further typo corrections

exercise/main.tex

@@ -1,7 +1,7 @@
 \documentclass[solution]{../course_template/exerciseClass}
 \title{Power Electronics}
 
-\includeonly{tex/exercise04}
+\includeonly{tex/exercise05}
 
 \begin{document}
     \include{tex/exercise01}

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
@@ -372,7 +372,7 @@
         i'(T_\mathrm{2,on}) = i'(T_\mathrm{2,off}) -\frac{ U_\mathrm{L,2}}{L}(T_\mathrm{2,on}-T_\mathrm{2,off}) = \SI{8.07}{\ampere} -\frac{\SI{306}{\volt}}{\SI{5}{\milli\henry}}\cdot (\SI{1040}{\micro\s}-\SI{960}{\micro\s}) = \SI{3.32}{\ampere}.
     \end{equation}
     The values determined in this task for mains current
-    $i'(t)$ must be entered in \autoref{fig:Current i_1 and control signal ex05}. This results in \autoref{fig:Complete Current i' control signal ex05 result}. Figure \autoref{fig:Reference signal $c(t)$ and duty cycle signal d(t)} shows that with this type of PWM current open-loop control, an unphysical duty cycle signal of $>1$ is described at the beginning. This results in a open-loop control deviation that can no longer be compensated for during the control process. An alternative to prevent this deviation would be to implement a feedback control system. This would keep the system continuously within the permissible range $(0,1)$ of the duty cycle. One type of feedback control for this problem would be a voltage time surface control. 
+    $i'(t)$ must be entered in \autoref{fig:Current i_1 and control signal ex05}. This results in \autoref{fig:Complete Current i' control signal ex05 result}. Figure \autoref{fig:Reference signal $c(t)$ and duty cycle signal d(t)} shows that with this type of PWM current open-loop control, a nonphysical duty cycle signal of $>1$ is described at the beginning. This results in an open-loop control deviation that can no longer be compensated for during the control process. An alternative to prevent this deviation would be to implement a feedback control system. This would keep the system continuously within the permissible range $(0,1)$ of the duty cycle. One type of feedback control for this problem would be a voltage time surface control. 
 \end{solutionblock}
 
 \input{fig/ex05/Fig_courses_i'_results}
@@ -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}
@@ -421,7 +421,7 @@
     \begin{equation}
        \Delta u_\mathrm{2}= \hat u_\mathrm{C} = \frac{P_\mathrm{2}}{2\omega C U_\mathrm{2}}.
     \end{equation}
-    And now the capacitance can be calculate with $\Delta u_{\mathrm{2}}<0.05  \hat u_{\mathrm{2}}$ as:
+    Next, the capacitance can be calculated with $\Delta u_{\mathrm{2}}<0.05  \hat u_{\mathrm{2}}$ as:
     \begin{equation}
         C = \frac{P_\mathrm{2}}{2\cdot 2\pi f  U_\mathrm{2}  \Delta u_\mathrm{2} }= \frac{U_\mathrm{2}I_\mathrm{2}}{2\cdot 2\pi f  U_\mathrm{2}\  0.05 U_\mathrm{2}} = \frac{\SI{400}{\volt}\cdot \SI{10}{\ampere}}{4 \pi \cdot\SI{50}{\hertz}\cdot\SI{400}{\volt}\cdot 0.05 \cdot \SI{400}{\volt}} = \SI{796}{\micro\farad}.  
     \end{equation}

exercise/tex/exercise07.tex

@@ -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}

lecture/tex/Lecture03.tex

@@ -1197,7 +1197,7 @@ \subsection{Forward converter}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \begin{frame}
     \frametitle{Forward converter: demagnetization via negative input voltage}
-    \vspace{-0.25cm}
+    \vspace{-0.3cm}
         \begin{figure}
             \begin{circuitikz}[]
                 %Asym. half-bridge
@@ -1217,7 +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_= $u_\mathrm{p}(t)\hspace{0.5cm}$, 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) {$T_2$};
 
@@ -1320,6 +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
@@ -1339,7 +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_= $u_\mathrm{p}(t)\hspace{0.5cm}$, 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) {$T_3$};