Electronic – How to use Luenberger observer for induction motor rotor flux estimation

I have implemented field oriented control algorithm of three phase induction motor with speed sensor. Now I would like to improve the rotor flux estimation algorithm with Luenberger observer usage. My idea was to use the state space model of the induction motor in following form:

$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}
\begin{bmatrix}
i_{s\alpha}\\ i_{s\beta}\\ \psi_{r\alpha}\\ \psi_{r\beta}
\end{bmatrix}
&=
\begin{bmatrix}
-\alpha & 0 & \beta & \gamma\cdot\omega_m \\
0 & -\alpha & -\gamma\cdot\omega_m & \beta \\
R_r\frac{L_h}{L_r} & 0 & -\frac{R_r}{L_r} & -\omega_m \\
0 & R_r\frac{L_h}{L_r} & \omega_m & -\frac{R_r}{L_r}
\end{bmatrix}
\cdot
\begin{bmatrix}
i_{s\alpha} \\ i_{s\beta} \\ \psi_{r\alpha} \\ \psi_{r\beta}
\end{bmatrix}
+
\begin{bmatrix}
\delta & 0 \\ 0 & \delta \\ 0 & 0 \\ 0 & 0 \\
\end{bmatrix}
\cdot
\begin{bmatrix}
u_{s\alpha} \\
u_{s\beta}
\end{bmatrix}
\\
\\
\dot{\mathbf{x}}_T
&=
\mathbf{A}_T\cdot \mathbf{x}_T + \textbf{B}_T\cdot \mathbf{u}
\\
\begin{bmatrix}
i_{s\alpha} \\ i_{s\beta}
\end{bmatrix}
&=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
i_{s\alpha} \\ i_{s\beta} \\ \psi_{r\alpha} \\ \psi_{r\beta}
\end{bmatrix}
\\
\mathbf{y}_T
&=
\mathbf{C}_T\cdot\mathbf{x}_T
\end{align}
$$
where

$$
\alpha = \frac{R_s + R_r\frac{L^2_h}{L^2_r}}{L_{s\sigma}+\frac{L_h}{L_r}L_{r\sigma}}
$$

$$
\beta = \frac{R_r\frac{L_h}{L^2_r}}{L_{s\sigma}+\frac{L_h}{L_r}L_{r\sigma}}
$$

$$
\gamma = \frac{\frac{L_h}{L_r}}{L_{s\sigma}+\frac{L_h}{L_r}L_{r\sigma}}\cdot p_p
$$

$$
\delta = \frac{1}{L_{s\sigma}+\frac{L_h}{L_r}L_{r\sigma}}
$$

System state consists of components of the space vector of the stator current in the stator reference frame along with the components of the space vector of the rotor flux in the stator reference frame. Input to the system consists of the components of the space vector of the stator voltage in the stator reference frame. Output of the system consists of the components of the space vector of the stator current in the stator reference frame.

Based on the inspection of the above mentioned system matrix it is obvious that this matrix is rotor speed dependent (\$\omega_m\$). This is not a problem because I have the information about the rotor speed (I have been using the foc algorithm with speed sensor). My intention is to use the standard Luenberger observer i.e. variant for linear time independent systems and in each sampling period evaluate the elements of the system matrix based on current value of the rotor mechanical speed at first. Please can anybody tell me whether this approach is correct or whether it is necessary to use the the extended Luenberger observer which is dedicated for non-linear systems? Thanks in advance for any ideas.

EDIT:

Transformation of the state space model from the T equivalent circuit into the inverse gamma equivalent circuit:

The goal is to have the state space model for the inverse gamma equivalent circuit. It means that it is necessary to transform the matrices \$\mathbf{A}_T, \mathbf{B}_T, \mathbf{C}_T\$ above with some trasformation matrix usage.

The transformation matrix can be found based on following relationship between the above mentioned state space model for the T equivalent circuit and the inverse gamma equivalent circuit:

$$
\hat{\mathbf{i}}_{s\Gamma^-1} = \hat{\mathbf{i}}_{sT} \\
\hat{\boldsymbol{\psi}}_{r\Gamma^{-1}}= \hat{\boldsymbol{\psi}}_{rT}\cdot\frac{L_h}{L_r}
$$

The previously mentioned formulas can be written in matrix form

$$
\begin{bmatrix}
\hat{\mathbf{i}}_{s\Gamma^{-1}} \\
\hat{\boldsymbol{\psi}}_{r\Gamma^{-1}}
\end{bmatrix}
=
\begin{bmatrix}
\mathbf{I} & 0 \\
0 & \frac{L_h}{L_r}\cdot\mathbf{I}
\end{bmatrix}
\cdot
\begin{bmatrix}
\hat{\mathbf{i}}_{sT} \\
\hat{\boldsymbol{\psi}}_{rT}
\end{bmatrix}
$$

In case we express the state vector for the T equivalent circuit from the last matrix equation, we have

$$
\mathbf{x}_T
=
\begin{bmatrix}
\hat{\mathbf{i}}_{sT} \\
\hat{\boldsymbol{\psi}}_{rT}
\end{bmatrix}
=
\begin{bmatrix}
\mathbf{I} & 0 \\
0 & \frac{L_r}{L_h}\cdot\mathbf{I}
\end{bmatrix}
\cdot
\begin{bmatrix}
\hat{\mathbf{i}}_{s\Gamma^{-1}} \\
\hat{\boldsymbol{\psi}}_{r\Gamma^{-1}}
\end{bmatrix}
=
\mathbf{T}
\cdot
\mathbf{x}_{\Gamma^{-1}}
$$

Now we can substitute the \$\mathbf{x}_T\$ into the state space model for the T equivalent circuit

$$
\dot{\mathbf{x}}_T = \mathbf{A}_T\cdot \mathbf{x}_T + \textbf{B}_T\cdot \mathbf{u}
$$

$$
\frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{T}\cdot\mathbf{x}_{\Gamma^{-1}}\right)
=
\mathbf{A}_T\cdot\left(\mathbf{T}\cdot\mathbf{x}_{\Gamma^{-1}}\right) + \mathbf{B}_T\cdot \mathbf{u}_T
$$

$$
\dot{\mathbf{x}}_{\Gamma^{-1}} = \mathbf{T}^{-1}\cdot\mathbf{A}_T\cdot\mathbf{T}\cdot\mathbf{x}_{\Gamma^{-1}} + \mathbf{T}^{-1}\cdot\mathbf{B}_T\cdot\mathbf{u}_T
$$

respectively

$$
\mathbf{y}_{T}
=
\mathbf{y}_{\Gamma^{-1}}
=
\mathbf{C}_T\cdot\mathbf{T}\cdot\mathbf{x}_{\Gamma^{-1}}
$$

Based on that we have know how to transform the matrices of the T state space model

$$
\mathbf{A}_{\Gamma^{-1}} = \mathbf{T}^{-1}\cdot\mathbf{A}_T\cdot\mathbf{T}
$$

$$
\mathbf{B}_{\Gamma^{-1}} = \mathbf{T}^{-1}\cdot\mathbf{B}_T
$$

$$
\mathbf{C}_{\Gamma^{-1}} = \mathbf{C}_T\cdot\mathbf{T}
$$

At this moment we can substitute into the transformation equations

$$
\mathbf{A}_{\Gamma^{-1}}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{L_h}{L_r} & 0 \\
0 & 0 & 0 & \frac{L_h}{L_r}
\end{bmatrix}
\cdot
\begin{bmatrix}
-\alpha & 0 & \beta & \gamma\cdot\omega_m \\
0 & -\alpha & -\gamma\cdot\omega_m & \beta \\
R_r\frac{L_h}{L_r} & 0 & \frac{-R_r}{L_r} & \omega_m \\
0 & R_r\frac{L_h}{L_r} & \omega_m & -\frac{R_r}{L_r}
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{L_r}{L_h} & 0 \\
0 & 0 & 0 & \frac{L_r}{L_h}
\end{bmatrix}
$$

$$
\mathbf{A}_{\Gamma^{-1}}
=
\begin{bmatrix}
-\alpha & 0 & \frac{L_r}{L_h}\cdot\beta & \frac{L_r}{L_h}\cdot\gamma\cdot\omega_m \\
0 & -\alpha & -\frac{L_r}{L_h}\cdot\gamma\cdot\omega_m & \frac{L_r}{L_h}\cdot\beta \\
\frac{L_h}{L_r}\cdot R_r\cdot\frac{L_h}{L_r} & 0 & -\frac{L_r}{L_h}\cdot\frac{L_h}{L_r}\cdot R_r & -\frac{L_r}{L_h}\cdot\frac{L_h}{L_r}\cdot\omega_m \\
0 & \frac{L_h}{L_r}\cdot R_r\cdot\frac{L_h}{L_r} & \frac{L_r}{L_h}\cdot\frac{L_h}{L_r}\cdot\omega_m & -\frac{L_r}{L_h}\cdot\frac{L_h}{L_r}\cdot R_r
\end{bmatrix}
$$

$$
\mathbf{B}_{\Gamma^{-1}}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{L_h}{L_r} & 0 \\
0 & 0 & 0 & \frac{L_h}{L_r}
\end{bmatrix}
\cdot
\begin{bmatrix}
\delta & 0 \\
0 & \delta \\
0 & 0 \\
0 & 0
\end{bmatrix}
=
\begin{bmatrix}
\delta & 0 \\
0 & \delta \\
0 & 0 \\
0 & 0
\end{bmatrix}
$$

$$
\mathbf{C}_{\Gamma^{-1}}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{L_h}{L_r} & 0 \\
0 & 0 & 0 & \frac{L_h}{L_r}
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$

Now we can substitute for the \$ \alpha, \beta, \gamma, \delta \$ into the matrices \$\mathbf{A}_{\Gamma^{-1}}, \mathbf{B}_{\Gamma^{-1}}, \mathbf{C}_{\Gamma^{-1}} \$. We also have take into account that the inverse gamma equivalent circuit contains following parameters: total leakage inductance \$L_L = L_{s\sigma} + \frac{L_h}{L_r}\cdot L_{r\sigma}\$, magnetizing inductance \$L_M = L_h\cdot\frac{L_h}{L_r}\$, stator resistance \$R_S = R_s\$ and rotor resistance \$R_R = R_r\cdot\left(\frac{L_h}{L_r}\right)^2\$. After simplification, we have

$$
\mathbf{A}_{\Gamma^{-1}}
=
\begin{bmatrix}
-\frac{R_S + R_R}{L_L} & 0 & \frac{R_R}{L_M\cdot L_L} & \frac{1}{L_L}\cdot\omega_m \\
0 & -\frac{R_S + R_R}{L_L} & -\frac{1}{L_L}\cdot\omega_m & \frac{R_R}{L_M\cdot L_L} \\
R_R & 0 & -\frac{R_R}{L_M} & -\omega_m \\
0 & R_R & \omega_m & -\frac{R_R}{L_M}
\end{bmatrix}
$$

$$
\mathbf{B}_{\Gamma^{-1}}
=
\begin{bmatrix}
\frac{1}{L_L} & 0 \\
0 & \frac{1}{L_L} \\
0 & 0 \\
0 & 0
\end{bmatrix}
$$

$$
\mathbf{C}_{\Gamma^{-1}}
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$

As far as the equation for the moment of the induction motor. We can use the equation for the moment based on the T equivalent circuit

$$
M_T = \frac{3}{2}\cdot p_p\cdot \frac{L_h}{L_r}\cdot\Re \{ \mathrm{j}\cdot\hat{\boldsymbol{\psi}}_{rT}\cdot \hat{\boldsymbol{i}}^{*}_{sT} \} = \frac{3}{2}\cdot p_p\cdot \frac{L_h}{L_r}\cdot(\psi_{rxT}\cdot i_{syT} – \psi_{ryT}\cdot i_{sxT})
$$

Then we can substitute for the space vectors \$\hat{\mathbf{i}}_{sT} = \hat{\mathbf{i}}_{s\Gamma^-1}\$ and \$\hat{\boldsymbol{\psi}}_{rT} = \frac{L_r}{L_h}\cdot\hat{\boldsymbol{\psi}}_{r\Gamma^{-1}}\$. After simplification, we have

$$
M_{\Gamma^{-1}} = \frac{3}{2}\cdot p_p\cdot(\psi_{rx\Gamma^{-1}}\cdot i_{sy\Gamma^{-1}} – \psi_{ry\Gamma^{-1}}\cdot i_{sx\Gamma^{-1}})
$$