磁动势不变约束

{ia=idcos⁡θ−iqsin⁡θib=idcos⁡(θ−120∘)−iqsin⁡(θ−120∘)ic=idcos⁡(θ+120∘)−iqsin⁡(θ+120∘)\left\{ \begin{aligned} &i_a=i_d\cos\theta-i_q\sin\theta\\ &i_b=i_d\cos(\theta-120^\circ)-i_q\sin(\theta-120^\circ)\\ &i_c=i_d\cos(\theta+120^\circ)-i_q\sin(\theta+120^\circ) \end{aligned} \right.

[iaibic]=[cos⁡θ−sin⁡θ1cos⁡(θ−120∘)−sin⁡(θ−120∘)1cos⁡(θ+120∘)−sin⁡(θ+120∘)1][idiqi0]\left[ \begin{matrix} i_a\\ i_b\\ i_c \end{matrix} \right] = \left[ \begin{array}{ccc} \cos\theta & -\sin\theta & 1\\ \cos(\theta-120^\circ) & -\sin(\theta-120^\circ) & 1\\ \cos(\theta+120^\circ) & -\sin(\theta+120^\circ) & 1 \end{array} \right] \left[ \begin{matrix} i_d\\ i_q\\ i_0 \end{matrix} \right]

[idiqi0]=23[cos⁡θcos⁡(θ−120∘)cos⁡(θ+120∘)−sin⁡θ−sin⁡(θ−120∘)−sin⁡(θ+120∘)1/21/21/2][iaibic]\left[ \begin{matrix} i_d\\ i_q\\ i_0 \end{matrix} \right] = \frac{2}{3} \left[ \begin{array}{ccc} \cos\theta & \cos(\theta-120^\circ) & \cos(\theta+120^\circ)\\ -\sin\theta & -\sin(\theta-120^\circ) & -\sin(\theta+120^\circ)\\ 1/2 & 1/2 & 1/2 \end{array} \right] \left[ \begin{matrix} i_a\\ i_b\\ i_c \end{matrix} \right]

功率不变约束

[iaibic]=23[cos⁡θ−sin⁡θ12cos⁡(θ−120∘)−sin⁡(θ−120∘)12cos⁡(θ+120∘)−sin⁡(θ+120∘)12][idiqi0]\left[ \begin{matrix} i_a\\ i_b\\ i_c \end{matrix} \right] = \sqrt\frac{2}{3} \left[ \begin{array}{ccc} \cos\theta & -\sin\theta & \sqrt\frac{1}{2}\\ \cos(\theta-120^\circ) & -\sin(\theta-120^\circ) & \sqrt\frac{1}{2}\\ \cos(\theta+120^\circ) & -\sin(\theta+120^\circ) & \sqrt\frac{1}{2} \end{array} \right] \left[ \begin{matrix} i_d\\ i_q\\ i_0 \end{matrix} \right]

[idiqi0]=23[cos⁡θcos⁡(θ−120∘)cos⁡(θ+120∘)−sin⁡θ−sin⁡(θ−120∘)−sin⁡(θ+120∘)121212][iaibic]\left[ \begin{matrix} i_d\\ i_q\\ i_0 \end{matrix} \right] = \sqrt\frac{2}{3} \left[ \begin{array}{ccc} \cos\theta & \cos(\theta-120^\circ) & \cos(\theta+120^\circ)\\ -\sin\theta & -\sin(\theta-120^\circ) & -\sin(\theta+120^\circ)\\ \sqrt\frac{1}{2} & \sqrt\frac{1}{2} & \sqrt\frac{1}{2} \end{array} \right] \left[ \begin{matrix} i_a\\ i_b\\ i_c \end{matrix} \right]

dq轴电感

iabc=Cidq0ψdq0=C−1ψabc=C−1Labciabc=C−1LabcCidq0=Ldq0idq0\boldsymbol i_{abc}=\boldsymbol C \boldsymbol i_{dq0}\\ \boldsymbol \psi_{dq0}=\boldsymbol C^{-1} \boldsymbol \psi_{abc}=\boldsymbol C^{-1} \boldsymbol L_{abc} \boldsymbol i_{abc}=\boldsymbol C^{-1} \boldsymbol L_{abc} \boldsymbol C \boldsymbol i_{dq0}=\boldsymbol L_{dq0} \boldsymbol i_{dq0}

C=23[cos⁡θ−sin⁡θ12cos⁡(θ−120∘)−sin⁡(θ−120∘)12cos⁡(θ+120∘)−sin⁡(θ+120∘)12]C−1=23[cos⁡θcos⁡(θ−120∘)cos⁡(θ+120∘)−sin⁡θ−sin⁡(θ−120∘)−sin⁡(θ+120∘)121212]\boldsymbol C = \sqrt\frac{2}{3} \left[ \begin{array}{ccc} \cos\theta & -\sin\theta & \sqrt\frac{1}{2}\\ \cos(\theta-120^\circ) & -\sin(\theta-120^\circ) & \sqrt\frac{1}{2}\\ \cos(\theta+120^\circ) & -\sin(\theta+120^\circ) & \sqrt\frac{1}{2} \end{array} \right] \\ \boldsymbol C^{-1} = \sqrt\frac{2}{3} \left[ \begin{array}{ccc} \cos\theta & \cos(\theta-120^\circ) & \cos(\theta+120^\circ)\\ -\sin\theta & -\sin(\theta-120^\circ) & -\sin(\theta+120^\circ)\\ \sqrt\frac{1}{2} & \sqrt\frac{1}{2} & \sqrt\frac{1}{2} \end{array} \right]

Ldq0=C−1LabcC\boldsymbol L_{dq0}=\boldsymbol C^{-1} \boldsymbol L_{abc} \boldsymbol C

To be continued.🍇