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

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	[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi Receiver^]]
	[[de:Aufgaben:3.~~12_Trellisdiagramm_für_2_Vorläufer~~]]		[[de:Aufgaben:Aufgabe 3.12: Trellisdiagramm für zwei Vorläufer]]

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	*Here, the minimum accumulated metric   ${\it \Gamma}_{\rm \nu}(01)$  is the smaller value obtained from the comparison of		*Here, the minimum accumulated metric   ${\it \Gamma}_{\rm \nu}(01)$  is the smaller value obtained from the comparison of
	:$$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm and}		:$$\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].$$
	\hspace{0.15cm}\big[{\it \Gamma}_{\nu-1}(10) + \varepsilon_{\nu}(101)\big].$$

	To calculate the minimum accumulated metric   ${\it \Gamma}_2(10)$  in subtasks '''(1)''' and '''(2)''',  assume the following numerical values:		To calculate the minimum accumulated metric   ${\it \Gamma}_2(10)$  in subtasks '''(1)''' and '''(2)''',  assume the following numerical values:
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	* minimum accumulated metric at time  $\nu = 1$:		* minimum accumulated metric at time  $\nu = 1$:
	:$${\it \Gamma}_{1}(00) = 0.0,\hspace{0.2cm}{\it \Gamma}_{1}(01) = 0.2, \hspace{0.2cm} {\it \Gamma}_{1}(10) = 0.6,\hspace{0.2cm}{\it \Gamma}_{1}(11) =		:$${\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) =1.2\hspace{0.05cm}.$$
	1.2
	\hspace{0.05cm}.$$

	The graph shows the simplified trellis diagram for time points   $\nu = 1$  to   $\nu = 8$.		The graph shows the simplified trellis diagram for time points   $\nu = 1$  to   $\nu = 8$.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The first metric is calculated as follows:		'''(1)'''  The first metric is calculated as follows:
	:$$\varepsilon_{2}(010) = [d_0 - 0 \cdot g_0 - 1 \cdot g_{-1}- 0 \cdot g_{-2}]^2= [0.2 -0.3]^2\hspace{0.15cm}\underline {=0.01}		:$$\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}.$$
	\hspace{0.05cm}.$$

	Correspondingly,  for the other metrics:		Correspondingly,  for the other metrics:
	:$$\varepsilon_{2}(011) \ = \ [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$		:$$\varepsilon_{2}(011) \ = \ [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$
	:$$\varepsilon_{2}(110) \ = \ [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$		:$$\varepsilon_{2}(110) \ = \ [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$
	:$$\varepsilon_{2}(111) \ = \ [0.2 -0.5- 0.3-0.2]^2\hspace{0.15cm}\underline {=0.64}		:$$\varepsilon_{2}(111) \ = \ [0.2 -0.5- 0.3-0.2]^2\hspace{0.15cm}\underline {=0.64}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


	'''(2)'''  The task is to find the minimum value of each of two comparison values:		'''(2)'''  The task is to find the minimum value of each of two comparison values:
	:$${\it \Gamma}_{2}(10) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(010),		:$${\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}\hspace{0.05cm},$$
	\hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(110)\right] = {\rm Min}\left[0.2+ 0.01, 1.2 + 0.36\right]\hspace{0.15cm}\underline {= 0.21}		:$${\it \Gamma}_{2}(11) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(011),\hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(111)\right] = {\rm Min}\left[0.2+ 0.09, 1.2 + 0.64\right]\hspace{0.15cm}\underline {= 0.29}\hspace{0.05cm}.$$
	\hspace{0.05cm},$$
	:$${\it \Gamma}_{2}(11) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(011),
	\hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(111)\right] = {\rm Min}\left[0.2+ 0.09, 1.2 + 0.64\right]\hspace{0.15cm}\underline {= 0.29}
	\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi Receiver^]]		[[Category:Digital Signal Transmission: Exercises\|^3.8 Viterbi Receiver^]]
			[[de:Aufgaben:3.12_Trellisdiagramm_für_2_Vorläufer]]

Guenter at 15:34, 4 July 2022

2022-07-04T15:34:06Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The first ~~error quantity~~ is calculated as follows:		'''(1)'''  The first metric is calculated as follows:
	:$$\varepsilon_{2}(010) = [d_0 - 0 \cdot g_0 - 1 \cdot g_{-1}- 0 \cdot g_{-2}]^2= [0.2 -0.3]^2\hspace{0.15cm}\underline {=0.01}		:$$\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}.$$		\hspace{0.05cm}.$$

	Correspondingly, for the other ~~error quantities~~:		Correspondingly,  for the other metrics:
	:$$\varepsilon_{2}(011) \ = \ [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$		:$$\varepsilon_{2}(011) \ = \ [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$$
	:$$\varepsilon_{2}(110) \ = \ [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$		:$$\varepsilon_{2}(110) \ = \ [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$$
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	'''(3)'''  The <u>first and last solutions</u> are correct:		'''(3)'''  The  <u>first and last solutions</u>  are correct:
	*The sequence $1011010$ can be recognized from the continuous path:     "red – blue – red – red – blue – red – blue".		*The sequence  "$1011010$"  can be recognized from the continuous path:     "red – blue – red – red – blue – red – 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,  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  $a_8 = 1$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 15:24, 4 July 2022

2022-07-04T15:24:55Z

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	[[File:P_ID1478__Dig_A_3_12.png\|right\|frame\|Trellis diagram for two precursors]]		[[File:P_ID1478__Dig_A_3_12.png\|right\|frame\|Trellis diagram for two precursors]]
	We assume the basic pulse values   $g_0$,  $g_{\rm –1}$  and  $g_{\rm –2}$:		We assume the basic pulse values   $g_0\ne 0$,  $g_{\rm –1}\ne 0$  and  $g_{\rm –2}\ne 0$:
	*This means that the decision on the symbol   $a_{\rm \nu}$  is also influenced by the subsequent coefficients   $a_{\rm \nu +1}$  and  $a_{\rm \nu +2}$.		*This means that the decision on the symbol  $a_{\rm \nu}$  is also influenced by the subsequent coefficients  $a_{\rm \nu +1}$  and  $a_{\rm \nu +2}$.
	*Thus, for each time point   $\nu$,  exactly eight ~~error quantities~~   $\varepsilon_{\rm \nu}$  have to be determined, from which the minimum ~~total error quantities~~   ${\it \Gamma}_{\rm \nu}(00)$,  ${\it \Gamma}_{\rm \nu}(01)$,  ${\it \Gamma}_{\rm \nu}(10)$  and  ${\it \Gamma}_{\rm \nu}(11)$  can be calculated.
	*~~Here, for~~ example,   ${\it \Gamma}_{\rm \nu}(01)$  provides information about the symbol   $a_{\rm \nu}$  under the assumption that   $a_{\rm \nu +1} = 0$  and  $a_{\rm \nu +2} = 1$  will be.		*Thus,  for each time point   $\nu$,  exactly eight  '''metrics'''   $\varepsilon_{\rm \nu}$  have to be determined, from which the  '''minimum accumulated metrics'''   ${\it \Gamma}_{\rm \nu}(00)$,  ${\it \Gamma}_{\rm \nu}(01)$,  ${\it \Gamma}_{\rm \nu}(10)$  and  ${\it \Gamma}_{\rm \nu}(11)$  can be calculated.
	*Here, the minimum ~~total error quantity~~   ${\it \Gamma}_{\rm \nu}(01)$  is the smaller value obtained from the comparison of
			*For example,   ${\it \Gamma}_{\rm \nu}(01)$  provides information about the symbol  $a_{\rm \nu}$  under the assumption that  $a_{\rm \nu +1} = 0$  and  $a_{\rm \nu +2} = 1$  will be.

			*Here, the minimum accumulated metric   ${\it \Gamma}_{\rm \nu}(01)$  is the smaller value obtained from the comparison of
	:$$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm and}		:$$\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].$$		\hspace{0.15cm}\big[{\it \Gamma}_{\nu-1}(10) + \varepsilon_{\nu}(101)\big].$$

	To calculate the minimum ~~total error quantity~~   ${\it \Gamma}_2(10)$  in subtasks '''(1)''' and '''(2)''', assume the following numerical values:		To calculate the minimum accumulated metric   ${\it \Gamma}_2(10)$  in subtasks '''(1)''' and '''(2)''',  assume the following numerical values:
	* unipolar amplitude coefficients:  $a_{\rm \nu} ∈ \{0, 1\}$,		* unipolar amplitude coefficients:  $a_{\rm \nu} ∈ \{0, 1\}$,

	* basic pulse values   $g_0 = 0.5$,  $g_{\rm –1} = 0.3$,  $g_{\rm –2} = 0.2$,		* basic pulse values   $g_0 = 0.5$,  $g_{\rm –1} = 0.3$,  $g_{\rm –2} = 0.2$,
	* applied detection sample:  $d_2 = 0.2$,
	* ~~Minimum total error quantities~~ at time  $\nu = 1$:		* applied noisy detection sample:  $d_2 = 0.2$,

			* minimum accumulated metric at time  $\nu = 1$:
	:$${\it \Gamma}_{1}(00) = 0.0,\hspace{0.2cm}{\it \Gamma}_{1}(01) = 0.2, \hspace{0.2cm} {\it \Gamma}_{1}(10) = 0.6,\hspace{0.2cm}{\it \Gamma}_{1}(11) =		:$${\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) =
	1.2		1.2
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	The graph shows the simplified trellis diagram for time points   $\nu = 1$  to   $\nu = 8$.		The graph shows the simplified trellis diagram for time points   $\nu = 1$  to   $\nu = 8$.
	*Blue branches come from either   ${\it \Gamma}_{\rm \nu –1}(00)$  or   ${\it \Gamma}_{\rm \nu –1}(01)$  and denote a hypothetical "$0$".		*Blue branches come from either   ${\it \Gamma}_{\rm \nu –1}(00)$   or   ${\it \Gamma}_{\rm \nu –1}(01)$   and denote a hypothetical  "$0$".
	*In contrast, all red branches – starting from the   ${\it \Gamma}_{\rm \nu –1}(10)$  or   ${\it \Gamma}_{\rm \nu –1}(11)$  states – indicate the symbol "$1$".
			*In contrast,  all red branches – starting from the   ${\it \Gamma}_{\rm \nu –1}(10)$  or   ${\it \Gamma}_{\rm \nu –1}(11)$  states – indicate the symbol  "$1$".



			Notes:

	''Notes:''
	*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Viterbi_Receiver\|"Viterbi Receiver"]].		*The exercise belongs to the chapter   [[Digital_Signal_Transmission/Viterbi_Receiver\|"Viterbi Receiver"]].

	* All quantities here are to be understood normalized.		* All quantities here are to be understood normalized.

	* Also, assume unipolar and equal probability amplitude coefficients:   ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$		* Also, assume unipolar and equal probability amplitude coefficients:   ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$
	* The topic is also covered in the interactive applet  [[Applets:Viterbi\|"Properties of the Viterbi Receiver"]].


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Calculate the following ~~error quantities~~:		{Calculate the following metrics:
	\|type="{}"}		\|type="{}"}
	$\varepsilon_2(010) \ = \ $ { 0.01 3% }		$\varepsilon_2(010) \ = \ $ { 0.01 3% }
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	$\varepsilon_2(111) \ = \ $ { 0.64 3% }		$\varepsilon_2(111) \ = \ $ { 0.64 3% }

	{Calculate the following minimum ~~total error quantities~~:		{Calculate the following minimum accumulated metrics:
	\|type="{}"}		\|type="{}"}
	${\it \Gamma}_2(10) \ = \ $ { 0.21 3% }		${\it \Gamma}_2(10) \ = \ $ { 0.21 3% }
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	{What are the symbols output by the Viterbi receiver?		{What are the symbols output by the Viterbi receiver?
	\|type="[]"}		\|type="[]"}
	+ The first seven symbols are  $1011010$.		+ The first seven symbols are   "$1011010$".
	- The first seven symbols are  $1101101$.		- The first seven symbols are   "$1101101$".
	- The last symbol  $a_8 = 1$  is safe.		- The last symbol  $a_8 = 1$  is safe.
	+ No definite statement can be made about the symbol  $a_8$.		+ No definite statement can be made about the symbol  $a_8$.

Hwang at 09:36, 8 June 2022

2022-06-08T09:36:45Z

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Guenter at 07:05, 1 June 2022

2022-06-01T07:05:10Z

← Older revision		Revision as of 09:05, 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^]]

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

← Older revision	Revision as of 09:00, 1 June 2022
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2021-05-28T14:27:43Z

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	In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte  $\nu = 1$  bis  $\nu = 8$  dargestellt.		In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte  $\nu = 1$  bis  $\nu = 8$  dargestellt.
	*Blaue Zweige kommen entweder von  ${\it \Gamma}_{\rm \nu –1}(00)$  oder von  ${\it \Gamma}_{\rm \nu –1}(01)$  und kennzeichnen eine hypothetische ~~&bdquo;~~$0$".		*Blaue Zweige kommen entweder von  ${\it \Gamma}_{\rm \nu –1}(00)$  oder von  ${\it \Gamma}_{\rm \nu –1}(01)$  und kennzeichnen eine hypothetische "$0$".
	*Dagegen weisen alle roten Zweige – ausgehend von den Zuständen  ${\it \Gamma}_{\rm \nu –1}(10)$  bzw.  ${\it \Gamma}_{\rm \nu –1}(11)$  – auf das Symbol ~~&bdquo;~~$1$" hin.		*Dagegen weisen alle roten Zweige – ausgehend von den Zuständen  ${\it \Gamma}_{\rm \nu –1}(10)$  bzw.  ${\it \Gamma}_{\rm \nu –1}(11)$  – auf das Symbol "$1$" hin.


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	'''(3)'''  Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:		'''(3)'''  Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:
	*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:     ~~&bdquo;~~Rot – Blau – Rot – Rot – Blau – Rot – Blau".		*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:     "Rot – Blau – Rot – Rot – Blau – Rot – Blau".
	*Dagegen kann über das Symbol $a_8$ zum Zeitpunkt $\nu = 8$ noch keine endgültige Aussage gemacht werden:		*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 $a_8 = 1$.		*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 $a_8 = 1$.

Javier: Text replacement - "”" to """

2021-05-28T14:27:21Z

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← Older revision		Revision as of 16:27, 28 May 2021
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	In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte  $\nu = 1$  bis  $\nu = 8$  dargestellt.		In der Grafik ist das vereinfachte Trellisdiagramm für die Zeitpunkte  $\nu = 1$  bis  $\nu = 8$  dargestellt.
	*Blaue Zweige kommen entweder von  ${\it \Gamma}_{\rm \nu –1}(00)$  oder von  ${\it \Gamma}_{\rm \nu –1}(01)$  und kennzeichnen eine hypothetische &bdquo;$0$~~”~~.		*Blaue Zweige kommen entweder von  ${\it \Gamma}_{\rm \nu –1}(00)$  oder von  ${\it \Gamma}_{\rm \nu –1}(01)$  und kennzeichnen eine hypothetische &bdquo;$0$".
	*Dagegen weisen alle roten Zweige – ausgehend von den Zuständen  ${\it \Gamma}_{\rm \nu –1}(10)$  bzw.  ${\it \Gamma}_{\rm \nu –1}(11)$  – auf das Symbol &bdquo;$1$~~”~~ hin.		*Dagegen weisen alle roten Zweige – ausgehend von den Zuständen  ${\it \Gamma}_{\rm \nu –1}(10)$  bzw.  ${\it \Gamma}_{\rm \nu –1}(11)$  – auf das Symbol &bdquo;$1$" hin.


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	'''(3)'''  Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:		'''(3)'''  Richtig sind der <u>erste und der letzte Lösungsvorschlag</u>:
	*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:     &bdquo;Rot – Blau – Rot – Rot – Blau – Rot – Blau~~”~~.		*Die Folge $1011010$ erkennt man aus dem durchgehenden Pfad:     &bdquo;Rot – Blau – Rot – Rot – Blau – Rot – Blau".
	*Dagegen kann über das Symbol $a_8$ zum Zeitpunkt $\nu = 8$ noch keine endgültige Aussage gemacht werden:		*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 $a_8 = 1$.		*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 $a_8 = 1$.