## Operating Principle
### Principle of Power Flow  
From Equation (4), we know that \(\overset{\rightharpoonup}{u_{s}}\) can be synthesized by \(\overset{\rightharpoonup}{u_{s1}}\) and \(\overset{\rightharpoonup}{u_{s2}}\), where \(\overset{\rightharpoonup}{u_{s1}}\) and \(\overset{\rightharpoonup}{u_{s2}}\) are generated by two independent SVPWM controllers separately. Thus, when \(\overset{\rightharpoonup}{u_{s}}\) is determined and \(V_{dc1}\), \(V_{dc2}\) are measured, the feasible region of voltage vector distribution can be obtained. As shown in Figure 3, \(O_{1}\) is the origin of vector \(\overset{\rightharpoonup}{u_{s}}\) and \(O_{2}\) is the end of \(\overset{\rightharpoonup}{u_{s}}\). Two hexagons indicating the modulation range of the two inverters can be acquired. Hexagon \(A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}\) centering on \(O_{1}\), having a side length of \(\sqrt{\frac{2}{3}}V_{dc1}\), gives the modulation range of inverter1. In the same way, hexagon \(A_{2}B_{2}C_{2}D_{2}E_{2}F_{2}\) centering on \(O_{2}\) with a side length of \(\sqrt{\frac{2}{3}}V_{dc2}\), gives the modulation range of inverter2. Vector \(\overset{\rightharpoonup}{u_{s1}}\) starts at \(O_{1}\) and has to end within the range of hexagon \(A_{1}B_{1}C_{1}D_{1}E_{1}F_{1}\), similarly vector \(\overset{\rightharpoonup}{u_{s2}}\) starts at \(O_{2}\) and has to end within the range of hexagon \(A_{2}B_{2}C_{2}D_{2}E_{2}F_{2}\). According to Equation (4), the end of \(\overset{\rightharpoonup}{u_{s1}}\) and the end of \(\overset{\rightharpoonup}{u_{s2}}\) must coincide at one point, assuming it is \(I\). The overlapping region of the two hexagons \(A_{2}B_{2}GD_{1}E_{1}H\) corresponds to the feasible region of \(I\), which also indicates the feasible region of voltage vector distribution. If the motor stator currents \(i_{AX}\), \(i_{BX}\) and \(i_{CX}\) are measured, the stator current vector of OW-PMSM \(\stackrel{{\rightharpoonup}}{{i_{s}}}\) can be obtained as follow:  
\[\stackrel{{\rightharpoonup}}{{i_{s}}}=\sqrt{\frac{2}{3}}(i_{AX}e ^{\mid 0}+i_{BY}e^{j\frac{2\pi}{3}}+i_{CZ}e^{j\frac{4\pi}{3}}) \tag{6}\]  
Then the power sharing of dual inverter can be determined as shown in Equation (7):  
\[\left\{ \begin{array}{l}P_{inv1}=\stackrel{{\rightharpoonup}}{{ i_{s1}}}\cdot\stackrel{{\rightharpoonup}}{{i_{s}}}\\ P_{inv2}=\stackrel{{\rightharpoonup}}{{i_{s2}}}\cdot\stackrel{{ \rightharpoonup}}{{-i_{s}}}=-\stackrel{{\rightharpoonup}}{{i_{s2 }}}\cdot\stackrel{{\rightharpoonup}}{{i_{s}}}\end{array} \right. \tag{7}\]  
where \(P_{inv1}\) and \(P_{inv2}\) are the output power of inverter1 and inverter2 respectively, positive value means the corresponding inverter is outputting active power and vice versa. It is important to notice the minus sign in the expression of \(P_{inv2}\). That is because the defined motor phase current polarities are flowing from inverter1 to inverter2, thus the direction of vector \(\stackrel{{\rightharpoonup}}{{i_{s}}}\) on inverter2's point of view is opposite to inverter1's. In Figure 3, \(P_{inv1}\) can be obtained from vector dot product of \(\stackrel{{\rightharpoonup}}{{u_{s1}}}\) and \(\stackrel{{\rightharpoonup}}{{i_{s}}}\) at point \(O_{1}\), similarly \(P_{inv2}\) can be obtained from vector dot product of \(\stackrel{{\rightharpoonup}}{{u_{s2}}}\) and \(\stackrel{{\rightharpoonup}}{{-i_{s}}}\) at point \(O_{2}\). Taking inverter1 for example, we have \(P_{inv1}=\stackrel{{\rightharpoonup}}{{u_{s1}}}\cdot\stackrel{{ \rightharpoonup}}{{i_{s}}}=|\stackrel{{\rightharpoonup}}{{i_{s}}} |\lvert O_{1}\rvert\), where \(\lvert O_{1}\rvert\)\(\mid\) is the projection of \(\stackrel{{\rightharpoonup}}{{u_{s1}}}\) on the direction of \(\stackrel{{\rightharpoonup}}{{i_{s}}}\). Thus all the available \(\stackrel{{\rightharpoonup}}{{-i_{s1}}}\) having the same projection \(\lvert O_{1}\rvert\)\(\mid\) obtain the same value of \(P_{inv1}\), which is indicated by line \(l_{1}\) through point \(J\) and orthogonal to \(\stackrel{{\rightharpoonup}}{{i_{s}}}\).