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Problem 3
Contemplation
The first step is to ponder the inductance matrix in $ \vec{\lambda}_{abs} = \mathbf{L}_s \vec{i}_{abs} $.
$ \begin{equation} \mathbf{L}_s = \begin{bmatrix} L_{asas} & L_{asbs} \\ L_{bsas} & L_{bsbs} \end{bmatrix} = \begin{bmatrix} 8 - 4\cos(2\theta_r) & -2\sin(2\theta_r) \\ -2\sin(2\theta_r) & 2 + \cos(2\theta_r) \end{bmatrix} \end{equation} $
Because the self-inductances change with rotor position $ \theta_r $, it is clear that some type of 2-phase, salient synchronous machine is the device under test. The self-inductances can be fit to a known general form.
$ \begin{align} L_{asas} &= L_{\ell as} + L_{Aas} + L_{Bas} \cos\left[2\left(\theta_r - 0\right)\right] = L_{\ell as} + L_{Aas} + L_{Bas} \cos\left(2\theta_r\right) \\ L_{bsbs} &= L_{\ell bs} + L_{Abs} + L_{Bbs} \cos\left[2\left(\theta_r - \frac{\pi}{2}\right)\right] = L_{\ell bs} + L_{Abs} - L_{Bbs} \cos\left(2\theta_r\right) \end{align} $
Phase Variable Referral
Something is amiss because the constant terms of the self-inductances are not equal: $ L_{\ell as} + L_{Aas} \neq L_{ell bs} + L_{Bbs} $ or $ 8 \, \textrm{H} \neq 2 \, \textrm{H} $. This would imply that the sinusoidal amplitudes of the winding functions are not equal for both windings. Before any calculations can proceed further, the machine has to be symmetrized through referred variables. The $ \mathit{bs} $-phase variables will be referred to the $ \mathit{as} $-phase variables.
$ \begin{align} \frac{N_a}{N_b} &= \sqrt{\frac{L_{\ell as} + L_{Aas}}{L_{\ell bs} + L_{Abs}}} = \sqrt{\frac{8}{2}} = 2 \\ \lambda_{bs}^{'} &= \frac{N_a}{N_b} \lambda_{bs} = 2 \lambda_{bs} \\ i_{bs}^{'} &= \frac{N_b}{N_a} i_{bs} = \frac{1}{2} i_{bs} \\ \mathbf{L}_s^{'} &= \begin{bmatrix} L_{asas} & L_{asbs}^{'} \\ L_{asbs}^{'} & L_{bsbs}^{'} \end{bmatrix} = \begin{bmatrix} L_{asas} & \frac{N_a}{N_b} L_{asbs} \\ \frac{N_a}{N_b} L_{asbs} & \left(\frac{N_a}{N_b}\right)^2 L_{bsbs} \end{bmatrix} = \begin{bmatrix} L_{asas} & 2 L_{asbs} \\ 2 L_{asbs} & 4 L_{bsbs} \end{bmatrix} \end{align} $
Thus, the referred inductance matrix is found. Notice that it is still a symmetric matrix and that the constant terms and the coefficient of the trigonometric function of $ 2\theta_r $ match respectively.
$ \begin{equation} \mathbf{L}_s^{'} = \begin{bmatrix} 8 - 4\cos(2\theta_r) & -4\sin(2\theta_r) \\ -4\sin(2\theta_r) & 8 + 4\cos(2\theta_r) \end{bmatrix} \end{equation} $
Drawing a Diagram
Now, the position-dependent behavior of the self-inductances is analyzed so that a diagram of the machine can be drawn with the $ \mathit{qs} $-axis and $ \mathit{ds} $-axis drawn correctly on the dogbone rotor and correct orientation at $ \theta_r = 0 $.
| $ \theta_r $ | $ 0^\circ $ | $ 45^\circ $ | $ 90^\circ $ | $ 135^\circ $ |
|---|---|---|---|---|
| $ L_{asas}(\theta_r) $ | min | ave | max | ave |
| $ L_{bsbs}(\theta_r) $ | max | ave | min | ave |
Orientation of the rotor can be determined from this table.
| $ \theta_r $ | $ 0^\circ $ | $ 45^\circ $ | $ 90^\circ $ | $ 135^\circ $ |
|---|---|---|---|---|
| $ \mathit{as} $-axis observes | mostly air | rotor corners | mostly steel | rotor corners |
| $ \mathit{bs} $-axis observes | mostly steel | rotor corners | mostly air | rotor corners |
It takes $ 90^\circ $ to go from mostly air to mostly steel, so there are only $ P = 2 $ poles on the machine, so the $ \mathit{ds} $-axis is $ \frac{2}{P} 90^\circ $ behind the $ \mathit{qs} $-axis. At $ \theta_r = 0^\circ $, the $ \mathit{qs} $-axis and $ \mathit{as} $-axis are aligned since the $ \mathit{qs} $-axis points out of the rotor where mostly air would be observed. A diagram can be made.
