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 bit in each  $T_{\rm R}$  frame is  $N_{1}$.  All parameters labelled "???" in the graphic should be calculated in the task.

The first blocks shown in the transmission chain are:

• 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:   $8.25 \ \ \rm bits)$  to a so called  Normal Burst.  The rate at the output is $R_{5}$.
• 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 GSM is $R_{\rm tot} = R_{7}$ .

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

Notes:

• The task belongs to the chapter  Similarities between GSM and 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  $T_{\rm R} = 20 \ \rm ms$.
• $N_{\rm tot} = 156.25$  is the number of bits after burst formation, related to the duration  $T_{\rm Z}$  of a TDMA time slot.  $N_{\rm Info} = 114$  of which 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:

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

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

(4)  In contrast, $N_{3} \hspace{0.15cm}\underline{= 456}$  follows by the specifed data rate  $R_{3} = 22.8 \ \rm kbit/s$.

• This means that from  $N_{3}' = 488 \ \rm bit$,  $N_{\rm P} = 32 \ \rm bit$ can be removed by puncturing.

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

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

(6)  The bit duration is  $T_{\rm B} = 1/R_{7} = 1/(0.270833 {\ \rm Mbit/s}) \approx 3.69 \ \rm µ s$.

• In every time slot  $T_{\rm Z}$  a burst of $156.25 \ \rm bit$  will be transmitted.
• This makes  $T_{\rm Z} \hspace{0.15cm}\underline{= 576.9 \ \rm µ s}$.

(7)  GSM has eight time slots, whereby each user is periodically assigned a time slot.

• 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 in the "normal burst" the portion of user data (including channel coding) is  $114/156.25$.

• The rate would be without consideration of the added signaling bits:

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

• The same result can be obtained if you consider that in GSM every thirteenth frame is reserved for "Common Control"  (signaling info):

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

• Thus the percentage of signaling bits is

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