Exercise 2.5Z: Range and Bit Rate with ADSL: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Examples_of_Communication_Systems/Methods_to_Reduce_the_Bit_Error_Rate_in_DSL | ||
}} | }} | ||
[[File: | [[File:EN_Bei_Z_2_5.png|right|frame|"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: | |||
:$$l_{\rm max}\,{\rm [in}\,\,{\rm km]} = \frac {20}{4 + R_{\rm ges}\,{\rm [in}\,\,{\rm Mbit/s]}} \hspace{0.05cm}.$$ | *All data applies to a balanced copper pair with $\text{0.4 mm}$ 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): | |||
:$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$ | :$${\rm a}_{\rm \star} = {\rm a}_{\rm K}(f = {R_{\rm B}}/{2}) = \alpha_{\rm K}(f = {R_{\rm B}}/{2}) \cdot l\hspace{0.05cm}.$$ | ||
The attenuation coefficient (noted as "alpha") is given for a $\text{0.4 mm}$ balanced copper pair as follows: | |||
:$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$ | :$$\alpha_{\rm K}(f ) = \left [ 5.1 + 14.3 \cdot \left ({f}/({\rm 1\,MHz})\right )^{0.59} \right ] {\rm dB}/{\rm km} \hspace{0.05cm}.$$ | ||
For the downlink of variant $\boldsymbol{\rm A}$ with $R_{\rm B} = 2 \ \rm Mbit/s$ thus results with $l = l_{\rm max} = 3.5 \ \rm km$: | |||
:$$\alpha_{\rm K}(f = {\rm 1\,MHz}) = \left [ 5.1 + 14.3 \right ] {\rm dB}/{\rm km} = 19.4\,{\rm dB}/{\rm km} | :$$\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 [[Examples_of_Communication_Systems/Methods_to_Reduce_the_Bit_Error_Rate_in_DSL|"Methods to Reduce the Bit Error Rate in DSL"]]. | |||
===Questions=== | |||
<quiz display=simple> | <quiz display=simple> | ||
{ | {For example, consider the system variant $\boldsymbol{\rm C}$. Which applies? | ||
|type=" | |type="()"} | ||
- | - The upstream bit rate is $13 \ \rm Mbit/s$. | ||
+ | + The downstream bit rate is $13 \rm Mbit/s$. | ||
{ | {Which of the drawn variants are ADSL or VDSL systems? | ||
|type=" | |type="()"} | ||
- | - The red dots indicate VDSL systems. | ||
+ | + The green dots indicate VDSL systems. | ||
{ | {What range would result from the given rule of thumb for $R_{\rm ges} = 1 \ \rm Gbit/s$. | ||
|type="{}"} | |type="{}"} | ||
$l_{\rm max} \ = \ ${ 20 3% } $ \ m$ | $l_{\rm max} \ = \ ${ 20 3% } $ \ \rm m$. | ||
{ | {Calculate the characteristic cable attenuations for the variant | ||
|type="{}"} | |type="{}"} | ||
$\boldsymbol{\rm B:} \ \ | $\boldsymbol{\rm B}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 64.9 3% } $ \ \rm dB$ | ||
$\boldsymbol{\rm C:} \ \ | $\boldsymbol{\rm C}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 48.2 3% } $ \ \rm dB$ | ||
$\boldsymbol{\rm D:} \ \ | $\boldsymbol{\rm D}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 56 3% } $ \ \rm dB$ | ||
$\boldsymbol{\rm E:} \ \ | $\boldsymbol{\rm E}\text{:} \hspace{0.4cm} {\rm a}_{\ast} \ = \ ${ 40.7 3% } $ \ \rm dB$ | ||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' Correct is <u>the second proposed solution</u>: | ||
*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 <u>the second proposed solution</u>: | |||
*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)''' | '''(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},$$ | :$${\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},$$ | :$${\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},$$ | :$${\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}.$$ | :$${\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}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Examples of Communication Systems: Exercises|^2.4 BER Lowering at DSL | ||
^]] | ^]] | ||
[[de:Aufgabe 2.5Z: Reichweite und Bitrate bei ADSL]] | |||
Latest revision as of 17:53, 16 March 2026

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
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.