Exercise 1.3: Frame Structure of ISDN

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

$T_{\rm R} \ = \ $

$\ \rm µ s$

$T_{\rm B} \ = \ $

$\ \rm µ s $

$R_{\rm gross} \ = \ $

$\ \rm kbit/s$

$N_{\rm CB} \ = \ $

$U_{10} \ = \ $

$\ \rm V $

$U_{11} \ = \ $

$\ \rm V $

$U_{12} \ = \ $

$\ \rm V $

$U_{48} \ = \ $

$\ \rm V $

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.