Answers and Discussions for

ECE Ph.D. Qualifying Exam ES-1 August 2009



Problem 3

The voltage equations of a symmetric, 2-phase machine share a common form that can be written instantly. Let $ r_s $ represent the stator winding resistance of a phase and $ \mathbf{I}_2 $ denote the 2x2 identity matrix.

$ \begin{equation} \boxed{\vec{v}_{abs} = r_s \mathbf{I}_2 \vec{i}_{abs} + \mathit{p}\vec{\lambda}_{abs}} \end{equation} $

To move from stator phase variables to the 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 rotor reference frame to stator phase variables, pre-multiply by $ \mathbf{K}_r^s = \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} $. The inverse matrix is found with the explicit formula the the inverse of a 2x2 matrix and its determinant as well. The Pythagorean identity $ \sin^2(x) + \cos^2(x) = 1 $ (not given explicitly) was used to simplify the determinant.

The flux linkage equations in the rotor reference frame are transformed into stator phase variables.

$ \begin{align} \vec{\lambda}_{abs} &= \mathbf{K}_r^s \vec{\lambda}_{qds}^r \\ \vec{\lambda}_{abs} &= \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r \vec{i}_{abs} \end{align} $

A detour is taken to simplify the matrix product.

$ \begin{align} \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r &= \begin{bmatrix} \cos(\theta_r) & -\sin(\theta_r) \\ +\sin(\theta_r) & \cos(\theta_r) \end{bmatrix} \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} \cos(\theta_r) & +\sin(\theta_r) \\ -\sin(\theta_r) & \cos(\theta_r) \end{bmatrix} \\ \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r &= \begin{bmatrix} \cos(\theta_r) & -\sin(\theta_r) \\ +\sin(\theta_r) & \cos(\theta_r) \end{bmatrix} \begin{bmatrix} 4 \cos(\theta_r) & 4\sin(\theta) \\ -2 \sin(\theta_r) & 2 \cos(\theta_r) \end{bmatrix} \\ \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r &= \begin{bmatrix} 4 \cos^2(\theta_r) + 2 \sin^2(\theta_r) & 4 \sin(\theta_r)\cos(\theta_r) - 2 \sin(\theta_r)\cos(\theta_r) \\ 4 \sin(\theta_r)\cos(\theta_r) - 2 \sin(\theta_r)\cos(\theta_r) & 4 \sin^2(\theta_r) + 2 \cos^2(\theta_r) \end{bmatrix} \\ \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r &= \begin{bmatrix} 2 + 2 \cos(2\theta_r) + 1 - \cos(2\theta_r) & \sin(2\theta_r) \\ \sin(2\theta_r) & 2 - 2\cos(2\theta_r) + 1 + \cos(2\theta_r) \end{bmatrix} \\ \mathbf{K}_r^s \mathbf{L}_s \mathbf{K}_s^r &= \begin{bmatrix} 3 + \cos(2\theta_r) & \sin(2\theta_r) \\ \sin(2\theta_r) & 3 - \cos(2\theta_r) \end{bmatrix} \end{align} $

The fourth line follows from the third using the Power Reduction Identity (given for sine and cosine) and the Double Angle Identity (given for sine). The vector flux linkage equation is finished.

$ \begin{equation} \boxed{\vec{\lambda}_{abs} = \begin{bmatrix} 3 + \cos(2\theta_r) & \sin(2\theta_r) \\ \sin(2\theta_r) & 3 - \cos(2\theta_r) \end{bmatrix} \vec{i}_{abs}} \end{equation} $


Discussion



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Sean Hu, ECE PhD 2009