Difference between revisions of "Exercise 2.5Z: Range and Bit Rate with ADSL"

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{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/Verfahren zur Senkung der Bitfehlerrate bei DSL
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{{quiz-Header|Buchseite=Examples_of_Communication_Systems/Methods_to_Reduce_the_Bit_Error_Rate_in_DSL
  
  
 
}}
 
}}
  
[[File:P_ID1981__Bei_Z_2_5.png|right|frame|Diagramm "Reichweite vs. Bitrate "]]
+
[[File:EN_Bei_Z_2_5.png|right|frame|"Range vs. bit rate" diagram]]
  
Die Entwicklung der xDSL–Technik begann 1995 mit dem ersten Standard für  $\rm ADSL$  (''Asymmetric Digital Subscriber Line''). Ab 2006 kam in Deutschland auch das schnellere  $\rm VDSL$  (''Very High Data Rate Digital Subscriber Line'') zum Einsatz.
+
The development of xDSL technology began in 1995 with the first standard for $\rm ADSL$  (''Asymmetric Digital Subscriber Line''). From 2006, the faster  $\rm VDSL$  (''Very High Data Rate Digital Subscriber Line'') also came into use in Germany.
  
Die Grafik zeigt fünf Systemvarianten in einem Diagramm, in dem die erreichbare Kabellänge  $l_{\rm max}$  in Abhängigkeit der Gesamtbitrate  $R_{\rm ges}$  aufgetragen ist:
+
The graph shows five system variants in a diagram in which the achievable cable length  $l_{\rm max}$  is plotted as a function of the total bit rate  $R_{\rm ges}$ :
*  $\boldsymbol{\rm A}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 2 \ {\rm Mbit/s}; \hspace{0.2cm} \text{      Kabellänge   } l_{\rm max} \approx 3.5 \ {\rm km},$
+
*  $\boldsymbol{\rm A}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 2 \ {\rm Mbit/s}; \hspace{0.2cm} \text{      cable length   } l_{\rm max} \approx 3.5 \ {\rm km},$
*  $\boldsymbol{\rm B}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 6 \ {\rm Mbit/s}; \hspace{0.2cm} \text{      Kabellänge   } l_{\rm max} \approx 2 \ {\rm km},$
+
*  $\boldsymbol{\rm B}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 6 \ {\rm Mbit/s}; \hspace{0.2cm} \text{      cable length   } l_{\rm max} \approx 2 \ {\rm km},$
*  $\boldsymbol{\rm C}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 13 \ {\rm Mbit/s}; \hspace{0.3cm} \text{      Kabellänge   }  l_{\rm max} \approx 1 \ {\rm km},$
+
*  $\boldsymbol{\rm C}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 13 \ {\rm Mbit/s}; \hspace{0.3cm} \text{      cable length   }  l_{\rm max} \approx 1 \ {\rm km},$
*  $\boldsymbol{\rm D}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 26 \ {\rm Mbit/s}; \hspace{0.3cm} \text{      Kabellänge   }  \ l_{\rm max} \approx 0.8 \ {\rm km},$
+
*  $\boldsymbol{\rm D}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 26 \ {\rm Mbit/s}; \hspace{0.3cm} \text{      cable length   }  \ l_{\rm max} \approx 0.8 \ {\rm km},$
*  $\boldsymbol{\rm E}\text{:} \hspace{0.3cm}\text{    Bitraten} \  2 \ {\rm Mbit/s} + 51 \ {\rm Mbit/s};  \hspace{0.35cm} \text{      Kabellänge   } l_{\rm max} \approx 0.4 \ {\rm km}.$
+
*  $\boldsymbol{\rm E}\text{:} \hspace{0.3cm}\text{    Bitraten} \  2 \ {\rm Mbit/s} + 51 \ {\rm Mbit/s};  \hspace{0.35cm} \text{      cable length   } l_{\rm max} \approx 0.4 \ {\rm km}.$
  
 
+
To this graphic is to be noted further:
Zu dieser Grafik ist weiter anzumerken:
+
*All data applies to a balanced copper pair with  $\text{0.4 mm}$  diameter.
* Alle Angaben gelten für eine Kupfer–Doppelader mit  $\text{0.4 mm}$  Durchmesser.
+
*One of the bit rates given here is for upstream, the other is for downstream.  
*Eine der hier angegebenen Bitraten bezieht sich auf den Upstream, die andere auf den Downstream.  
+
*The total bitrate is the sum of the two portions.  
*Die Gesamtbitrate ist die Summe der beiden Anteile.  
+
*Which bit rate refers to the upstream and which to the downstream is asked in the subtask '''(1)'''.
*Welche Bitrate sich auf den Upstream bezieht und welche auf den Downstream, wird in der Teilaufgabe '''(1)''' abgefragt.
+
*The colored differentiation of the drawn points refers to the subdivision into ADSL and VDSL. This is referred to in the subtask '''(2)'''.  
*Die farbliche Unterscheidung der eingezeichneten Punkte bezieht sich auf die Unterteilung in ADSL und VDSL. Hierauf wird in der Teilaufgabe '''(2)''' Bezug genommen.  
+
*The curve drawn in blue shows a rule of thumb that approximates the relationship between range and total bit rate:
*Die blau eingezeichnete Kurve zeigt eine Faustformel, die den Zusammenhang zwischen Reichweite und Gesamtbitrate annähert:
 
 
:$$l_{\rm max}\,{\rm \big [in}\,\,{\rm km \big]} = \frac {20}{4 + R_{\rm ges}\,{\rm \big[in}\,\,{\rm Mbit/s\big]}} \hspace{0.05cm}.$$
 
:$$l_{\rm max}\,{\rm \big [in}\,\,{\rm km \big]} = \frac {20}{4 + R_{\rm ges}\,{\rm \big[in}\,\,{\rm Mbit/s\big]}} \hspace{0.05cm}.$$
*Gestrichelt eingezeichnet sind Abweichungen hiervon um  $\pm 25\%$.
+
*Dashed are deviations from this by  $\pm 25\%$.
  
  
  
Häufig charakterisiert man ein leitungsgebundenes Übertragungssystem anhand der Kabeldämpfung bei der halben Bitrate (beachten Sie bitte das „a” bei der Dämpfung):
+
One often characterizes a wireline transmission system by the cable attenuation at half the bit rate (note the "a" in the attenuation):
 
:$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$
 
:$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$
  
Das Dämpfungsmaß (mit „alpha” notiert) ist für eine   $\text{0.4 mm}$ Kupfer–Doppelader wie folgt gegeben:
+
The attenuation coefficient (noted as "alpha") is given for a $\text{0.4 mm}$ balanced copper pair as follows:
 
:$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$
 
:$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$
  
Für den Downlink von Variante  $\boldsymbol{\rm A}$  mit  $R_{\rm B} = 2 \ \rm Mbit/s$  ergibt sich somit mit  $l = l_{\rm max} = 3.5 \ \rm km$:
+
For the downlink of variant  $\boldsymbol{\rm A}$  with  $R_{\rm B} = 2 \ \rm Mbit/s$  thus results with  $l = l_{\rm max} = 3.5 \ \rm km$:
 
:$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km}\hspace{0.3cm}
 
:$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km}\hspace{0.3cm}
 
\Rightarrow \hspace{0.3cm} {\rm a}_{\rm \star} = 19.4\,{\rm dB}/{\rm km} \cdot 3.5\,{\rm km} = 67.9\,{\rm dB}\hspace{0.05cm}.$$
 
\Rightarrow \hspace{0.3cm} {\rm a}_{\rm \star} = 19.4\,{\rm dB}/{\rm km} \cdot 3.5\,{\rm km} = 67.9\,{\rm dB}\hspace{0.05cm}.$$
  
Die Werte für die anderen Systemvarianten sollen in der Teilaufgabe '''(4)''' ermittelt werden.
+
The values for the other system variants are to be determined in subtask '''(4)'''.
  
  
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''Hinweis:''
+
Hint:  
  
*Die Aufgabe gehört zum  Kapitel  [[Examples_of_Communication_Systems/Verfahren_zur_Senkung_der_Bitfehlerrate_bei_DSL|Verfahren zur Senkung der Bitfehlerrate bei DSL]].
+
*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/Methods_to_Reduce_the_Bit_Error_Rate_in_DSL|"Methods to Reduce the Bit Error Rate in DSL"]].  
 
   
 
   
  
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===Fragebogen===
+
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Betrachten Sie beispielsweise die Systemvariante&nbsp; $\boldsymbol{\rm C}$. Was trifft zu?
+
{For example, consider the system variant&nbsp; $\boldsymbol{\rm C}$. Which applies?
 
|type="()"}
 
|type="()"}
- Die Upstream–Bitrate beträgt&nbsp; $13 \ \rm Mbit/s$.
+
- The upstream bit rate is&nbsp; $13 \ \rm Mbit/s$.
+ Die Downstream–Bitrate beträgt&nbsp; $13 \ \rm Mbit/s$.
+
+ The downstream bit rate is&nbsp; $13 \rm Mbit/s$.
  
  
{Welche der eingezeichneten Varianten sind ADSL– bzw. VDSL–Systeme?
+
{Which of the drawn variants are ADSL or VDSL systems?
 
|type="()"}
 
|type="()"}
- Die roten Punkte kennzeichnen VDSL–Systeme.
+
- The red dots indicate VDSL systems.
+ Die grünen Punkte kennzeichnen VDSL–Systeme.
+
+ The green dots indicate VDSL systems.
  
  
{Welche Reichweite ergäbe sich aus der angegebenen Faustformel für&nbsp; $R_{\rm ges} = 1 \ \rm Gbit/s$.
+
{What range would result from the given rule of thumb for&nbsp; $R_{\rm ges} = 1 \ \rm Gbit/s$.
 
|type="{}"}
 
|type="{}"}
$l_{\rm max} \ = \ ${ 20 3% } $ \ \rm m$
+
$l_{\rm max} \ = \ ${ 20 3% } $ \ \rm m$.
  
{Berechnen Sie die charakteristischen Kabeldämpfungen für die Variante
+
{Calculate the characteristic cable attenuations for the variant
 
|type="{}"}
 
|type="{}"}
 
$\boldsymbol{\rm B}\text{:} \hspace{0.4cm}  {\rm a}_{\ast} \ = \ ${ 64.9 3% } $ \ \rm dB$
 
$\boldsymbol{\rm B}\text{:} \hspace{0.4cm}  {\rm a}_{\ast} \ = \ ${ 64.9 3% } $ \ \rm dB$
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig ist <u>der zweite Lösungsvorschlag</u>:  
+
'''(1)'''&nbsp; Correct is <u>the second proposed solution</u>:  
*Bei allen xDSL–Varianten wird der Downstream mit höherer Bitrate betrieben als der Upstream.  
+
*In all xDSL variants, the downstream is operated at a higher bit rate than the upstream.  
*Dieses Prinzip ergibt sich aus dem Nutzerverhalten. Man holt sehr viel mehr Daten zum Rechner (Downstream) als in der umgekehrten Richtung.
+
*This principle results from user behavior. One fetches much more data to the computer (downstream) than in the reverse direction.
  
  
  
'''(2)'''&nbsp; Richtig ist <u>der zweite Lösungsvorschlag</u>:  
+
'''(2)'''&nbsp; Correct is <u>the second proposed solution</u>:  
*VDSL bietet höhere Datenraten an.  
+
*VDSL offers higher data rates.  
*Eine hohe Datenrate ist allerdings nur bei relativ kurzen Leitungslängen möglich.
+
*High data rate, however, is only possible with relatively short line lengths.
  
  
  
'''(3)'''&nbsp; Die Reichweite eines solchen Gbit/s–Systems über Zweidrahtleitung wäre etwa $20/1000  
+
'''(3)'''&nbsp; The range of such a Gbit/s system over two-wire line would be about $20/1000  
  \ {\rm km} \underline{= 20 \ \rm Meter}$.  
+
  \ {\rm km} \underline{= 20 \ \rm meters}$.  
*Betrachten Sie diese Teilaufgabe eher als akademisch.
+
*Consider this subtask rather academic.
  
  
  
'''(4)'''&nbsp; Hier ergeben sich folgende charakteristische Kabeldämpfungen. Für
+
'''(4)'''&nbsp; Here the following characteristic cable attenuations result. For
  
*Variante $\boldsymbol{\rm B} \ (R_{\rm B}/2 = 3 {\rm \ Mbit/s}, \ l_{\rm max} = 2 {\rm \ km})\text{:}$
+
*variant $\boldsymbol{\rm B} \ (R_{\rm B}/2 = 3 {\rm \ Mbit/s}, \ l_{\rm max} = 2 {\rm \ km})\text{:}$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 3^{0.59} \right ] \cdot 2\,{\rm dB}\hspace{0.15cm}\underline{ \approx 64.9\,{\rm dB}}\hspace{0.05cm},$$  
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 3^{0.59} \right ] \cdot 2\,{\rm dB}\hspace{0.15cm}\underline{ \approx 64.9\,{\rm dB}}\hspace{0.05cm},$$  
*Variante $\boldsymbol{\rm C} \ (R_{\rm B}/2 = 6.5 {\rm \ Mbit/s}, \ l_{\rm max} = 1 {\rm \ km})\text{:}$
+
*variant $\boldsymbol{\rm C} \ (R_{\rm B}/2 = 6.5 {\rm \ Mbit/s}, \ l_{\rm max} = 1 {\rm \ km})\text{:}$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 6.5^{0.59} \right ] \cdot 1\,{\rm dB} \hspace{0.15cm}\underline{\approx 48.2\,{\rm dB}}\hspace{0.05cm},$$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 6.5^{0.59} \right ] \cdot 1\,{\rm dB} \hspace{0.15cm}\underline{\approx 48.2\,{\rm dB}}\hspace{0.05cm},$$
*Variante $\boldsymbol{\rm D} \ (R_{\rm B}/2 = 13 {\rm \ Mbit/s}, \ l_{\rm max} = 0.8 {\rm \ km})\text{:}$
+
*variant $\boldsymbol{\rm D} \ (R_{\rm B}/2 = 13 {\rm \ Mbit/s}, \ l_{\rm max} = 0.8 {\rm \ km})\text{:}$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 13^{0.59} \right ] \cdot 0.8\,{\rm dB}\hspace{0.15cm}\underline{ \approx 56\,{\rm dB}}\hspace{0.05cm},$$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 13^{0.59} \right ] \cdot 0.8\,{\rm dB}\hspace{0.15cm}\underline{ \approx 56\,{\rm dB}}\hspace{0.05cm},$$
*Variante $\boldsymbol{\rm E} \ (R_{\rm B}/2 = 25.5 {\rm \ Mbit/s}, \ l_{\rm max} = 0.4 {\rm \ km})\text{:}$
+
*variant $\boldsymbol{\rm E} \ (R_{\rm B}/2 = 25.5 {\rm \ Mbit/s}, \ l_{\rm max} = 0.4 {\rm \ km})\text{:}$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 25.5^{0.59} \right ] \cdot 0.4\,{\rm dB}\hspace{0.15cm}\underline{ \approx 40.7\,{\rm dB}}\hspace{0.05cm}.$$
 
:$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 25.5^{0.59} \right ] \cdot 0.4\,{\rm dB}\hspace{0.15cm}\underline{ \approx 40.7\,{\rm dB}}\hspace{0.05cm}.$$
  
Weiter ist anzumerken:
+
Further, it should be noted:
*Die charakteristische Kabeldämpfung ${\rm a}_{\ast}$ von ADSL–Systemen liegt im Bereich $65 \ \rm dB$ ... $68 \ \rm dB$.  
+
*The characteristic cable attenuation ${\rm a}_{\ast}$ of ADSL systems is in the range $65 \ \rm dB$ ... $68 \ \rm dB$.  
*Die VDSL–Varianten liefern charakteristische Kabeldämpfungen zwischen $40 \ \rm dB$ und $56 \ \rm dB$.  
+
*The VDSL variants provide characteristic cable attenuations between $40 \ \rm dB$ and $56 \ \rm dB$.  
*Zu beachten ist allerdings, dass dieser bei herkömmlicher binärer Basisbandübertragung wichtige Systemparameter ${\rm a}_{\ast}$ die Gegebenheiten bei ''OFDM'' bzw. ''Discrete Multitone Transmission'' nicht ausreichend gut wiedergibt.
+
*It should be noted, however, that this system parameter ${\rm a}_{\ast}$, which is important in conventional binary baseband transmission, does not reflect the conditions in ''OFDM'' or ''Discrete Multitone Transmission'' sufficiently well.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
[[Category:Examples of Communication Systems: Exercises|^2.4 BER lowering at DSL
+
[[Category:Examples of Communication Systems: Exercises|^2.4 BER Lowering at DSL
  
  
 
^]]
 
^]]

Latest revision as of 18:34, 25 March 2023

"Range vs. bit rate" diagram

The development of xDSL technology began in 1995 with the first standard for $\rm ADSL$  (Asymmetric Digital Subscriber Line). From 2006, the faster  $\rm VDSL$  (Very High Data Rate Digital Subscriber Line) also came into use in Germany.

The graph shows five system variants in a diagram in which the achievable cable length  $l_{\rm max}$  is plotted as a function of the total bit rate  $R_{\rm ges}$ :

  • $\boldsymbol{\rm A}\text{:} \hspace{0.3cm} \text{ Bitraten} \ 0.2 \ {\rm Mbit/s} + 2 \ {\rm Mbit/s}; \hspace{0.2cm} \text{ cable length } l_{\rm max} \approx 3.5 \ {\rm km},$
  • $\boldsymbol{\rm B}\text{:} \hspace{0.3cm} \text{ Bitraten} \ 0.2 \ {\rm Mbit/s} + 6 \ {\rm Mbit/s}; \hspace{0.2cm} \text{ cable length } l_{\rm max} \approx 2 \ {\rm km},$
  • $\boldsymbol{\rm C}\text{:} \hspace{0.3cm} \text{ Bitraten} \ 2 \ {\rm Mbit/s} + 13 \ {\rm Mbit/s}; \hspace{0.3cm} \text{ cable length } l_{\rm max} \approx 1 \ {\rm km},$
  • $\boldsymbol{\rm D}\text{:} \hspace{0.3cm} \text{ Bitraten} \ 2 \ {\rm Mbit/s} + 26 \ {\rm Mbit/s}; \hspace{0.3cm} \text{ cable length } \ l_{\rm max} \approx 0.8 \ {\rm km},$
  • $\boldsymbol{\rm E}\text{:} \hspace{0.3cm}\text{ Bitraten} \ 2 \ {\rm Mbit/s} + 51 \ {\rm Mbit/s}; \hspace{0.35cm} \text{ cable length } l_{\rm max} \approx 0.4 \ {\rm km}.$

To this graphic is to be noted further:

  • All data applies to a balanced copper pair with  $\text{0.4 mm}$  diameter.
  • One of the bit rates given here is for upstream, the other is for downstream.
  • The total bitrate is the sum of the two portions.
  • Which bit rate refers to the upstream and which to the downstream is asked in the subtask (1).
  • The colored differentiation of the drawn points refers to the subdivision into ADSL and VDSL. This is referred to in the subtask (2).
  • The curve drawn in blue shows a rule of thumb that approximates the relationship between range and total bit rate:
$$l_{\rm max}\,{\rm \big [in}\,\,{\rm km \big]} = \frac {20}{4 + R_{\rm ges}\,{\rm \big[in}\,\,{\rm Mbit/s\big]}} \hspace{0.05cm}.$$
  • Dashed are deviations from this by  $\pm 25\%$.


One often characterizes a wireline transmission system by the cable attenuation at half the bit rate (note the "a" in the attenuation):

$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$

The attenuation coefficient (noted as "alpha") is given for a $\text{0.4 mm}$ balanced copper pair as follows:

$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$

For the downlink of variant  $\boldsymbol{\rm A}$  with  $R_{\rm B} = 2 \ \rm Mbit/s$  thus results with  $l = l_{\rm max} = 3.5 \ \rm km$:

$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm a}_{\rm \star} = 19.4\,{\rm dB}/{\rm km} \cdot 3.5\,{\rm km} = 67.9\,{\rm dB}\hspace{0.05cm}.$$

The values for the other system variants are to be determined in subtask (4).





Hint:



Questions

1

For example, consider the system variant  $\boldsymbol{\rm C}$. Which applies?

The upstream bit rate is  $13 \ \rm Mbit/s$.
The downstream bit rate is  $13 \rm Mbit/s$.

2

Which of the drawn variants are ADSL or VDSL systems?

The red dots indicate VDSL systems.
The green dots indicate VDSL systems.

3

What range would result from the given rule of thumb for  $R_{\rm ges} = 1 \ \rm Gbit/s$.

$l_{\rm max} \ = \ $

$ \ \rm m$.

4

Calculate the characteristic cable attenuations for the variant

$\boldsymbol{\rm B}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ $

$ \ \rm dB$
$\boldsymbol{\rm C}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ $

$ \ \rm dB$
$\boldsymbol{\rm D}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ $

$ \ \rm dB$
$\boldsymbol{\rm E}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ $

$ \ \rm dB$


Solution

(1)  Correct is the second proposed solution:

  • In all xDSL variants, the downstream is operated at a higher bit rate than the upstream.
  • This principle results from user behavior. One fetches much more data to the computer (downstream) than in the reverse direction.


(2)  Correct is the second proposed solution:

  • VDSL offers higher data rates.
  • High data rate, however, is only possible with relatively short line lengths.


(3)  The range of such a Gbit/s system over two-wire line would be about $20/1000 \ {\rm km} \underline{= 20 \ \rm meters}$.

  • Consider this subtask rather academic.


(4)  Here the following characteristic cable attenuations result. For

  • variant $\boldsymbol{\rm B} \ (R_{\rm B}/2 = 3 {\rm \ Mbit/s}, \ l_{\rm max} = 2 {\rm \ km})\text{:}$
$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 3^{0.59} \right ] \cdot 2\,{\rm dB}\hspace{0.15cm}\underline{ \approx 64.9\,{\rm dB}}\hspace{0.05cm},$$
  • variant $\boldsymbol{\rm C} \ (R_{\rm B}/2 = 6.5 {\rm \ Mbit/s}, \ l_{\rm max} = 1 {\rm \ km})\text{:}$
$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 6.5^{0.59} \right ] \cdot 1\,{\rm dB} \hspace{0.15cm}\underline{\approx 48.2\,{\rm dB}}\hspace{0.05cm},$$
  • variant $\boldsymbol{\rm D} \ (R_{\rm B}/2 = 13 {\rm \ Mbit/s}, \ l_{\rm max} = 0.8 {\rm \ km})\text{:}$
$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 13^{0.59} \right ] \cdot 0.8\,{\rm dB}\hspace{0.15cm}\underline{ \approx 56\,{\rm dB}}\hspace{0.05cm},$$
  • variant $\boldsymbol{\rm E} \ (R_{\rm B}/2 = 25.5 {\rm \ Mbit/s}, \ l_{\rm max} = 0.4 {\rm \ km})\text{:}$
$${\rm a}_{\rm \star} = \left [ 5.1 + 14.3 \cdot 25.5^{0.59} \right ] \cdot 0.4\,{\rm dB}\hspace{0.15cm}\underline{ \approx 40.7\,{\rm dB}}\hspace{0.05cm}.$$

Further, it should be noted:

  • The characteristic cable attenuation ${\rm a}_{\ast}$ of ADSL systems is in the range $65 \ \rm dB$ ... $68 \ \rm dB$.
  • The VDSL variants provide characteristic cable attenuations between $40 \ \rm dB$ and $56 \ \rm dB$.
  • It should be noted, however, that this system parameter ${\rm a}_{\ast}$, which is important in conventional binary baseband transmission, does not reflect the conditions in OFDM or Discrete Multitone Transmission sufficiently well.