永磁同步电机无传感器之龙博格观测器(Luenberger Observer)离散化推导及建模

一、龙博格观测器简介

龙博格观测器，一种典型的全维状态观测器，依赖系统的输出状态与搭建的状态误差收敛状态对状态进行观测

假设一个系统为：
{ x ˙ = A x + B u y = C x \left\{ \begin{aligned} \dot{x} &= A x + B u \\ y &= C x \end{aligned} \right.{x˙y​=Ax+Bu=Cx​
根据框图格式，构建观测器如下（K为增益）：
{ x ^ ˙ = A x ^ + B u + K ( y − y ^ ) y ^ = C x ^ \left\{ \begin{aligned} \dot{\hat{x}} &= A \hat{x} + B u+K(y-\hat{y}) \\ \hat{y} &= C \hat{x} \end{aligned} \right.{x^˙y^​​=Ax^+Bu+K(y−y^​)=Cx^​
进一步，即状态x的观测值为：
x ^ ˙ = ( A − K C ) x ^ + B u + K y （式 1 ） \ \dot{\hat{x}}= (A-KC) \hat{x} + B u+Ky \ （式1）x^˙=(A−KC)x^+Bu+Ky（式1）
进一步，离散化后状态x为：
x ^ ( k + 1 ) = [ ( A − K C ) x ^ ( k ) + B u ( k ) + K y ( k ) ] ∗ T s + x ^ ( k ) \ {\hat{x}(k+1)}=[ (A-KC) \hat{x}(k) + B u(k)+Ky (k) ]*T_s+\hat{x}(k)\x^(k+1)=[(A−KC)x^(k)+Bu(k)+Ky(k)]∗Ts​+x^(k)
在PMSM控制中，我们通过对输出状态i a \ i_aia​和i β \ i_βiβ​的追踪实现对反电动势e a \ e_aea​与e a \ e_aea​的观测，进而通过PLL可以提取PMSM的转子信息

二、状态变量推导

表贴式PMSM两相静止坐标系下电压方程：
{ u α = R s i α + L s d i α d t + e α u β = R s i β + L s d i β d t + e β \left\{ \begin{aligned} u_{\alpha} &= R_s i_{\alpha} + L_s \frac{di_{\alpha}}{dt} + e_{\alpha} \\ u_{\beta} &= R_s i_{\beta} + L_s \frac{di_{\beta}}{dt} + e_{\beta} \end{aligned} \right.⎩⎨⎧​uα​uβ​​=Rs​iα​+Ls​dtdiα​​+eα​=Rs​iβ​+Ls​dtdiβ​​+eβ​​
其中，反电动势为：
{ e α = − ω r ψ f sin ⁡ ( θ r ) e β = ω r ψ f cos ⁡ ( θ r ) \left\{ \begin{aligned} e_{\alpha} = -\omega_{r} \psi_{f} \sin \left( \theta_{r} \right) \\ e_{\beta} = \omega_{r} \psi_{f} \cos \left( \theta_{r} \right) \end{aligned} \right.{eα​=−ωr​ψf​sin(θr​)eβ​=ωr​ψf​cos(θr​)​
其中，ω r \ \omega_{r}ωr​为电角度、θ r \ \theta_{r}θr​为转子位置、ψ f \ \psi_{f}ψf​为永磁体磁链

由假设近似：ω r ˙ = 0 \ \dot{ \omega_{r} }=0ωr​˙​=0，d θ r d t = ω r \ \frac{d\theta_r}{dt} = \omega_rdtdθr​​=ωr​从电压方程解出电流导数和反电势导数：
d i α d t = 1 L s ( u α − R s i α − e α ) \frac{di_{\alpha}}{dt} = \frac{1}{L_s}(u_{\alpha} - R_s i_{\alpha} - e_{\alpha})dtdiα​​=Ls​1​(uα​−Rs​iα​−eα​)
d i β d t = 1 L s ( u β − R s i β − e β ) \frac{di_{\beta}}{dt} = \frac{1}{L_s}(u_{\beta} - R_s i_{\beta} - e_{\beta})dtdiβ​​=Ls​1​(uβ​−Rs​iβ​−eβ​)
d e α d t = d d t ( − ω r ψ f sin ⁡ ( θ r ) ) = − ω r d d t ( ψ f sin ⁡ ( θ r ) ) ≈ − ω r e β \frac{de_{\alpha}}{dt} = \frac{d}{dt} \Bigl(-\omega_{r} \psi_{f} \sin(\theta_{r}) \Bigr) = -\omega_{r} \frac{d}{dt} \Bigl( \psi_{f} \sin(\theta_{r}) \Bigr) \approx -\omega_{r} e_{\beta}dtdeα​​=dtd​(−ωr​ψf​sin(θr​))=−ωr​dtd​(ψf​sin(θr​))≈−ωr​eβ​
d e β d t = d d t ( ω r ψ f cos ⁡ ( θ r ) ) = ω r d d t ( ψ f cos ⁡ ( θ r ) ) ≈ ω r e α \frac{de_{\beta}}{dt} = \frac{d}{dt} \Bigl( \omega_{r} \psi_{f} \cos(\theta_{r}) \Bigr) = \omega_{r} \frac{d}{dt} \Bigl( \psi_{f} \cos(\theta_{r}) \Bigr) \approx \omega_{r} e_{\alpha}dtdeβ​​=dtd​(ωr​ψf​cos(θr​))=ωr​dtd​(ψf​cos(θr​))≈ωr​eα​
故而状态空间方程构建如下：
状态空间方程为：d d t [ i α i β e α e β ] = [ − R s L s 0 − 1 L s 0 0 − R s L s 0 − 1 L s 0 0 0 − ω r 0 0 ω r 0 ] [ i α i β e α e β ] + [ 1 L s 0 0 1 L s 0 0 0 0 ] [ u α u β 0 0 ] \frac{d}{d t}\left[\begin{array}{c} i_{\alpha} \\ i_{\beta} \\ e_{\alpha} \\ e_{\beta} \end{array}\right] = \left[\begin{array}{cccc} -\frac{R_{s}}{L_{s}} & 0 & -\frac{1}{L_{s}} & 0 \\ 0 & -\frac{R_{s}}{L_{s}} & 0 & -\frac{1}{L_{s}} \\ 0 & 0 & 0 & -\omega_{r} \\ 0 & 0 & \omega_{r} & 0 \end{array}\right] \left[\begin{array}{c} i_{\alpha} \\ i_{\beta} \\ e_{\alpha} \\ e_{\beta} \end{array}\right] + \left[\begin{array}{cc} \frac{1}{L_{s}} & 0 \\ 0 & \frac{1}{L_{s}} \\ 0 & 0 \\ 0 & 0 \end{array}\right] \left[\begin{array}{c} u_{\alpha} \\ u_{\beta} \\ 0 \\ 0 \end{array}\right]dtd​​iα​iβ​eα​eβ​​​=​−Ls​Rs​​000​0−Ls​Rs​​00​−Ls​1​00ωr​​0−Ls​1​−ωr​0​​​iα​iβ​eα​eβ​​​+​Ls​1​000​0Ls​1​00​​​uα​uβ​00​​

输出方程为：[ i α i β ] = [ 1 0 0 0 0 1 0 0 ] [ i α i β e α e β ] \left[\begin{array}{l} i_{\alpha} \\ i_{\beta} \end{array}\right] = \left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right] \left[\begin{array}{l} i_{\alpha} \\ i_{\beta} \\ e_{\alpha} \\ e_{\beta} \end{array}\right][iα​iβ​​]=[10​01​00​00​]​iα​iβ​eα​eβ​​​

三、实现过程

仿照一、中，对上面给出的状态方程设计Lunberger观测器如下：
d d t [ i ^ α i ^ β e ^ α e ^ β ] = A [ i ^ α i ^ β e ^ α e ^ β ] + B [ u α u β 0 0 ] + K ( [ i α i β 0 0 ] − [ i ^ α i ^ β 0 0 ] ) ( 式 2 ) \frac{d}{d t}\left[\begin{array}{c} \hat{i}_{\alpha} \\ \hat{i}_{\beta} \\ \hat{e}_{\alpha} \\ \hat{e}_{\beta} \end{array}\right] = \mathbf{A}\left[\begin{array}{c} \hat{i}_{\alpha} \\ \hat{i}_{\beta} \\ \hat{e}_{\alpha} \\ \hat{e}_{\beta} \end{array}\right] + \mathbf{B}\left[\begin{array}{c} u_{\alpha} \\ u_{\beta} \\ 0 \\ 0 \end{array}\right] + \mathbf{K}\left( \left[\begin{array}{c} i_{\alpha} \\ i_{\beta}\\0 \\0 \end{array}\right] - \left[\begin{array}{c} \hat{i}_{\alpha} \\ \hat{i}_{\beta} \\0 \\0 \end{array}\right] \right)(式2)dtd​​i^α​i^β​e^α​e^β​​​=A​i^α​i^β​e^α​e^β​​​+B​uα​uβ​00​​+K​​iα​iβ​00​​−​i^α​i^β​00​​​(式2)
其中，系数矩阵A、B、C为：
A = [ − R s L s 0 − 1 L s 0 0 − R s L s 0 − 1 L s 0 0 0 − ω r 0 0 ω r 0 ] \mathbf{A}=\left[\begin{array}{cccc} -\frac{R_{s}}{L_{s}} & 0 & -\frac{1}{L_{s}} & 0 \\ 0 & -\frac{R_{s}}{L_{s}} & 0 & -\frac{1}{L_{s}} \\ 0 & 0 & 0 & -\omega_{r} \\ 0 & 0 & \omega_{r} & 0 \end{array}\right]A=​−Ls​Rs​​000​0−Ls​Rs​​00​−Ls​1​00ωr​​0−Ls​1​−ωr​0​​

B = [ 1 L s 0 0 1 L s 0 0 0 0 ] , C = [ 1 0 0 0 0 1 0 0 ] \mathbf{B}=\left[\begin{array}{cc} \frac{1}{L_{s}} & 0 \\ 0 & \frac{1}{L_{s}} \\ 0 & 0 \\ 0 & 0 \end{array}\right],\quad \mathbf{C}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right]B=​Ls​1​000​0Ls​1​00​​,C=[10​01​00​00​]
增益矩阵为：
K = [ K 1 0 0 0 0 K 1 0 0 K 2 0 0 0 0 K 2 0 0 ] \mathbf{K} = \left[ \begin{array}{cc} K_1 & 0 & 0& 0\\ 0 & K_1 & 0& 0\\ K_2 & 0 & 0& 0\\ 0 & K_2& 0& 0\\ \end{array} \right]K=​K1​0K2​0​0K1​0K2​​0000​0000​​
其中，K1是对电流的观测增益，K2是对反电动势的观测增益
状态变量，输入矩阵，输出矩阵分别为：

x = [ i α i β e α e α ] , u = [ u α u β 0 0 ] , y = [ i α i β 0 0 ] \mathbf{x}=\left[\begin{array}{c} i_{\alpha} \\ i_{\beta} \\ e_{\alpha} \\ e_{\alpha} \end{array}\right] ,\mathbf{u}=\left[\begin{array}{c} u_{\alpha} \\ u_{\beta} \\ 0 \\ 0 \end{array}\right] ,\mathbf{y}=\left[\begin{array}{c} i_{\alpha} \\ i_{\beta} \\ 0 \\ 0 \end{array}\right]x=​iα​iβ​eα​eα​​​,u=​uα​uβ​00​​,y=​iα​iβ​00​​
对(式2)离散化后，得到反电动势的龙博格观测器为：
i ^ α ( k + 1 ) = i ^ α ( k ) + T [ − R s L s i ^ α ( k ) − 1 L s e ^ α ( k ) + 1 L s u α ( k ) + K 1 ( i α ( k ) − i ^ α ( k ) ) ] i ^ β ( k + 1 ) = i ^ β ( k ) + T [ − R s L s i ^ β ( k ) − 1 L s e ^ β ( k ) + 1 L s u β ( k ) + K 1 ( i β ( k ) − i ^ β ( k ) ) ] e ^ α ( k + 1 ) = e ^ α ( k ) + T [ − ω ^ e e ^ β ( k ) + K 2 ( i α ( k ) − i ^ α ( k ) ) ] e ^ β ( k + 1 ) = e ^ β ( k ) + T [ ω ^ e e ^ α ( k ) + K 2 ( i β ( k ) − i ^ β ( k ) ) ] \begin{aligned} \hat{i}_{\alpha}(k+1) &= \hat{i}_{\alpha}(k) + T\bigg[-\frac{R_s}{L_s}\hat{i}_{\alpha}(k) - \frac{1}{L_s}\hat{e}_{\alpha}(k) + \frac{1}{L_s}u_{\alpha}(k) + K_1\big(i_{\alpha}(k) - \hat{i}_{\alpha}(k)\big)\bigg] \\ \hat{i}_{\beta}(k+1) &= \hat{i}_{\beta}(k) + T\bigg[-\frac{R_s}{L_s}\hat{i}_{\beta}(k) - \frac{1}{L_s}\hat{e}_{\beta}(k) + \frac{1}{L_s}u_{\beta}(k) + K_1\big(i_{\beta}(k) - \hat{i}_{\beta}(k)\big)\bigg] \\ \hat{e}_{\alpha}(k+1) &= \hat{e}_{\alpha}(k) + T\bigg[-\hat{\omega}_e \hat{e}_{\beta}(k) + K_2\big(i_{\alpha}(k) - \hat{i}_{\alpha}(k)\big)\bigg] \\ \hat{e}_{\beta}(k+1) &= \hat{e}_{\beta}(k) + T\bigg[\,\hat{\omega}_e \hat{e}_{\alpha}(k) + K_2\big(i_{\beta}(k) - \hat{i}_{\beta}(k)\big)\bigg] \end{aligned}i^α​(k+1)i^β​(k+1)e^α​(k+1)e^β​(k+1)​=i^α​(k)+T[−Ls​Rs​​i^α​(k)−Ls​1​e^α​(k)+Ls​1​uα​(k)+K1​(iα​(k)−i^α​(k))]=i^β​(k)+T[−Ls​Rs​​i^β​(k)−Ls​1​e^β​(k)+Ls​1​uβ​(k)+K1​(iβ​(k)−i^β​(k))]=e^α​(k)+T[−ω^e​e^β​(k)+K2​(iα​(k)−i^α​(k))]=e^β​(k)+T[ω^e​e^α​(k)+K2​(iβ​(k)−i^β​(k))]​
T为采样时间

四、仿真

simulink搭建仿真验证如下：

选取合适增益后运行，用示波器查看i_α和hat(i_α)波形发现观测收敛，如下：

反电动势为：