Ex07 Task2: Add missing correction and update solution.

exercise/fig/ex07/Fig_standardization_to_fudamental_freq.tex

@@ -27,5 +27,5 @@
     }
 \end{tikzpicture}
 \caption{Normalization to the amplitude of the fundamental oscillation.}
-\label{fig:Normalization to the amplitude of the fundamental oscillation}
+\label{fig:fig:NormalizationToTheAmplitude}
 \end{solutionfigure}

exercise/tex/exercise07.tex

@@ -77,9 +77,9 @@
         \toprule
         Input voltages: & $U_\mathrm{1}=\SI{510}{\volt}$ \\
         Internal voltages: & $u_{\mathrm{2ai}}(t) = \sqrt{2} \cdot \SI{220}{\volt} \cdot \sin(\omega_1t)$ \\
-        Angular load frequency: & $\omega_1 = \SI{2 \pi \cdot 30}{\frac{1}{\second}}$ \\ 
+        Angular load frequency: & $\omega_1 = \SI{2 \pi \cdot 30}{\frac{1}{\second}}$ \\ 11
         Inductance per phase: & $L= \SI{10}{\milli \henry}$ \\
-        Phase angle between  $u_{\mathrm{2ai}}(t)$ and $i_{\mathrm{2ai}}^\mathrm{(1)}(t)$ & $\alpha=\SI{30}{\degree}$ \\
+        Phase angle between  $u_{\mathrm{2ai}}(t)$ and $i_{\mathrm{2ai}}^\mathrm{(1)}(t)$ & $\varphi_{\mathrm{2ai}}^\mathrm{(1)}=\SI{30}{\degree}$ \\
         \bottomrule
     \end{tabular}
     \caption{Parameters of three-phase inverter in six-step mode.}  
@@ -285,6 +285,7 @@
     \end{equation}   
     The amplitudes are depicted in \autoref{fig:NormalizationToTheAmplitude},
     \input{fig/ex07/Fig_standardization_to_fudamental_freq.tex}
+    \FloatBarrier
     The relation between fundamental und harmonic amplitude is calculated by
     \begin{equation}
         \begin{split}        
@@ -301,14 +302,17 @@
 \begin{solutionblock}
     According \eqref{eq:Ex07T2_FundamentelVoltage} the fundamental voltage  is calculated as:
     \begin{equation}
-        \hat{u}_\mathrm{2a0,1}(t)=\frac{2U_{\mathrm{1}}}{1\pi}=\frac{2\cdot \SI{510}{\volt}}{1\pi}=\SI{324,68}{\volt}
+        \hat{u}_\mathrm{2a0,1}(t)=\frac{2U_{\mathrm{1}}}{1\pi}=\frac{2\cdot \SI{510}{\volt}}{1\pi}=\SI{324,68}{\volt}.
     \end{equation}
-    The amplitude of $u_\mathrm{2ae}(t)$ is given with
+    The amplitude of $u_\mathrm{2ai}(t)$ is given with
     \begin{equation}
-        \hat{u}_\mathrm{2ae}=\sqrt{2} \cdot \SI{220}{\volt}=\SI{311,13}{\volt}
+        \hat{u}_\mathrm{2ai}=\sqrt{2} \cdot \SI{220}{\volt}=\SI{311,13}{\volt}.
     \end{equation}
-
-    A triangle is formed by the inverter voltage $u_\mathrm{2a}^\mathrm{(1)}(t)$, the voltage $u_\mathrm{2ae}(t)$ 
+    The angle between $u_\mathrm{2ai}(t)$ and the voltage drop at the inductor is calculated by
+    \begin{equation}
+        \alpha=\SI{90}{\degree}+\varphi_{\mathrm{2ai}}^\mathrm{(1)}=\SI{90}{\degree}+\SI{30}{\degree}=\SI{120}{\degree}.  
+    \end{equation}
+    A triangle is formed by the inverter voltage $u_\mathrm{2a}^\mathrm{(1)}(t)$, the voltage $u_\mathrm{2ai}(t)$ 
     and the voltage drop across the inductance $L$. Two sides and one angle are known. By applying the sine theorem results in
     \begin{equation}
         \begin{split}        
@@ -332,12 +336,12 @@
     \input{fig/ex07/Fig_ trigonometric_approach_triangle.tex}
     In a symmetrical three-phase system, the active power is:
     \begin{equation}
-        P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\phi)
+        P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\varphi)
         \label{eq:Ex07T2_EffPowergen}
     \end{equation}
 
     $U_{\mathrm{L-L}}$ corresponds to the effective value of the line-to-line voltage and $IU_{\mathrm{L}}$ is the effective value
-    of the line current and $\phi$ is the phase angle between voltage and current. For this case $U_{\mathrm{L-L}}$ is calculated by
+    of the line current and $\varphi$ is the phase angle between voltage and current. For this case $U_{\mathrm{L-L}}$ is calculated by
     \begin{equation}
         U_{\mathrm{L-L}}=\sqrt{3}\frac{\hat{u}_\mathrm{2a}^\mathrm{(1)}}{\sqrt{2}}
         = \sqrt{\frac{3}{2}} \frac{2U_\mathrm{1}}{\pi} = \sqrt{3}\sqrt{2}\frac{U_\mathrm{1}}{\pi}
@@ -350,18 +354,18 @@
         = \frac{\SI{12.37}{\ampere}}{\sqrt{2}}=\SI{8.75}{\ampere}
         \label{eq:Ex07T2_EffPowercurrent}
     \end{equation}    
-    The angle $\phi$ results in
+    The angle $\varphi$ results in
     \begin{equation}
         \begin{split}        
-            &\phi=\SI{30}{\degree}+ \gamma= \SI{30}{\degree}+\SI{3.9}{\degree}= \SI{33.9}{\degree}\phi=\SI{30}{\degree}+ \gamma= \SI{30}{\degree}+\SI{3.9}{\degree}= \SI{33.9}{\degree} \\
-            &\cos(\phi)=\cos(\SI{33.9}{\degree})=0.83
+            &\varphi=\SI{30}{\degree}+ \gamma= \SI{30}{\degree}+\SI{3.9}{\degree}= \SI{33.9}{\degree} \\
+            &\cos(\varphi)=\cos(\SI{33.9}{\degree})=0.83
         \end{split}          
         \label{eq:Ex07T2_EffPowerangle}
     \end{equation}
     Using \eqref{eq:Ex07T2_EffPowervoltage},  \eqref{eq:Ex07T2_EffPowercurrent} and  \eqref{eq:Ex07T2_EffPowerangle}
     in \eqref{eq:Ex07T2_EffPowergen} leads to
     \begin{equation}
-        P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\phi)
+        P=\sqrt{3} U_{\mathrm{L-L}} I_{\mathrm{L}} \cos(\varphi)
         = \sqrt{3} \SI{397.6}{\volt} \cdot \SI{8.75}{\ampere} \cdot 0.83= \SI{5}{\kilo\watt}
     \end{equation}
 \end{solutionblock}