Aufgaben:Exercise 3.10: Metric Calculation - Revision history

Guenter at 13:53, 18 November 2022

2022-11-18T13:53:14Z

Guenter at 13:29, 18 November 2022

2022-11-18T13:29:49Z

Hwang at 19:10, 1 November 2022

2022-11-01T19:10:49Z

Noah at 19:54, 18 October 2022

2022-10-18T19:54:20Z

Noah at 19:18, 18 October 2022

2022-10-18T19:18:41Z

Noah at 19:18, 18 October 2022

2022-10-18T19:18:31Z

Guenter at 12:16, 10 October 2022

2022-10-10T12:16:19Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.10: Fehlergrößenberechnung to Aufgaben:Exercise 3.10: Metric Calculation

2022-10-10T12:09:09Z

Guenter moved page Aufgabe 3.10: Fehlergrößenberechnung to Exercise 3.10: Metric Calculation

Javier: Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

2021-03-23T12:48:22Z

Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

Javier: Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

2020-07-09T12:54:30Z

Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

@@ Line 78: / Line 78: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; At all nodes $S_{\mu}$ a decision must be made between the two incoming branches. The branch that led to the (minimum) error metric ${\it \Gamma}_5(S_{\mu})$ is then selected in each case. With $\underline{y}_5 = (01)$ one obtains:
+'''(1)'''&nbsp; At all nodes&nbsp; $S_{\mu}$&nbsp; a decision must be made between the two incoming branches.&nbsp; The branch that led to the (minimum) error metric&nbsp; ${\it \Gamma}_5(S_{\mu})$&nbsp; is then selected in each case.&nbsp; With&nbsp; $\underline{y}_5 = (01)$&nbsp; one obtains:
 :$${\it \Gamma}_5(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
 :$${\it \Gamma}_5(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
@@ Line 84: / Line 84: @@
 :$${\it \Gamma}_5(S_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_1) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_3) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+0\hspace{0.05cm},\hspace{0.05cm} 2+2 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
-The left graph shows the final evaluated ${\it \Gamma}_i(S_{\mu})$ trellis.
+[[File:P_ID2682__KC_A_3_10c_neu.png|right|frame|Evaluated trellis diagrams]]
+<br><br><br><br>
-[[File:P_ID2682__KC_A_3_10c_neu.png|center|frame|Evaluated trellis diagrams]]
+The left sketch in the graph shows the final evaluated ${\it \Gamma}_i(S_{\mu})$ trellis.
+<br clear=all>
+'''(2)'''&nbsp; At time&nbsp; $i = 6$&nbsp; the termination is already effective and there are only two branch metrics left.&nbsp; For these one obtains with&nbsp; $\underline{y}_6 = (01)$:
 :$${\it \Gamma}_6(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
 :$${\it \Gamma}_6(S_2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_1) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_3) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
@@ Line 97: / Line 96: @@
 :$${\it \Gamma}_7(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{6}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (11) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{6}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (11) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
-In the BSC model, one can infer from ${\it \Gamma}_7(S_{\mu}) = 3$ that three transmission errors occurred &nbsp; &#8658; &nbsp; <u>solutions 1 and 3</u>.
+*With the BSC model,&nbsp; one can infer from&nbsp; ${\it \Gamma}_7(S_{\mu}) = 3$&nbsp; that three transmission errors occurred &nbsp; &#8658; &nbsp; <u>solutions 1 and 3</u>.
-'''(4)'''&nbsp; Correct are <u>statements 1 and 2</u>:
+*Also,&nbsp; the surviving and deleted branches are identical in both graphs.
-*Maximizing the branch metrics ${\it \Lambda}_i(S_{\mu})$ according to the right sketch in the above graph gives the same result as minimizing the branch metrics ${\it \Gamma}_i(S_{\mu})$ shown on the left. Also, the surviving and deleted branches are identical in both graphs.
-*The given equation is also correct, which is shown here only on the example $i = 7$:
+*The given equation is also correct,&nbsp; which is shown here only on the example&nbsp; $i = 7$:
 :$${\it \Lambda}_7(S_0)) =  2 \cdot \big [i - {\it \Gamma}_7(S_0) \big ] = 2 \cdot \big [7 - 3 \big ] \hspace{0.15cm}\underline{= 8}\hspace{0.05cm}.$$
-*The last statement is false. Rather applies&nbsp; $&#9001;x_i', \, y_i&#9002; &#8712; \{&ndash;2, \, 0, \, +2\}$.
+*The last statement is false.&nbsp; Rather applies&nbsp; $&#9001;x_i', \, y_i&#9002; &#8712; \{&ndash;2, \, 0, \, +2\}$.
-Hints: In [[Aufgaben:Exercise_3.11:_Viterbi_Path_Finding| "Exercise 3.11"]], path finding is demonstrable for the same example, assuming ${\it \Lambda}_i(S_{\mu})$&ndash;metrics as shown in the right graph.
+*In&nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Path_Finding| $\text{Exercise 3.11}$]],&nbsp; the path finding will be demonstrated for the same example,&nbsp; assuming ${\it \Lambda}_i(S_{\mu})$&nbsp; metrics as shown in the right graph.
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 [[File:P_ID2681__KC_A_3_10.png|right|frame|Only partially evaluated trellis]]
-In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$&nbsp; between the possible code words&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
+In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| $\text{theory section}$]]&nbsp; of this chapter,&nbsp; the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail,&nbsp; based on the Hamming distance &nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$ &nbsp; between
-The exercise deals exactly with this topic. In the adjacent graph
+*and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$.
-For example, the branch metric&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; with&nbsp; $\underline{y}_4 = (01)$&nbsp; as the minimum of the two comparison values
+The exercise deals exactly with this topic.&nbsp;  In the adjacent graph
-* ${\it \Gamma}_3(S_0) + d_{\rm H}((00), \ (01)) = 3 + 1 = 4$, and
+* the considered trellis is shown&nbsp; &ndash; valid for the code with rate&nbsp; $R = 1/2$, &nbsp; memory&nbsp; $m = 2$ &nbsp; and&nbsp;
-* ${\it \Gamma}_3(S_2) + d_{\rm H}((11), \ (01)) = 2 + 1 = 3$.
+:$$\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2),$$
-In the subtask '''(4)''' the relationship between
+For example,&nbsp; the branch metric&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; with&nbsp; $\underline{y}_4 = (01)$&nbsp; as the minimum of the two comparison values
-*the&nbsp; ${\it \Gamma}_i(S_{\mu})$ minimization and
+* ${\it \Gamma}_3(S_0) + d_{\rm H}((00), \ (01)) = 3 + 1 = 4$,&nbsp; and
-Hints:
+<u>Hints:</u>
 * The exercise refers to the chapter&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes| "Decoding Convolutional Codes"]].
-* For the time being, the search of surviving paths is not considered. This will be dealt with for the same example in the later&nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Path_Finding| "Exercise 3.11"]].
@@ Line 62: / Line 69: @@
 + This final value suggests three transmission errors.
-{Which statements are true for the&nbsp; ${\it \Lambda}_i(S_{\mu})$ evaluation?
+{Which statements are true for the&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; evaluation?
 |type="[]"}
-+ The metrics&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; provide the same information as&nbsp; ${\it \Gamma}_i(S_{\mu})$.
++ The correlation metrics&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; provide the same information as&nbsp; ${\it \Gamma}_i(S_{\mu})$.
 + For all nodes,&nbsp; ${\it \Lambda}_i(S_{\mu}) = 2 \cdot \big [i \, &ndash;{\it \Gamma}_i(S_{\mu})\big ]$.
 - For the metric increments,&nbsp; $&#9001; \underline{x}_i', \, \underline{y}_i &#9002; &#8712; \{0, \, 1, \, 2\}$.

@@ Line 2: / Line 2: @@
 [[File:P_ID2681__KC_A_3_10.png|right|frame|Only partially evaluated trellis]]
-In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$&nbsp; between the possible codewords&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
+In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$&nbsp; between the possible code words&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
 The exercise deals exactly with this topic. In the adjacent graph

@@ Line 44: / Line 44: @@
 ===Questions===
 <quiz display=simple>
-{Wie lauten die Fehlergrößen für den Zeitpunkt&nbsp; $i = 5$?
+{What are the branch metrics for time&nbsp; $i = 5$?
 |type="{}"}
 ${\it \Gamma}_5(S_0) \ = \ ${ 3 3% }
@@ Line 51: / Line 51: @@
 ${\it \Gamma}_5(S_3) \ = \ ${ 3 3% }
-{Wie lauten die Fehlergrößen für den Zeitpunkt&nbsp; $i = 6$?
+{What are the branch metrics for time&nbsp; $i = 6$?
 |type="{}"}
 ${\it \Gamma}_6(S_0) \ = \ ${ 3 3% }
 ${\it \Gamma}_6(S_2) \ = \ ${ 3 3% }
-{Welcher Endwert ergibt sich bei diesem Trellis, basierend auf&nbsp; ${\it \Gamma}_i(S_{\mu})$?
+{What is the final value of this trellis based on&nbsp; ${\it \Gamma}_i(S_{\mu})$?
 |type="[]"}
-+ Es gilt&nbsp; ${\it \Gamma}_7(S_0) = 3$.
++ It holds&nbsp; ${\it \Gamma}_7(S_0) = 3$.
-- Dieser Endwert lässt auf eine fehlerfreie Übertragung schließen.
+- This final value suggests one error-free transmission.
-+ Dieser Endwert lässt auf drei Übertragungsfehler schließen.
++ This final value suggests three transmission errors.
-{Welche Aussagen sind für die&nbsp; ${\it \Lambda}_i(S_{\mu})$&ndash;Auswertung zutreffend?
+{Which statements are true for the&nbsp; ${\it \Lambda}_i(S_{\mu})$ evaluation?
 |type="[]"}
-+ Die Metriken&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; liefern gleiche Informationen wie&nbsp; ${\it \Gamma}_i(S_{\mu})$.
++ The metrics&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; provide the same information as&nbsp; ${\it \Gamma}_i(S_{\mu})$.
-+ Für alle Knoten gilt&nbsp; ${\it \Lambda}_i(S_{\mu}) = 2 \cdot \big [i \, &ndash;{\it \Gamma}_i(S_{\mu})\big ]$.
++ For all nodes,&nbsp; ${\it \Lambda}_i(S_{\mu}) = 2 \cdot \big [i \, &ndash;{\it \Gamma}_i(S_{\mu})\big ]$.
-- Für die Metrikzuwächse gilt&nbsp; $&#9001; \underline{x}_i', \, \underline{y}_i &#9002; &#8712; \{0, \, 1, \, 2\}$.
+- For the metric increments,&nbsp; $&#9001; \underline{x}_i', \, \underline{y}_i &#9002; &#8712; \{0, \, 1, \, 2\}$.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Bei allen Knoten $S_{\mu}$ muss eine Entscheidung zwischen den beiden ankommenden Zweigen getroffen werden. Ausgewählt wird dann jeweils der Zweig, der zur (minimalen) Fehlergröße ${\it \Gamma}_5(S_{\mu})$ geführt hat. Mit $\underline{y}_5 = (01)$ erhält man:
+'''(1)'''&nbsp; At all nodes $S_{\mu}$ a decision must be made between the two incoming branches. The branch that led to the (minimum) error metric ${\it \Gamma}_5(S_{\mu})$ is then selected in each case. With $\underline{y}_5 = (01)$ one obtains:
 :$${\it \Gamma}_5(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
 :$${\it \Gamma}_5(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
@@ Line 77: / Line 77: @@
 :$${\it \Gamma}_5(S_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_1) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_3) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+0\hspace{0.05cm},\hspace{0.05cm} 2+2 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
-Die linke Grafik zeigt das endgültig ausgewertete ${\it \Gamma}_i(S_{\mu})$&ndash;Trellis.
+The left graph shows the final evaluated ${\it \Gamma}_i(S_{\mu})$ trellis.
-[[File:P_ID2682__KC_A_3_10c_neu.png|center|frame|Ausgewertete Trellisdiagramme]]
+[[File:P_ID2682__KC_A_3_10c_neu.png|center|frame|Evaluated trellis diagrams]]
-'''(2)'''&nbsp; Zum Zeitpunk $i = 6$ ist bereits die Terminierung wirksam und es gibt nur noch zwei Fehlergrößen. Für diese erhält man mit $\underline{y}_6 = (01)$:
+'''(2)'''&nbsp; At time $i = 6$ the termination is already effective and there are only two branch metrics left. For these one obtains with $\underline{y}_6 = (01)$:
 :$${\it \Gamma}_6(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$
 :$${\it \Gamma}_6(S_2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_1) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_3) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; Der Endwert ergibt sich zu
+'''(3)'''&nbsp; The final value results to
 :$${\it \Gamma}_7(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{6}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (11) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{6}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (11) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$
-Beim BSC&ndash;Modell kann man aus ${\it \Gamma}_7(S_{\mu}) = 3$ darauf schließen, dass drei Übertragungsfehler aufgetreten sind &nbsp; &#8658; &nbsp; <u>Lösungsvorschläge 1 und 3</u>.
+In the BSC model, one can infer from ${\it \Gamma}_7(S_{\mu}) = 3$ that three transmission errors occurred &nbsp; &#8658; &nbsp; <u>solutions 1 and 3</u>.
-'''(4)'''&nbsp; Richtig sind die <u>Aussagen 1 und 2</u>:
+'''(4)'''&nbsp; Correct are <u>statements 1 and 2</u>:
-*Die Maximierung der Metriken ${\it \Lambda}_i(S_{\mu})$ entsprechend der rechten Skizze in obiger Grafik liefert das gleiche Ergebnis wie die links dargestellte Minimierung der Fehlergrößen ${\it \Gamma}_i(S_{\mu})$. Auch die überlebenden und gestrichenen Zweige sind in beiden Grafiken identisch.
+*Maximizing the branch metrics ${\it \Lambda}_i(S_{\mu})$ according to the right sketch in the above graph gives the same result as minimizing the branch metrics ${\it \Gamma}_i(S_{\mu})$ shown on the left. Also, the surviving and deleted branches are identical in both graphs.
-*Die angegebene Gleichung ist ebenfalls richtig, was hier nur am Beispiel $i = 7$ gezeigt wird:
+*The given equation is also correct, which is shown here only on the example $i = 7$:
 :$${\it \Lambda}_7(S_0)) =  2 \cdot \big [i - {\it \Gamma}_7(S_0) \big ] = 2 \cdot \big [7 - 3 \big ] \hspace{0.15cm}\underline{= 8}\hspace{0.05cm}.$$
-*Die letzte Aussage ist falsch. Vielmehr gilt&nbsp; $&#9001;x_i', \, y_i&#9002; &#8712; \{&ndash;2, \, 0, \, +2\}$.
+*The last statement is false. Rather applies&nbsp; $&#9001;x_i', \, y_i&#9002; &#8712; \{&ndash;2, \, 0, \, +2\}$.
-''Hinweis:'' In der [[Aufgaben:Aufgabe_3.11:_Viterbi–Pfadsuche| Aufgabe 3.11]] wird für das gleiche Beispiel die Pfadsuche demonstiert, wobei von den ${\it \Lambda}_i(S_{\mu})$&ndash;Metriken gemäß der rechten Grafik ausgegangen wird.
+Hints: In [[Aufgaben:Exercise_3.11:_Viterbi_Path_Finding| "Exercise 3.11"]], path finding is demonstrable for the same example, assuming ${\it \Lambda}_i(S_{\mu})$&ndash;metrics as shown in the right graph.
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 [[File:P_ID2681__KC_A_3_10.png|right|frame|Only partially evaluated trellis]]
-In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0. 05cm}', \ \underline{y}_i)$&nbsp; between the possible codewords&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
+In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$&nbsp; between the possible codewords&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
 The exercise deals exactly with this topic. In the adjacent graph

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Kanalcodierung/Decodierung von Faltungscodes}}
+{{quiz-Header|Buchseite=Channel_Coding/Decoding_of_Convolutional_Codes}}
-[[File:P_ID2681__KC_A_3_10.png|right|frame|Nur teilweise ausgewertetes Trellis]]
+[[File:P_ID2681__KC_A_3_10.png|right|frame|Only partially evaluated trellis]]
-Im&nbsp; [[Channel_Coding/Decodierung_von_Faltungscodes#Vorbemerkungen_zu_den_nachfolgenden_Decodierbeispielen| Theorieteil]]&nbsp; zu diesem Kapitel wurde die Berechnung der Fehlergrößen&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; ausführlich behandelt, die auf der Hamming&ndash;Distanz&nbsp; $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$&nbsp; zwischen den möglichen Codeworten&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; und dem zum Zeitpunkt&nbsp; $i$&nbsp; empfangenen 2&ndash;Bit&ndash;Worten&nbsp; $\underline{y}_i$&nbsp; basiert.
+In the&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes#Preliminary_remarks_on_the_following_decoding_examples| "theory section"]]&nbsp; of this chapter, the calculation of the branch metrics&nbsp; ${\it \Gamma}_i(S_{\mu})$&nbsp; has been discussed in detail, based on the Hamming distance&nbsp; $d_{\rm H}(\underline{x}\hspace{0. 05cm}', \ \underline{y}_i)$&nbsp; between the possible codewords&nbsp; $\underline{x}\hspace{0.05cm}' &#8712; \{00, \, 01, \, 10, \, 11\}$&nbsp; and the 2&ndash;bit&ndash;words&nbsp; $\underline{y}_i$&nbsp; received at time&nbsp; $i$&nbsp; is based.
-Die Aufgabe beschäftigt sich genau mit dieser Thematik. In nebenstehender Grafik
+The exercise deals exactly with this topic. In the adjacent graph
-* ist das betrachtete Trellis dargestellt &ndash; gültig für den Code mit Rate&nbsp; $R = 1/2$,&nbsp; Gedächtnis&nbsp; $m = 2$&nbsp; sowie&nbsp; $\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$,
+* the considered trellis is shown&ndash; valid for the code with rate&nbsp; $R = 1/2$,&nbsp; memory&nbsp; $m = 2$&nbsp; and&nbsp; $\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$,
-* sind die Empfangsworte&nbsp; $\underline{y}_1 = (01), \hspace{0.05cm}\text{ ...} \hspace{0.05cm} , \ \underline{y}_7 = (11)$&nbsp; in den Rechtecken angegeben,
+* are the received words&nbsp; $\underline{y}_1 = (01), \hspace{0.05cm}\text{ ...} \hspace{0.05cm} , \ \underline{y}_7 = (11)$&nbsp; indicated in the rectangles,
-* sind alle Fehlergrößen&nbsp; ${\it \Gamma}_0(S_{\mu}), \hspace{0.05cm}\text{ ...} \hspace{0.05cm}  , \ {\it \Gamma}_4(S_{\mu})$&nbsp;  bereits eingetragen.
+* are all branch metrics&nbsp; ${\it \Gamma}_0(S_{\mu}), \hspace{0.05cm}\text{ ...} \hspace{0.05cm}  , \ {\it \Gamma}_4(S_{\mu})$&nbsp; already entered.
-Beispielsweise ergibt sich die Fehlergröße&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; mit&nbsp; $\underline{y}_4 = (01)$&nbsp; als das Minimum der beiden Vergleichswerte
+For example, the branch metric&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; with&nbsp; $\underline{y}_4 = (01)$&nbsp; as the minimum of the two comparison values
-* ${\it \Gamma}_3(S_0) + d_{\rm H}((00), \ (01)) = 3 + 1 = 4$, und
+* ${\it \Gamma}_3(S_0) + d_{\rm H}((00), \ (01)) = 3 + 1 = 4$, and
 * ${\it \Gamma}_3(S_2) + d_{\rm H}((11), \ (01)) = 2 + 1 = 3$.
-Der überlebende Zweig &ndash; hier von&nbsp; ${\it \Gamma}_3(S_2)$&nbsp; nach&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; &ndash; ist durchgezogen gezeichnet, der eliminierte Zweig von&nbsp; ${\it \Gamma}_3(S_0)$&nbsp; nach&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; punktiert. Rote Pfeile stehen für das Informationsbit $u_i = 0$, blaue Pfeile für $u_i = 1$.
+The surviving branch &ndash; here from&nbsp; ${\it \Gamma}_3(S_2)$&nbsp; to&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; &ndash; is drawn solid, the eliminated branch from&nbsp; ${\it \Gamma}_3(S_0)$&nbsp; to&nbsp; ${\it \Gamma}_4(S_0)$&nbsp; dotted. Red arrows represent the information bit $u_i = 0$, blue arrows $u_i = 1$.
-In der Teilaufgabe '''(4)''' soll der Zusammenhang zwischen
+In the subtask '''(4)''' the relationship between
-*der&nbsp;  ${\it \Gamma}_i(S_{\mu})$&ndash;Minimierung und
+*the&nbsp; ${\it \Gamma}_i(S_{\mu})$ minimization and
-*der&nbsp; ${\it \Lambda}_i(S_{\mu})$&ndash;Maximierung
+*the&nbsp; ${\it \Lambda}_i(S_{\mu})$ maximization.
-herausgearbeitet werden. Hierbei bezeichnet man die Knoten&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; als <i>Metriken</i>, wobei sich der Metrikzuwachs gegenüber den Vorgängerknoten aus dem Korrelationswert&nbsp; $&#9001;\underline{x}_i\hspace{0.05cm}', \, \underline{y}_i &#9002;$&nbsp; ergibt. Näheres zu dieser Thematik finden Sie auf den folgenden Theorieseiten:
+shall be worked out. Here, we refer to the nodes&nbsp; ${\it \Lambda}_i(S_{\mu})$&nbsp; as <i>metrics</i>, where the metric increment over the predecessor nodes results from the correlation value&nbsp; $&#9001;\underline{x}_i\hspace{0.05cm}', \, \underline{y}_i &#9002;$&nbsp;. For more details on this topic, see the following theory pages:
-* [[Channel_Coding/Decodierung_von_Faltungscodes#Zusammenhang_zwischen_Hamming.E2.80.93Distanz_und_Korrelation| Zusammenhang zwischen Hamming&ndash;Distanz und Korrelation]],
+* [[Channel_Coding/Decoding_of_Convolutional_Codes#Relationship_between_Hamming_distance_and_correlation| "Relationship between Hamming distance and correlation"]]
-* [[Channel_Coding/Decodierung_von_Faltungscodes#Viterbi.E2.80.93Algorithmus.2C_basierend_auf_Korrelation_und_Metriken| Viterbi&ndash;Algorithmus, basierend auf Korrelation und Metriken]],
+* [[Channel_Coding/Decoding_of_Convolutional_Codes#Viterbi_algorithm_based_on_correlation_and_metrics| "Viterbi algorithm based on correlation and metrics"]]
-* [[Channel_Coding/Decodierung_von_Faltungscodes#Viterbi.E2.80.93Entscheidung_bei_nicht.E2.80.93terminierten_Faltungscodes| Viterbi&ndash;Entscheidung bei nicht&ndash;terminierten Faltungscodes]].
+* [[Channel_Coding/Decoding_of_Convolutional_Codes#Viterbi_decision_for_non-terminated_convolutional_codes| "Viterbi decision for non&ndash;terminated convolutional codes"]].
@@ Line 34: / Line 34: @@
-''Hinweise:''
+Hints:
-* Die Aufgabe bezieht sich auf das Kapitel&nbsp; [[Channel_Coding/Decodierung_von_Faltungscodes| Decodierung von Faltungscodes]].
+* The exercise refers to the chapter&nbsp; [[Channel_Coding/Decoding_of_Convolutional_Codes| "Decoding Convolutional Codes"]].
-* Vorerst nicht betrachtet wird die Suche der überlebenden Pfade. Damit beschäftigt sich für das gleiche Beispiel die spätere&nbsp; [[Aufgaben:3.11_Viterbi%E2%80%93Pfadsuche| Aufgabe 3.11]].
+* For the time being, the search of surviving paths is not considered. This will be dealt with for the same example in the later&nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Path_Finding| "Exercise 3.11"]].
@@ Line 42: / Line 42: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
 {Wie lauten die Fehlergrößen für den Zeitpunkt&nbsp; $i = 5$?

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−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^3.4 Decodierung von Faltungscodes^]]	+	[[Category:Channel Coding: Exercises\|^3.4 Decodierung von Faltungscodes^]]