Exercise 2.3Z: xDSL Frequency Band

xDSL frequency band allocation

The figure shows the frequency band allocation of a common $\rm xDSL$ system:

The ISDN band is located at the bottom.
Two bands follow $\rm A$ and $\rm B$, representing downstream and upstream.
Nothing is said about the order of the two bands. This is the question for subtask (2).

Further it is standardized with xDSL/DMT that.

$4000$ frames are transmitted per second,
a synchronization frame is inserted after every $68$ data frames,
the symbol duration must be shortened by the factor $16/17$ because of the cyclic prefix,
each data frame is encoded to a DMT symbol.

This also determines the integration duration $T$ which is evaluated at the receiver for detection, and at the same time also represents the fundamental frequency $f_{0} = 1/T$ of the DMT (Discrete Multitone Transmission) method considered here.

Hint:

This exercise refers to the chapter "xDSL as Transmission Technology".
For information on the cyclic prefix, refer to the chapter "Methods to Reduce the Bit Error Rate in DSL".

Questions

	ADSL,
	ADSL2+,
	VDSL.

	$\rm A$ identifies the upstream and $\rm B$ the downstream.
	$\rm A$ denotes the downstream and $\rm B$ the upstream.

$T \ = \ $

$\ \rm µ s$

$ f_{0} \ = \ $

$\ \rm kHz$

$ K_{\rm max} \ = \ $

$K_{\rm down} \ = \ $

$ {\rm E}\big [ \hspace{0.05cm} b \hspace{0.05cm}\big ] \ = \ $

$\ \rm bit$

Solution

(1) Correct is the second proposed solution:

For ADSL2+, the frequency band ends at $2208 \rm kHz$ as shown in the sketch.
For ADSL, the frequency band already ends at $1104 \rm kHz$.
VDSL has a much larger bandwidth, depending on the band plan, with upstream and downstream bands alternating in each case.

(2) Correct is the first proposed solution:

The upstream was assigned the better (lower) frequencies, since a loss of the fewer upstream channels has a less favorable percentage effect than a loss of a downstream channel.

(3) Without taking into account the synchronization frames (after every $68$ of frames occupied with user data) and the guard interval, the frame duration would result in

$$T = 1/(4000/{\rm s}) = 250 \ \rm µ s.$$

With this overhead taken into account, the symbol duration is shorter by a factor of $68/69 \cdot 16/17$:

$$T = \frac{68}{69} \cdot \frac{16}{17} \cdot 250\, {\rm \mu s} \hspace{0.15cm}\underline{ \approx 232\, {\rm µ s}} \hspace{0.05cm}.$$

(4) The subcarriers lie at DMT at all multiples of $f_0$, where must hold:

$$f_0 = \frac{1}{T} \hspace{0.15cm}\underline{= 4.3125 \, {\rm kHz}}.$$

In fact, the time windowing corresponds to the multiplication of the cosine carrier signals by a square wave function of duration $T$.
In the frequency domain, this results in the convolution with the si function.
If the system quantities $T$ and $f_0 = 1/T$ were not tuned to each other, a de-orthogonalization of the individual DMT channels and thus intercarrier interference would occur.

(5) Ignoring ISDN/upstream reservation, we get $K_{\rm max} = 2208/4.3125 \underline{= 512}.$

(6) The lower $276/4.3125 = 64$ channels are reserved for ISDN and upstream in the "ADSL2+" system considered here.

This leaves $K_{\rm down} = 512 - 64\hspace{0.15cm} \underline{= 448}$ usable channels.

(7) For the bitrate holds.

$$R_{\rm B} = 4000 \, \,\frac {\rm frame}{\rm s} \cdot K \cdot b \hspace{0.05cm}.$$

This results in the (average) bit allocation per bin:

$${\rm E}\big [ \hspace{0.05cm} b \hspace{0.05cm}\big ] = \frac{R_{\rm B}}{ 4000 \, \, {\rm frame}/{\rm s} \cdot K} = \frac{25 \cdot 10^6 \,\, {\rm bit/s}}{ 4000 \, \, {1}/{\rm s} \cdot 448} \hspace{0.15cm}\underline{= 13.95 \, \, {\rm bit}}\hspace{0.05cm}.$$