diff --git a/ex05/kn05.pdf b/ex05/kn05.pdf
index e38add0..887842d 100644
--- a/ex05/kn05.pdf
+++ b/ex05/kn05.pdf
Binary files differ
diff --git a/ex05/kn05.tex b/ex05/kn05.tex
index 8f59ae8..79a4e56 100644
--- a/ex05/kn05.tex
+++ b/ex05/kn05.tex
@@ -113,7 +113,10 @@
     549 ~kHz & 549000 ~Hz &  546.45 ~m & \text{Low Frequency LF}
   \end{array}
   $
-\end{center} \vspace{1cm}
+  \begin{corr}
+      Anwendung fehlt
+  \end{corr}
+\end{center} \pagebreak
 \Aufgabe{Frequency-Division Multiplexing (FDM)}{6+6+6+6}\\
 The available analogue satellite channel with bandwidth $B_{STA} = 4MHz$ is divided into the required $5$
 channels, of which each has now $4MHz / 5 = 0.8MHz$ for transferring $N = 6Mbit/s$. Each digital channel is
@@ -121,14 +124,19 @@
 is demultiplexed and again demodulated with QAM. This leads to 5 output channel with the original 6Mbit/s.
 \begin{enumerate}
   \item The required bandwidth for each channel is $B_{QAM} = (1 + d) \cdot S$, with $d = 0$ and $S = 0.8MHz$ this leads to
-  $B_{QAM} = 0.8MHz$.
+  $B_{QAM} = 0.8MHz$.\begin{corr}
+      B=B, not S
+  \end{corr}
   \item The Nyquist theorem states, that the maximum bit rate $R_{max}$ is given as $R_{max} = 2 \cdot \log{2}{L}$.
   Since $\log{2}{L}$ is the number of bits per Baud, the maximum Baud per second $Bd_{max}$ is:
   \begin{align*}
     Bd_{max} = 2 \cdot B_{channel} = 2 \cdot 0.8 MHz = 1.6 MHz
     \end{align*}
-  \item The required number of bits per symbol is $r = \frac{N}{S} = \frac{\enot{6}{6}bit/s}{\enot{0.8}{6}1/s} = 7.5 Bits$
-  \item The required number of symbols for QAM is $L = 2^{\lceil r\rceil} = 2^{\lceil 7.5\rceil}=2^8 = 256$.
+    \begin{corr}
+        $S=0.8MBd$, since $B=(1+0)S$
+    \end{corr}
+  \item The required number of bits per symbol is $r = \frac{N}{S} = \frac{\enot{6}{6}bit/s}{\enot{0.8}{6}1/s} = 7.5 Bits$\ok
+  \item The required number of symbols for QAM is $L = 2^{\lceil r\rceil} = 2^{\lceil 7.5\rceil}=2^8 = 256$.\ok
 \end{enumerate} \vspace{1cm}
 \Aufgabe{Analog and Digital Hierarchy}{14+14}
 \begin{enumerate}
@@ -138,6 +146,14 @@
   \textbf{analog} signals are multiplexed into a single link using \textbf{Frequency Division Multiplexing}. In digital hierarchy, a bunch
   of \textbf{digital} signals are multiplexed into a single link using \textbf{Time Division Multiplexing}. It is also to be noted, that
   the digital signals can carry analog information (e.g. speech signals).
+  \begin{corr}
+  \begin{itemize}
+      \item FDM $\leftrightarrow$ TDM
+      \item digital vs. analog
+      \item Stufenaufbau etwas verschieden
+      \item Hierachie
+  \end{itemize}
+  \end{corr}
   \item All three stand for the lowest level of their hierarchy.
   \begin{enumerate}
     \item STS-1, OC-1\\
@@ -146,6 +162,9 @@
     \item DS-0\\
         has 64kBit/s (PCM-modulated human voice) and was used by phone companies. It is a single voice channel and was used to transport one phone call.\\
         DS is used for digital signals.
+    \item \begin{corr}
+        Bitraten(nicht gefragt), Stufen fehlen
+    \end{corr}
   \end{enumerate}
 \end{enumerate}
 \end{document}