Difference between revisions of "Aufgaben:Exercise 1.4Z: Modified MS43 Code"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Examples_of_Communication_Systems/ISDN_Basic_Access |
}} | }} | ||
− | [[File: | + | [[File:EN_Bei_Z_1_4.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 "$\rm U_{\rm K0}$" interface $($transmission path between the exchange and the NTBA$)$. | |
− | + | The abbreviation "MMS43" stands for "'''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 $l$ 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: | ||
+ | |||
+ | *The exercise belongs to the chapter [[Examples_of_Communication_Systems/ISDN_Basic_Access|"ISDN Basic Access"]]. | ||
+ | |||
+ | *Information about the MMS43 code can be found in the chapter [[Digital_Signal_Transmission/Block_Coding_with_4B3T_Codes|"Block Coding with 4B3T Codes"]] of the book "Digital signal transmission". | ||
+ | |||
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What are the reasons for using the 4B3T code instead of the redundancy-free binary code in ISDN? |
|type="[]"} | |type="[]"} | ||
− | - 4B3T | + | - 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. |
− | { | + | {Encode the binary sequence "$1100\hspace{0.08cm} 0100 \hspace{0.08cm} 0110 \hspace{0.08cm} 1010$" 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 } | ||
− | { | + | {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? |
|type="{}"} | |type="{}"} | ||
− | ${\rm Pr}(\ | + | ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 0 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=0) \ = \ $ { 0.375 3% } |
− | ${\rm Pr}(\ | + | ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 2 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=0) \ = \ $ { 0.1875 3% } |
− | ${\rm Pr}(\ | + | ${\rm Pr}({\it \Sigma}\hspace{0.05cm}_{l+1} = 0 \ | \ {\it \Sigma}\hspace{0.05cm}_{l}=2) \ = \ $ { 0 3% } |
− | { | + | {What properties follow from the Markov diagram? |
|type="[]"} | |type="[]"} | ||
− | - | + | - 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 $2$. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' <u>Statements 2 and 3</u> 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 $a_{12}\hspace{0.15cm} \underline{ = \ –1}$. | |
− | |||
+ | [[File:P_ID1341_Dig_A_2_6c.png|right|frame|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: | ||
− | |||
− | |||
:$${\rm Pr}({\it \Sigma}_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ 6/16 \underline{ \ = \ 0.375},$$ | :$${\rm Pr}({\it \Sigma}_{l+1} = 0 \ | \ {\it \Sigma}_{l}=0) \ = \ 6/16 \underline{ \ = \ 0.375},$$ | ||
:$${\rm Pr}({\it \Sigma}_{l+1} = 2 \ | \ {\it \Sigma}_{l}=0) \ = \ 3/16 \underline{ \ = \ 0.1875},$$ | :$${\rm Pr}({\it \Sigma}_{l+1} = 2 \ | \ {\it \Sigma}_{l}=0) \ = \ 3/16 \underline{ \ = \ 0.1875},$$ | ||
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− | '''(4)''' | + | |
− | * | + | '''(4)''' <u>Statements 2 and 3</u> 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 | + | 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: | ||
:$${\rm Pr}(0)= \frac{6}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(1)+ \frac{1}{16} \cdot {\rm Pr}(3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{9}{16} \cdot {\rm Pr}(0)= \frac{4}{16} \cdot {\rm Pr}(1).$$ | :$${\rm Pr}(0)= \frac{6}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(1)+ \frac{1}{16} \cdot {\rm Pr}(3)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\frac{9}{16} \cdot {\rm Pr}(0)= \frac{4}{16} \cdot {\rm Pr}(1).$$ | ||
− | + | *From the further condition ${\rm Pr}(0) + {\rm Pr}(1) = 1/2$ follows further: | |
:$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$ | :$${\rm Pr}(0)= {\rm Pr}(3)= \frac{9}{26}\hspace{0.05cm}, \hspace{0.2cm} {\rm Pr}(1)= {\rm Pr}(2)= \frac{4}{26}\hspace{0.05cm}.$$ | ||
− | + | :This calculation is based on the <u>sum of the incoming arrows in the "0" condition</u>. | |
+ | |||
+ | *One could also give equations for the other three states, but they all give the same result: | ||
:$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$ | :$${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$ | ||
:$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$ | :$$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$ | ||
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− | [[Category: | + | [[Category:Examples of Communication Systems: Exercises|^1.2 ISDN Basic Access^]] |
Latest revision as of 16:55, 24 October 2022
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:
- The exercise belongs to the chapter "ISDN Basic Access".
- Information about the MMS43 code can be found in the chapter "Block Coding with 4B3T Codes" of the book "Digital signal transmission".
Questions
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 $a_{12}\hspace{0.15cm} \underline{ = \ –1}$.
(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:
- $${\rm Pr}(1) \ = \ \frac{6}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{3}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
- $$ {\rm Pr}(2) \ = \ \frac{3}{16} \cdot {\rm Pr}(0) + \frac{6}{16} \cdot {\rm Pr}(1)+ \frac{6}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm},$$
- $$ {\rm Pr}(3) \ = \ \frac{1}{16} \cdot {\rm Pr}(0) + \frac{4}{16} \cdot {\rm Pr}(2)+\frac{6}{16} \cdot {\rm Pr}(3)\hspace{0.05cm}.$$