Answers and Discussions for
Problem 3
Contemplation
The first step is to ponder the inductance matrix in $ \vec{\lambda}_{abs} = \mathbf{L}_s \vec{i}_{abs} $.
$ \begin{equation} \mathbf{L}_s = \begin{bmatrix} L_{asas} & L_{asbs} \\ L_{bsas} & L_{bsbs} \end{bmatrix} \end{equation} $
Because the self-inductances change with rotor position $ \theta_r $, it is clear that the reluctance machine is salient (a vital operational principle of all reluctance machines). The self-inductances and mutual fit a known form as given. All leakage inductance terms are zero based on that given form. Note that $ L_{Aas} = L_{Abs} = L_A $ and $ L_{Bas} = L_{Bbs} = L_{Basbs} = -L_B $ in this symmetric machine.
$ \begin{align} L_{asas} &= \cancel{L_{\ell as}} + L_{Aas} + L_{Bas} \cos\left[2\left(\theta_r - 0\right)\right] = L_{A} - L_{B} \cos\left(2\theta_r\right) \\ L_{bsbs} &= \cancel{L_{\ell bs}} + L_{Abs} + L_{Bbs} \cos\left[2\left(\theta_r - \frac{\pi}{2}\right)\right] = L_{A} + L_{B} \cos\left(2\theta_r\right) \\ L_{asbs} &= L_{bsas} = +L_{Basbs} \sin\left(2\theta_r\right) = -L_{B} \sin\left(2\theta_r\right) \end{align} $
Rotor Reference Frame Transformation
To move from stator phase variables to the derived rotor reference frame, pre-multiply by $ \mathbf{K}_s^r =\begin{bmatrix} +\cos(\theta_r) & \sin(\theta_r) \\ \sin(\theta_r) & -\cos(\theta_r) \end{bmatrix} $. To do the opposite and move from the derived rotor reference frame to stator phase variables, pre-multiply by $ \left(\mathbf{K}_s^r\right)^{-1} = \frac{1}{\cancelto{-1}{-\cos^2(\theta_r) - \sin^2(\theta_r)}} \begin{bmatrix} -\cos(\theta_r) & -\sin(\theta_r) \\ -\sin(\theta_r) & +\cos(\theta_r) \end{bmatrix} = \begin{bmatrix} +\cos(\theta_r) & \sin(\theta_r) \\ \sin(\theta_r) & -\cos(\theta_r) \end{bmatrix} = \mathbf{K}_s^r $. The inverse matrix is found with the explicit formula the the inverse of a 2x2 matrix and its determinant as well. (By coincidence, $ \mathbf{K}_s^r $ is an involutary matrix.)
The transformation of the flux linkage equations proceeds.
$ \begin{align} \vec{\lambda}_{qds}^{r} &= \mathbf{K}_s^r \vec{\lambda}_{abs}^{'} \\ \vec{\lambda}_{qds}^{r} &= \mathbf{K}_s^r \mathbf{L}_s \left(\mathbf{K}_s^r\right)^{-1} \vec{i}_{qds}^{r} \\ \vec{\lambda}_{qds}^{r} &= \begin{bmatrix} +\cos(\theta_r) & \sin(\theta_r) \\ \sin(\theta_r) & -\cos(\theta_r) \end{bmatrix} \begin{bmatrix} L_A - L_B \cos(2\theta_r) & -L_B \sin(2\theta_r) \\ -L_B \sin(2\theta_r) & L_A + L_B \cos(2\theta_r) \end{bmatrix} \begin{bmatrix} +\cos(\theta_r) & \sin(\theta_r) \\ \sin(\theta_r) & -\cos(\theta_r) \end{bmatrix} \vec{i}_{qds}^{r} \\ \vec{\lambda}_{qds}^{r} &= \begin{bmatrix} L_A \cos(\theta_r) - L_B \cos(\theta_r)\cos(2\theta_r) - L_B \sin(\theta_r)\sin(2\theta_r) & -L_B \cos(\theta_r)\sin(2\theta_r) + L_A \sin(\theta_r) + L_B \sin(\theta_r)\cos(2\theta_r) \\ L_A \sin(\theta_r) - L_B \sin(\theta_r)\cos(2\theta_r) + L_B \cos(\theta_r)\sin(2\theta_r) & -L_B \sin(\theta_r)\sin(2\theta_r) - L_A \cos(\theta_r) - L_B \cos(\theta_r)\cos(2\theta_r) \end{bmatrix} \begin{bmatrix} +\cos(\theta_r) & \sin(\theta_r) \\ \sin(\theta_r) & -\cos(\theta_r) \end{bmatrix} \vec{i}_{qds}^{r} \end{align} $
The matrix has gotten so out of hand that reverting back to separated flux linkage equations helps fit the equation on the display.
$ \begin{align} \lambda_{qs}^r &= \begin{split} &{} \left[L_A \cos^2(\theta_r) - L_B \cos^2(\theta_r)\cos(2\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\sin(2\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\sin(2\theta_r) + L_A \sin^2(\theta_r) + L_B \sin^2(\theta_r)\cos(2\theta_r)\right] i_{qs}^r \\ &{}+ \left[L_A \sin(\theta_r)\cos(\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\cos(2\theta_r) - L_B \sin^2(\theta_r)\sin(2\theta_r) + L_B \cos^2(\theta_r)\sin(2\theta_r) - L_A \sin(\theta_r)\cos(\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\cos(2\theta_r)\right] i_{ds}^r \end{split} \\ \lambda_{qs}^r &= \begin{split} &{} \left[L_A \left(\sin^2(\theta_r) + \cos^2(\theta_r)\right) - L_B \left(2\sin(\theta_r)\cos(\theta_r)\right)\sin(2\theta_r) - L_B \left(\cos^2(\theta_r) - \sin^2(\theta_r)\right)\cos(2\theta_r)\right] i_{qs}^r \\ &{}+ \left[-L_B \left(2\sin(\theta_r)\cos(\theta_r)\right)\cos(2\theta_r) + L_B \left(\cos^2(\theta_r) - \sin^2(\theta_r)\right)\sin(2\theta_r)\right] i_{ds}^r \end{split} \\ \lambda_{qs}^r &= \left[L_A - L_B \sin(2\theta_r)\sin(2\theta_r) - L_B \cos(2\theta_r)\cos(2\theta_r)\right] i_{qs}^r + \left[-L_B \sin(2\theta_r)\cos(2\theta_r) - L_B \cos(2\theta_r)\sin(2\theta_r)\right] i_{ds}^r \\ \lambda_{qs}^r &= \left[L_A - L_B \left(\sin^2(2\theta_r) + \cos^2(2\theta_r)\right)\right] i_{qs}^r + \left[0\right] i_{ds}^r \\ \lambda_{qs}^r &= \left[L_A - L_B\right] i_{qs}^r \end{align} $
$ \begin{align} \lambda_{ds}^r &= \begin{split} &{} \left[L_A \sin(\theta_r)\cos(\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\cos(2\theta_r) + L_B \cos^2(\theta_r)\sin(2\theta_r) - L_B \sin^2(\theta_r)\sin(2\theta_r) - L_A \sin(\theta_r)\cos(\theta_r) - L_B \sin(\theta_r)\cos(\theta_r)\cos(2\theta_r)\right] i_{qs}^r \\ &{}+ \left[L_A \sin^2(\theta_r) - L_B \sin^2(\theta_r)\cos(2\theta_r) + L_B \sin(\theta_r)\cos(\theta_r)\sin(2\theta_r) + L_B \sin(\theta_r)\cos(\theta_r)\sin(2\theta_r) + L_A \cos^2(\theta_r) + L_B\cos^2(\theta_r)\cos(2\theta_r)\right] i_{ds}^r \end{split} \\ \lambda_{ds}^r &= \left[0\right] i_{qs}^r + \left[L_A \left(\sin^2(\theta_r) + \cos^2(\theta_r)\right) + L_B \left(2\sin(\theta_r)\cos(\theta_r)\right)\sin(2\theta_r) + L_B \left(\cos^2(\theta_r) - \sin^2(\theta_r)\right)\cos(2\theta_r)\right] i_{ds}^r \\ \lambda_{ds}^r &= \left[L_A + L_B \sin(2\theta_r)\sin(2\theta_r) + L_B \cos(2\theta_r)\cos(2\theta_r)\right] i_{ds}^r \\ \lambda_{ds}^r &= \left[L_A + L_B \left(\sin^2(2\theta_r) + \cos^2(2\theta_r)\right)\right] i_{ds}^r \\ \lambda_{ds}^r &= \left[L_A + L_B\right] i_{ds}^r \end{align} $
These trigonometric simplifications rely on the Pythagorean Identity of $ \sin^2(x) + \cos^2(x) = 1 $ (not given), the Double Angle Identity for sine of $ \sin(2x) = 2\sin(x)\cos(x) $ (derived from given), and the Double Angle Identity for cosine of $ \cos(2x) = \cos^2(x) - \sin^2(x) $ (derived from given). All rotor position dependence has been removed from the equations as would be expected from a reference frame transformation. The vector flux linkage equation is finished. The result is consistent with equations for $ L_{mq} $ and $ L_{md} $ in a 2-phase machine with the given form of inductance matrix.
$ \begin{equation} \boxed{\mathbf{K}_s^r \mathbf{L}_s \left(\mathbf{K}_s^r\right)^{-1} = \begin{bmatrix} L_A - L_B & 0 \\ 0 & L_A + L_B \end{bmatrix}} \end{equation} $
