Aufgaben:Exercise 3.3: GSM Frame Structure - Revision history

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	[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]		[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]
	[[de:Aufgaben:3.~~3_GSM–Rahmenstruktur~~]]		[[de:Aufgaben:Aufgabe 3.3: GSM–Rahmenstruktur]]

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	[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]		[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]
			[[de:Aufgaben:3.3_GSM–Rahmenstruktur]]

Guenter at 16:02, 16 January 2023

2023-01-16T16:02:28Z

← Older revision		Revision as of 18:02, 16 January 2023
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	[[File:~~EN_Bei_A_3_3~~.png\|right\|frame\|GSM frame structure ~~'''Korrektur'''~~]]		[[File:EN_Bei_A_3_3_v2.png\|right\|frame\|GSM frame structure]]
	In the 2G cellular mobile communication standard  $\rm GSM$  the following frame structure is specified:		In the 2G cellular mobile communication standard  $\rm GSM$  the following frame structure is specified:
	*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.		*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.

Guenter at 12:14, 11 January 2023

2023-01-11T12:14:04Z

← Older revision		Revision as of 14:14, 11 January 2023
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	$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$		$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$

	{How long does a burst last?		{How long does a burst  $($one time slot$)$  last?
	\|type="{}"}		\|type="{}"}
	$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm µ s$		$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm µ s$

	{At what intervals  $\Delta T_{\rm burst}$  is a user assigned ~~timeslots~~?		{At what intervals  $\Delta T_{\rm burst}$  is a user assigned time slots  $($bursts$)$ ?
	\|type="{}"}		\|type="{}"}
	$\Delta T_{\rm burst} \ = \ ${ 4.615 3% } $ \ \rm ms$		$\Delta T_{\rm burst} \ = \ ${ 4.615 3% } $ \ \rm ms$
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	{What is the gross data rate of a user?		{What is the gross data rate of a user?
	\|type="{}"}		\|type="{}"}
	$R_{\rm ~~Gross~~} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$		$R_{\rm gross} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$

	{What is the net data rate of one user?		{What is the net data rate of one user?
	\|type="{}"}		\|type="{}"}
	$R_{\rm ~~Netto~~} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$		$R_{\rm net} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$


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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  A superframe consists of 51 multiframes with respective durations $T_{\rm MF} = 120 \rm ms$. From this follows:		'''(1)'''  A superframe consists of  $51$  multiframes with respective durations $T_{\rm MF} = 120 \rm ms$.  From this follows:
	:$$T_{\rm SF} = 51 \cdot T_{\rm MF} \hspace{0.15cm} \underline {= 6.12\,{\rm s}}\hspace{0.05cm}.$$		:$$T_{\rm SF} = 51 \cdot T_{\rm MF} \hspace{0.15cm} \underline {= 6.12\,{\rm s}}\hspace{0.05cm}.$$


	'''(2)'''  Each multiframe is divided into $26$ TDMA frames according to the specification. Therefore:		'''(2)'''  Each multiframe is divided into  $26$  TDMA frames  $\rm TFs$  according to the specification.  Therefore:
	:$$T_{\rm R} = \frac{ T_{\rm MF}}{26} = \frac{ 120\,{\rm ms}}{26} \hspace{0.15cm} \underline {= 4.615\,{\rm ms}}\hspace{0.05cm}.$$		:$$T_{\rm TF} = \frac{ T_{\rm MF}}{26} = \frac{ 120\,{\rm ms}}{26} \hspace{0.15cm} \underline {= 4.615\,{\rm ms}}\hspace{0.05cm}.$$


	'''(3)'''  A TDMA frame consists of $8$ ~~time slots~~. Therefore		'''(3)'''  A TDMA frame consists of  $8$  bursts.  Therefore
	:$$T_{\rm Z} = \frac{ T_{\rm R}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm µ s}}\hspace{0.05cm}.$$		:$$T_{\rm burst} = \frac{ T_{\rm TF}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm µ s}}\hspace{0.05cm}.$$


	'''(4)'''  The spacing of time slots allocated for a user is.		'''(4)'''  The spacing of time slots allocated for a user is
	:$$\Delta T_{\rm Z} = T_{\rm R} \underline{= 4.615 \ \rm ms}.$$		:$$\Delta T_{\rm burst} = T_{\rm TF} \underline{= 4.615 \ \rm ms}.$$


	'''(5)'''  Each burst consists - considering the guard period - of $156.25 \ \rm bits$, which must be transmitted within the time duration $T_{\rm Z} = 576.9 \ \rm \mu s$. This results in:		'''(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:
	:$$T_{\rm B} = \frac{ T_{\rm Z}}{156.25} = \frac{ 576.9\,{\rm µ s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm µ s}}\hspace{0.05cm}.$$		:$$T_{\rm B} = \frac{ T_{\rm burst}}{156.25} = \frac{ 576.9\,{\rm µ s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm µ s}}\hspace{0.05cm}.$$


	'''(6)'''  For example, the bit rate can be calculated as the reciprocal of the bit duration:		'''(6)'''  For example,  the bit rate can be calculated as the reciprocal of the bit duration:
	:$$R_{\rm B} = \frac{ 1}{T_{\rm B}} = \frac{ 1}{3.69216\,{\rm µ s}} \hspace{0.15cm} \underline {= 270.833\,{\rm kbit/s}}\hspace{0.05cm}.$$		:$$R_{\rm B} = \frac{ 1}{T_{\rm B}} = \frac{ 1}{3.69216\,{\rm µ s}} \hspace{0.15cm} \underline {= 270.833\,{\rm kbit/s}}\hspace{0.05cm}.$$


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


	'''(8)'''  For the net data rate, according to the specifications:		'''(8)'''  For the net data rate,  according to the specifications:
	:$$R_{\rm ~~Netto~~} = \frac{ 114}{156.25} \cdot R_{\rm ~~Brutto~~} - 1.9\,{\rm kbit/s} \hspace{0.15cm} \underline {= 22.8\,{\rm kbit/s}}\hspace{0.05cm}.$$		:$$R_{\rm net} = \frac{ 114}{156.25} \cdot R_{\rm B} - 1.9\,{\rm kbit/s} \hspace{0.15cm} \underline {= 22.8\,{\rm kbit/s}}\hspace{0.05cm}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 11:58, 11 January 2023

2023-01-11T11:58:15Z

← Older revision		Revision as of 13:58, 11 January 2023
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	<quiz display=simple>		<quiz display=simple>

	{How long does a superframe last?		{How long does a superframe  $\rm (SF)$  last?
	\|type="{}"}		\|type="{}"}
	$T_{\rm SF} \ = \ ${ 6.12 3% } $ \ \rm s$		$T_{\rm SF} \ = \ ${ 6.12 3% } $ \ \rm s$

	{What is the duration of a TDMA frame?		{What is the duration of a TDMA frame  $\rm (TF)$?
	\|type="{}"}		\|type="{}"}
	$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$		$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$

	{How long does a ~~time slot~~ last?		{How long does a burst last?
	\|type="{}"}		\|type="{}"}
	$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm µ s$		$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm µ s$
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	{At what intervals  $\Delta T_{\rm burst}$  is a user assigned timeslots?		{At what intervals  $\Delta T_{\rm burst}$  is a user assigned timeslots?
	\|type="{}"}		\|type="{}"}
	$\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$		$\Delta T_{\rm burst} \ = \ ${ 4.615 3% } $ \ \rm ms$

	{What is the bit duration?		{What is the bit duration?
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	$R_{\rm Gross} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$		$R_{\rm Gross} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$

	{What is the net data rate of a user?		{What is the net data rate of one user?
	\|type="{}"}		\|type="{}"}
	$R_{\rm Netto} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$		$R_{\rm Netto} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$

Guenter at 11:53, 11 January 2023

2023-01-11T11:53:32Z

← Older revision		Revision as of 13:53, 11 January 2023
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	[[File:EN_Bei_A_3_3.png\|right\|frame\|GSM frame structure '''Korrektur''']]		[[File:EN_Bei_A_3_3.png\|right\|frame\|GSM frame structure '''Korrektur''']]
	In the 2G cellular standard  $\rm GSM$  the following frame structure is specified:		In the 2G cellular mobile communication standard  $\rm GSM$  the following frame structure is specified:
	*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.		*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 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  $1.9 \rm kbit/s$ .

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

	~~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~~.		*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:		<u>Hints:</u>

	*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/Radio_Interface\|"Radio Interface"]].		*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/Radio_Interface\|"Radio Interface"]].
	*Reference is made in particular to the ~~page~~  [[Examples_of_Communication_Systems/Radio_Interface#GSM_frame_structure\|"GSM frame structure"]].
			*Reference is made in particular to the sectione  [[Examples_of_Communication_Systems/Radio_Interface#GSM_frame_structure\|"GSM frame structure"]].


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	{What is the duration of a TDMA frame?		{What is the duration of a TDMA frame?
	\|type="{}"}		\|type="{}"}
	$T_{\rm R} \ = \ ${ 4.615 3% } $ \ \rm ms$		$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$

	{How long does a time slot last?		{How long does a time slot last?
	\|type="{}"}		\|type="{}"}
	$ T_{\rm Z} \ = \ ${ 576.9 3% } $ \ \rm µ s$		$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm µ s$

	{At what intervals  $\Delta T_{\rm Z}$  is a user assigned timeslots?		{At what intervals  $\Delta T_{\rm burst}$  is a user assigned timeslots?
	\|type="{}"}		\|type="{}"}
	$\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$		$\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$

Guenter at 16:37, 3 January 2023

2023-01-03T16:37:11Z

← Older revision		Revision as of 18:37, 3 January 2023
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	}}		}}

	[[File:~~P_ID1224__Bei_A_3_3~~.png\|right\|frame\|GSM frame structure '''Korrektur''']]		[[File:EN_Bei_A_3_3.png\|right\|frame\|GSM frame structure '''Korrektur''']]
	In the 2G cellular standard  $\rm GSM$  the following frame structure is specified:		In the 2G cellular standard  $\rm GSM$  the following frame structure is specified:
	*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.		*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.

Guenter at 14:03, 2 January 2023

2023-01-02T14:03:34Z

← Older revision		Revision as of 16:03, 2 January 2023
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	}}		}}

	[[File:P_ID1224__Bei_A_3_3.png\|right\|frame\|GSM frame structure]]		[[File:P_ID1224__Bei_A_3_3.png\|right\|frame\|GSM frame structure '''Korrektur''']]
	In the 2G cellular standard  $\rm GSM$  the following frame structure is specified:		In the 2G cellular standard  $\rm GSM$  the following frame structure is specified:
	*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.		*A superframe consists of  $51$  multiframes and has duration  $T_{\rm SF}$.

Guenter at 16:34, 31 December 2022

2022-12-31T16:34:50Z

← Older revision		Revision as of 18:34, 31 December 2022
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	[[Category:Examples of Communication Systems: Exercises\|^3.2 ~~Funkschnittstelle~~^]]		[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 3.3: GSM–Rahmenstruktur to Aufgaben:Exercise 3.3: GSM Frame Structure

2022-12-31T16:25:59Z

Guenter moved page Aufgaben:Aufgabe 3.3: GSM–Rahmenstruktur to Aufgaben:Exercise 3.3: GSM Frame Structure

← Older revision	Revision as of 18:25, 31 December 2022
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