Aufgaben:Exercise 4.6Z: ISDN Supply Lines

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Main bundle, basic bundle and star quad KORREKTUR: four-wire line  ⇒   star quad

In  $\rm ISDN$  ("Integrated Services Digital Networks"),  the terminal (near the subscriber) is connected to a local exchange ("OVSt") by a copper pair,  with two pairs twisted into a so-called  "star quad".

• Several such star quad pairs are twisted to form a basic bundle,
• several basic bundles are combined to form a main bundle (see diagram).

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

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

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

Here  $l$  denotes the line length.

Notes:

• The exercise belongs to the chapter  Properties of Balanced Copper Pairs.
• For more information on the attenuation behavior of copper lines, see the page  Four-wire and two-wire transmission  in the book "Examples of Communication Systems".

• You can use the  (German language)  interactive   HTML5/JS applet  "Attenuation of copper cables"  to check your results.
• [PW95]  denotes the following literature reference:  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 via the main cable shown at the beginning?

$N \ = \ $

2

What are the consequences of two-wire transmission?

	The two transmission directions can interfere with each other.
	Crosstalk interference can 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

What is the attenuation and phase 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 therefore equal to the number of pairs in the main cable:

$$N = 5 \cdot 5 \cdot 2 \hspace{0.15cm}\underline{= 50}.$$

(2)  Solutions 1 and 2  are correct:

• In two-wire transmission,  a direction separation method is required,  namely the so-called  "hybrid coil".  This has the task that at receiver  $\rm A$  only arrives the transmitted signal of subscriber  $\rm B$,  but not the own transmitted signal.  This is generally quite successful with narrow-band signals – for example speech – but not completely  ⇒  solution 1 is correct.
• Solution 2 is also correct.  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 correct.  Intersymbol interference   ⇒   the mutual interfering influence of neighboring symbols can certainly occur,  but it is not related to two-wire transmission.  The reason for such intersymbol interference is rather linear distortion,  caused by a non-ideal attenuation or phase response.

(3)  The DC 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  $\alpha_0 =5.1 \ \rm dB/km$,  this gives the line length  $l = 12.04/5.1\; \underline{= 2.36 \ \rm km}$.

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

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

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