Aufgaben:Exercise 2.6: Modified MS43 Code - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^2.3 Blockwise 4B3T Coding^]]		[[Category:Digital Signal Transmission: Exercises\|^2.3 Blockwise 4B3T Coding^]]
	[[de:Aufgaben:2.~~6_Modifizierter_MS43~~-Code]]		[[de:Aufgaben:Aufgabe 2.6: Modifizierter MS43-Code]]

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	[[Category:Digital Signal Transmission: Exercises\|^2.3 Blockwise 4B3T Coding^]]		[[Category:Digital Signal Transmission: Exercises\|^2.3 Blockwise 4B3T Coding^]]
			[[de:Aufgaben:2.6_Modifizierter_MS43-Code]]

Guenter at 15:56, 24 October 2022

2022-10-24T15:56:23Z

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	[[File:~~EN_Dig_A_2_6_v2~~.png\|right\|frame\|Code table of the MMS43 code ~~'''Korrektur''' Strich~~]]		[[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):		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):
	:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$		:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$

Guenter at 16:00, 24 August 2022

2022-08-24T16:00:40Z

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	[[File:~~P_ID1340__Dig_A_2_6~~.png\|right\|frame\|Code table of the MMS43 code '''Korrektur''']]		[[File:EN_Dig_A_2_6_v2.png\|right\|frame\|Code table of the MMS43 code '''Korrektur''' Strich]]
	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):		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):
	:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$		:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$

Guenter at 14:57, 19 May 2022

2022-05-19T14:57:01Z

← Older revision		Revision as of 16:57, 19 May 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Statements 2 and 3</u> are correct:		'''(1)'''  <u>Statements 2 and 3</u>  are correct:
	*The first statement, on the other hand, is not true: For example, the AWGN (~~''Additives Weißes Gaußsches Rauschen''~~) channel with a 4B3T code results in a much larger 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 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 $25 \%$ smaller ~~step~~ 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.
			*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.


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

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

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

	:$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} \|\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 6/16 \ \underline{= 0.375},$$		:$$ {\rm Pr}({\it \Sigma}_{l+1} = 0\hspace{0.05cm} \|\hspace{0.05cm}{\it \Sigma}_{l} = 0) = 6/16 \ \underline{= 0.375},$$
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	'''(4)'''  <u>Statements 2 and 3</u> are correct:		'''(4)'''  <u>Statements 2 and 3</u>  are correct:
	*Statement 1 is false, recognizable by the asymmetries in the Markov diagram.		*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.		*On the other hand,  there are symmetries with respect to the states  "$0$"  and  "$3$"  as well as between  "$1$"  and  "$2$".
	*Taking advantage of the property ${\Pr}(3) = {\Pr}(0)$ and ${\Pr}(2) = {\Pr}(1)$, we obtain from the Markov diagram:
			*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:
	:$${\rm Pr}(0)= {6}/{16} \cdot {\rm Pr}(0) +{4}/{16} \cdot {\rm Pr}(1)+ {1}/{16} \cdot {\rm Pr}(3)\hspace{0.15cm} \Rightarrow \hspace{0.15cm}{9}/{16} \cdot {\rm Pr}(0)= {4}/{16} \cdot {\rm Pr}(1)$$		:$${\rm Pr}(0)= {6}/{16} \cdot {\rm Pr}(0) +{4}/{16} \cdot {\rm Pr}(1)+ {1}/{16} \cdot {\rm Pr}(3)\hspace{0.15cm} \Rightarrow \hspace{0.15cm}{9}/{16} \cdot {\rm Pr}(0)= {4}/{16} \cdot {\rm Pr}(1)$$
	From the further condition ${\Pr}(0) + {\Pr}(1) = 1/2$ it follows further:		*From the further condition  ${\Pr}(0) + {\Pr}(1) = 1/2$  it follows further:
	:$${\rm Pr}(0)= {\rm Pr}(3)= {9}/{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= {4}/{26}\hspace{0.05cm}.$$		:$${\rm Pr}(0)= {\rm Pr}(3)= {9}/{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= {4}/{26}\hspace{0.05cm}.$$
	This calculation is based on the "sum of incoming arrows in state $0$".		: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:		One could also give equations for the other three states,  but they all give the same result:
	:$${\rm Pr}(1) = \ {6}/{16} \cdot {\rm Pr}(0) + {6}/{16} \cdot {\rm Pr}(1)+ {6}/{16} \cdot {\rm Pr}(2)+{3}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$		:$${\rm Pr}(1) = \ {6}/{16} \cdot {\rm Pr}(0) + {6}/{16} \cdot {\rm Pr}(1)+ {6}/{16} \cdot {\rm Pr}(2)+{3}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
	:$$ {\rm Pr}(2) = \ {3}/{16} \cdot {\rm Pr}(0) +{6}/{16} \cdot {\rm Pr}(1)+{6}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$		:$$ {\rm Pr}(2) = \ {3}/{16} \cdot {\rm Pr}(0) +{6}/{16} \cdot {\rm Pr}(1)+{6}/{16} \cdot {\rm Pr}(2)+{6}/{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$

Guenter at 14:39, 19 May 2022

2022-05-19T14:39:21Z

← Older revision		Revision as of 16:39, 19 May 2022
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	[[File:P_ID1340__Dig_A_2_6.png\|right\|frame\|Code table of the MMS43 code]]		[[File:P_ID1340__Dig_A_2_6.png\|right\|frame\|Code table of the MMS43 code '''Korrektur''']]
	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)		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):
	:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu$$		:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu.$$
	${\it \Sigma}_{0} = 0$  is used for initialization.		${\it \Sigma}_{0} = 0$  is used for initialization.

	The colorings in the table mean:		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.		*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.

			*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:''		Notes:
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes\|"Block Coding with 4B3T Codes"]].		*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 ~~'''~~L~~'''~~ ("Low") and ~~'''~~H~~'''~~ ("High") in this learning tutorial. 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~~'''~~.		*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 "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~~  [[Applets:4B3T-Codes\|"Principle of 4B3T coding"]].		*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"]].


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

	{Code the binary sequence  $\rm HHLL\hspace{0.08cm} LHLL\hspace{0.08cm} LHHL\hspace{0.08cm} HLHL$  $($with  ${\it \Sigma}_{0} = 0)$ . <br>What is the amplitude coefficient of the third ternary symbol of the fourth block?		{Code the binary sequence  $\rm HHLL\hspace{0.08cm} LHLL\hspace{0.08cm} LHHL\hspace{0.08cm} HLHL$  $($with  ${\it \Sigma}_{0} = 0)$ . <br>What is the amplitude coefficient of the third ternary symbol of the fourth block?
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	{Determine the Markov diagram for the transition from  ${\it \Sigma}_{l}$  to  ${\it \Sigma}_{l+1}$. What are the transition probabilities?		{Determine the Markov diagram for the transition from  ${\it \Sigma}_{l}$  to  ${\it \Sigma}_{l+1}$.  What are the transition probabilities?
	\|type="{}"}		\|type="{}"}
	$ {\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} = 0\hspace{0.05cm} \|\hspace{0.05cm}{\it \Sigma}_{l} = 0) \ = \ $ { 0.375 3% }
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	\|type="[]"}		\|type="[]"}
	- Equal probabilities:   ${\rm Pr}({\it \Sigma}_{l} = 0) = \text{...} = {\rm Pr}(\Sigma_{l} = 3)$.		- 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.		+   ${\rm Pr}({\it \Sigma}_{l} = 0) = {\rm Pr}({\it \Sigma}_{l} = 3)$   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  ${\it \Sigma}_{l} = 2)$.		+ The extreme values  $({\it \Sigma}_{l} = 0$  and  ${\it \Sigma}_{l} = 3)$  occur less frequently than the inner values  $({\it \Sigma}_{l} = 1$  and  ${\it \Sigma}_{l} = 2)$.

	</quiz>		</quiz>

Hwang at 15:45, 16 May 2022

2022-05-16T15:45:57Z

← Older revision		Revision as of 17:45, 16 May 2022
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	''Notes:''		''Notes:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes\|Block Coding with 4B3T Codes]].		*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 '''L''' ("Low") and '''H''' ("High") in this learning tutorial. 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'''.		*The binary symbols are denoted by '''L''' ("Low") and '''H''' ("High") in this learning tutorial. 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.		*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  [[Applets:4B3T-Codes\|Principle of 4B3T coding]].		*You can check the results with the interaction module  [[Applets:4B3T-Codes\|"Principle of 4B3T coding"]].

Guenter at 14:12, 5 May 2022

2022-05-05T14:12:49Z

← Older revision		Revision as of 16:12, 5 May 2022
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	[[Category:Digital Signal Transmission: Exercises\|^2.3 ~~Blockweise~~ 4B3T~~-Codierung~~^]]		[[Category:Digital Signal Transmission: Exercises\|^2.3 Blockwise 4B3T Coding^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 2.6: Modifizierter MS43-Code to Aufgaben:Exercise 2.6: Modified MS43 Code

2022-05-05T14:10:01Z

Guenter moved page Aufgaben:Aufgabe 2.6: Modifizierter MS43-Code to Aufgaben:Exercise 2.6: Modified MS43 Code

← Older revision	Revision as of 16:10, 5 May 2022
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Hwang at 13:22, 20 April 2022

2022-04-20T13:22:23Z

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