Add change

exercise/tex/exercise05.tex

@@ -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 $d(t) > 1$ is necessary at the beginning of the period. This results in a open-loop control deviation, that is, a mismatch between the reference and the actual current cruves. An alternative to prevent this deviation would be to implement a feedback control system. This closed loop allows the deviation caused by $d(t) > 1$ to be compensated for downstream.
+    $i'(t)$ 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 nonphysical duty cycle signal of $d(t) > 1$ is necessary at the beginning of the period. This results in a open-loop control deviation, that is, a mismatch between the reference and the actual current cruves. An alternative to prevent this deviation would be to implement a feedback control system. This closed loop allows the deviation caused by $d(t) > 1$ to be compensated for downstream.
 \end{solutionblock}
 
 \input{fig/ex05/Fig_courses_i'_results}