Aufgaben:Exercise 3.3: GSM Frame Structure - Revision history

Guenter at 16:02, 16 January 2023

2023-01-16T16:02:28Z

Guenter at 12:14, 11 January 2023

2023-01-11T12:14:04Z

Guenter at 11:58, 11 January 2023

2023-01-11T11:58:15Z

Guenter at 11:53, 11 January 2023

2023-01-11T11:53:32Z

Guenter at 16:37, 3 January 2023

2023-01-03T16:37:11Z

Guenter at 14:03, 2 January 2023

2023-01-02T14:03:34Z

Guenter at 16:34, 31 December 2022

2022-12-31T16:34:50Z

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 Aufgabe 3.3: GSM–Rahmenstruktur to Exercise 3.3: GSM Frame Structure

Noah at 19:31, 30 December 2022

2022-12-30T19:31:31Z

Javier: Text replacement - "Category:Aufgaben zu Beispiele von Nachrichtensystemen" to "Category:Examples of Communication Systems: Exercises"

2021-03-23T12:52:29Z

Text replacement - "Category:Aufgaben zu Beispiele von Nachrichtensystemen" to "Category:Examples of Communication Systems: Exercises"

@@ Line 3: / Line 3: @@
 }}
-[[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&nbsp; $\rm GSM$&nbsp; the following frame structure is specified:
 *A superframe consists of&nbsp; $51$&nbsp; multiframes and has duration&nbsp; $T_{\rm SF}$.

@@ Line 44: / Line 44: @@
 $T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$
-{How long does a burst last?
+{How long does a burst&nbsp; $($one time slot$)$&nbsp; last?
 |type="{}"}
 $ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm &micro; s$
-{At what intervals&nbsp; $\Delta T_{\rm burst}$&nbsp; is a user assigned timeslots?
+{At what intervals&nbsp; $\Delta T_{\rm burst}$&nbsp; is a user assigned time slots&nbsp; $($bursts$)$&nbsp;?
 |type="{}"}
 $\Delta T_{\rm burst} \ = \ ${ 4.615 3% } $ \ \rm ms$
@@ Line 62: / Line 62: @@
 {What is the gross data rate of a user?
 |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?
 |type="{}"}
-$R_{\rm Netto} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$
+$R_{\rm net} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$
@@ Line 73: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; A superframe consists of 51 multiframes with respective durations $T_{\rm MF} = 120 \rm ms$. From this follows:
+'''(1)'''&nbsp; A superframe consists of&nbsp; $51$&nbsp; multiframes with respective durations $T_{\rm MF} = 120 \rm ms$.&nbsp; From this follows:
 :$$T_{\rm SF} = 51 \cdot T_{\rm MF} \hspace{0.15cm} \underline {= 6.12\,{\rm s}}\hspace{0.05cm}.$$
-'''(2)'''&nbsp; Each multiframe is divided into $26$ TDMA frames according to the specification. Therefore:
+'''(2)'''&nbsp; Each multiframe is divided into&nbsp; $26$&nbsp; TDMA frames&nbsp; $\rm TFs$&nbsp; according to the specification.&nbsp; 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)'''&nbsp; A TDMA frame consists of $8$ time slots. Therefore
+'''(3)'''&nbsp; A TDMA frame consists of&nbsp; $8$&nbsp; bursts.&nbsp; Therefore
-:$$T_{\rm Z} = \frac{ T_{\rm R}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm &micro; s}}\hspace{0.05cm}.$$
+:$$T_{\rm burst} = \frac{ T_{\rm TF}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm &micro; s}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; The spacing of time slots allocated for a user is.
+'''(4)'''&nbsp; 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)'''&nbsp; 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)'''&nbsp; Each burst consists - considering the guard period - of $156.25 \ \rm bits$,&nbsp; which must be transmitted within the time duration $T_{\rm burst} = 576.9 \ \rm \mu s$.&nbsp; This results in:
-:$$T_{\rm B} = \frac{ T_{\rm Z}}{156.25} = \frac{ 576.9\,{\rm &micro; s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm &micro; s}}\hspace{0.05cm}.$$
+:$$T_{\rm B} = \frac{ T_{\rm burst}}{156.25} = \frac{ 576.9\,{\rm &micro; s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm &micro; s}}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; For example, the bit rate can be calculated as the reciprocal of the bit duration:
+'''(6)'''&nbsp; For example,&nbsp; the bit rate can be calculated as the reciprocal of the bit duration:
 :$$R_{\rm B} = \frac{ 1}{T_{\rm B}} = \frac{ 1}{3.69216\,{\rm &micro; s}} \hspace{0.15cm} \underline {= 270.833\,{\rm kbit/s}}\hspace{0.05cm}.$$
-'''(7)'''&nbsp; 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)'''&nbsp; In each time slot,&nbsp; the data rate&nbsp; $R_{\rm B} \approx 271 \rm kbit/s$.&nbsp; However,&nbsp; since each user is assigned only one of the eight time slots,&nbsp; 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)'''&nbsp; For the net data rate, according to the specifications:
+'''(8)'''&nbsp; For the net data rate,&nbsp; 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}}

@@ Line 36: / Line 36: @@
 <quiz display=simple>
-{How long does a superframe last?
+{How long does a superframe&nbsp; $\rm (SF)$&nbsp; last?
 |type="{}"}
 $T_{\rm SF} \ = \ ${ 6.12 3% } $ \ \rm s$
-{What is the duration of a TDMA frame?
+{What is the duration of a TDMA frame&nbsp; $\rm (TF)$?
 |type="{}"}
 $T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$
-{How long does a time slot last?
+{How long does a burst last?
 |type="{}"}
 $ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm &micro; s$
@@ Line 50: / Line 50: @@
 {At what intervals&nbsp; $\Delta T_{\rm burst}$&nbsp; is a user assigned timeslots?
 |type="{}"}
-$\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$
+$\Delta T_{\rm burst} \ = \ ${ 4.615 3% } $ \ \rm ms$
 {What is the bit duration?
@@ Line 64: / Line 64: @@
 $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="{}"}
 $R_{\rm Netto} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$

@@ Line 4: / Line 4: @@
 [[File:EN_Bei_A_3_3.png|right|frame|GSM frame structure '''Korrektur''']]
-In the 2G cellular standard&nbsp; $\rm GSM$&nbsp; the following frame structure is specified:
+In the 2G cellular mobile communication standard&nbsp; $\rm GSM$&nbsp; the following frame structure is specified:
 *A superframe consists of&nbsp; $51$&nbsp; multiframes and has duration&nbsp; $T_{\rm SF}$.
 *Each multiframe has&nbsp; $26$&nbsp; TDMA frames and lasts a total of&nbsp; $T_{\rm MF} = 120 \rm ms$.
-It should also be noted that, in addition to the described multiframe structure with&nbsp; $26$&nbsp; TDMA frames, there are also multiframes with&nbsp; $51$&nbsp; TDMA frames, but these are used almost exclusively for the transmission of signaling information.
+*For example,&nbsp; in such a time slot,&nbsp; a&nbsp; "Normal Burst"&nbsp; with&nbsp; $156.25$&nbsp; bits is transmitted.
-Hints:
+<u>Hints:</u>
 *This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/Radio_Interface|"Radio Interface"]].
-*Reference is made in particular to the page&nbsp; [[Examples_of_Communication_Systems/Radio_Interface#GSM_frame_structure|"GSM frame structure"]].
@@ Line 40: / Line 42: @@
 {What is the duration of a TDMA frame?
 |type="{}"}
-$T_{\rm R} \ = \ ${ 4.615 3% } $ \ \rm ms$
+$T_{\rm TF} \ = \ ${ 4.615 3% } $ \ \rm ms$
 {How long does a time slot last?
 |type="{}"}
-$ T_{\rm Z} \ = \ ${ 576.9 3% } $ \ \rm &micro; s$
+$ T_{\rm burst} \ = \ ${ 576.9 3% } $ \ \rm &micro; s$
-{At what intervals&nbsp; $\Delta T_{\rm Z}$&nbsp; is a user assigned timeslots?
+{At what intervals&nbsp; $\Delta T_{\rm burst}$&nbsp; is a user assigned timeslots?
 |type="{}"}
 $\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$

@@ Line 3: / Line 3: @@
 }}
-[[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&nbsp; $\rm GSM$&nbsp; the following frame structure is specified:
 *A superframe consists of&nbsp; $51$&nbsp; multiframes and has duration&nbsp; $T_{\rm SF}$.

← Older revision		Revision as of 16:34, 31 December 2022
Line 106:		Line 106:


−	[[Category:Examples of Communication Systems: Exercises\|^3.2 ~~Funkschnittstelle~~^]]	+	[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]

← Older revision	Revision as of 16:25, 31 December 2022
(No difference)

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/Funkschnittstelle
+{{quiz-Header|Buchseite=Examples_of_Communication_Systems/Radio_Interface
 }}
-[[File:P_ID1224__Bei_A_3_3.png|right|frame|Zur GSM-Rahmenstruktur]]
+[[File:P_ID1224__Bei_A_3_3.png|right|frame|GSM frame structure]]
-Beim 2G&ndash;Mobilfunkstandard&nbsp; $\rm GSM$&nbsp; ist folgende Rahmenstruktur spezifiziert:
+In the 2G cellular standard&nbsp; $\rm GSM$&nbsp; the following frame structure is specified:
-*Ein Superframe besteht aus&nbsp; $51$&nbsp; Multiframes und hat die Zeitdauer&nbsp; $T_{\rm SF}$.
+*A superframe consists of&nbsp; $51$&nbsp; multiframes and has duration&nbsp; $T_{\rm SF}$.
-*Jeder Multiframe hat&nbsp; $26$&nbsp; TDMA–Rahmen und dauert insgesamt&nbsp; $T_{\rm MF} = 120 \ \rm ms$.
+*Each multiframe has&nbsp; $26$&nbsp; TDMA frames and lasts a total of&nbsp; $T_{\rm MF} = 120 \rm ms$.
-*Jeder TDMA–Rahmen hat die Dauer&nbsp; $T_{\rm R}$&nbsp; und ist eine Abfolge von acht Zeitschlitzen mit Dauer&nbsp; $T_{\rm Z}$.
+*Each TDMA frame has duration&nbsp; $T_{\rm R}$&nbsp; and is a sequence of eight time slots with duration&nbsp; $T_{\rm Z}$.
-*In einem solchen Zeitschlitz wird zum Beispiel ein&nbsp; ''Normal Burst''&nbsp; mit&nbsp; $156.25$&nbsp; Bit übertragen.
+*For example, in such a time slot, a&nbsp; ''Normal Burst''&nbsp; with&nbsp; $156.25$&nbsp; bits is transmitted.
-*Davon sind jedoch nur&nbsp; $114$&nbsp; Datenbits. Weitere Bits werden benötigt für die so genannte ''Guard Period'', die Signalisierung, Synchronisation und Kanalschätzung.
+*Of these, however, only&nbsp; $114$&nbsp; are data bits. Further bits are needed for the so called ''Guard Period'', signaling, synchronization and channel estimation.
-*Weiter ist bei der Berechnung der Netto–Datenrate zu berücksichtigen, dass die logischen Kanäle SACCH und IDLE insgesamt&nbsp; $1.9 \ \rm kbit/s$&nbsp; benötigen.
+*Further, when calculating the net data rate, it must be taken into account that the logical channels SACCH and IDLE require a total of&nbsp; $1.9 \rm kbit/s$&nbsp;.
-Anzumerken ist ferner, dass es neben der beschriebenen Multiframe–Struktur mit&nbsp; $26$&nbsp; TDMA–Rahmen auch Multiframes mit&nbsp; $51$&nbsp; TDMA–Rahmen gibt, die jedoch fast ausschließlich zur Übertragung von Signalisierungsinformation benutzt werden.
+It should also be noted that, in addition to the described multiframe structure with&nbsp; $26$&nbsp; TDMA frames, there are also multiframes with&nbsp; $51$&nbsp; TDMA frames, but these are used almost exclusively for the transmission of signaling information.
@@ Line 22: / Line 22: @@
-''Hinweise:''
+Hints:
-*Diese Aufgabe gehört zum Kapitel&nbsp;  [[Examples_of_Communication_Systems/Funkschnittstelle|Funkschnittstelle]].
+*This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/Radio_Interface|"Radio Interface"]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Examples_of_Communication_Systems/Funkschnittstelle#GSM.E2.80.93Rahmenstruktur|GSM&ndash;Rahmenstruktur]]
+*Reference is made in particular to the page&nbsp; [[Examples_of_Communication_Systems/Radio_Interface#GSM_frame_structure|"GSM frame structure"]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lange dauert ein Superframe?
+{How long does a superframe last?
 |type="{}"}
 $T_{\rm SF} \ = \ ${ 6.12 3% } $ \ \rm s$
-{Welche Dauer hat ein TDMA–Rahmen?
+{What is the duration of a TDMA frame?
 |type="{}"}
 $T_{\rm R} \ = \ ${ 4.615 3% } $ \ \rm ms$
-{Wie lange dauert ein Zeitschlitz?
+{How long does a time slot last?
 |type="{}"}
 $ T_{\rm Z} \ = \ ${ 576.9 3% } $ \ \rm &micro; s$
-{In welchen Zeitabständen&nbsp; $\Delta T_{\rm Z}$&nbsp; bekommt ein Benutzer Zeitschlitze zugewiesen?
+{At what intervals&nbsp; $\Delta T_{\rm Z}$&nbsp; is a user assigned timeslots?
 |type="{}"}
 $\Delta T_{\rm Z} \ = \ ${ 4.615 3% } $ \ \rm ms$
-{Wie groß ist die Bitdauer?
+{What is the bit duration?
 |type="{}"}
 $T_{\rm B} \ = \ ${ 3.692 3% } $ \ \rm &micro; s$
-{Wie groß ist die Gesamtbitrate des GSM?
+{What is the total bit rate of the GSM?
 |type="{}"}
 $R_{\rm B} \ = \ ${ 270.833 3% } $ \ \rm kbit/s $
-{Wie groß ist die Brutto–Datenrate eines Benutzers?
+{What is the gross data rate of a user?
 |type="{}"}
-$R_{\rm Brutto} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$
+$R_{\rm Gross} \ = \ ${ 33.854 3% } $ \ \rm kbit/s$
-{Wie groß ist die Netto–Datenrate eines Benutzers?
+{What is the net data rate of a user?
 |type="{}"}
 $R_{\rm Netto} \ = \ ${ 22.8 3% } $ \ \rm kbtit/s$
@@ Line 69: / Line 69: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Ein Superframe besteht aus 51 Multiframes mit jeweiliger Zeitdauer $T_{\rm MF} = 120 \ \rm ms$. Daraus folgt:
+'''(1)'''&nbsp; 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)'''&nbsp; Jeder Multiframe ist entsprechend der Angabe in $26$ TDMA–Rahmen unterteilt. Deshalb gilt:
+'''(2)'''&nbsp; 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)'''&nbsp; Ein TDMA–Rahmen besteht aus $8$ Zeitschlitzen. Deshalb ist
+'''(3)'''&nbsp; A TDMA frame consists of $8$ time slots. Therefore
 :$$T_{\rm Z} = \frac{ T_{\rm R}}{8} = \frac{ 4.615\,{\rm ms}}{8} \hspace{0.15cm} \underline {= 576.9\,{\rm &micro; s}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Der Abstand der für einen Benutzer zugewiesenen Zeitschlitze ist
+'''(4)'''&nbsp; The spacing of time slots allocated for a user is.
 :$$\Delta T_{\rm Z} = T_{\rm R} \underline{= 4.615 \ \rm ms}.$$
-'''(5)'''&nbsp;  Ein jeder Burst besteht – unter Berücksichtigung der Guard Period – aus $156.25 \ \rm Bit$, die innerhalb der Zeitdauer $T_{\rm Z} = 576.9 \ \rm \mu s$ übertragen werden müssen. Daraus ergibt sich:
+'''(5)'''&nbsp; 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:
 :$$T_{\rm B} = \frac{ T_{\rm Z}}{156.25} = \frac{ 576.9\,{\rm &micro; s}}{156.25} \hspace{0.15cm} \underline {= 3.69216\,{\rm &micro; s}}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; Die Bitrate kann beispielsweise als Kehrwert der Bitdauer berechnet werden:
+'''(6)'''&nbsp; For example, the bit rate can be calculated as the reciprocal of the bit duration:
 :$$R_{\rm B} = \frac{ 1}{T_{\rm B}} = \frac{ 1}{3.69216\,{\rm &micro; s}} \hspace{0.15cm} \underline {= 270.833\,{\rm kbit/s}}\hspace{0.05cm}.$$
-'''(7)'''&nbsp; In jedem Zeitschlitz beträgt die Datenrate $R_{\rm B} \approx 271 \ \rm kbit/s$. Da jedem Benutzer jedoch nur einer von acht Zeitschlitzen zugewiesen wird, beträgt die Brutto–Datenrate eines Benutzers
+'''(7)'''&nbsp; 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
-:$$R_{\rm Brutto} = \frac{ R_{\rm B}}{8} = \frac{ 270.833\,{\rm kbit/s}}{8} \hspace{0.15cm} \underline {= 33.854\,{\rm kbit/s}}\hspace{0.05cm}.$$
+:$$R_{\rm gross} = \frac{ R_{\rm B}}{8} = \frac{ 270.833\,{\rm kbit/s}}{8} \hspace{0.15cm} \underline {= 33.854\,{\rm kbit/s}}\hspace{0.05cm}.$$
-'''(8)'''&nbsp; Für die Netto–Datenrate gilt entsprechend den Angaben:
+'''(8)'''&nbsp; For the net data rate, according to the specifications:
 :$$R_{\rm Netto} = \frac{ 114}{156.25} \cdot R_{\rm Brutto} - 1.9\,{\rm kbit/s} \hspace{0.15cm} \underline {= 22.8\,{\rm kbit/s}}\hspace{0.05cm}.$$

← Older revision		Revision as of 12:52, 23 March 2021
Line 106:		Line 106:


−	[[Category:~~Aufgaben zu Beispiele von Nachrichtensystemen~~\|^3.2 Funkschnittstelle^]]	+	[[Category:Examples of Communication Systems: Exercises\|^3.2 Funkschnittstelle^]]