Aufgaben:Exercise 1.5: HDB3 Coding

• Return to book

Signals with HDB3 coding

The ISDN primary rate interface 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 (8 bits) of all 32 channels. The duration of such a frame 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 range  $M = 3$, 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 block of four in 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  oder  "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, which you are to determine in the course of this exercise, is denoted by  $c(t)$. 

Notes:

• The exercise belongs to the chapter  "ISDN Primary Multiplex Connection" .
• Information about the pseudo-ternary codes can be found in the  "Symbolwise Coding with Pseudo-Ternary Codes"  of "Digital Signal Transmission".

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  coded?

$c_{6} \ = \ $

$c_{7} \ = \ $

$c_{8} \ = \ $

$c_{9} \ = \ $

$c_{10} \ = \ $

4

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

$c_{14} \ = \ $

$c_{15} \ = \ $

$c_{16} \ = \ $

$c_{17} \ = \ $

5

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

$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 by 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: $\underline{c_{24} = 0}$.