Exercise 3.3: GSM Frame Structure

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GSM frame structure

In the 2G cellular mobile communication standard  $\rm GSM$  the following frame structure is specified:

• A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.

• Each multiframe has  $26$  TDMA frames and lasts a total of  $T_{\rm MF} = 120 \rm ms$.

• Each TDMA frame has duration  $T_{\rm TF}$  and is a sequence of eight time slots with duration  $T_{\rm burst}$.

• For example,  in such a time slot,  a  "Normal Burst"  with  $156.25$  bits is transmitted.

• Of these,  however,  only  $114$  are data bits.  Further bits are needed for the so called  "Guard Period"  $\rm (GP)$,  signaling,  synchronization and channel estimation.

• Further,  when calculating the net data rate,  it must be taken into account that the logical channels SACCH and IDLE require a total of  $1.9 \rm kbit/s$.

It should also be noted that,  in addition to the described multiframe structure with  $26$  TDMA frames,  there are also multiframes with  $51$  TDMA frames,  but these are used almost exclusively for the transmission of signaling information.

Hints:

• This exercise belongs to the chapter  "Radio Interface".

• Reference is made in particular to the sectione  "GSM frame structure".

Questions

1

How long does a superframe  $\rm (SF)$  last?

$T_{\rm SF} \ = \ $ $ \ \rm s$

2

What is the duration of a TDMA frame  $\rm (TF)$?

$T_{\rm TF} \ = \ $ $ \ \rm ms$

3

How long does a burst  $($one time slot$)$  last?

$ T_{\rm burst} \ = \ $ $ \ \rm µ s$

4

At what intervals  $\Delta T_{\rm burst}$  is a user assigned time slots  $($bursts$)$ ?

$\Delta T_{\rm burst} \ = \ $ $ \ \rm ms$

5

What is the bit duration?

$T_{\rm B} \ = \ $ $ \ \rm µ s$

6

What is the total bit rate of the GSM?

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

7

What is the gross data rate of a user?

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

8

What is the net data rate of one user?

$R_{\rm net} \ = \ $ $ \ \rm kbtit/s$

Solution

Solution

(1)  A superframe consists of  $51$  multiframes with respective durations $T_{\rm MF} = 120 \rm ms$.  From this follows:

$$T_{\rm SF} = 51 \cdot T_{\rm MF} \hspace{0.15cm} \underline {= 6.12\,{\rm s}}\hspace{0.05cm}.$$

(2)  Each multiframe is divided into  $26$  TDMA frames  $\rm TFs$  according to the specification.  Therefore:

$$T_{\rm TF} = \frac{ T_{\rm MF}}{26} = \frac{ 120\,{\rm ms}}{26} \hspace{0.15cm} \underline {= 4.615\,{\rm ms}}\hspace{0.05cm}.$$

(3)  A TDMA frame consists of  $8$  bursts.  Therefore

$$T_{\rm burst} = \frac{ T_{\rm TF}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm µ s}}\hspace{0.05cm}.$$

(4)  The spacing of time slots allocated for a user is

$$\Delta T_{\rm burst} = T_{\rm TF} \underline{= 4.615 \ \rm ms}.$$

(5)  Each burst consists - considering the guard period - of $156.25 \ \rm bits$,  which must be transmitted within the time duration $T_{\rm burst} = 576.9 \ \rm \mu s$.  This results in:

$$T_{\rm B} = \frac{ T_{\rm burst}}{156.25} = \frac{ 576.9\,{\rm µ s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm µ s}}\hspace{0.05cm}.$$

(6)  For example,  the bit rate can be calculated as the reciprocal of the bit duration:

$$R_{\rm B} = \frac{ 1}{T_{\rm B}} = \frac{ 1}{3.69216\,{\rm µ s}} \hspace{0.15cm} \underline {= 270.833\,{\rm kbit/s}}\hspace{0.05cm}.$$

(7)  In each time slot,  the data rate  $R_{\rm B} \approx 271 \rm kbit/s$.  However,  since each user is assigned only one of the eight time slots,  the gross data rate of a user is

$$R_{\rm gross} = \frac{ R_{\rm B}}{8} = \frac{ 270.833\,{\rm kbit/s}}{8} \hspace{0.15cm} \underline {= 33.854\,{\rm kbit/s}}\hspace{0.05cm}.$$

(8)  For the net data rate,  according to the specifications:

$$R_{\rm net} = \frac{ 114}{156.25} \cdot R_{\rm B} - 1.9\,{\rm kbit/s} \hspace{0.15cm} \underline {= 22.8\,{\rm kbit/s}}\hspace{0.05cm}.$$