Aufgaben:Exercise 3.2: GSM Data Rates - Revision history

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

2021-03-23T12:38:14Z

Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

Guenter at 14:23, 20 January 2021

2021-01-20T14:23:45Z

Guenter at 14:05, 20 January 2021

2021-01-20T14:05:55Z

Guenter at 17:32, 18 January 2021

2021-01-18T17:32:39Z

Javier: Text replacement - "Mobile Kommunikation/Gemeinsamkeiten von GSM und UMTS" to "Mobile_Communications/Similarities_Between_GSM_and_UMTS"

2020-12-15T14:02:29Z

Text replacement - "Mobile Kommunikation/Gemeinsamkeiten von GSM und UMTS" to "Mobile_Communications/Similarities_Between_GSM_and_UMTS"

Javier: Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

2020-07-09T09:12:41Z

Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

Javier: Javier moved page Exercises:Exercise 3.2: GSM Data Rates to Aufgaben:Exercise 3.2: GSM Data Rates

2020-06-30T07:39:30Z

Javier moved page Exercises:Exercise 3.2: GSM Data Rates to Exercise 3.2: GSM Data Rates

Oezdemir at 19:17, 25 June 2020

2020-06-25T19:17:13Z

Oezdemir at 19:16, 25 June 2020

2020-06-25T19:16:02Z

Oezdemir at 19:14, 25 June 2020

2020-06-25T19:14:43Z

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^3.2 Similarities between GSM and UMTS	+	[[Category:Mobile Communications: Exercises\|^3.2 Similarities between GSM and UMTS
	^]]		^]]

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 {{ML-Kopf}}
-'''(1)'''&nbsp; The following applies $N_{1} = R_{1} \cdot T_{\rm R} = 9.6  {\ \rm kbit/s} \cdot 20  {\ \rm ms} \hspace{0.15cm} \underline{= 192 \ \rm Bit}$.
+'''(1)'''&nbsp; The following applies:&nbsp;
-'''(2)'''&nbsp;  Analogous to subtask '''(1)''' applies:
+'''(2)'''&nbsp;  Analogous to subtask&nbsp; '''(1)'''&nbsp; applies:
-:$$R_2= \frac{N_2}{T_{\rm R}} = \frac{244\,{\rm Bit}}{20\,{\rm ms}}\hspace{0.15cm} \underline { = 12.2\,{\rm kbit/s}}\hspace{0.05cm}.$$
+:$$R_2= \frac{N_2}{T_{\rm R}} = \frac{244\,{\rm bit}}{20\,{\rm ms}}\hspace{0.15cm} \underline { = 12.2\,{\rm kbit/s}}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; 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 input bits $N_{2} = 244$ .
+'''(3)'''&nbsp; The convolutional encoder of rate&nbsp; $1/2$&nbsp; alone would generate exactly&nbsp; $N_{3}\hspace{0.01cm}' \hspace{0.15cm}\underline{= 488}$&nbsp; output bits from the&nbsp; $N_{2} = 244$&nbsp; input bits.
-'''(4)'''&nbsp;  In contrast, $N_{3} \hspace{0.15cm}\underline{= 456}$ is followed by the specifed data rate $R_{3} = 22.8 \ \rm kbit/s$
+'''(4)'''&nbsp;  In contrast, $N_{3} \hspace{0.15cm}\underline{= 456}$&nbsp; follows by the specifed data rate&nbsp; $R_{3} = 22.8 \ \rm kbit/s$.
-*This means that from $N_{3}' = 488 \ \rm Bit$ ,$N_{\rm P} = 32 \ \rm Bit$ can be removed by puncturing.
+*This means that from&nbsp; $N_{3}' = 488 \ \rm bit$,&nbsp; $N_{\rm P} = 32 \ \rm bit$ can be removed by puncturing.
+'''(5)'''&nbsp; Both the interleaving and the encryption are "data neutral".&nbsp; Thus the following applies:
 :$$R_{4} = R_{3} \hspace{0.15cm}\underline{= 22.8 \ {\rm kbit/s}}  \Rightarrow  N_{4} = N_{3} = 456.$$
+'''(6)'''&nbsp; The bit duration is&nbsp; $T_{\rm B} = 1/R_{7} = 1/(0.270833 {\ \rm Mbit/s}) \approx 3.69 \ \rm &micro; s$.
-'''(6)'''&nbsp; The bit duration is $T_{\rm B} = 1/R_{7} = 1/(0.270833 {\ \rm Mbit/s}) \approx 3.69 \ \rm &micro; s$.
+*In every time slot&nbsp; $T_{\rm Z}$&nbsp; a  burst of $156.25 \ \rm bit$&nbsp; will be transmitted.
-*In every time slot $T_{\rm Z}$ a  Burst of $156.25 \ \rm Bit$ – will be transmitted.
+*This makes&nbsp; $T_{\rm Z} \hspace{0.15cm}\underline{= 576.9 \ \rm &micro; s}$.
 '''(7)'''&nbsp; GSM has eight time slots, whereby each user is periodically assigned a time slot.
-*The gross data rate for each user is $R_{6} = R_{7}/8 \hspace{0.15cm}\underline{ \approx 33.854 \ \rm kbit/s}$.
+*The gross data rate for each user is&nbsp; $R_{6} = R_{7}/8 \hspace{0.15cm}\underline{ \approx 33.854 \ \rm kbit/s}$.
-'''(8)'''&nbsp; 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:
+'''(8)'''&nbsp; Considering that in the "normal burst" the portion of user data (including channel coding) is&nbsp; $114/156.25$.
-:$$R_5 = \frac{n_{\rm tot} }{n_{\rm Info} } \cdot R_4 = \frac{156.25}{114} \cdot 22.8\,{\rm kbit/s}\hspace{0.15cm} \underline { = 31,250\,{\rm kbit/s}}\hspace{0.05cm}.$$
+* The rate would be without consideration of the added signaling bits:
-*The same result can be obtained if you consider that in GSM every 13th frame is reserved for ''Common Control'' (signaling info):
+:$$R_5 = \frac{n_{\rm tot} }{n_{\rm Info} } \cdot R_4 = \frac{156.25}{114} \cdot 22.8\,{\rm kbit/s}\hspace{0.15cm} \underline { = 31.250\,{\rm kbit/s}}\hspace{0.05cm}.$$
-:$$R_5 = \frac{12}{13} \cdot 33,854\,{\rm kbit/s} ={ 31,250\,{\rm kbit/s}}\hspace{0.05cm}.$$
+*The same result can be obtained if you consider that in GSM every thirteenth frame is reserved for "Common Control"&nbsp; (signaling info):
 *Thus the percentage of signaling bits is
 :$$\alpha_{\rm SB} = \frac{33.854 - 31.250}{33.854 } { \approx 7.7\%}\hspace{0.05cm}.$$

@@ Line 7: / Line 7: @@
 The first blocks shown in the transmission chain are:
-*the outer coder (block code including four tail bits) with&nbsp; $N_{2} = 244 \ \rm Bit$&nbsp; per frame&nbsp; $(T_{\rm R} = 20 \ \ \rm ms)$ &nbsp; &rArr; &nbsp; Rate&nbsp; $R_{2}$&nbsp; is to be determined,
+*the outer coder (block code including four tail bits) with&nbsp; $N_{2} = 244 \ \rm bit$&nbsp; per frame&nbsp; $(T_{\rm R} = 20 \ \ \rm ms)$ &nbsp; &rArr; &nbsp; Rate&nbsp; $R_{2}$&nbsp; is to be determined,
 *the convolutional coder with the code rate&nbsp; $1/2$, and subsequent puncturing $($waiver of&nbsp; $N_{\rm P} \ \rm bit)$&nbsp; &nbsp; &rArr; &nbsp; Rate $R_{3} = 22.8 \ \rm kbit/s$,
 *interleaving and encryption, both rate-neutral.&nbsp; At the output of this block the rate&nbsp; $R_4$&nbsp; occurs.
@@ Line 13: / Line 13: @@
 The further signal processing is basically as follows:
-*Each&nbsp; $114$&nbsp; (coded, scrambled, encrypted) data bits are combined together with&nbsp; $34$&nbsp; control bits (for training sequence, tail bits, guard period) and a pause $($Duration: &nbsp; $8.25 \ \ \rm Bit)$&nbsp; to a so called&nbsp; ''Normal  Burst''.&nbsp; The rate at the output is $R_{5}$.
+*Each&nbsp; $114$&nbsp; (coded, scrambled, encrypted) data bits are combined together with&nbsp; $34$&nbsp; control bits (for training sequence, tail bits, guard period) and a pause $($Duration: &nbsp; $8.25 \ \ \rm bits)$&nbsp; to a so called&nbsp; ''Normal  Burst''.&nbsp; The rate at the output is $R_{5}$.
 *Additionally, further bursts (''Frequency Correction Burst, Synchronisation Burst, Dummy Burst, Access Bursts'') are added for signalling.&nbsp; The rate after this block is $R_{6}$.
 *Finally the TDMA multiplexing equipment follows, so that the total gross data rate of GSM is $R_{\rm tot} = R_{7}$&nbsp;.
@@ Line 26: / Line 26: @@
 *The task belongs to the chapter&nbsp; [[Mobile_Communications/Similarities_Between_GSM_and_UMTS|Similarities between GSM and UMTS]].
-*The graphic above summarizes the present description and defines the data rates used.
+*The graphic above summarizes the present description and defines the data rates used.&nbsp; All rates are given in&nbsp; $ \rm kbit/s$.
 *$N_{1},&nbsp; N_{2},&nbsp; N_{3}$&nbsp; and&nbsp; $N_{4}$&nbsp; denote the respective number of bits at the corresponding points of the above block diagram within a time frame of duration&nbsp; $T_{\rm R} = 20 \ \rm ms$.
 *$N_{\rm tot} = 156.25$&nbsp; is the number of bits after burst formation, related to the duration&nbsp; $T_{\rm Z}$&nbsp; of a TDMA time slot.&nbsp; $N_{\rm Info} = 114$&nbsp; of which are information bits including channel coding.
@@ Line 39: / Line 38: @@
 {How many bits are provided by the source in each frame?
 |type="{}"}
-$N_{1} \ = \ $ { 192 3% } $\ \ \rm Bit$
+$N_{1} \ = \ $ { 192 3% } $\ \ \rm bit$
@@ Line 49: / Line 48: @@
 {How many bits would the convolutional coder deliver alone (without dotting)?
 |type="{}"}
-$N_{3}\hspace{0.01cm}' \ = \ $ { 488 3% } $\ \ \rm Bit$
+$N_{3}\hspace{0.01cm}' \ = \ $ { 488 3% } $\ \ \rm bit$
 {How many bits does the dotted convolutional coder actually emit?
 |type="{}"}
-$N_{3} \ = \ $ { 456 3% } $\ \ \rm Bit$
+$N_{3} \ = \ $ { 456 3% } $\ \ \rm bit$

@@ Line 4: / Line 4: @@
 [[File:EN_Mob_A_3_2.png|right|frame|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&nbsp; $T_{\rm R} = 20 \ \rm ms$&nbsp; of a voice frame as a temporal reference in the following calculations. The input data rate is&nbsp; $R_{1} = 9.6 \ \rm kbit/s$. The number of input bits in each $T_{\rm R}$ frame is&nbsp; $N_{1}$. All parameters labelled "???" in the graphic should be calculated in the task.
+In this task, the data transmission with GSM is considered.&nbsp; However, since this system was mainly specified for voice transmission, we usually use the duration&nbsp; $T_{\rm R} = 20 \ \rm ms$&nbsp; of a voice frame as a temporal reference in the following calculations.&nbsp; The input data rate is&nbsp; $R_{1} = 9.6 \ \rm kbit/s$.&nbsp; The number of input bit in each&nbsp; $T_{\rm R}$&nbsp; frame is&nbsp; $N_{1}$.&nbsp; All parameters labelled "'''???'''" in the graphic should be calculated in the task.
-The first blocks are shown in the transmission chain shown:
+The first blocks shown in the transmission chain are:
 *the outer coder (block code including four tail bits) with&nbsp; $N_{2} = 244 \ \rm Bit$&nbsp; per frame&nbsp; $(T_{\rm R} = 20 \ \ \rm ms)$ &nbsp; &rArr; &nbsp; Rate&nbsp; $R_{2}$&nbsp; is to be determined,
 *the convolutional coder with the code rate&nbsp; $1/2$, and subsequent puncturing $($waiver of&nbsp; $N_{\rm P} \ \rm bit)$&nbsp; &nbsp; &rArr; &nbsp; Rate $R_{3} = 22.8 \ \rm kbit/s$,
-*Interleaving and encryption, both rate-neutral At the output of this block the rate&nbsp; $R_4$&nbsp; occurs.
+*interleaving and encryption, both rate-neutral.&nbsp; At the output of this block the rate&nbsp; $R_4$&nbsp; occurs.
 The further signal processing is basically as follows:
-*Each&nbsp; $114$&nbsp; (coded, scrambled, encrypted) data bits are combined together with&nbsp; $34$&nbsp; control bits (for training sequence, tail bits, guard period) and a pause $($Duration: &nbsp; $8.25 \ \ \rm Bit)$&nbsp; to a so called&nbsp; ''Normal \ Burst''&nbsp; . The rate at the output is called&nbsp; $R_{5}$&nbsp;.
+*Each&nbsp; $114$&nbsp; (coded, scrambled, encrypted) data bits are combined together with&nbsp; $34$&nbsp; control bits (for training sequence, tail bits, guard period) and a pause $($Duration: &nbsp; $8.25 \ \ \rm Bit)$&nbsp; to a so called&nbsp; ''Normal  Burst''.&nbsp; 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&nbsp; $R_{6}$.
+*Additionally, further bursts (''Frequency Correction Burst, Synchronisation Burst, Dummy Burst, Access Bursts'') are added for signalling.&nbsp; The rate after this block is $R_{6}$.
-*Finally the TDMA multiplexing equipment follows, so that the total gross data rate of the GSM is&nbsp; $R_{\rm tot} = R_{7}$&nbsp;.
+*Finally the TDMA multiplexing equipment follows, so that the total gross data rate of GSM is $R_{\rm tot} = R_{7}$&nbsp;.
 The total gross digital data rate&nbsp; $R_{\rm tot} = 270,833 \ \rm kbit/s$&nbsp; (for eight users) is assumed to be known.
@@ Line 27: / Line 25: @@
 ''Notes:''
-*The task belongs to the chapter&nbsp; [[Mobile_Communications/Gemeinsamkeiten_von_GSM_und_UMTS|Gemeinsamkeiten von GSM und
+*The task belongs to the chapter&nbsp; [[Mobile_Communications/Similarities_Between_GSM_and_UMTS|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$".
+*All rates are given in&nbsp; $ \rm kbit/s$.
 *$N_{1},&nbsp; N_{2},&nbsp; N_{3}$&nbsp; and&nbsp; $N_{4}$&nbsp; denote the respective number of bits at the corresponding points of the above block diagram within a time frame of duration&nbsp; $T_{\rm R} = 20 \ \rm ms$.
-*$N_{\rm tot} = 156.25$&nbsp; is the number of bits after burst formation, related to the duration&nbsp; $T_{\rm Z}$&nbsp; of a TDMA time slot. Of which&nbsp; $N_{\rm Info} = 114$&nbsp; are information bits including channel coding.
+*$N_{\rm tot} = 156.25$&nbsp; is the number of bits after burst formation, related to the duration&nbsp; $T_{\rm Z}$&nbsp; of a TDMA time slot.&nbsp; $N_{\rm Info} = 114$&nbsp; of which are information bits including channel coding.

← Older revision		Revision as of 14:02, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Gemeinsamkeiten von GSM und UMTS~~	+	{{quiz-Header\|Buchseite=Mobile_Communications/Similarities_Between_GSM_and_UMTS

	}}		}}

@@ Line 15: / Line 15: @@
 *Each&nbsp; $114$&nbsp; (coded, scrambled, encrypted) data bits are combined together with&nbsp; $34$&nbsp; control bits (for training sequence, tail bits, guard period) and a pause $($Duration: &nbsp; $8.25 \ \ \rm Bit)$&nbsp; to a so called&nbsp; ''Normal \ Burst''&nbsp;. The rate at the output is called&nbsp; $R_{5}$&nbsp;.
 *Additionally, further bursts (''Frequency Correction Burst, Synchronisation Burst, Dummy Burst, Access Bursts'') are added for signalling. The rate after this block is&nbsp; $R_{6}$.
-*Finally the TDMA multiplexing equipment follows, so that the total gross data rate of the GSM is&nbsp; $R_{\rm ges} = R_{7}$&nbsp;.
+*Finally the TDMA multiplexing equipment follows, so that the total gross data rate of the GSM is&nbsp; $R_{\rm tot} = R_{7}$&nbsp;.
-The total gross digital data rate&nbsp; $R_{\rm ges} = 270,833 \ \rm kbit/s$&nbsp; (for eight users) is assumed to be known.
+The total gross digital data rate&nbsp; $R_{\rm tot} = 270,833 \ \rm kbit/s$&nbsp; (for eight users) is assumed to be known.
@@ Line 32: / Line 32: @@
 *All rates are given in "$ \rm kbit/s$".
 *$N_{1},&nbsp; N_{2},&nbsp; N_{3}$&nbsp; and&nbsp; $N_{4}$&nbsp; denote the respective number of bits at the corresponding points of the above block diagram within a time frame of duration&nbsp; $T_{\rm R} = 20 \ \rm ms$.
-*$N_{\rm ges} = 156.25$&nbsp; is the number of bits after burst formation, related to the duration&nbsp; $T_{\rm Z}$&nbsp; of a TDMA time slot. Of which&nbsp; $N_{\rm Info} = 114$&nbsp; are information bits including channel coding.
+*$N_{\rm tot} = 156.25$&nbsp; is the number of bits after burst formation, related to the duration&nbsp; $T_{\rm Z}$&nbsp; of a TDMA time slot. Of which&nbsp; $N_{\rm Info} = 114$&nbsp; are information bits including channel coding.
@@ Line 80: / Line 80: @@
 </quiz>
-=== sample solution===
+=== Sample Solution===
 {{ML-Kopf}}
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-'''(5)'''&nbsp;  Sowohl das Interleaving als auch die Verschlüsselung erfolgt sozusagen „datenneutral”. Damit gilt:
+'''(5)'''&nbsp; Both the interleaving and the encryption are "data neutral" so to speak. Thus the following applies:
 :$$R_{4} = R_{3} \hspace{0.15cm}\underline{= 22.8 \ {\rm kbit/s}}  \Rightarrow  N_{4} = N_{3} = 456.$$
-'''(8)'''&nbsp; Berücksichtigt man, dass beim ''Normal Burst'' der Anteil der Nutzdaten (inkl. Kanalcodierung) $114/156.25$ beträgt, so wäre die Rate ohne Berücksichtigung der zugefügten Signalisierungsbits:
+'''(7)'''&nbsp; GSM has eight time slots, whereby each user is periodically assigned a time slot.
-:$$R_5 = \frac{n_{\rm ges} }{n_{\rm Info} } \cdot R_4 = \frac{156.25 }{114} \cdot 22.8\,{\rm kbit/s}\hspace{0.15cm} \underline { = 31.250\,{\rm kbit/s}}\hspace{0.05cm}.$$
+*The gross data rate for each user is $R_{6} = R_{7}/8 \hspace{0.15cm}\underline{ \approx 33.854 \ \rm kbit/s}$.
-*Zum gleichen Ergebnis kommt man, wenn man berücksichtigt, dass bei GSM jeder 13. Rahmen für ''Common Control'' (Signalisierungs–Info) reserviert ist:
-:$$R_5 = \frac{12 }{13 } \cdot 33.854\,{\rm kbit/s} ={ 31.250\,{\rm kbit/s}}\hspace{0.05cm}.$$
-*Damit beträgt der prozentuale Anteil der Signalisierungsbits:
 :$$\alpha_{\rm SB} = \frac{33.854 - 31.250}{33.854 } { \approx 7.7\%}\hspace{0.05cm}.$$
 {{ML-Fuß}}