minor typo corrections to ex04

exercise/main.tex

@@ -1,7 +1,7 @@
 \documentclass[solution]{../course_template/exerciseClass}
 \title{Power Electronics}
 
-\includeonly{tex/exercise07}
+\includeonly{tex/exercise04}
 
 \begin{document}
     \include{tex/exercise01}

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}