## 3 Analytical Investigation and Autonomous Amplitude-Phase Equations
\begin{array}{ll}\big{(}D_{0}^{2}+\omega^{2}\big{)}y_{1}&=[-2i\omega(D_{1}B) -2i\hat{\mu}\omega B+3\hat{\alpha}_{1}B^{2}\overline{B}+2\hat{\alpha}_{2}BA \overline{A}+\hat{\alpha}_{2}\overline{B}A^{2}+i\hat{\alpha}_{3}\omega B^{2} \overline{B}+2i\hat{\alpha}_{4}\omega BA\overline{A}\\ &-i\hat{\alpha}_{4}\omega\overline{B}A^{2}+2\hat{\alpha}_{5}\omega^{2}BA \overline{A}-\hat{\alpha}_{5}\omega^{2}\overline{B}A^{2}+\hat{\alpha}_{6} \omega^{2}B^{2}\overline{B}+i\hat{\alpha}_{7}\omega\overline{B}A^{2}]e^{i \omega T_{0}}+\big{[}\hat{\alpha}_{1}B^{3}\\ &+\hat{\alpha}_{2}BA^{2}+i\hat{\alpha}_{3}\omega B^{3}+i\hat{\alpha}_{4} \omega BA^{2}-\hat{\alpha}_{5}\omega^{2}BA^{2}-\hat{\alpha}_{6}\omega^{2}B^{3 }+i\hat{\alpha}_{7}\omega BA^{2}\big{]}e^{3i\omega T_{0}}\\ &-\frac{1}{2}i\Omega^{2}\hat{f}e^{i\Omega T_{0}}\end{array}  \tag{21}\]  
To obtain a bounded solution for Equations (20) and (21), the possible resonant conditions should be determined first, which are the primary (i.e., \(\Omega=\omega\)) and subharmonic (i.e., \(\Omega=3\omega\)) resonance cases. Accordingly, to investigate the system dynamics at primary response condition, the detuning parameter \(\sigma\) is introduced to distinguish the closeness of the rotor angular speed to the twelve-pole system natural frequency as follows:  
\[\Omega=\omega+\sigma=\omega+\hat{e}\sigma \tag{22}\]  
By introducing Equation (22) into Equations (20) and (21), we get  
\[\begin{array}{ll}\big{(}D_{0}^{2}+\omega^{2}\big{)}x_{1}&=[-2i\omega(D_{1}A) -2i\hat{\mu}\omega A+3\hat{\alpha}_{1}A^{2}\overline{A}+2\hat{\alpha}_{2}AB \overline{B}+\hat{\alpha}_{2}\overline{A}B^{2}+i\hat{\alpha}_{3}\omega A^{2} \overline{A}\\ &+2i\hat{\alpha}_{4}\omega AB\overline{B}-i\hat{\alpha}_{4}\omega\overline{A}B ^{2}+2\hat{\alpha}_{5}\omega^{2}AB\overline{B}-\hat{\alpha}_{5}\omega^{2} \overline{A}B^{2}+\hat{\alpha}_{6}\omega^{2}A^{2}\overline{A}+i\hat{\alpha}_{7} \omega\overline{A}B^{2}\\ &+\frac{1}{2}(\omega+\sigma)^{2}\hat{f}e^{i\omega T_{0}}\big{]}e^{i\omega T_{0 }}+\big{[}\hat{\alpha}_{1}A^{3}+\hat{\alpha}_{2}AB^{2}+i\hat{\alpha}_{3} \omega A^{3}+i\hat{\alpha}_{4}\omega AB^{2}-\hat{\alpha}_{5}\omega^{2}AB^{2} \]  
\[ \begin{array}{ll}\big{(}D_{0}^{2}+\omega^{2}\big{)}y_{1}&=[-2i\omega(D_{1}B) -2i\mu\omega B+3\hat{\alpha}_{1}B^{2}\overline{B}+2\hat{\alpha}_{2}BA\overline{A }+\hat{\alpha}_{2}\overline{B}A^{2}+i\hat{\alpha}_{3}\omega B^{2}\overline{B} \\ &+2i\hat{\alpha}_{4}\omega BA\overline{A}-i\hat{\alpha}_{4}\omega\overline{B}A^{2 }+2\hat{\alpha}_{5}\omega^{2}BA\overline{A}-\hat{\alpha}_{5}\omega^{2} \overline{B}A^{2}\ +\hat{\alpha}_{6}\omega^{2}B^{2}\overline{B}+i\hat{\alpha}_{7}\omega \overline{B}A^{2}\\ &-\frac{1}{2}i(\omega+\sigma)^{2}\hat{f}e^{i\omega T_{0}}\big{]}e^{i\omega T_{0 }}+\big{[}\hat{\alpha}_{1}B^{3}+\hat{\alpha}_{2}BA^{2}+i\hat{\alpha}_{3} \omega B^{3}+i\hat{\alpha}_{4}\omega BA^{2}-\hat{\alpha}_{5}\omega^{2}BA^{2} \\ &-\hat{\alpha}_{6}\omega^{2}B^{3}+i\hat{\alpha}_{7}\omega BA^{2}\big{]}e^{3i \omega T_{0}}\end{array}  \tag{24}\]  
To have a bounded solution for Equations (23) and (24), the coefficients of \(e^{i\omega T_{0}}\) should vanish. So, the solvability conditions of Equations (23) and (24) are:  
\[\begin{array}{ll}-2i\omega(D_{1}A)-2i\hat{\mu}\omega A+3\hat{\alpha}_{1}A^{2} \overline{A}+2\hat{\alpha}_{2}AB\overline{B}+i\hat{\alpha}_{2}\overline{A}B^{2 }+i\hat{\alpha}_{3}\omega A^{2}\overline{A}+2i\hat{\alpha}_{4}\omega AB \overline{B}-i\hat{\alpha}_{4}\omega\overline{A}B^{2}\\ +2\hat{\alpha}_{5}\omega^{2}AB\overline{B}-\hat{\alpha}_{5}\omega^{2}\overline{A}B ^{2}+\hat{\alpha}_{6}\omega^{2}A^{2}\overline{A}+i\hat{\alpha}_{7}\omega \overline{A}B^{2}+\frac{1}{2}(\omega+\sigma)^{2}\hat{f}e^{i\omega T_{1}}=0\\ -2i\omega(D_{1}B)-2i\hat{\mu}\omega B+3\hat{\alpha}_{1}B^{2}\overline{B}+2\hat{ \alpha}_{2}BA\overline{A}+\hat{\alpha}_{2}\overline{B}A^{2}+i\hat{\alpha}_{3} \omega B^{2}\overline{BInserting Equation (27) into Equations (25) and (26) and reverting to the original system parameters (i.e., \(t=\frac{T_{1}}{\varepsilon},\hat{\sigma}=\frac{\varepsilon}{\varepsilon}\).