Aufgaben:Exercise 1.1: ISDN Supply Lines

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Main bundle, basic bundle, and star quad

In ISDN  (Integrated Services Digital Network)  the final branch (near the subscriber) is connected to a local exchange (OVSt) 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 line 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 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 line 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:

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

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