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

$N_{1} \ = \ $

$\ \ \rm Bit$

$R_{2} \ = \ $

$\ \ \rm kbit/s$

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

$\ \ \rm Bit$

$N_{3} \ = \ $

$\ \ \rm Bit$

$R_{4} \ = \ $

$\ \ \rm kbit/s$

$T_{\rm Z} \ = \ $

$\ \ \rm µ s$

$R_{6} \ = \ $

$\ \ \rm kbit/s$

$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}.$$

Category:Exercises for Mobile Communications|^3.2 Similarities between GSM and UMTS ^]]