Difference between revisions of "Aufgaben:Exercise 1.4Z: Modified MS43 Code"

From LNTwww
 
(15 intermediate revisions by 4 users not shown)
Line 1: Line 1:
  
{{quiz-Header|Buchseite=Beispiele von Nachrichtensystemen/ISDN-Basisanschluss
+
{{quiz-Header|Buchseite=Examples_of_Communication_Systems/ISDN_Basic_Access
 
}}
 
}}
  
[[File:P_ID1583__Bei_Z_1_4.png|right|frame|Codetabelle des MMS43-Codes]]
+
[[File:EN_Bei_Z_1_4.png|right|frame|Code table of the MMS43 code]]
Bei der ISDN–Datenübertragung wird in Deutschland und Belgien auf der so genannten $\rm U_{\rm K0}$–Schnittstelle (Übertragungsstrecke zwischen der Vermittlungsstelle und dem NTBA) der MMS43–Code eingesetzt. Die Abkürzung „MMS43” steht für '''M'''odified '''M'''onitored '''S'''um '''4'''B'''3'''T.
+
For ISDN data transmission,  the MMS43 code is used in Germany and Belgium on the so-called  "$\rm U_{\rm K0}$"  interface  $($transmission path between the exchange and the NTBA$)$. 
  
Es handelt sich hierbei um einen 4B3T–Blockcode mit den vier in der Grafik gezeigten Codetabellen, die gemäß der so genannten „Laufenden Digitalen Summe” (nach $l$–Blöcken)
+
The abbreviation  "MMS43"  stands for  "'''M'''odified '''M'''onitored '''S'''um '''4'''B'''3'''T".
:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.05cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu$$
 
zur Codierung benutzt werden. Zur Initialisierung wird ${\it \Sigma}_{0} = 0$ verwendet.
 
  
 +
This is a 4B3T block code with the four code tables shown in the graphic,  which are used for coding according to the so-called "running digital sum"  $($after  $l$  blocks$)$:
 +
:$${\it \Sigma}\hspace{0.05cm}_l = \sum_{\nu = 1}^{3 \hspace{0.05cm}\cdot \hspace{0.05cm} l}\hspace{0.02cm} a_\nu$$
 +
For initialization:  ${\it \Sigma}_{0} = 0$  is used.
  
Die Farbgebungen in der Grafik bedeuten:
 
*Ändert sich die laufende digitale Summe nicht $({\it \Sigma}_{l+1} = {\it \Sigma} _{l})$, so ist ein Feld grau hinterlegt.
 
*Eine Zunahme $({\it \Sigma}_{l+1} > {\it \Sigma}_{l})$ ist rot hinterlegt, eine Abnahme $({\it \Sigma}_{l+1} < {\it \Sigma} _{l})$ blau.
 
*Je intensiver diese Farben sind, um so größer ist die Änderung der laufenden digitalen Summe.
 
  
 +
The colorings in the graph mean:
 +
*If the running digital sum does not change &nbsp; $({\it \Sigma}\hspace{0.05cm}_{l+1} = {\it \Sigma}\hspace{0.05cm} _{l})$,&nbsp; a field is grayed out.
  
 +
*An increase &nbsp; $({\it \Sigma}\hspace{0.05cm}_{l+1} > {\it \Sigma}\hspace{0.05cm}_{l})$ &nbsp; is highlighted in red,&nbsp; a decrease &nbsp; $({\it \Sigma}\hspace{0.05cm}_{l+1} < {\it \Sigma}\hspace{0.05cm} _{l})$ &nbsp; in blue.
  
''Hinweise:''
+
*The more intense these colors are,&nbsp; the greater the change in the running digital sum.
  
*Die Aufgabe gehört zum Kapitel [[Beispiele_von_Nachrichtensystemen/ISDN-Basisanschluss|ISDN-Basisanschluss]].  
+
 
*Angaben zum MMS43–Code finden Sie im Kapitel [[Digitalsignalübertragung/Blockweise_Codierung_mit_4B3T-Codes|Blockweise Codierung mit 4B3T-Codes]] des Buches „Digitalsignalübertragung”.
+
 
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
+
 
 +
 
 +
Notes:
 +
 
 +
*The exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/ISDN_Basic_Access|"ISDN Basic Access"]].
 +
 +
*Information about the MMS43 code can be found in the chapter&nbsp; [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes|"Block Coding with 4B3T Codes"]]&nbsp; of the book&nbsp; "Digital signal transmission".
 +
  
  
 
   
 
   
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Was sind Gründe dafür, dass bei ISDN statt des redundanzfreien Binärcodes ein 4B3T–Code verwendet wird?
+
{What are the reasons for using the 4B3T code instead of the redundancy-free binary code in ISDN?
 
|type="[]"}
 
|type="[]"}
- 4B3T ist prinzipiell besser als der redundanzfreie Binärcode.
+
- 4B3T is in principle better than the redundancy-free binary code.
+ Das Sendesignal sollte gleichsignalfrei sein wenn $H_{\rm K}(f = 0) = 0$ ist.  
+
+ The transmitted signal should be free of DC signals if the channel frequency response&nbsp; $H_{\rm K}(f = 0) = 0$.&nbsp;
+ Eine kleine Symbolrate $(1/T)$ ermöglicht größere Kabellänge.
+
+ A small symbol rate&nbsp; $(1/T)$&nbsp; allows a longer cable length.
  
{Codieren Sie die Binärfolge $1100\hspace{0.08cm} 0100 \hspace{0.08cm} 0110 \hspace{0.08cm} 1010$ gemäß der Tabelle. Wie lautet der Koeffizient des dritten Ternärsymbols des vierten Blocks?
+
{Encode the binary sequence&nbsp; "$1100\hspace{0.08cm} 0100 \hspace{0.08cm} 0110 \hspace{0.08cm} 1010$"&nbsp; according to the table. <br>What is the coefficient of the third ternary symbol of the fourth block?
 
|type="{}"}
 
|type="{}"}
 
$a_{12} \ = \ $ { -1.03--0.97 }
 
$a_{12} \ = \ $ { -1.03--0.97 }
  
{Ermitteln Sie das Markovdiagramm für den Übergang von $\Sigma_{l}$ auf $\Sigma_{l+1}$. Welche Übergangswahrscheinlichkeiten ergeben sich?
+
{Determine the Markov diagram for the transition from&nbsp; ${\it \Sigma}\hspace{0.05cm}_{l}$&nbsp; to&nbsp; ${\it \Sigma}\hspace{0.05cm}_{l+1}$.&nbsp; What are the transition probabilities?
 
|type="{}"}
 
|type="{}"}
${\rm Pr}(\Sigma_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ $ { 0.375 3% }
+
${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 0 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=0) \ = \ $ { 0.375 3% }
${\rm Pr}(\Sigma_{l+1} = 2 \ | \ {\it \Sigma}_{l}=0) \ = \ $ { 0.1875 3% }
+
${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 2 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=0) \ = \ $ { 0.1875 3% }
${\rm Pr}(\Sigma_{l+1} = 0 \ | \ {\it \Sigma}_{l}=2) \ = \ $ { 0 3% }
+
${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 0 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=2) \ = \ $ { 0 3% }
  
{Welche Eigenschaften folgen aus dem Markovdiagramm?
+
{What properties follow from the Markov diagram?
 
|type="[]"}
 
|type="[]"}
- Die Wahrscheinlichkeiten ${\rm Pr}({\it \Sigma}_{l} = 0), \text{ ...} \  , {\rm Pr}({\it \Sigma}_{l} = 3)$ sind gleich.
+
- The probabilities &nbsp; ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0), \text{ ...} \  , {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$ &nbsp; are equal.
+ Es gilt ${\rm Pr}(\Sigma_{l} = 0) = {\rm Pr}({\it \Sigma}_{l} = 3)$ und ${\rm Pr}(\Sigma_{l} = 1) = {\rm Pr}({\it \Sigma}_{l} = 2)$.
+
+ ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$ &nbsp; and &nbsp; ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 1) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 2)$&nbsp;  are valid.
+ Die Extremwerte ($0$ bzw. $3$) treten seltener auf als $1$ oder $2$.
+
+ The extreme values&nbsp; $(0$ or $3)$&nbsp; occur less frequently than&nbsp; $1$&nbsp; or&nbsp; $2$.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
  
'''(1)'''&nbsp; Richtig sind die <u>Aussagen 2 und 3</u>:
+
'''(1)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
*Die erste Aussage trifft nicht zu: Beispielsweise ergibt sich beim AWGN–Kanal (additives weißes Gaußsches Rauschen) mit einem 4B3T–Code im Vergleich zum redundanzfreien Binärcode eine deutlich größere Fehlerwahrscheinlichkeit aufgrund der ternären Entscheidung.  
+
*The first statement is not true:&nbsp; For example,&nbsp; the AWGN channel&nbsp; ("additive white Gaussian noise") with a 4B3T code results in a much larger error probability due to the ternary decision compared to the redundancy-free binary code.
*Der wesentliche Grund für die Verwendung eines redundanten Übertragungscodes ist vielmehr, dass über einen „Telefonkanal” kein Gleichsignalanteil übertragen werden kann.
+
 
*Auch die um $25 \%$ kleinere Schrittgeschwindigkeit $(1/T)$ des 4B3T–Codes kommt den Übertragungseigenschaften von Kupferleitungen (starker Dämpfungsanstieg mit der Frequenz) entgegen.  
+
*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".
*Bei gegebener Leitungsdämpfung lässt sich deshalb mit dem 4B3T–Code eine größere Länge überbrücken als mit einem redundanzfreien Binärsignal.  
+
 
 +
*The&nbsp; $25 \%$&nbsp; smaller symbol rate&nbsp; $(1/T)$&nbsp; of the 4B3T code also accommodates the transmission characteristics of copper lines&nbsp; (strong increase in attenuation with frequency).
 +
 
 +
*For a given line attenuation,&nbsp; therefore,&nbsp; a greater length can be bridged with the 4B3T code than with a redundancy-free binary signal.  
 +
 
 +
 
 +
 
 +
'''(2)'''&nbsp; With the initial value&nbsp; ${\it \Sigma}_{0} = 0$,&nbsp; the 4B3T coding results in:
  
 +
* '''1100''' &nbsp; &rArr; &nbsp; "+  +  +" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
  
'''(2)'''&nbsp; Die 4B3T–Codierung ergibt mit dem Initialwert ${\it \Sigma}_{0} = 0$:
+
* '''0100''' &nbsp; &rArr; &nbsp; " –  +  '''0'''" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
  
* '''1100''' &nbsp; &rArr; &nbsp; &bdquo;+ + +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
+
* '''0110''' &nbsp; &rArr; &nbsp; "– +" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
  
* '''0100''' &nbsp; &rArr; &nbsp; &bdquo; –  +  '''0'''&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
+
* '''1010''' &nbsp; &rArr; &nbsp; "+  –  –" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.
  
* '''0110''' &nbsp; &rArr; &nbsp; &bdquo;–  –  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
 
  
* '''1010''' &nbsp; &rArr; &nbsp; &bdquo;+  –  –&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.
+
Thus,&nbsp; the amplitude coefficient we are looking for is&nbsp; $a_{12}\hspace{0.15cm} \underline{ = \ –1}$.
  
  
Der gesuchte Amplitudenkoeffizient ist somit $a_{12}\ \underline{ = –1}$.
 
  
 +
[[File:P_ID1341_Dig_A_2_6c.png|right|frame|Markov diagram for the MMS43 code]]
 +
'''(3)'''&nbsp; From the coloring of the given code table,&nbsp; one can determine the following Markov diagram.
 +
*From it,&nbsp; the transition probabilities we are looking for can be read:
  
'''(3)'''&nbsp; Aus der Farbgebung der vorgegebenen Codetabelle kann man das folgende Markovdiagramm ermitteln. Daraus können die gesuchten Übergangswahrscheinlichkeiten abgelesen werden:
 
[[File:P_ID1341_Dig_A_2_6c.png|right|frame|Markovdiagramm für den MMS43-Code]]
 
 
:$${\rm Pr}({\it \Sigma}_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ 6/16 \underline{ \ = \ 0.375},$$  
 
:$${\rm Pr}({\it \Sigma}_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ 6/16 \underline{ \ = \ 0.375},$$  
 
:$${\rm Pr}({\it \Sigma}_{l+1} = 2 \ | \ {\it \Sigma}_{l}=0) \ = \ 3/16 \underline{ \ = \ 0.1875},$$
 
:$${\rm Pr}({\it \Sigma}_{l+1} = 2 \ | \ {\it \Sigma}_{l}=0) \ = \ 3/16 \underline{ \ = \ 0.1875},$$
Line 84: Line 98:
  
  
'''(4)'''&nbsp; Richtig sind <u>die Aussagen 2 und 3</u>:
+
 
*Die erste Aussage ist falsch, was man an den Asymmetrien im Markovdiagramm erkennt.  
+
'''(4)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
*Dagegen gibt es Symmetrien bezüglich der Zustände „0” und „3” und zwischen „1” und „2”.
+
*The first statement is false,&nbsp; which can be seen from 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; and between&nbsp; "1"&nbsp; and&nbsp; "2".
  
  
In der folgenden Berechnung schreiben wir anstelle von ${\rm Pr}({\it \Sigma}_{l} = 0)$ vereinfachend ${\rm Pr}(0)$. Unter Ausnutzung der Eigenschaft ${\rm Pr}(3) = {\rm Pr}(0)$ und ${\rm Pr}(2) = {\rm Pr}(1)$ ergeben sich folgende Gleichungen aus dem Markovdiagramm:
+
In the following calculation,&nbsp; instead of&nbsp; ${\rm Pr}({\it \Sigma}_{l} = 0)$,&nbsp; we write&nbsp; ${\rm Pr}(0)$&nbsp; in a simplified way.
 +
* Taking advantage of the properties&nbsp; ${\rm Pr}(3) = {\rm Pr}(0)$&nbsp; and&nbsp; ${\rm Pr}(2) = {\rm Pr}(1)$,&nbsp; we get the following equations from the Markov diagram:
 
:$${\rm Pr}(0)= \frac{6}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(1)+ \frac{1}{16} \cdot {\rm Pr}(3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{9}{16} \cdot {\rm Pr}(0)= \frac{4}{16} \cdot {\rm Pr}(1).$$
 
:$${\rm Pr}(0)= \frac{6}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(1)+ \frac{1}{16} \cdot {\rm Pr}(3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{9}{16} \cdot {\rm Pr}(0)= \frac{4}{16} \cdot {\rm Pr}(1).$$
Aus der weiteren Bedingung ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$ folgt weiter:
+
*From the further condition&nbsp; ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$&nbsp; follows further:
 
:$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$
 
:$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$
Diese Berechnung basiert auf der <u>Summe der ankommenden Pfeile im Zustand &bdquo;0”</u>. Man könnte auch Gleichungen für die drei anderen Zustände angeben, die aber alle zum gleichen Ergebnis führen:
+
:This calculation is based on the&nbsp; <u>sum of the incoming arrows in the "0" condition</u>.  
 +
 
 +
*One could also give equations for the other three states,&nbsp; but they all give the same result:
 
:$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
 
:$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
 
:$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
 
:$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
Line 103: Line 122:
  
  
[[Category:Aufgaben zu Beispiele von Nachrichtensystemen|^1.2 ISDN-Basisanschluss^]]
+
[[Category:Examples of Communication Systems: Exercises|^1.2 ISDN Basic Access^]]

Latest revision as of 16:55, 24 October 2022

Code table of the MMS43 code

For ISDN data transmission,  the MMS43 code is used in Germany and Belgium on the so-called  "$\rm U_{\rm K0}$"  interface  $($transmission path between the exchange and the NTBA$)$. 

The abbreviation  "MMS43"  stands for  "Modified Monitored Sum 4B3T".

This is a 4B3T block code with the four code tables shown in the graphic,  which are used for coding according to the so-called "running digital sum"  $($after  $l$  blocks$)$:

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

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


The colorings in the graph mean:

  • If the running digital sum does not change   $({\it \Sigma}\hspace{0.05cm}_{l+1} = {\it \Sigma}\hspace{0.05cm} _{l})$,  a field is grayed out.
  • An increase   $({\it \Sigma}\hspace{0.05cm}_{l+1} > {\it \Sigma}\hspace{0.05cm}_{l})$   is highlighted in red,  a decrease   $({\it \Sigma}\hspace{0.05cm}_{l+1} < {\it \Sigma}\hspace{0.05cm} _{l})$   in blue.
  • The more intense these colors are,  the greater the change in the running digital sum.



Notes:



Questions

1

What are the reasons for using the 4B3T code instead of the redundancy-free binary code in ISDN?

4B3T is in principle better than the redundancy-free binary code.
The transmitted signal should be free of DC signals if the channel frequency response  $H_{\rm K}(f = 0) = 0$. 
A small symbol rate  $(1/T)$  allows a longer cable length.

2

Encode the binary sequence  "$1100\hspace{0.08cm} 0100 \hspace{0.08cm} 0110 \hspace{0.08cm} 1010$"  according to the table.
What is the coefficient of the third ternary symbol of the fourth block?

$a_{12} \ = \ $

3

Determine the Markov diagram for the transition from  ${\it \Sigma}\hspace{0.05cm}_{l}$  to  ${\it \Sigma}\hspace{0.05cm}_{l+1}$.  What are the transition probabilities?

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

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

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

4

What properties follow from the Markov diagram?

The probabilities   ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0), \text{ ...} \ , {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$   are equal.
${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$   and   ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 1) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 2)$  are valid.
The extreme values  $(0$ or $3)$  occur less frequently than  $1$  or  $2$.


Solution

(1)  Statements 2 and 3  are correct:

  • The first statement is not true:  For example,  the AWGN channel  ("additive white Gaussian noise") 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  $25 \%$  smaller symbol rate  $(1/T)$  of the 4B3T code also accommodates the transmission characteristics of copper lines  (strong increase in attenuation with frequency).
  • For a given line attenuation,  therefore,  a greater length can be bridged with the 4B3T code than with a redundancy-free binary signal.


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

  • 1100   ⇒   "+ + +"   ⇒   ${\it \Sigma}_{1} = 3$,
  • 0100   ⇒   " – + 0"   ⇒   ${\it \Sigma}_{2} = 3$,
  • 0110   ⇒   "– – +"   ⇒   ${\it \Sigma}_{3} = 2$,
  • 1010   ⇒   "+ – –"   ⇒   ${\it \Sigma}_{4} = 1$.


Thus,  the amplitude coefficient we are looking for is  $a_{12}\hspace{0.15cm} \underline{ = \ –1}$.


Markov diagram for the MMS43 code

(3)  From the coloring of the given code table,  one can determine the following Markov diagram.

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


(4)  Statements 2 and 3  are correct:

  • The first statement is false,  which can be seen from the asymmetries in the Markov diagram.
  • On the other hand,  there are symmetries with respect to the states  "0"  and  "3"  and 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 properties  ${\rm Pr}(3) = {\rm Pr}(0)$  and  ${\rm Pr}(2) = {\rm Pr}(1)$,  we get the following equations from the Markov diagram:
$${\rm Pr}(0)= \frac{6}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(1)+ \frac{1}{16} \cdot {\rm Pr}(3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{9}{16} \cdot {\rm Pr}(0)= \frac{4}{16} \cdot {\rm Pr}(1).$$
  • From the further condition  ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$  follows further:
$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$
This calculation is based on the  sum of the incoming arrows in the "0" condition.
  • One could also give equations for the other three states,  but they all give the same result:
$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
$$ {\rm Pr}(3) \ = \ \frac{1}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm}.$$