Aufgaben:Exercise 3.11Z: Metric and Accumutated Metric - Revision history

Guenter at 11:19, 13 July 2022

2022-07-13T11:19:18Z

Guenter: Guenter moved page Aufgaben:Exercise 3.11Z: Maximum Likelihood Error Variables to Aufgaben:Exercise 3.11Z: Metric and Accumutated Metric

2022-07-13T11:18:32Z

Guenter moved page Exercise 3.11Z: Maximum Likelihood Error Variables to Exercise 3.11Z: Metric and Accumutated Metric

Hwang at 13:45, 11 July 2022

2022-07-11T13:45:19Z

Hwang at 13:44, 11 July 2022

2022-07-11T13:44:57Z

Guenter at 13:01, 5 July 2022

2022-07-05T13:01:13Z

Guenter at 15:09, 4 July 2022

2022-07-04T15:09:12Z

Guenter at 13:50, 4 July 2022

2022-07-04T13:50:55Z

Guenter at 13:28, 4 July 2022

2022-07-04T13:28:44Z

Guenter at 13:21, 4 July 2022

2022-07-04T13:21:45Z

Hwang at 09:22, 8 June 2022

2022-06-08T09:22:21Z

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← Older revision		Revision as of 11:19, 13 July 2022
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	[[File:P_ID1476__Dig_Z_3_11.png\|right\|frame\|Calculation of the minimum accumulated metrics]]		[[File:P_ID1476__Dig_Z_3_11.png\|right\|frame\|Calculation of the minimum accumulated metrics]]
−	~~'''KORREKTUR: Titel - error variables -> metrics?'''~~
	For the maximum likelihood constellation with bipolar amplitude coefficient   $a_{\rm \nu} ∈ \{+1, –1\}$  discussed in   [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram\|"Exercise 3.11"]],  the metrics   $\varepsilon_{\rm \nu}(i)$  and the minimum accumulated metrics   ${\it \Gamma}_{\rm \nu}(–1)$ and ${\it \Gamma}_{\rm \nu}(+1)$  are to be determined.		For the maximum likelihood constellation with bipolar amplitude coefficient   $a_{\rm \nu} ∈ \{+1, –1\}$  discussed in   [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram\|"Exercise 3.11"]],  the metrics   $\varepsilon_{\rm \nu}(i)$  and the minimum accumulated metrics   ${\it \Gamma}_{\rm \nu}(–1)$ and ${\it \Gamma}_{\rm \nu}(+1)$  are to be determined.

@@ Line 3: / Line 3: @@
 [[File:P_ID1476__Dig_Z_3_11.png|right|frame|Calculation of the minimum accumulated metrics]]
-'''KORREKTUR: Titel'''
+'''KORREKTUR: Titel - error variables -> metrics?'''
 For the maximum likelihood constellation with bipolar amplitude coefficient &nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; discussed in &nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram|"Exercise 3.11"]],&nbsp;  the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; and the minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(&ndash;1)$ and ${\it \Gamma}_{\rm \nu}(+1)$&nbsp; are to be determined.

← Older revision		Revision as of 13:44, 11 July 2022
Line 3:		Line 3:

	[[File:P_ID1476__Dig_Z_3_11.png\|right\|frame\|Calculation of the minimum accumulated metrics]]		[[File:P_ID1476__Dig_Z_3_11.png\|right\|frame\|Calculation of the minimum accumulated metrics]]
		+	'''KORREKTUR: Titel'''
	For the maximum likelihood constellation with bipolar amplitude coefficient   $a_{\rm \nu} ∈ \{+1, –1\}$  discussed in   [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram\|"Exercise 3.11"]],  the metrics   $\varepsilon_{\rm \nu}(i)$  and the minimum accumulated metrics   ${\it \Gamma}_{\rm \nu}(–1)$ and ${\it \Gamma}_{\rm \nu}(+1)$  are to be determined.		For the maximum likelihood constellation with bipolar amplitude coefficient   $a_{\rm \nu} ∈ \{+1, –1\}$  discussed in   [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram\|"Exercise 3.11"]],  the metrics   $\varepsilon_{\rm \nu}(i)$  and the minimum accumulated metrics   ${\it \Gamma}_{\rm \nu}(–1)$ and ${\it \Gamma}_{\rm \nu}(+1)$  are to be determined.

← Older revision		Revision as of 13:01, 5 July 2022
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	*Also assume bipolar and equal probability amplitude coefficients:   ${\rm Pr} (a_\nu = -\hspace{-0.05cm}1) = {\rm Pr} (a_\nu = +1)= 0.5.$		*Also assume bipolar and equal probability amplitude coefficients:   ${\rm Pr} (a_\nu = -\hspace{-0.05cm}1) = {\rm Pr} (a_\nu = +1)= 0.5.$

−	* The topic is also covered in the interactive SWF applet   [[Applets:Viterbi\|"Properties of the Viterbi Receiver"]].

@@ Line 5: / Line 5: @@
 For the maximum likelihood constellation with bipolar amplitude coefficient &nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; discussed in &nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram|"Exercise 3.11"]],&nbsp;  the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; and the minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(&ndash;1)$ and ${\it \Gamma}_{\rm \nu}(+1)$&nbsp; are to be determined.
-The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp; These,&nbsp; as well as the detection samples &nbsp; $d_0$&nbsp; and&nbsp; $d_1$,&nbsp; can be taken from the following calculations for the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; at times &nbsp; $\nu = 0$&nbsp; and&nbsp; $\nu = 1$.&nbsp; Note that the symbol &nbsp;$a_0 = 0$&nbsp; is always sent before the actual message &nbsp; $(a_1$,&nbsp; $a_2$,&nbsp; $a_3)$.&nbsp;
+#The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp;
 For time &nbsp;$\nu = 0$&nbsp; holds:
 :$$\varepsilon_{0}(+1) \ = \ \big[-0.4- 0.4\big]^2=0.64 \hspace{0.05cm},$$

@@ Line 71: / Line 71: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From the equations on the information section one can see $d_0 = \underline{&ndash;0.4}$ and $d_1 = \underline {&ndash;0.8}$.
+'''(1)'''&nbsp; From the equations on the information section one can see&nbsp; $d_0 = \underline{&ndash;0.4}$&nbsp; and&nbsp; $d_1 = \underline {&ndash;0.8}$.
-'''(2)'''&nbsp; The metrics $\varepsilon_0(i)$ include the basic pulse value $g_{\rm &ndash;1}$, which is used to establish the relationship between the amplitude coefficient $a_1$ and the detection sample $d_0$ ($g_0$ is not included in these equations).
+'''(2)'''&nbsp; The metric&nbsp; $\varepsilon_0(i)$&nbsp; include the basic pulse value&nbsp; $g_{\rm &ndash;1}$,&nbsp; which is used to establish the relationship between the amplitude coefficient&nbsp; $a_1$&nbsp; and the detection sample&nbsp; $d_0$&nbsp; $(g_0$&nbsp; is not included in these equations$)$.
-*One can see $g_{\rm &ndash;1}\  \underline {= 0.4}$.
+*One can see&nbsp; $g_{\rm &ndash;1}\  \underline {= 0.4}$.
 *From the equations for $\nu = 1$, the main value $g_0 \ \underline {= 0.6}$ can be read.
-'''(3)'''&nbsp; The correct solutions are <u>1 and 4</u>:
+'''(3)'''&nbsp; The correct solutions are&nbsp; <u>1 and 4</u>:
-*The possible useful samples are $\pm g_0 \pm g_{\rm &ndash;1} = \pm 0.6 \pm0.4$, i.e. $\underline {&plusmn;0.2}$ and $\underline {&plusmn;1.0}$.
+*The possible useful samples are&nbsp; $\pm g_0 \pm g_{\rm &ndash;1} = \pm 0.6 \pm0.4$,&nbsp; i.e.&nbsp; $\underline {&plusmn;0.2}$&nbsp; and&nbsp; $\underline {&plusmn;1.0}$.
-*In contrast, unipolar signaling &nbsp; &rArr; &nbsp; $a_\nu \in \{0, \hspace{0.05cm} 1\}$ would result in values of $0, \ 0.4, \ 0.6$ and $1$.
-*The relationship between bipolar values $b_i$ and unipolar equivalents $u_i$ is generally: &nbsp; $b_i = 2 \cdot u_i - 1  \hspace{0.05cm}.$
+*In contrast,&nbsp; unipolar signaling &nbsp; &rArr; &nbsp; $a_\nu \in \{0, \hspace{0.05cm} 1\}$&nbsp; would result in values of&nbsp; $0, \ 0.4, \ 0.6$&nbsp; and&nbsp; $1$.
 '''(4)'''&nbsp; The metrics are obtained for $\nu = 2$ considering the result from (3) as follows:
 :$$\varepsilon_{2}(+1, +1)  \ = \ [0.1 - 1.0]^2=0.81,\hspace{0.2cm}
    \varepsilon_{2}(-1, +1)  = [0.1 +0.2]^2=0.09
@@ Line 95: / Line 98: @@
    \hspace{0.05cm}.$$
-Thus, the minimum accumulated metrics are:
+Thus,&nbsp; the minimum accumulated metrics are:
 :$${\it \Gamma}_{2}(+1) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(+1) + \varepsilon_{2}(+1, +1),
   \hspace{0.2cm}{\it \Gamma}_{1}(-1) + \varepsilon_{2}(-1, +1)\right] =
@@ Line 105: / Line 108: @@
 [[File:P_ID1480__Dig_Z_3_11d.png|right|frame|Calculation of the minimum accumulated metrics]]
-<br>In the adjacent trellis diagram, the state "$1$" is to be interpreted as "$+1$" and "$0$" as "$&ndash;1$".
+<br>In the adjacent trellis diagram,&nbsp; the state&nbsp; "$1$"&nbsp; is to be interpreted as&nbsp; "$+1$"&nbsp; and&nbsp; "$0$"&nbsp; as&nbsp; "$&ndash;1$".
 Then holds:
-*${\it \Gamma}_2(+1) = 0.13$ is the minimum total error quantity under the hypothesis that the following symbol will be $a_3 = +1$.
+*${\it \Gamma}_2(+1) = 0.13$&nbsp; is the minimum accumulated metric under the hypothesis that the following symbol will be&nbsp;  $a_3 = +1$.
 *Under this assumption, $a_2 = \ &ndash;1$ is more likely than $a_2 = +1$, as shown in the trellis diagram (the incoming path is blue).
-*A realistic alternative to the combination "$a_2 = \ &ndash;1, a_3 = +1$" is "$a_2 = +1, a_3 = \ &ndash;1$", which lead to the minimum total error quantity ${\it \Gamma}_2(&ndash;1) = 0.37$. Here, the incoming path is red.
+*Under this assumption,&nbsp; $a_2 = \ &ndash;1$&nbsp; is more likely than&nbsp; $a_2 = +1$,&nbsp; as shown in the trellis diagram&nbsp; (the incoming path is blue).
-'''(5)'''&nbsp; For time $\nu = 3$, the following equations hold:
+'''(5)'''&nbsp; For time&nbsp; $\nu = 3$,&nbsp; the following equations hold:
 :$$\varepsilon_{3}(+1, +1)  \ = \ [0.5 - 1.0]^2=0.25,\hspace{0.2cm}
    \varepsilon_{3}(-1, +1)  = [0.5 +0.2]^2=0.49
@@ Line 125: / Line 131: @@
 .22}  \hspace{0.05cm}.$$
-*In both equations, the first term in each case is the smaller, with ${\it \Gamma}_2(+1) = 0.13$ included in each case.
+*In both equations,&nbsp; the first term in each case is the smaller,&nbsp; with&nbsp; ${\it \Gamma}_2(+1) = 0.13$&nbsp; included in each case.
-If we follow the continuous path in the trellis diagram, the other amplitude coefficients are also fixed by fixing $a_3 = +1$:
+*If we follow the continuous path in the trellis diagram from the right to the left,&nbsp; the other amplitude coefficients are also fixed by fixing&nbsp; $a_3 = +1$:
 :$$a_1 = a_2 = \ &ndash;1.$$
 {{ML-Fuß}}

@@ Line 5: / Line 5: @@
 For the maximum likelihood constellation with bipolar amplitude coefficient &nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; discussed in &nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram|"Exercise 3.11"]],&nbsp;  the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; and the minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(&ndash;1)$ and ${\it \Gamma}_{\rm \nu}(+1)$&nbsp; are to be determined.
-The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp; These, as well as the detection samples &nbsp; $d_0$&nbsp; and&nbsp; $d_1$,&nbsp; can be taken from the following calculations for the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; at the times &nbsp; $\nu = 0$&nbsp; and&nbsp; $\nu = 1$.&nbsp; Note that the symbol &nbsp;$a_0 = 0$&nbsp; is always sent before the actual message &nbsp; $(a_1$, $a_2$, $a_3)$.&nbsp;
+The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp; These,&nbsp; as well as the detection samples &nbsp; $d_0$&nbsp; and&nbsp; $d_1$,&nbsp; can be taken from the following calculations for the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; at times &nbsp; $\nu = 0$&nbsp; and&nbsp; $\nu = 1$.&nbsp; Note that the symbol &nbsp;$a_0 = 0$&nbsp; is always sent before the actual message &nbsp; $(a_1$,&nbsp; $a_2$,&nbsp; $a_3)$.&nbsp;
-For the time &nbsp;$\nu = 0$&nbsp; holds:
+For time &nbsp;$\nu = 0$&nbsp; holds:
 :$$\varepsilon_{0}(+1) \ = \ \big[-0.4- 0.4\big]^2=0.64 \hspace{0.05cm},$$
 :$$\varepsilon_{0}(-1) \ = \ \big[-0.4+ 0.4\big]^2=0.00 \hspace{0.05cm}.$$
@@ Line 23: / Line 23: @@
    \hspace{0.05cm}.$$
-The minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(-\hspace{-0.07cm}1)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(+1)$ that can be calculated with these six metrics are already plotted in the graph. The other detection samples are &nbsp; $d_{2}=0.1 \hspace{0.05cm},\hspace{0.1cm}
+The minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(-\hspace{-0.07cm}1)$ &nbsp; and &nbsp; ${\it \Gamma}_{\rm \nu}(+1)$ &nbsp; that can be calculated with these six metrics are already plotted in the graph.&nbsp; The other detection samples are &nbsp; $d_{2}=0.1 \hspace{0.05cm},\hspace{0.1cm}
 d_{3}=0.5  \hspace{0.05cm}.$
 Notes:
 *The exercise belongs to the chapter &nbsp;  [[Digital_Signal_Transmission/Viterbi_Receiver|"Viterbi Receiver"]].
-* All quantities are to be understood normalized here.
+* All quantities here are to be understood normalized.
 *Also assume bipolar and equal probability amplitude coefficients: &nbsp; ${\rm Pr} (a_\nu = -\hspace{-0.05cm}1) = {\rm Pr} (a_\nu = +1)= 0.5.$
 * The topic is also covered in the interactive applet &nbsp; [[Applets:Viterbi|"Properties of the Viterbi Receiver"]].

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Viterbi_Receiver}}
-[[File:P_ID1476__Dig_Z_3_11.png|right|frame|Calculation of the minimum total error quantities]]
+[[File:P_ID1476__Dig_Z_3_11.png|right|frame|Calculation of the minimum accumulated metrics]]
-For the maximum likelihood constellation with bipolar amplitude coefficient &nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; discussed in &nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram|"Exercise 3.11"]],&nbsp; behandelte Maximum&ndash;Likelihood&ndash;Konstellation mit bipolaren Amplitudenkoeffizient&nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; the error quantities &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; and the minimum total error quantities &nbsp; ${\it \Gamma}_{\rm \nu}(&ndash;1)$ and ${\it \Gamma}_{\rm \nu}(+1)$&nbsp; are to be determined.
+For the maximum likelihood constellation with bipolar amplitude coefficient &nbsp; $a_{\rm \nu} &#8712; \{+1, &ndash;1\}$&nbsp; discussed in &nbsp; [[Aufgaben:Exercise_3.11:_Viterbi_Receiver_and_Trellis_Diagram|"Exercise 3.11"]],&nbsp;  the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; and the minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(&ndash;1)$ and ${\it \Gamma}_{\rm \nu}(+1)$&nbsp; are to be determined.
-The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp; These, as well as the detection samples &nbsp; $d_0$&nbsp; and&nbsp; $d_1$,&nbsp; can be taken from the following calculations for the error quantities &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; at the times &nbsp; $\nu = 0$&nbsp; and&nbsp; $\nu = 1$.&nbsp; Note that the symbol &nbsp;$a_0 = 0$&nbsp; is always sent before the actual message &nbsp; $(a_1$, $a_2$, $a_3)$.&nbsp;
+The basic pulse is given by the two values &nbsp; $g_0$&nbsp; and&nbsp; $g_{\rm &ndash;1}$.&nbsp; These, as well as the detection samples &nbsp; $d_0$&nbsp; and&nbsp; $d_1$,&nbsp; can be taken from the following calculations for the metrics &nbsp; $\varepsilon_{\rm \nu}(i)$&nbsp; at the times &nbsp; $\nu = 0$&nbsp; and&nbsp; $\nu = 1$.&nbsp; Note that the symbol &nbsp;$a_0 = 0$&nbsp; is always sent before the actual message &nbsp; $(a_1$, $a_2$, $a_3)$.&nbsp;
 For the time &nbsp;$\nu = 0$&nbsp; holds:
@@ Line 13: / Line 13: @@
 From this it could be concluded already at time &nbsp; $\nu = 0$&nbsp; that with high probability &nbsp; $a_1 =  -\hspace{-0.05cm}1$.&nbsp;
-For time &nbsp; $\nu = 1$,&nbsp; the following error quantities result:
+For time &nbsp; $\nu = 1$,&nbsp; the following metrics result:
 :$$\varepsilon_{1}(+1, +1) \ = \ \big[-0.8- 0.6 -0.4\big]^2=3.24
   \hspace{0.05cm},$$
@@ Line 23: / Line 23: @@
    \hspace{0.05cm}.$$
-The minimum total error quantities &nbsp; ${\it \Gamma}_{\rm \nu}(-\hspace{-0.07cm}1)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(+1)$ that can be calculated with these six error quantities are already plotted in the graph. The other detection samples are &nbsp; $d_{2}=0.1 \hspace{0.05cm},\hspace{0.1cm}
+The minimum accumulated metrics &nbsp; ${\it \Gamma}_{\rm \nu}(-\hspace{-0.07cm}1)$&nbsp; and&nbsp; ${\it \Gamma}_{\rm \nu}(+1)$ that can be calculated with these six metrics are already plotted in the graph. The other detection samples are &nbsp; $d_{2}=0.1 \hspace{0.05cm},\hspace{0.1cm}
 d_{3}=0.5  \hspace{0.05cm}.$
@@ Line 58: / Line 58: @@
 + $&plusmn;1.0.$
-{Give the minimum total error quantities for time &nbsp; $\nu = 2$&nbsp; &nbsp;$(d_2 = 0.1)$.
+{Give the minimum accumulated metrics for time &nbsp; $\nu = 2$&nbsp; &nbsp;$(d_2 = 0.1)$.
 |type="{}"}
 ${\it \Gamma}_2(+1)\ = \ $ { 0.13 3% }
 ${\it \Gamma}_2(-\hspace{-0.05cm}1)\ = \ $ { 0.37 3% }
-{Calculate the minimum total error quantities for time &nbsp; $\nu = 3$&nbsp; $(d_3 = 0.5)$.
+{Calculate the minimum accumulated metric for time &nbsp; $\nu = 3$&nbsp; $(d_3 = 0.5)$.
 |type="{}"}
 ${\it \Gamma}_3(+1) \ = \ $ { 0.38 3% }
@@ Line 74: / Line 74: @@
-'''(2)'''&nbsp; The error quantities $\varepsilon_0(i)$ include the basic pulse value $g_{\rm &ndash;1}$, which is used to establish the relationship between the amplitude coefficient $a_1$ and the detection sample $d_0$ ($g_0$ is not included in these equations).
+'''(2)'''&nbsp; The metrics $\varepsilon_0(i)$ include the basic pulse value $g_{\rm &ndash;1}$, which is used to establish the relationship between the amplitude coefficient $a_1$ and the detection sample $d_0$ ($g_0$ is not included in these equations).
 *One can see $g_{\rm &ndash;1}\  \underline {= 0.4}$.
 *From the equations for $\nu = 1$, the main value $g_0 \ \underline {= 0.6}$ can be read.
@@ Line 87: / Line 87: @@
-'''(4)'''&nbsp; The error quantities are obtained for $\nu = 2$ considering the result from (3) as follows:
+'''(4)'''&nbsp; The metrics are obtained for $\nu = 2$ considering the result from (3) as follows:
 :$$\varepsilon_{2}(+1, +1)  \ = \ [0.1 - 1.0]^2=0.81,\hspace{0.2cm}
    \varepsilon_{2}(-1, +1)  = [0.1 +0.2]^2=0.09
@@ Line 95: / Line 95: @@
    \hspace{0.05cm}.$$
-Thus, the minimum total error quantities are:
+Thus, the minimum accumulated metrics are:
 :$${\it \Gamma}_{2}(+1) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(+1) + \varepsilon_{2}(+1, +1),
   \hspace{0.2cm}{\it \Gamma}_{1}(-1) + \varepsilon_{2}(-1, +1)\right] =
@@ Line 104: / Line 104: @@
   \hspace{0.05cm}.$$
-[[File:P_ID1480__Dig_Z_3_11d.png|right|frame|Calculation of the minimum total error quantities]]
+[[File:P_ID1480__Dig_Z_3_11d.png|right|frame|Calculation of the minimum accumulated metrics]]
 <br>In the adjacent trellis diagram, the state "$1$" is to be interpreted as "$+1$" and "$0$" as "$&ndash;1$".