Aufgaben:Exercise 1.3: Frame Structure of ISDN

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Frame structure of the  $\rm S_{0}$ interface

The graphic shows the frame structure of the  $\rm S_{0}$ interface. Each frame of the duration  $T_{\rm R}$  contains $48$ bits, among them:

• $16$ bits for the Bearer Channel   $\rm B1$ (light blue),
• $16$ bits for the Bearer Channel   $\rm B2$ (dark blue),
• $4$ bits for the Data Channel   $\rm D$ (green).

The required control bits are shown in yellow.

For this exercise, it is specified that each of the two base channels  $\rm B1$  and  $\rm B2$  should provide a net data rate of  $R_{\rm B} = 64 \ \rm kbit/s$. 

It should also be noted that the bit duration  $T_{\rm B}$  of the uncoded binary signal simultaneously indicates the symbol duration of the (modified) AMI code, which assigns each binary  "$1$"  to the voltage level  $0 \ \rm V$  and alternately represents each binary  "$0$"  with  $+0.75 \ \rm V$  and  $–0.75 \ \rm V$. 

The numerical values in the graphic (marked in red) indicate an example sequence which is to be converted into voltage levels in subtask  (5)  according to the modified AMI code.

• Bit number $48$ contains the so-called  L bit.
• This is to be set in subtask  (6)  in such a way that the signal  $s(t)$  becomes free of equal signals.

Notes:

• This exercise is part of the chapter  "ISDN Basic Access".
• The AMI code is described in detail in the chapter  "Properties of the AMI code"  of the book "Digital Signal Transmission".
• It should also be noted that the first $47$ bits contain exactly $22$ "zeros".

Questions

1

What is the frame duration  $T_{\rm R}$?

$T_{\rm R} \ = \ $ $\ \rm µ s$

2

What is the bit duration  $T_{\rm B}$? Note:   This is equal to the symbol duration after AMI coding.

$T_{\rm B} \ = \ $ $\ \rm µ s $

3

What is the total gross data rate  $R_{\rm ges}$?

$R_{\rm ges} \ = \ $ $\ \rm kbit/s$

4

How many control bits  $(N_{\rm St})$  are transmitted per frame?

$N_{\rm St} \ = \ $

5

With which voltage values  $(0 \ {\rm V}, \ +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$  are bits 10, 11 and 12 (gray shaded block) represented?

$U_{10} \ = \ $ $\ \rm V $

$U_{11} \ = \ $ $\ \rm V $

$U_{12} \ = \ $ $\ \rm V $

6

What is the voltage value  $(0 \ {\rm V}, \ +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$  of the L bit at the end?

$U_{48} \ = \ $ $\ \rm V $

Solution

Solution

(1) 

• In each frame, 16 bits of the base channels B1 and B2 are transmitted.
• With the frame duration $T_{\rm R}$, the bit rate $(R_{\rm B} = 64 \ \rm kbit/s)$ of each frame is thus:

$$R_{\rm B} = \frac{16\,\,{\rm bit}}{T_{\rm R}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} T_{\rm R} = \frac{16\,\,{\rm bit}}{64 \cdot 10^3\,\,{\rm bit/s}} \hspace{0.15cm}\underline{= 250 \,{\rm µ s}} \hspace{0.05cm}.$$

(2) 

• Thus, the following time duration is available for each of the 48 bits.

$$T_{\rm B} = \frac{T_{\rm R}}{48} = \frac{250 \,{\rm µ s}}{48} \hspace{0.15cm}\underline{ = 5.208 \,{\rm µ s}}$$

• Since in (modified) AMI encoding each binary symbol is replaced by a ternary symbol of the same duration, the symbol duration after AMI encoding is also equal to $T_{\rm B}$.

(3)  The gross data rate is equal to the reciprocal of the bit duration:

$$R_{\rm ges} = \frac{1}{T_{\rm B}} \hspace{0.15cm}\underline{= 192 \,{\rm kbit/s}} \hspace{0.05cm}.$$

(4)  The number of control bits is:

$$N_{\rm St} = 48 - 2 \cdot 16 -4 \hspace{0.15cm}\underline{= 12} \hspace{0.05cm}.$$

• These are marked in yellow in the graph.
• Thus, the total gross data rate calculated in the last subquestion is composed as follows:

$$R_{\rm ges} = 2 \cdot {R_{\rm B}} + {R_{\rm D}} + {R_{\rm St}} = 2 \cdot 64 \,{\rm kbit/s} + 16 \,{\rm kbit/s} + 48 \,{\rm kbit/s} = 192 \,{\rm kbit/s} \hspace{0.05cm}.$$

(5)  Note that the first "0" is coded with positive polarity and all following alternating with $±0.75 \ {\rm V}$:

• $U_{1} = U_{5} = U_{9} = U_{12} =\text{ ...} = +0.75 \ {\rm V},$
• $ U_{2} = U_{7} = U_{10} = U_{13} = \text{ ...} = -0.75 \ {\rm V}$.

It follows further:

• Bit $b_{10} = 0$ is represented by $U_{10} \underline{= -0.75 \ \rm V}$,
• Bit $b_{11} = 1$ by $U_{11} \underline{= 0 \ \rm V}$ and
• Bit $b_{12} = 0$ by $U_{12} \underline{= +0.75 \ \rm V}$.

(6) 

• The L bit has the task of keeping the AMI encoded signal (over all 48 ternary symbols) equal signal free.
• Since the binary symbol "0" has occurred 22 times (i.e. 11 times each the voltage values $+0.75 \ \rm V$ and $-0.75 \ \rm V$) and correspondingly 27 times the binary symbol "1" (voltage value $0 \ \rm V$), $U_{48}\hspace{0.15cm}\underline{=0 \ \rm V}$ must be set.