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  • 坐标变换的艺术—PMSM高频注入法公式推导
    2022-04-30 15:31:22

    坐标变换的艺术—高频注入法公式推导

    PMSM无位置传感器技术总体而言可分为两个部分,第一部分获知位置误差信号然后输入位置观测器进行处理,使得位置信号收敛,第二部分是极性判断。本文着重讲解第一部分中位置误差信号的获取,至于后续的观测器、极性判断等环节,以及另外的感应电机的无速度传感器技术会在文末给出学习资料。

    PMSM高频模型的由来

    PMSM在旋转轴系的电压为
    [ u d u q ] = R [ i d i q ] + d d t [ ψ d ψ q ] + ω e [ − ψ q ψ d ] (1) \tag{1} \left[ \begin{array}{c} u_d\\ u_q\\ \end{array} \right] =R\left[ \begin{array}{c} i_d\\ i_q\\ \end{array} \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} \psi _d\\ \psi _q\\ \end{array} \right] +\omega _e\left[ \begin{array}{c} -\psi _q\\ \psi _d\\ \end{array} \right] [uduq]=R[idiq]+dtd[ψdψq]+ωe[ψqψd](1)
    在零低速工况,基于一些假设,公式 ( 1 ) (1) (1)被进一步简化,习惯将简化后的方程称之为高频模型。通常,电机的电路模型视为R-L电路,控制回路中注入高频信号,此时电路中占据主导地位的是电感项,又因为电机运行于零低速,因此旋转电压方程中的第一项和第三项可忽略,则PMSM的高频电压模型为
    [ u d h u q h ] = d d t [ ψ d ψ q ] = [ L d 0 0 L q ] d d t [ i d h i q h ] (2) \tag{2} \left[ \begin{array}{c} u_{dh}\\ u_{qh}\\ \end{array} \right] =\frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} \psi _d\\ \psi _q\\ \end{array} \right] =\left[ \begin{matrix} L_d& 0\\ 0& L_q\\ \end{matrix} \right] \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{dh}\\ i_{qh}\\ \end{array} \right] [udhuqh]=dtd[ψdψq]=[Ld00Lq]dtd[idhiqh](2)
    高频电压模型是高频注入法的核心所在,由此衍生出了旋转高频注入法、脉振高频注入法(脉振正弦和脉振方波)。

    公式推导

    为了便于公式推导,将式 ( 2 ) (2) (2)转换成电流微分表达式,可得:
    d d t [ i d h i q h ] = [ 1 L d 0 0 1 L q ] [ u d h u q h ] (3) \tag{3} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{dh}\\ i_{qh}\\ \end{array} \right] =\left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{array}{c} u_{dh}\\ u_{qh}\\ \end{array} \right] dtd[idhiqh]=[Ld100Lq1][udhuqh](3)
    所有的高频方波注入法遵循坐标变换理论,各坐标系间的关系如下图所示,

    估计轴系至旋转轴系的坐标变换矩阵满足
    P γ δ / d q = [ cos ⁡ θ ~ sin ⁡ θ ~ − sin ⁡ θ ~ cos ⁡ θ ~ ] (4) \tag{4} P_{\gamma \delta /dq}=\left[ \begin{matrix} \cos \tilde{\theta}& \sin \tilde{\theta}\\ -\sin \tilde{\theta}& \cos \tilde{\theta}\\ \end{matrix} \right] Pγδ/dq=[cosθ~sinθ~sinθ~cosθ~](4)

    静止轴系至旋转轴系的坐标变换矩阵满足

    P α β / d q = [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] (5) \tag{5} P_{\alpha \beta /dq}=\left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] Pαβ/dq=[cosθesinθesinθecosθe](5)

    上述 ( 4 ) (4) (4) ( 5 ) (5) (5)是单位正交矩阵,满足
    P − 1 = P T (6) \tag{6} P^{-1}=P^T P1=PT(6)

    为了读者更好的理解下文各高频注入法的公式推导过程,现将各方法的信号注入、响应信号提取轴系罗列成表,作为推导线索。

    算法名称高频电压信号注入轴系高频电流提取轴系
    旋转高频注入法两相静止轴系两相静止轴系
    脉振高频正弦注入法估计轴系估计轴系
    脉振高频方波注入法估计轴系静止轴系
    旋转高频注入法

    旋转高频注入法在静止轴系完成信号注入、响应信号提取过程,依据坐标变换理论,可得:
    [ i d h i q h ] = [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] [ i α i β ] (7) \tag{7} \left[ \begin{array}{c} i_{dh}\\ i_{qh}\\ \end{array} \right] =\left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{array}{c} i_{\alpha}\\ i_{\beta}\\ \end{array} \right] [idhiqh]=[cosθesinθesinθecosθe][iαiβ](7)

    [ u d h u q h ] = [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] [ u α u β ] (8) \tag{8} \left[ \begin{array}{c} u_{dh}\\ u_{qh}\\ \end{array} \right] =\left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{array}{c} u_{\alpha}\\ u_{\beta}\\ \end{array} \right] [udhuqh]=[cosθesinθesinθecosθe][uαuβ](8)

    结合式 ( 7 ) (7) (7),对式 ( 3 ) (3) (3)中等式左边的微分项化简,可得:
    d d t [ i d h i q h ] = d d t { [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] [ i α i β ] } = [ − sin ⁡ θ e cos ⁡ θ e − cos ⁡ θ e − sin ⁡ θ e ] ω e [ i α i β ] + [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] d d t [ i α i β ] (9) \tag{9} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{dh}\\ i_{qh}\\ \end{array} \right] =\frac{\mathrm{d}}{\mathrm{d}t}\left\{ \left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{array}{c} i_{\alpha}\\ i_{\beta}\\ \end{array} \right] \right\} =\left[ \begin{matrix} -\sin \theta _e& \cos \theta _e\\ -\cos \theta _e& -\sin \theta _e\\ \end{matrix} \right] \omega _e\left[ \begin{array}{c} i_{\alpha}\\ i_{\beta}\\ \end{array} \right] +\left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha}\\ i_{\beta}\\ \end{array} \right] dtd[idhiqh]=dtd{[cosθesinθesinθecosθe][iαiβ]}=[sinθecosθecosθesinθe]ωe[iαiβ]+[cosθesinθesinθecosθe]dtd[iαiβ](9)
    分析式 ( 9 ) (9) (9),展开式中的第一项含有速度项,因电机运行于零低速工况,故第一项略去,保留第二项,可得:
    d d t [ i d h i q h ] = [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] d d t [ i α i β ] (10) \tag{10} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{dh}\\ i_{qh}\\ \end{array} \right] =\left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha}\\ i_{\beta}\\ \end{array} \right] dtd[idhiqh]=[cosθesinθesinθecosθe]dtd[iαiβ](10)
    ( 8 ) (8) (8) ( 10 ) (10) (10)带入式 ( 3 ) (3) (3),结合式 ( 6 ) (6) (6),式 ( 3 ) (3) (3)可重写为
    d d t [ i α h i β h ] = [ cos ⁡ θ e − sin ⁡ θ e sin ⁡ θ e cos ⁡ θ e ] [ 1 L d 0 0 1 L q ] [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] [ u α h u β h ] (11) \tag{11} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] =\left[ \begin{matrix} \cos \theta _e& -\sin \theta _e\\ \sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{array}{c} u_{\alpha h}\\ u_{\beta h}\\ \end{array} \right] dtd[iαhiβh]=[cosθesinθesinθecosθe][Ld100Lq1][cosθesinθesinθecosθe][uαhuβh](11)
    对式 ( 11 ) (11) (11)中等式右边的矩阵乘积因子化简,可得:
    [ cos ⁡ θ e − sin ⁡ θ e sin ⁡ θ e cos ⁡ θ e ] [ 1 L d 0 0 1 L q ] [ cos ⁡ θ e sin ⁡ θ e − sin ⁡ θ e cos ⁡ θ e ] ⇒ [ 1 2 ( 1 L d + 1 L q ) + 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ e 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ e 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ e 1 2 ( 1 L d + 1 L q ) − 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ e ] (12) \tag{12} \begin{aligned} &\left[ \begin{matrix} \cos \theta _e& -\sin \theta _e\\ \sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{matrix} \cos \theta _e& \sin \theta _e\\ -\sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \\ \Rightarrow& \left[ \begin{matrix} \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) +\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\theta _e& \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\theta _e\\ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\theta _e& \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\theta _e\\ \end{matrix} \right] \end{aligned} [cosθesinθesinθecosθe][Ld100Lq1][cosθesinθesinθecosθe]21(Ld1+Lq1)+21(Ld1Lq1)cos2θe21(Ld1Lq1)sin2θe21(Ld1Lq1)sin2θe21(Ld1+Lq1)21(Ld1Lq1)cos2θe(12)
    注入的高频信号满足下式
    [ u α h u β h ] = [ V h cos ⁡ ω h t V h sin ⁡ ω h t ] (13) \tag{13} \left[ \begin{array}{c} u_{\alpha h}\\ u_{\beta h}\\ \end{array} \right] =\left[ \begin{array}{c} V_h\cos \omega _ht\\ V_h\sin \omega _ht\\ \end{array} \right] [uαhuβh]=[VhcosωhtVhsinωht](13)
    将式 ( 12 ) (12) (12) ( 13 ) (13) (13)带入式 ( 11 ) (11) (11),可得:
    d d t [ i α h i β h ] = [ 1 2 ( 1 L d + 1 L q ) + 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ e 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ e 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ e 1 2 ( 1 L d + 1 L q ) − 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ e ] [ V h cos ⁡ ω h t V h sin ⁡ ω h t ] (14) \tag{14} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] =\left[ \begin{matrix} \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) +\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\theta _e& \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\theta _e\\ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\theta _e& \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\theta _e\\ \end{matrix} \right] \left[ \begin{array}{c} V_h\cos \omega _ht\\ V_h\sin \omega _ht\\ \end{array} \right] dtd[iαhiβh]=21(Ld1+Lq1)+21(Ld1Lq1)cos2θe21(Ld1Lq1)sin2θe21(Ld1Lq1)sin2θe21(Ld1+Lq1)21(Ld1Lq1)cos2θe[VhcosωhtVhsinωht](14)

    通过积分运算,两相静止轴系的高频响应电流为:
    [ i α h i β h ] = [ 1 2 ( 1 L d + 1 L q ) V h ω h sin ⁡ ω h t − 1 2 ( 1 L d − 1 L q ) V h ω h sin ⁡ ( − ω h t + 2 θ e ) − 1 2 ( 1 L d + 1 L q ) V h ω h cos ⁡ ω h t + 1 2 ( 1 L d − 1 L q ) V h ω h cos ⁡ ( 2 θ e − ω h t ) ] (15) \tag{15} \left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] =\left[ \begin{array}{c} \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \omega _ht-\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \left( -\omega _ht+2\theta _e \right)\\ -\frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\cos \omega _ht+\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\cos \left( 2\theta _e-\omega _ht \right)\\ \end{array} \right] [iαhiβh]=21(Ld1+Lq1)ωhVhsinωht21(Ld1Lq1)ωhVhsin(ωht+2θe)21(Ld1+Lq1)ωhVhcosωht+21(Ld1Lq1)ωhVhcos(2θeωht)(15)

    将高频电流变换至高频轴系,可得:
    [ i d h ′ i q h ′ ] = [ cos ⁡ ω h t sin ⁡ ω h t − sin ⁡ ω h t cos ⁡ ω h t ] [ i α h i β h ] (16) \tag{16} \left[ \begin{array}{c} i_{dh}^{'}\\ i_{qh}^{'}\\ \end{array} \right] =\left[ \begin{matrix} \cos \omega _ht& \sin \omega _ht\\ -\sin \omega _ht& \cos \omega _ht\\ \end{matrix} \right] \left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] [idhiqh]=[cosωhtsinωhtsinωhtcosωht][iαhiβh](16)
    则高频轴系的响应电流为
    [ i d h ′ i q h ′ ] = [ − 1 2 ( 1 L d − 1 L q ) V h ω h sin ⁡ ( − 2 ω h t + 2 θ e ) − 1 2 ( 1 L d + 1 L q ) V h ω h + 1 2 ( 1 L d − 1 L q ) V h ω h cos ⁡ ( − 2 ω h t + 2 θ e ) ] (17) \tag{17} \left[ \begin{array}{c} i_{dh}^{'}\\ i_{qh}^{'}\\ \end{array} \right] =\left[ \begin{array}{c} -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \left( -2\omega _ht+2\theta _e \right)\\ -\frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) \frac{V_h}{\omega _h}+\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\cos \left( -2\omega _ht+2\theta _e \right)\\ \end{array} \right] [idhiqh]=21(Ld1Lq1)ωhVhsin(2ωht+2θe)21(Ld1+Lq1)ωhVh+21(Ld1Lq1)ωhVhcos(2ωht+2θe)(17)

    在高频轴系,使用高通滤波器滤波,然后电流变换至两相静止轴系,可得:
    [ i α h ∗ i β h ∗ ] = [ cos ⁡ ω h t − sin ⁡ ω h t sin ⁡ ω h t cos ⁡ ω h t ] H P F { [ i d h ′ i q h ′ ] } (18) \tag{18} \left[ \begin{array}{c} i_{\alpha h}^{*}\\ i_{\beta h}^{*}\\ \end{array} \right] =\left[ \begin{matrix} \cos \omega _ht& -\sin \omega _ht\\ \sin \omega _ht& \cos \omega _ht\\ \end{matrix} \right] HPF\left\{ \left[ \begin{array}{c} i_{dh}^{'}\\ i_{qh}^{'}\\ \end{array} \right] \right\} [iαhiβh]=[cosωhtsinωhtsinωhtcosωht]HPF{[idhiqh]}(18)
    此时,两相静止轴系的高频电流变为
    [ i α h ∗ i β h ∗ ] = [ − 1 2 ( 1 L d − 1 L q ) V h ω h sin ⁡ ( − ω h t + 2 θ e ) 1 2 ( 1 L d − 1 L q ) V h ω h cos ⁡ ( − ω h t + 2 θ e ) ] (19) \tag{19} \left[ \begin{array}{c} i_{\alpha h}^{*}\\ i_{\beta h}^{*}\\ \end{array} \right] =\left[ \begin{array}{c} -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \left( -\omega _ht+2\theta _e \right)\\ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\cos \left( -\omega _ht+2\theta _e \right)\\ \end{array} \right] [iαhiβh]=21(Ld1Lq1)ωhVhsin(ωht+2θe)21(Ld1Lq1)ωhVhcos(ωht+2θe)(19)
    通过外差法可得位置误差信号
    i α h ∗ cos ⁡ ( − ω h t + 2 θ ^ e ) + i β h ∗ sin ⁡ ( − ω h t + 2 θ ^ e ) ⇒ − 1 2 ( 1 L d − 1 L q ) V h ω h sin ⁡ ( − ω h t + 2 θ e ) cos ⁡ ( − ω h t + 2 θ ^ e ) + 1 2 ( 1 L d − 1 L q ) V h ω h cos ⁡ ( − ω h t + 2 θ e ) sin ⁡ ( − ω h t + 2 θ ^ e ) ⇒ − 1 2 ( 1 L d − 1 L q ) V h ω h sin ⁡ ( 2 θ e − 2 θ ^ e ) ≈ − 1 2 ( 1 L d − 1 L q ) V h ω h ( 2 θ e − 2 θ ^ e ) = ( 1 L q − 1 L d ) V h ω h ( θ e − θ ^ e ) (20) \tag{20} \begin{aligned} &i_{\alpha h}^{*}\cos \left( -\omega _ht+2\hat{\theta}_e \right) +i_{\beta h}^{*}\sin \left( -\omega _ht+2\hat{\theta}_e \right) \\ \Rightarrow& -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \left( -\omega _ht+2\theta _e \right) \cos \left( -\omega _ht+2\hat{\theta}_e \right) +\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\cos \left( -\omega _ht+2\theta _e \right) \sin \left( -\omega _ht+2\hat{\theta}_e \right) \\ \Rightarrow& -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \left( 2\theta _e-2\hat{\theta}_e \right) \approx -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\left( 2\theta _e-2\hat{\theta}_e \right) =\left( \frac{1}{L_q}-\frac{1}{L_d} \right) \frac{V_h}{\omega _h}\left( \theta _e-\hat{\theta}_e \right) \end{aligned} iαhcos(ωht+2θ^e)+iβhsin(ωht+2θ^e)21(Ld1Lq1)ωhVhsin(ωht+2θe)cos(ωht+2θ^e)+21(Ld1Lq1)ωhVhcos(ωht+2θe)sin(ωht+2θ^e)21(Ld1Lq1)ωhVhsin(2θe2θ^e)21(Ld1Lq1)ωhVh(2θe2θ^e)=(Lq1Ld1)ωhVh(θeθ^e)(20)

    脉振高频正弦注入法

    脉振高频正弦注入法在估计轴系完成信号注入、响应信号提取过程,参考旋转高频注入法(可结合前文的各轴系的空间关系图理解),可得:

    d d t [ i γ h i δ h ] = [ cos ⁡ θ ~ − sin ⁡ θ ~ sin ⁡ θ ~ cos ⁡ θ ~ ] [ 1 L d 0 0 1 L q ] [ cos ⁡ θ ~ sin ⁡ θ ~ − sin ⁡ θ ~ cos ⁡ θ ~ ] [ u γ h u δ h ] (21) \tag{21} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\gamma h}\\ i_{\delta h}\\ \end{array} \right] =\left[ \begin{matrix} \cos \tilde{\theta}& -\sin \tilde{\theta}\\ \sin \tilde{\theta}& \cos \tilde{\theta}\\ \end{matrix} \right] \left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{matrix} \cos \tilde{\theta}& \sin \tilde{\theta}\\ -\sin \tilde{\theta}& \cos \tilde{\theta}\\ \end{matrix} \right] \left[ \begin{array}{c} u_{\gamma h}\\ u_{\delta h}\\ \end{array} \right] dtd[iγhiδh]=[cosθ~sinθ~sinθ~cosθ~][Ld100Lq1][cosθ~sinθ~sinθ~cosθ~][uγhuδh](21)

    注入的信号满足下式
    [ u γ h u δ h ] = [ V h cos ⁡ ω h t 0 ] (22) \tag{22} \left[ \begin{array}{c} u_{\gamma h}\\ u_{\delta h}\\ \end{array} \right] =\left[ \begin{array}{c} V_h\cos \omega _ht\\ 0\\ \end{array} \right] [uγhuδh]=[Vhcosωht0](22)
    将式 ( 22 ) (22) (22)带入 ( 21 ) (21) (21),参考式 ( 14 ) (14) (14),化简结果可写为
    d d t [ i γ h i δ h ] = [ 1 2 ( 1 L d + 1 L q ) + 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ ~ 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ ~ 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ ~ 1 2 ( 1 L d + 1 L q ) − 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ ~ ] [ V h cos ⁡ ω h t 0 ] (23) \tag{23} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\gamma h}\\ i_{\delta h}\\ \end{array} \right] =\left[ \begin{matrix} \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) +\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\tilde{\theta}& \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\tilde{\theta}\\ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\tilde{\theta}& \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) -\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\tilde{\theta}\\ \end{matrix} \right] \left[ \begin{array}{c} V_h\cos \omega _ht\\ 0\\ \end{array} \right] dtd[iγhiδh]=21(Ld1+Lq1)+21(Ld1Lq1)cos2θ~21(Ld1Lq1)sin2θ~21(Ld1Lq1)sin2θ~21(Ld1+Lq1)21(Ld1Lq1)cos2θ~[Vhcosωht0](23)
    高频响应电流为
    [ i γ h i δ h ] = [ 1 2 ( 1 L d + 1 L q ) V h ω h sin ⁡ ω h t + 1 2 ( 1 L d − 1 L q ) cos ⁡ 2 θ ~ V h ω h sin ⁡ ω h t 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ ~ V h ω h sin ⁡ ω h t ] (24) \tag{24} \left[ \begin{array}{c} i_{\gamma h}\\ i_{\delta h}\\ \end{array} \right] =\left[ \begin{array}{c} \frac{1}{2}\left( \frac{1}{L_d}+\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin \omega _ht+\frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \cos 2\tilde{\theta}\frac{V_h}{\omega _h}\sin \omega _ht\\ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\tilde{\theta}\frac{V_h}{\omega _h}\sin \omega _ht\\ \end{array} \right] [iγhiδh]=21(Ld1+Lq1)ωhVhsinωht+21(Ld1Lq1)cos2θ~ωhVhsinωht21(Ld1Lq1)sin2θ~ωhVhsinωht(24)

    分析式 ( 24 ) (24) (24)可知, γ \gamma γ轴的电流分量存在位置误差信号,因此,通过幅值解调技术可得位置误差信号为
    L P F ( i δ h × sin ⁡ ω h t ) = L P F [ 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ ~ V h ω h sin ⁡ ω h t sin ⁡ ω h t ] = L P F [ 1 2 ( 1 L d − 1 L q ) sin ⁡ 2 θ ~ V h ω h 1 − cos ⁡ 2 ω h t 2 ] = 1 4 ( 1 L d − 1 L q ) V h ω h sin ⁡ 2 θ ~ ≈ 1 2 ( 1 L d − 1 L q ) V h ω h θ ~ (25) \tag{25} \begin{aligned} LPF\left( i_{\delta h}\times \sin \omega _ht \right) &=LPF\left[ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\tilde{\theta}\frac{V_h}{\omega _h}\sin \omega _ht\sin \omega _ht \right] \\ &=LPF\left[ \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \sin 2\tilde{\theta}\frac{V_h}{\omega _h}\frac{1-\cos 2\omega _ht}{2} \right] \\ &=\frac{1}{4}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\sin 2\tilde{\theta}\approx \frac{1}{2}\left( \frac{1}{L_d}-\frac{1}{L_q} \right) \frac{V_h}{\omega _h}\tilde{\theta} \end{aligned} LPF(iδh×sinωht)=LPF[21(Ld1Lq1)sin2θ~ωhVhsinωhtsinωht]=LPF[21(Ld1Lq1)sin2θ~ωhVh21cos2ωht]=41(Ld1Lq1)ωhVhsin2θ~21(Ld1Lq1)ωhVhθ~(25)

    脉振高频方波注入法

    脉振高频方波注入法在估计轴系注入信号、两相静止轴系提取响应信号,式 ( 3 ) (3) (3)可写为

    d d t [ i α h i β h ] = [ cos ⁡ θ e − sin ⁡ θ e sin ⁡ θ e cos ⁡ θ e ] [ 1 L d 0 0 1 L q ] [ cos ⁡ θ ~ sin ⁡ θ ~ − sin ⁡ θ ~ cos ⁡ θ ~ ] [ u γ h u δ h ] (26) \tag{26} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] =\left[ \begin{matrix} \cos \theta _e& -\sin \theta _e\\ \sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{matrix} \cos \tilde{\theta}& \sin \tilde{\theta}\\ -\sin \tilde{\theta}& \cos \tilde{\theta}\\ \end{matrix} \right] \left[ \begin{array}{c} u_{\gamma h}\\ u_{\delta h}\\ \end{array} \right] dtd[iαhiβh]=[cosθesinθesinθecosθe][Ld100Lq1][cosθ~sinθ~sinθ~cosθ~][uγhuδh](26)

    对上式中的矩阵乘积因子化简,可得:
    [ cos ⁡ θ e − sin ⁡ θ e sin ⁡ θ e cos ⁡ θ e ] [ 1 L d 0 0 1 L q ] [ cos ⁡ θ ~ sin ⁡ θ ~ − sin ⁡ θ ~ cos ⁡ θ ~ ] ⇒ [ 1 L d cos ⁡ θ e cos ⁡ θ ~ + 1 L q sin ⁡ θ e sin ⁡ θ ~ 1 L d cos ⁡ θ e sin ⁡ θ ~ − 1 L q sin ⁡ θ e cos ⁡ θ ~ 1 L d sin ⁡ θ e cos ⁡ θ ~ − 1 L q cos ⁡ θ e sin ⁡ θ ~ 1 L d sin ⁡ θ e sin ⁡ θ ~ + 1 L q cos ⁡ θ e cos ⁡ θ ~ ] (27) \tag{27} \begin{aligned} &\left[ \begin{matrix} \cos \theta _e& -\sin \theta _e\\ \sin \theta _e& \cos \theta _e\\ \end{matrix} \right] \left[ \begin{matrix} \frac{1}{L_d}& 0\\ 0& \frac{1}{L_q}\\ \end{matrix} \right] \left[ \begin{matrix} \cos \tilde{\theta}& \sin \tilde{\theta}\\ -\sin \tilde{\theta}& \cos \tilde{\theta}\\ \end{matrix} \right] \\ \Rightarrow& \left[ \begin{matrix} \frac{1}{L_d}\cos \theta _e\cos \tilde{\theta}+\frac{1}{L_q}\sin \theta _e\sin \tilde{\theta}& \frac{1}{L_d}\cos \theta _e\sin \tilde{\theta}-\frac{1}{L_q}\sin \theta _e\cos \tilde{\theta}\\ \frac{1}{L_d}\sin \theta _e\cos \tilde{\theta}-\frac{1}{L_q}\cos \theta _e\sin \tilde{\theta}& \frac{1}{L_d}\sin \theta _e\sin \tilde{\theta}+\frac{1}{L_q}\cos \theta _e\cos \tilde{\theta}\\ \end{matrix} \right] \end{aligned} [cosθesinθesinθecosθe][Ld100Lq1][cosθ~sinθ~sinθ~cosθ~][Ld1cosθecosθ~+Lq1sinθesinθ~Ld1sinθecosθ~Lq1cosθesinθ~Ld1cosθesinθ~Lq1sinθecosθ~Ld1sinθesinθ~+Lq1cosθecosθ~](27)

    将式 ( 27 ) (27) (27)带入式 ( 26 ) (26) (26),可得:
    d d t [ i α h i β h ] = [ 1 L d cos ⁡ θ e cos ⁡ θ ~ + 1 L q sin ⁡ θ e sin ⁡ θ ~ 1 L d cos ⁡ θ e sin ⁡ θ ~ − 1 L q sin ⁡ θ e cos ⁡ θ ~ 1 L d sin ⁡ θ e cos ⁡ θ ~ − 1 L q cos ⁡ θ e sin ⁡ θ ~ 1 L d sin ⁡ θ e sin ⁡ θ ~ + 1 L q cos ⁡ θ e cos ⁡ θ ~ ] [ u γ h u δ h ] (28) \tag{28} \frac{\mathrm{d}}{\mathrm{d}t}\left[ \begin{array}{c} i_{\alpha h}\\ i_{\beta h}\\ \end{array} \right] =\left[ \begin{matrix} \frac{1}{L_d}\cos \theta _e\cos \tilde{\theta}+\frac{1}{L_q}\sin \theta _e\sin \tilde{\theta}& \frac{1}{L_d}\cos \theta _e\sin \tilde{\theta}-\frac{1}{L_q}\sin \theta _e\cos \tilde{\theta}\\ \frac{1}{L_d}\sin \theta _e\cos \tilde{\theta}-\frac{1}{L_q}\cos \theta _e\sin \tilde{\theta}& \frac{1}{L_d}\sin \theta _e\sin \tilde{\theta}+\frac{1}{L_q}\cos \theta _e\cos \tilde{\theta}\\ \end{matrix} \right] \left[ \begin{array}{c} u_{\gamma h}\\ u_{\delta h}\\ \end{array} \right] dtd[iαhiβh]=[Ld1cosθecosθ~+Lq1sinθesinθ~Ld1sinθecosθ~Lq1cosθesinθ~Ld1cosθesinθ~Lq1sinθecosθ~Ld1sinθesinθ~+Lq1cosθecosθ~][uγhuδh](28)

    对于脉振高频方波注入法,可得电流增量信息为
    [ Δ i α h Δ i β h ] = { [ ( 1 L d cos ⁡ θ e cos ⁡ θ ~ + 1 L q sin ⁡ θ e sin ⁡ θ ~ ) V h T s ( 1 L d sin ⁡ θ e cos ⁡ θ ~ − 1 L q cos ⁡ θ e sin ⁡ θ ~ ) V h T s ] , V i n > 0 [ − ( 1 L d cos ⁡ θ e cos ⁡ θ ~ + 1 L q sin ⁡ θ e sin ⁡ θ ~ ) V h T s − ( 1 L d sin ⁡ θ e cos ⁡ θ ~ − 1 L q cos ⁡ θ e sin ⁡ θ ~ ) V h T s ] , V i n < 0 (29) \tag{29} \left[ \begin{array}{c} \Delta i_{\alpha h}\\ \Delta i_{\beta h}\\ \end{array} \right] =\begin{cases} \left[ \begin{array}{c} \left( \frac{1}{L_d}\cos \theta _e\cos \tilde{\theta}+\frac{1}{L_q}\sin \theta _e\sin \tilde{\theta} \right) V_hT_s\\ \left( \frac{1}{L_d}\sin \theta _e\cos \tilde{\theta}-\frac{1}{L_q}\cos \theta _e\sin \tilde{\theta} \right) V_hT_s\\ \end{array} \right] ,V_{in}>0\\ \left[ \begin{array}{c} -\left( \frac{1}{L_d}\cos \theta _e\cos \tilde{\theta}+\frac{1}{L_q}\sin \theta _e\sin \tilde{\theta} \right) V_hT_s\\ -\left( \frac{1}{L_d}\sin \theta _e\cos \tilde{\theta}-\frac{1}{L_q}\cos \theta _e\sin \tilde{\theta} \right) V_hT_s\\ \end{array} \right] ,V_{in}<0\\ \end{cases} [ΔiαhΔiβh]=(Ld1cosθecosθ~+Lq1sinθesinθ~)VhTs(Ld1sinθecosθ~Lq1cosθesinθ~)VhTs,Vin>0(Ld1cosθecosθ~+Lq1sinθesinθ~)VhTs(Ld1sinθecosθ~Lq1cosθesinθ~)VhTs,Vin<0(29)

    符号化处理,即乘以 s i g n ( V i n ) sign(V_{in}) sign(Vin),上式可简化为
    [ Δ i α h Δ i β h ] = [ ( 1 L d cos ⁡ θ e cos ⁡ θ ~ + 1 L q sin ⁡ θ e sin ⁡ θ ~ ) V h T s ( 1 L d sin ⁡ θ e cos ⁡ θ ~ − 1 L q cos ⁡ θ e sin ⁡ θ ~ ) V h T s ] (30) \tag{30} \left[ \begin{array}{c} \Delta i_{\alpha h}\\ \Delta i_{\beta h}\\ \end{array} \right] =\left[ \begin{array}{c} \left( \frac{1}{L_d}\cos \theta _e\cos \tilde{\theta}+\frac{1}{L_q}\sin \theta _e\sin \tilde{\theta} \right) V_hT_s\\ \left( \frac{1}{L_d}\sin \theta _e\cos \tilde{\theta}-\frac{1}{L_q}\cos \theta _e\sin \tilde{\theta} \right) V_hT_s\\ \end{array} \right] [ΔiαhΔiβh]=(Ld1cosθecosθ~+Lq1sinθesinθ~)VhTs(Ld1sinθecosθ~Lq1cosθesinθ~)VhTs(30)

    位置误差足够小时,此时 θ ~ ≈ 0 \tilde{\theta}\approx 0 θ~0,可得:
    [ Δ i α h Δ i β h ] = [ V h T s 1 L d cos ⁡ θ e V h T s 1 L d sin ⁡ θ e ] (31) \tag{31} \left[ \begin{array}{c} \Delta i_{\alpha h}\\ \Delta i_{\beta h}\\ \end{array} \right] =\left[ \begin{array}{c} V_hT_s\frac{1}{L_d}\cos \theta _e\\ V_hT_s\frac{1}{L_d}\sin \theta _e\\ \end{array} \right] [ΔiαhΔiβh]=[VhTsLd1cosθeVhTsLd1sinθe](31)

    位置误差信号为
    − Δ i α h sin ⁡ θ ^ e + Δ i β h cos ⁡ θ ^ e ⇒ − V h T s 1 L d cos ⁡ θ e sin ⁡ θ ^ e + V h T s 1 L d sin ⁡ θ e cos ⁡ θ ^ e ⇒ V h T s 1 L d sin ⁡ ( θ e − θ ^ e ) ≈ V h T s L d ( θ e − θ ^ e ) (32) \tag{32} \begin{aligned} &-\Delta i_{\alpha h}\sin \hat{\theta}_e+\Delta i_{\beta h}\cos \hat{\theta}_e \\ \Rightarrow& -V_hT_s\frac{1}{L_d}\cos \theta _e\sin \hat{\theta}_e+V_hT_s\frac{1}{L_d}\sin \theta _e\cos \hat{\theta}_e \\ \Rightarrow& V_hT_s\frac{1}{L_d}\sin \left( \theta _e-\hat{\theta}_e \right) \approx \frac{V_hT_s}{L_d}\left( \theta _e-\hat{\theta}_e \right) \end{aligned} Δiαhsinθ^e+Δiβhcosθ^eVhTsLd1cosθesinθ^e+VhTsLd1sinθecosθ^eVhTsLd1sin(θeθ^e)LdVhTs(θeθ^e)(32)

    参考文献

    【1】袁雷. 现代永磁同步电机控制原理及 MATLAB 仿真[M]. 北京航空航天大学出版社, 2016.

    【2】Yoon Y D, Sul S K, Morimoto S, et al. High-bandwidth sensorless algorithm for AC machines based on square-wave-type voltage injection[J]. IEEE transactions on Industry Applications, 2011, 47(3): 1361-1370.

    拓展阅读

    观测器、极性判断123

    sensorless control of IM(感应电机)45


    1. Jeong Y, Lorenz R D, Jahns T M, et al. Initial rotor position estimation of an interior permanent-magnet synchronous machine using carrier-frequency injection methods[J]. IEEE Transactions on Industry Applications, 2005, 41(1): 38-45. ↩︎

    2. 刘颖, 周波, 李帅, 等. 转子磁钢表贴式永磁同步电机转子初始位置检测[J]. 中国电机工程学报, 2011, 31(18): 48-54. ↩︎

    3. Holtz J. Acquisition of position error and magnet polarity for sensorless control of PM synchronous machines[J]. IEEE Transactions on Industry Applications, 2008, 44(4): 1172-1180. ↩︎

    4. Holtz J. Sensorless control of induction motor drives[J]. Proceedings of the IEEE, 2002, 90(8): 1359-1394. ↩︎

    5. Holtz J. Sensorless control of induction machines—With or without signal injection?[J]. IEEE Transactions on Industrial Electronics, 2006, 53(1): 7-30. ↩︎

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