Aufgaben:Exercise 3.2: GSM Data Rates

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Block diagram of GSM

In this task, the data transmission with GSM is considered. However, since this system was mainly specified for voice transmission, we usually use the duration  $T_{\rm R} = 20 \ \rm ms$  of a voice frame as a temporal reference in the following calculations. The input data rate is  $R_{1} = 9.6 \ \rm kbit/s$. The number of input bits in each $T_{\rm R}$ frame is  $N_{1}$. All parameters labelled "???" in the graphic should be calculated in the task.

The first blocks are shown in the transmission chain shown:

• the outer coder (block code including four tail bits) with  $N_{2} = 244 \ \rm Bit$  per frame  $(T_{\rm R} = 20 \ \ \rm ms)$   ⇒   Rate  $R_{2}$  is to be determined,
• the convolutional coder with the code rate  $1/2$, and subsequent puncturing $($waiver of  $N_{\rm P} \ \rm bit)$    ⇒   Rate $R_{3} = 22.8 \ \rm kbit/s$,
• Interleaving and encryption, both rate-neutral At the output of this block the rate  $R_4$  occurs.

The further signal processing is basically as follows:

• Each  $114$  (coded, scrambled, encrypted) data bits are combined together with  $34$  control bits (for training sequence, tail bits, guard period) and a pause $($Duration:   $8.25 \ \ \rm Bit)$  to a so called  Normal \ Burst . The rate at the output is called  $R_{5}$ .
• Additionally, further bursts (Frequency Correction Burst, Synchronisation Burst, Dummy Burst, Access Bursts) are added for signalling. The rate after this block is  $R_{6}$.
• Finally the TDMA multiplexing equipment follows, so that the total gross data rate of the GSM is  $R_{\rm ges} = R_{7}$ .

The total gross digital data rate  $R_{\rm ges} = 270,833 \ \rm kbit/s$  (for eight users) is assumed to be known.

Notes:

• The task belongs to the chapter  Gemeinsamkeiten von GSM und UMTS.
• The graphic above summarizes the present description and defines the data rates used.
• All rates are given in "$ \rm kbit/s$".
• $N_{1},  N_{2},  N_{3}$  and  $N_{4}$  denote the respective number of bits at the corresponding points of the above block diagram within a time frame of duration  $T_{\rm R} = 20 \ \rm ms$.
• $N_{\rm ges} = 156.25$  is the number of bits after burst formation, related to the duration  $T_{\rm Z}$  of a TDMA time slot. Of which  $N_{\rm Info} = 114$  are information bits including channel coding.

Questionnaire

1

How many bits are provided by the source in each frame?

$N_{1} \ = \ $ $\ \ \rm Bit$

2

What is the data rate after the outer coder?

$R_{2} \ = \ $ $\ \ \rm kbit/s$

3

How many bits would the convolutional coder deliver alone (without dotting)?

$N_{3}\hspace{0.01cm}' \ = \ $ $\ \ \rm Bit$

4

How many bits does the dotted convolutional coder actually emit?

$N_{3} \ = \ $ $\ \ \rm Bit$

5

What is the data rate after Interleaver and encryption?

$R_{4} \ = \ $ $\ \ \rm kbit/s$

6

How long does a time slot last?

$T_{\rm Z} \ = \ $ $\ \ \rm µ s$

7

What is the gross data rate for each individual TDMA user?

$R_{6} \ = \ $ $\ \ \rm kbit/s$

8

What gross data rate would be without signaling bits?

$R_{5} \ = \ $ $\ \ \rm kbit/s$

sample solution

Solution

(1)  The following applies $N_{1} = R_{1} \cdot T_{\rm R} = 9.6 {\ \rm kbit/s} \cdot 20 {\ \rm ms} \hspace{0.15cm} \underline{= 192 \ \rm Bit}$.

(2)  Analogous to subtask (1) applies:

$$R_2= \frac{N_2}{T_{\rm R}} = \frac{244\,{\rm Bit}}{20\,{\rm ms}}\hspace{0.15cm} \underline { = 12.2\,{\rm kbit/s}}\hspace{0.05cm}.$$

Please note:   For a redundancy-free binary source (but only on this one), there is no difference between "$\rm Bit$" and "$\rm bit$".

(3)  Der Faltungscoder der Rate $1/2$ allein würde aus seinen $N_{2} = 244$ Eingangsbits genau $N_{3}\hspace{0.01cm}' \hspace{0.15cm}\underline{= 488}$ Ausgangsbits pro Rahmen generieren.

(4)  Aus der angegebenen Datenrate $R_{3} = 22.8 \ \rm kbit/s$ folgt dagegen $N_{3} \hspace{0.15cm}\underline{= 456}$.

• Das bedeutet, dass von den $N_{3}' = 488 \ \rm Bit$ durch die Punktierung $N_{\rm P} = 32 \ \rm Bit$ entfernt werden.

(5)  Sowohl das Interleaving als auch die Verschlüsselung erfolgt sozusagen „datenneutral”. Damit gilt:

$$R_{4} = R_{3} \hspace{0.15cm}\underline{= 22.8 \ {\rm kbit/s}} \Rightarrow N_{4} = N_{3} = 456.$$

(6)  Für die Bitdauer gilt $T_{\rm B} = 1/R_{7} = 1/(0.270833 {\ \rm Mbit/s}) \approx 3.69 \ \rm µ s$.

• In jedem Zeitschlitz der Dauer $T_{\rm Z}$ wird ein Burst – bestehend aus $156.25 \ \rm Bit$ – übertragen.
• Daraus ergibt sich $T_{\rm Z} \hspace{0.15cm}\underline{= 576.9 \ \rm µ s}$.

(7)  Bei GSM gibt es acht Zeitschlitze, wobei jedem Nutzer periodisch ein Zeitschlitz zugewiesen wird.

• Damit beträgt die Bruttodatenrate für jeden Nutzer $R_{6} = R_{7}/8 \hspace{0.15cm}\underline{ \approx 33.854 \ \rm kbit/s}$.

(8)  Berücksichtigt man, dass beim Normal Burst der Anteil der Nutzdaten (inkl. Kanalcodierung) $114/156.25$ beträgt, so wäre die Rate ohne Berücksichtigung der zugefügten Signalisierungsbits:

$$R_5 = \frac{n_{\rm ges} }{n_{\rm Info} } \cdot R_4 = \frac{156.25 }{114} \cdot 22.8\,{\rm kbit/s}\hspace{0.15cm} \underline { = 31.250\,{\rm kbit/s}}\hspace{0.05cm}.$$

• Zum gleichen Ergebnis kommt man, wenn man berücksichtigt, dass bei GSM jeder 13. Rahmen für Common Control (Signalisierungs–Info) reserviert ist:

$$R_5 = \frac{12 }{13 } \cdot 33.854\,{\rm kbit/s} ={ 31.250\,{\rm kbit/s}}\hspace{0.05cm}.$$

• Damit beträgt der prozentuale Anteil der Signalisierungsbits:

$$\alpha_{\rm SB} = \frac{33.854 - 31.250}{33.854 } { \approx 7.7\%}\hspace{0.05cm}.$$