【正文】 cally, 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 )s i n()c os ( )s i n()c os ( 22 1121 ?? ?? 1 ?1 = 0 is assumed to correspond with the magic axis of phase a of motor 1 and similarly for motor 2. so that when ? 2 ——? 1 ? 0, vd1, vd2 ? ? and consequently, the currents id1, id2 can be large near the singularity. To avoid the singularity, the control scheme proposed here is to physically offset the angular position of the rotors of the two motors by (π/2)/np. Then, as both 外文翻譯（ 原 文） 11 nominally track the same trajectory (except for the position offset of (π/2)/np), a tight trajectory tracking control loop would keep np(? 2 ——? 1) close to π/2 and thus keep the system away from the singularity. . Simulations A basic simulation was run using the one inverter two motor controller. The speed profile for the two motors is shown in Fig. 1 and the position profiles are shown in Fig. 2. These plots show that the doors open in s. The speed tracking error is less than 5? 104m/s and therefore is not shown . The difference in position shown in Fig. 2 is due to the fact that x1(0) = 0 and x2（ 0） = req2?np=. The uncontrolled currents id1, id2 are shown in Fig. 3. The difference in the trajectories of the two currents can be explained by the fact the initial angular position of the two motors are different by (π/2)/np to avoid the singularity in the control. The consequence of not being able to control the direct current to zero results in the wasted power RSi2d because in a oneinverter/onemotor configuration, this current would typically be zero. The quadrature currents iq1, iq2 are controlled and, as shown, are on top of each other. The linear force F = Keq iq / req is required to make the move. Finally, the phase voltage vS1 is shown .where it is noted that the maximum of the voltage is just under 100V. . Series connection In the series connection, the current in each motor is the same in their corresponding phases. To analyze this situation, consider the twophase equivalent models of the two synchronous motors: 111 )s i n( sapeqsaSsa vnKiRdtdiL ???? ?? 外文翻譯（ 原 文） 12 111 )s i n( sbpeqsbSsb vnKiRdtdiL ???? ?? 222 )s i n( sapeqsaSsa vnKiRdtdiL ???? ?? 222 )s i n( sbpeqsbSsb vnKiRdtdiL ???? ?? The control input is vsa = vsa1 + vsa2 vsb = vsb1 + vsb2. Using highgain current feedback, vsa = KP (isaref + isa) vsb = Kp(isaref ——isa). one can force isa ? isa_ref, isb ? isb_ref fast enough that isa, isb can be considered as the ‖inputs‖ . Note that the highgain feedback does not have an integrator. This is due to the fact that the currents are sinusoids and will be of high frequency at high speeds and an integrator can have trouble tracking such a fast varying signal .With τ1, τ2 the torq