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 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  $+0.75 \ \rm V$  resp.  $–0.75 \ \rm V$. 

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  "$\rm L$  bit.
• This is to be set in subtask  (6)  in such a way that the signal  $s(t)$  becomes DC–free.

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  $($German:  "Rahmendauer"   ⇒   subscript "R"$)$  $T_{\rm R}$?

$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  $R_{\rm gross}$?

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

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

$N_{\rm CB} \ = \ $

5 With which voltage values  $(0 \ {\rm V}, +0.75 \ {\rm V}, \ –0.75 \ {\rm V})$  are the 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  $\rm L$  bit at the end?

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

Solution

Solution

(1)  In each frame,  $16$  bits of the base channels  $\rm B1$  and  $\rm B2$  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  $T_{\rm B}$.

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

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

(4)  The number of control bits  $\rm (CB)$  is:

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

• These are marked in yellow in the graph.

• Thus,  the total gross data rate calculated in the last subquestion is composed as follows:

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

(5)  Note that the first  "0"  is encoded 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  $U_{11} \underline{= 0 \ \rm V}$,

• bit  $b_{12} = 0$  by  $U_{12} \underline{= +0.75 \ \rm V}$.

(6) 

• The  $\rm L$  bit has the task of keeping the AMI encoded signal  $($over all  $48$  ternary symbols$)$  DC-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.