Aufgaben:Exercise 1.3: Frame Structure of ISDN - Revision history

Guenter at 15:51, 24 October 2022

2022-10-24T15:51:23Z

Guenter at 14:21, 18 October 2022

2022-10-18T14:21:52Z

Guenter at 13:46, 18 October 2022

2022-10-18T13:46:15Z

Hwang at 12:53, 9 October 2022

2022-10-09T12:53:08Z

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Guenter at 15:26, 9 September 2022

2022-09-09T15:26:34Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.3: ISDN–Rahmenstruktur to Aufgaben:Exercise 1.3: Frame Structure of ISDN

2022-09-09T15:22:03Z

Guenter moved page Aufgabe 1.3: ISDN–Rahmenstruktur to Exercise 1.3: Frame Structure of ISDN

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

2021-03-23T12:49:16Z

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

Javier: Text replacement - "[[Beispiele_von_Nachrichtensystemen" to "[[Examples_of_Communication_Systems"

2020-07-09T12:53:33Z

Text replacement - "[[Beispiele_von_Nachrichtensystemen" to "[[Examples_of_Communication_Systems"

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:44:09Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

Guenter at 14:33, 15 July 2019

2019-07-15T14:33:52Z

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1581__Bei_A_1_3_neu.png|right|frame|Frame structure of the&nbsp; $\rm S_{0}$ interface]]
+[[File:EN_Bei_A_1_3.png|right|frame|Frame structure of the&nbsp; $\rm S_{0}$ interface]]
 The graphic shows the frame structure of the&nbsp; $\rm S_{0}$ interface.&nbsp; Each frame of duration&nbsp; $T_{\rm R}$&nbsp; contains&nbsp; $48$&nbsp; bits, among them:
 *$16$&nbsp; bits for the bearer channel&nbsp;  $\rm B1$&nbsp; $($light blue$)$,

@@ Line 75: / Line 75: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp;
+'''(1)'''&nbsp; In each frame,&nbsp; $16$&nbsp; bits of the base channels&nbsp; $\rm B1$&nbsp; and&nbsp; $\rm B2$&nbsp; are transmitted.
-*In each frame, 16 bits of the base channels B1 and B2 are transmitted.
+*With the frame duration&nbsp; $T_{\rm R}$,&nbsp; the bit rate&nbsp; $(R_{\rm B} = 64 \ \rm kbit/s)$&nbsp; of each frame is thus:
 :$$R_{\rm B} = \frac{16\,\,{\rm bit}}{T_{\rm R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm R} = \frac{16\,\,{\rm bit}}{64 \cdot 10^3\,\,{\rm bit/s}} \hspace{0.15cm}\underline{= 250 \,{\rm &micro; s}} \hspace{0.05cm}.$$
-'''(2)'''&nbsp;
+'''(2)'''&nbsp; Thus,&nbsp; the following time duration is available for each of the&nbsp; $48$&nbsp; bits:
 :$$T_{\rm B} = \frac{T_{\rm R}}{48} = \frac{250 \,{\rm &micro; s}}{48} \hspace{0.15cm}\underline{ = 5.208 \,{\rm &micro; s}}$$
-*Since in (modified) AMI encoding each binary symbol is replaced by a ternary symbol of the same duration, the symbol duration after AMI encoding is also equal to $T_{\rm B}$.
+*Since in&nbsp; (modified)&nbsp; AMI encoding each binary symbol is replaced by a ternary symbol of the same duration,&nbsp; the symbol duration after AMI encoding is also&nbsp; $T_{\rm B}$.
 '''(3)'''&nbsp; The gross data rate is equal to the reciprocal of the bit duration:
-:$$R_{\rm ges} = \frac{1}{T_{\rm B}} \hspace{0.15cm}\underline{= 192 \,{\rm kbit/s}} \hspace{0.05cm}.$$
+:$$R_{\rm gross} = \frac{1}{T_{\rm B}} \hspace{0.15cm}\underline{= 192 \,{\rm kbit/s}} \hspace{0.05cm}.$$
-'''(4)'''&nbsp; The number of control bits is:
+'''(4)'''&nbsp; The number of control bits&nbsp; $\rm (CB)$&nbsp; is:
-:$$N_{\rm St} = 48 - 2 \cdot 16 -4 \hspace{0.15cm}\underline{= 12} \hspace{0.05cm}.$$
+:$$N_{\rm CB} = 48 - 2 \cdot 16 -4 \hspace{0.15cm}\underline{= 12} \hspace{0.05cm}.$$
 *These are marked in yellow in the graph.
-'''(5)'''&nbsp; Note that the first "0" is coded with positive polarity and all following alternating with $±0.75 \ {\rm V}$:
 *$U_{1} = U_{5} = U_{9} = U_{12} =\text{ ...} = +0.75 \ {\rm V},$
 *$ U_{2} = U_{7} = U_{10} = U_{13} = \text{ ...}  = -0.75 \ {\rm V}$.
 It follows further:
-*Bit $b_{10} = 0$ is represented by $U_{10} \underline{= -0.75 \ \rm V}$,
+*Bit&nbsp; $b_{10} = 0$&nbsp; is represented by&nbsp; $U_{10} \underline{= -0.75 \ \rm V}$,
-*Bit $b_{11} = 1$ by $U_{11} \underline{= 0 \ \rm V}$ and
-*Bit $b_{12} = 0$ by $U_{12} \underline{= +0.75 \ \rm V}$.
+*bit&nbsp; $b_{11} = 1$&nbsp; by&nbsp; $U_{11} \underline{= 0 \ \rm V}$,
 '''(6)'''&nbsp;
-*The '''L''' bit has the task of keeping the AMI encoded signal (over all 48 ternary symbols) equal signal free.
+*The&nbsp; $\rm L$&nbsp; bit has the task of keeping the AMI encoded signal&nbsp; $($over all&nbsp; $48$&nbsp; ternary symbols$)$&nbsp; DC-free.
-*Since the binary symbol "0" has occurred 22 times (i.e. 11 times each the voltage values $+0.75 \ \rm V$ and $-0.75 \ \rm V$) and correspondingly 27 times the binary symbol "1" (voltage value $0 \ \rm V$), $U_{48}\hspace{0.15cm}\underline{=0 \ \rm V}$ must be set.

@@ Line 4: / Line 4: @@
 [[File:P_ID1581__Bei_A_1_3_neu.png|right|frame|Frame structure of the&nbsp; $\rm S_{0}$ interface]]
-The graphic shows the frame structure of the&nbsp; $\rm S_{0}$ interface. Each frame of the duration&nbsp; $T_{\rm R}$&nbsp; contains $48$ bits, among them:
+The graphic shows the frame structure of the&nbsp; $\rm S_{0}$ interface.&nbsp; Each frame of duration&nbsp; $T_{\rm R}$&nbsp; contains&nbsp; $48$&nbsp; bits, among them:
-*$16$ bits for the ''Bearer Channel'' &nbsp;  $\rm B1$ (light blue),
+*$16$&nbsp; bits for the bearer channel&nbsp;  $\rm B1$&nbsp; $($light blue$)$,
-*$16$ bits for the ''Bearer Channel'' &nbsp; $\rm B2$ (dark blue),
-*$4$ bits for the ''Data Channel'' &nbsp; $\rm D$ (green).
+*$16$&nbsp; bits for the bearer channel&nbsp; $\rm B2$&nbsp; $($dark blue$)$,
 The required control bits are shown in yellow.
-For this exercise, it is specified that each of the two base channels&nbsp; $\rm B1$&nbsp; and&nbsp; $\rm B2$&nbsp; should provide a net data rate of&nbsp; $R_{\rm B} = 64 \ \rm kbit/s$.&nbsp;
+For this exercise,&nbsp; it is specified that each of the two base channels&nbsp; $\rm B1$&nbsp; and&nbsp; $\rm B2$&nbsp; should provide a net data rate of&nbsp; $R_{\rm B} = 64 \ \rm kbit/s$.&nbsp;
-It should also be noted that the bit duration&nbsp; $T_{\rm B}$&nbsp; of the uncoded binary signal simultaneously indicates the symbol duration of the (modified) AMI code, which assigns each binary&nbsp; "$1$"&nbsp; to the voltage level&nbsp; $0 \ \rm V$&nbsp; and alternately represents each binary&nbsp; "$0$"&nbsp; with&nbsp; $+0.75 \ \rm V$&nbsp; and&nbsp; $–0.75 \ \rm V$.&nbsp;
+It should also be noted that the bit duration&nbsp; $T_{\rm B}$&nbsp; of the uncoded binary signal simultaneously indicates the symbol duration of the&nbsp; $($modified$)$&nbsp; AMI code,&nbsp;
-The numerical values in the graphic (marked in red) indicate an example sequence which is to be converted into voltage levels in subtask&nbsp; '''(5)'''&nbsp; according to the modified AMI code.
+*alternately represents each binary&nbsp; "$0$"&nbsp; with&nbsp; $+0.75 \ \rm V$&nbsp; resp.&nbsp; $–0.75 \ \rm V$.&nbsp;
-''Notes:''
+*This exercise is part of the chapter&nbsp; [[Examples_of_Communication_Systems/ISDN_Basic_Access|"ISDN Basic Access"]].
-*This exercise is part of the chapter&nbsp; [[Examples_of_Communication_Systems/ISDN_Basic_Access|"ISDN Basic Access"]].
+*It should also be noted that the first&nbsp; $47$&nbsp; bits contain exactly&nbsp; $22$&nbsp; "zeros".
@@ Line 40: / Line 45: @@
 <quiz display=simple>
-{What is the frame duration&nbsp; $T_{\rm R}$?
+{What is the frame duration&nbsp; $($German:&nbsp; "Rahmendauer" &nbsp; &rArr; &nbsp; subscript "R"$)$&nbsp;  $T_{\rm R}$?
 |type="{}"}
 $T_{\rm R} \ = \ $ { 250 3% } $\ \rm &micro; s$
-{What is the bit duration&nbsp; $T_{\rm B}$? ''Note:'' &nbsp; This is equal to the symbol duration after AMI coding.
+{What is the bit duration&nbsp; $T_{\rm B}$?&nbsp; Note:&nbsp; This is equal to the symbol duration after AMI coding.
 |type="{}"}
 $T_{\rm B} \ = \ $ { 5.208 3% } $\ \rm &micro; s $
-{What is the total gross data rate&nbsp; $R_{\rm ges}$?
+{What is the total gross data rate&nbsp; $R_{\rm gross}$?
 |type="{}"}
-$R_{\rm ges} \ = \ $ { 192 3% } $\ \rm kbit/s$
+$R_{\rm gross} \ = \ $ { 192 3% } $\ \rm kbit/s$
-{How many control bits&nbsp; $(N_{\rm St})$&nbsp; are transmitted per frame?
+{How many control bits&nbsp; $(N_{\rm CB})$&nbsp; are transmitted per frame?
 |type="{}"}
-$N_{\rm St} \ = \ $ { 12 3% }
+$N_{\rm CB} \ = \ $ { 12 3% }
-{With which voltage values&nbsp; $(0 \ {\rm V}, \ +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$&nbsp; are bits 10, 11 and 12 (gray shaded block) represented?
+{With which voltage values&nbsp; $(0 \ {\rm V},  +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$&nbsp; are the bits 10, 11 and 12&nbsp; (gray shaded block)&nbsp; represented?
 |type="{}"}
 $U_{10} \ = \ $ { -0.8025--0.6975 } $\ \rm V $
@@ Line 62: / Line 67: @@
 $U_{12} \ = \ $ { 0.75 3% } $\ \rm V $
-{What is the voltage value&nbsp; $(0 \ {\rm V}, \ +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$&nbsp; of the '''L bit''' at the end?
+{What is the voltage value&nbsp; $(0 \ {\rm V},  +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$&nbsp; of the&nbsp; $\rm L$&nbsp; bit at the end?
 |type="{}"}
 $U_{48} \ = \ $ { 0. } $\ \rm V $

← Older revision		Revision as of 15:26, 9 September 2022
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−	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN~~-Basisanschluss~~^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN Basic Access^]]

← Older revision		Revision as of 12:49, 23 March 2021
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−	[[Category:~~Aufgaben zu Beispiele von Nachrichtensystemen~~\|^1.2 ISDN-Basisanschluss^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN-Basisanschluss^]]

@@ Line 29: / Line 29: @@
 ''Hinweise:''
-*Die Aufgabe gehört zum Kapitel&nbsp; [[Beispiele_von_Nachrichtensystemen/ISDN-Basisanschluss|ISDN-Basisanschluss]].
+*Die Aufgabe gehört zum Kapitel&nbsp; [[Examples_of_Communication_Systems/ISDN-Basisanschluss|ISDN-Basisanschluss]].
 *Der AMI–Code wird ausführlich im Kapitel&nbsp; [[Digital_Signal_Transmission/Symbolweise_Codierung_mit_Pseudoternärcodes#Eigenschaften_des_AMI-Codes|Eigenschaften des AMI-Codes]]&nbsp; des Buches „Digitalsignalübertragung” beschrieben.
 *Anzumerken ist ferner, dass die ersten $47$ Bit genau $22$ „Nullen” enthalten.

@@ Line 71: / Line 71: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; In jedem Rahmen werden jeweils 16 Bit der Basiskanäle B1 und B2 übertragen. Mit der Rahmendauer $T_{\rm R}$ gilt somit für die Bitrate $(R_{\rm B} = 64 \ \rm kbit/s)$ eines jeden Rahmens:
+'''(1)'''&nbsp;
-:$$R_{\rm B} = \frac{16\,\,{\rm bit}}{T_{\rm R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm R} = \frac{16\,\,{\rm bit}}{64 \cdot 10^3\,\,{\rm bit/s}} \hspace{0.15cm}\underline{= 250 \,{\rm \mu s}} \hspace{0.05cm}.$$
+*In jedem Rahmen werden jeweils 16 Bit der Basiskanäle B1 und B2 übertragen.
@@ Line 85: / Line 90: @@
 '''(4)'''&nbsp; Die Anzahl der Steuerbit beträgt:
 :$$N_{\rm St} = 48 - 2 \cdot 16 -4 \hspace{0.15cm}\underline{= 12} \hspace{0.05cm}.$$
-Diese sind in der Grafik gelb markiert. Die in der letzten Teilfrage berechnete Gesamt–Bruttodatenrate setzt sich somit wie folgt zusammen:
+*Diese sind in der Grafik gelb markiert.
 :$$R_{\rm ges} = 2 \cdot {R_{\rm B}} + {R_{\rm D}} + {R_{\rm St}} = 2 \cdot 64 \,{\rm kbit/s} + 16 \,{\rm kbit/s} + 48 \,{\rm kbit/s} = 192 \,{\rm kbit/s} \hspace{0.05cm}.$$
 '''(5)'''&nbsp;
 *Das Bit $b_{10} = 0$ wird dargestellt durch $U_{10} \underline{= -0.75 \ \rm V}$,
 *Das Bit $b_{11} = 1$ durch $U_{11} \underline{= 0 \ \rm V}$ und
@@ Line 94: / Line 106: @@
-Zu beachten ist, dass die erste „0” mit positiver Polarität codiert wird und alle folgenden alternierend mit $±0.75 \ {\rm V}$:
+'''(6)'''&nbsp;
-*$U_{1} = U_{5} = U_{9} = U_{12} =\text{ ...} = +0.75 \ {\rm V},$
+*Das '''L'''–Bit hat die Aufgabe, das AMI–codierte Signal (über alle 48 Ternärsymbole) gleichsignalfrei zu halten.
 *Da 22 mal das Binärsymbol „0” aufgetreten ist (also je 11 mal die Spannungswerte $+0.75 \ \rm V$ und $-0.75 \ \rm V$) und dementsprechend 27 mal das Binärsymbol „1” (Spannungswert $0 \ \rm V$), ist $U_{48}\hspace{0.15cm}\underline{=0 \ \rm V}$ zu setzen.