 and QPCV 2 flowing into the tank. There are also check valves between the cylinder and the pump, so that the flow from the cylinder does not directly enter the pump. By designing the hydraulic circuit in this way, the hydraulic fluids circulate in the system but can be viewed as an open circuit. So, the mathematical modeling is simplified. Also, if the motor is not driven, the pilot-operated check valve is not opened, and the cylinder can maintain a constant force because of the check valve. Figure 7b shows the actual hydraulic circuit; the circuit was modularized with a compact size of 130×× 100 70 mm3 . Because of this, the hydraulic hose between the valves was eliminated. So, the circuit has been simplified, and maintenance of leaks has become Appl. Sci. 2019, 9, 4295 9 of 17 easier. The oil tank was made of acrylic, and its size is 70×× 70 110 mm3 . The cylinder’s maximum pressure was 3.5 MPa, the inner diameter was 20 mm, and the stroke length was 120 mm.

(a) (b)

FigureFigure 7. 7. TheThe hydraulic hydraulic fluids ﬂuids circulate in thethe system:system: ((aa)) isis thethe proposedproposed hydraulichydraulic circuit, circuit, and and ( b(b)) is is a amanifold manifold of of the the hydraulic hydraulic circuit. circuit. 4. Modeling of the EHA System 4. Modeling of the EHA System By setting each chamber of the single rod hydraulic cylinder to the control volume and applying By setting each chamber of the single rod hydraulic cylinder to the control volume and applying the continuity equation for the pressure change, it is derived as follows [27]: the continuity equation for the pressure change, it is derived as follows [27]:

dV1 V1 dP1 Q1 = + Cip(P1 P2) dt βe dt − − . (10) dV2 V2 dP2 Q2 = + + Cip(P P2) CepP2 − dt βe dt 1 − −

V1 and V2 are the cylinder volumes, P1 and P2 are the cylinder pressures, Cip is the internal leakage coefficient of the piston, and Cep is the external leakage coefficient of the piston. β is the effective bulk modulus of the hydraulic fluid. After rewriting Equation (10) with respect to the change in pressure and arranging it, it is the same as Equation (11): dP1 βe dV1 = + Q1 Cip(P1 P2) dt V1 − dt − − . (11) dP2 βe dV2 = Q2 + Cip(P1 P2) CepP2 dt V2 − dt − − −

The volumes V1 and V2 of both chambers of the cylinder can be expressed as Equation (12) with respect to the initial volume: V = V + A x 1 01 1 L (12) V = V A xL 2 02 − 2 dP1 βe . = A xL + Q C (P P ) dt V01+A1xL 1 1 ip 1 2 .− − − . (13) dP2 βe ( ) dt = V A x A2xL Q2 + Cip P1 P2 CepP2 02− 2 L − − − when the cylinder moves forward, that is, when the electric motor rotates forward, the total ﬂow rate of the hydraulic circuit can be expressed as follows: ( QP = QT = Q Q 1 − R1 . (14) Q2 = QPCV2

In practice, the relief valves were not considered in this model, because they are safety devices and stay below the valve set during operation of the EHA system. So, QP = QT = Q1. Additionally, Appl. Sci. 2019, 9, 4295 10 of 17

QT can be expressed as Dωm k (P P ). D is the displacement of the pump, ωm is the motor − leakage 1 − 2 speed, and kleakage is the leakage coefficient of the pump. The flow rate QPCV through the pilot-operated check valve is expressed as shown in Equation (15), and the flow rate varies depending on the area Acv. As a result, the pressure–flow rate equation for the hydraulic circuit becomes a nonlinear function. The flow rate QPCV is determined according to the following equation [28]: s 2 ∆P Q = C A , (15) PCV D cv 1/4 · ρ · 2 2 ∆P + Pcr

where   Acv,leak for Pe Pmax  ≤ A (∆P) =  A + k (P P ) for P < P < P cv  cv,leak e crack crack e max  · − . Acv,leak for Pe Pmax Acv,max A ≥ ∆P = P P , k = − cv,leak 1 2 Pmax P − − crack CD is the flow discharge coefficient, Acv is the instantaneous orifice passage area, Aleak is the closed valve leakage area, Amax is the fully open valve passage area, Pcr is the minimum pressure for turbulent flow, Pe is the equivalent pressure differential across the control member, Pcrack is the valve cracking pressure, and Pmax is the pressure needed to fully open the valve. Using the above Equations (12) and (13) and the Laplace transform, we can obtain Equation (16): ! V s + C P (s) = A sXL(s) + Dω(s). (16) 2β ip 1 − 1 when the left chamber in the single rod is equal to the right chamber, V is the total volume and the volume V/2 is V01 + A1xL = V02 + A2xL. By referencing the hydraulic circuit of the EHA system shown in Figure7, the equation of motion of the load system can be described as

. .. A P A P BxL FL = MxL. (17) 1 1 − 2 2 − − when the cylinder moves forward, the pilot-operated check valve on the opposite side is opened by the pressure P1. Because the ﬂow rate QPCV2 is drained into the tank, we do not need to consider pressure P2. Assuming that there is no disturbance, Equation (17) is rearranged by the Laplace transform and Equation (18) is obtained as:

1 1 1 XL(s) = A P (s) = A P (S) . (18) 1 1 Ms2 + Bs 1 1 Ms + B · s

After arranging Equation (16) to determine the pressure and substituting it in Equation (18), the transfer function for the entire system can be determined, as shown in Equation (19). The block diagram is shown in Figure8:

XL(s) 1 A = D 1 . (19) A2 2 ω(s) V 1 · Ms + Bs 2β s + Cip + Ms+B

The hydraulic system has nonlinearity, as shown in Equations (10) and (15). Also, the temperature of the hydraulic ﬂuid changes because of the friction and pressure acting on the hydraulic cylinder and the hydraulic pump. Nonlinearity occurs because of the uncertainty of the modeling, the viscosity change, and the leakage. So, the EHA system must be robustly controlled against these nonlinearities and uncertainties [26,29]. Appl. Sci. 2019, 9, x FOR PEER REVIEW 10 of 17

A for PP≤  cv,max leak e ()Δ= +⋅− ( ) << AcvPA cv,max leak kPP e crack for P crack PP e  ≥ . Acv,max leak for PP e A − A Δ=PPP −, k = cv,max cv , leak 12 − PPmax crack

CD is the flow discharge coefficient, Acv is the instantaneous orifice passage area, Aleak is the

closed valve leakage area, Amax is the fully open valve passage area, Pcr is the minimum pressure

for turbulent flow, Pe is the equivalent pressure differential across the control member, Pcrack is the

valve cracking pressure, and Pmax is the pressure needed to fully open the valve. Using the above Equations (12) and (13) and the Laplace transform, we can obtain Equation (16):

V +=−+ω sCip Ps11() AsXs L () D () s. (16) 2β

when the left chamber in the single rod is equal to the right chamber, V is the total volume and the +=+ volume V /2 is VAxVAx01 1LL 02 2 . By referencing the hydraulic circuit of the EHA system shown in Figure 7, the equation of motion of the load system can be described as −−−= AP11 AP 22 BxLL F Mx L. (17)

when the cylinder moves forward, the pilot-operated check valve on the opposite side is opened by

the pressure P1 . Because the flow rate QPCV 2 is drained into the tank, we do not need to consider

pressure P2 . Assuming that there is no disturbance, Equation (17) is rearranged by the Laplace transform and Equation (18) is obtained as: 111 Xs()==⋅ APs () APS () . (18) L 11MsBs2 + 11 MsBs+ After arranging Equation (16) to determine the pressure and substituting it in Equation (18), the Appl. Sci. 2019, 9, 4295 11 of 17 transfer function for the entire system can be determined, as shown in Equation (19). The block diagram is shown in Figure 8:

Figure 8. The block diagram of the entire system.

5. The Position and Force Control Experiments Xs() 1 A L =⋅D 1 . 5.1. Design of the Sliding Mode Controlω()s A22Ms+ Bs (19) V ++ 1 sCip In order to apply a sliding mode control2β to theMs EHA+ B system, the nonlinear dynamic system is defined as follows: The hydraulic system has nonlinearity, as shown in Equations (10) and (15). Also, the x(n) = f (x) + b(x)u, (20) temperature of the hydraulic fluid changes because of the friction and pressure acting on the wherehydraulic the cylinder scalar andx isthe the hydrau outputlic pump. of interest, Nonlinearity the occurs scalar be ucause is the of the control uncertainty input, of andthe T h . (n 1) i x = x x ... x − is the state vector. One of the typical methods for a nonlinear system is sliding mode control. First, a sliding surface is defined, and the state (x) slides along the surface of the desired state (xd). Once on the sliding surface, s (t) = 0 is satisfied and the desired state is maintained. For the nonlinear control of the EHA system, the sliding surface can be defined as [30]:

n 1 d − s(x, t) = dt + λ ex n 1 , (21) d d − = c1e1 + c2 dt e2 + + cn n 1 en ··· dt − where, e = ex = x xd is the tracking error, and λ and c are the positive constants. In this study, because − the position, velocity, and acceleration of the single rod of the EHA system should be tracked, n is set to 3, and the sliding surface is deﬁned as follows:

. .. s = c1e + c2e + e. (22)

. . For s 0 to occur, the Lyapunov function V is set to V = 1 s2, and when V = ss 0, the current → 2 ≤ state does not deviate from the sliding surface. The control input u consists of the equivalent control input ueq for entering the sliding surface and the switching control input sgn(s) for not going out of the surface at the sliding surface (s = 0). However, the function sgn(s), which is the nonlinear function, generates chattering because of the discontinuous response. To avoid this, we replaced it with a saturation function. So, the control input is ( s, s < 1 u = ueq Ks sat(s), sat(s) = | | . (23) − · sgn(s), s 1 | | ≥ Assuming that the cylinder rod moves forward, the equation diﬀerentiating Equation (16) can be rearranged with respect to time for x as follows: ! ... 1 . .. dF x = AP Bx d . (24) M − − dt

By substituting Equation (13) in Equation (24), we can obtain Equation (25):

2 ... B .. 2A β . 2Aβ 2Aβ x = x x + Dω C P + d. (25) −M − MV MV − MV ip Appl. Sci. 2019, 9, 4295 12 of 17 Appl. Sci. 2019, 9, x FOR PEER REVIEW 12 of 17

2Aβ If ω in this equation is replaced by u, and summarized2 ββ as , then the equation is as follows: =−BA −22 + A()MV − x xxDud  e , (26) MMVMV2 ... B .. 2A β . 2Aβ x = x x + D(u de), (26) where, de includes the disturbance −d,M volumetric− MV elasticMV modulus,− area, volume, and so on. where, de includes the disturbance d, volumetric=++= elastic modulus, area, volume, and so on. scecee12    0 , (27) ...... = + + = where the control input u satisfyings Equationc1e c2e (26)e is the0, equivalent control input ueq . It can (27) be whereobtained the through control inputEquationsu satisfying (26) and Equation (27). (26) is the equivalent control input ueq. It can be obtained through Equations=++ (26) and (27). scecee12 . 2 βββ...... (28) s =−+−++−()()   222ABADAD= c e + c e + e + +=  cx12dd x cx x x1 2 x u eqed d x 0 2 . . .. .. MV2A β . MB .. MV2AβD 2 MVAβD ... (28) = c x x + c x x + x + x ueq + de + x = 0 1 d − 2 d − MV M − MV MV d MV2 A2 β  B ucxcxdcxcxx=++++−−+( 2 !      ,) eqMVβ 2A β 1212. B .. e. d.. d... d (29) ueq = 2AD MV+ c1 x +  M+ c2 x + de c1xd c2xd + x d , (29) 2AβD MV M − − × where dde is is the the disturbance disturbance that that cannot cannot actuallyactually bebe measured and is is set set by by KKep e because because the state e p × value mustmust gogo intointo thethe sliding sliding surface. surface. Finally, Finally, the the sliding sliding mode mode control control law law is shown is shown by Equationby Equation (28) which(28) which was was designed designed using using Equations Equations (22) and(22) (23):and (23):

( 2 2 ! ) MV MV2Aβ2 Aβ . B  B ...... . .. u = ucxcxKecxcxxKsatcecee=++++−−+−⋅+++ c x + + c x + K pe c x c x + x Ks sat ()c e + c e + e .. (30) A D β MV1 121212M 2 pddds1 d 2 d d 1 2 (30) 2 β2AD MV M − − − · 5.2. Experiment Configuration and Control Experiments 5.2. Experiment Configuration and Control Experiments This chapter explains the experiment configuration and control experiments. The experiment configurationThis chapter consisted explains of thethe hydraulic experiment pump, configur the manifoldation and of control the hydraulic experiments. circuit, The a double experiment acting configurationsingle rod cylinder, consisted a linear of the variablehydraulic di pump,fferential the transformermanifold of the (LVDT), hydraulic a load circuit, cell, a double motor driver,acting singleand an rod embedded cylinder, system a linear for variable control anddifferential data acquisition. transformer The (LVDT), experimental a load setup cell, isa motor shown driver, in Figure and9. Thean embedded embedded system system for consisted control ofand the data control acquisition. board and The the experimental data acquisition setup board. is shown Each in board Figure was 9. Theequipped embedded with system controller consisted area network of the control (CAN) board communication, and the data the acquisition serial peripheral board. Each interface board (SPI)was equippedcommunication, with controller amplifier, area analog network filter, and(CAN) the 200commun MHzication, microcontroller the serial unit peripheral (MCU). interface The single (SPI) rod communication,cylinder for the experiment amplifier, analog had a 120filter, mm and stroke the 200 length. MHz The microcontroller inner diameter unit was (MCU). 20 mm, The and single the rod cylinderdiameter for was the 10 experiment mm. The LVDT, had awhich 120 mm measures stroke le thength. position The inner of the diameter cylinder was rod, 20 had mm, a stroke and the length rod diameterof 150 mm. was The 10 loadmm. cellThe thatLVDT, measures which themeasures reaction the force position of the of cylinderthe cylinder could rod, measure had a stroke up to 100length kg ofand 150 was mm. installed The load at thecell endthat of measures the cylinder. the reaction force of the cylinder could measure up to 100 kg and was installed at the end of the cylinder.

Figure 9. All experimentsexperiments used used a loada load cell, cell, hydraulic hydraulic cylinder, cylinder, motor, motor, pump, pump, manifold, manifold, and control and system.control system. Appl. Sci. 2019, 9, 4295 13 of 17

Appl. Sci. 2019, 9, x FOR PEER REVIEW 13 of 17 Meanwhile, when the exoskeleton robot operates in the stance phase, the robot performs position Meanwhile, when the exoskeleton robot operates in the stance phase, the robot performs control, and when operating in the swing phase, the robot grasps the wearer’s intention and performs position control, and when operating in the swing phase, the robot grasps the wearer’s intention and force control. And speed control loop is generally contained within the position control loop. performs force control. And speed control loop is generally contained within the position control So, the veriﬁcation of the hydraulic pump and the hydraulic circuit were performed through position loop. So, the verification of the hydraulic pump and the hydraulic circuit were performed through and force tracking. Also, the designed sliding mode controller was veriﬁed in comparison with a position and force tracking. Also, the designed sliding mode controller was verified in comparison proportional integral derivative (PID) controller. with a proportional integral derivative (PID) controller. The reference input for tracking of the position was a sine wave, and the PID control and sliding The reference input for tracking of the position was a sine wave, and the PID control and sliding mode control were compared through experiments. First, we simulated them with Matlab Simulink, mode control were compared through experiments. First, we simulated them with Matlab Simulink, as shown in Figure 10, and the results are shown in Figure 11. Table3 shows the parameters of the as shown in Figure 10, and the results are shown in Figure 11. Table 3 shows the parameters of the cylinder and pump, and Table4 shows the PID control and sliding mode control parameter values for cylinder and pump, and Table 4 shows the PID control and sliding mode control parameter values each experiment. for each experiment.

Appl. Sci. 2019, 9, x FOR PEER REVIEWFigure 10. Control simulation with Matlab Simulink. 14 of 17 Figure 10. Control simulation with Matlab Simulink.

Table 3. The parameters of the cylinder and pump.

Parameter Unit Value 2 AA1 , cm 3.1416 Kg M 0.2 Kg D cc/rev 1.5 B N/(cm/s) 0.1 β 2 9 Kg/cm 1.7× 10 3 VVt , cc, cm 32.32 C ()() ip cm3/s / N / cm 2 0.001 (a) (b)

FigureFigure 11.11. SimulationSimulation result result of theof slidingthe sliding mode mode control control scheme: scheme: (a) is the ( resulta) is ofthe position result performance,of position performance,and (b) is the and speed (b) ofTable is the the motor. 4.speed The parametersof the motor. of the control for each experiment.

Figure 12 gives a comparisonPositionTable Control 3. Theof the parameters SMC (slidi of theng cylinder mode andForcecontrol) pump. Control and the PID control for the SMC1 KK==200, 15 SMC1 KK==20, 15 experimental equipment.Parameter In theps simulation, the position Unit reference inputps value Value was the sine wave. At == == this time, the SMC2control periodKK wasps100, set to 1015 ms. As in2 SMC2 the simulation, KKps c20, and c1.5 were set to be the A1, A cm 1 3.1416 2 KKKM==200, 0.02, ==0.1Kg = KK==40, 0.2 1.5 = same, and in FigurePID 12a,pi SMC 1 was set to dKp 200 SMC3 and Ks 15 , psSMC 2 was set to Kp 100 and D cc/rev 1.5 = = = PID1 = KK==20, 0.001, K =1 Ks 15 , and PID was setB to Kp 200 , Ki 0.02N/(cm, and/s) Kd 0.1pi. This shows 0.1 that the d error increased 2 9 β Kg/cm KK==30, 1.70.001,10 K =1 at 1 s and 5 s for all three controls. This is because the3PID2 moving pidistance of the× cylinder d per time was Vt, V cc, cm 32.32 12bit − 3 2 short and the resolutionC (ip120 mm/2 1 )cm of the/s / LVDTN/cm with the stroke length0.001 of 120 mm was low. In otherAs shownwords, inthe Figure LVDT 11a, error we occurred can see becausethat the ofcy thelinder characteristics followed the of sinethe analogwave well. system Figure (system 11b noiseshows and the the speed measurement of the motor. noise), Because which the increased initial position the rate of of the change cylinder in a wasvery 0 short mm, time.the initial Thus, rpm the controlof the motor response was was too alarge little toslow, follo andw the the reference position trackinginput. After was thenot currentgood at state1, 3, andreached 5 s. To the improve sliding thesurface, control we performance,can see that control it would value be switchedbetter to usein succession high-resolution for continuous LVDT or tracking. to increase the control period of the system.

(a) (b)

Figure 12. Comparisons of the experimental results: (a) is the position control performance of the SMC (sliding motor control) and the PID control, (b) is a comparison of errors.

On the other hand, when the moving speed of the cylinder was high, this problem did not occur in the sliding mode control and PID control. However, PID had a larger error than sliding mode control when tracking the position (Figure 12b). The PID controls showed good follow-up, but it is unstable and slower than the designed SMC 1. In case of PID control, the error value has some positive offset in 0–1 s, 3–5 s, and 7–8 s intervals. This is because the applied cylinder is a single rod, not a double rod. In other words, because the volumes of both chambers are different, when the same Appl. Sci. 2019, 9, x FOR PEER REVIEW 14 of 17

Appl. Sci. 2019, 9, 4295 14 of 17

Table 4. The parameters of the control for each experiment.

Position Control Force Control

SMC1 Kp = 200, Ks = 15 SMC1 Kp = 20, Ks = 15 SMC2 Kp = 100, Ks = 15 SMC2 Kp = 20, Ks = 1.5 PID Kp = 200, Ki = 0.02, Kd = 0.1 SMC3 Kp = 40, Ks = 1.5 PID1 Kp = 20, Ki = 0.001, Kd = 1 PID2 Kp = 30, Ki = 0.001, Kd = 1

(a) (b) As shown in Figure 11a, we can see that the cylinder followed the sine wave well. Figure 11b showsFigure the speed11. Simulation of the motor. result Because of the sliding the initial mode position control of scheme: the cylinder (a) is wasthe 0result mm, of the position initial rpm of theperformance, motor was and too ( largeb) is the to speed follow of the the referencemotor. input. After the current state reached the sliding surface, we can see that control value switched in succession for continuous tracking. Figure 12 gives a comparison of the SMC (sliding mode control) and the PID control for the Figure 12 gives a comparison of the SMC (sliding mode control) and the PID control for the experimental equipment. In the simulation, the position reference input value was the sine wave. At experimental equipment. In the simulation, the position reference input value was the sine wave. this time, the control period was set to 10 ms. As in the simulation, c1 and c2 were set to be the At this time, the control period was set to 10 ms. As in the simulation, c1 and c2 were set to be the same, = = = andsame, in and Figure in Figure 12a, SMC 12a, 1SMC was 1 set was to Ksetp =to 200Kp and 200Ks and= 15 ,K SMCs 15 2, wasSMC set 2 was to K pset= to100 Kandp 100Ks = and15, and= PID was set to Kp = 200, Ki = 0.02, and=Kd = 0.1. This= shows that the error increased at 1 s and Ks 15 , and PID was set to Kp 200 , Ki 0.02 , and Kd 0.1 . This shows that the error increased 5 s for all three controls. This is because the moving distance of the cylinder per time was short and theat 1 resolutions and 5 s for (120 all mmthree/2 12bitcontrols.1) ofThis the is LVDT becaus withe the the moving stroke distance length of of 120 the mmcylinder was low.per time In other was − 12bit words,short and the the LVDT resolution error occurred ( 120 mm/2 because− of 1 ) theof the characteristics LVDT withof the the stroke analog length system of (system120 mm noisewas low. and theIn other measurement words, the noise), LVDT which error occurred increased because the rate of of the change characteristics in a very shortof the time.analog Thus, system the (system control responsenoise and wasthe measurement a little slow, and noise), the which position increased tracking the was rate not of goodchange at in 1, a 3, very and short 5 s. To time. improve Thus, the control performance,response was ita little would slow, be better and the to useposition high-resolution tracking was LVDT not good or to at increase 1, 3, and the 5 controls. To improve period ofthe the control system. performance, it would be better to use high-resolution LVDT or to increase the control period of the system.

(a) (b)

Figure 12. ComparisonsComparisons of the experimental results: ( a) is is the the position position control control performance of the SMC (sliding motormotor control)control) andand thethe PIDPID control,control, ((bb)) isis aa comparisoncomparison ofof errors.errors.

On the otherother hand,hand, whenwhen thethe moving moving speed speed of of the the cylinder cylinder was was high, high, this this problem problem did did not not occur occur in thein the sliding sliding mode mode control control and and PID control.PID control. However, However, PID had PID a had larger a larger error than error sliding than sliding mode control mode whencontrol tracking when tracking the position the position (Figure 12(Figureb). The 12b). PID Th controlse PID controls showed goodshowed follow-up, good follow-up, but it is unstablebut it is andunstable slower and than slower the designed than the SMCdesigned 1. In caseSMC of 1. PID In control,case of PID the errorcontrol, value the has error some value positive has osomeffset inpositive 0–1 s, offset 3–5 s, in and 0–1 7–8 s, s3–5 intervals. s, and 7–8 This s intervals. is because This the is applied because cylinder the applied is a singlecylinder rod, is nota single a double rod, rod.not a Indouble other rod. words, In other because words, the because volumes the of volumes both chambers of both chambers are different, are different, when the when same the amount same of flow rate flows into the cylinder, a speed difference of the forward and backward movement of the rod occurs. In order to follow the sine wave, the motor must rotate at high speed in a section of 0–1, 3–5, 7–8 s. At this time, the limit of the PID controller was shown in this section. In terms of the Appl. Sci. 2019, 9, x FOR PEER REVIEW 15 of 17 amount of flow rate flows into the cylinder, a speed difference of the forward and backward movement of the rod occurs. In order to follow the sine wave, the motor must rotate at high speed in a section of 0–1, 3–5, 7–8 s. At this time, the limit of the PID controller was shown in this section. In terms of the performance of the controllers, SMC 1 was more accurate than SMC 2 in terms of Appl. Sci. 2019, 9, 4295 15 of 17 following the desired input into the sliding surface. When the state variables enter the sliding surface, the control input sgn(s ) is switched to the magnitude of Ks . If the larger disturbance enters the performance of the controllers, SMC 1 was more accurate than SMC 2 in terms of following the desired system, it is temporarily outside of the surface boundary. Theoretically, a larger K value can be input into the sliding surface. When the state variables enter the sliding surface, thes control input given to prevent this, but when applied to the actual system, a large vibration is generated. In the sgn(s) is switched to the magnitude of Ks. If the larger disturbance enters the system, it is temporarily experiment, the average error of SMC 1 was 0.16 mm, which is a satisfactory value for a practical outside of the surface boundary. Theoretically, a larger Ks value can be given to prevent this, but when system.applied to the actual system, a large vibration is generated. In the experiment, the average error of SMCFigure 1 was 0.1613 shows mm, whichthe results is a satisfactoryof the cylinder value trac forking a practical the desired system. force through the sliding mode controlFigure and 13the shows PID control. the results The of cylinder the cylinder followed tracking forces the of desired 60, 90, forceand 120 through N at the2 s slidingintervals. mode As = = = showncontrol andin Figure the PID 13a, control. SMC The1 was cylinder set to followed Kp 20 forces andof K 60,s 90,15 and, SMC 120 2 N was at 2 sset intervals. to Kp As20 shown and in Figure= 13a, SMC 1 was set to Kp == 20 and Ks = =15, SMC 2 was set to Kp = 20 and Ks = 1.5, and= SMC Ks 1.5 , and SMC 3 was set to Kp 40 and Ks 1.5 . Overshoot occurred in the system at Kp 40 3 was set to Kp = 40 and Ks = 1.5. Overshoot occurred in the system at Kp = 40, and the system became , and the system became unstable. Also, chattering occurred when K = 15 even with the same K unstable. Also, chattering occurred when Ks = 15 even with the same Ksp value. Therefore, the cylinderp = = value.was most Therefore, stable when the cylinderKp = 20 andwasK mosts = 1.5 stable(SMC when 2). Figure Kp 1320b showsand K as comparison1.5 (SMC between2). Figure SMC 13b 2 showsand the a PID1,2comparison control. between The PID SMC control 2 and showed the PID1,2 good control. follow-up The butPID was control unstable showed and good less accurate follow- upthan but the was designed unstable SMC and 2. less accurate than the designed SMC 2.

(a) (b)

Figure 13. ComparisonsComparisons of of the the experimental experimental results: results: ( (aa)) is is the the force force control control pe performancerformance of of SMC, SMC, and and ((b) is a comparison of SMC and PID control.

6. Discussion Discussion InIn thisthis study,study, we we developed developed the the EHA EHA system system which which combines combines the advantages the adva ofntages hydraulic of hydraulic actuators actuatorsand electric and actuators. electric actuators. The pump The design pump of thedesi EHAgn of system the EHA was system determined was determined through human through gait humananalysis gait and analysis design and of an design exoskeleton of an exoskeleton robot. In addition,robot. In addition, the number the ofnu pistonsmber of waspistons set was to nine, set toin nine, consideration in consideration of the pulsation of the pulsation rate and rate pump and pump size. Atsize. this At time, this time, the pump the pump was was designed designed and andmanufactured manufactured by a by bidirectional a bidirectional swash swash piston piston pump. pump. In addition, In additi severalon, several valves valves were were made made into into one onemanifold, manifold, and theand circuit the circuit was simplified was simplified to make to itmake smaller it smaller and lighter. and Thelighter. controller The controller was designed was designedbased on thebased mathematical on the mathematical modeling ofmodeling the EHA of system. the EHA Because system. hydraulic Because systems hydraulic have systems nonlinearity have nonlinearityand system uncertaintyand system characteristics,uncertainty characteristic the slidings, mode the sliding controller mode was controller designed. was We designed. verified We the verifiedperformance the performance of the developed of the developed EHA system EHA through system positionthrough andposition force and control. force control. So, in thisSo, in study, this the control parameters were set to Kp = 200=and Ks = 15 in= order to enter the sliding surface when study, the control parameters were set to Kp 200 and Ks 15 in order to enter the sliding surface tracking the position with respect to the sine wave. To track the desired force, the control parameters when tracking the position with respect to the sine wave. To track the desired force, the control were set to Kp = 20 and Ks = 1.5. It can be seen that the overshoot occurs as the control parameter Kp = = parametersincreases in were each controlset to K experiment.p 20 and AsKsKs 1.5is larger,. It can state be seen variable that doesthe overshoot not deviate occurs the sliding as the surface,control but the system becomes unstable. Rather than enlarging the control parameters for the fast response of the system, it should be determined considering the stability of the system. Furthermore, compared with the PID control, we found that the sliding mode control performed better in a nonlinear system. The EHA was operated by controlling the speed and force of the cylinder. When the actual cylinder is not moving, the motor is not driven and the forward and backward movement of the cylinder is Appl. Sci. 2019, 9, 4295 16 of 17 controlled by the rotation of the motor. This exoskeleton robot with a conventional unidirectional pump can increase the efficiency by overcoming the disadvantage that the motor must continue to operate regardless of the operation of the cylinder. In addition, when the knee joint of the robot is bent in the swing phase, it is possible to reduce the burden on the wearer by driving the motor instead of the wearer’s force. The hydraulic circuit was designed with consideration of the walking operation of the robot and compacted by integrating them. Therefore, the proposed EHA system applied to the exoskeleton robot can solve the disadvantages of the existing robot which is used by a conventional unidirectional pump. The design and control of the exoskeleton robot to which this system is applied should be continuously researched.

Author Contributions: D.L. designed the pump, EHA system, and controller, and wrote the paper; D.L. and B.S. contributed to conceptualization, and performed the experiments; B.S. and S.Y.P. analyzed the data and reviewed and edited the paper; Y.S.B. contributed funding acquisition and project administration. Funding: This research was supported by the Basic Science Research Program grant funded by the National Research Foundation of Korea (NRF) (No. 2017R1A2B4009265). Conﬂicts of Interest: The authors declare no conﬂict of interest.

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