Exercise 2.5Z: Range and Bit Rate with ADSL

"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.