# Exercise 1.5: HDB3 Coding

Signals with HDB3 coding

The ISDN primary rate interface  $\rm (PRI)$  is based on the  $\rm PCM\ system \ 30/32$  and offers

• $30$  full-duplex basic channels,
• plus a signaling channel
• and a synchronization channel.

Each of these channels,  which are transmitted in time division multiplex,  has a data rate of  $64 \ \rm kbit/s$.  A frame consists of one byte  $\rm (8$  bits$)$  of all  $32$  channels.  The duration of such a frame  $($German:  "Rahmen"$)$  is denoted by  $T_{\rm R}$,  while  $T_{\rm B}$  indicates the bit duration.

On both the  $\rm S_{\rm 2M}$ and  $\rm U_{\rm K2}$ interfaces of the ISDN system under consideration,  the   HDB3 code   is used,  which is derived from the AMI code.  This is a pseudo-ternary code  $($symbol set size  $M = 3$,  symbol duration  $T = T_{\rm B})$,  that differs from the AMI code in that long zero sequences are avoided by deliberately violating the AMI coding rule.  The following applies:

If four consecutive  "0"  symbols occur in the AMI-encoded signal  $a(t)$,  these are replaced by four other ternary symbols:

• If an even number of  "+1"  occurred before this four-symbol block the signal  $a(t)$  and the last pulse is positive,  "0 0 0 0"  is replaced by  "– 0 0 –".  If the last pulse is negative,  "0 0 0 0"  is replaced by  "+ 0 0 +".
• On the other hand,  if there is an odd number of  "ones"  before this  "0 0 0 0" block,  "0 0 0 +"  $($if last pulse positive$)$  or  "0 0 0 –"  $($if last pulse negative$)$  are selected.

The graph above shows the binary signal  $q(t)$  and the signal  $a(t)$  after AMI coding.  The HDB3 signal is denoted by  $c(t)$.

Notes:

### Questions

1

What is the total data rate of the ISDN rate interface?

 $R_{\rm B} \ = \$ $\ \rm Mbit/s$

2

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

 $T_{\rm B} \ = \$ $\ \rm µ s$ $T_{\rm R} \ = \$ $\ \rm µ s$

3

How is the zero block between bit  6  and bit  10  encoded?
Possible input values are  $0$,  $+1$  and  $–1$.

 $c_{6} \ = \$ $c_{7} \ = \$ $c_{8} \ = \$ $c_{9} \ = \$ $c_{10} \ = \$

4

How is the zero block between bit  14  and bit  17  encoded?

 $c_{14} \ = \$ $c_{15} \ = \$ $c_{16} \ = \$ $c_{17} \ = \$

5

How is the zero block between bit  20  and bit  24  encoded?

 $c_{20} \ = \$ $c_{21} \ = \$ $c_{22} \ = \$ $c_{23} \ = \$ $c_{24} \ = \$

### Solution

#### Solution

(1)  The total data rate of the  $32$  channels at  $64 \ \rm kbit/s$  each results in

$$R_{\rm B} \underline{ = 2.048 \ \rm Mbit/s}.$$

(2)  The bit duration is  $T_{\rm B} = 1/R_{\rm B} \underline{ = 0.488 \ \rm µ s}$.

• One byte (8 bits) of each channel is transmitted per frame.  It follows that:
$$T_{\rm R} = 32 \cdot 8 \cdot T_{\rm B} \hspace{0.15cm}\underline{= 125 \,{\rm µ s}}\hspace{0.05cm}.$$

(3)  By time  $t = 6T$,  a  "+1"  has occurred exactly once in the AMI-encoded signal  $a(t)$.

Relationship between AMI code and HDB3 code
• Because of  $a_{5} = –1$,  in the HDB3 code  "0 0 0 0"  is replaced by  (see diagram)
$$\underline{c_{6} = 0, \hspace{0.2cm}c_{7} = 0, \hspace{0.2cm}c_{8} = 0, \hspace{0.2cm}c_{9} = -1} \hspace{0.05cm}.$$
• In contrast,  $\underline{c_{10} = a_{10} = 0}$  is not changed bythe HDB3 coding.

(4)  Up to and including  $a_{13}$,  there are three times a  "+1"   ⇒   odd number.  Because of  $a_{12} = +1$,  this zero block is replaced as follows:

$$\underline{c_{14} = 0, \hspace{0.2cm}c_{15} = 0, \hspace{0.2cm}c_{16} = 0, \hspace{0.2cm}c_{17} = +1} \hspace{0.05cm}.$$

(5)  In the AMI-encoded signal,  "+1"  occurs exactly four times up to and including  $a_{19}$    ⇒   even number.

• Because of  $a_{19} = +1$,  the substitution according to rule 2 in the information section is:
$$\underline{c_{20} = -1, \hspace{0.2cm}c_{21} = 0, \hspace{0.2cm}c_{22} = 0, \hspace{0.2cm}c_{23} = -1} \hspace{0.05cm}.$$
• The zero symbol  $a_{24}$  remains unchanged: $\underline{c_{24} = 0}$.