## 3 Principles and Modeling of an Adaptive Control Method for PMSMs Based on Super-Helical SMC
### MRAS-Based PMSM Position Sensorless Control System
By substituting matrix \(\mathrm{A}\) and the new matrix \(\mathrm{P}\) into Equation (12), we obtain:  
\[\mathbf{Q}=-(\mathbf{P}\mathbf{A}+\mathbf{A}^{\mathrm{T}}\mathbf{P})= \begin{bmatrix}2\frac{\mathbf{R} }{L_{d}}&0\\ 0&2\frac{\mathbf{R}}{L_{d}}\end{bmatrix}  \tag{13}\]  
It is evident that \(\mathrm{Q}\) is a symmetric positive definite matrix. Therefore, for permanent magnet synchronous motor, when the linear compensator matrix \(\mathrm{C}\) is chosen as given in Equation (4), the transfer function matrix \(\mathrm{G}(\mathrm{s})\) of the linear time-invariant forward channel is a strictly positive real matrix. (2) The nonlinear time-varying feedback channel satisfies the Popov integral inequality:  
\[\eta(0,t_{1})=\int_{0}^{t_{1}}\mathbf{V}^{\mathrm{T}}\mathbf{W}\mathrm{d}t\geq-\gamma _{0}^{2}\;\;(\forall t_{1}\geq 0) \tag{14}\]  
The design of the adaptive law in the nonlinear time-varying feedback channel must ensure that the nonlinear feedback channel satisfies Inequality (14). By substituting \(\mathrm{V}=\mathrm{C}\mathrm{e}\) and \(\mathrm{W}\) into Equation (14), we obtain:  
\[\eta(0,t_{1})=\int_{0}^{t_{1}}\mathbf{e}^{\mathrm{T}}\mathbf{C}\mathbf{f}(\hat{\omega}_{ \mathrm{e}}-\omega_{\mathrm{e}})\hat{\mathbf{i}}^{\gamma}\mathrm{d}t\geq-\gamma_{ 0}^{2},\;\;\forall t_{1}\geq 0 \tag{15}\]  
In the proportional-integral form of the model reference adaptive system, \(\hat{\omega}_{\mathrm{e}}\) can be expressed as:  
\[\hat{\omega}_{\mathrm{e}}=\int_{0}^{t}F_{1}(v,t,\tau)\mathrm{d}\tau+F_{2}(v,t )+\hat{\omega}_{\mathrm{e}}(0) \tag{16}\]  
In the equation, \(\hat{\omega}_{\mathrm{e}}(0)\) represents the estimated initial value of the speed. Substituting Equation (16) into Equation (15) yields:  
\[ \begin{split}\eta(0,t_{1})&=\int_{0}^{t_{1}}\Big{[}\int_ {0}^{t}F_{\rm I}(v,t,\tau)\mathrm{d}\tau+F_{2}(v,t)+\hat{\omega}_{\rm e}(0)- \omega_{\rm e}\Big{]}e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^{\gamma}\mathrm{d}t\\ &=\eta_{1}(0,t_{1})+\eta_{2}(0,t_{1})\end{split}  \tag{17}\]  
To ensure that \(\eta(0,t_{1})\geq-\gamma_{0}^{2}\), it is necessary to satisfy the following conditions individually:  
\[\eta_{1}(0,t_{1})=\int_{0}^{t_{1}}\Big{[}\int_{0}^{t}F_{\rm I}(v,t,\tau)\mathrm{ d}\tau+\hat{\omega}_{\rm e}(0)-\omega_{\rm e}\Big{]}e^{\intercal}C\mathbf{j}\hat{ \mathbf{i}}^{\gamma}\mathrm{d}t\geq-\gamma_{1}^{2} \tag{18}\]  
\[\eta_{2}(0,t_{1})=\int_{0}^{t_{1}}F_{2}(v,t)e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^ {\gamma}\mathrm{d}t\geq-\gamma_{2}^{2} \tag{19}\]  
In the equation, \(\gamma_{1}^{2}\) and \(\gamma_{2}^{2}\) are both finite positive numbers. To construct a function \(f(t)\) that satisfies Inequality (18), we can define \(f(t)\) as follows:  
\[\left\{ \begin{array}{l}K_{\rm i}f(t)=\int_{0}^{t}F_{\rm I}(v,t,\tau)\mathrm{ d}\tau+\hat{\omega}_{\rm e}(0)-\omega_{\rm e}\\ \frac{\mathrm{d}f(t)}{\mathrm{d}t}=e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^{\gamma} \end{array} \right. \tag{20}\]  
In the equation, \(K_{i}>0\). Substituting Equation (20) into Equation (18) yields the following:  
\[ \begin{split}\eta_{1}(0,t_{1})&=\int_{0}^{t_{1}}\frac{ \mathrm{d}f(t)}{\mathrm{d}t}K_{\rm i}f(t)\mathrm{d}t=\frac{K_{\rm i}}{2}\big{[} f^{2}(t)-f^{2}(0)\big{]}\\ &\geq-\frac{1}{2}K_{\rm i}f^{2}(0)\geq-\gamma_{1}^{2}\end{split}  \tag{21}\]  
By differentiating the first equation in Equation (20) and using the second equality in Equation (20), we obtain the following:  
\[F_{\rm I}(v,t,\tau)=K_{\rm i}e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^{\gamma}\ \ \ (K_{\rm i }>0) \tag{22}\]  
If the integrand in Equation (19) is positive, then the inequality is guaranteed to hold. Thus, we choose the following equation:  
\[F_{2}(v,t)=K_{\rm p}e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^{\gamma}\ \ (K_{\rm P}>0) \tag{23}\]  
Substituting the above expression into Equation (19) yields the following:  
\[\eta_{2}(0,t_{1})=\int_{0}^{t_{1}}K_{\rm P}(e^{\intercal}C\mathbf{j}\hat{\mathbf{i}}^ {\gamma})^{2}\mathrm{d}t\geq 0\geq-\gamma_{2}^{2} \tag{24}\]  
Thus, the proof is complete.