Answers and Discussions for

## Problem 2

To move from stator phase variables to the weird reference frame, pre-multiply by $\mathbf{K}_s^w = \begin{bmatrix} 1 & -t^2 \\ +t^2 & 1 \end{bmatrix}$. To do the opposite and move from the weird reference frame to stator phase variables, pre-multiply by $\mathbf{K}_w^s = \left(\mathbf{K}_s^w\right)^{-1} = \frac{1}{1 + t^4} \begin{bmatrix} 1 & +t^2 \\ -t^2 & 1 \end{bmatrix}$. The inverse matrix is found with the explicit formula the the inverse of a 2x2 matrix and its determinant as well.

The voltage equations in stator phase variables are put in vector form.

$$$\vec{v}_{abs} = \mathbf{r}_s \vec{i}_{abs} + \mathit{p}\vec{\lambda}_{abs} = r_s \mathbf{I}_2 \vec{i}_{abs} + \mathit{p}\vec{\lambda}_{abs}$$$

The transformation proceeds next.

\begin{align} \vec{v}_{qds}^{w} &= \mathbf{K}_s^w \vec{v}_{abs} \\ \vec{v}_{qds}^{w} &= \mathbf{K}_s^w r_s \mathbf{I}_2 \left(\mathbf{K}_w^s \vec{i}_{qds}^{w}\right) + \mathbf{K}_s^w \mathit{p} \left(\mathbf{K}_w^s \vec{\lambda}_{qds}^{w}\right) \\ \vec{v}_{qds}^{w} &= r_s \cancelto{\mathbf{I}_2}{\mathbf{K}_s^w \mathbf{I}_2 \mathbf{K}_w^s} \vec{i}_{qds}^{w} + \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) \vec{\lambda}_{qds}^{w} + \cancelto{\mathbf{I}_2}{\mathbf{K}_s^w \mathbf{K}_w^s} \mathit{p}\vec{\lambda}_{qds}^{w} \end{align}

The Product Rule is applied for $\mathit{p} \left(\mathbf{K}_w^s \vec{\lambda}_{qds}^{w}\right)$ to produce two terms, the latter of which also simplifies due to $\mathbf{K}_s^w$ and $\mathbf{K}_w^s$ being matrix inverses. A detour is needed to simplify the middle term.

\begin{align} \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) &= \begin{bmatrix} 1 & -t^2 \\ +t^2 & 1 \end{bmatrix} \mathit{p} \left(\frac{1}{1 + t^4} \begin{bmatrix} 1 & +t^2 \\ -t^2 & 1 \end{bmatrix}\right) \\ \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) &= \begin{bmatrix} 1 & -t^2 \\ +t^2 & 1 \end{bmatrix} \frac{-4t^3}{\left(1 + t^4\right)^2} \begin{bmatrix} 1 & +t^2 \\ -t^2 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -t^2 \\ +t^2 & 1 \end{bmatrix} \frac{1}{1 + t^4} \begin{bmatrix} 0 & +2t \\ -2t & 0 \end{bmatrix} \\ \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) &= \frac{-4t^3}{\left(1 + t^4\right)^2} \begin{bmatrix} 1 + t^4 & 0 \\ 0 & 1 + t^4 \end{bmatrix} + \frac{1}{1 + t^4} \begin{bmatrix} +2t^3 & +2t \\ -2t & +2t^3 \end{bmatrix} \\ \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) &= \frac{1}{1 + t^4} \begin{bmatrix} -4t^3 & 0 \\ 0 & -4t^3 \end{bmatrix} + \frac{1}{1 + t^4} \begin{bmatrix} +2t^3 & +2t \\ -2t & +2t^3 \end{bmatrix} \\ \mathbf{K}_s^w \left(\mathit{p}\mathbf{K}_w^s\right) &= \frac{1}{1 + t^4} \begin{bmatrix} -2t^3 & +2t \\ -2t & -2t^3 \end{bmatrix} \end{align}

Recall by the Quotient Rule or the Chain Rule + Power Rule that $\frac{d}{dt} \frac{a}{\sum_{k=0}^K b_k t^k} = \frac{-a\sum_{k=1}^K k b_k t^{k - 1}}{\left(\sum_{k=0}^K b_k t^k\right)^2}$ for $a, b_k, K \in \mathbb{R}$. The vector voltage equation is finished.

$$$\vec{v}_{qds}^{w} = r_s \vec{i}_{qds}^{w} + \frac{1}{1 + t^4} \begin{bmatrix} -2t^3 & +2t \\ -2t & -2t^3 \end{bmatrix} \vec{\lambda}_{qds}^{w} + \mathit{p}\vec{\lambda}_{qds}^{w}$$$

The weird reference frame voltage equations can be expressed separately.

$$$\boxed{v_{qs}^w = r_s i_{qs}^w - \frac{2t^3}{1 + t^4} \lambda_{qs}^w + \frac{2t}{1 + t^4} \lambda_{ds}^w + \mathit{p}\lambda_{qs}^w}$$$

$$$\boxed{v_{ds}^w = r_s i_{ds}^w - \frac{2t}{1 + t^4} \lambda_{qs}^w - \frac{2t^3}{1 + t^4} \lambda_{ds}^w + \mathit{p}\lambda_{ds}^w}$$$

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