diff --git a/ex05/kn05.pdf b/ex05/kn05.pdf
index 3e34477..e38add0 100644
--- a/ex05/kn05.pdf
+++ b/ex05/kn05.pdf
Binary files differ
diff --git a/ex05/kn05.tex b/ex05/kn05.tex
index 0a1d227..8f59ae8 100644
--- a/ex05/kn05.tex
+++ b/ex05/kn05.tex
@@ -114,21 +114,21 @@
   \end{array}
   $
 \end{center} \vspace{1cm}
-\Aufgabe{Frequency-Division Multiplexing (FDM)}{6+6+6+6}
+\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 $5MHz / 5 = 0.8MHz$ for transferring $N = 6Mbit/s$. Each digital channel is
+channels, of which each has now $4MHz / 5 = 0.8MHz$ for transferring $N = 6Mbit/s$. Each digital channel is
 modulated by QAM and sent over the analogue satellite link using the reserved bandwidth. At the receiver, the signal
 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
   $B_{QAM} = 0.8MHz$.
-  \item The nyquist theorem states, that the maximum bit rate $R_{max}$ is given as $R_{max} = 2 \cdot \log{2}{L}$.
+  \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 $L = 2^r = 2^8 = 256$.
+  \item The required number of symbols for QAM is $L = 2^{\lceil r\rceil} = 2^{\lceil 7.5\rceil}=2^8 = 256$.
 \end{enumerate} \vspace{1cm}
 \Aufgabe{Analog and Digital Hierarchy}{14+14}
 \begin{enumerate}
@@ -138,11 +138,14 @@
   \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).
-  \item
+  \item All three stand for the lowest level of their hierarchy.
   \begin{enumerate}
-    \item STS-1
-    \item OC-1
-    \item DS-0
+    \item STS-1, OC-1\\
+        STS-1 is the frame format for OC-1 (optical carrier) in SONET\\
+        SONET is used for optical signals.
+    \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.
   \end{enumerate}
 \end{enumerate}
 \end{document}