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==Problem 1, part a==
 
==Problem 1, part a==
−
===<small>Coenergy Calculation</small>===
+
===<small>Voltage Calculation</small>===
  
−
Sample text<!--Coenergy will be calculated in steps using a sequential ramping process. Before any steps can be completed, the contribution to coenergy of fixing the mechanical system should be documented (zero unless the mechanical system can store energy in its position).
+
The voltage equation for these five identical, magnetically coupled inductors may be written down in matrix form (resistance is neglected).
−
 
+
−
<math>\begin{equation}
+
−
W_{c,0} = 0
+
−
\end{equation}</math>
+
−
 
+
−
The first step in the coenergy calculation will ramp dummy variable <math>i_{1}'</math> from <math>0</math> to its final value of <math>i_{1}</math> while dummy variable <math>i_{2}' = 0</math> is held at its initial value. Recall that <math>\int \frac{au}{\sqrt{b + u^2}} \, du = a \sqrt{b + u^2} + C</math> where <math>C</math> is an arbitrary constant.
+
  
 
<math>\begin{align}
 
<math>\begin{align}
−
W_{c,1} &= \left.\int_{i_{1}' = 0}^{i_{1}} \lambda_{1}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{1}'\right|_{i_{2}' = 0} + \left.\int_{i_{1}' = 0}^{i_{1}} \lambda_{2}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{1}'\right|_{i_{2}' = 0} \\
+
\vec{v}_{12345s} &= \cancelto{\mathbf{0}}{\mathbf{r}_s} \vec{i}_{12345s} + p \vec{\lambda}_{12345s} \\
−
W_{c,1} &= \int_{i_{1}' = 0}^{i_{1}} \left[10i_{1}' + 2 \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1}' + \cancelto{0}{i_{2}'}}{\sqrt{1 + (2i_{1}' + \cancelto{0}{i_{2}'})^2}}\right] \, di_{1}' + \int_{i_{1}' = 0}^{i_{1}} \left[5\cancelto{0}{i_{2}'} + \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1}' + \cancelto{0}{i_{2}'}}{\sqrt{1 + (2i_{1}' + \cancelto{0}{i_{2}'})^2}}\right] \, di_{1}' \\
+
\vec{v}_{12345s} &= p \vec{\lambda}_{12345s}
−
W_{c,1} &= 5 i_{1}^2 + 2 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} - 2 \left(5 + 2\sin(2\theta_{rm})\right) + \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} - \left(5 + 2\sin(2\theta_{rm})\right) \\
+
−
W_{c,1} &= 5 i_{1}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right)
+
 
\end{align}</math>
 
\end{align}</math>
  
−
The evaluation of the previous integral has terms for <math>i_{1}' = 0</math> (lower bound evaluation) that must not be forgotten. The second step in the coenergy calculation will ramp dummy variable <math>i_{2}'</math> from <math>0</math> to its final value of <math>i_{2}</math> while dummy variable <math>i_{1}' = i_{1}</math> is held at its final value.
+
The flux linkage equation needs to be written (linear magnetics are assumed).
  
 
<math>\begin{align}
 
<math>\begin{align}
−
W_{c,2} &= \left.\int_{i_{2}' = 0}^{i_{2}} \lambda_{1}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{2}'\right|_{i_{1}' = i_{1}} + \left.\int_{i_{2}' = 0}^{i_{2}} \lambda_{2}\left(i_{1}', i_{2}', \theta_{rm}\right) \, di_{2}'\right|_{i_{1}' = i_{1}} \\
+
\vec{\lambda}_{12345s} &= \mathbf{L}_{ss} \vec{i}_{12345s} \\
−
W_{c,2} &= \int_{i_{2}' = 0}^{i_{2}} \left[10i_{1} + 2 \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1} + i_{2}'}{\sqrt{1 + (2i_{1} + i_{2}')^2}}\right] \, di_{2}' + \int_{i_{2}' = 0}^{i_{2}} \left[5i_{2}' + \left(5 + 2\sin(2\theta_{rm})\right) \frac{2i_{1} + i_{2}'}{\sqrt{1 + (2i_{1} + i_{2}')^2}}\right] \, di_{2}' \\
+
\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \\ \lambda_5 \end{bmatrix} &=
−
W_{c,2} &= 10 i_{1} i_{2} + 2 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 2 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} + \frac{5}{2} i_{2}^2 + \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} \\
+
\begin{bmatrix}
−
W_{c,2} &= 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2}
+
  L_s & M_{ns} & M_{fs} & M_{fs} & M_{ns} \\
 +
  M_{ns} & L_s & M_{ns} & M_{fs} & M_{fs} \\
 +
  M_{fs} & M_{ns} & L_s & M_{ns} & M_{fs} \\
 +
  M_{fs} & M_{fs} & M_{ns} & L_s & M_{ns} \\
 +
  M_{ns} & M_{fs} & M_{fs} & M_{ns} & L_s
 +
\end{bmatrix}
 +
\begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \end{bmatrix}
 
\end{align}</math>
 
\end{align}</math>
  
−
The evaluation of the previous integral has terms for <math>i_{2}' = 0</math> (lower bound evaluation) that must not be forgotten. The final step is the sum all the individual coenergy contributions <math>W_c(i_{1}, i_{2}, \theta_{rm}) = \sum_{k = 0}^{2} W_{c,k}(i_{1}, i_{2}, \theta_{rm})</math>.
+
Because the inductors are identical to each other, only three unique, '''constant''' entries exist in the inductance matrix <math>\mathbf{L}_{ss}</math>. The self-inductance <math>L_s > 0</math> combines the effects of leakage inductance and magnetizing inductance. The mutual inductance between inductors ''near'' each other, <math>\frac{2\pi}{5} \, \text{rad}</math> apart in other words, is denoted <math>M_{ns}</math>. The mutual inductance between inductors ''far'' from each other, <math>\frac{4\pi}{5} \, \text{rad}</math> apart in other words, is denoted <math>M_{fs}</math>. Because inductors with spacing <math> -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2}</math> produce a positive value of cosine, the mutual inductance of near conductors is positive <math>M_{ns} > 0</math>. The inductors with spacing <math>-\pi < \Delta\theta < -\frac{\pi}{2}</math> or <math>+\frac{\pi}{2} < \Delta\theta < \pi</math> produce a negative value of cosine, leading to a negative mutual inductance <math>M_{fs} < 0</math>.
  
−
<math>\begin{align}
+
===<small>Solving Imbalanced Conditions</small>===
−
W_c(i_{1}, i_{2}, \theta_{rm}) &= W_{c,0} + W_{c,1} + W_{c,2} \\
+
−
W_c(i_{1}, i_{2}, \theta_{rm}) &=
+
−
\begin{split}
+
−
  &{}0 + 5 i_{1}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \\
+
−
  &{}+ 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1})^2} \\
+
−
\end{split} \\
+
−
W_c(i_{1}, i_{2}, \theta_{rm}) &= 5 i_{1}^2 + 10 i_{1} i_{2} + \frac{5}{2} i_{2}^2 + 3 \left(5 + 2\sin(2\theta_{rm})\right) \sqrt{1 + (2i_{1} + i_{2})^2} - 3 \left(5 + 2\sin(2\theta_{rm})\right)
+
−
\end{align}</math>
+
−
<!--
+
−
This calculation is consistent with the known result of a magnetically linear system having a coenergy of <math>W_c(i_{as}, i_{bs}, \theta_{rm}) = \frac{1}{2} L_{asas}(\theta_{rm}) i_{as}^2 + L_{asbs}(\theta_{rm}) i_{as} i_{bs} + \frac{1}{2} L_{bsbs}(\theta_{rm}) i_{bs}^2 = \frac{1}{2} \sum_{j = 1}^2 \sum_{k = 1}^2 \lambda_j(i, \theta_{rm}) i_k</math>.
+
−
--><!--
+
−
===<small>Electromechanical Torque Calculation</small>===
+
  
−
Electromechanical torque is just the partial derivative of coenergy with respect to mechanical rotor position.
+
The current through the circuit is given knowing that four the the five inductors are open-circuited and carry no current.
 +
 
 +
<math>\begin{equation}
 +
\begin{bmatrix} i_1(t) \\ i_2(t) \\ i_3(t) \\ i_4(t) \\ i_5(t) \end{bmatrix} = \begin{bmatrix} 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t\right) \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \text{A}
 +
\end{equation}</math>
 +
 
 +
The following, partially complete information is measured from the circuit.
 +
 
 +
<math>\begin{equation}
 +
\begin{bmatrix} v_1(t) \\ v_2(t) \\ v_3(t) \\ v_4(t) \\ v_5(t) \end{bmatrix} = \begin{bmatrix} 40 \cos(\omega_1 t + \phi_1) \\ 10 \cos(\omega_2 t + \phi_2) \\ v_3(t) \\ 1 \cos(\omega_4 t + \phi_4) \\ v_5(t) \end{bmatrix} \, \text{V}
 +
\end{equation}</math>
 +
 
 +
Because of the linear magnetics, the frequency of the voltage response must match the frequency of the current excitation: <math>\omega_j = 100 \, \frac{\text{rad}}{\text{s}} \, \forall j \in \{1, 2, 3, 4, 5\}</math>.
 +
 
 +
<math>\begin{equation}
 +
\boxed{\omega_1 = 100 \, \frac{\text{rad}}{\text{s}}}
 +
\end{equation}</math>
 +
 
 +
<math>\begin{equation}
 +
\boxed{\omega_2 = 100 \, \frac{\text{rad}}{\text{s}}}
 +
\end{equation}</math>
 +
 
 +
<math>\begin{equation}
 +
\boxed{\omega_4 = 100 \, \frac{\text{rad}}{\text{s}}}
 +
\end{equation}</math>
 +
 
 +
By analyzing the voltage equation and flux linkage equation, the phase shift of the voltage of each coil becomes known. The first coil is analyzed.
  
 
<math>\begin{align}
 
<math>\begin{align}
−
T_e &= \frac{\partial W_c(i_{1}, i_{2}, \theta_{rm})}{\partial \theta_{rm}} \\
+
v_1(t) &= p \lambda_1(t) \\
−
T_e &= 0 + 0 + 0 + 12\cos(2\theta_{rm}) \sqrt{1 + (2i_{1} + i_{2})^2} - 12\cos(2\theta_{rm})
+
v_1(t) &= p \left[L_s i_1(t)\right] \\
 +
v_1(t) &= L_s \frac{d}{dt} \left[10 \cos\left(\omega_1 t\right) \, \text{A}\right] \\
 +
v_1(t) &= L_s \left[-10 \omega_1 \sin\left(\omega_1 t\right) \, \text{A}\right] \\
 +
v_1(t) &= 10^3 L_s \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\
 +
v_2(t) &= 10^3 M_{ns} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\
 +
v_4(t) &= 10^3 M_{fs} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\
 +
v_4(t) &= 10^3 |M_{fs}| \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \, \text{V}
 
\end{align}</math>
 
\end{align}</math>
  
−
Thus, the electromechanical torque equation is obtained for this device.
+
Similar calculations unfold for <math>v_2(t)</math> and <math>v_4(t)</math>, just with different amplitudes. The voltages all have a phase shift <math>\phi_j = -\frac{\pi}{2} \, \text{rad} \, j \in \{1, 2, 5\}</math> and <math>\phi_j = +\frac{\pi}{2} \, \text{rad} \, j \in \{3, 4\}</math>. In other words, the current lags the voltage with quadrature phase shift for inductors shifted <math> -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2}</math>, and the current leads the voltage with quadrature phase shift for the remaining inductors.
  
 
<math>\begin{equation}
 
<math>\begin{equation}
−
\boxed{T_e = 12\cos(2\theta_{rm}) \left[\sqrt{1 + (2i_{1} + i_{2})^2} - 1\right]}
+
\boxed{\phi_1 = -\frac{\pi}{2} \, \text{rad}}
−
\end{equation}</math>-->
+
\end{equation}</math>
 +
 
 +
<math>\begin{equation}
 +
\boxed{\phi_2 = -\frac{\pi}{2} \, \text{rad}}
 +
\end{equation}</math>
 +
 
 +
<math>\begin{equation}
 +
\boxed{\phi_4 = +\frac{\pi}{2} \, \text{rad}}
 +
\end{equation}</math>
  
 
----
 
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Latest revision as of 14:22, 4 August 2018


Answers and Discussions for

ECE Ph.D. Qualifying Exam PE-1 August 2014

Problem 1, part a

Voltage Calculation

The voltage equation for these five identical, magnetically coupled inductors may be written down in matrix form (resistance is neglected).

$ \begin{align} \vec{v}_{12345s} &= \cancelto{\mathbf{0}}{\mathbf{r}_s} \vec{i}_{12345s} + p \vec{\lambda}_{12345s} \\ \vec{v}_{12345s} &= p \vec{\lambda}_{12345s} \end{align} $

The flux linkage equation needs to be written (linear magnetics are assumed).

$ \begin{align} \vec{\lambda}_{12345s} &= \mathbf{L}_{ss} \vec{i}_{12345s} \\ \begin{bmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \lambda_4 \\ \lambda_5 \end{bmatrix} &= \begin{bmatrix} L_s & M_{ns} & M_{fs} & M_{fs} & M_{ns} \\ M_{ns} & L_s & M_{ns} & M_{fs} & M_{fs} \\ M_{fs} & M_{ns} & L_s & M_{ns} & M_{fs} \\ M_{fs} & M_{fs} & M_{ns} & L_s & M_{ns} \\ M_{ns} & M_{fs} & M_{fs} & M_{ns} & L_s \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ i_3 \\ i_4 \\ i_5 \end{bmatrix} \end{align} $

Because the inductors are identical to each other, only three unique, constant entries exist in the inductance matrix $ \mathbf{L}_{ss} $. The self-inductance $ L_s > 0 $ combines the effects of leakage inductance and magnetizing inductance. The mutual inductance between inductors near each other, $ \frac{2\pi}{5} \, \text{rad} $ apart in other words, is denoted $ M_{ns} $. The mutual inductance between inductors far from each other, $ \frac{4\pi}{5} \, \text{rad} $ apart in other words, is denoted $ M_{fs} $. Because inductors with spacing $ -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2} $ produce a positive value of cosine, the mutual inductance of near conductors is positive $ M_{ns} > 0 $. The inductors with spacing $ -\pi < \Delta\theta < -\frac{\pi}{2} $ or $ +\frac{\pi}{2} < \Delta\theta < \pi $ produce a negative value of cosine, leading to a negative mutual inductance $ M_{fs} < 0 $.

Solving Imbalanced Conditions

The current through the circuit is given knowing that four the the five inductors are open-circuited and carry no current.

$ \begin{equation} \begin{bmatrix} i_1(t) \\ i_2(t) \\ i_3(t) \\ i_4(t) \\ i_5(t) \end{bmatrix} = \begin{bmatrix} 10 \cos\left(100 \, \frac{\text{rad}}{\text{s}} t\right) \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \, \text{A} \end{equation} $

The following, partially complete information is measured from the circuit.

$ \begin{equation} \begin{bmatrix} v_1(t) \\ v_2(t) \\ v_3(t) \\ v_4(t) \\ v_5(t) \end{bmatrix} = \begin{bmatrix} 40 \cos(\omega_1 t + \phi_1) \\ 10 \cos(\omega_2 t + \phi_2) \\ v_3(t) \\ 1 \cos(\omega_4 t + \phi_4) \\ v_5(t) \end{bmatrix} \, \text{V} \end{equation} $

Because of the linear magnetics, the frequency of the voltage response must match the frequency of the current excitation: $ \omega_j = 100 \, \frac{\text{rad}}{\text{s}} \, \forall j \in \{1, 2, 3, 4, 5\} $.

$ \begin{equation} \boxed{\omega_1 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $

$ \begin{equation} \boxed{\omega_2 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $

$ \begin{equation} \boxed{\omega_4 = 100 \, \frac{\text{rad}}{\text{s}}} \end{equation} $

By analyzing the voltage equation and flux linkage equation, the phase shift of the voltage of each coil becomes known. The first coil is analyzed.

$ \begin{align} v_1(t) &= p \lambda_1(t) \\ v_1(t) &= p \left[L_s i_1(t)\right] \\ v_1(t) &= L_s \frac{d}{dt} \left[10 \cos\left(\omega_1 t\right) \, \text{A}\right] \\ v_1(t) &= L_s \left[-10 \omega_1 \sin\left(\omega_1 t\right) \, \text{A}\right] \\ v_1(t) &= 10^3 L_s \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_2(t) &= 10^3 M_{ns} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_4(t) &= 10^3 M_{fs} \cos\left(100 \, \frac{\text{rad}}{\text{s}} t - \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \\ v_4(t) &= 10^3 |M_{fs}| \cos\left(100 \, \frac{\text{rad}}{\text{s}} t + \frac{\pi}{2} \, \text{rad}\right) \, \text{V} \end{align} $

Similar calculations unfold for $ v_2(t) $ and $ v_4(t) $, just with different amplitudes. The voltages all have a phase shift $ \phi_j = -\frac{\pi}{2} \, \text{rad} \, j \in \{1, 2, 5\} $ and $ \phi_j = +\frac{\pi}{2} \, \text{rad} \, j \in \{3, 4\} $. In other words, the current lags the voltage with quadrature phase shift for inductors shifted $ -\frac{\pi}{2} < \Delta\theta < +\frac{\pi}{2} $, and the current leads the voltage with quadrature phase shift for the remaining inductors.

$ \begin{equation} \boxed{\phi_1 = -\frac{\pi}{2} \, \text{rad}} \end{equation} $

$ \begin{equation} \boxed{\phi_2 = -\frac{\pi}{2} \, \text{rad}} \end{equation} $

$ \begin{equation} \boxed{\phi_4 = +\frac{\pi}{2} \, \text{rad}} \end{equation} $

Discussion

Back to PE-1, August 2014

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