Merge branch 'main' of https://github.com/IAS-Uni-Siegen/PE_course

exercise/fig/ex04/Fig_currentI1I2PeriodTask1.tex

@@ -49,6 +49,6 @@
             };
         \end{axis}
         \end{tikzpicture} 
-    \caption{Display of the current $i_\mathrm{1}(t)$.}
+    \caption{Input and output currents.}
     \label{fig:currentSecondarySideTask1}
     \end{solutionfigure}

exercise/fig/ex04/Fig_voltageTransistorUsPeriodTask1.tex

@@ -49,7 +49,7 @@
             };
         \end{axis}
         \end{tikzpicture} 
-    \caption{Display of the voltage $u_\mathrm{T}(t)$ and $u_\mathrm{s}(t)$.}
+    \caption{Display of the voltages $u_\mathrm{T}(t)$ and $u_\mathrm{s}(t)$.}
     \label{fig:voltageTransistorPeriodTask1}
 
 

exercise/fig/ex04/sFigDia_Ex04SignalsSingledEndedForwardConverter.tex

@@ -59,7 +59,7 @@
                 \node[black, fill=white, inner sep = 1pt, anchor = south] at (axis cs:0.55,200) {$U_{\mathrm{1}}=\SI{240}{\volt}$};                    
             \end{axis}     
     \end{tikzpicture}
-    \caption{Voltage at transistor.}
+    \caption{Voltage at the transistor.}
     \label{fig:ex04_VoltageAtTransistor}
     \begin{tikzpicture}
         \begin{axis}[

exercise/tex/exercise04.tex

@@ -65,17 +65,17 @@
 
 \end{solutionblock}
 
-\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. Calculate and sketch the following voltage and current curves for this operating points over two cycle periods: $u_\mathrm{T}(t), u_\mathrm{s}(t), i_\mathrm{2}(t), i_\mathrm{1}(t)$. Here, $u_\mathrm{T}(t)$ is the transistor voltage and $u_\mathrm{s}(t)$ is the voltage on the secondary side of the transformer.}
+\subtask{The input voltage is  $U_\mathrm{1}=\SI{382}{\volt}$ at nominal load. Calculate and sketch the following voltage and current curves for this operating point over two cycle periods: $u_\mathrm{T}(t), u_\mathrm{s}(t), i_\mathrm{2}(t), i_\mathrm{1}(t)$. Here, $u_\mathrm{T}(t)$ is the transistor voltage and $u_\mathrm{s}(t)$ is the voltage on the secondary side of the transformer.}
 
 \begin{solutionblock}
-    Because of a different input voltage, the duty cycle $D$ has changed:
+    Because of the different input voltage, the duty cycle $D$ has changed:
     \begin{equation}
         \Delta i_\mathrm{m,max}= \frac{T_\mathrm{s} \cdot U_1}{L_\mathrm{m}} = \frac{\frac{1}{\SI{50}{\kilo\hertz}}\cdot \SI{382}{\volt}}{\SI{760}{\micro\henry}}=\SI{10.05}{\ampere}.
     \end{equation}
     \begin{equation}
         D = \sqrt{\frac{2U_2\overline{i}_2}{U_1\Delta i_\mathrm{m,max}}} = \sqrt{\frac{2\cdot \SI{15}{\volt}\cdot\SI{30}{\watt}}{\SI{382}{\volt}\cdot\SI{15}{\volt}\cdot\SI{10.05}{\ampere}}} = 0.125.
     \end{equation}
-    Trough this result $T_\mathrm{on}$ can be calculated as:
+    Based on this result, $T_\mathrm{on}$ can be calculated as:
     \begin{equation}
         T_\mathrm{on} = D T_\mathrm{s} = 0.125 \cdot \frac{1}{\SI{50}{\kilo\hertz}} = \SI {2.5}{\micro\s}.
     \end{equation}
@@ -104,7 +104,7 @@
         u_\mathrm{T} = \SI{382}{\volt} + \frac{60}{12}\cdot\SI{15}{\volt} = \SI{457}{\volt}.
     \end{equation}
     Because of the conducting diode, for the voltage $U_\mathrm{2}$ is following $U_\mathrm{2} = u_\mathrm{s} = \SI{15}{\volt}$.
-    The third interval is $T_\mathrm{on}+T'_\mathrm{off}$ < t < $T_\mathrm{s}$. In this interval the diode is not conducting and the transistor blocks. From this follows $u_\mathrm{T} = U_1$ and $u_\mathrm{s}=0$.
+    The third interval is $T_\mathrm{on}+T'_\mathrm{off} < t < T_\mathrm{s}$. In this interval the diode is not conducting and the transistor blocks. From this follows $u_\mathrm{T} = U_1$ and $u_\mathrm{s}=0$.
     Looking at the current $i_\mathrm{1}(t)$ over the entire course, the current can only be present when the transistor is switched on. When the transistor is switched on, the current value rises to the peak value $\hat i_\mathrm{1}= \SI{1.257}{\ampere}$. Current $i_\mathrm{2}(t)$ is the discharge current of the secondary winding when the diode is conducting and the transistor blocks. The discharge current starts at the peak value of the secondary side $\hat i_\mathrm{2}= \SI{6.28}{\ampere}$.
     \input{./fig/ex04/Fig_voltageTransistorUsPeriodTask1.tex}
     \input{./fig/ex04/Fig_currentI1I2PeriodTask1.tex}
@@ -205,9 +205,9 @@
 
 \subtask{Calculate the peak value $\hat{i}_\mathrm{m}$ of the magnetizing current $i_\mathrm{m}$.}
 \begin{solutionblock}
-    The duty cycle is less than $0.5$, i.e. the magnetizing current $i_\mathrm{m}$ increase 
+    The duty cycle is less than $0.5$, i.e., the magnetizing current $i_\mathrm{m}$ increase 
     while the transistors are active and decrease to $\SI{0}{\ampere}$ before the next period starts.
-    This means that
+    This leads to 
     \begin{equation}
         \hat{i}_\mathrm{m}=\Delta i_\mathrm{m,max}= \frac{D \cdot U_\mathrm{1}}{L_\mathrm{m} \cdot f_\mathrm{s}} =
         \frac{0.4615 \cdot \SI{325}{\volt}}{\SI{2}{\milli\henry} \cdot \SI{50}{\kilo\hertz}} = \SI{1.5}{\ampere}.
@@ -399,7 +399,7 @@
         I_\mathrm{2}=P_\mathrm{2}/U_\mathrm{2}=\frac{\SI{125}{\watt}}{\SI{5}{\volt}}=\SI{25}{\ampere}
         \label{eq:ex04ouputcurrent}.
     \end{equation}
-    This is corresponds to the current at the primary side of
+    This corresponds to the current at the primary side of
     \begin{equation}
         I_\mathrm{2s}'=I_\mathrm{2}\frac{N_\mathrm{2}}{N_\mathrm{1}}.
     \end{equation}

exercise/tex/exercise05.tex

@@ -7,7 +7,7 @@
 %% Task 1: B2U topology with capacitive filtering %%
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
-\task{B2U topology with capactive filtering}
+\task{B2U topology with capacitive filtering}
 An uncontrolled single-phase, two-pulse rectifier circuit with capacitive filtering 
 is shown in \autoref{fig:B2U_Topology_Cap_Filtering}.
 All components, including the diodes, are assumed to be ideal. On the input side, 
@@ -120,7 +120,7 @@
     \input{fig/ex05/sFig_VoltageU2AndCurrenti1_ic}  
 \end{solutionblock}
 
-\subtask{Calculate the currents $i_\mathrm{1}(\omega t)$  and $i_\mathrm{C}(\omega t)$ and and add them to the previous plot.}
+\subtask{Calculate the currents $i_\mathrm{1}(\omega t)$  and $i_\mathrm{C}(\omega t)$ and add them to the previous plot.}
 \begin{solutionblock}
     Two cases are to consider for the currents of $i_\mathrm{1}(\omega t)$ and  $i_\mathrm{C}(\omega t)$. If $\alpha<\omega t<\alpha+\beta$
     and $\pi+\alpha<\omega t<\pi+\alpha+\beta$, the capacitor supplies the load current $I_{\mathrm{0}}$.
@@ -315,7 +315,7 @@
     \end{equation}
 \end{solutionblock}
 
-\subtask{Complete the current curve for a switching frequency $f_\mathrm{s2} = \SI{2}{\kilo\hertz}$ and a inductance $L = \SI{5}{\milli\henry}$ in \autoref{fig:Current i_1 and control signal ex05}. Note: At the time $t =0$ is $i'=0$. The switch-on and switch-off times are determined by the control signal of the transistor $T$ and are summarized for the first 4 switching times in \autoref{tab:switching_times}.}
+\subtask{Complete the current curve for a switching frequency $f_\mathrm{s2} = \SI{2}{\kilo\hertz}$ and an inductance $L = \SI{5}{\milli\henry}$ in \autoref{fig:Current i_1 and control signal ex05}. Note: At the time $t =0$ is $i'=0$. The switch-on and switch-off times are determined by the control signal of the transistor $T$ and are summarized for the first 4 switching times in \autoref{tab:switching_times}.}
 
 \begin{table}[ht]
     \centering
@@ -409,7 +409,7 @@
     \begin{equation}
         i_\mathrm{C}(t)=C \frac{\mathrm{d}u_\mathrm{c}}{\mathrm{d}t} = \frac{\Delta p_\mathrm{2}}{U_\mathrm{2}} = \frac{P_\mathrm{2}\cos(2\omega t)}{U_\mathrm{2}}.\label{eq:ex056currentic}
     \end{equation}
-    The following follows from \eqref{eq:ex056currentic} for $U_\mathrm{C}$:
+    This follows from \eqref{eq:ex056currentic} for $U_\mathrm{C}$:
     \begin{equation}
         \frac{\mathrm{d}u_\mathrm{C}}{\mathrm{d}t} = \frac{P_\mathrm{2}\cos (2\omega t)}{C U_\mathrm{2}},
     \end{equation}

exercise/tex/exercise07.tex

@@ -64,7 +64,7 @@
         i^\mathrm{(1)}_\mathrm{2}(t) = \frac{u^\mathrm{(1)}_\mathrm{2}(t) - u_{2\mathrm{i}}(t)}{j \omega_\mathrm{2} L},
         \label{7.1.2:eq:i_2_fund}         
     \end{equation}
-    which can be representred in phasor as 
+    which can be represented in phasor as 
     \begin{equation}
         i^\mathrm{(1)}_\mathrm{2}(t) = \frac{\hat{u}^\mathrm{(1)}_\mathrm{2} e^{j \cdot 0} - \hat{u}_{2\mathrm{i}} e^{j \cdot \frac{\pi}{6}}}{j \omega_\mathrm{2} L} = \frac{\hat{u}^\mathrm{(1)}_\mathrm{2} - \hat{u}_{2\mathrm{i}} \cos(\frac{\pi}{6}) + j \hat{u}_{2\mathrm{i}} \sin(\frac{\pi}{6})}{j \omega_\mathrm{2} L}.
         \label{7.1.2:eq:i_2_fund_phasor}         
@@ -105,7 +105,7 @@
 
 Hint:  The current harmonics $i^{(k)}_\mathrm{2}(t)$ are free of any bias.}
 \begin{solutionblock}
-    The harmonics $u^{(k)}_\mathrm{2}(t)$ can be calculated from the output voltage $u_\mathrm{2}(t)$ and the fundamnetal component $u^{(1)}_\mathrm{2}(t)$ with
+    The harmonics $u^{(k)}_\mathrm{2}(t)$ can be calculated from the output voltage $u_\mathrm{2}(t)$ and the fundamental component $u^{(1)}_\mathrm{2}(t)$ with
     \begin{equation}
         u^{(k)}_\mathrm{2}(t) = u_\mathrm{2}(t) - u^{(1)}_\mathrm{2}(t),
         \label{7.1.2:eq:u_2_harm}         
@@ -217,7 +217,7 @@
     \end{cases}
 \bigskip    
 \end{align*}
-Sketch the switching states in the correct chronological order for one periode.
+Sketch the switching states in the correct chronological order for one period.
 Calculate and sketch the voltages $u_\mathrm{2a0}(t)$, $u_\mathrm{2b0}(t)$ and $u_\mathrm{2c0}(t)$ 
 depending on these switching states.}
 \begin{solutionblock}
@@ -266,8 +266,7 @@
     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.
-    The voltages $u_\mathrm{2a0}(\mathrm{\omega t})$, $u_\mathrm{2b0}(\mathrm{\omega t})$, $u_\mathrm{2c0}(\mathrm{\omega t})$
-    ,$u_\mathrm{2ab}(\mathrm{\omega t})$ and $u_\mathrm{2bc}(\mathrm{\omega t})$ are displayed in \autoref{sfig:voltage_u2a0_u2b0_u2c0}.
+    The voltages $u_\mathrm{2a0}(\mathrm{\omega t})$, $u_\mathrm{2b0}(\mathrm{\omega t})$, $u_\mathrm{2c0}(\mathrm{\omega t})$ ,$u_\mathrm{2ab}(\mathrm{\omega t})$ and $u_\mathrm{2bc}(\mathrm{\omega t})$ are displayed in \autoref{sfig:voltage_u2a0_u2b0_u2c0}.
 
     \input{fig/ex07/sFig_Voltage_u2a0_u2b0_u2c0}
 \end{solutionblock}
@@ -410,7 +409,7 @@
     The amplitudes are depicted in \autoref{sfig:NormalizationToTheAmplitude}.
     \input{fig/ex07/sFig_standardization_to_fudamental_freq.tex}
     \FloatBarrier
-    The relation between fundamental und harmonic amplitude is calculated by
+    The relation between fundamental and harmonic amplitude is calculated by
     \begin{equation}
         \begin{split}        
             \frac{\hat{u}_\mathrm{2a0,1}}{\hat{u}_\mathrm{2a0,k}} &= \frac{1}{k} \\
@@ -456,7 +455,6 @@
     Using the result for $\beta$ leads to
     \begin{equation}
         \gamma=\SI{180}{\degree}-\alpha-\beta = \SI{180}{\degree}-\SI{120}{\degree}-\SI{56.1}{\degree}= \SI{3.9}{\degree}.
-        \label{eq:Ex07T2_sin_beta}
     \end{equation}
     In \autoref{sfig:IllustrationForUsingSineTheorem} the triangle is depicted.
     \input{fig/ex07/sFig_ trigonometric_approach_triangle.tex}

lecture/main.ist

@@ -1,5 +1,5 @@
 % makeindex style file created by the glossaries package
-% for document 'main' on 2025-1-28
+% for document 'main' on 2025-2-2
 actual '?'
 encap '|'
 level '!'

lecture/tex/Lecture03.tex

@@ -1197,6 +1197,7 @@ \subsection{Forward converter}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
     \frametitle{Forward converter: demagnetization via negative input voltage}
+    \vspace{-0.3cm}
         \begin{figure}
             \begin{circuitikz}[]
                 %Asym. half-bridge
@@ -1216,7 +1217,7 @@ \subsection{Forward converter}
                 to [short, -o] (B)
                 (F) to [short,*-] ++(1,0) coordinate (I)
                 to [short] ++(1,0) coordinate (H)
-                (J) to [open,v^= $u_\mathrm{p}(t)\hspace{1.6cm}$, voltage = straight] (I);
+                (J) to [open,v_= $u_\mathrm{p}(t)$, voltage = straight] (I);
                 \draw let \p1 = (npn1.B) in node[anchor=east] at (\x1,\y1) {$T_1$};
                 \draw let \p1 = (npn2.B) in node[anchor=east] at (\x1,\y1) {$T_2$};
 
@@ -1319,6 +1320,7 @@ \subsection{Forward converter}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
     \frametitle{Forward converter: demagnetization via negative input voltage (cont.)}
+    \vspace{-0.3cm}
         \begin{figure}
             \begin{circuitikz}[]
                 %full-bridge
@@ -1338,7 +1340,7 @@ \subsection{Forward converter}
                 to [short, -o] (B)
                 (F) to [short,*-] ++(1,0) coordinate (I)
                 to [short] ++(1,0) coordinate (H)
-                (J) to [open,v^= $u_\mathrm{p}(t)\hspace{1.6cm}$, voltage = straight] (I);
+                (J) to [open,v_= $u_\mathrm{p}(t)$, voltage = straight] (I);
                 \draw let \p1 = (npn1.B) in node[anchor=east] at (\x1,\y1) {$T_1$};
                 \draw let \p1 = (npn2.B) in node[anchor=east] at (\x1,\y1) {$T_2$};
                 \draw let \p1 = (npn3.B) in node[anchor=east] at (\x1,\y1) {$T_3$};