Aufgaben:Exercise 2.6: Modified MS43 Code: Difference between revisions

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{{quiz-Header|Buchseite=Digitalsignalübertragung/2.3 Blockweise 4B3T-Codierung
{{quiz-Header|Buchseite=Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes
}}


[[File:|right|]]
[[File:EN_Bei_Z_1_4.png|right|frame|Code table of the MMS43 code ]]
For ISDN data transmission,  the  '''MMS43 code'''  ("'''M'''odified '''M'''onitored '''S'''um '''4'''B'''3'''T")  is used in Germany and Belgium on the so-called  $\rm {U_{K0}}$ interface,  which describes the transmission path between the exchange and home.  This is a 4B3T code with four code tables,  which are used for coding according to the running digital sum  (after  $l$ blocks):
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$
 ${\it \Sigma}_{0} = 0$  is used for initialization.


The colorings in the table mean:
*If the running digital sum does not change  $({\it \Sigma}_{l+1} = {\it \Sigma}_{l})$,  a field is grayed out.


===Fragebogen===
*An increase &nbsp;$({\it \Sigma}_{l+1} > {\it \Sigma}_{l})$&nbsp; is highlighted in red,&nbsp; a decrease &nbsp;$({\it \Sigma}_{l+1} < {\it \Sigma}_{l})$&nbsp; in blue.
 
*The more intense the coloring,&nbsp; the larger the change.
 
 
 
Notes:
*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes|"Block Coding with 4B3T Codes"]].
*The binary symbols are denoted by&nbsp; $\rm L$&nbsp; ("Low")&nbsp; and&nbsp; $\rm H$&nbsp; ("High")&nbsp; in this learning tutorial.&nbsp; Often you can find the binary symbols&nbsp; $\rm L$&nbsp; and&nbsp; $\rm 0$&nbsp; $($instead of&nbsp; $\rm H)$&nbsp; in the literature.&nbsp; Sometimes,&nbsp; however,&nbsp;  $\rm L$&nbsp; corresponds to our&nbsp; $\rm H$&nbsp; and&nbsp; $\rm 0$&nbsp; to &nbsp; $\rm L$.
 
*To avoid such confusion and to prevent the&nbsp; $\rm 0$&nbsp; from appearing in both alphabets&nbsp; (binary and ternary)&nbsp; - in addition with different meanings - we have used the nomenclature which admittedly takes some getting used to.&nbsp; We are well aware that our nomenclature will also confuse some readers.
 
*You can check the results with the&nbsp; (German language)&nbsp; SWF applet &nbsp;[[Applets:4B3T-Codes|"Principle of 4B3T coding"]].&nbsp;
 
 
 
===Questions===


<quiz display=simple>
<quiz display=simple>
{Multiple-Choice Frage
 
{Why is a 4B3T code used for ISDN instead of the redundancy-free binary code?
|type="[]"}
|type="[]"}
- Falsch
- 4B3T is always better than the redundancy-free binary code.
+ Richtig
+ The transmitted signal should be DC signal free because of &nbsp;$H_{\rm K}(f=0) = 0$.&nbsp;
+ A smaller baud rate allows longer cable length.


{Code the binary sequence &nbsp;$\rm HHLL\hspace{0.08cm} LHLL\hspace{0.08cm} LHHL\hspace{0.08cm} HLHL$&nbsp; $($with &nbsp;${\it \Sigma}_{0} = 0)$&nbsp;. <br>What is the amplitude coefficient of the third ternary symbol of the fourth block?
|type="{}"}
$a_{12} \ = \ $  { -1.03--0.97 }


{Input-Box Frage
 
{Determine the Markov diagram for the transition from &nbsp;${\it \Sigma}_{l}$&nbsp; to &nbsp;${\it \Sigma}_{l+1}$.&nbsp; What are the transition probabilities?
|type="{}"}
|type="{}"}
$\alpha$ = { 0.3 }
$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) \ = \ $ { 0.375 3% }
 
$ {\rm Pr}({\it \Sigma}_{l+1} = 2\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) \ = \ $ { 0.1875 3% }
$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 2) \ = \ $ { 0 3% }


{What properties follow from the Markov diagram?
|type="[]"}
- Equal probabilities: &nbsp; ${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}(\Sigma_{l} = 3)$.
+ &nbsp; ${\rm Pr}({\it \Sigma}_{l} = 0) =  {\rm Pr}({\it \Sigma}_{l} = 3)$ &nbsp; and &nbsp; ${\rm Pr}({\it \Sigma}_{l} = 1) =  {\rm Pr}({\it \Sigma}_{l} = 2)$&nbsp; are valid.
+ The extreme values &nbsp;$({\it \Sigma}_{l} = 0$&nbsp; and&nbsp;  ${\it \Sigma}_{l} = 3)$&nbsp; occur less frequently than the inner values &nbsp;$({\it \Sigma}_{l} = 1$&nbsp;  and &nbsp;${\it \Sigma}_{l} = 2)$.


</quiz>
</quiz>


===Musterlösung===
===Solution===
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'''(1)'''&nbsp;
'''(1)'''&nbsp; <u>Statements 2 and 3</u>&nbsp;  are correct:
'''(2)'''&nbsp;
*The first statement,&nbsp; on the other hand,&nbsp; is not true:&nbsp; For example,&nbsp; the AWGN ("Additive White Gaussian Noise")&nbsp; channel with a 4B3T code results in a much larger symbol error probability due to the ternary decision compared to the redundancy-free binary code.&nbsp;
'''(3)'''&nbsp;
'''(4)'''&nbsp;
*The essential reason for the use of a redundant transmission code is rather that no DC signal component can be transmitted via a&nbsp; "telephone channel".
'''(5)'''&nbsp;
 
'''(6)'''&nbsp;
*The&nbsp; $25 \%$&nbsp; smaller baud rate&nbsp; $(1/T)$&nbsp; of the 4B3T code compared to the redundancy-free binary code also accommodates the transmission characteristics of symmetrical copper lines&nbsp; (strong increase in attenuation with frequency).&nbsp;
*For a given line attenuation,&nbsp; a greater length can be bridged with the 4B3T code than with a redundancy-free binary signal.
 
 
[[File:P_ID1341__Dig_A_2_6c.png|right|frame|Markov diagram for the MMS43 code]]
'''(2)'''&nbsp; The 4B3T coding results with the initial value ${\it\Sigma}_{0} = 0$:
 
*$\rm HHLL$ &nbsp; &rArr; &nbsp; $+ + + \hspace{0.2cm}({\it\Sigma}_{1} = 3)$,
*$\rm LHLL$ &nbsp; &rArr; &nbsp; $– + 0 \hspace{0.2cm}({\it\Sigma}_{2} = 3)$,
*$\rm LHHL$ &nbsp; &rArr; &nbsp; $– – + \hspace{0.2cm}({\it\Sigma}_{3} = 2)$,
*$\rm HLHL$ &nbsp; &rArr; &nbsp; $+ – – \hspace{0.2cm}({\it\Sigma}_{4} = 1)$.
 
Thus,&nbsp; the amplitude coefficient we are looking for is&nbsp; $a_{12} \ \underline{=  -\hspace{-0.05cm}1}$.
 
 
'''(3)'''&nbsp; From the coloring of the code table,&nbsp; the Markov diagram can be obtained.&nbsp;  
*From it,&nbsp; one can read the transition probabilities we are looking for:
 
:$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 6/16 \ \underline{= 0.375},$$
:$$ {\rm Pr}({\it \Sigma}_{l+1} = 2\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 0)= 3/16 \ \underline{= 0.1875},$$
:$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 2)  \ \underline{= 0}.$$
 
 
'''(4)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
*Statement 1 is false,&nbsp; recognizable by the asymmetries in the Markov diagram.
 
*On the other hand,&nbsp; there are symmetries with respect to the states&nbsp; "$0$"&nbsp; and&nbsp; "$3$"&nbsp; as well as between&nbsp; "$1$"&nbsp; and&nbsp; "$2$".
*In the following calculation,&nbsp; instead of&nbsp; $ {\rm Pr}({\it \Sigma}_{l} = 0)$&nbsp; we write $ {\rm Pr}(0)$&nbsp; in a simplified way.
*Taking advantage of the property&nbsp; ${\Pr}(3) = {\Pr}(0)$&nbsp; and&nbsp; ${\Pr}(2) = {\Pr}(1)$,&nbsp; we obtain from the Markov diagram:
:$${\rm Pr}(0)= {6}/{16} \cdot {\rm Pr}(0) +{4}/{16} \cdot {\rm Pr}(1)+ {1}/{16} \cdot {\rm Pr}(3)\hspace{0.15cm} \Rightarrow \hspace{0.15cm}{9}/{16} \cdot {\rm Pr}(0)= {4}/{16} \cdot {\rm Pr}(1)$$
*From the further condition&nbsp; ${\Pr}(0) + {\Pr}(1) = 1/2$&nbsp; it follows further:
:$${\rm Pr}(0)= {\rm Pr}(3)= {9}/{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= {4}/{26}\hspace{0.05cm}.$$
:This calculation is based on the&nbsp; "sum of incoming arrows in state&nbsp; $0$".
 
One could also give equations for the other three states,&nbsp; but they all give the same result:
:$${\rm Pr}(1)  = \ {6}/{16} \cdot {\rm Pr}(0) + {6}/{16} \cdot {\rm Pr}(1)+ {6}/{16} \cdot {\rm Pr}(2)+{3}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
:$$ {\rm Pr}(2) = \ {3}/{16} \cdot {\rm Pr}(0) +{6}/{16} \cdot {\rm Pr}(1)+{6}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
:$${\rm Pr}(3) = \ {1}/{16} \cdot {\rm Pr}(0) + {4}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm}.$$
 


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[[Category:Aufgaben zu Digitalsignalübertragung|^2.3 Blockweise 4B3T-Codierung^]]
[[Category:Digital Signal Transmission: Exercises|^2.3 Blockwise 4B3T Coding^]]
[[de:Aufgaben:Aufgabe 2.6: Modifizierter MS43-Code]]

Latest revision as of 17:56, 16 March 2026


Code table of the MMS43 code

For ISDN data transmission,  the  MMS43 code  ("Modified Monitored Sum 4B3T")  is used in Germany and Belgium on the so-called  $\rm {U_{K0}}$ interface,  which describes the transmission path between the exchange and home.  This is a 4B3T code with four code tables,  which are used for coding according to the running digital sum  (after  $l$ blocks):

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

 ${\it \Sigma}_{0} = 0$  is used for initialization.

The colorings in the table mean:

  • If the running digital sum does not change  $({\it \Sigma}_{l+1} = {\it \Sigma}_{l})$,  a field is grayed out.
  • An increase  $({\it \Sigma}_{l+1} > {\it \Sigma}_{l})$  is highlighted in red,  a decrease  $({\it \Sigma}_{l+1} < {\it \Sigma}_{l})$  in blue.
  • The more intense the coloring,  the larger the change.


Notes:

  • 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.


Questions

1 Why is a 4B3T code used for ISDN instead of the redundancy-free binary code?

4B3T is always better than the redundancy-free binary code.
The transmitted signal should be DC signal free because of  $H_{\rm K}(f=0) = 0$. 
A smaller baud rate allows longer cable length.

2 Code the binary sequence  $\rm HHLL\hspace{0.08cm} LHLL\hspace{0.08cm} LHHL\hspace{0.08cm} HLHL$  $($with  ${\it \Sigma}_{0} = 0)$ .
What is the amplitude coefficient of the third ternary symbol of the fourth block?

$a_{12} \ = \ $

3 Determine the Markov diagram for the transition from  ${\it \Sigma}_{l}$  to  ${\it \Sigma}_{l+1}$.  What are the transition probabilities?

$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) \ = \ $
$ {\rm Pr}({\it \Sigma}_{l+1} = 2\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) \ = \ $
$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 2) \ = \ $

4 What properties follow from the Markov diagram?

Equal probabilities:   ${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}(\Sigma_{l} = 3)$.
  ${\rm Pr}({\it \Sigma}_{l} = 0) = {\rm Pr}({\it \Sigma}_{l} = 3)$   and   ${\rm Pr}({\it \Sigma}_{l} = 1) = {\rm Pr}({\it \Sigma}_{l} = 2)$  are valid.
The extreme values  $({\it \Sigma}_{l} = 0$  and  ${\it \Sigma}_{l} = 3)$  occur less frequently than the inner values  $({\it \Sigma}_{l} = 1$  and  ${\it \Sigma}_{l} = 2)$.


Solution

(1)  Statements 2 and 3  are correct:

  • The first statement,  on the other hand,  is not true:  For example,  the AWGN ("Additive White Gaussian Noise")  channel with a 4B3T code results in a much larger symbol error probability due to the ternary decision compared to the redundancy-free binary code. 
  • The essential reason for the use of a redundant transmission code is rather that no DC signal component can be transmitted via a  "telephone channel".
  • The  $25 \%$  smaller baud rate  $(1/T)$  of the 4B3T code compared to the redundancy-free binary code also accommodates the transmission characteristics of symmetrical copper lines  (strong increase in attenuation with frequency). 
  • For a given line attenuation,  a greater length can be bridged with the 4B3T code than with a redundancy-free binary signal.


Markov diagram for the MMS43 code

(2)  The 4B3T coding results with the initial value ${\it\Sigma}_{0} = 0$:

  • $\rm HHLL$   ⇒   $+ + + \hspace{0.2cm}({\it\Sigma}_{1} = 3)$,
  • $\rm LHLL$   ⇒   $– + 0 \hspace{0.2cm}({\it\Sigma}_{2} = 3)$,
  • $\rm LHHL$   ⇒   $– – + \hspace{0.2cm}({\it\Sigma}_{3} = 2)$,
  • $\rm HLHL$   ⇒   $+ – – \hspace{0.2cm}({\it\Sigma}_{4} = 1)$.

Thus,  the amplitude coefficient we are looking for is  $a_{12} \ \underline{= -\hspace{-0.05cm}1}$.


(3)  From the coloring of the code table,  the Markov diagram can be obtained. 

  • From it,  one can read the transition probabilities we are looking for:
$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 6/16 \ \underline{= 0.375},$$
$$ {\rm Pr}({\it \Sigma}_{l+1} = 2\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 0)= 3/16 \ \underline{= 0.1875},$$
$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} |\hspace{0.05cm}{\it \Sigma}_{l} = 2) \ \underline{= 0}.$$


(4)  Statements 2 and 3  are correct:

  • Statement 1 is false,  recognizable by the asymmetries in the Markov diagram.
  • On the other hand,  there are symmetries with respect to the states  "$0$"  and  "$3$"  as well as between  "$1$"  and  "$2$".
  • In the following calculation,  instead of  $ {\rm Pr}({\it \Sigma}_{l} = 0)$  we write $ {\rm Pr}(0)$  in a simplified way.
  • Taking advantage of the property  ${\Pr}(3) = {\Pr}(0)$  and  ${\Pr}(2) = {\Pr}(1)$,  we obtain from the Markov diagram:
$${\rm Pr}(0)= {6}/{16} \cdot {\rm Pr}(0) +{4}/{16} \cdot {\rm Pr}(1)+ {1}/{16} \cdot {\rm Pr}(3)\hspace{0.15cm} \Rightarrow \hspace{0.15cm}{9}/{16} \cdot {\rm Pr}(0)= {4}/{16} \cdot {\rm Pr}(1)$$
  • From the further condition  ${\Pr}(0) + {\Pr}(1) = 1/2$  it follows further:
$${\rm Pr}(0)= {\rm Pr}(3)= {9}/{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= {4}/{26}\hspace{0.05cm}.$$
This calculation is based on the  "sum of incoming arrows in state  $0$".

One could also give equations for the other three states,  but they all give the same result:

$${\rm Pr}(1) = \ {6}/{16} \cdot {\rm Pr}(0) + {6}/{16} \cdot {\rm Pr}(1)+ {6}/{16} \cdot {\rm Pr}(2)+{3}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
$$ {\rm Pr}(2) = \ {3}/{16} \cdot {\rm Pr}(0) +{6}/{16} \cdot {\rm Pr}(1)+{6}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
$${\rm Pr}(3) = \ {1}/{16} \cdot {\rm Pr}(0) + {4}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm}.$$