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 you will see 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 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 interleaving 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 result without signaling bits?

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

sample solution

{{{ML Kopf}

(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 the 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 this one), there is no difference between "$\rm bit$" and "$\rm bit$".

(3)  The convolutional encoder of rate $1/2$ alone would generate exactly $N_{3}\hspace{0.01cm}' from its $N_{2} = 244$ input bits. \hspace{0.15cm}\underline{= 488}$ generate output bits per frame.

(4)  Contrary, $N_{3} follows from the specified data rate $R_{3} = 22.8 \ \rm kbit/s$ \hspace\underline $456.

• This means that from $N_{3}' = 488 \ \ \rm Bit$ are removed by the dotting $N_{\rm P} = 32 \ \ \rm Bit$.

(5)  Both the interleaving and the encryption are "data neutral" so to speak. Thus the following applies:

$$R_{4} = R_{3} \hspace{0.15cm}\underline{= 22.8 \ {\rm kbit/s}} '''(6)'''  The bit duration is $T_{\rm B} = 1/R_{7} = 1/(0.270833 {\ \ \rm Mbit/s}) \approx 3.69 \ \rm µ s$. *A burst - consisting of $156.25 \ \ \rm Bit$ - is transmitted in each time slot of duration $T_{\rm Z}$. *This results in $T_{\rm Z} \hspace{0.15cm}\underline{= 576.9 \ \rm µ s}$. '''(7)'''  GSM has eight time slots, with one time slot being periodically assigned to each user. *The gross data rate for each user is $R_{6} = R_{7}/8 \hspace{0.15cm}\underline{ \approx 33.854 \ \rm kbit/s}$. '''(8)'''  Considering that for the ''Normal Burst'' the amount of user data (including channel coding) is $114/156.25$, the rate would be without consideration of the added signaling bits: :$$R_5 = \frac{n_{\rm ges} }{\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}.$$ *The same result can be obtained if you consider that in GSM every 13th frame is reserved for ''Common Control'' (signalling info): :$$R_5 = \frac{12}{13} \cdot 33,854\,{\rm kbit/s} ={ 31,250\,{\rm kbit/s}}\hspace{0.05cm}.$$ *Consequently , the percentage of signaling bits is: $$\alpha_{\rm SB} = \frac{33.854 - 31.250}{33.854} { \approx 7.7\%}\hspace{0.05cm}.$$

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