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Ex07 Task2: Add calculations until subtask 4.
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@SevenOfNinePE
SevenOfNinePE committed Jan 24, 2025
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21 changes: 13 additions & 8 deletions exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex
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\begin{circuitikz}
% Add voltage U1p
\draw (0,0) coordinate (U1p) to [open, o-o, v = $U_1p\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (Gnd)
(Gnd) to [short,o-o] ++(1,0)
(Gnd) to [short,o-o] ++(0.4,0)
(Gnd) to [open, -o, v = $U_1m\hspace{0.5cm}$, voltage = straight] ++(0,-2.5) coordinate (U1m)
% Add current
(U1p) to [short, o-, i=$i_1(t)$] ++(2,0) coordinate (jT1c)
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% Add u2a inductor
(ju2ax) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2ae)
% Add u2ae
(ju2ae) to [sV=$u_\mathrm{1ae}$] ++(1.5,0) coordinate (ju2an)
(ju2ae) to [sV=$u_\mathrm{2ae}$] ++(1.5,0) coordinate (ju2an)
% Add u2b inductor
(ju2b) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2be)
% Add u2be
(ju2be) to [sV=$u_\mathrm{1be}$] ++(1.5,0) coordinate (ju2bn)
(ju2be) to [sV=$u_\mathrm{2be}$] ++(1.5,0) coordinate (ju2bn)
% Add connection to u2c inductor
(ju2c) to [short,-] ++(0,-2) coordinate (ju2cx)
% Add u2a inductor
(ju2cx) to [L, l=$L$, name = L] ++(2,0) coordinate (ju2ce)
% Add u2ce
(ju2ce) to [sV=$u_\mathrm{1ce}$] ++(1.5,0) coordinate (ju2cn)
(ju2ce) to [sV=$u_\mathrm{2ce}$] ++(1.5,0) coordinate (ju2cn)
% Add connection of u2in
(ju2an) to [short,-*] (ju2bn) to [short,-] (ju2cn);

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(i2b) node[anchor=north,color=black]{$i_\mathrm{2b}(t)$}
(jT6c) ++(1,0) node[currarrow](i2c){}
(i2c) node[anchor=north,color=black]{$i_\mathrm{2c}(t)$}
% Add voltage arrows u2an, u2bn and u2cn
% Add voltage arrow u2an, u2bn and u2cn
(ju2ax) ++(0,-0.8) to [open,v^=$u_\mathrm{2a}(t)$, voltage = straight] ++(3.8,0)
(ju2b) ++(0,-0.8) to [open,v^=$u_\mathrm{2b}(t)$,voltage = straight] ++(3.8,0)
(ju2cx) ++(0,-0.8) to [open,v^=$u_\mathrm{2c}(t)$,voltage = straight] ++(3.8,0)
% Add voltage arrows u2ab
% Add voltage arrow u2ab
(ju2ax) ++(0.2,0) to [open,v^=$$,voltage = straight] ++(0,-2.5)
(ju2ax) ++ (0.9,-1) node[anchor=north,color=black]{$u_\mathrm{2ab}(t)$}
% Add voltage arrows u2bc
% Add voltage arrow u2bc
(ju2b) ++(0.2,0) to [open,v^=$$,voltage = straight] ++(0,-2.5)
(ju2b) ++ (0.9,-1) node[anchor=north,color=black]{$u_\mathrm{2bc}(t)$};
(ju2b) ++ (0.9,-1) node[anchor=north,color=black]{$u_\mathrm{2bc}(t)$}
% Add voltage arrow ua0
(Gnd) ++(2,0) to [open,v^=$$,voltage = straight] ++(-1.6,0)
(Gnd) ++ (1.2,0.7) node[anchor=north,color=black]{$u_\mathrm{2a,0}(t)$};
\end{circuitikz}
\end{center}
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124 changes: 120 additions & 4 deletions exercise/tex/exercise07.tex
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\centering % Center the table
\begin{tabular}{ll}
\toprule
Input voltages: & $U_{\mathrm{1p},i}=\SI{255}{\volt}$ \quad $U_{\mathrm{1m},i}=\SI{255}{\volt}$ \\
Internal voltages: & $u_{\mathrm{1ae}(t)} = \sqrt{2} \cdot \SI{220}{\volt} \cdot \sin(\omega_1t)$ \\
Input voltages: & $U_\mathrm{1p}=\SI{255}{\volt}$ \quad $U_\mathrm{1m}=\SI{255}{\volt}$ \\
Internal voltages: & $u_{\mathrm{2ae}}(t) = \sqrt{2} \cdot \SI{220}{\volt} \cdot \sin(\omega_1t)$ \\
Circular frequency: & $\omega_1 = \SI{2 \pi \cdot 30}{\frac{1}{\second}}$ \\
Inductivity per phase: & $L= \SI{10}{\milli \henry}$ \\
Phase angle between $u_{\mathrm{1ae}(t)}$ and $i_{\mathrm{1ae}^\mathrm{(1)}(t)}$ & $\alpha=\SI{30}{\degree}$ \\
Phase angle between $u_{\mathrm{2ae}}(t)$ and $i_{\mathrm{2ae}^\mathrm{(1)}}(t)$ & $\alpha=\SI{30}{\degree}$ \\
\bottomrule
\end{tabular}
\caption{Parameters of three-phase inverter in six-step mode.}
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Calculate and sketch the voltages $u_\mathrm{a,0}(t)$, $u_\mathrm{b,0}(t)$ and $u_\mathrm{c,0}(t)$ depending on these switching states.
}
\begin{solutionblock}
Each half bridge has got the 2 states '+1' and '-1', which results in $2^3 = 8$ combinations according table \autoref{stable:ex07_Task2_Switchingstates}.
The correct chronological order is displayed in table \autoref{stable:ex07_Task2_UsedSwitchingStates}.
\bigskip
\FloatBarrier

\begin{minipage}{0.3\textwidth}
\begin{tabular}{|c|c|c|} % Each column is separated by a line
\hline
\bfseries $s_\mathrm{a}(t)$ & \bfseries $s_\mathrm{b}(t)$ & \bfseries $s_\mathrm{c}(t)$ \\ \hline
-1 & -1 & -1 \\ \hline
-1 & -1 & +1 \\ \hline
-1 & +1 & -1 \\ \hline
-1 & +1 & +1 \\ \hline
+1 & -1 & -1 \\ \hline
+1 & -1 & +1 \\ \hline
+1 & +1 & -1 \\ \hline
+1 & +1 & +1 \\ \hline
\end{tabular}
\noindent
\captionof{table}{Possible switching states.}
\label{stable:ex07_Task2_Switchingstates}
\end{minipage}
\hfill
\begin{minipage}{0.55\textwidth}
\begin{tabular}{|c|c|c|c|c|c|} % Each column is separated by a line
\hline
\bfseries $s_\mathrm{a}(t)$ & \bfseries $s_\mathrm{b}(t)$ & \bfseries $s_\mathrm{c}(t)$
& \bfseries $+U_\mathrm{2a,0}$ & \bfseries $+U_\mathrm{2b,0}$ & \bfseries $+U_\mathrm{2c,0}$ \\ \hline
+1 & -1 & +1 & $U_\mathrm{1p}$ & $-U_\mathrm{1m}$ & $U_\mathrm{1p}$ \\ \hline
+1 & -1 & -1 & $U_\mathrm{1p}$ & $-U_\mathrm{1m}$ & $-U_\mathrm{1m}$ \\ \hline
+1 & +1 & -1 & $U_\mathrm{1p}$ & $U_\mathrm{1p}$ & $-U_\mathrm{1m}$ \\ \hline
-1 & +1 & -1 & $-U_\mathrm{1m}$ & $U_\mathrm{1p}$ & $-U_\mathrm{1m}$ \\ \hline
-1 & +1 & +1 & $-U_\mathrm{1m}$ & $U_\mathrm{1p}$ & $U_\mathrm{1p}$ \\ \hline
-1 & -1 & +1 & $-U_\mathrm{1m}$ & $-U_\mathrm{1m}$ & $U_\mathrm{1p}$ \\ \hline
\end{tabular}
\captionof{table}{Used switching states and voltages.}
\label{stable:ex07_Task2_UsedSwitchingStates}
\end{minipage}
\bigskip
\FloatBarrier
The switching state (-1,-1,-1) and (+1,+1,+1) are not used. In this case the chained voltages are zero.
This additional degree of freedom is applied at a higher switching frequency in order to reduce the amplitude of the
output voltage on average. In case of block switching, these switching states are not used, since
the switching only occurs twice per period. This results in the maximum possible voltage (square wave) at the output.
\end{solutionblock}

\subtask{The internal voltages $u_\mathrm{ea}(t)$, $u_\mathrm{eb}(t)$ and $u_\mathrm{ec}(t)$ are a symmetrical voltage system,
i.e. the following always applies: $u_\mathrm{ea}(t)+u_\mathrm{eb}(t)+u_\mathrm{ec}(t)=0V$.
Show that this equation is also applicable for the voltages $u_\mathrm{a}(t)$, $u_\mathrm{b}(t)$ and $u_\mathrm{c}(t)$ under the same conditions.
}
\begin{solutionblock}
In the case of a symmetrical three-phase consumer where the current sum at the consumer star point is zero,
the following results:
\begin{equation}
u_{\mathrm{2a}(t)} + u_{\mathrm{2b}(t)} + u_{\mathrm{2c}(t)} = \SI{0}{\volt} \quad
i_{\mathrm{2a}(t)} + i_{\mathrm{2b}(t)} + i_{\mathrm{2c}(t)} = \SI{0}{\ampere}.
\label{eq:u2_i2_symgen}
\end{equation}
This leads to
\begin{equation}
u_{\mathrm{2a}}(t) = L \frac{\mathrm{d}i_{\mathrm{2a}}(t)}{\mathrm{d}t}+u_{\mathrm{2ae}}(t)
\quad u_{\mathrm{2a}}(t) = L \frac{\mathrm{d}i_{\mathrm{2a}}(t)}{\mathrm{d}t}+u_{\mathrm{2ae}}(t)
\quad u_{\mathrm{2a}}(t) = L \frac{\mathrm{d}i_{\mathrm{2a}}(t)}{\mathrm{d}t}+u_{\mathrm{2ae}}(t).
\label{eq:u2_i2_symL}
\end{equation}
Using \eqref{eq:u2_i2_symgen} leads to
\begin{equation}
u_{\mathrm{2a}(t)} + u_{\mathrm{2b}(t)} + u_{\mathrm{2c}(t)}
= L \frac{\mathrm{d}}{\mathrm{d}t} \left( i_{\mathrm{2a}}(t)+i_{\mathrm{2b}}(t)+i_{\mathrm{2c}}(t) \right)
+ \left( u_{\mathrm{2ae}}(t) + u_{\mathrm{2be}}(t) + u_{\mathrm{2ce}}(t)\right)=\SI{0}{\volt}.
\label{eq:u2_i2_symres}
\end{equation}
This derivation is valid under following conditions:
\begin{itemize}
\item The induction $L$ is constant.
\item The internal voltages $u_{\mathrm{2ae}}(t)$, $u_{\mathrm{2be}}(t)$ and $u_{\mathrm{2ce}}(t)$
are purely sinusoidal (no harmonics), symmetrical (sum equals zero) and independent of the currents
$i_{\mathrm{2a}}(t)$, $i_{\mathrm{2b}}(t)$ and $i_{\mathrm{2c}}(t)$.
\end{itemize}
\end{solutionblock}

\subtask{Calculate and sketch the voltages $u_\mathrm{ab}(t)$, $u_\mathrm{bc}(t)$, $u_\mathrm{a}(t)$ and $u_\mathrm{a,0}(t)$
\subtask{Calculate and sketch the voltages $u_\mathrm{2ab}(t)$, $u_\mathrm{2bc}(t)$, $u_\mathrm{2a}(t)$ and $u_\mathrm{2a,0}(t)$
depending on these switching states.}
\begin{solutionblock}
The voltage $u_{\mathrm{2ab}(t)}$ is calculated by
\begin{equation}
u_{\mathrm{2ab}(t)} = u_{\mathrm{2a,0}(t)} - u_{\mathrm{2b,0}(t)}.
\label{eq:u2ab_gen}
\end{equation}
In similar way the voltage $u_{\mathrm{2bc}(t)}$ is calculated by
\begin{equation}
u_{\mathrm{2bc}(t)} = u_{\mathrm{2b,0}(t)} - u_{\mathrm{2c,0}(t)}.
\label{eq:u2bc_gen}
\end{equation}
The voltage $u_{\mathrm{2a}(t)}$ is obtained by
\begin{equation}
u_{\mathrm{2a}(t)} = u_{\mathrm{2ab}(t)} + u_{\mathrm{2b,0}(t)}.
\label{eq:u2a_1}
\end{equation}
Additional voltage $u_{\mathrm{2a}(t)}$ is obtained by
\begin{equation}
u_{\mathrm{2a}(t)} = u_{\mathrm{2ab}(t)} + u_{\mathrm{2bc}(t)} + u_{\mathrm{2c}(t)}.
\label{eq:u2a_2}
\end{equation}
The addition of \eqref{eq:u2a_1} and \eqref{eq:u2a_2} results in
\begin{equation}
2u_{\mathrm{2a}(t)} = 2u_{\mathrm{2ab}(t)} + u_{\mathrm{2bc}(t)}
+ \left( u_{\mathrm{2a}(t)} + u_{\mathrm{2b}(t)} + u_{\mathrm{2v}(t)}\right)
- u_{\mathrm{2a}(t)}.
\label{eq:u2a_gen}
\end{equation}
Solving \eqref{eq:u2a_gen} with respect to $u_{\mathrm{2a}(t)}$ leads to
\begin{equation}
u_{\mathrm{2a}(t)} = \frac{2}{3} u_{\mathrm{2ab}(t)} + \frac{1}{3} u_{\mathrm{2bc}(t)}
\end{equation}
The voltage $u_{\mathrm{0,n}(t)}$ is obtained by
\begin{equation}
u_{\mathrm{0,n}(t)} = u_{\mathrm{2a,0}(t)} - u_{\mathrm{2a}(t)}
= u_{\mathrm{2a,0}(t)} - \frac{2}{3} u_{\mathrm{2ab}(t)} - \frac{1}{3} u_{\mathrm{2bc}(t)}
\end{equation}
Using \eqref{eq:u2ab_gen} and \eqref{eq:u2bc_gen} leads to
\begin{equation}
\begin{split}
u_{\mathrm{0,n}(t)} &= u_{\mathrm{2a,0}(t)} - \frac{2}{3} \left( u_{\mathrm{2a,0}(t)} - u_{\mathrm{2b,0}(t)} \right)
- \frac{1}{3} \left( u_{\mathrm{2b,0}(t)} - u_{\mathrm{2c,0}(t)} \right) \\
u_{\mathrm{0,n}(t)} &= \frac{1}{3} \left( u_{\mathrm{2a,0}(t)} + u_{\mathrm{2b,0}(t)} + u_{\mathrm{2c,0}(t)} \right).
\end{split}
\end{equation}
\end{solutionblock}

\subtask{Decompose the voltage $u_\mathrm{a}(t)$ into a Fourier series and sketch the spectral lines related to the
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