Examples of Communication Systems/ISDN Primary Multiplex Connection

1. Examples of Communication Systems
2. ISDN – Integrated Services Digital Network
3. ISDN Primary Multiplex Connection

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$\text{Preliminary remark:}$  Again,  we would like to point out that the content of this chapter no longer fully reflects the current state of the art  $($2018$)$.  Therefore,  consider the following text as a historical treatise,  even if parts of it are still relevant in practice today.

General description

First of all,  it should be explained why an ISDN rate interface is  $($or rather was$)$  needed.  This was only offered as a   system connection   $($"point-to-point"$)$. This means that only one device could be connected to the network termination:  Onr speaks of a  "telecommunications system",  abbreviated to  "TC system"  in the following.

There were many reasons for using a system connection:

1. Companies,  government agencies or hospitals often need a central number and a block of extension numbers. 
2. Most often,  the extension number of the central office is  "0".
3. The central call number has  $3$  to  $5$  digits,  an extension number thereafter has  "2"  to  "5"  digits.  This allows direct dialing of a call partner from the outside.
4. Telephone calls between employees  – i.e.,  an internal connection –  should be free of charge.

$\text{Example 1:}$  Let's consider a company in Munich whose head office can be reached from outside via  "089/4711 - 0"  and internally with  "0".  Employee  $X$  can be reached from outside at a charge by dialing extension  "089/4711 - 432"  and internally without charge by dialing  "432".

Larger companies usually work with a   primary rate interface   $\rm (PRI)$,  to which the telecommunications or data processing equipment is connected by a four-wire line.  The primary rate interface according to the adjacent diagram offers:

ISDN primary rate interface

• $30$  full-duplex basic channels with  $\text{64 kbit/s}$  each,

• one signaling channel  $($"data channel"$)$  with  $\text{64 kbit/s}$,

• one synchronization channel  (also with  $\text{64 kbit/s})$,  and accordingly

• a  gross data rate  of  $32 · 64 \hspace{0.15cm}\underline{ = 2048 \ \rm kbit/s}$.

It follows some general information about the primary rate interface:

1. The  $30$  user channels are implemented with the  "PCM-30"  multiplex system.  In contrast to the basic rate interface,  only a point-to-point connection is possible here.  This means that a second system cannot be connected to the same line as with a bus.
2. The telecommunications system is connected to the local exchange via the network termination equipment  $($"Network Termination for Primary Rate Multiplex Access"   ⇒   "$\rm NTPM$"$)$.  This connection is four-wire   ⇒   both transmission directions are separated.  Thus,  no direction separation procedures  $($fork circuit,  echo cancellation,  etc.$)$  are required in the NTPM and in the local exchange.
3. The  "reference point"  $\rm U$  between the network termination and the local exchange is designated  $\rm U_{K2}$ in the case of a primary multiplex connection if a copper cable  $\rm (K)$  is used; the  $\rm (2)$  stands for the transmission rate of  $\text{2 Mbit/s}$.  In the case of a fiber optic connection,  this point is called  $\rm U_{G2}$.
4. Accordingly,  the connection between the network termination and the TC system is generally referred to as the  "$\rm S_{2M}$"  interface.  Technically, however,  there is not much difference between the  "$\rm U_{K2}$"  and the  "$\rm S_{2M}$"  interfaces.

Frame structure of S_2M and U_K2 interface

The  $\rm S_{2M}$ interface represents the connection between the telecommunications system and the network termination  $\rm (NTPM)$,  which is implemented with two copper pairs.  Since only point-to-point operation is possible here,  the  $\rm S_{2M}$ interface is not designed as a bus like the  $\rm S_{0}$ interface in the basic connection,  and therefore no collision detection method is required here.

Frame structure of the  $\rm S_{2M}$ interface

The graphic shows the frame structure of the  $\rm S_{2M}$ interface. It can be seen:

• In time division multiplex,  a TDMA frame is transmitted every  $125\ \rm µs$.  However,  each of the  $32$  channels occupies the TDMA frame only for the duration of  $125\ \rm µs/32 = 3.906 \ \rm µs$.

• Eight bits are transmitted per channel and TDMA frame; the bit duration is 

$$T_\text{B} = 3.906 \ \rm µs/8 = 0.488 \ \rm µs.$$

• The reciprocal of this is the gross data rate 

$$R_\text{B} \hspace{0.15cm}\underline{= 2.048 \ \rm Mbit/s}.$$

• The channels  1  to  15  and  17  to  31  represent the bearer  ($\rm B)$  channels,  all of which are operated independently of each other at  $64 \ \rm kbit/s$.

• The data channel  16  $($marked red in the graph$)$  provides control of these user channels and the entire telephone system.

• The synchronization channel  0  $($marked in blue$)$  is used for frame detection in the case of odd frames  $($with number  $1, 3, 5,$ ...$)$,  while the even frames  ($2, 4, 6,$ ...$)$  are used for maintenance purposes and for error handling.  Both are done with the help of the  $\rm CRC4$  method,  which is described in more detail in the next section.

The  $\rm U_{K2}$ interface has exactly the same properties as the  $\rm S_{2M}$ interface and thus also has the same frame structure.

Frame synchronization

Synchronization is implemented in the  synchronization channel  (channel  0)  of a frame in the primary rate interface. The  Cyclic Redundancy Check  $\rm (CRC4)$ is used for this purpose, which can be illustrated briefly as follows:

• Channel  0  of each odd time frame  (number 1, 3, ... , 15)  transmits the so-called frame password  (FP), while each even frame  (number 2, 4, ... , 16)  of channel  0  contains the message word  (MW).
• Based on the frame password with the fixed bit pattern "$\rm X001\hspace{0.08cm} 1011$", the synchronization between the transmission and the reception direction is established. The first bit  $\rm X ∈ {0, 1}$ is determined by the CRC4 method.
• The message word is "$\rm X1DN\hspace{0.08cm} YYYY$". Error messages  are signaled via the D bit and N bit. The four  $\rm Y$ bits are reserved for service functions.  The  $\rm X$ bit is again obtained by the CRC4 method.
• The CRC4 method requires 16  $\rm X$ bits and thus 16 consecutive pulse frames, which are divided into two multiple frames. The length of a multiple frame is therefore  $8 · 256 = 2048$  bits and the time duration is  $8 · 0.125 = 1$  millisecond.
• The CRC4 checksum is formed as a sequence of four bits  $(\rm C0$, ... , $\rm C3)$  in each multiple frame and provides the first bit  $\rm (X)$  for each of four consecutive frame identifiers.

The table shows the respective frame assignment of synchronization channel  0  for one cycle of the CRC4 method.

Frame assignment of the synchronization channel

$\text{Example 2:}$  The procedure of the CRC4 method shall be explained by an example, where from the generator polynomial 

$$D^4 + D + 1$$

is assumed. In the binary representation this is:  $10011$.

Example for the CRC4 method

The graphic shows the extraction of the CRC4 checksum (left) and its evaluation at the receiver (right). You can see:

• The CRC4 checksum at the transmitter results as the remainder of the division of a data block with a total of twelve bits  (eight useful bits, in the example  $1000\hspace{0.05cm} 1100$, to which  $0000$ is appended)  by the generator polynomial in binary notation  $(10011)$. In polynomial notation, the remainder of the division results in  $(D^{11} + D^7 + D^6 ) : ( D^4 + D + 1)$  to  $R(D) = D^3 + 1$    ⇒   binary $1001$.
• The division is realized by a  modulo-2 addition  (bitwise XOR operation). In the example, the division yields the remainder  $1001$. These four bits  $(\rm C0$, ... , $\rm C3)$  of the CRC checksum are then transmitted to the receiver in different frames of the synchronization channel (see frame assignment in the above graphic).
• After the receiver has received these twelve bits  (data block and CRC4 checksum),  it also divides this 12-digit binary word by the generator polynomial. In the example, this division  $1000\hspace{0.05cm} 1100\hspace{0.05cm} 1001$  divided by  $10011$  gives the remainder zero. This result indicates that no transmission errors have occurred.
• If the division remainder is not zero, the result indicates a transmission error. In this case, the data must be requested again from the transmitter.

Telecommunications aspects

With the ISDN primary rate interface, the so-called  HDB3 line code  (High Density Bipolar 3ary) is used on the  $\rm S_{2M}$ interface and also on the  $\rm U_{K2}$ interface. Compared to the modified AMI code on the  $\rm S_{0}$ interface of the base connection

• the occurrence of long zero sequences is avoided, thus
• providing the receiver a more reliable clock recovery and synchronization.

AMI code and HDB3 code

HDB3 line coding works as follows:

• As in the AMI code, each binary  "0"  is assigned the signal level  $\rm 0\hspace{0.09cm} V$,  while the binary "1" is alternately represented by the values  $+s_0$  and  $–s_0$. 
• If four consecutive  „0”bits occur in the AMI coded signal, they are replaced by a sequence of four other bits that violate the AMI coding rule.
• If, as in the above graphic, the number of ones is even or zero and the last pulse before these four bits is negative,  "0 0 0 0"  is replaced by the sequence  "+ 0 0 +".  On the other hand, if the last pulse before these four bits were positive, "0 0 0 0" would be replaced by "– 0 0 –".
• If there were an odd number of ones before this  "0 0 0 0" block, on the other hand,  "0 0 0 +"  (if last pulse positive) or  "0 0 0 –"  (if last pulse negative) would be selected as replacements. The DC freedom is maintained by these measures.
• In all four cases, the decoder can detect the violation of the AMI rule and replace this block again with  "0 0 0 0". 

Exercises for the chapter

Exercise 1.5: HDB3 Coding

Exercise 1.6: Cyclic Redundancy Check