Answers and Discussions for
Problem 3
To move from stator phase variables to the "new" reference frame, pre-multiply by $ \mathbf{K}_s^n = \begin{bmatrix} 2 & +1 \\ -1 & 2 \end{bmatrix} $. To do the opposite and move from the "new" reference frame to stator phase variables, pre-multiply by $ \left(\mathbf{K}_s^n\right)^{-1} = \frac{1}{5} \begin{bmatrix} 2 & -1 \\ +1 & 2 \end{bmatrix} $. The inverse matrix is found with the explicit formula the the inverse of a 2x2 matrix and its determinant as well. To move from rotor phase variables to the "new" reference frame, pre-multiply by $ \mathbf{K}_r^n = \begin{bmatrix} 4 & +4 \\ +2 & 4 \end{bmatrix} $. To do the opposite and move from the "new" reference frame to rotor phase variables, pre-multiply by $ \left(\mathbf{K}_r^n\right)^{-1} = \frac{1}{8} \begin{bmatrix} 4 & -4 \\ -2 & 4 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -1 & 2 \end{bmatrix} $.
The stator flux linkage equations in stator phase variables and rotor phase variables are put in vector form.
$ \begin{equation} \vec{\lambda}_{abs} = \mathbf{L}_{ss} \vec{i}_{abs} + \mathbf{L}_{sr} \vec{i}_{abr} = \begin{bmatrix} 5 & 0 \\ -1 & 5 \end{bmatrix} \vec{i}_{abs} + \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \vec{i}_{abr} \end{equation} $
The transformation proceeds next.
$ \begin{align} \vec{\lambda}_{qds}^{n} &= \mathbf{K}_s^n \vec{\lambda}_{abs} \\ \vec{\lambda}_{qds}^{n} &= \mathbf{K}_s^n \mathbf{L}_{ss} \left(\mathbf{K}_s^n\right)^{-1} \vec{i}_{qds}^{n} + \mathbf{K}_s^n \mathbf{L}_{sr} \left(\mathbf{K}_r^n\right)^{-1} \vec{i}_{qdr}^{n} \\ \vec{\lambda}_{qds}^{n} &= \begin{bmatrix} 2 & +1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ -1 & 5 \end{bmatrix} \frac{1}{5} \begin{bmatrix} 2 & -1 \\ +1 & 2 \end{bmatrix} \vec{i}_{qds}^{n} + \begin{bmatrix} 2 & +1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -1 & 2 \end{bmatrix} \vec{i}_{qdr}^{n} \\ \vec{\lambda}_{qds}^{n} &= \frac{1}{5} \begin{bmatrix} 9 & 5 \\ -7 & 10 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ +1 & 2 \end{bmatrix} \vec{i}_{qds}^{n} + \frac{1}{4} \begin{bmatrix} -2 & -3 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -1 & 2 \end{bmatrix} \vec{i}_{qdr}^{n} \\ \vec{\lambda}_{qds}^{n} &= \frac{1}{5} \begin{bmatrix} 23 & 1 \\ -4 & 27 \end{bmatrix} \vec{i}_{qds}^{n} + \frac{1}{4} \begin{bmatrix} -1 & -2 \\ 3 & -4 \end{bmatrix} \vec{i}_{qdr}^{n} \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.
The "new" reference frame stator flux linkage equations can be expressed separately.
$ \begin{equation} \boxed{\lambda_{qs}^n = \frac{23}{5} i_{qs}^n + \frac{1}{5} i_{ds}^n - \frac{1}{4} i_{qr}^n - \frac{1}{2} i_{dr}^n} \end{equation} $
$ \begin{equation} \boxed{\lambda_{ds}^n = -\frac{4}{5} i_{qs}^n + \frac{27}{5} i_{ds}^n + \frac{3}{4} i_{qr}^n - i_{dr}^n} \end{equation} $
