Mechatronic Servo System Control - M. Nakamura S. Goto and N. Kyura Part 3 doc

2.14 th OrderM od el of One Axis in aM ec hatronic Serv oS ystem 21 K J ssN N K J s s D p M L L L - K v T θ U L L g G G 1111 1 1 S e rvo c ontroller 2 m a ss model Fig. 2.2. Block diagram of 4th order mo del of industrial mechatronic servosystem T M ( s )=K g v [ K p { U ( s ) − θ M ( s ) }−sθ M ( s )] (2.7) where T M ( s )inequation (2.7)denotes the torque generated fr om the motor. The first item of (2.7)isthe transferfunctionofthe servocontroller.The second item expressesthe influence of thereactionforce T L ( s ). U ( s )isthe angle input to the motor. K p is position loop gain. K g v is velocityamplifier gain. The transfer fu nction from the angle input U ( s )for themotor of the whole mechatronic servosystem to the angle output θ L ( s )ofthe load can be written as below, when deriving the relation equation between U ( s )and θ L ( s ) by eliminating θ M ( s ), T L ( s ), T M ( s )fromfourrelationequation (2.4) ∼ (2.7) with fivevariables U ( s ), θ L ( s ), θ M ( s ), T L ( s ), T M ( s )( refert oF ig. 2.2). G ( s )= a 0 N G ( s 4 + a 3 s 3 + a 2 s 2 + a 1 s + a 0 ) (2.8) a 0 = K L K p K g v J L J M a 1 = K L K g v J L J M + D L K p K g v J L J M + D L K L N 2 G J L J M a 2 = K L J L + D L K g v J L J M + K p K g v J M + K L N 2 G J M a 3 = D L J L + K g v J M . This 4th order mo del of am ec hatronic serv os ystem can be effectiv ely adopted in the developmentofservoparameterdeterminationorcontrol strategy. In theactual mechatronic servosystem, for changingvelocitycontroller as PI controller, it is as shownstrictly in the block diagram of Fig. 1.1. To this controller, in the 4th order model of Fig. 2.2, velocitycontroller is expressed by an equivalent Pcontrol. Theintegral(I) actioninvelocitycontroller in the actualmechatronic servosystem is performedfor torqu edisturbance com- pensation.The time shift of output response is nominatedbythe gain of P 22 2M athematical Mo del Construction of aM ec hatronic Serv oS ystem control. On above way ,the ratiogain K s v of PI controlinthe general motion of an actualsystem is not the velocityamplifier gain K g v in the model of Fig. 2.2, butisexpressed by the ratio gain when PI Controller is equivalenttothe P control. (2) Normalized 4th Order Mo del for Servo Pa rameter Determination The parameters of the serv oc on troller in the4 th order mo de (2.8)a re po sition lo op gain K p andv elo cit ya mplifier gain K g v .C oncerning the ve lo cit ya mplifier gain K g v ,the totalinertialmoment transformed from the motoraxis with a rigid connection is assumed as J T = J M + J L N 2 G . (2.9) K v is defined as the velo cityloopgain by using this J T as K v = K g v J T . (2.10) This velocityloopgain is regarded as aservoparameter. Hence,position loop gain K p andvelocityloopgain K v hasthe same order forusing later. In addition, in equation (2.8), by viscous friction coefficient D L ,spring constant K L andload momentofinertia J L ,the naturalangularfrequency ω L and damping factor ζ L expressed by the features of mechanism part is written as ω L =  K L J L (2.11 a ) ζ L = D L 2 √ J L K L . (2.11 b ) When expressing the general features of the mechanism part, for convenient expression by naturalangularfrequency ω L anddampingfactor ζ L with vis- cous friction coefficient D L ands pring constan t K L , ω L and ζ L area dopted as theparameters of themechanism part. The 4th order model derived in the last part is determined by the natur al angular frequency ω L anddampingfact or ζ L as thefeatures of the mecha- nism part, as well as the servoparameter K p , K v .H owe ve r, since the natural angular frequencyofthe mechanismparthas astrong dependence on its size or mass, it is expected that the standard determination of servoparameters is notbasedonthe naturalangularfrequencyofthe mechanismpart. Therefore, theposition loop gain K p andvelocityloopgain K v areexpressed as below by using the naturalangularfrequency ω L of themechanism part as K p = c p ω L (2.12 a ) K v = c v ω L . (2.12 b ) 2.14 th OrderM od el of One Axis in aM ec hatronic Serv oS ystem 23 It is thetransformation of equation (2.8) using c p , c v in equation (2.12 a )and (2.12 b ). When we put equation (2.11b ) ∼ (2.12 b )into(2.8),the normalized4th order model without dependence on natural angular frequency ω L is derived G c ( s )= b 0 N G ( s 4 + b 3 s 3 + b 2 s 2 + b 1 s + b 0 ) (2.13) b 0 =( 1+ N L ) c p c v b 1 =(1+N L )(c v +2c p c v ζ L )+2 N L ζ L b 2 =( 1+ N L )(1 +2 c v ζ L + c p c v ) b 3 =2ζ L +(1+N L ) c v where N L = J L N 2 G J M (2.14) is the ratio between the inertial moment andmotor axisequivalentinertial moment of themechanism part. By usingthis normalized4th order model (2.13), the commondiscussion on the arbitrary natural angular frequency ω L of themechanism part can be carried out. 2.1.3Determination Method of Servo Parameters Using a Mathematical Model (1)C on trol Pe rformance Required in an IndustrialM ec hatronic Servo System The response characteristicofanindustrialmechatronic servosystem is re- quired to have afast response in the system withinthe regionwhere there is no generation of oscil lation andovershoot (refer to 1.1.2item 3). Previously, the servoparameters aredetermined by satisfyingthe requirement basedon the test error or experience. The prop er determination methodcan be derived by anormalized 4th order model (2.15) here In an industrialmechatronic servosystem, the following conditions are successful: • The motorisselected when the momentofinertia J M of themotor is satisfying 3 ≤ N L ≤ 10 fromthe moment of inertia J L of themechanism part and gearratio; • The dampingfactor ζ L of mechanismpartis0≤ ζ L ≤ 0 . 02. Forthe latter condition, since the damping factor ζ L is very small in an industrialmechatronic servosystem, then ζ L =0.However, ζ L =0is existed in the situation of continuous oscillation generationwhichisthe most difficult to control. Thenthis assumptionissufficient forthis situation.When put ζ L =0into equation (2.13), it can be as 24 2M athematical Mo del Construction of aM ec hatronic Serv oS ystem G c ( s ) ≈ 1 N G  s 4 (1 + N L ) c p c v + s 3 c p + (1 + c p c v ) s 2 c p c v + s c p +1  . (2.15) From thecurrentutilizationofanindustrialmechatronic servosystem, therea re the follow ing conditions for serv op arameters determination satisfy- ing the desiredc on trol pe rformances 1. Thereare two realpolesand onecomplexconjugate root in thenormalized 4th order model (2.15) (conditionA) 2. Ther esp onse comp onen to ft he complex conjugate ro ot is smaller than theresponse componentofthe principalroot(condition B). 3. The response componentofthe complex conjugate ro ot is morequickly converged thanthe response componentofthe principalroot(condition C). 4. If satisfying the above three conditions, the servoparameters K p , K v can be determinedfor afaster response. (2) Ramp Response of the Normalized4th Order Model Fordetermining the servoparameters satisfying the requiredcontrol perfor- mance intro ducedin2.1.3(1), the ramp response of the normalized 4th order model (2.15) should be worked out. Thereasonfor using aramp response is that, the ramp input canbeadoptedineachaxis of an industrial mechatronic servosystem in almost all contour control (refer to 1.1.2 item 8). Forthe ramp response of thenor malized 4th order model,ramp input is u ( t )=vt.FromconditionA,there aregiven two poles as − τ 1 , − τ 2 ( τ 1 <τ 2 ) andone complex conjugate root − σ + jρ, − σ − jρ,and the ramp response is calculatedas(refertoappendixA.2) y 4 ( t )=  t − K 0 + K 1 e − τ 1 t + K 2 e − τ 2 t + K 3 e − σt sin(ρt +2φ 1 − φ 2 − φ 3 )  v (2.16) K 0 = ( τ 1 + τ 2 )(σ 2 + ρ 2 )+2στ 1 τ 2 τ 1 τ 2 ( σ 2 + ρ 2 ) K 1 = τ 2 ( σ 2 + ρ 2 ) τ 1 ( τ 2 − τ 1 )(τ 2 1 − 2 στ 1 + σ 2 + ρ 2 ) K 2 = τ 1 ( σ 2 + ρ 2 ) τ 2 ( τ 1 − τ 2 )(τ 2 2 − 2 στ 2 + σ 2 + ρ 2 ) K 3 = τ 1 τ 2 ρ  ((τ 1 − σ ) 2 + ρ 2 )((τ 2 − σ ) 2 + ρ 2 ) where φ 1 =tan − 1 ( ρ/σ), φ 2 =tan − 1 ( ρ/( τ 1 − σ )), φ 3 =tan − 1 ( ρ/( τ 2 − σ )), K 0 steady-state velocitydeviationofthe 4th order model, K 1 ,K 2 response componentoftwo realpoles, K 3 respo nse comp onen to fc omplexc onjugate root. 2.14 th OrderM od el of One Axis in aM ec hatronic Serv oS ystem 25 (3)Relation between Servo Parameters andCharacteristicRoot By usingthe ramp response of thenormalized 4th order model,the relation between servoparameters andcharacteristic root is investigated.The moment of inertiar atioi sg iv en as N L =3,whose value is alwaysadoptedinindustrial mechatr onic servosystems. The region of c p and c v satisfying conditions A, B, Cisillustrated in Fig. 2.3(a),(b),(c), respectively.Fig. 2.3(d) shows the equivalentheightline about the region of c p and c v satisfying conditions A, B, Cand principalroot τ 1 .When the regionofthe response componentofthe complex conjugate root of conditionBis very small, K 3 K 1 ≤ 0 . 1(2.17) is given.When the regionofthe response componentofthe complex conjugate root of conditionCis converged quickly σ τ 1 ≥ 2 . 0(2.18) is given. Forreference, the calculated ratio of the response component K 1 of prin- cipalrootwhen changingparameters c p and c v ,and response component K 3 of thecomplexconjugate root is showninFig. 2.4(a). The calculated ratio of the principal root − τ 1 andt he realp art − σ of thec omplexc onjugate ro ot 0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 Cv Cp A 0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 Cv Cp B (a) Condition A(b) Condition B 0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 Cv Cp C 0.5 1 1.5 2 0.1 0.2 0.3 0.4 0.5 Cv Cp A ∩ B ∩ C 0.82 0.24 τ 1 =-0.1 τ 1 =-0.2 τ 1 =-0.3 τ 1 =-0.492 τ 1 =-0.6 τ 1 =-0.7 τ 1 =-0.4 (c) Condition C(d) Condition ABC Fig. 2.3. Relation of c p and c v for various conditions 26 2M athematical Mo del Construction of aM ec hatronic Serv oS ystem 0.1 0.2 0.3 0 0.05 0.1 0.15 Cv=0.4 Cv=0.6 Cv=0.9 Cv=0.8 Cv=1.2 Cv=1.4 Cv=1.6 Cv=1.8 Cv=2.0 K 3 / K 1 Cp Cv=0.82 0.1 0.2 0.3 0 1 2 3 4 5 Cv=0.4 Cv=0.6 Cv=0.8 Cv=0.9 Cv=1.2 Cv=1.4 Cv=1.6 Cv=1.8 Cv=2.0 Cp σ / τ 1 Cv=0.82 (a) Relationof K 3 /K 1 and c p (b) Relationof σ/τ 1 and c p Fig. 2.4. Relation of various parameters for various c v is shown in Fig. 2.4(b). From Fig.(a), when c v is fixed and c p is increased, K 3 /K 1 becomes big. That is, the response componentofthe complex conju- gate root cannotbeneglected. In Fig.(b), when c v is fixed and c p is increased, σ/τ 1 becomes small. That is, the declinationofthe response componentof complex conjugate root is delayed. (4) Determination Method of Servo Parameters Based on Control Performance Fr om thes erv op arameterd eterminationc onditions of 2.1.3(1),t he serv op a- rameters c p and c v aredetermined in order to obtain the fast response when satisfying equation (2.17) in 2.1.3(3)and equation (2.18), i.e.,the principal root τ 1 is small. Accordingtothe equivalent heightline of principal root τ 1 shown in Fig. 2.3(d), when the servoparame ters are c p =0. 24 and c v =0. 82, the minimal va lue is τ 1 = − 0 . 492. This is thegeneral result whichisnot de- pendentonthe naturalangularfrequency ω L of them ec hanism part in the normalized 4th order mo del (2.15). In order to verify the obtained servoparameterresults, the results of ramp resp onse calculated by equation (2.16) are illustrated in Fig. 2.5.F ig.(a)s ho ws the results when N L =3.Fig.(b) shows the results when N L =10. In thecom- mon velocityresponse of Fig.(a)and Fig.(b), the conditions of faster response in the regionofnooscillationorovershoot generation are c p =0. 24 and c v =0. 82. In addition, by comparing the results of Fig.(a)and Fig.(b), the po sition and ve lo cit ya re almost thes ame.W ith the general industrial field condition 3 ≤ N L ≤ 10, theconditions of faster response in velocityresponse without oscillation or overshoot gener ation are c p =0. 24 and c v =0. 82. From theseresults, the servoparameters K p , K v arecalculatedbythe naturalangularfrequency ω L of themechanism in experiment. In equation (2.12 a )and (2.12 b ) c p =0. 24 (2.19a ) c v =0. 82 (2.19b ) 2.14 th OrderM od el of One Axis in aM ec hatronic Serv oS ystem 27 0.92 0.94 0.96 0.98 1 Cp=0.24, Cv=0.82 Cp=0.3, Cv=0.82 Cp=0.2, Cv=0.82 Cp=0.24, Cv=1.2 Cp=0.24, Cv=0.7 Objective trajectory Position[1] 55 60 65 0 0.01 0.02 Velocity[1/s] Time[s] 0.92 0.94 0.96 0.98 1 Cp=0.24, Cv=0.82 Cp=0.3, Cv=0.82 Cp=0.2, Cv=0.82 Cp=0.24, Cv=1.2 Cp=0.24, Cv=0.7 Objective trajectory Position[1] 55 60 65 0 0.01 0.02 Velocity[1/s] Time[s] (a) N L =3 (b) N L =10 Fig. 2.5. Simulation results of normalized 4th order model as equation (2.15) with various c p andc v c p =0. 24, c v =0. 82; c p =0. 3, c v =0. 82; c p =0. 2, c v =0. 82; c p =0. 24, c v =1. 2; c p =0. 24, c v =0. 7, (a) N L =3,(b) N L =10. are given.The regulation of amechatronic servosystem, for fast response without oscillation or overshoot, can be carried out. 2.1.4E xp eriment Ve rificationo ft he Mathematical Mo del (1)Simulation and Experiment The appropriation of the determination methodfor theservoparameterof industrial mec hatronic serv os ystem, deriv ed in the former part, is ve rified by the experimentofDEC-1(refertoexperimentdevice E.1). The sampling time interval of the experimentisgiven as 1[ms] (refer to 3.1). The value of po sition lo op gain K p can be changed in th ecomputerprogram. Thevalue of velocityloopgain K v needs the equivalent value when K s v in Fig. 1.1 is adjusted by alteringt he va riable resistance.T he concrete metho di st hat, whenthe position loop is at the outside and the step signal of velocityis given,the time constantcorresponding to thisresponse wave is worked ou t and K v is calculated by its in ve rse va lue. When ch angingt he va lue of va riable resistance,t he va riable resistance,a st he regulation va lue, whic hi sc onsisten t with thedetermined K v value by the above experimentwith the methodof 2.1.3(4),isadopted. With thismethod, the ratio gain K s v of thePIcontroller of theactual velocitycontroller,corresponding to the optimal gain K v of P controller of velocitycontrol in the4th order model, can also be worked out. Themotion velocityofthe mechatronic servosystem serves as the op- eration ve lo cit yi nt he general industrialfi eld. With ab out 1/10 of motor rated speed u ( t )=10t [rad/s] as well as two conditions (a) K p =22.6[1/s], 28 2M athematical Mo del Construction of aM ec hatronic Serv oS ystem 0 5 1 0 P o s i t ion[ r a d ] Objec t i v e tra jec t o ry S imu l a t ion E x per iment 0 1 2 0 5 1 0 T ime[ s ] V eloc i ty[ r a d /s] 00.5 11.5 8.5 9 9.5 1 0 x [ r a d ] y [ r a d ] Objec t i v elo c us S imu l a t ion E x per iment (a) K p =22.6[1/s], K v =77.24[1/s] 0 5 1 0 P o s i t ion[ r a d ] Objec t i v e tra jec t o ry S imu l a t ion E x per iment 0 1 2 0 5 1 0 T ime[ s ] V eloc i ty[ r a d /s] 00.5 11.5 8.5 9 9.5 1 0 x [ r a d ] y [ r a d ] Objec t i v elo c us S imu l a t ion E x per iment (b) K p =50[1/s], K v =50[1/s] Fig. 2.6. Experimental results by using DEC-1 experimentdevice and comparison with simulation results by using 4th order model K v =77.24[1/s], (b) K p =50[1/s], the exp erimen tw as carriedo ut. Condition( a) is theappropriate servoparametercalculatedbyputting c p =0. 24, c v =0. 82 and ω L =9 4 . 2[rad/s] in to equation (2.12 a )a nd (2.12 b ). Condition (b) is the deviation of the servoparameterfromthe propervalue. Theseexperimental results and simulation results are illustrated in Fig. 2.6.However, for grasping visually the influence giv en to con tour con trol pe rformance,F ig. 2.6 sho ws the expansion graph of the angular part of the contour control results when same experimental results were usedtwice for the positions of the x axisand the y axis. Forproperservoparameters andservoparameters completewith errors, the simulation results based on the 4th order model of amechatronic servo system are almost identical to the actual experimental results. Therefore, it verifiedthatthe 4th order model is the correctexpression of the dyn amic ch aracteristico fa ni ndustrialm ec hatronic serv os ystem. The va lidation of 2.2R educed Order Mo del of One Axis in aM ec hatronic Serv oS ystem 29 adaptationofthe 4th order model in the designofaservocontroller is also shown. Moreover, in the simulationand experimentalresult of condition (a), the desired response characteristics withoutoscillationorovershoot at all in both position response and velocityresponse is illustrated. However, in condition (b),the position response is near to the objectivetrajectory comparing with thatof(a). Butoscillationisgenerated both in the position response andve- locityresponse. Additionally,incontourcontrol, theovershoot hasoccur red and the control performance hasdeteriorated. Since this overshoot must be avo ided in the con tour con trol in the industrial field,t his conditionc annotb e adoptedi nt he cont ourc on trol. Based on thea bo ve explanation, the effectiv e- ness of the pr oposed determination methodofservoparameterwas verified by experimental results. In an industrial mechatronic servosystem, for regulating eachaxis charac- teristicwith consistence, this methodisadaptedfor all axesofthe mechatronic servosystem and the high-precision contour control of industrial mechatronic servosystems can be realized. 2.2Reduced Order ModelofOne Axis in aMechatronic ServoSystem The expressiono fa mec hatronic serv os ystem by ar educed order mo del cor- resp ondin gtothe movementvelocityconditionisdesired from the simple controller design. Accordingt ot he 4th order mo del,t he mo del appro ximation errori sd efined and the linear 1st order equation (2.23) and the linear 2nd order equation (2.29) are constructed. The relationbetween the model parameters of the 4th order mo del and the mo del parameters of the reduced order mo del is giv en in equations (2.24), (2.30) and (2.37). The 1st order model for expressing the lowspeed operationofthe mecha- tronic serv os ystem (v elo cit yb elo w1 /20 rateds pe ed) and the 2nd order mo del for expressing the middle speed operation(velocitybelow1/5 ratedspeed) trace the experience of one of the authors. The significance of these reduced order mo dels has be en prove d. The effective usage of them od el for serv o controller design is also verifiedbyexample. 2.2.1Necessary Conditionsofthe Reduced Order Model As introduced in section2.1, one axis of mechatronic servosystem is con- structed by manyblocks(parts). These blocks(parts)have respectively at least one or two order transferfunctions. From block diagrams expressing cor- rectlytheseblo cks, it is very difficulttograsp quickly and entirely the features of the serv os ystem. In an industrial field,t hesem ec hatronic serv os ystems 30 2M athematical Mo del Construction of aM ec hatronic Serv oS ystem are previouslyregarded as asimple 1st order system (refer to 1.2.1(1)). How- ever, since these are the approximated judgmentfromthe movementofthe mechatr onic servosystem, it is hard to saythatthis possesses thedistinctly theoretical ground. In this section, consider ing the selection methodofthe servomotor firstly, the necessity of thereduced order model of the mechatronic servosystem is arranged as below. 1. In the mechanism part determined from the operation purpose (the fea- tures of mechanism part are expressed by naturalangularfrequencyand damping rate), the serv om otor is setu pa ccordingt ot he motors election metho d [8] .When controlling thisservomotor by the servocontroller,the actualmechanism is established according to the whole features of the servosystem and the entire servosystem is known before regulation. 2. Forunderstanding the entire features, the exchange of themechanism part is needed and also the revision of motorselection should be judged. 3. From this feature,itshould judgehow long to followthe current as- sumedoperation pattern (Generally, trapezoidal wave of velocityisalways adoptedinthe positioning control). 4. In the contour control, the trace of actual trajectory in term of command should be judged andthe properaction should be briefly known. Next,the important factorsinthe reduced order model arelisted below. 1. The features of the main structureblocksofthe mechatronic servosystem (suchasnaturalangularfrequencyofthe mechanismpart, properties of damping rate andmotor,etc) should be reflected. 2. Thegeneral regulationconditionofthe servosystem (overshoot is not absolutely generatednot only in theposition loop but alsointhe velocity loop) should be reflected. 3. Theaction conditions of the servosystem (e.g., the instruction is the ramp input of eachindependent axis, thetrajectory speedinthe contourcontrol is below1/5 of maximum velocity, etc) should be reflected. 4. Thereduced order is adopted for modelingand onemodel can be usedfor oneaction status. The reduced order model of mechatronic servosystems satisfying the above conditions is the 1st order mo del in lo ws pe ed con tour con trol, i.e., the ch arac- teristicparameterisonly K p 1 ;the 2ndorder model in middle contourcontrol, i.e.,the characteristicparameters are K p 2 , K v 2 .The detailed explanation is as below. 2.2.2StructureStandard of Model With the4th order model (2.13) as standard,for thecontourcontrol of indus- trial mechatronic servosystems, lowspeed 1st order model expressing pr operly [...]... Motor and mechanism part 1 s Y(s) Position loop Fig 2.7 Low speed 1st order model of industrial mechatronic servo system 2.2 Reduced Order Model of One Axis in a Mechatronic Servo System 33 Mechatronic servo system Motor and mechanism part Servo controller U(s) + - + K p2 - K v2 1 s 1 s Y(s) Velocity loop Position loop Fig 2.8 Middle speed 2nd order model of industrial mechatronic servo system Here, the... industrial mechatronic servo system from 1/20 to 1/5 of rated speed can be derived This middle speed 2nd order model is the 2nd order system The whole mechatronic servo system is as d2 y(t) dy(t) − cp2 cv2 y(t) + cp2 cv2 u(t) = −cv2 2 dt dt (2.25) and the model expressing by transfer function is as Gc2 (s) = s2 cv2 cp2 + cv2 s + cv2 cp2 (2.26) Mechatronic servo system Servo controller U(s) + - K p1 Motor and. .. expressing properly the industrial mechatronic servo system can be derived This low speed 1st order model is expressed as a 1st order system In the mechanism part, the inertial moment of the load is transformed into the motor axis Considering both the whole inertial moment of the mechatronic servo system and the electric characteristic of the servo motor, the whole mechatronic servo system is as dy(t) = −cp1... Axis in a Mechatronic Servo System Table 2.1 Evaluation of reduced order model (rated speed VM = 104[rad/s], ωL = 94.2[rad/s], servo parameter of low speed 1st order model Kp1 = 23. 6[1/s], servo parameter of middle speed 2nd order model Kp2 = 23. 6[1/s], Kv2 = 84.8[1/s]) Velocity [rad/s] 5.02(= VM /20) 20.1(= VM /5) 34 .0(= VM /3) Low velocity eq(2. 23) [rad2 ] 7.07 × 10−5 1. 13 × 10 3 × 7.07 × 10 3 × Middle... (2 .30 ) In the mechatronic servo system regulated properly, cp = 0.24 is given From equation (2 .30 ) and (2 .37 ), cp2 = 0.24 and cv2 = 0.96 are given 2.2.5 Evaluation of the Low Speed 1st Order Model and the Middle Speed 2nd Order Model Through the respective movement velocities of the low speed 1st order model and the middle speed 2nd order model derived in 2.2 .3 and 2.2.4, the appropriate modeling mechatronic. .. modeling mechatronic servo system is illustrated In the contour control of an industrial mechatronic servo system, ramp input is always adopted As the performance standard of the reduced order model, the error squared integral of the ramp response error between the 4th order model and the reduced order model is adopted In the contour control, the ramp input of mechatronic servo system is 1/20 of the... 8 .30 × 10−5 5.18 × 10−4 × 2 2 (τ1 + τ2 )(K1 τ2 + K2 τ1 ) + 4K1 K2 τ1 τ2 2τ1 τ2 (τ1 + τ2 ) 13 2K1 (τ1 + 3cp2 ) 2K2 (τ2 + 3cp2 ) + v2 3 − c (τ + 2c )2 − c (τ + 2c )2 32 cp2 p2 1 p2 p2 2 p2 J2min = (2 .36 ) From the above discussion, cv2 satisfying conditions can be derived for the minimum by cv2 = 4cp2 ≈ 4cp (2 .37 ) The approximation equation (2 .30 ) is as same as (2.24) The approximation equation (2 .37 )... between the normalized 4th order model and the 2nd order model is calculated with the differential about ζ2 by equation (2 .33 ) as dJ2 = dζ2 3 2 16K1 c4 ζ2 2ζ2 − 1 p2 + 5 2 2 8c3 ζ2 (cp2 τ1 ζ2 + 4c2 (τ1 + cp2 ) 3 )2 p2 p2 + 3 16K2 c4 ζ2 p2 2 3 (cp2 τ2 ζ2 + 4c2 (τ2 + cp2 )ζ2 )2 p2 v2 (2 .34 ) This value is often positive if ζ2 > 1 That is, since J2 (ζ2 ) is the mono-increase function in the scale of ζ2... response (2.16) of relationship ω2 = 2ζ2 cp2 and normalized 4th model, and the 2nd order model is given as J2 = 2 2 (τ1 + τ2 )(K1 τ2 + K2 τ1 ) + 4K1 K2 τ1 τ2 2τ1 τ2 (τ1 + τ2 ) 3 2 4 16ζ2 − 4ζ2 + 1 2K1 ((τ1 − cp2 )ζ2 + 4cp2 ζ2 ) − + 3 ζ4 2 ζ + 4c2 (τ + c )ζ 3 32cp2 2 cp2 τ1 2 p2 2 p2 1 − 3 2K2 ((τ2 − cp2 )ζ2 + 4cp2 ζ2 ) 2 ζ + 4c2 (τ + c )ζ 3 cp2 τ2 2 p2 2 p2 2 v2 (2 .33 ) The squared integral of the output... derived with the condition 1 of 2.2.2 and for agreement with the steady-state velocity deviation as b0 ≈ cp , (2.24) cp1 = b1 Here, the final approximation equation in (2.24) is the results approximated with ζL ≈ 0 for very small damping rate from 0 to 0.02 of the mechanism part in the industrial mechatronic servo system When given cp = 0.24 in the mechatronic servo system regulated properly, cp1 = 0.24 . Model (1)C on trol Pe rformance Required in an IndustrialM ec hatronic Servo System The response characteristicofanindustrialmechatronic servosystem is re- quired to have afast response in the system withinthe regionwhere. a1st order system. In the mechanism part, the inertial momentofthe load is trans- formedintothe motoraxis. Considering both the whole inertial momentof the mechatronic servosystem and the electric. the servocontroller,the actualmechanism is established according to the whole features of the servosystem and the entire servosystem is known before regulation. 2. Forunderstanding the entire