Aufgaben:Exercise 2.6Z: 4B3T Code according to Jessop and Waters

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Code tables for the 4B3T code according to Jessop/Waters

The graphic shows the two code tables for the 4B3T code according to Jessop and Waters. 

Depending on the current value of the running digital sum

$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.05cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}$$

there are for each binary input tuple  $\rm LLLL$ ... $\rm \ HHHH$  two different ternary code sequences.

• In the table,  "$+$"  and  "$-$"  stand for the amplitude coefficients  $a_{\nu} = +1$  and  $a_{\nu} = -1$.

• The index  $l$  identifies the individual blocks.

• In the exercise,  the following six input blocks are assumed:

$$\rm LLHL\hspace{0.1cm} HLLH \hspace{0.1cm}LHHH \hspace{0.1cm}HLLH \hspace{0.1cm}HLHH \hspace{0.1cm}HHLH.$$

• The running digital sum is initialized to  ${\it \Sigma}_{0} = 0$  in subtasks up to and including (2) or to  ${\it \Sigma}_{0} = 5$  in subtask (5).

Notes:

• The exercise belongs to the chapter  "Block Coding with 4B3T Codes".

• The binary symbols are denoted by  $\rm L$  ("Low")  and  $\rm H$  ("High")  in this learning tutorial.  Often you can find the binary symbols  $\rm L$  and  $\rm 0$  $($instead of  $\rm H)$  in the literature.  Sometimes,  however,  $\rm L$  corresponds to our  $\rm H$  and  $\rm 0$  to   $\rm L$.

• To avoid such confusion and to prevent the  $\rm 0$  from appearing in both alphabets  (binary and ternary)  - in addition with different meanings - we have used the nomenclature which admittedly takes some getting used to.  We are well aware that our nomenclature will also confuse some readers.

• You can check the results with the  (German language)  SWF applet  "Principle of 4B3T coding". 

Questions

1

Code the input sequence   $\rm LLHL\hspace{0.1cm} HLLH \hspace{0.1cm}LHHH \hspace{0.1cm}HLLH \hspace{0.1cm}HLHH \hspace{0.1cm}HHLH$   starting from the initial value  ${\it \Sigma}_{0} = 0$.
What is the ternary output sequence?

	$ \text{0 – +} \hspace{0.4cm} \text{– + +} \hspace{0.5cm} \text {– – –} \hspace{0.65cm} \text{– ++} \hspace{0.4cm} \text{+ 0 0} \hspace{0.4cm} \text{0 0 +}\hspace{0.1cm}, $
	$ \text{0 – +} \hspace{0.4cm} \text{– + +} \hspace{0.4cm} \text{+ + +} \hspace{0.4cm} \text{+ – –} \hspace{0.5cm} \text{– 0 0} \hspace{0.4cm} \text{0 0 +} \hspace{0.1cm},$
	$ \text{0 – +} \hspace{0.4cm} \text{+ – –} \hspace{0.5cm} \text{– – –} \hspace{0.65cm} \text{– + +} \hspace{0.4cm} \text{+ 0 0} \hspace{0.4cm} \text{0 0 –}\hspace{0.1cm}. $

2

What is the value of the running digital sum after coding the six blocks?

${\it \Sigma}_{6} \ = \ $

3

What is the maximum number of ternary values  $+1$  that can succeed each other?

$K_{+1} \ = \ $

4

What is the maximum number of ternary values  $0$  that can succeed each other?

$K_{0} \ = \ $

5

What is the value of the running digital sum after coding the six blocks,  assuming  ${\it \Sigma}_{0} = 5$? 

${\it \Sigma}_{6} \ = \ $

Solution

Solution

1  The  second solution  is correct. 

• The first ternary sequence would result with  ${\it \Sigma}_{0} = 2$, 
• the last with  ${\it \Sigma}_{0} = 5$.

2  Starting from  ${\it \Sigma}_{0} = 0$,  the following values result for the running digital sum:

    ${\it \Sigma}_{1} = 0,$     ${\it \Sigma}_{2} = 1,$     ${\it \Sigma}_{3} = 4,$     ${\it \Sigma}_{4}= 3,$     ${\it \Sigma}_{5} = 2,$     ${\it \Sigma}_{6} \ \underline{= 3}.$

3  $K_{+1}\hspace{0.15cm}\underline{ = 6}$ holds.   Also in the coded sequence of this exercise,

$$ \text{0 – +} \hspace{0.4cm} \text{– + +} \hspace{0.4cm} \text{+ + +} \hspace{0.4cm} \text{+ – –} \hspace{0.5cm} \text{– 0 0} \hspace{0.4cm} \text{0 0 +} \hspace{0.1cm},$$

one recognizes six consecutive plus signs coming from a total of three blocks:

• Two at the end of the second block,
• then three  "$+1$"  in block  $3$,  and
• finally one  "$+1$"  at the beginning of the fourth block.

Similarly,  $K_{-1} = 6$  (see solution suggestion 3 in the first subtask).

4  If  ${\it \Sigma}_{l} = 2$,  the binary sequence  "$\rm HLHH\hspace{0.1cm} HHLH$"  leads to the ternary sequence  "$+ 0 0 \hspace{0.1cm}0 0 –$".

• More than  $K_{0}\ \underline{ = 4}$  consecutive zeros are not possible.

5  The ternary sequence here is:   "$ \text{0 – +} \hspace{0.4cm} \text{+ – –} \hspace{0.5cm} \text{– – –} \hspace{0.65cm} \text{– + +} \hspace{0.4cm} \text{+ 0 0} \hspace{0.4cm} \text{0 0 –} $".

• The running digital sum builds up as follows:   ${\it \Sigma}_{1} = 5,$     ${\it \Sigma}_{2} = 4,$     ${\it \Sigma}_{3} = 1,$     ${\it \Sigma}_{4}= 2,$     ${\it \Sigma}_{5} = 3,$     ${\it \Sigma}_{6} \ \underline{= 2}.$