Ex07 Task2: Additional update of task description to simplify solution

exercise/fig/ex07/Fig_ThreePhaseInverter_6StepMode.tex

@@ -86,6 +86,9 @@
                 (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 arrow u1
+                (U1p) ++(0.3,0.5) to [open,v^=$$,voltage = straight] ++(0,-6)coordinate (Uges)
+                (Uges) ++ (0.3,3) node[anchor=north,color=black]{$U_\mathrm{1}$}
                 % 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)

exercise/tex/exercise07.tex

@@ -28,7 +28,7 @@
     \centering  % Center the table
     \begin{tabular}{ll}
         \toprule
-        Input voltages: & $U_\mathrm{1p}=\SI{255}{\volt}$ \quad $U_\mathrm{1m}=\SI{255}{\volt}$ \\
+        Input voltages: & $U_\mathrm{1}=\SI{510}{\volt}$ \quad $U_\mathrm{1p}=U_\mathrm{1m}=U_\mathrm{1}/2$ \\
         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: & $L= \SI{10}{\milli \henry}$ \\
@@ -190,6 +190,25 @@
 \label{sub:DecomposeVoltage}
 }
 \begin{solutionblock}
+    In the case of odd and alternating functions corresponding to $f(x)=-f(x+\pi)$ the Fourier coefficients are:
+    \begin{equation}
+        \begin{split}
+        a_\mathrm{k} &= 0 \\
+        a_\mathrm{k} &= \frac{4}{\pi} \int_0^{\pi/2} f(x)\sin(x) \mathrm{d}x \quad k=\mathrm{odd} \\
+        f(x) &= \sum_{k}^{} \left( b_k \sin(kx) \right).
+        \end{split}
+    \end{equation}
+    The coefficients $b_k$ are the amplitudes of the respective harmonic. The voltage $u_{\mathrm{2a}}(t)$ needs 
+    only to be integrated up to $\pi/2$. Only the terms with odd order numbers are taken into account.
+    \begin{equation}
+        \begin{split}
+        a_\mathrm{k} &= \frac{4}{\pi} \int_0^{\pi/3} f(x)\sin(x) \mathrm{d}x \quad k=\mathrm{odd} \\
+        f(x) &= \sum_{k}^{} \left( b_k \sin(kx) \right).
+        \end{split}
+    \end{equation}
+
+
+
     \input{fig/ex07/Fig_Voltage_U_um_excerpt}   
     \input{fig/ex07/Fig_graphic_solutions_cos_terms}
     \input{fig/ex07/Fig_standardization_to_fudamental_freq.tex}