Aufgaben:Exercise 1.7: Coding for Broadband ISDN

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HDB3 and 1T2B coding

For conventional ISDN over copper lines the HDB3 code is used – see  "Exercise 1.5":

This was derived from the so-called AMI code,

• is like the latter a pseudo-ternary code,
• but avoids more than three consecutive  "$0$" symbols,
• by deliberately violating the stricter AMI coding rule for longer zero sequences.

The graph shows the HDB3 encoded signal  $c(t)$ resulting from the binary redundancy-free source signal  $q(t)$.  Since there are no more than three consecutive zeros in the source signal,  $c(t)$  is identical to the AMI-encoded signal.

The broadband ISDN planned for the late 1990s was to provide data rates of up to 155 Mbit/s compared with 144 kbit/s of conventional ISDN with two B channels and one D channel. To achieve this higher data rate, it was necessary to

• newer technology (ATM) had to be used,
• secondly, the transmission medium had to be changed from copper to fiber optics.

However, since the HDB3-encoded signal  $c(t) ∈ \{–1, \ 0, +1\}$  cannot be transmitted by means of light, a second encoding was required. The  1T2B code  provided for this purpose replaces each ternary symbol with two binary symbols. The diagram below shows an example of the binary signal  $b(t) ∈ \{0, 1\}$, which results from the signal  $c(t)$  after this 1T2B coding.

For this exercise, assume that the bit rate of the redundancy-free source signal  $q(t)$  is equal to  $R_{q} = 2.048 \ \rm Mbit/s$.  The respective symbol durations of the signals  $q(t),  c(t)$  and  $b(t)$  are denoted by  $T_{q}$,  $T_{c}$  and  $T_{b}$. 

The equivalent bit rate of the pseudo-ternary signal  $c(t)$  is  $R_{c} = {\rm log_2}(3)/T_{c}$, from which the bit rate  $R_{q} = 1/T_{q}$  of the source signal can be used to calculate the relative redundancy of the AMI or HDB3 code:

$$r_{\rm HDB3} = \frac{R_c - R_q}{R_c}= 1 - \frac{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_c)} \hspace{0.05cm}.$$

A similar equation can be established for the 1T2B code, as well as for the two codes in combination.

Notes:

• The exercise belongs to the chapter  "Further Developments of ISDN".
• Redundancy is defined and illustrated with examples in the chapter  "Basics of Coded Transmission"  of the book "Digital Signal Transmission".

Questions

1

What is the assignment of the  1T2B code?

	$c(t) = +1 \Rightarrow b(t) = 10, \hspace{1cm}c(t) = 0 \Rightarrow b(t) = 00, \hspace{1cm}c(t) = -1 \Rightarrow b(t) = 01,$
	$c(t) = +1 \Rightarrow b(t) = 11, \hspace{1cm}c(t) = 0 \Rightarrow b(t) = 01, \hspace{1cm}c(t) = -1 \Rightarrow b(t) = 00,$
	$c(t) = +1 \Rightarrow b(t) = 01, \hspace{1cm}c(t) = 0 \Rightarrow b(t) = 11, \hspace{1cm}c(t) = -1 \Rightarrow b(t) = 10.$

2

What are the symbol durations of  $q(t),  c(t)$  and  $b(t)$?

$T_{q} \ = \ $ $\ \rm µ s$

$T_{c} \ = \ $ $\ \rm µ s$

$T_{b} \ = \ $ $\ \rm µ s$

3

Calculate the relative redundancy of the  HDB3 code.

$r_{\rm HDB3} \ = \ $ $\ \%$

4

Calculate the relative redundancy of the  1T2B code.

$r_{\rm 1T2B} \ = \ $ $\ \%$

5

What is the relative redundancy of the signal  $b(t)$, i.e. the  combination  of HDB3 code and 1T2B code?

$r_{\rm HDB3+1T2B} \ = \ $ $\ \%$

Solution

Solution

(1)  Solution 2 is correct, as a comparison of the signal characteristics $c(t)$ and $b(t)$ shows.

(2)  The symbol duration (bit duration) of $q(t)$ is   $T_{q} \hspace{0.15cm}\underline{ = 1/R_{q} = 0.488 \ \rm µ s}$.

• The symbol duration of the AMI code (and the HDB3 code) is exactly the same:   $T_{c} \hspace{0.15cm}\underline{ = 0.488 \ \rm µ s}$.
• In contrast, the symbol duration (bit duration) after 1T2B encoding is only half as large: $T_{b} = T_{c}/2 \hspace{0.15cm}\underline{= 0.244 \ \rm µ s}$.

(3)  Using the given equation, with $M_{q} = 2, M_{c} = 3$ and $T_{c} = T_{q}$, we get:

$$r_{\rm HDB3} = 1 - \frac{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_c)} = 1 - \frac{1}{{\rm log_2}\hspace{0.1cm}(3)} \hspace{0.15cm}\underline{= 36.9\,\%} \hspace{0.05cm}.$$

(4)  Fitting the equation to the 1T2B code, we obtain with $M_{c} = 3, M_{b} = 2, T_{b} = T_{c}/2$:

$$r_{\rm 1T2B} = 1 - \frac{T_b \cdot {\rm log_2}\hspace{0.1cm}(M_c)}{T_c \cdot {\rm log_2}\hspace{0.1cm}(M_b)} = 1 - \frac{{\rm log_2}\hspace{0.1cm}(3)}{2} \hspace{0.15cm}\underline{= 20.7\,\%} \hspace{0.05cm}.$$

(5)  The resulting redundancy of both codes is obtained by relating the given equation to the input signal $q(t)$ and the output signal $c(t)$. With $M_{q} = M_{b} = 2$ and $T_{b} = T_{q}/2$ it follows:

$$r_{\rm HDB3+1T2B} = 1 - \frac{T_b \cdot {\rm log_2}\hspace{0.1cm}(M_q)}{T_q \cdot {\rm log_2}\hspace{0.1cm}(M_b)} = 1 - \frac{T_b}{T_q} \hspace{0.15cm}\underline{= 50\,\%} \hspace{0.05cm}.$$

The same result is obtained by the calculation

$$1-r_{\rm HDB3+1T2B} \ = \ (1-r_{\rm HDB3}) \cdot (1-r_{\rm 1T2B}) =(1- 1 +\frac{1}{{\rm log_2}\hspace{0.1cm}(3)}) \cdot (1-1+ \frac{{\rm log_2}\hspace{0.1cm}(3)}{2}) = 50\,\% \hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}r_{\rm HDB3+1T2B}= 50\,\% \hspace{0.05cm}.$$