Answers and Discussions for
Problem 2
It should be known that the arbitrary reference frame transformation has the following form.
$ \begin{align} \vec{f}_{qd0s} &= \mathbf{K}_s(\theta) \vec{f}_{abcs} \\ \mathbf{K}_s(\theta) &= \frac{2}{3} \begin{bmatrix} \cos\left(\theta\right) & \cos\left(\theta - \frac{2\pi}{3}\right) & \cos\left(\theta + \frac{2\pi}{3}\right) \\ \sin\left(\theta\right) & \sin\left(\theta - \frac{2\pi}{3}\right) & \sin\left(\theta + \frac{2\pi}{3}\right) \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} \end{align} $
The stationary reference frame transformation is a linear transformation represented by the matrix $ \mathbf{K}_s^s $ obtained from the arbitrary reference frame transformation matrix.
$ \begin{equation} \mathbf{K}_s^s = \left.\mathbf{K}_s(\theta)\right|_{\theta = 0} = \frac{2}{3} \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & -\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix} = \mathbf{C} \end{equation} $
The matrices of the stationary reference reference frame transformation and Clarke's transformation are identical: $ \mathbf{K}_s^s = \mathbf{I}_3 \mathbf{C} $. Therefore, $ \vec{f}_{qd0s}^s = \mathbf{I}_3 \vec{f}_{\alpha\beta0} $.
This identity may be expressed in scalar equations.
$ \begin{equation} \boxed{f_\alpha = f_{qs}^s} \end{equation} $
$ \begin{equation} \boxed{f_\beta = f_{ds}^s} \end{equation} $
$ \begin{equation} \boxed{f_0 = f_{0s}} \end{equation} $
