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

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:

This exercise belongs to the chapter "Methods to Reduce the Bit Error Rate in DSL".

Questions

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

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

$l_{\rm max} \ = \ $

$ \ \rm m$.

$\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.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/Verfahren zur Senkung der Bitfehlerrate bei DSL
+{{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|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 ADSL (''Asymmetric Digital Subscriber Line)''. Ab 2006 kommt in Deutschland auch das schnellere VDSL (''Very High Data Rate Digital Subscriber Line'') zum Einsatz.
+The development of xDSL technology began in 1995 with the first standard for $\rm ADSL$&nbsp; (''Asymmetric Digital Subscriber Line''). From 2006, the faster&nbsp; $\rm VDSL$&nbsp; (''Very High Data Rate Digital Subscriber Line'') also came into use in Germany.
-Die Grafik zeigt 5 Systemvarianten in einem Diagramm, in dem die erreichbare Kabellänge $l_{\rm max}$ in Abhängigkeit der Gesamtbitrate $R_{\rm ges}$ aufgetragen ist:
-*  $\boldsymbol{\rm A:} \ \  0.2 \ {\rm Mbit/s} + 2 \ {\rm Mbit/s}; \ l_{\rm max} \approx 3.5 \ {\rm km},$
-*  $\boldsymbol{\rm B:} \ \  0.2 \ {\rm Mbit/s} + 6 \ {\rm Mbit/s};\approx \ l_{\rm max} \approx 2 \ {\rm km},$
-*  $\boldsymbol{\rm C:} \ \  2 \ {\rm Mbit/s} + 13 \ {\rm Mbit/s};\approx \ l_{\rm max} \approx 1 \ {\rm km},$
-*  $\boldsymbol{\rm D:} \ \  2 \ {\rm Mbit/s} + 26 \ {\rm Mbit/s};\approx \ l_{\rm max} \approx 0.8 \ {\rm km},$
-*  $\boldsymbol{\rm E:} \ \  2 \ {\rm Mbit/s} + 51 \ {\rm Mbit/s}; \ l_{\rm max} \approx 0.4 \ {\rm km}.$
-Eine der hier angegebenen Bitraten bezieht sich auf den Upstream, die andere auf den Downstream. Die Gesamtbitrate ist die Summe der beiden Anteile. Welche Bitrate sich auf den Upstream bezieht und welche auf den Downstream, wird in der Teilaufgabe 1) abgefragt. Alle Angaben gelten für eine Kupfer–Doppelader mit $0.4 \rm mm$ Durchmesser.
+The graph shows five system variants in a diagram in which the achievable cable length&nbsp; $l_{\rm max}$&nbsp; is plotted as a function of the total bit rate&nbsp; $R_{\rm ges}$&nbsp;:
+*  $\boldsymbol{\rm A}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 2 \ {\rm Mbit/s}; \hspace{0.2cm} \text{       cable length   } l_{\rm max} \approx 3.5 \ {\rm km},$
+*  $\boldsymbol{\rm B}\text{:} \hspace{0.3cm} \text{    Bitraten} \  0.2 \ {\rm Mbit/s} + 6 \ {\rm Mbit/s}; \hspace{0.2cm} \text{       cable length   } l_{\rm max} \approx 2 \ {\rm km},$
+*  $\boldsymbol{\rm C}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 13 \ {\rm Mbit/s}; \hspace{0.3cm} \text{       cable length   }  l_{\rm max} \approx 1 \ {\rm km},$
+*  $\boldsymbol{\rm D}\text{:} \hspace{0.3cm} \text{    Bitraten} \  2 \ {\rm Mbit/s} + 26 \ {\rm Mbit/s}; \hspace{0.3cm} \text{       cable length   }  \ l_{\rm max} \approx 0.8 \ {\rm km},$
+*  $\boldsymbol{\rm E}\text{:} \hspace{0.3cm}\text{    Bitraten} \  2 \ {\rm Mbit/s} + 51 \ {\rm Mbit/s};  \hspace{0.35cm} \text{       cable length   } l_{\rm max} \approx 0.4 \ {\rm km}.$
-Die farbliche Unterscheidung der eingezeichneten Punkte bezieht sich auf die Unterteilung in ADSL und VDSL. Hierauf wird in der Teilaufgabe 2) Bezug genommen. Die blau eingezeichnete Kurve zeigt eine Faustformel, die den Zusammenhang zwischen Reichweite und Gesamtbitrate annähert:
+To this graphic is to be noted further:
-:$$l_{\rm max}\,{\rm [in}\,\,{\rm km]} = \frac {20}{4 + R_{\rm ges}\,{\rm [in}\,\,{\rm Mbit/s]}} \hspace{0.05cm}.$$
+*All data applies to a balanced copper pair with&nbsp; $\text{0.4 mm}$&nbsp; 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&nbsp; $\pm 25\%$.
-Gestrichelt eingezeichnet sind Abweichungen hiervon um $\pm 25%$.
-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):
 :$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$
-Das Dämpfungsmaß (mit „alpha” notiert) ist für eine $0.4 \rm mm$ Kupfer–Doppelader wie folgt gegeben:
+The attenuation coefficient (noted as "alpha") is given for a $\text{0.4 mm}$ balanced copper pair as follows:
 :$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$
-Für den Downlink von Variante $\boldsymbol{\rm A} (R_{\rm B} = 2 \ \rm Mbit/s)$ ergibt sich somit mit $l = l_{\rm  max} = 3.5 \ \rm km$:
+For the downlink of variant&nbsp; $\boldsymbol{\rm A}$&nbsp; with&nbsp; $R_{\rm B} = 2 \ \rm Mbit/s$&nbsp; thus results with&nbsp; $l = l_{\rm max} = 3.5 \ \rm km$:
-:$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km}$$
+:$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km}\hspace{0.3cm}
-:$$\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}.$$
+\Rightarrow \hspace{0.3cm} {\rm a}_{\rm \star} = 19.4\,{\rm dB}/{\rm km} \cdot 3.5\,{\rm km} = 67.9\,{\rm dB}\hspace{0.05cm}.$$
+The values for the other system variants are to be determined in subtask '''(4)'''.
+Hint:
-Die Werte für die anderen Systemvarianten sollen in der Teilaufgabe 4) ermittelt werden.
+*This exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/Methods_to_Reduce_the_Bit_Error_Rate_in_DSL|"Methods to Reduce the Bit Error Rate in DSL"]].
-''Hinweis:''
-Die Aufgabe bezieht sich auf das Kapitel [[Beispiele_von_Nachrichtensystemen/Verfahren_zur_Senkung_der_Bitfehlerrate_bei_DSL|Verfahren zur Senkung der Bitfehlerrate bei DSL]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Betrachten Sie beispielsweise die Systemvariante $\boldsymbol{\rm C}$. Was trifft zu?
+{For example, consider the system variant&nbsp; $\boldsymbol{\rm C}$. Which applies?
-|type="[]"}
+|type="()"}
-- Die Upstream–Bitrate beträgt $13 \ \rm Mbit/s$.
+- The upstream bit rate is&nbsp; $13 \ \rm Mbit/s$.
-+ Die Downstream–Bitrate beträgt $13 \ \rm Mbit/s$.
++ 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 Faustformel für $R_{\rm ges} = 1 \ \rm Gbit/s$.
+{What range would result from the given rule of thumb for&nbsp; $R_{\rm ges} = 1 \ \rm Gbit/s$.
 |type="{}"}
-$l_{\rm max} \ = \ ${ 20 3% } $ \ m$
+$l_{\rm max} \ = \ ${ 20 3% } $ \ \rm m$.
-{Berechnen Sie die charakteristischen Kabeldämpfungen für die Variante
+{Calculate the characteristic cable attenuations for the variant
 |type="{}"}
-$\boldsymbol{\rm B:} \ \ a_{\ast} \ = \ ${ 64.9 3% } $ \ \rm dB$
+$\boldsymbol{\rm B}\text{:} \hspace{0.4cm}  {\rm a}_{\ast} \ = \ ${ 64.9 3% } $ \ \rm dB$
-$\boldsymbol{\rm C:} \ \ a_{\ast} \ = \ ${ 48.2 3% } $ \ \rm dB$
+$\boldsymbol{\rm C}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 48.2 3% } $ \ \rm dB$
-$\boldsymbol{\rm D:} \ \ a_{\ast} \ = \ ${ 56 3% } $ \ \rm dB$
+$\boldsymbol{\rm D}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 56 3% } $ \ \rm dB$
-$\boldsymbol{\rm E:} \ \ a_{\ast} \ = \ ${ 40.7 3% } $ \ \rm dB$
+$\boldsymbol{\rm E}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 40.7 3% } $ \ \rm dB$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;
+'''(1)'''&nbsp; Correct is <u>the second proposed solution</u>:
-'''(2)'''&nbsp;
+*In all xDSL variants, the downstream is operated at a higher bit rate than the upstream.
-'''(3)'''&nbsp;
+*This principle results from user behavior. One fetches much more data to the computer (downstream) than in the reverse direction.
-'''(4)'''&nbsp;
-'''(5)'''&nbsp;
+'''(2)'''&nbsp; Correct is <u>the second proposed solution</u>:
+*VDSL offers higher data rates.
+*High data rate, however, is only possible with relatively short line lengths.
+'''(3)'''&nbsp; 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)'''&nbsp; 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.
 {{ML-Fuß}}
-[[Category:Aufgaben zu Beispiele von Nachrichtensystemen|^2.4 Verfahren zur Senkung der Bitfehlerrate bei DSL
+[[Category:Examples of Communication Systems: Exercises|^2.4 BER Lowering at DSL
 ^]]