【正文】 ary motor satisfies 2preq = np2dp ? req = . The torque constant of an equivalent threephase rotary motor is found from the linear force constant by setting Km = req KM = ()(32)Nm/A = and the moment of inertia is J =r2eqm =() =? 103kgm2. The parameters LS, M, RS, np are the same as for the linear motor. Here x = 0 for the linear motor corresponds to the magic axis of its rotor phase a being lined up with the magic axis of stator phase a and similarly for the equivalent rotary motor. The corresponding equivalent twophase parameters are then L = LS + M = , RS = ? , Keq = 23 Km=, Imax (continuous) = 23 imax =,Vmax= 23 vmax= linear force put out by this motor is F = Keq iq / req= 23 Km iq / req . 外文翻譯（ 原 文） 7 . Door model The door model is from the technical report of He [4] and is of the form dx/dt = Ax+bu y =Cx where A ? R8? 8 , b?R8 , C?R8? 8 . The values of the triple {A,b,C} are given in [4]. Here x1 is the door position, x2 is the door speed and the input u to the door is the linear force F = Keq iq / req put out by the motor. The state variables x1, x2 are the two measured/puted state variables so that the output matrix is simply ??????? 00000010 00000001C The mass of the door is denoted by Mc so that the total mass of the door/motor bination is Mc + m. The observability matrix ? ?765432 CACACACACACACAC has rank 4 while the controllability matrix ? ?b765432 AbAbAbAbAbAAbb has rank 5. However, A is stable. The control approach is to feed back x, ?(? = x/req,ω=?/req) treating the transfer function from input u = F to x as a double integrator. The resolution of the linear position feedback from the wall to the door control system is . The maximum door speed is ?max = 1m=s, the maximum acceleration is αmax = , and the jerk rate is limited to j max = distance traveled by each door is 555mm. . Standard controller 外文翻譯（ 原 文） 8 A straightforward way to do servo control of this motor for the standard controller in which there is one inverter for each motor is to choose the linear force as ))()()(( 0012 dtxxKxxKxvKMu t r e fr e fr e fr e fc ???????? ? dtiiKiiKv t qqr e fIqqr e fpq ? ???? 0 )()( dtiiKiiKv t ddr e fIddr e fpq ? ???? 0 )()( where the reference trajectories xref, vref, αref, iqref are as shown in Section and idref is typically taken to be zero [1–3]. 4. Two motors and one inverter The interest here is to control the motor using a single inverter and the approach to independently control the quadrature current in each motor which in turn requires leaving the direct currents uncontrolled. One approach to controlling two PM synchronous motors using one inverter would be to just control the two motors identically. Specifically, as they nominally follow identical trajectories, just set vd1 = vd2, vq1 = vq2 so that ? 1 = ? 2, ω1 = ω2, id1 = id2, iq1 = iq2. This is a standard approach for torque/speed control of induction motor propulsion systems (light rail vehicles, subway cars, etc.) in which two traction (induction) motors are driven by a single inverter. In this case (torque control), the induction motors require only the rotor speed (to estimate the rotor fluxes) so that the average speed of the two motors can be used in the flux estimator and for speed control. However, in the case of synchronous motors, the position of the rotor is required for control as (7), (8) show. The external disturbances τL1, τL2 on the individual motors are not necessarily equal and so the rotors will misalign, ., ? 1 ? ? 2. In this situation, a control scheme based on putting the same input into both motors is not able to recover after misalignment occurs, that is, to realign? 1 = ? 2. Again, the objective here is to use one inverter to control two motors. (If there was an inverter for each motor, then the feedback controller given in (6) would suffice). To develop a controller using a single inverter for the two motors, two cases 外文翻譯（ 原 文） 9 are considered: (1) the motors are connected in parallel (2) the motors are connected in series. . Parallel connection First, consider the motors connected in parallel, that is, the applied voltage to 111 Lqeq iKdtdJ ?? ?? 11 ?? ?dtd each phase of the motors are the same. The model of the two motors in the dq coordinate system are then and To run these two motors off of a single inverter, one must take into account how the voltages are manded to the motor. The same three voltages vS1, vS2, vS3, or equivalently, the same two phase equivalent voltages va, vb are manded to both motors. The dq voltages for the two motors are given by ???????????????????? sbsapp ppqd iinn nnii )c os ()s i n( )s i n()c os ( ?? ?? ???????????????????? sbsapp ppqd vvnn nnvv )c os ()s i n( )s i n()c os ( ?? ?? 111 Lqeq iKdtdJ ?? ?? 11 ?? ?dtd where ? 1,? 2 are the angular position of motor 1 and motor 2, respectively 1. As the controller (6) indicates, this can be done by specifying the quadrature 外文翻譯（ 原 文） 10 ???????????? ????????? sbsapp ppqd vvnn nnvv )c os ()s i n( )s i n()c os ( ?? ?? voltage vq of each motor to control the torque producing current iq of each motor. To do so, Eqs. (7) and (8) show this requires choosing va, vb such that Clearly there is a singularity in the inverse at (np(? 2? 1 ))mod? =0. The reference trajectories for the two linear motors are designed to maintain an angular separation of the two motors. A control scheme for the two motors is then eqr e fr e fr e fr e fq K KKJi )()( 111111 ???? ?? ?????? dtiiKiiKv t qr e fqIqr e fqpq ? ???? 0 11111 )()( eqr e fr e fr e fr e fq K KKJi )()( 222222 ???? ?? ?????? dtiiKiiKv t qr e fqIqr e fqpq ? ???? 0 22222 )()( The direct voltages are determined by the quadrature voltages given by (9). Specifically, substitute (9) into ??????????????????? bapp ppdd vvnn nnvv