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
通过i_α和i_β估计反电势e_α和e_β
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,通过搭建依赖系统的可测输出状态(y)与搭建的观测输出状态(hat_(y))误差收敛实现对输入状态(x)的观测

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
假设输出测定值为
y
\ y
y,输出的观测值为
y
^
\ \hat{y}
y^,定义输出观测误差
y
−
y
^
\ y-\hat{y}
y−y^,对其求导并令其指数收敛即有:
y
˙
−
y
^
˙
=
−
K
1
(
y
−
y
^
)
\ \dot{y}-\dot{\hat{y}}=-K_1(\ y-\hat{y})
y˙−y^˙=−K1( y−y^),将
y
˙
=
C
x
˙
,
y
^
˙
=
C
x
^
˙
\dot{y} = C \dot{x},\dot{\hat{y}} = C \dot{\hat{x}}
y˙=Cx˙,y^˙=Cx^˙ ,解出带入状态方程
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_a
ia 和
i
β
\ i_β
iβ 的追踪实现对反电动势
e
a
\ e_a
ea 与
e
a
\ e_a
ea 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.
⎩
⎨
⎧uαuβ=Rsiα+Lsdtdiα+eα=Rsiβ+Lsdtdiβ+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=−ωrψfsin(θr)eβ=ωrψfcos(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} }=0
ωr˙=0 ,
d
θ
r
d
t
=
ω
r
\ \frac{d\theta_r}{dt} = \omega_r
dtdθr=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=Ls1(uα−Rsiα−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=Ls1(uβ−Rsiβ−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
[
ψ
f
cos
(
θ
r
)
ω
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) = -\omega_{r}[\psi_{f} \cos(\theta_{r})\omega_{r}] \approx -\omega_{r} e_{\beta}
dtdeα=dtd(−ωrψfsin(θr))=−ωrdtd(ψfsin(θr))=−ωr[ψfcos(θr)ωr]≈−ωreβ
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(ωrψfcos(θr))=ωrdtd(ψfcos(θr))≈ωreα
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
]
(式
2
)
\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](式2)
dtd
iαiβeαeβ
=
−LsRs0000−LsRs00−Ls100ωr0−Ls1−ωr0
iαiβeαeβ
+
Ls10000Ls100
uαuβ00
(式2)
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] [iαiβ]=[10010000] iαiβeαeβ
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
i
^
α
i
^
β
]
−
[
i
α
i
β
i
α
i
β
]
)
(
式
3
)
\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} \hat{i}_{\alpha} \\ \hat{i}_{\beta} \\\hat{i}_{\alpha} \\ \hat{i}_{\beta} \end{array}\right]-\left[\begin{array}{c} i_{\alpha} \\ i_{\beta}\\i_{\alpha} \\ i_{\beta} \end{array}\right] \right)(式3)
dtd
i^αi^βe^αe^β
=A
i^αi^βe^αe^β
+B
uαuβ00
−K
i^αi^βi^αi^β
−
iαiβiαiβ
(式3)
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=
−LsRs0000−LsRs00−Ls100ωr0−Ls1−ωr0
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=
Ls10000Ls100
,C=[10010000]
增益矩阵为:
K
=
[
K
1
0
0
0
0
K
1
0
0
0
0
K
2
0
0
0
0
K
2
]
\mathbf{K} = \left[ \begin{array}{cc} K_1 & 0 & 0& 0\\ 0 & K_1 & 0& 0\\ 0 & 0 & K_2& 0\\ 0 &0& 0&K_2\\ \end{array} \right]
K=
K10000K10000K20000K2
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= iαiβeαeα ,u= uαuβ00 ,y= iαiβ00
将(式3)和(式2)做差得到观测误差方程为:
[
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
]
−
[
K
1
0
0
0
0
K
1
0
0
0
0
K
2
0
0
0
0
K
2
]
[
i
~
α
i
~
β
e
~
α
e
~
β
]
\left[\begin{array}{c} \dot{\widetilde{i}}_{\alpha} \\\dot{\widetilde{i}}_{\beta} \\ \dot{\widetilde{e}}_{\alpha}\\ \dot{\widetilde{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} \widetilde{i}_{\alpha} \\ \widetilde{i}_{\beta} \\ \widetilde{e}_{\alpha} \\ \widetilde{e}_{\beta} \end{array}\right]- \left[ \begin{array}{cc} K_1 & 0 & 0& 0\\ 0 & K_1 & 0& 0\\ 0 & 0 & K_2& 0\\ 0 &0& 0&K_2\\ \end{array} \right] \left[\begin{array}{c} {\widetilde{i}}_{\alpha} \\{\widetilde{i}}_{\beta} \\{\widetilde{e}}_{\alpha}\\ {\widetilde{e}}_{\beta} \end{array}\right]
i
˙αi
˙βe
˙αe
˙β
=
−LsRs0000−LsRs00−Ls100ωr0−Ls1−ωr0
i
αi
βe
αe
β
−
K10000K10000K20000K2
i
αi
βe
αe
β
选取正确增益观测器可以收敛
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
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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(\hat{i}_{\alpha}(k) - {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( \hat{i}_{\beta}(k) -{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( \hat{i}_{\alpha}(k) -{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(\hat{i}_{\beta}(k) - {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[−LsRsi^α(k)−Ls1e^α(k)+Ls1uα(k)+K1(i^α(k)−iα(k))]=i^β(k)+T[−LsRsi^β(k)−Ls1e^β(k)+Ls1uβ(k)+K1(i^β(k)−iβ(k))]=e^α(k)+T[−ω^ee^β(k)+K2(i^α(k)−iα(k))]=e^β(k)+T[ω^ee^α(k)+K2(i^β(k)−iβ(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:

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