# A PINN-Based Nonlinear PMSM Electromagnetic Model Using Differential Inductance Theory
## 2 The Nonlinear PMSM Electromagnetic Model Based on Differential Inductance
### PMSM Equations in the dq0 Frame  
Under magnetic saturation and dq-axis cross-coupling effects, the mathematical model in the dq0 reference frame transforms AC quantities into DC quantities. However, decoupling of the inductance matrix becomes unattainable under these conditions. Define the dq0-frame variables as follows:  
\[\left\{\begin{array}{l}\mathbf{i}_{dq0}=\mathbf{C}_{dq0}^{-1}\mathbf{i}_{abc }\\ \Psi_{dq0\_{\text{AM}}}=\mathbf{C}_{dq0}^{-1}\Psi_{abc\_{\text{AM}}}\\ \Psi_{dq0\_{\text{PM}}}=\mathbf{C}_{dq0}^{-1}\Psi_{abc\_{\text{PM}}}\\ \mathbf{L}_{dq0}=\mathbf{C}_{dq0}^{-1}\mathbf{L}_{abc}\mathbf{C}_{dq0}\end{array}\right. \tag{20}\]  
where \(\mathbf{C}_{dq0}\) is the coordinate transformation matrix from ABC to dq0:  
\[\mathbf{C}_{dq0}=\left[\begin{array}{ccc}\cos\theta_{e}&-\sin\theta_{e}&1\\ \cos(\theta_{e}-\frac{2}{3}\pi)&-\sin(\theta_{e}-\frac{2}{3}\pi)&1\\ \cos(\theta_{e}+\frac{2}{3}\pi)&-\sin(\theta_{e}+\frac{2}{3}\pi)&1\end{array}\right] \tag{21}\]  
Its inverse matrix \(\mathbf{C}_{dq0}^{-1}\) is given by:  
\[\mathbf{C}_{dq0}^{-1}=\frac{2}{3}\begin{bmatrix}\cos\theta_{e}&\cos(\theta_{ e}-\frac{2}{3}\pi)&\cos(\theta_{e}+\frac{2}{3}\pi)\\ -\sin\theta_{e}&-\sin(\theta_{e}-\frac{2}{3}\pi)&-\sin(\theta_{e}+\frac{2}{3} \pi)\\ \frac{1}{2}&\frac{1}{2}&\frac{1}{2}\end{bmatrix} \tag{22}\]  
Substituting (20) into (2) yields the equation for the flux linkage equation in dq0 frame:  
\[\Psi_{dq0}|_{\theta_{e}}=\left(\Psi_{dq0\_{\text{PM}}}+\Psi_{dq0\_{\text{AM} }}\right)|_{\theta_{e}}=\left(\Psi_{dq0\_{\text{PM}}}+\int_{0}^{i_{abc}}\mathbf{ L}_{dq0}\text{di}\mathbf{i}_{dq0}\right)|_{\theta_{e}} \tag{23}\]After transformation, \(L_{4d0}\) is an asymmetric matrix. Let:  
\[\left\{\begin{array}{l}\widetilde{\mathbf{L}}_{4d0}(=\mathbf{\rho}_{4d0} \mathbf{L}_{d0}\\ \widetilde{\Psi}_{4d0\_AM}=\mathbf{\rho}_{4d0}\Psi_{4d0\_AM}\\ \widetilde{\Psi}_{4d0\_PM}=\mathbf{\rho}_{4d0}\Psi_{4d0\_PM}\\ \widetilde{\Psi}_{4d0}=\mathbf{\rho}_{4d0}\Psi_{4d0}\end{array}\right. \tag{24}\]  
\(\mathbf{\rho}_{4d0}(=\mathbf{C}_{4d0}^{T}\mathbf{C}_{4d0})\) is the metric tensor, ensuring that the length of vectors remains invariant under coordinate transformations. After the transformation, \(\widetilde{\mathbf{L}}_{4d0}\) is still a real symmetric array. Thus, (23) can be rewritten as:  
\[\widetilde{\Psi}_{4d0}=\left(\widetilde{\Psi}_{4d0\_PM}+\widetilde{\Psi}_{4d0 \_AM}\right)|_{\theta_{e}}=\left(\widetilde{\Psi}_{4d0\_PM}+\int_{0}^{\mathbf{ i}_{4d0}}\widetilde{\mathbf{L}}_{4d0}\mathbf{di}_{4d0}\right)|_{\theta_{e}} \tag{25}\]  
Substituting (20) into (13) yields the magnetic energy equation in dq0 frame:  
\[W_{m}|_{\theta_{e}}=\left(W_{m\_PM}+\int_{0}^{\mathbf{i}_{4d0}}\mathbf{i}_{d0 }^{T}\widetilde{\mathbf{L}}_{4d0}\mathbf{di}_{4d0}\right)|_{\theta_{e}} \tag{26}\]  
Combining (10), (25) and (26), the torque equation in the dq0 frame can be derived:  
\[\begin{array}{l}\gamma_{s}=\mathbf{C}_{4d0}^{-1}\frac{d\mathbf{C}_{4d0}}{ \partial\theta_{e}}\\ T_{e}=n_{p}\mathbf{i}_{dd0}^{2}\mathbf{\rho}_{4d0}(\gamma_{s}\Psi_{4d0}+\frac{ \partial\Psi_{4d0}}{\partial\theta_{e}})-n_{p}\frac{\partial W_{m}}{\partial \theta_{e}}\end{array} \tag{27}\]  
The field equation becomes:  
\[\
abla_{\mathbf{i}_{4d0}}W_{c\_AM}=\widetilde{\Psi}_{4d0\_AM} \tag{28}\]