## 2 Materials and Methods
\begin{bmatrix}L_{1o}&M_{m}\cos \frac{2\pi}{3}&M_{m}\cos\frac{4\pi}{3}\\ M_{m}\cos\frac{4\pi}{3}&L_{1o}&M_{m}\cos\frac{2\pi}{3}\\ M_{m}\cos\frac{2\pi}{3}&M_{m}\cos\frac{4\pi}{3}&L_{1o}\end{bmatrix} = \begin{bmatrix} L_{1o}&-\frac{M_{m}}{2}&-\frac{M_{m}}{2}\\ -\frac{M_{m}}{2}&L_{1o}&-\frac{M_{m}}{2}\\ -\frac{M_{m}}{2}&-\frac{M_{m}}{2}&L_{1o}\end{bmatrix} , \tag{7}\]  
\[[L_{2}]= \begin{bmatrix}L_{2o}&M_{2A2B}&M_{2A2C}\\ M_{2B2A}&L_{2o}&M_{2B2C}\\ M_{2C2A}&M_{2C2B}&L_{2o}\end{bmatrix} = \begin{bmatrix}L_{2o}&M_{m}\cos\frac{2 \pi}{3}&M_{m}\cos\frac{4\pi}{3}\\ M_{m}\cos\frac{4\pi}{3}&L_{2o}&M_{m}\cos\frac{2\pi}{3}\\ M_{m}\cos\frac{2\pi}{3}&M_{m}\cos\frac{4\pi}{3}&L_{2o}\end{bmatrix} = \begin{bmatrix} L_{2o}&-\frac{M_{m}}{2}&-\frac{M_{m}}{2}\\ -\frac{M_{m}}{2}&L_{2o}&-\frac{M_{m}}{2}\\ -\frac{M_{m}}{2}&-\frac{M_{m}}{2}&L_{2o}\end{bmatrix} , \tag{8}\]  
where \(M_{m}\) is the maximum value of mutual inductance arising from the alignment of phases. The assumption made earlier about the uniformity of the air gap allows us to conclude that all the eigeninductances of phases do not depend on the angular position of the rotor and are determined by Equation (9). \[L_{1o}=L_{1 co}+L_{1e0};\;L_{2o}=L_{2 co}+L_{2e0}; \tag{9}\]  
where \(L_{1 co}\) is stator phase inductance from the main magnetic flux;  
\(L_{1e0}\) is inductance from stator magnetic flux dissipation;  
\(L_{2e0}\) is the inductance of the rotor phase reduced to the stator from the main magnetic flux;  
\(L_{2e0}\) is inductance reduced to the stator from the magnetic flux of the rotor dissipation. For a given IM, the maximum value of mutual inductance \(M_{m}\) is equal to the inductance from the main magnetic flux of the phase and is related to the inductance of the magnetizing circuit by equality (10):  
\[L_{1 co}=L_{2 co}=M_{m}=\frac{2}{3}L_{o} \tag{10}\]  
Let us write the matrices of mutual inductances as Equations (11) and (12):  
\[[M_{12}]= \begin{bmatrix}M_{1A2A}&M_{1A2B}&M_{1A2C}\\ M_{1B2A}&M_{1B2B}&M_{1B2C}\\ M_{1C2A}&M_{1C2B}&M_{1C2C}\end{bmatrix} =M_{m}\times \begin{bmatrix}\cos\alpha& \cos\bigl{(}\alpha+\frac{2\pi}{3}\bigr{)}&\cos\bigl{(}\alpha+\frac{4\pi}{3} \bigr{)}\\ \cos\bigl{(}\alpha+\frac{4\pi}{3}\bigr{)}&\cos\alpha&\cos\bigl{(}\alpha+\frac {2\pi}{3}\bigr{)}\\ \cos\bigl{(}\alpha+\frac{2\pi}{3}\bigr{)}&\cos\bigl{(}\alpha+\frac{4\pi}{3} \bigr{)}&\cos\alpha\end{bmatrix} ; \tag{11}\]  
\[[M_{21}]= \begin{bmatrix}M_{2A1A}&M_{2A1B}&M_{2A1C}\\ M_{2B1A}&M_{2B1B}&M_{2B1C}\\ M_{2C1A}&M_{2C1B}&M_{2C1C}\end{bmatrix} =[M_{12}]^{T}=M_{m}\times \begin{bmatrix} \cos\alpha&\cos\bigl{(}\alpha+\frac{4\pi}{3}\bigr{)}&\cos\bigl{(}\alpha+\frac {2\pi}{3}\bigr{)}\\ \cos\bigl{(}\alpha+\frac{2\pi}{3}\bigr{)}&\cos\alpha&\cos\bigl{(}\alpha+\frac {4\pi}{3}\bigr{)}\\ \cos\bigl{(}\alpha+\frac{4\pi}{3}\bigr{)}&\cos\bigl{(}\alpha+\frac{2\pi}{3} \bigr{)}&\cos\alpha\end{bmatrix} , \tag{12}\]  
where T is the transpose sign of the matrix. From Equation (13), it is possible to determine the electric angle of rotation of the rotor:  
\[\alpha(t)=\alpha(0)+\int_{0}^{t}\omega(t)dt, \tag{13}\]  
where \(\omega\) is the angular velocity of the rotor, el. rad/s. Since the elements of matrices \([L_{1}]\) and \([L_{2}]\) according to Equations (7) and (8) are independent on the angular position of the rotor, by taking partial derivatives of matrix elements by rotor rotation angle \(\alpha\), we obtain zero matrices of size \(3\times 3\). Taking partial derivatives of the elements of matrices \([M_{12}]\) and \([M_{21}]\) on the rotor rotation angle \(\alpha\), we obtain Equations (14) and (15):  
\[\frac{\partial[M_{12}]}{\partial\alpha}=-M_{m}\times \begin{bmatrix}\sin\alpha& \sin\alpha\left(\alpha+\frac{2\pi}{3}\right)&\sin\alpha\left(\alpha+\frac{4\pi }{3}\right)\\ \sin\alpha\left(\alpha+\frac{4\pi}{3}\right)&\sin\alpha&\sin\alpha\left(\alpha+ \frac{2\pi}{3}\right)\\ \sin\alpha\left(\alpha+\frac{2\pi}{3}\right)&\sin\alpha\left(\alpha+\frac{4 \pi}{3}\right)&\sin\alpha\end{bmatrix} ; \tag{14}\]  
\[\frac{\partial[M_{21}]}{\partial\alpha}=-M_{m}\times