Difference between revisions of "Aufgaben:Exercise 1.4: Maximum Likelihood Decision"

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{{quiz-Header|Buchseite=Kanalcodierung/Kanalmodelle und Entscheiderstrukturen
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{{quiz-Header|Buchseite=Channel Coding/Channel Models and Decision Structures
  
 
}}
 
}}
  
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[[File:P_ID2384__KC_A_1_4.png|right|frame|Modell zur Maximum–Likelihood–Decodierung]]
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[[File:EN_KC_A_1_4_neu.png|right|frame|Maximum likelihood decoding model]]
  
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Wir betrachten das digitale Übertragungssystem entsprechend der Grafik. Berücksichtigt sind dabei:
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We consider the digital transmission system according to the graph.  Considered are:
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*ein systematischer  $(5, 2)$–Blockcode  $\mathcal{C}$  mit den Codeworten
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*a systematic  $(5, 2)$  block code  $\mathcal{C}$  with the code words
 
:$$\underline{x}_{0} \hspace{-0.1cm} \ =  \  \hspace{-0.1cm} (0, 0, 0, 0, 0) \hspace{0.05cm},$$ $$\underline{x}_{1} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (0, 1, 0, 1, 0) \hspace{0.05cm},$$ $$\underline{x}_{2} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (1, 0, 1, 0, 1) \hspace{0.05cm},$$ $$\underline{x}_{3} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (1, 1, 1, 1, 1) \hspace{0.05cm};$$
 
:$$\underline{x}_{0} \hspace{-0.1cm} \ =  \  \hspace{-0.1cm} (0, 0, 0, 0, 0) \hspace{0.05cm},$$ $$\underline{x}_{1} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (0, 1, 0, 1, 0) \hspace{0.05cm},$$ $$\underline{x}_{2} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (1, 0, 1, 0, 1) \hspace{0.05cm},$$ $$\underline{x}_{3} \hspace{-0.1cm} \  =  \  \hspace{-0.1cm} (1, 1, 1, 1, 1) \hspace{0.05cm};$$
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*ein digitales (binäres) Kanalmodell, das den Vektor  $\underline{x} \in {\rm GF} (2^{5})$  in den Vektor  $\underline{y} \in {\rm GF} (2^{5})$  verfälscht;
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*a digital (binary) channel model that changes the vector  $\underline{x} \in {\rm GF} (2^{5})$  over into the vector  $\underline{y} \in {\rm GF} (2^{5})$,
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*ein  [[Kanalcodierung/Kanalmodelle_und_Entscheiderstrukturen#Maximum-Likelihood.E2.80.93Entscheidung_beim_BSC.E2.80.93Kanal|Maximum–Likelihood–Decoder]]  mit der Entscheidungsregel
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*a  [[Channel_Coding/Channel_Models_and_Decision_Structures#Maximum_likelihood_decision_at_the_BSC_channel|Maximum likelihood decoder]]  (short:   ML decoder)  with the decision rule
 
:$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm}|\hspace{0.05cm} \underline{y} ) = {\rm arg} \min_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm H}(\underline{y} \hspace{0.05cm}, \hspace{0.1cm}\underline{x}_{\hspace{0.03cm}i}).$$
 
:$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm}|\hspace{0.05cm} \underline{y} ) = {\rm arg} \min_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm H}(\underline{y} \hspace{0.05cm}, \hspace{0.1cm}\underline{x}_{\hspace{0.03cm}i}).$$
  
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Hier bezeichnet  $d_{\rm H} (\underline{y}, \ \underline{x_{i}})$  die  [[Kanalcodierung/Zielsetzung_der_Kanalcodierung#Einige_wichtige_Definitionen_zur_Blockcodierung|Hamming–Distanz]]  zwischen dem Empfangswort  $\underline{y}$  und dem (möglicherweise) gesendeten Codewort  $\underline{x_{i}}$.
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Here,  $d_{\rm H} (\underline{y}, \ \underline{x_{i}})$  is the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|Hamming distance]]  between the received word  $\underline{y}$  and the  (possibly)  sent code word  $\underline{x_{i}}$.
  
  
  
  
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Note:  This exercise belongs to the chapter  [[Channel_Coding/Channel_Models_and_Decision_Structures|"Channel Models and Decision Structures"]].  
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''Hinweis'':
 
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*Die Aufgabe gehört zum Kapitel  [[Kanalcodierung/Kanalmodelle_und_Entscheiderstrukturen|Kanalmodelle und  Entscheiderstrukturen]].  
 
 
   
 
   
  
  
  
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===Fragebogen===
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===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
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{Es sei&nbsp; $\underline{y} = (1, 0, 0, 0, 1)$. Welche Entscheidungen erfüllen das Maximum–Likelihood–Kriterium?
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{Let&nbsp; $\underline{y} = (1, 0, 0, 0, 1)$.&nbsp; Which decisions satisfy the maximum likelihood criterion?
 
|type="[]"}
 
|type="[]"}
 
- $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
 
- $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
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{Es sei&nbsp; $\underline{y} = (0, 0, 0, 1, 0)$. Welche Entscheidungen erfüllen das Maximum–Likelihood–Kriterium?
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{Let&nbsp; $\underline{y} = (0, 0, 0, 1, 0)$.&nbsp; Which decisions satisfy the maximum likelihood criterion?
 
|type="[]"}
 
|type="[]"}
 
+ $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
 
+ $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
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- $\underline{z} = \underline{x}_{3} = (1, 1, 1, 1, 1)$.
 
- $\underline{z} = \underline{x}_{3} = (1, 1, 1, 1, 1)$.
  
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{Welche Entscheidung trifft der ML–Decoder für&nbsp; $\underline{y} = (1, 0, 1, 1, 1)$, wenn ihm mitgeteilt wird, dass die beiden letzten Symbole unsicher sind?
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{What decision does the maximum likelihood decoder make for&nbsp; $\underline{y} = (1, 0, 1, 1, 1)$ when it is told that the last two symbols are uncertain?
 
|type="[]"}
 
|type="[]"}
 
- $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
 
- $\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
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{Zu welchem Informationswort&nbsp; $v = (v_{1}, v_{2})$&nbsp; führt die Entscheidung gemäß der letzten Teilaufgabe?
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{What information word&nbsp; $v = (v_{1}, v_{2})$&nbsp; does the decision lead to according to the last subtask?
 
|type="{}"}
 
|type="{}"}
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$v_{1} \ = \ $ { 1 }
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$v_{1} \ = \ $ { 1 }
 
$v_{2} \ = \ $ { 0. }
 
$v_{2} \ = \ $ { 0. }
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</quiz>
 
</quiz>
  
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
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'''(1)'''&nbsp; Richtig ist die <u>Antwort 3</u>:
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'''(1)'''&nbsp; Correct&nbsp; <u>answer 3</u> &nbsp; &rArr; &nbsp;  $\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$:
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*Die Hamming–Distanzen zwischen dem spezifischen Empfangswort $\underline{y} = (1, 0, 0, 0, 1)$ und den vier möglichen Codeworten $\underline{x}_{i}$ ergeben sich wie folgt:
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*The Hamming distances between the specific received word $\underline{y} = (1, 0, 0, 0, 1)$ and the four possible code words $\underline{x}_{i}$ are as follows:
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 3\hspace{0.05cm}.$$
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 3\hspace{0.05cm}.$$
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*Entschieden wird sich für die Folge mit der geringsten Hamming–Distanz $d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1$.
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*A decision is made for the sequence with the smallest Hamming distance $d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1$.
  
  
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'''(2)'''&nbsp; Für $\underline{y} = (0, 0, 0, 1, 0)$ sind die <u>Antworten 1 und 2</u> richtig, wie die folgende Rechnung zeigt:
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 +
'''(2)'''&nbsp; For&nbsp; $\underline{y} = (0, 0, 0, 1, 0)$&nbsp; the&nbsp; <u>answers 1 and 2</u>&nbsp; are correct,&nbsp; as the following calculation shows:
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 4\hspace{0.05cm}.$$
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 4\hspace{0.05cm}.$$
  
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'''(3)'''&nbsp; Richtig ist die <u>Antwort 3</u>:
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*Entsprechend der Hamming–Distanz wäre eine Entscheidung zugunsten von $x_{2}$ genau so möglich wie für $x_{3}$, wenn der Vektor $\underline{y} = (1, 0, 1, 1, 1)$ empfangen wird:
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'''(3)'''&nbsp; Correct <u>answer 3</u> &nbsp; &rArr; &nbsp;  $\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$:
 +
*According to the Hamming distance,&nbsp;  a decision in favor of&nbsp; $x_{2}$&nbsp; would be just as possible as for&nbsp; $x_{3}$&nbsp; if the vector&nbsp; $\underline{y} = (1, 0, 1, 1, 1)$&nbsp; is received:
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 1\hspace{0.05cm}.$$
 
:$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 1\hspace{0.05cm}.$$
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*Der Empfangsvektor $\underline{y}$ unterscheidet sich aber von $x_{2}$ bezüglich des vierten Bits und von $x_{3}$ im zweiten Bit. Da das vierte Bit unsicherer ist als das zweite, wird er sich für $x_{2}$ entscheiden .
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*But the received vector&nbsp; $\underline{y}$&nbsp; is different from&nbsp; $x_{2}$&nbsp; with respect to the fourth bit and from&nbsp; $x_{3}$&nbsp; in the second bit.
 +
* Since the fourth bit is more uncertain than the second,&nbsp; it will choose&nbsp; $x_{2}$ .
 +
 
  
  
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'''(4)'''&nbsp; Da es sich hier um einen systematischen Code handelt, ist die Entscheidung für $\underline{z} = (1, 0, 1, 0, 1)$ gleichbedeutend mit der Entscheidung
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'''(4)'''&nbsp; Since this is a systematic code,&nbsp; the decision for&nbsp; $\underline{z} = (1, 0, 1, 0, 1)$&nbsp; is equivalent to the decision
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:$$v_{1}  \ \underline{ = 1}, \ v_{2} \ \underline{= 0}.$$
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:$$v_{1}  \ \underline{ = 1}, \ v_{2} \ \underline{= 0}.$$
  
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Es ist nicht sicher, dass $\underline{u} = (1, 0)$ tatsächlich gesendet wurde, aber die Wahrscheinlichkeit ist angesichts des Empfangsvektors $\underline{y} = (1, 0, 1, 1, 1)$ hierfür am größten.
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*It is not certain that&nbsp; $\underline{u} = (1, 0)$&nbsp; was actually sent.
 +
*But the probability is highest for this given the received vector $\underline{y} = (1, 0, 1, 1, 1)$.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu  Kanalcodierung|^1.2 Kanal und Entscheiderstrukturen
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[[Category:Channel Coding: Exercises|^1.2 Channel and Decision Device^]]
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^]]
 

Latest revision as of 17:54, 1 November 2022

Maximum likelihood decoding model

We consider the digital transmission system according to the graph.  Considered are:

  • a systematic  $(5, 2)$  block code  $\mathcal{C}$  with the code words
$$\underline{x}_{0} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (0, 0, 0, 0, 0) \hspace{0.05cm},$$ $$\underline{x}_{1} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (0, 1, 0, 1, 0) \hspace{0.05cm},$$ $$\underline{x}_{2} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (1, 0, 1, 0, 1) \hspace{0.05cm},$$ $$\underline{x}_{3} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} (1, 1, 1, 1, 1) \hspace{0.05cm};$$
  • a digital (binary) channel model that changes the vector  $\underline{x} \in {\rm GF} (2^{5})$  over into the vector  $\underline{y} \in {\rm GF} (2^{5})$,
  • a  Maximum likelihood decoder  (short:   ML decoder)  with the decision rule
$$\underline{z} = {\rm arg} \max_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} {\rm Pr}( \underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm}|\hspace{0.05cm} \underline{y} ) = {\rm arg} \min_{\underline{x}_{\hspace{0.03cm}i} \hspace{0.05cm} \in \hspace{0.05cm} \mathcal{C}} \hspace{0.1cm} d_{\rm H}(\underline{y} \hspace{0.05cm}, \hspace{0.1cm}\underline{x}_{\hspace{0.03cm}i}).$$

Here,  $d_{\rm H} (\underline{y}, \ \underline{x_{i}})$  is the  Hamming distance  between the received word  $\underline{y}$  and the  (possibly)  sent code word  $\underline{x_{i}}$.



Note:  This exercise belongs to the chapter  "Channel Models and Decision Structures".



Questions

1

Let  $\underline{y} = (1, 0, 0, 0, 1)$.  Which decisions satisfy the maximum likelihood criterion?

$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1, 1)$.

2

Let  $\underline{y} = (0, 0, 0, 1, 0)$.  Which decisions satisfy the maximum likelihood criterion?

$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1, 1)$.

3

What decision does the maximum likelihood decoder make for  $\underline{y} = (1, 0, 1, 1, 1)$ when it is told that the last two symbols are uncertain?

$\underline{z} = \underline{x}_{0} = (0, 0, 0, 0, 0)$,
$\underline{z} = \underline{x}_{1} = (0, 1, 0, 1, 0)$,
$\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$,
$\underline{z} = \underline{x}_{3} = (1, 1, 1, 1, 1)$.

4

What information word  $v = (v_{1}, v_{2})$  does the decision lead to according to the last subtask?

$v_{1} \ = \ $

$v_{2} \ = \ $


Solution

Solution

(1)  Correct  answer 3   ⇒   $\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$:

  • The Hamming distances between the specific received word $\underline{y} = (1, 0, 0, 0, 1)$ and the four possible code words $\underline{x}_{i}$ are as follows:
$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 3\hspace{0.05cm}.$$
  • A decision is made for the sequence with the smallest Hamming distance $d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1$.


(2)  For  $\underline{y} = (0, 0, 0, 1, 0)$  the  answers 1 and 2  are correct,  as the following calculation shows:

$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 4\hspace{0.05cm}.$$


(3)  Correct answer 3   ⇒   $\underline{z} = \underline{x}_{2} = (1, 0, 1, 0, 1)$:

  • According to the Hamming distance,  a decision in favor of  $x_{2}$  would be just as possible as for  $x_{3}$  if the vector  $\underline{y} = (1, 0, 1, 1, 1)$  is received:
$$d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_0) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_1) = 4\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_2) = 1\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{y}, \hspace{0.05cm}\underline{x}_3) = 1\hspace{0.05cm}.$$
  • But the received vector  $\underline{y}$  is different from  $x_{2}$  with respect to the fourth bit and from  $x_{3}$  in the second bit.
  • Since the fourth bit is more uncertain than the second,  it will choose  $x_{2}$ .


(4)  Since this is a systematic code,  the decision for  $\underline{z} = (1, 0, 1, 0, 1)$  is equivalent to the decision

$$v_{1} \ \underline{ = 1}, \ v_{2} \ \underline{= 0}.$$
  • It is not certain that  $\underline{u} = (1, 0)$  was actually sent.
  • But the probability is highest for this given the received vector $\underline{y} = (1, 0, 1, 1, 1)$.

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