Answers and Discussions for
Problem 2
To move from stator phase variables to the arbitrary reference frame, pre-multiply by $ \mathbf{K}_s = \begin{bmatrix} \cos(\theta) & +\sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} $. To do the opposite and move from the arbitrary reference frame to stator phase variables, pre-multiply by $ \left(\mathbf{K}_s\right)^{-1} = \frac{1}{\cancelto{1}{\cos^2(\theta) + \sin^2(\theta)}} \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \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 arbitrary reference frame, pre-multiply by $ \mathbf{K}_r = \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} $ where $ \theta_r $ is the electrical rotor angular position. To do the opposite and move from the arbitrary reference frame to rotor phase variables, pre-multiply by $ \left(\mathbf{K}_r\right)^{-1} = \frac{1}{\cancelto{1}{\cos^2(\theta - \theta_r) + \sin^2(\theta - \theta_r)}} \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} = \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} $.
The rotor flux linkage equations in stator phase variables and rotor phase variables are put in vector form without referral.
$ \begin{equation} \vec{\lambda}_{abr} = \mathbf{L}_{rs} \vec{i}_{abs} + \mathbf{L}_{rr} \vec{i}_{abr} = L_{sr} \begin{bmatrix} \cos(\theta_r) & +\sin(\theta_r) \\ -\sin(\theta_r) & \cos(\theta_r) \end{bmatrix} \vec{i}_{abs} + L_{rr} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{i}_{abr} \end{equation} $
The transformation proceeds next.
$ \begin{align} \vec{\lambda}_{qdr} &= \mathbf{K}_r \vec{\lambda}_{abr} \\ \vec{\lambda}_{qdr} &= \mathbf{K}_r \mathbf{L}_{rs} \left(\mathbf{K}_s\right)^{-1} \vec{i}_{qds} + \mathbf{K}_r \mathbf{L}_{rr} \left(\mathbf{K}_r\right)^{-1} \vec{i}_{qdr} \\ \vec{\lambda}_{qdr} &= \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} L_{sr} \begin{bmatrix} \cos(\theta_r) & +\sin(\theta_r) \\ -\sin(\theta_r) & \cos(\theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \end{bmatrix} \vec{i}_{qds} + \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} L_{rr} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \vec{i}_{qdr} \\ \vec{\lambda}_{qdr} &= \begin{split} &L_{sr} \begin{bmatrix} \cos(\theta_r) \cos(\theta - \theta_r) - \sin(\theta_r) \sin(\theta - \theta_r) & \sin(\theta_r) \cos(\theta - \theta_r) + \cos(\theta_r) \sin(\theta - \theta_r) \\ -\cos(\theta_r) \sin(\theta - \theta_r) - \sin(\theta_r) \cos(\theta - \theta_r) & \cos(\theta_r) \cos(\theta - \theta_r) - \sin(\theta_r) \sin(\theta - \theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \end{bmatrix} \vec{i}_{qds} \\ &+ L_{rr} \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \vec{i}_{qdr} \end{split} \\ \vec{\lambda}_{qdr} &= \begin{split} &\frac{L_{sr}}{2} \begin{bmatrix} \cos(\theta) + \cos(\theta - 2\theta_r) + \cos(\theta) - \cos(\theta - 2\theta_r) & +\sin(\theta) - \sin(\theta - 2\theta_r) + \sin(\theta) + \sin(\theta - 2\theta_r) \\ -\sin(\theta) + \sin(\theta - 2\theta_r) - \sin(\theta) - \sin(\theta - 2\theta_r) & \cos(\theta) + \cos(\theta - 2\theta_r) + \cos(\theta) - \cos(\theta - 2\theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \end{bmatrix} \vec{i}_{qds} \\ &+ L_{rr} \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \vec{i}_{qdr} \end{split} \\ \vec{\lambda}_{qdr} &= \begin{split} &\frac{L_{sr}}{2} \begin{bmatrix} 2\cos(\theta) & +2\sin(\theta) \\ -2\sin(\theta) & 2\cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ +\sin(\theta) & \cos(\theta) \end{bmatrix} \vec{i}_{qds} \\ &+ L_{rr} \begin{bmatrix} \cos(\theta - \theta_r) & +\sin(\theta - \theta_r) \\ -\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \begin{bmatrix} \cos(\theta - \theta_r) & -\sin(\theta - \theta_r) \\ +\sin(\theta - \theta_r) & \cos(\theta - \theta_r) \end{bmatrix} \vec{i}_{qdr} \end{split} \\ \vec{\lambda}_{qdr} &= L_{sr} \begin{bmatrix} \cos^2(\theta) + \sin^2(\theta) & -\sin(\theta) \cos(\theta) + \sin(\theta) \cos(\theta) \\ -\sin(\theta) \cos(\theta) + \sin(\theta) \cos(\theta) & \sin^2(\theta) + \cos^2(\theta) \end{bmatrix} \vec{i}_{qds} + L_{rr} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{i}_{qdr} \\ \vec{\lambda}_{qdr} &= L_{sr} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{i}_{qds} + L_{rr} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{i}_{qdr} \\ \vec{\lambda}_{qdr} &= L_{sr} \vec{i}_{qds} + L_{rr} \vec{i}_{qdr} \end{align} $
The fifth line follows from the fourth using the Product-to-Sum Identities (Prosthaphaeresis Formulas given for cosine-cosine, sine-sine, and sine-cosine). The sixth line takes advantage of the fact that $ \mathbf{K}_r \left(\mathbf{K}_r\right)^{-1} = \mathbf{I}_2 $ because one matrix is the inverse of the other when moving to the seventh line. The Pythagorean Identity $ \sin^2(x) + \cos^2(x) = 1 $ (not given explicitly) was used to simplify the seventh line. The arbitrary reference frame rotor flux linkage equations can be expressed separately.
$ \begin{equation} \boxed{\lambda_{qr} = L_{sr} i_{qs} + L_{rr} i_{qr}} \end{equation} $
$ \begin{equation} \boxed{\lambda_{dr} = L_{sr} i_{ds} + L_{rr} i_{dr}} \end{equation} $
