Aufgaben:Exercise 3.3: GSM Frame Structure

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GSM frame structure Korrektur

In the 2G cellular 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  $T_{\rm R}$  and is a sequence of eight time slots with duration  $T_{\rm Z}$.
• 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, 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 page  "GSM frame structure".

Questions

1

How long does a superframe last?

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

2

What is the duration of a TDMA frame?

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

3

How long does a time slot last?

$ T_{\rm Z} \ = \ $ $ \ \rm µ s$

4

At what intervals  $\Delta T_{\rm Z}$  is a user assigned timeslots?

$\Delta T_{\rm Z} \ = \ $ $ \ \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 a user?

$R_{\rm Netto} \ = \ $ $ \ \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 according to the specification. Therefore:

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

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

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

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

$$\Delta T_{\rm Z} = T_{\rm R} \underline{= 4.615 \ \rm ms}.$$

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

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

(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 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:

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