永磁同步电机模型推导（静止坐标系+旋转坐标系）

参考文献

1. 电机传动系统控制-薛承基
2. 交流电机动态分析-汤蕴缪
3. Sychronous Machines Theory and Performances – Charles Concordia

1. 静止坐标系建模

$$V_{abc}=R_s{i}_{abc}+\frac{d}{dt}{\psi }_{abc}$$
其中：
$$\begin{gathered} V_{abc}={\left[\begin{array}{ccc}{V}_a& V_b& V_c\end{array}\right]}^T \\ {i}_{abc}={\left[\begin{array}{ccc}{i}_a& i_b& i_c\end{array}\right]}^T \\ {\psi }_{abc}={\left[\begin{array}{ccc}{\psi }_a& {\psi }_b& {\psi }_c\end{array}\right]}^T \end{gathered}$$

自感计算方法

单独推导了自感的计算方法，见单独的超链接如下。
自感计算推导
$$\begin{gathered} L_{aa}=L_{ls}+L_A+L_B\cos \left(2\theta \right) \\ {L}_{bb}=L_{ls}+L_A+L_B\cos \left(\frac{4}{3}\pi -2\theta \right) \\ {L}_{bb}=L_{ls}+L_A+L_B\cos \left(\frac{4}{3}\pi +2\theta \right) \end{gathered}$$

相间互感的计算方法

单独推导了相间互感计算方法，见单独的超链接如下。
相间互感计算方法
$$\begin{gathered} L_{ab}=-\frac{L_A}{2}+L_B\cos \left(2\theta -\frac{2}{3}\pi \right) \\ {L}_{ac}=-\frac{L_A}{2}+L_B\cos \left(2\theta +\frac{2}{3}\pi \right) \\ {L}_{bc}=-\frac{L_A}{2}+L_B\cos \left(2\theta \right) \end{gathered}$$
$L_A$表示静止坐标系下，每相自感的平均值；$L_B$表示静止坐标系下每项自感波动分量的最大值；

定子绕组交链转子磁链的部分

$$\begin{gathered} {\psi }_{af}={\psi }_f\cos \left(\theta \right) \\ {\psi }_{bf}={\psi }_f\cos \left(\theta -\frac{2}{3}\pi \right) \\ {\psi }_{cf}={\psi }_f\cos \left(\theta +\frac{2}{3}\pi \right) \end{gathered}$$

定子绕组磁链（矩阵形式）

ψ ⃗ a b c = [ L a a L a b L a c L b a L b b L b c L c a L c b L c c ] [ i a i b i c ] + [ cos ⁡ θ cos ⁡ ( 2 3 π − θ ) cos ⁡ ( 2 3 π + θ ) ] ψ f \vec{\psi}_{abc}=\left[ \begin{matrix} L_{aa}& L_{ab}& L_{ac}\\ L_{ba}& L_{bb}& L_{bc}\\ L_{ca}& L_{cb}& L_{cc}\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] +\left[ \begin{array}{c} \cos \theta\\ \cos \left( \frac{2}{3}\pi -\theta \right)\\ \cos \left( \frac{2}{3}\pi +\theta \right)\\ \end{array} \right] \psi _f ψ ​abc​= ​Laa​Lba​Lca​​Lab​Lbb​Lcb​​Lac​Lbc​Lcc​​ ​ ​ia​ib​ic​​ ​+ ​cosθcos(32​π−θ)cos(32​π+θ)​ ​ψf​

转化为空间复矢量

计算定子电流产生的磁链复矢量

L s = [ L l s + L A + L B cos ⁡ ( 2 θ a ) − 1 2 L A + L B cos ⁡ ( 2 θ − 2 3 π ) − L A 2 + L B cos ⁡ ( 2 θ + 2 3 π ) − 1 2 L A + L B cos ⁡ ( 2 θ − 2 3 π ) L l s + L A + L B cos ⁡ ( 2 θ + 2 3 π ) − L A 2 + L B cos ⁡ ( 2 θ ) − L A 2 + L B cos ⁡ ( 2 θ + 2 3 π ) − L A 2 + L B cos ⁡ ( 2 θ ) L l s + L A + L B cos ⁡ ( 2 θ − 2 3 π ) ] L_s=\left[ \begin{matrix} L_{ls}+L_A+L_B\cos \left( 2\theta _a \right)& -\frac{1}{2}L_A+L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& -\frac{L_A}{2}+L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)\\ -\frac{1}{2}L_A+L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_{ls}+L_A+L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& -\frac{L_A}{2}+L_B\cos \left( 2\theta \right)\\ -\frac{L_A}{2}+L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& -\frac{L_A}{2}+L_B\cos \left( 2\theta \right)& L_{ls}+L_A+L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)\\ \end{matrix} \right] Ls​= ​Lls​+LA​+LB​cos(2θa​)−21​LA​+LB​cos(2θ−32​π)−2LA​​+LB​cos(2θ+32​π)​−21​LA​+LB​cos(2θ−32​π)Lls​+LA​+LB​cos(2θ+32​π)−2LA​​+LB​cos(2θ)​−2LA​​+LB​cos(2θ+32​π)−2LA​​+LB​cos(2θ)Lls​+LA​+LB​cos(2θ−32​π)​ ​
将电感矩阵可以分拆为三个电感矩阵以便计算：
L s = L 1 + L 2 + L 3 L 1 = [ L l s 0 0 0 L l s 0 0 0 L l s ] ; L 2 = [ L A − 1 2 L A − 1 2 L A − 1 2 L A L A − 1 2 L A − 1 2 L A − 1 2 L A L A ] ; L 3 = [ L B cos ⁡ ( 2 θ a ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ − 2 3 π ) ] L_s=L_1+L_2+L_3 \\ L_1=\left[ \begin{matrix} L_{ls}& 0& 0\\ 0& L_{ls}& 0\\ 0& 0& L_{ls}\\ \end{matrix} \right] ;L_2=\left[ \begin{matrix} L_A& -\frac{1}{2}L_A& -\frac{1}{2}L_A\\ -\frac{1}{2}L_A& L_A& -\frac{1}{2}L_A\\ -\frac{1}{2}L_A& -\frac{1}{2}L_A& L_A\\ \end{matrix} \right] ; \\ L_3=\left[ \begin{matrix} L_B\cos \left( 2\theta _a \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)\\ L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)\\ L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)\\ \end{matrix} \right] Ls​=L1​+L2​+L3​L1​= ​Lls​00​0Lls​0​00Lls​​ ​;L2​= ​LA​−21​LA​−21​LA​​−21​LA​LA​−21​LA​​−21​LA​−21​LA​LA​​ ​;L3​= ​LB​cos(2θa​)LB​cos(2θ−32​π)LB​cos(2θ+32​π)​LB​cos(2θ−32​π)LB​cos(2θ+32​π)LB​cos(2θ)​LB​cos(2θ+32​π)LB​cos(2θ)LB​cos(2θ−32​π)​ ​

计算磁链复矢量：
ψ ⃗ s = 2 3 [ 1 a a 2 ] [ ψ a ψ b ψ c ] ψ ⃗ s = 2 3 [ 1 a a 2 ] [ L 1 + L 2 + L 3 ] [ i a i b i c ] \vec{\psi}_s=\frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{array}{c} \psi _a\\ \psi _b\\ \psi _c\\ \end{array} \right] \\ \vec{\psi}_s=\frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ L_1+L_2+L_3 \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] ψ ​s​=32​[1​a​a2​] ​ψa​ψb​ψc​​ ​ψ ​s​=32​[1​a​a2​][L1​+L2​+L3​] ​ia​ib​ic​​ ​
带入L1,L2,L3可得：
ψ ⃗ s = 2 3 [ 1 a a 2 ] [ L l s 0 0 0 L l s 0 0 0 L l s ] [ i a i b i c ] + 2 3 [ 1 a a 2 ] [ L A − 1 2 L A − 1 2 L A − 1 2 L A L A − 1 2 L A − 1 2 L A − 1 2 L A L A ] [ i a i b i c ] + 2 3 [ 1 a a 2 ] [ L B cos ⁡ ( 2 θ a ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ − 2 3 π ) ] [ i a i b i c ] \vec{\psi}_s=\frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{matrix} L_{ls}& 0& 0\\ 0& L_{ls}& 0\\ 0& 0& L_{ls}\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] +\frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{matrix} L_A& -\frac{1}{2}L_A& -\frac{1}{2}L_A\\ -\frac{1}{2}L_A& L_A& -\frac{1}{2}L_A\\ -\frac{1}{2}L_A& -\frac{1}{2}L_A& L_A\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] \\ +\frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{matrix} L_B\cos \left( 2\theta _a \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)\\ L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)\\ L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] ψ ​s​=32​[1​a​a2​] ​Lls​00​0Lls​0​00Lls​​ ​ ​ia​ib​ic​​ ​+32​[1​a​a2​] ​LA​−21​LA​−21​LA​​−21​LA​LA​−21​LA​​−21​LA​−21​LA​LA​​ ​ ​ia​ib​ic​​ ​+32​[1​a​a2​] ​LB​cos(2θa​)LB​cos(2θ−32​π)LB​cos(2θ+32​π)​LB​cos(2θ−32​π)LB​cos(2θ+32​π)LB​cos(2θ)​LB​cos(2θ+32​π)LB​cos(2θ)LB​cos(2θ−32​π)​ ​ ​ia​ib​ic​​ ​
前两项可以可以容易计算得到：
$$L_{ls}{i}_s+\frac{3}{2}{L}_A{i}_s$$
下面着重计算第三项：
2 3 [ 1 a a 2 ] [ L B cos ⁡ ( 2 θ a ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ − 2 3 π ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ + 2 3 π ) L B cos ⁡ ( 2 θ ) L B cos ⁡ ( 2 θ − 2 3 π ) ] [ i a i b i c ] \frac{2}{3}\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{matrix} L_B\cos \left( 2\theta _a \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)\\ L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)\\ L_B\cos \left( 2\theta +\frac{2}{3}\pi \right)& L_B\cos \left( 2\theta \right)& L_B\cos \left( 2\theta -\frac{2}{3}\pi \right)\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] 32​[1​a​a2​] ​LB​cos(2θa​)LB​cos(2θ−32​π)LB​cos(2θ+32​π)​LB​cos(2θ−32​π)LB​cos(2θ+32​π)LB​cos(2θ)​LB​cos(2θ+32​π)LB​cos(2θ)LB​cos(2θ−32​π)​ ​ ​ia​ib​ic​​ ​
根据欧拉公式有：
$$a=e^{j\frac{2\pi }{3}};a^2=e^{-j\frac{2\pi }{3}};\cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2}$$

因此可化简为：
2 3 L B [ 1 a a 2 ] [ cos ⁡ ( 2 θ a ) cos ⁡ ( 2 θ a − 2 3 π ) cos ⁡ ( 2 θ a + 2 3 π ) cos ⁡ ( 2 θ a − 2 3 π ) cos ⁡ ( 2 θ a + 2 3 π ) cos ⁡ ( 2 θ a ) cos ⁡ ( 2 θ a + 2 3 π ) cos ⁡ ( 2 θ a ) cos ⁡ ( 2 θ a − 2 3 π ) ] [ i a i b i c ] = 1 3 L B [ 1 e j 2 π 3 e − j 2 π 3 ] [ e j 2 θ a e j ( 2 θ a − 2 3 π ) e j ( 2 θ a + 2 3 π ) e j ( 2 θ a − 2 3 π ) e j ( 2 θ a + 2 3 π ) e j 2 θ a e j ( 2 θ a + 2 3 π ) e j 2 θ a e j ( 2 θ a − 2 3 π ) ] [ i a i b i c ] + 1 3 L B [ 1 e j 2 π 3 e − j 2 π 3 ] [ e − j 2 θ a e − j ( 2 θ a − 2 3 π ) e − j ( 2 θ a + 2 3 π ) e j ( − 2 θ a + 2 3 π ) e − j ( 2 θ a + 2 3 π ) e − j 2 θ a e j ( − 2 θ a − 2 3 π ) e − j 2 θ a e − j ( 2 θ a − 2 3 π ) ] [ i a i b i c ] \frac{2}{3}L_B\left[ \begin{matrix} 1& a& a^2\\ \end{matrix} \right] \left[ \begin{matrix} \cos \left( 2\theta _a \right)& \cos \left( 2\theta _a-\frac{2}{3}\pi \right)& \cos \left( 2\theta _a+\frac{2}{3}\pi \right)\\ \cos \left( 2\theta _a-\frac{2}{3}\pi \right)& \cos \left( 2\theta _a+\frac{2}{3}\pi \right)& \cos \left( 2\theta _a \right)\\ \cos \left( 2\theta _a+\frac{2}{3}\pi \right)& \cos \left( 2\theta _a \right)& \cos \left( 2\theta _a-\frac{2}{3}\pi \right)\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] \\ =\frac{1}{3}L_B\left[ \begin{matrix} 1& e^{j\frac{2\pi}{3}}& e^{-j\frac{2\pi}{3}}\\ \end{matrix} \right] \left[ \begin{matrix} e^{j2\theta _a}& e^{j\left( 2\theta _a-\frac{2}{3}\pi \right)}& e^{j\left( 2\theta _a+\frac{2}{3}\pi \right)}\\ e^{j\left( 2\theta _a-\frac{2}{3}\pi \right)}& e^{j\left( 2\theta _a+\frac{2}{3}\pi \right)}& e^{j2\theta _a}\\ e^{j\left( 2\theta _a+\frac{2}{3}\pi \right)}& e^{j2\theta _a}& e^{j\left( 2\theta _a-\frac{2}{3}\pi \right)}\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] \\ +\frac{1}{3}L_B\left[ \begin{matrix} 1& e^{j\frac{2\pi}{3}}& e^{-j\frac{2\pi}{3}}\\ \end{matrix} \right] \left[ \begin{matrix} e^{-j2\theta _a}& e^{-j\left( 2\theta _a-\frac{2}{3}\pi \right)}& e^{-j\left( 2\theta _a+\frac{2}{3}\pi \right)}\\ e^{j\left( -2\theta _a+\frac{2}{3}\pi \right)}& e^{-j\left( 2\theta _a+\frac{2}{3}\pi \right)}& e^{-j2\theta _a}\\ e^{j\left( -2\theta _a-\frac{2}{3}\pi \right)}& e^{-j2\theta _a}& e^{-j\left( 2\theta _a-\frac{2}{3}\pi \right)}\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] 32​LB​[1​a​a2​] ​cos(2θa​)cos(2θa​−32​π)cos(2θa​+32​π)​cos(2θa​−32​π)cos(2θa​+32​π)cos(2θa​)​cos(2θa​+32​π)cos(2θa​)cos(2θa​−32​π)​ ​ ​ia​ib​ic​​ ​=31​LB​[1​ej32π​​e−j32π​​] ​ej2θa​ej(2θa​−32​π)ej(2θa​+32​π)​ej(2θa​−32​π)ej(2θa​+32​π)ej2θa​​ej(2θa​+32​π)ej2θa​ej(2θa​−32​π)​ ​ ​ia​ib​ic​​ ​+31​LB​[1​ej32π​​e−j32π​​] ​e−j2θa​ej(−2θa​+32​π)ej(−2θa​−32​π)​e−j(2θa​−32​π)e−j(2θa​+32​π)e−j2θa​​e−j(2θa​+32​π)e−j2θa​e−j(2θa​−32​π)​ ​ ​ia​ib​ic​​ ​
= 1 3 L B [ 3 e j 2 θ a 3 e j ( 2 θ a − 2 3 π ) 3 e j ( 2 θ a + 2 3 π ) ] [ i a i b i c ] + 1 3 L B [ 0 0 0 ] [ i a i b i c ] = 3 2 L B i ⃗ s ∗ e j 2 θ a i ⃗ s ∗ 为共轭电流矢量,因此可得定子电流产生的定子磁链分量为： ψ ⃗ s 1 = ( L l s + 3 2 L A ) i ⃗ s + 3 2 L B i ⃗ s ∗ e j 2 θ a =\frac{1}{3}L_B\left[ \begin{matrix} 3e^{j2\theta _a}& 3e^{j\left( 2\theta _a-\frac{2}{3}\pi \right)}& 3e^{j\left( 2\theta _a+\frac{2}{3}\pi \right)}\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] +\frac{1}{3}L_B\left[ \begin{matrix} 0& 0& 0\\ \end{matrix} \right] \left[ \begin{array}{c} i_a\\ i_b\\ i_c\\ \end{array} \right] \\ =\frac{3}{2}L_B\vec{i}_{s}^{*}e^{j2\theta _a} \\ \vec{i}_{s}^{*}\text{为共轭电流矢量,} 因此可得定子电流产生的定子磁链分量为： \\ \vec{\psi}_{s1}=\left( L_{ls}+\frac{3}{2}L_A \right) \vec{i}_s+\frac{3}{2}L_B\vec{i}_{s}^{*}e^{j2\theta _a} =31​LB​[3ej2θa​​3ej(2θa​−32​π)​3ej(2θa​+32​π)​] ​ia​ib​ic​​ ​+31​LB​[0​0​0​] ​ia​ib​ic​​ ​=23​LB​i s∗​ej2θa​i s∗​为共轭电流矢量,因此可得定子电流产生的定子磁链分量为：ψ ​s1​=(Lls​+23​LA​)i s​+23​LB​i s∗​ej2θa​

计算转子磁场在定子产生的磁链

ψ ⃗ s f = 2 3 [ 1 e j 2 π 3 e − j 2 π 3 ] [ cos ⁡ θ a cos ⁡ ( 2 3 π − θ a ) cos ⁡ ( 2 3 π + θ a ) ] ψ f = 1 3 [ 1 e j 2 π 3 e − j 2 π 3 ] [ e j θ a e j ( θ a − 2 3 π ) e j ( θ a + 2 3 π ) ] ψ f + 1 3 [ 1 e j 2 π 3 e − j 2 π 3 ] [ e − j θ a e − j ( θ a − 2 3 π ) e − j ( θ a + 2 3 π ) ] ψ f = ψ f e j θ a \vec{\psi}_{sf}=\frac{2}{3}\left[ \begin{matrix} 1& e^{j\frac{2\pi}{3}}& e^{-j\frac{2\pi}{3}}\\ \end{matrix} \right] \left[ \begin{array}{c} \cos \theta _a\\ \cos \left( \frac{2}{3}\pi -\theta _a \right)\\ \cos \left( \frac{2}{3}\pi +\theta _a \right)\\ \end{array} \right] \psi _f \\ =\frac{1}{3}\left[ \begin{matrix} 1& e^{j\frac{2\pi}{3}}& e^{-j\frac{2\pi}{3}}\\ \end{matrix} \right] \left[ \begin{array}{c} e^{j\theta _a}\\ e^{j\left( \theta _a-\frac{2}{3}\pi \right)}\\ e^{j\left( \theta _a+\frac{2}{3}\pi \right)}\\ \end{array} \right] \psi _f+\frac{1}{3}\left[ \begin{matrix} 1& e^{j\frac{2\pi}{3}}& e^{-j\frac{2\pi}{3}}\\ \end{matrix} \right] \left[ \begin{array}{c} e^{-j\theta _a}\\ e^{-j\left( \theta _a-\frac{2}{3}\pi \right)}\\ e^{-j\left( \theta _a+\frac{2}{3}\pi \right)}\\ \end{array} \right] \psi _f \\ =\psi _fe^{j\theta _a} ψ ​sf​=32​[1​ej32π​​e−j32π​​] ​cosθa​cos(32​π−θa​)cos(32​π+θa​)​ ​ψf​=31​[1​ej32π​​e−j32π​​] ​ejθa​ej(θa​−32​π)ej(θa​+32​π)​ ​ψf​+31​[1​ej32π​​e−j32π​​] ​e−jθa​e−j(θa​−32​π)e−j(θa​+32​π)​ ​ψf​=ψf​ejθa​

定子总磁链

$${\psi }_s=\left(L_{ls}+\frac{3}{2}{L}_A\right)i_s+\frac{3}{2}{L}_B{i}_s^{\ast }{e}^{j2{\theta }_a}+{\psi }_f{e}^{j{\theta }_a}$$

电压方程的复失量形式

$$\begin{gathered} u_s=R_s{i}_s+\frac{d}{dt}{\psi }_s \\ {u}_s=R_s{i}_s+\frac{d}{dt}\left[\left(L_{ls}+\frac{3}{2}{L}_A\right)i_s+\frac{3}{2}{L}_B{i}_s^{\ast }{e}^{j2{\theta }_a}+{\psi }_f{e}^{j{\theta }_a}\right] \\ {u}_s{e}^{-j{\theta }_a}=R_s{i}_s{e}^{-j{\theta }_a}+e^{-j{\theta }_a}\frac{d}{dt}\left[\left(L_{ls}+\frac{3}{2}{L}_A\right)\left(i_{dqs}{e}^{j{\theta }_a}\right)+\frac{3}{2}{L}_B{i}_s^{\ast }{e}^{j2{\theta }_a}+{\psi }_f{e}^{j{\theta }_a}\right] \\ \text{根据以下公式：} \\ \frac{3}{2}{L}_B{i}_s^{\ast }{e}^{j2{\theta }_a}=\frac{3}{2}{L}_B{i}_{dqs}^{\ast }{e}^{j{\theta }_a} \\ {u}_{dqs}=R_s{i}_{dqs}+\left(L_{ls}+\frac{3}{2}{L}_A\right)\frac{d}{dt}{i}_{dqs}+j{\omega }_r\left(L_{ls}+\frac{3}{2}{L}_A\right)i_{dqs}+\frac{3}{2}{L}_B\frac{d}{dt}{i}_{dqs}^{\ast }+j{\omega }_r\frac{3}{2}{L}_B{i}_{dqs}^{\ast }+j{\omega }_r{\Psi }_f \end{gathered}$$
$$\begin{gathered} \text{根据以下公式：}{i}_{dqs}^{\ast }=i_d-j{i}_q,i_{dqs}=i_d+j{i}_q \\ \text{可以化简得到电压方程：} \\ {u}_{sd}=R_s{i}_{sd}+\left(L_{ls}+\frac{3}{2}{L}_A\right)\frac{d}{dt}{i}_{sd}+\frac{3}{2}{L}_B\frac{d}{dt}{i}_{sd}-{\omega }_r\left(L_{ls}+\frac{3}{2}{L}_A\right)i_{sq}+{\omega }_r\frac{3}{2}{L}_B{i}_{sq} \\ {u}_{sd}=R_s{i}_{sq}+\left(L_{ls}+\frac{3}{2}{L}_A\right)\frac{d}{dt}{i}_{sq}-\frac{3}{2}{L}_B\frac{d}{dt}{i}_{sd}+{\omega }_r\left(L_{ls}+\frac{3}{2}{L}_A\right)i_{sd}+{\omega }_r\frac{3}{2}{L}_B{i}_{sd}+{\omega }_r{\Psi }_f \\ \text{令：}{L}_d=L_{ls}+\frac{3}{2}\left(L_A+L_B\right);L_q=L_{ls}+\frac{3}{2}\left(L_A-L_B\right) \\ \text{化简可得：}{u}_{sd}=R_s{i}_{sd}+L_d\frac{d{i}_{sd}}{dt}-{\omega }_r{L}_q{i}_{sq} \\ \text{同理可得：}{u}_{sd}=R_s{i}_{sq}+L_q\frac{d{i}_{sq}}{dt}+{\omega }_r{L}_d{i}_{sd}+{\omega }_r{\Psi }_f \end{gathered}$$

静止两相坐标系模型推导

静止坐标系下推导磁链矢量：
$$\begin{gathered} {\psi }_s=\left(L_{ls}+\frac{3}{2}{L}_A\right)i_s+\frac{3}{2}{L}_B{i}_s^{\ast }{e}^{j2{\theta }_a}+{\psi }_f{e}^{j{\theta }_a} \\ {\psi }_s=\left(L_{ls}+\frac{3}{2}{L}_A\right)\left(i_{\alpha }+j{i}_{\beta }\right)+\frac{3}{2}{L}_B\left(i_{\alpha }-j{i}_{\beta }\right)\left(\cos 2{\theta }_r+j\sin 2{\theta }_r\right)+{\psi }_f\left(\cos {\theta }_r+j\sin {\theta }_r\right) \\ {\psi }_s=\left[\left(L_{ls}+\frac{3}{2}{L}_A\right)+\frac{3}{2}{L}_B\cos 2{\theta }_r\right]i_{\alpha }+\frac{3}{2}{L}_B\sin 2{\theta }_r{i}_{\beta }+{\psi }_f\cos {\theta }_r \\ +j\left[\left(L_{ls}+\frac{3}{2}{L}_A\right)-\frac{3}{2}{L}_B\cos 2{\theta }_r\right]i_{\beta }+j\frac{3}{2}{L}_B\sin 2{\theta }_r{i}_{\alpha }+j{\psi }_f\sin {\theta }_r \\ {\psi }_s=\left[\begin{array}{cc}\left(L_{ls}+\frac{3}{2}{L}_A\right)+\frac{3}{2}{L}_B\cos 2{\theta }_r& \frac{3}{2}{L}_B\sin 2{\theta }_r\\ \frac{3}{2}{L}_B\sin 2{\theta }_r& \left(L_{ls}+\frac{3}{2}{L}_A\right)-\frac{3}{2}{L}_B\cos 2{\theta }_r\end{array}\right]\left[\begin{array}{c}{i}_{\alpha }\\ i_{\beta }\end{array}\right]+\left[\begin{array}{c}{\psi }_f\cos {\theta }_r\\ {\psi }_f\sin {\theta }_r\end{array}\right] \end{gathered}$$
进行变量代换：
$$\begin{gathered} \text{根据电感定义：} \\ {L}_d=L_{ls}+\frac{3}{2}{L}_A+\frac{3}{2}{L}_B \\ {L}_q=L_{ls}+\frac{3}{2}{L}_A-\frac{3}{2}{L}_B \\ \text{可得：}{L}_{ls}+\frac{3}{2}{L}_A=\frac{L_d+L_q}{2},\frac{3}{2}{L}_B=\frac{L_d-L_q}{2} \\ \text{分别令：}{L}_{ls}+\frac{3}{2}{L}_A=L_1\text{，}\frac{3}{2}{L}_B=L_2 \\ \text{因此：} \\ {\psi }_s=\left[\begin{array}{cc}{L}_1+L_2\cos 2{\theta }_r& L_2\sin 2{\theta }_r\\ L_2\sin 2{\theta }_r& L_1-L_2\cos 2{\theta }_r\end{array}\right]\left[\begin{array}{c}{i}_{\alpha }\\ i_{\beta }\end{array}\right]+\left[\begin{array}{c}{\psi }_f\cos {\theta }_r\\ {\psi }_f\sin {\theta }_r\end{array}\right] \end{gathered}$$
因此可得静止两相坐标系下的电压方程：
$$\begin{gathered} u_s=R_s{i}_s+\frac{d}{dt}{\psi }_s \\ {\psi }_s=[\begin{array}{c}{L}_1+L_2\cos 2{\theta }_r\\ L_2\end{array} \end{gathered}$$