Aufgaben:Exercise 1.1: ISDN Supply Lines

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Main bundle, basic bundle, and star quad Korrehtur auch das LZI Bild

In ISDN  ("Integrated Services Digital Network")  the final branch  (near the subscriber)  is connected to a  "local exchange"  $\rm (LE)$  by a copper twisted pair,  whereby two twisted pairs are twisted into a so-called  "star quad".  Several such star quads are then combined to form a  "basic bundle",  and several basic bundles are combined to form a  "main bundle"  (see graphic).

In the network of  "Deutsche Telekom"  (formerly:  "Deutsche Bundespost"),  mostly copper lines with  $0.4$  mm core diameter are found,  for whose attenuation and phase function the following equations are given in  [PW95]: 

$$\frac{a_{\rm K}(f)}{\rm dB} = \left [ 5.1 + 14.3 \cdot \left (\frac{f}{\rm MHz}\right )^{0.59}\right ]\cdot\frac{l}{\rm km} \hspace{0.05cm},$$

$$\frac{b_{\rm K}(f)}{\rm rad} = \left [ 32.9 \cdot \frac{f}{\rm MHz} + 2.26 \cdot \left (\frac{f}{\rm MHz}\right )^{0.5}\right ]\cdot\frac{l}{\rm km} \hspace{0.05cm}.$$

Here  $l$  denotes the cable length.

Notes:

• The exercise belongs to the chapter  "General Description of ISDN".

• In particular,  reference is made to the section  "Network infrastructure for ISDN".

• Further information on the attenuation of copper lines can be found in the chapter  "Properties of Electrical Cables"  of the book  "Linear and Time Invariant Systems".

• [PW95]  refers to the following  (German language)  publication:  Pollakowski, P.; Wellhausen, H.-W.: Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz. Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.

Questions

1

How many subscribers  $(N)$  can be connected to an ISDN local exchange through the main cable shown?

$N \ = \ $

2

What are the consequences of two-wire transmission?

	The two transmission directions interfere with each other.
	Crosstalk noise may occur.
	Intersymbol interference occurs.

3

A DC signal is attenuated by a factor of  $4$.  What is the cable length  $l$ ?

$l \ = \ $ $\ \rm km$

4

Which attenuation and phase value results from this for the frequency  $f = 120 \ \rm kHz$ ?

$a_{\rm K}(f = 120 \ \rm kHz) \ = \ $ $\ \rm dB$

$b_{\rm K}(f = 120 \ \rm kHz) \ = \ $ $\ \rm rad$

Solution

Solution

(1)  Two-wire transmission is used in the connection area.  The possible connections are equal to the number of pairs in the main cable:  

$$\underline{N = 50}.$$

(2)  Solutions 1 and 2  are correct:

• Two-wire transmission requires a directional separation method,  namely the so-called  "fork circuit".  This has the task that at receiver  $\rm A$  only the transmitted signal of subscriber  $\rm B$  arrives,  but not the own transmitted signal.  This is generally quite successful with narrowband signals – for example,  speech – but not completely.

• Due to inductive and capacitive couplings,  crosstalk can occur from the twin wire located in the same star quad,  whereby  "near-end crosstalk"   $($i.e. the interfering transmitter and the interfered receiver are located together$)$  leads to greater impairments than  "far-end crosstalk".

• On the other hand,  the last solution is not applicable.  Intersymbol interference – i.e. the mutual interference of neighboring symbols – can certainly occur,  but it is not related to two-wire transmission.  The reason for this are rather  (linear)  distortions due to the specific attenuation and phase curves.

(3)  The DC signal attenuation by a factor of  $4$  can be expressed as follows:

$$a_{\rm K}(f = 0) = 20 \cdot {\rm lg}\,\,(4) = 12.04\,{\rm dB}\hspace{0.05cm}.$$

• With the given coefficient  $\text{5.1 dB/km}$,  this gives the cable length  $l = 12.04/5.1\hspace{0.15cm}\underline{ = 2.36 \ \rm km}$.

(4)  Using the given equations and  $ l = 2.36 \ \rm km$,  we obtain:

$$a_{\rm K}(f = 120\,{\rm kHz})= (5.1 + 14.3 \cdot 0.12^{\hspace{0.05cm}0.59}) \cdot 2.36\,{\rm dB} \hspace{0.15cm}\underline{\approx 21.7\,{\rm dB}}\hspace{0.05cm},$$

$$b_{\rm K}(f = 120\,{\rm kHz}) = (32.9 \cdot 0.12 + 2.26 \cdot 0.12^{\hspace{0.05cm}0.5}) \cdot 2.36\,{\rm rad}\hspace{0.15cm}\underline{ \approx 11.2\,{\rm rad}}\hspace{0.05cm}.$$