Aufgaben:Exercise 1.1: ISDN Supply Lines - Revision history

Guenter at 14:18, 26 October 2022

2022-10-26T14:18:44Z

Guenter at 15:47, 24 October 2022

2022-10-24T15:47:51Z

Guenter at 16:03, 14 October 2022

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Guenter at 15:51, 14 October 2022

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Hwang at 09:15, 9 October 2022

2022-10-09T09:15:17Z

Hwang at 07:47, 2 October 2022

2022-10-02T07:47:52Z

Hwang at 20:14, 1 October 2022

2022-10-01T20:14:51Z

Guenter at 15:19, 9 September 2022

2022-09-09T15:19:29Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.1: ISDN–Versorgungsleitungen to Aufgaben:Exercise 1.1: ISDN Supply Lines

2022-09-09T15:17:16Z

Guenter moved page Aufgabe 1.1: ISDN–Versorgungsleitungen to Exercise 1.1: ISDN Supply Lines

Javier: Text replacement - "„" to """

2021-05-28T14:19:48Z

Text replacement - "„" to """

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_ISDN
 }}
-[[File:EN_Bei_A_1_1.png|right|frame|Main bundle, basic bundle, and star quad '''Korrehtur''' auch das LZI Bild]]
+[[File:EN_Bei_A_1_1.png|right|frame|Main bundle, basic bundle, and star quad]]
 In ISDN&nbsp; ("Integrated Services Digital Network")&nbsp; the final branch&nbsp; (near the subscriber)&nbsp; is connected to a&nbsp; "local exchange"&nbsp; $\rm (LE)$&nbsp; by a copper twisted pair,&nbsp; whereby two twisted pairs are twisted into a so-called&nbsp; "star quad".&nbsp; Several such star quads are then combined to form a&nbsp; "basic bundle",&nbsp; and several basic bundles are combined to form a&nbsp; "main bundle"&nbsp; (see graphic).

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_ISDN
 }}
-[[File:EN_LZI_Z_4_6_neu.png|right|frame|Main bundle, basic bundle, and star quad '''Korrehtur''' auch das LZI Bild]]
+[[File:EN_Bei_A_1_1.png|right|frame|Main bundle, basic bundle, and star quad '''Korrehtur''' auch das LZI Bild]]
 In ISDN&nbsp; ("Integrated Services Digital Network")&nbsp; the final branch&nbsp; (near the subscriber)&nbsp; is connected to a&nbsp; "local exchange"&nbsp; $\rm (LE)$&nbsp; by a copper twisted pair,&nbsp; whereby two twisted pairs are twisted into a so-called&nbsp; "star quad".&nbsp; Several such star quads are then combined to form a&nbsp; "basic bundle",&nbsp; and several basic bundles are combined to form a&nbsp; "main bundle"&nbsp; (see graphic).

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_ISDN
 }}
-[[File:EN_LZI_Z_4_6_neu.png|right|frame|Main bundle, basic bundle, and star quad]]
+[[File:EN_LZI_Z_4_6_neu.png|right|frame|Main bundle, basic bundle, and star quad '''Korrehtur''' auch das LZI Bild]]
 In ISDN&nbsp; ("Integrated Services Digital Network")&nbsp; the final branch&nbsp; (near the subscriber)&nbsp; is connected to a&nbsp; "local exchange"&nbsp; $\rm (LE)$&nbsp; by a copper twisted pair,&nbsp; whereby two twisted pairs are twisted into a so-called&nbsp; "star quad".&nbsp; Several such star quads are then combined to form a&nbsp; "basic bundle",&nbsp; and several basic bundles are combined to form a&nbsp; "main bundle"&nbsp; (see graphic).
@@ Line 54: / Line 54: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Two-wire transmission is used in the connection area. The possible connections are equal to the number of pairs in the main cable: &nbsp; $\underline{N = 50}$.
+'''(1)'''&nbsp; Two-wire transmission is used in the connection area.&nbsp; The possible connections are equal to the number of pairs in the main cable: &nbsp;
-'''(2)'''&nbsp; <u>Solutions 1 and 2</u> are correct:
+'''(2)'''&nbsp; <u>Solutions 1 and 2</u>&nbsp; are correct:
-*Two-wire transmission requires a directional separation method, namely the so-called fork circuit. This has the task that at receiver &nbsp;$\rm A$&nbsp; only the transmitted signal of subscriber &nbsp;$\rm B$&nbsp; arrives, but not the own transmitted signal. This is generally quite successful with narrowband signals – for example, speech – but not completely.
+*Two-wire transmission requires a directional separation method,&nbsp; namely the so-called&nbsp; "fork circuit".&nbsp; This has the task that at receiver &nbsp;$\rm A$&nbsp; only the transmitted signal of subscriber &nbsp;$\rm B$&nbsp; arrives,&nbsp; but not the own transmitted signal.&nbsp; This is generally quite successful with narrowband signals – for example,&nbsp; speech – but not completely.
 *Due to inductive and capacitive couplings, crosstalk can occur from the twin wire located in the same star quad, whereby near-end crosstalk (i.e. the interfering transmitter and the interfered receiver are located together) leads to greater impairments than far-end crosstalk.
-*On the other hand, the last solution is not applicable. Intersymbol interference – i.e. the mutual interference of neighboring symbols – can certainly occur, but it is not related to two-wire transmission. The reason for this are rather (linear) distortions due to the specific attenuation and phase curves.
+*Due to inductive and capacitive couplings,&nbsp; crosstalk can occur from the twin wire located in the same star quad,&nbsp; whereby&nbsp; "near-end crosstalk"&nbsp;&nbsp; $($i.e. the interfering transmitter and the interfered receiver are located together$)$&nbsp; leads to greater impairments than&nbsp; "far-end crosstalk".
@@ Line 66: / Line 69: @@
 '''(3)'''&nbsp; The DC signal attenuation by a factor of&nbsp; $4$&nbsp; can be expressed as follows:
 :$$a_{\rm K}(f = 0) = 20 \cdot {\rm lg}\,\,(4) = 12.04\,{\rm dB}\hspace{0.05cm}.$$
-*With the given coefficient&nbsp; $\text{5.1 dB/km}$,&nbsp; this gives the line length $l = 12.04/5.1\hspace{0.15cm}\underline{ = 2.36 \ \rm km}$.
+*With the given coefficient&nbsp; $\text{5.1 dB/km}$,&nbsp; this gives the cable length&nbsp; $l = 12.04/5.1\hspace{0.15cm}\underline{ = 2.36 \ \rm km}$.
-'''(4)'''&nbsp; Using the given equations and&nbsp; $ l = 2.36 \ \rm km$, we obtain:
+'''(4)'''&nbsp; Using the given equations and&nbsp; $ l = 2.36 \ \rm km$,&nbsp; we obtain:
 :$$a_{\rm K}(f = 120\,{\rm kHz})= (5.1 + 14.3 \cdot 0.12^{\hspace{0.05cm}0.59}) \cdot 2.36\,{\rm dB} \hspace{0.15cm}\underline{\approx 21.7\,{\rm dB}}\hspace{0.05cm},$$
 :$$b_{\rm K}(f = 120\,{\rm kHz}) = (32.9 \cdot 0.12 + 2.26 \cdot 0.12^{\hspace{0.05cm}0.5}) \cdot 2.36\,{\rm rad}\hspace{0.15cm}\underline{ \approx 11.2\,{\rm rad}}\hspace{0.05cm}.$$

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Examples_of_Communication_Systems/General_Description_of_ISDN
 }}
-[[File:P_ID1577__Bei_A_1_1.png|right|frame|Main bundle, basic bundle, and star quad]]
+In the network of&nbsp; "Deutsche Telekom"&nbsp; (formerly:&nbsp; "Deutsche Bundespost"),&nbsp; mostly copper lines with&nbsp; $0.4$&nbsp; mm core diameter are found,&nbsp; for whose attenuation and phase function the following equations are given in&nbsp; '''[PW95]''':&nbsp;
 :$$\frac{a_{\rm K}(f)}{\rm dB} = \left [ 5.1 + 14.3 \cdot \left (\frac{f}{\rm MHz}\right )^{0.59}\right ]\cdot\frac{l}{\rm km}
      \hspace{0.05cm},$$
 :$$\frac{b_{\rm K}(f)}{\rm rad} = \left [ 32.9 \cdot \frac{f}{\rm MHz} + 2.26 \cdot \left (\frac{f}{\rm MHz}\right )^{0.5}\right ]\cdot\frac{l}{\rm km}    \hspace{0.05cm}.$$
-Here&nbsp; $l$&nbsp; denotes the line length.
+Here&nbsp; $l$&nbsp; denotes the cable length.
@@ Line 16: / Line 15: @@
-''Notes:''
+*In particular,&nbsp; reference is made to the section&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN#Network_infrastructure_for_ISDN|"Network infrastructure for ISDN"]].
-*The exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN|"General Description of ISDN"]].
-*In particular, reference is made to the section&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN#Network_infrastructure_for_ISDN|"Network infrastructure for ISDN"]].
+*Further information on the attenuation of copper lines can be found in the chapter&nbsp; "Properties of Electrical Cables"&nbsp; of the book&nbsp; [[Lineare zeitinvariante Systeme|"Linear and Time Invariant Systems"]].
 *Further information on the attenuation of copper lines can be found in the chapter "Properties of Electrical Cables" of the book [[Lineare zeitinvariante Systeme|"Linear and Time Invariant Systems"]].
-*[PW95]&nbsp; refers to the following publication: Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
+*'''[PW95]'''&nbsp; refers to the following&nbsp; (German language)&nbsp;  publication:&nbsp; Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
@@ Line 30: / Line 31: @@
 <quiz display=simple>
-{How many subscribers&nbsp; ($N$)&nbsp; can be connected to an ISDN local exchange through the main cable shown?
+{How many subscribers&nbsp; $(N)$&nbsp; can be connected to an ISDN local exchange through the main cable shown?
 |type="{}"}
 $N \ = \ $ { 50 3% }
@@ Line 40: / Line 41: @@
 - Intersymbol interference occurs.
-{A DC signal is attenuated by a factor of $4$.&nbsp; What is the cable length&nbsp; $l$&nbsp;?
+{A DC signal is attenuated by a factor of&nbsp; $4$.&nbsp; What is the cable length&nbsp; $l$&nbsp;?
 |type="{}"}
 $l \ = \ $ { 2.36 3% } $\ \rm km$

@@ Line 4: / Line 4: @@
 [[File:P_ID1577__Bei_A_1_1.png|right|frame|Main bundle, basic bundle, and star quad]]
-In ISDN&nbsp; (''Integrated Services Digital Network'')&nbsp; the final branch (near the subscriber) is connected to a local exchange (OVSt) by a copper twisted pair, whereby two twisted pairs are twisted into a so-called star quad. Several such star quads are then combined to form a basic bundle, and several basic bundles are combined to form a main bundle (see graphic).
+In ISDN&nbsp; (''Integrated Services Digital Network'')&nbsp; the final branch (near the subscriber) is connected to a local exchange (LE) by a copper twisted pair, whereby two twisted pairs are twisted into a so-called star quad. Several such star quads are then combined to form a basic bundle, and several basic bundles are combined to form a main bundle (see graphic).
 In the network of Deutsche Telekom (formerly: &nbsp;Deutsche Bundespost), mostly copper lines with 0.4 mm core diameter are found, for whose attenuation and phase function the following equations are given in&nbsp; [PW95]:&nbsp;

@@ Line 10: / Line 10: @@
      \hspace{0.05cm},$$
 :$$\frac{b_{\rm K}(f)}{\rm rad} = \left [ 32.9 \cdot \frac{f}{\rm MHz} + 2.26 \cdot \left (\frac{f}{\rm MHz}\right )^{0.5}\right ]\cdot\frac{l}{\rm km}    \hspace{0.05cm}.$$
-Hierbei bezeichnet&nbsp; $l$&nbsp; die Leitungslänge.
+Here&nbsp; $l$&nbsp; denotes the line length.
@@ Line 17: / Line 17: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN|"General Description of ISDN"]].
+*The exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN|"General Description of ISDN"]].
-*Insbesondere wird Bezug genommen auf die Seite&nbsp; [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN#Netzinfrastruktur_f.C3.BCr_das_ISDN|Netzinfrastruktur für das ISDN]].
+*In particular, reference is made to the section&nbsp; [[Examples_of_Communication_Systems/General_Description_of_ISDN#Network_infrastructure_for_ISDN|"Network infrastructure for ISDN"]].
-*Weitere Informationen zur Dämpfung von Kupferleitungen finden Sie im Kapitel "Eigenschaften elektrischer Leitungen"  des Buches [[Lineare zeitinvariante Systeme]].
+*Further information on the attenuation of copper lines can be found in the chapter "Properties of Electrical Cables" of the book [[Lineare zeitinvariante Systeme|"Linear and Time Invariant Systems"]].
-*[PW95]&nbsp; bezieht sich auf die folgende Veröffentlichung: Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
+*[PW95]&nbsp; refers to the following publication: Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.
@@ Line 27: / Line 27: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wieviele Teilnehmer&nbsp; ($N$)&nbsp; können durch das dargestellte Hauptkabel an eine ISDN–Ortsvermittlungsstelle angeschlossen werden?
+{How many subscribers&nbsp; ($N$)&nbsp; can be connected to an ISDN local exchange through the main cable shown?
 |type="{}"}
 $N \ = \ $ { 50 3% }
-{Welche Konsequenzen ergeben sich aus der Zweidrahtübertragung?
+{What are the consequences of two-wire transmission?
 |type="[]"}
-+ Die beiden Übertragungsrichtungen stören sich gegenseitig.
++ The two transmission directions interfere with each other.
-+ Es kann zu Nebensprechstörungen kommen.
++ Crosstalk noise may occur.
-- Es treten Impulsinterferenzen auf.
+- Intersymbol interference occurs.
-{Ein Gleichsignal wird um den Faktor $4$&nbsp; gedämpft. Wie groß ist die Kabellänge&nbsp; $l$&nbsp;?
+{A DC signal is attenuated by a factor of $4$.&nbsp; What is the cable length&nbsp; $l$&nbsp;?
 |type="{}"}
 $l \ = \ $ { 2.36 3% } $\ \rm km$
-{Welcher Dämpfungs– und Phasenwert ergibt sich daraus für die Frequenz&nbsp; $f = 120 \ \rm kHz$&nbsp;?
+{Which attenuation and phase value results from this for the frequency&nbsp; $f = 120 \ \rm kHz$&nbsp;?
 |type="{}"}
 $a_{\rm K}(f = 120 \ \rm kHz) \ = \ $ { 21.7 3% } $\ \rm dB$
@@ Line 51: / Line 51: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Im Anschlussbereich wird eine Zweidrahtübertragung verwendet. Die möglichen Anschlüsse sind gleich der Anzahl der Doppeladern im Hauptkabel: &nbsp; $\underline{N = 50}$.
+'''(1)'''&nbsp; Two-wire transmission is used in the connection area. The possible connections are equal to the number of pairs in the main cable: &nbsp; $\underline{N = 50}$.
-'''(2)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 2</u>:
+'''(2)'''&nbsp; <u>Solutions 1 and 2</u> are correct:
-*Bei Zweidrahtübertragung ist ein Richtungstrennungsverfahren erforderlich, nämlich die so genannte Gabelschaltung. Diese hat die Aufgabe, dass beim Empfänger &nbsp;$\rm A$&nbsp; nur das Sendesignal von Teilnehmer &nbsp;$\rm B$&nbsp; ankommt, nicht jedoch das eigene Sendesignal. Dies gelingt bei schmalbandigen Signalen – zum Beispiel Sprache – im allgemeinen recht gut, jedoch nicht vollständig.
+*Two-wire transmission requires a directional separation method, namely the so-called fork circuit. This has the task that at receiver &nbsp;$\rm A$&nbsp; only the transmitted signal of subscriber &nbsp;$\rm B$&nbsp; arrives, but not the own transmitted signal. This is generally quite successful with narrowband signals – for example, speech – but not completely.
-*Aufgrund von induktiven und kapazitiven Kopplungen kann es zu Übersprechen von der im gleichen Sternvierer befindlichen Doppelader kommen, wobei Nahnebensprechen (das heißt: der störende Sender und der gestörte Empfänger liegen örtlich zusammen) zu größeren Beeinträchtigungen führt als Fernnebensprechen.
+*Due to inductive and capacitive couplings, crosstalk can occur from the twin wire located in the same star quad, whereby near-end crosstalk (i.e. the interfering transmitter and the interfered receiver are located together) leads to greater impairments than far-end crosstalk.
-*Nicht zutreffend ist dagegen der letzte Lösungsvorschlag. Impulsinterferenzen – also die gegenseitige störende Beeinflussung benachbarter Symbole – können zwar durchaus auftreten, hängen aber nicht mit der Zweidrahtübertragung zusammen. Der Grund hierfür sind vielmehr (lineare) Verzerrungen aufgrund des spezifischen Dämpfungs– und Phasenverlaufs.
+*On the other hand, the last solution is not applicable. Intersymbol interference – i.e. the mutual interference of neighboring symbols – can certainly occur, but it is not related to two-wire transmission. The reason for this are rather (linear) distortions due to the specific attenuation and phase curves.
-'''(3)'''&nbsp; Die Gleichsignal–Dämpfung um den Faktor&nbsp; $4$&nbsp; kann wie folgt ausgedrückt werden:
+'''(3)'''&nbsp; The DC signal attenuation by a factor of&nbsp; $4$&nbsp; can be expressed as follows:
 :$$a_{\rm K}(f = 0) = 20 \cdot {\rm lg}\,\,(4) = 12.04\,{\rm dB}\hspace{0.05cm}.$$
-*Mit dem angegebenen Koeffizienten&nbsp; $\text{5.1 dB/km}$&nbsp; ergibt sich somit die Leitungslänge $l = 12.04/5.1\hspace{0.15cm}\underline{ = 2.36 \ \rm km}$.
+*With the given coefficient&nbsp; $\text{5.1 dB/km}$,&nbsp; this gives the line length $l = 12.04/5.1\hspace{0.15cm}\underline{ = 2.36 \ \rm km}$.
-'''(4)'''&nbsp; Mit den angegebenen Gleichungen und&nbsp; $ l = 2.36 \ \rm km$ erhält man:
+'''(4)'''&nbsp; Using the given equations and&nbsp; $ l = 2.36 \ \rm km$, we obtain:
 :$$a_{\rm K}(f = 120\,{\rm kHz})= (5.1 + 14.3 \cdot 0.12^{\hspace{0.05cm}0.59}) \cdot 2.36\,{\rm dB} \hspace{0.15cm}\underline{\approx 21.7\,{\rm dB}}\hspace{0.05cm},$$
 :$$b_{\rm K}(f = 120\,{\rm kHz}) = (32.9 \cdot 0.12 + 2.26 \cdot 0.12^{\hspace{0.05cm}0.5}) \cdot 2.36\,{\rm rad}\hspace{0.15cm}\underline{ \approx 11.2\,{\rm rad}}\hspace{0.05cm}.$$

← Older revision		Revision as of 15:19, 9 September 2022
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−	[[Category:Examples of Communication Systems: Exercises\|^1.1 ~~Allgemeine Beschreibung von~~ ISDN^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.1 General Description of ISDN^]]