## 2 Materials and Methods
\begin{bmatrix}\sin\alpha& \sin\alpha\left(\alpha+\frac{4\pi}{3}\right)&\sin\alpha\left(\alpha+\frac{2 \pi}{3}\right)\\ \sin\alpha\left(\alpha+\frac{2\pi}{3}\right)&\sin\alpha&\sin\alpha\left(\alpha+ \frac{4\pi}{3}\right)\\ \sin\alpha\left(\alpha+\frac{4\pi}{3}\right)&\sin\alpha\left(\alpha+\frac{2 \pi}{3}\right)&\sin\alpha\end{bmatrix} . \tag{15}\]  
By multiplying the vectors and matrices in Equation (6), we obtain a system of differential Equations (16)-(21). \[ \begin{array}{l}u_{1A}=R_{10}i_{1A}+L_{10}\frac{d_{1A}}{dt}-\frac{M_{m}}{2} \left(\frac{d_{1B}}{dt}+\frac{d_{1C}}{dt}\right)+\\ +M_{m}\left(\frac{d_{1A}}{dt}\cos\alpha+\frac{4\pi}{3}\right)+\frac{d_{1B}}{ dt}\cos\left(\alpha+\frac{2\pi}{3}\right)+\frac{d_{1C}}{dt}\cos\left(\alpha+ \frac{4\pi}{3}\right)\right)-\\ -\omega M_{m}\left(i_{2A}\sin\alpha+i_{2B}\sin\left(\alpha+\frac{2\pi}{3} \right)+i_{2C}\sin\left(\alpha+\frac{4\pi}{3}\right)\right);\end{array}  \tag{16}\]  
\[ \begin{array}{l}u_{1B}=R_{10}i_{1B}+L_{10}\frac{d_{1B}}{dt}-\frac{M_{m}}{2} \left(\frac{d_{1A}}{dt}+\frac{d_{1B}}{dt}\right)+\\ +M_{m}\left(\frac{d_{1A}}{dt}\cos\left(\alpha+\frac{2\pi}{3}\right)+\frac{d_{1 B}}{dt}\cos\left(\alpha+\frac{4\pi}{3}\right)+\frac{d_{1C}}{dt}\cos\alpha\right)-\\ -\omega M_{m}\left(i_{2A}\sin\left(\alpha+\frac{2\pi}{3}\right)+i_{2B}\sin \left(\alpha+\frac{4\pi}{3}\right)+i_{2C}\sin\alpha\right);\end{array}  \tag{17}\]  
\[ \begin{array}{l}0=R_{20}i_{2A}+L_{20}\frac{d_{21A}}{dt}-\frac{M_{m}}{2} \left(\frac{d_{2B}}{dt}+\frac{d_{2C}}{dt}\right)+\\ +M_{m}\left(\frac{d_{1A}}{dt}\cos\alpha+\frac{d_{1B}}{dt}\cos\left(\alpha+ \frac{4\pi}{3}\right)+\frac{d_{1C}}{dt}\cos\left(\alpha+\frac{2\pi}{3}\right) \right)-\\ -\omega M_{m}\left(i_{1A}\sin\alpha+i_{1B}\sin\left(\alpha+\frac{4\pi}{3} \right)+i_{1C}\sin\left(\alpha+\frac{2\pi}{3}\right)\right);\end{array}  \tag{19}\]  
\[ \begin{array}{l}0=R_{20}i_{2B}+L_{20}\frac{d_{21A}}{dt}-\frac{M_{m}}{2} \left(\frac{d_{21A}}{dt}+\frac{d_{2C}}{dt}\right)+\\ +M_{m}\left(\frac{d_{1A}}{dt}\cos\left(\alpha+\frac{2\pi}{3}\right)+\frac{d_{1 B}}{dt}\cos\alpha+\frac{d_{1C}}{dt}\cos\left(\alpha+\frac{4\pi}{3}\right) \right)-\\ -\omega M_{m}\left(i_{1A}\sin\left(\alpha+\frac{2\pi}{3}\right)+i_{1B}\sin \left(\alpha+\frac{4\pi}{3}\right)\right);\end{array}  \tag{20}\]  
\[ \begin{array}{l}0=R_{20}i_{2C}+L_{20}\frac{d_{21C}}{dt}-\frac{M_{m}}{2} \left(\frac{d_{21A}}{dt}+\frac{d_{21B}}{dt}\right)+\\ +M_{m}\left(\frac{d_{1A}}{dt}\cos\left(\alpha+\frac{4\pi}{3}\right)+\frac{d_{1 B}}{dt}\cos\left(\alpha+\frac{2\pi}{3}\right)+\frac{d_{1C}}{dt}\cos\left(\alpha- \frac{4\pi}{3}\right)\right)-\\ -\omega M_{m}\left(i_{1A}\sin\left(\alpha+\frac{2\pi}{3}\right)+i_{1B}\sin \left(\alpha+\frac{2\pi}{3}\right)+i_{1C}\sin\alpha\right);\end{array}  \tag{21}\]  
We find the electromagnetic torque through the electromagnetic energy \(W_{y^{\prime}}\) concentrated in the air gap of the IM Equation (22):  
\[M_{y^{\prime}}=\frac{\partial W_{y^{\prime}}}{\partial\alpha_{\bar{a}}}=\frac{ 1}{2}[i]\partial\frac{[L]}{\partial\alpha_{\bar{a}}}, \tag{22}\]  
where \(\alpha_{\bar{a}}\) is the geometric angle of the rotor rotation. By multiplying the matrices included in Equation (22), we find an analytical expression of the electromagnetic moment of the three-phase IM Equation (23):\[ \begin{split} M_{y^{\prime}}&=p_{i}M_{m}\big{[}\sin\big{(} \alpha+\frac{\pi}{3}\big{)}(i_{1A}i_{2C}+i_{1B}i_{2A}+i_{1C}i_{2B})-\\ -\sin\alpha(i_{1A}i_{2A}+i_{1B}i_{2B}+i_{1C}i_{2C})+\sin\big{(} \alpha-\frac{\pi}{3}\big{)}(i_{1A}i_{2B}+i_{1B}i_{2C}+i_{1C}i_{2A})\big{]}, \end{split}  \tag{23}\]  
where \(p_{i}\) is the pole pairs number. The motion equation has the form (24)  
\[J_{\Sigma}\frac{d\omega}{dt}=p_{i}\Big{(}M_{y^{\prime}}-M_{c}\Big{)} \tag{24}\]  
where \(M_{c}\) is the static moment; \(J_{\Sigma}\) is the total inertia moment of the moving parts.