Aufgaben:Exercise 1.4Z: Modified MS43 Code - Revision history

Guenter at 15:55, 24 October 2022

2022-10-24T15:55:24Z

Guenter at 10:12, 20 October 2022

2022-10-20T10:12:49Z

Guenter at 10:08, 20 October 2022

2022-10-20T10:08:39Z

Guenter at 10:00, 20 October 2022

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Hwang at 16:03, 9 October 2022

2022-10-09T16:03:59Z

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Guenter at 12:37, 10 September 2022

2022-09-10T12:37:43Z

Guenter: Guenter moved page Aufgaben:Aufgabe 1.4Z: Modifizierter MS43–Code to Aufgaben:Exercise 1.4Z: Modified MS43 Code

2022-09-09T15:25:17Z

Guenter moved page Aufgabe 1.4Z: Modifizierter MS43–Code to Exercise 1.4Z: Modified MS43 Code

Javier: Text replacement - "”" to """

2021-05-28T14:28:57Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:21:25Z

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Javier: Text replacement - "Category:Aufgaben zu Beispiele von Nachrichtensystemen" to "Category:Examples of Communication Systems: Exercises"

2021-03-23T12:49:59Z

Text replacement - "Category:Aufgaben zu Beispiele von Nachrichtensystemen" to "Category:Examples of Communication Systems: Exercises"

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Dig_A_2_6_v2.png|right|frame|Code table of the MMS43 code]]
+[[File:EN_Bei_Z_1_4.png|right|frame|Code table of the MMS43 code]]
 For ISDN data transmission,&nbsp; the MMS43 code is used in Germany and Belgium on the so-called&nbsp; "$\rm U_{\rm K0}$"&nbsp; interface &nbsp;$($transmission path between the exchange and the NTBA$)$.&nbsp;

@@ Line 100: / Line 100: @@
 '''(4)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; are correct:
-*The first statement is false, which can be seen from the asymmetries in the Markov diagram.
+*The first statement is false,&nbsp; which can be seen from the asymmetries in 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 ${\rm Pr}(3) = {\rm Pr}(0)$ 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 ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$ t follows further:
+*From the further condition&nbsp; ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$&nbsp; follows further:
 :$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$
-This calculation is based on the <u>sum of the incoming arrows in the "0" condition</u>.
+: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, but they all give the same result:
+*One could also give equations for the other three states,&nbsp; but they all give the same result:
 :$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
 :$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$

@@ Line 39: / Line 39: @@
 |type="[]"}
 - 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&nbsp; $H_{\rm K}(f = 0) = 0$.&nbsp;
++ The transmitted signal should be free of DC signals if the channel frequency response&nbsp; $H_{\rm K}(f = 0) = 0$.&nbsp;
 + A small symbol rate&nbsp; $(1/T)$&nbsp; allows a longer cable length.
@@ Line 63: / Line 63: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Statements 2 and 3</u> are correct:
+'''(1)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; 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 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.
-'''(2)'''&nbsp; With the initial value ${\it \Sigma}_{0} = 0$, the 4B3T coding results in:
+*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.
 * '''1100''' &nbsp; &rArr; &nbsp; "+  +  +" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
@@ Line 82: / Line 85: @@
-Thus, the amplitude coefficient we are looking for is $a_{12}\hspace{0.15cm} \underline{ = \ –1}$.
+Thus,&nbsp; the amplitude coefficient we are looking for is&nbsp; $a_{12}\hspace{0.15cm} \underline{ = \ –1}$.
 [[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, one can determine the following Markov diagram.
+'''(3)'''&nbsp; From the coloring of the given code table,&nbsp; one can determine the following Markov diagram.
-*From it, the transition probabilities we are looking for can be read:
+*From it,&nbsp; the transition probabilities we are looking for can be read:
 :$${\rm Pr}({\it \Sigma}_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ 6/16 \underline{ \ = \ 0.375},$$
@@ Line 96: / Line 99: @@
-'''(4)'''&nbsp; <u>Statements 2 and 3</u> are correct:
+'''(4)'''&nbsp; <u>Statements 2 and 3</u>&nbsp; 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".

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1583__Bei_Z_1_4.png|right|frame|Code table of the MMS43 code]]
+[[File:EN_Dig_A_2_6_v2.png|right|frame|Code table of the MMS43 code]]
-For ISDN data transmission, the MMS43 code is used in Germany and Belgium on the so-called&nbsp; $\rm U_{\rm K0}$ interface &nbsp;(transmission path between the exchange and the NTBA).&nbsp;
+For ISDN data transmission,&nbsp; the MMS43 code is used in Germany and Belgium on the so-called&nbsp; "$\rm U_{\rm K0}$"&nbsp; interface &nbsp;$($transmission path between the exchange and the NTBA$)$.&nbsp;
-The abbreviation "MMS43" stands for '''M'''odified '''M'''onitored '''S'''um '''4'''B'''3'''T.
+The abbreviation&nbsp; "MMS43"&nbsp; stands for&nbsp; "'''M'''odified '''M'''onitored '''S'''um '''4'''B'''3'''T".
-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&nbsp; $l$&nbsp; blocks)
+This is a 4B3T block code with the four code tables shown in the graphic,&nbsp; which are used for coding according to the so-called "running digital sum"&nbsp; $($after&nbsp; $l$&nbsp; 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 wird&nbsp; ${\it \Sigma}_{0} = 0$&nbsp; is used.
+For initialization:&nbsp; ${\it \Sigma}_{0} = 0$&nbsp; is used.
 The colorings in the graph mean:
-*If the running digital sum does not change&nbsp; $({\it \Sigma}\hspace{0.05cm}_{l+1} = {\it \Sigma}\hspace{0.05cm} _{l})$, a field is grayed out.
+*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.
@@ Line 24: / Line 24: @@
 *The exercise belongs to the chapter&nbsp; [[Examples_of_Communication_Systems/ISDN_Basic_Access|"ISDN Basic Access"]].
-''Notes:''
+*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".
@@ Line 37: / Line 36: @@
 <quiz display=simple>
-{What are the reasons for using a 4B3T code instead of the redundancy-free binary code in ISDN?
+{What are the reasons for using the 4B3T code instead of the redundancy-free binary code in ISDN?
 |type="[]"}
 - 4B3T is in principle better than the redundancy-free binary code.
-+ The transmitted signal should be free of equal signals if the channel frequency response&nbsp; $H_{\rm K}(f = 0) = 0$.&nbsp;
++ The transmitted signal should be free of DC-signals if the channel frequency response&nbsp; $H_{\rm K}(f = 0) = 0$.&nbsp;
 + A small symbol rate&nbsp; $(1/T)$&nbsp; allows a longer cable length.
-{Code 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?
+{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="{}"}
 $a_{12} \ = \ $ { -1.03--0.97 }
-{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?
+{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="{}"}
 ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 0 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=0) \ = \ $ { 0.375 3% }
@@ Line 55: / Line 54: @@
 {What properties follow from the Markov diagram?
 |type="[]"}
-- 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.
+- 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.
-+ &nbsp; ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$&nbsp; and&nbsp; ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 1) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 2)$ are valid.
++ ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 0) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 3)$ &nbsp; and &nbsp; ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 1) = {\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l} = 2)$&nbsp;  are valid.
 + The extreme values&nbsp; $(0$ or $3)$&nbsp; occur less frequently than&nbsp; $1$&nbsp; or&nbsp; $2$.

← Older revision		Revision as of 12:37, 10 September 2022
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−	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN~~-Basisanschluss~~^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN Basic Access^]]

← Older revision	Revision as of 15:25, 9 September 2022
(No difference)

@@ Line 74: / Line 74: @@
 '''(2)'''&nbsp; Die 4B3T–Codierung ergibt mit dem Initialwert ${\it \Sigma}_{0} = 0$:
-* '''1100''' &nbsp; &rArr; &nbsp; "+  +  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
+* '''1100''' &nbsp; &rArr; &nbsp; "+  +  +" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
-* '''0100''' &nbsp; &rArr; &nbsp; " –  +  '''0'''&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
+* '''0100''' &nbsp; &rArr; &nbsp; " –  +  '''0'''" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
-* '''0110''' &nbsp; &rArr; &nbsp; "–  –  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
+* '''0110''' &nbsp; &rArr; &nbsp; "–  –  +" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
-* '''1010''' &nbsp; &rArr; &nbsp; "+  –  –&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.
+* '''1010''' &nbsp; &rArr; &nbsp; "+  –  –" &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.

@@ Line 74: / Line 74: @@
 '''(2)'''&nbsp; Die 4B3T–Codierung ergibt mit dem Initialwert ${\it \Sigma}_{0} = 0$:
-* '''1100''' &nbsp; &rArr; &nbsp; &bdquo;+  +  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
+* '''1100''' &nbsp; &rArr; &nbsp; "+  +  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{1} = 3$,
-* '''0100''' &nbsp; &rArr; &nbsp; &bdquo; –  +  '''0'''&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
+* '''0100''' &nbsp; &rArr; &nbsp; " –  +  '''0'''&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{2} = 3$,
-* '''0110''' &nbsp; &rArr; &nbsp; &bdquo;–  –  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
+* '''0110''' &nbsp; &rArr; &nbsp; "–  –  +&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{3} = 2$,
-* '''1010''' &nbsp; &rArr; &nbsp; &bdquo;+  –  –&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.
+* '''1010''' &nbsp; &rArr; &nbsp; "+  –  –&rdquo; &nbsp; &rArr; &nbsp; ${\it \Sigma}_{4} = 1$.
@@ Line 106: / Line 106: @@
 Aus der weiteren Bedingung ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$ folgt weiter:
 :$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$
-Diese Berechnung basiert auf der <u>Summe der ankommenden Pfeile im Zustand &bdquo;0”</u>.
+Diese Berechnung basiert auf der <u>Summe der ankommenden Pfeile im Zustand "0”</u>.
 Man könnte auch Gleichungen für die drei anderen Zustände angeben, die aber alle zum gleichen Ergebnis führen:

← Older revision		Revision as of 12:49, 23 March 2021
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−	[[Category:~~Aufgaben zu Beispiele von Nachrichtensystemen~~\|^1.2 ISDN-Basisanschluss^]]	+	[[Category:Examples of Communication Systems: Exercises\|^1.2 ISDN-Basisanschluss^]]