Exercise 3.2: GSM Data Rates

• Return to book

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

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