Difference between revisions of "Aufgaben:Exercise 2.6Z: 4B3T Code according to Jessop and Waters"
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[[File:EN_Dig_Z_2_6.png|right|frame|Code tables for the 4B3T code according to Jessop/Waters]] | [[File:EN_Dig_Z_2_6.png|right|frame|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 | + | 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 | ||
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.05cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}$$ | :$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.05cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}$$ | ||
there are for each binary input tuple $\rm LLLL$ ... $\rm \ HHHH$ two different ternary code sequences. | 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 $a_{\nu} = | + | *In the table, "$+$" and "$-$" stand for the amplitude coefficients $a_{\nu} = +1$ and $a_{\nu} = -1$. |
+ | |||
*The index $l$ identifies the individual blocks. | *The index $l$ identifies the individual blocks. | ||
− | *In the exercise, the following six input blocks are assumed: | + | |
+ | *In the exercise, the following six input blocks are assumed: | ||
:$$\rm LLHL\hspace{0.1cm} HLLH \hspace{0.1cm}LHHH \hspace{0.1cm}HLLH \hspace{0.1cm}HLHH \hspace{0.1cm}HHLH.$$ | :$$\rm LLHL\hspace{0.1cm} HLLH \hspace{0.1cm}LHHH \hspace{0.1cm}HLLH \hspace{0.1cm}HLHH \hspace{0.1cm}HHLH.$$ | ||
*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)'''. | *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)'''. | ||
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+ | Notes: | ||
+ | *The exercise belongs to the chapter [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes|"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 [[Applets:4B3T-Codes|"Principle of 4B3T coding"]]. | ||
Revision as of 17:04, 19 May 2022
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".
Notes:
- The exercise belongs to the chapter "Block Coding with 4B3T Codes".
- In this tutorial, the binary symbols are denoted by L ("Low") and H ("High"). Often you can find the binary symbols L and 0 (instead of H) in the literature. Sometimes, however, L corresponds to our H and 0 to L.
- To avoid such confusion and to prevent the "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 interaction module "Principle of 4B3T coding".
Questions
Solution
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 $K_{+1}\underline{ = 6}$ holds. Also in the coded sequence of this exercise, 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 –}\hspace{0.1cm}. $.
- The running digital sum builds up as follows:
${\it \Sigma}_{1} = 5,$ ${\it \Sigma}_{2} = 4,$ ${\it \Sigma}_{3} = 1,$ ${\it \Sigma}_{4}= 2,$ ${\it \Sigma}_{5} = 3,$ ${\it \Sigma}_{6} \ \underline{= 2}.$