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Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2011



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} $

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} $


Discussion



Back to PE-1, August 2011

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Correspondence Chess Grandmaster and Purdue Alumni

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