# Exercise 2.3Z: xDSL Frequency Band

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

### 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:

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