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\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}
 
\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}</math>
 
\end{align}</math>
 
Recall by the Quotient Rule or the Chain Rule + Power Rule that <math>\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}</math> for <math>a, b_k, K \in \mathbb{R}</math>. The vector voltage equation is finished.
 
  
 
The "new" reference frame stator flux linkage equations can be expressed separately.
 
The "new" reference frame stator flux linkage equations can be expressed separately.

Revision as of 18:01, 26 January 2018


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



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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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