Exercise 2.3Z: xDSL Frequency Band

• Return to book

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

1

What  $\rm xDSL$ system is it?

	ADSL,
	ADSL2+,
	VDSL.

2

What is the order of upstream and downstream?

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

3

What symbol duration  $T$  results for the DMT system?

$T \ = \ $ $\ \rm µ s$

4

What is the fundamental frequency  $ f_{0}$  underlying the DMT process?

$ f_{0} \ = \ $ $\ \rm kHz$

5

How many channels  $(\hspace{-0.03cm}K_{\rm max}\hspace{-0.03cm})$  could be transmitted in  $2208 \ \rm kHz$ ?

$ K_{\rm max} \ = \ $

6

How many downstream channels  $(\hspace{-0.03cm}K_{\rm down}\hspace{-0.03cm})$  result in this system, given the omission of lower frequencies?

$K_{\rm down} \ = \ $

7

With how many bits  $(b)$  would the bins have to be occupied on average   ⇒   ${\rm E}\big [ \hspace{0.05cm} b \hspace{0.05cm}\big ]$  so that the bit rate  $R_{\rm B} = 25 \ \rm Mbit/s$  is?

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

Solution

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:

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