Aufgaben:Exercise 3.12: Trellis Diagram for Two Precursors - Revision history

Guenter at 15:34, 4 July 2022

2022-07-04T15:34:06Z

Guenter at 15:24, 4 July 2022

2022-07-04T15:24:55Z

Hwang at 09:36, 8 June 2022

2022-06-08T09:36:45Z

Guenter at 07:05, 1 June 2022

2022-06-01T07:05:10Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.12: Trellisdiagramm für zwei Vorläufer to Aufgaben:Exercise 3.12: Trellis Diagram for Two Precursors

2022-06-01T07:00:11Z

Guenter moved page Aufgabe 3.12: Trellisdiagramm für zwei Vorläufer to Exercise 3.12: Trellis Diagram for Two Precursors

Javier: Text replacement - "„" to """

2021-05-28T14:27:43Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:27:21Z

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Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:46:45Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Guenter at 10:10, 8 March 2019

2019-03-08T10:10:01Z

Guenter at 10:06, 8 March 2019

2019-03-08T10:06:22Z

@@ Line 66: / Line 66: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The first error quantity is calculated as follows:
+'''(1)'''&nbsp; The first metric is calculated as follows:
 :$$\varepsilon_{2}(010)  = [d_0 - 0 \cdot g_0 - 1 \cdot g_{-1}- 0 \cdot g_{-2}]^2= [0.2 -0.3]^2\hspace{0.15cm}\underline {=0.01}
    \hspace{0.05cm}.$$
-Correspondingly, for the other error quantities:
+Correspondingly,&nbsp; for the other metrics:
 :$$\varepsilon_{2}(011) \ = \  [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$
 :$$\varepsilon_{2}(110) \ = \  [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$
@@ Line 86: / Line 86: @@
-'''(3)'''&nbsp; The <u>first and last solutions</u> are correct:
+'''(3)'''&nbsp; The&nbsp; <u>first and last solutions</u>&nbsp; are correct:
-*The sequence $1011010$ can be recognized from the continuous path: &nbsp; &nbsp; "red &ndash; blue &ndash; red &ndash; red &ndash; blue &ndash; red &ndash; blue".
+*The sequence&nbsp; "$1011010$"&nbsp; can be recognized from the continuous path: &nbsp; &nbsp; "red &ndash; blue &ndash; red &ndash; red &ndash; blue &ndash; red &ndash; blue".
 *On the other hand, no final statement can be made about the symbol $a_8$ at time $\nu = 8$:
-*Only under the hypothesis $a_9 = 1$ <u>and</u> $a_{\rm 10} = 1$ one would decide for $a_8 = 0$, under other hypotheses for $a_8 = 1$.
+*On the other hand,&nbsp; no final statement can be made about the symbol&nbsp; $a_8$&nbsp; at time&nbsp; $\nu = 8$:
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID1478__Dig_A_3_12.png|right|frame|Trellis diagram for two precursors]]
-We assume the basic pulse values &nbsp; $g_0$,&nbsp; $g_{\rm &ndash;1}$&nbsp; and&nbsp; $g_{\rm &ndash;2}$:&nbsp;
+We assume the basic pulse values &nbsp; $g_0\ne 0$,&nbsp; $g_{\rm &ndash;1}\ne 0$&nbsp; and&nbsp; $g_{\rm &ndash;2}\ne 0$:&nbsp;
-*This means that the decision on the symbol &nbsp; $a_{\rm \nu}$&nbsp; is also influenced by the subsequent coefficients &nbsp; $a_{\rm \nu +1}$&nbsp; and&nbsp; $a_{\rm \nu +2}$.&nbsp;
+*This means that the decision on the symbol&nbsp; $a_{\rm \nu}$&nbsp; is also influenced by the subsequent coefficients&nbsp; $a_{\rm \nu +1}$&nbsp; and&nbsp; $a_{\rm \nu +2}$.&nbsp;
-*Thus, for each time point &nbsp; $\nu$,&nbsp; exactly eight error quantities &nbsp; $\varepsilon_{\rm \nu}$&nbsp; have to be determined, from which the minimum total error quantities &nbsp; ${\it \Gamma}_{\rm \nu}(00)$,&nbsp; ${\it \Gamma}_{\rm \nu}(01)$,&nbsp; ${\it \Gamma}_{\rm \nu}(10)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(11)$&nbsp; can be calculated.
-*Here, for example, &nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; provides information about the symbol &nbsp; $a_{\rm \nu}$&nbsp; under the assumption that &nbsp; $a_{\rm \nu +1} = 0$&nbsp; and&nbsp; $a_{\rm \nu +2} = 1$&nbsp; will be.
+*Thus,&nbsp; for each time point &nbsp; $\nu$,&nbsp; exactly eight&nbsp; '''metrics''' &nbsp; $\varepsilon_{\rm \nu}$&nbsp; have to be determined, from which the&nbsp; '''minimum accumulated metrics''' &nbsp; ${\it \Gamma}_{\rm \nu}(00)$,&nbsp; ${\it \Gamma}_{\rm \nu}(01)$,&nbsp; ${\it \Gamma}_{\rm \nu}(10)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(11)$&nbsp; can be calculated.
-*Here, the minimum total error quantity &nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; is the smaller value obtained from the comparison of
 :$$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm and}
   \hspace{0.15cm}\big[{\it \Gamma}_{\nu-1}(10) + \varepsilon_{\nu}(101)\big].$$
-To calculate the minimum total error quantity &nbsp; ${\it \Gamma}_2(10)$&nbsp; in subtasks '''(1)''' and '''(2)''', assume the following numerical values:
+To calculate the minimum accumulated metric &nbsp; ${\it \Gamma}_2(10)$&nbsp; in subtasks '''(1)''' and '''(2)''',&nbsp; assume the following numerical values:
 * unipolar amplitude coefficients:&nbsp; $a_{\rm \nu} &#8712; \{0, 1\}$,
 * basic pulse values &nbsp; $g_0 = 0.5$,&nbsp; $g_{\rm &ndash;1} = 0.3$,&nbsp; $g_{\rm &ndash;2} = 0.2$,
 * applied detection sample:&nbsp; $d_2 = 0.2$,
-* Minimum total error quantities at time&nbsp; $\nu = 1$:
+* applied noisy detection sample:&nbsp; $d_2 = 0.2$,
 :$${\it \Gamma}_{1}(00) = 0.0,\hspace{0.2cm}{\it \Gamma}_{1}(01) = 0.2, \hspace{0.2cm} {\it \Gamma}_{1}(10) = 0.6,\hspace{0.2cm}{\it \Gamma}_{1}(11) =
 .2
@@ Line 21: / Line 27: @@
 The graph shows the simplified trellis diagram for time points &nbsp; $\nu = 1$&nbsp; to &nbsp; $\nu = 8$.&nbsp;
-*Blue branches come from either &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$&nbsp; or &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$&nbsp; and denote a hypothetical "$0$".
+*Blue branches come from either &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$ &nbsp; or &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$ &nbsp; and denote a hypothetical&nbsp; "$0$".
 *In contrast, all red branches &ndash; starting from the &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(10)$&nbsp; or &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(11)$&nbsp; states &ndash; indicate the symbol "$1$".
 Notes:
 *The exercise belongs to the chapter &nbsp;  [[Digital_Signal_Transmission/Viterbi_Receiver|"Viterbi Receiver"]].
 * All quantities here are to be understood normalized.
 * Also, assume unipolar and equal probability amplitude coefficients: &nbsp; ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$
@@ Line 39: / Line 44: @@
 ===Questions===
 <quiz display=simple>
-{Calculate the following error quantities:
+{Calculate the following metrics:
 |type="{}"}
 $\varepsilon_2(010) \ = \ $ { 0.01 3% }
@@ Line 46: / Line 51: @@
 $\varepsilon_2(111) \ = \ $ { 0.64 3% }
-{Calculate the following minimum total error quantities:
+{Calculate the following minimum accumulated metrics:
 |type="{}"}
 ${\it \Gamma}_2(10) \ = \ $ { 0.21 3% }
@@ Line 53: / Line 58: @@
 {What are the symbols output by the Viterbi receiver?
 |type="[]"}
-+ The first seven symbols are &nbsp;$1011010$.
++ The first seven symbols are &nbsp; "$1011010$".
-- The first seven symbols are &nbsp;$1101101$.
+- The first seven symbols are &nbsp; "$1101101$".
 - The last symbol &nbsp;$a_8 = 1$&nbsp; is safe.
 + No definite statement can be made about the symbol &nbsp;$a_8$.&nbsp;

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Viterbi–Empfänger}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Viterbi_Receiver}}
-[[File:P_ID1478__Dig_A_3_12.png|right|frame|Trellisdiagramm für zwei Vorläufer]]
+[[File:P_ID1478__Dig_A_3_12.png|right|frame|Trellis diagram for two precursors]]
-Wir gehen von den Grundimpulswerten&nbsp; $g_0$,&nbsp; $g_{\rm &ndash;1}$&nbsp; und&nbsp; $g_{\rm &ndash;2}$&nbsp; aus:
+We assume the basic pulse values &nbsp; $g_0$,&nbsp; $g_{\rm &ndash;1}$&nbsp; and&nbsp; $g_{\rm &ndash;2}$:&nbsp;
-*Das bedeutet, dass die Entscheidung über das Symbol&nbsp; $a_{\rm \nu}$&nbsp; auch durch die nachfolgenden Koeffizienten&nbsp; $a_{\rm \nu +1}$&nbsp; und&nbsp; $a_{\rm \nu +2}$&nbsp; beeinflusst wird.
+*This means that the decision on the symbol &nbsp; $a_{\rm \nu}$&nbsp; is also influenced by the subsequent coefficients &nbsp; $a_{\rm \nu +1}$&nbsp; and&nbsp; $a_{\rm \nu +2}$.&nbsp;
-*Damit sind für jeden Zeitpunkt&nbsp; $\nu$&nbsp; genau acht Fehlergrößen&nbsp; $\varepsilon_{\rm \nu}$&nbsp; zu bestimmen, aus denen die minimalen Gesamtfehlergrößen&nbsp; ${\it \Gamma}_{\rm \nu}(00)$,&nbsp; ${\it \Gamma}_{\rm \nu}(01)$,&nbsp; ${\it \Gamma}_{\rm \nu}(10)$&nbsp; und&nbsp; ${\it \Gamma}_{\rm \nu}(11)$&nbsp; berechnet werden können.
+*Thus, for each time point &nbsp; $\nu$,&nbsp; exactly eight error quantities &nbsp; $\varepsilon_{\rm \nu}$&nbsp; have to be determined, from which the minimum total error quantities &nbsp; ${\it \Gamma}_{\rm \nu}(00)$,&nbsp; ${\it \Gamma}_{\rm \nu}(01)$,&nbsp; ${\it \Gamma}_{\rm \nu}(10)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(11)$&nbsp; can be calculated.
-*Hierbei liefert beispielsweise&nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; Information über das Symbol&nbsp; $a_{\rm \nu}$&nbsp; unter der Annahme, dass&nbsp; $a_{\rm \nu +1} = 0$&nbsp; und&nbsp; $a_{\rm \nu +2} = 1$&nbsp; sein werden.
+*Here, for example, &nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; provides information about the symbol &nbsp; $a_{\rm \nu}$&nbsp; under the assumption that &nbsp; $a_{\rm \nu +1} = 0$&nbsp; and&nbsp; $a_{\rm \nu +2} = 1$&nbsp; will be.
-*Die minimale Gesamtfehlergröße&nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; ist hierbei der kleinere Wert aus dem Vergleich von
+*Here, the minimum total error quantity &nbsp; ${\it \Gamma}_{\rm \nu}(01)$&nbsp; is the smaller value obtained from the comparison of
-:$$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm und}
+:$$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm and}
   \hspace{0.15cm}\big[{\it \Gamma}_{\nu-1}(10) + \varepsilon_{\nu}(101)\big].$$
-Zur Berechnung der minimalen Gesamtfehlergröße&nbsp; ${\it \Gamma}_2(10)$&nbsp; in den Teilaufgaben '''(1)''' und '''(2)''' soll von folgenden Zahlenwerten ausgegangen werden:
+To calculate the minimum total error quantity &nbsp; ${\it \Gamma}_2(10)$&nbsp; in subtasks '''(1)''' and '''(2)''', assume the following numerical values:
-* unipolare Amplitudenkoeffizienten:&nbsp; $a_{\rm \nu} &#8712; \{0, 1\}$,
+* unipolar amplitude coefficients:&nbsp; $a_{\rm \nu} &#8712; \{0, 1\}$,
-* Grundimpulswerte&nbsp; $g_0 = 0.5$,&nbsp; $g_{\rm &ndash;1} = 0.3$,&nbsp; $g_{\rm &ndash;2} = 0.2$,
+* basic pulse values &nbsp; $g_0 = 0.5$,&nbsp; $g_{\rm &ndash;1} = 0.3$,&nbsp; $g_{\rm &ndash;2} = 0.2$,
-* anliegender Detektionsabtastwert:&nbsp; $d_2 = 0.2$,
+* applied detection sample:&nbsp; $d_2 = 0.2$,
-* Minimale Gesamtfehlergrößen zum Zeitpunkt&nbsp; $\nu = 1$:
+* Minimum total error quantities at time&nbsp; $\nu = 1$:
 :$${\it \Gamma}_{1}(00) = 0.0,\hspace{0.2cm}{\it \Gamma}_{1}(01) = 0.2, \hspace{0.2cm} {\it \Gamma}_{1}(10) = 0.6,\hspace{0.2cm}{\it \Gamma}_{1}(11) =
 .2
   \hspace{0.05cm}.$$
-In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte&nbsp; $\nu = 1$&nbsp; bis&nbsp; $\nu = 8$&nbsp; dargestellt.
+The graph shows the simplified trellis diagram for time points &nbsp; $\nu = 1$&nbsp; to &nbsp; $\nu = 8$.&nbsp;
-*Blaue Zweige kommen entweder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$&nbsp; oder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$&nbsp; und kennzeichnen eine hypothetische "$0$".
+*Blue branches come from either &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$&nbsp; or &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$&nbsp; and denote a hypothetical "$0$".
-*Dagegen weisen alle roten Zweige &ndash; ausgehend von den Zuständen&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(10)$&nbsp; bzw.&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(11)$&nbsp; &ndash; auf das Symbol "$1$" hin.
+*In contrast, all red branches &ndash; starting from the &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(10)$&nbsp; or &nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(11)$&nbsp; states &ndash; indicate the symbol "$1$".
@@ Line 28: / Line 28: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Digitalsignal%C3%BCbertragung/Viterbi%E2%80%93Empf%C3%A4nger|Viterbi&ndash;Empfänger]].
+*The exercise belongs to the chapter &nbsp;  [[Digital_Signal_Transmission/Viterbi_Receiver|"Viterbi Receiver"]].
-* Alle Größen sind hier normiert zu verstehen.
+* All quantities here are to be understood normalized.
-*Gehen Sie zudem von unipolaren und gleichwahrscheinlichen Amplitudenkoeffizienten aus: &nbsp; ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$
+* Also, assume unipolar and equal probability amplitude coefficients: &nbsp; ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$
-* Die Thematik wird auch im interaktiven Applet&nbsp;  [[Applets:Viterbi|Eigenschaften des Viterbi&ndash;Empfängers]]&nbsp; behandelt.
+* The topic is also covered in the interactive applet&nbsp;  [[Applets:Viterbi|"Properties of the Viterbi Receiver"]].&nbsp;
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die folgenden Fehlergrößen:
+{Calculate the following error quantities:
 |type="{}"}
 $\varepsilon_2(010) \ = \ $ { 0.01 3% }
@@ Line 46: / Line 46: @@
 $\varepsilon_2(111) \ = \ $ { 0.64 3% }
-{Berechnen Sie die folgenden minimalen Gesamtfehlergrößen:
+{Calculate the following minimum total error quantities:
 |type="{}"}
 ${\it \Gamma}_2(10) \ = \ $ { 0.21 3% }
 ${\it \Gamma}_2(11) \ = \ $ { 0.29 3% }
-{Wie lauten die vom Viterbi&ndash;Empfänger ausgegebene Symbole?
+{What are the symbols output by the Viterbi receiver?
 |type="[]"}
-+ Die ersten sieben Symbole sind &nbsp;$1011010$.
++ The first seven symbols are &nbsp;$1011010$.
-- Die ersten sieben Symbole sind &nbsp;$1101101$.
+- The first seven symbols are &nbsp;$1101101$.
-- Das letzte Symbol &nbsp;$a_8 = 1$&nbsp; ist sicher.
+- The last symbol &nbsp;$a_8 = 1$&nbsp; is safe.
-+ Über das Symbol &nbsp;$a_8$&nbsp; ist noch keine endgültige Aussage möglich.
++ No definite statement can be made about the symbol &nbsp;$a_8$.&nbsp;
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die erste Fehlergröße wird wie folgt berechnet:
+'''(1)'''&nbsp; The first error quantity is calculated as follows:
 :$$\varepsilon_{2}(010)  = [d_0 - 0 \cdot g_0 - 1 \cdot g_{-1}- 0 \cdot g_{-2}]^2= [0.2 -0.3]^2\hspace{0.15cm}\underline {=0.01}
    \hspace{0.05cm}.$$
-Entsprechend gilt für die weiteren Fehlergrößen:
+Correspondingly, for the other error quantities:
 :$$\varepsilon_{2}(011) \ = \  [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$
 :$$\varepsilon_{2}(110) \ = \  [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$
@@ Line 72: / Line 72: @@
-'''(2)'''&nbsp; Die Aufgabe ist, jeweils den minimalen Wert von zwei Vergleichswerten zu finden:
+'''(2)'''&nbsp; The task is to find the minimum value of each of two comparison values:
 :$${\it \Gamma}_{2}(10) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(010),
   \hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(110)\right] =  {\rm Min}\left[0.2+ 0.01, 1.2 + 0.36\right]\hspace{0.15cm}\underline {= 0.21}
@@ Line 81: / Line 81: @@
-'''(3)'''&nbsp; Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:
+'''(3)'''&nbsp; The <u>first and last solutions</u> are correct:
-*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:  &nbsp; &nbsp; "Rot &ndash; Blau &ndash; Rot &ndash; Rot &ndash; Blau &ndash; Rot &ndash; Blau".
+*The sequence $1011010$ can be recognized from the continuous path: &nbsp; &nbsp; "red &ndash; blue &ndash; red &ndash; red &ndash; blue &ndash; red &ndash; blue".
-*Dagegen kann über das Symbol $a_8$ zum Zeitpunkt $\nu = 8$ noch keine endgültige Aussage gemacht werden:
+*On the other hand, no final statement can be made about the symbol $a_8$ at time $\nu = 8$:
-*Nur unter der Hypothese $a_9 = 1$ <u>und</u> $a_{\rm 10} = 1$ würde man sich für $a_8 = 0$ entscheiden, bei anderen Hypothesen für $a_8 = 1$.
+*Only under the hypothesis $a_9 = 1$ <u>and</u> $a_{\rm 10} = 1$ one would decide for $a_8 = 0$, under other hypotheses for $a_8 = 1$.
 {{ML-Fuß}}

← Older revision		Revision as of 07:05, 1 June 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi~~-Empfänger~~^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi Receiver^]]

@@ Line 21: / Line 21: @@
 In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte&nbsp; $\nu = 1$&nbsp; bis&nbsp; $\nu = 8$&nbsp; dargestellt.
-*Blaue Zweige kommen entweder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$&nbsp; oder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$&nbsp; und kennzeichnen eine hypothetische &bdquo;$0$".
+*Blaue Zweige kommen entweder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(00)$&nbsp; oder von&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(01)$&nbsp; und kennzeichnen eine hypothetische "$0$".
-*Dagegen weisen alle roten Zweige &ndash; ausgehend von den Zuständen&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(10)$&nbsp; bzw.&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(11)$&nbsp; &ndash; auf das Symbol &bdquo;$1$" hin.
+*Dagegen weisen alle roten Zweige &ndash; ausgehend von den Zuständen&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(10)$&nbsp; bzw.&nbsp; ${\it \Gamma}_{\rm \nu &ndash;1}(11)$&nbsp; &ndash; auf das Symbol "$1$" hin.
@@ Line 82: / Line 82: @@
 '''(3)'''&nbsp; Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:
-*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:  &nbsp; &nbsp; &bdquo;Rot &ndash; Blau &ndash; Rot &ndash; Rot &ndash; Blau &ndash; Rot &ndash; Blau".
+*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:  &nbsp; &nbsp; "Rot &ndash; Blau &ndash; Rot &ndash; Rot &ndash; Blau &ndash; Rot &ndash; Blau".
 *Dagegen kann über das Symbol $a_8$ zum Zeitpunkt $\nu = 8$ noch keine endgültige Aussage gemacht werden:
 *Nur unter der Hypothese $a_9 = 1$ <u>und</u> $a_{\rm 10} = 1$ würde man sich für $a_8 = 0$ entscheiden, bei anderen Hypothesen für $a_8 = 1$.

← Older revision	Revision as of 07:00, 1 June 2022
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← Older revision		Revision as of 12:46, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^3.8 Viterbi-Empfänger^]]	+	[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi-Empfänger^]]