Difference between revisions of "Aufgaben:Exercise 5.7: McCullough and Gilbert-Elliott Parameters"

Latest revision as of 13:25, 19 October 2022

Gilbert-Elliott and McCullough model

In "Exercise 5.6" and "Exercise 5.6Z" the GE model with the parameter values

$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001, \hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm} {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.1, \hspace{0.2cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

are examined in more detail. Compared to these exercises, the transition probabilities are now renamed, for example $p(\rm B\hspace{0.05cm}|\hspace{0.05cm}G)$ is written instead of $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G)$. In the upper graph this renaming is already done.

The lower graphic shows the MC model of McCullough. This has exactly the same structure as the GE model, but now all probabilities are denoted by $q$ instead of $p$.

For example, in the MC model $q\rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$ denotes the transition probability from state $\rm G$ to state $\rm B$ under the assumption that an error has just occurred in state $\rm G$. In contrast, the GE parameter $p \rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$ denotes this transition probability without any additional condition.

The parameters of the GE model ⇒ $p_{\rm G}, \hspace{0.1cm} p_{\rm B}, \hspace{0.1cm} p({\rm B\hspace{0.05cm}|\hspace{0.05cm}G}), \hspace{0.1cm} p({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$ can be converted to the corresponding MC parameters $q_{\rm G}, \hspace{0.1cm} q_{\rm B},\hspace{0.1cm} q({\rm B\hspace{0.05cm}|\hspace{0.05cm}G})$ and $q({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$ in such a way that an error sequence identical in its statistical properties to the GE model is produced, but not the identical sequence.

The conversion equations are:

$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1-\beta_{\rm G}\hspace{0.05cm}, \hspace{0.2cm}q_{\rm B} = 1-\beta_{\rm B}\hspace{0.05cm},\hspace{0.5cm} q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{\alpha_{\rm B} \cdot \big [{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) + {\rm Pr}( G\hspace{0.05cm}|\hspace{0.05cm} B )\big ]}{\alpha_{\rm G} \cdot {\it q}_{\rm B} + \alpha_{\rm B} \cdot {\it q}_{\rm G}} \hspace{0.05cm},\hspace{0.5cm} \hspace{0.2cm}{\it q}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot {\it q}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )\hspace{0.05cm}.$$

Here, the following auxiliary quantities are used:

$$u_{\rm GG} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\rm G}) \hspace{0.05cm},\hspace{0.2cm} {\it u}_{\rm GB} ={\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\hspace{0.03cm} \rm G}) \hspace{0.05cm},$$

$$u_{\rm BB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} B ) \cdot (1-{\it p}_{\hspace{0.03cm}\rm B}) \hspace{0.05cm},\hspace{0.29cm} {\it u}_{\rm BG} ={\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) \cdot (1-{\it p}_{\hspace{0.03cm}\rm B})\hspace{0.05cm}$$

$$\Rightarrow \hspace{0.3cm} \beta_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm GG} + u_{\rm BB} + \sqrt{(u_{\rm GG} - u_{\rm BB})^2 + 4 \cdot u_{\rm GB}\cdot u_{\rm BG}}}{2} \hspace{0.05cm},$$

$$\beta_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm GG} + u_{\rm BB} - \sqrt{(u_{\rm GG} - u_{\rm BB})^2 + 4 \cdot u_{\rm GB}\cdot u_{\rm BG}}}{2}\hspace{0.05cm},$$

$$x_{\rm G} =\frac{u_{\rm BG}}{\beta_{\rm G}-u_{\rm BB}} \hspace{0.05cm},\hspace{0.2cm} x_{\rm B} =\frac{u_{\rm BG}}{\beta_{\rm B}-u_{\rm BB}}$$

$$\Rightarrow \hspace{0.3cm} \alpha_{\rm G} = \frac{(w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}\cdot x_{\rm G})( x_{\rm B}-1)}{p_{\rm M} \cdot( x_{\rm B}-x_{\rm G})} \hspace{0.05cm}, \hspace{0.2cm}\alpha_{\rm B} = 1-\alpha_{\rm G}\hspace{0.05cm}.$$

$w_{\rm G}$ and $w_{\rm B}$ are the state probabilities for "GOOD" and "BAD" of the GE model. In "Exercise 5.6Z", these were calculated as follows:

$$w_{\rm G} = {10}/{11}\hspace{0.05cm}, \hspace{0.2cm}w_{\rm B} = {1}/{11}\hspace{0.05cm}.$$

The corresponding state probabilities of the MC model are $\alpha_{\rm G}$ and $\alpha_{\rm B}$.

Notes:

The exercise belongs to the chapter "Burst Error Channels".
In the following "Exercise 5.7Z" the most important descriptive quantities are calculated directly from the MC parameters:
- Error correlation function,
- correlation duration,
- mean error probability and
- error distance distribution

Questions

$u_{\rm GG}\hspace{0.05cm} = \ $

$u_{\rm BG}\ = \ $

$u_{\rm GB}\ = \ $

$u_{\rm BB}\ = \ $

$\beta_{\rm G}\hspace{0.3cm} = \ $

$\beta_{\rm B}\hspace{0.32cm} = \ $

$q_{\rm G} \hspace{0.08cm} = \ $

$q_{\rm B} \ = \ $

$x_{\rm G} \ = \ $

$x_{\rm B} \ = \ $

$\alpha_{\rm G} \hspace{0.05cm} = \ $

$\alpha_{\rm B} \ = \ $

$q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)\ = \ $

$q(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)\ = \ $

Solution

(1) For the $u$ auxiliary variables:

$$u_{\rm GG} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\rm G}) = 0.99 \cdot (1 -0.001) \hspace{0.15cm}\underline {\approx 0.98901} \hspace{0.05cm},$$

$$u_{\rm BG} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) \cdot (1-{\it p}_{\rm B})= 0.1 \cdot (1 -0.1) \hspace{0.15cm}\underline {\approx 0.09} \hspace{0.05cm},$$

$$u_{\rm GB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\rm G})= 0.01 \cdot (1 -0.001) \hspace{0.15cm}\underline {\approx 0.00999} \hspace{0.05cm},$$

$$u_{\rm BB} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} B ) \cdot (1-{\it p}_{\rm B})= 0.9 \cdot (1 -0.1) \hspace{0.15cm}\underline {\approx 0.81}\hspace{0.05cm}.$$

It follows for the $\beta$ auxiliary quantities:

$$\beta_{\rm G} =\frac{u_{\rm GG} + u_{\rm BB} + \sqrt{(u_{\rm GG} - u_{\rm BB})^2 + 4 \cdot u_{\rm GB}\cdot u_{\rm BG}}}{2} =\frac{0.98901 + 0.81 + \sqrt{(0.98901 - 0.81)^2 + 4 \cdot 0.00999\cdot 0.09}}{2} $$

$$\Rightarrow \hspace{0.3cm}\beta_{\rm G}= \frac{1.79901 + \sqrt{0.03204 + 0.003596}}{2} = \frac{1.79901 + 0.18877}{2} \hspace{0.15cm}\underline {= 0.9939} \hspace{0.05cm},$$

$$\beta_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm GG} + u_{\rm BB} - \sqrt{(u_{\rm GG} - u_{\rm BB})^2 + 4 \cdot u_{\rm GB}\cdot u_{\rm BG}}}{2}\hspace{0.05cm}, = \ \hspace{-0.1cm}\text{...} = \frac{1.79901 - 0.18877}{2} \hspace{0.15cm}\underline {= 0.8051} \hspace{0.05cm}.$$

(2) Using the result of subtask (1), we obtain:

$$q_{\rm G} = 1-\beta_{\rm G} = 1 - 0.9939 \hspace{0.15cm}\underline {= 0.0061}\hspace{0.05cm}, \hspace{0.3cm}q_{\rm B} = 1-\beta_{\rm B}= 1 - 0.8051 \hspace{0.15cm}\underline {= 0.1949}\hspace{0.05cm}.$$

(3) According to the data section, the following is to be applied here

$$x_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm BG}}{\beta_{\rm G}-u_{\rm BB}}= \frac{0.0999}{0.9939-0.81}\hspace{0.15cm}\underline {= 0.5432}\hspace{0.05cm},$$

$$ x_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm BG}}{\beta_{\rm B}-u_{\rm BB}}= \frac{0.0999}{0.8051-0.81}\hspace{0.15cm}\underline {= -20.388}\hspace{0.05cm},$$

$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{(w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}\cdot x_{\rm G})( x_{\rm B}-1)}{p_{\rm M} \cdot( x_{\rm B}-x_{\rm G})} = \frac{(0.9091 \cdot 0.001 + 0.0909 \cdot 0.1\cdot 0.5432)( -20.388-1)}{0.01 \cdot( -20.388 -0.5432)} \hspace{0.15cm}\underline {= 0.5975} \hspace{0.05cm},$$

$$\alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1-\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$

(4) According to the given equations:

$$q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{\alpha_{\rm B} \cdot[{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )]}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdot q_{\rm G}} =\frac{0.4025 \cdot[0.1 + 0.01]}{0.5975 \cdot 0.1949 + 0.4025 \cdot 0.0061}\hspace{0.15cm}\underline {= 0.3724} \hspace{0.05cm},$$

$$q(\rm G\hspace{0.05cm}|\hspace{0.05cm} B ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )= \frac{0.5975}{0.4025} \cdot 0.3724 \hspace{0.15cm}\underline {= 0.5528}\hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Bündelfehlerkanäle}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Burst_Error_Channels}}
-[[File:P_ID1844__Dig_A_5_7.png|right|frame|Gilbert-Elliott– und McCullough–Modell]]
+[[File:EN Dig A 5 7.png|right|frame|Gilbert-Elliott and McCullough model]]
-In [[Aufgaben:5.6:_Fehlerkorrelationsdauer| Aufgabe 5.6]] und [[Aufgaben:5.6Z_GE-Modelleigenschaften| Aufgabe 5.6Z]] wurden jeweils das GE&ndash;Modell mit den Parameterwerten
+In&nbsp; [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration| "Exercise 5.6"]]&nbsp; and&nbsp; [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model| "Exercise 5.6Z"]]&nbsp; the GE model with the parameter values
 :$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,
 \hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}
@@ Line 11: / Line 11: @@
 B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
-genauer untersucht. Gegenüber diesen Aufgaben werden nun die Übergangswahrscheinlichkeiten umbenannt, beispielsweise wird $p(\rm B\hspace{0.05cm}|\hspace{0.05cm}G)$ anstelle von $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G)$ geschrieben. In der oberen Grafik ist diese Umbenennung bereits vorgenommen.
+are examined in more detail. Compared to these exercises, the transition probabilities are now renamed, for example&nbsp; $p(\rm B\hspace{0.05cm}|\hspace{0.05cm}G)$&nbsp; is written instead of &nbsp;$\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G)$.&nbsp; In the upper graph this renaming is already done.
-Die untere Grafik zeigt das MC&ndash;Modell von McCullough. Dieses besitzt die genau gleiche Struktur wie das GE&ndash;Modell, doch werden nun alle Wahrscheinlichkeiten mit $q$ anstelle von $p$ bezeichnet.
+The lower graphic shows the MC model of McCullough. This has exactly the same structure as the GE model, but now all probabilities are denoted by&nbsp; $q$&nbsp; instead of&nbsp; $p$.&nbsp;
-Beispielsweise bezeichnet beim MC&ndash;Modell $q\rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$ die Übergangswahrscheinlichkeit von Zustand $\rm G$ in den Zustand$\rm B$ unter der Voraussetzung, dass im Zustand $\rm G$ gerade ein Fehler aufgetreten ist. Der GE&ndash;Parameter $p \rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$ kennzeichnet dagegen diese Übergangswahrscheinlichkeit ohne Zusatzbedingung.
+For example, in the MC model&nbsp; $q\rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$&nbsp; denotes the transition probability from state&nbsp; $\rm G$&nbsp; to state&nbsp; $\rm B$&nbsp; under the assumption that an error has just occurred in state&nbsp; $\rm G$.&nbsp; In contrast, the GE parameter&nbsp; $p \rm (B\hspace{0.05cm}|\hspace{0.05cm}G)$&nbsp; denotes this transition probability without any additional condition.
-Die Parameter des GE&ndash;Modells &nbsp; &#8658; &nbsp; $p_{\rm G}, p_{\rm B}, p({\rm B\hspace{0.05cm}|\hspace{0.05cm}G}), p({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$ können so in die entsprechenden MC&ndash;Parameter $q_{\rm G}, q_{\rm B}, q({\rm B\hspace{0.05cm}|\hspace{0.05cm}G})$ und $q({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$ umgerechnet werden, dass eine in ihren statistischen Eigenschaften gleiche Fehlerfolge wie beim GE&ndash;Modell erzeugt wird, allerdings nicht die identische Folge.
+The parameters of the GE model &nbsp; &#8658; &nbsp; $p_{\rm G}, \hspace{0.1cm} p_{\rm B}, \hspace{0.1cm} p({\rm B\hspace{0.05cm}|\hspace{0.05cm}G}), \hspace{0.1cm} p({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$ &nbsp; can be converted to the corresponding MC parameters&nbsp; $q_{\rm G}, \hspace{0.1cm} q_{\rm B},\hspace{0.1cm} q({\rm B\hspace{0.05cm}|\hspace{0.05cm}G})$&nbsp; and &nbsp;$q({\rm G\hspace{0.05cm}|\hspace{0.05cm}B})$&nbsp; in such a way that an  error sequence identical in its statistical properties to the GE model is produced, but not the identical sequence.
-Die Umrechnungsgleichungen lauten:
+The conversion equations are:
 :$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1-\beta_{\rm
 G}\hspace{0.05cm}, \hspace{0.2cm}q_{\rm
 B} = 1-\beta_{\rm B}\hspace{0.05cm},\hspace{0.5cm}
 q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \hspace{-0.1cm} \ = \
-\hspace{-0.1cm}\frac{\alpha_{\rm B} \cdot[{\rm Pr}(\rm
+\hspace{-0.1cm}\frac{\alpha_{\rm B} \cdot \big [{\rm Pr}(\rm
-B\hspace{0.05cm}|\hspace{0.05cm} G ) + {\rm Pr}(\rm
+B\hspace{0.05cm}|\hspace{0.05cm} G ) + {\rm Pr}(
-G\hspace{0.05cm}|\hspace{0.05cm} B )]}{\alpha_{\rm G} \cdot q_{\rm
+G\hspace{0.05cm}|\hspace{0.05cm} B )\big ]}{\alpha_{\rm G} \cdot {\it q}_{\rm
-B} + \alpha_{\rm B} \cdot q_{\rm G}} \hspace{0.05cm},\hspace{0.5cm}
+B} + \alpha_{\rm B} \cdot {\it q}_{\rm G}} \hspace{0.05cm},\hspace{0.5cm}
-\hspace{0.2cm}q(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )\hspace{-0.1cm} \ = \
+\hspace{0.2cm}{\it q}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )\hspace{-0.1cm} \ = \
 \hspace{-0.1cm}
-\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm
+\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot {\it q}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G )\hspace{0.05cm}.$$
-Hierbei sind die folgenden Hilfsgrößen verwendet:
+Here, the following auxiliary quantities are used:
 :$$u_{\rm GG} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\rm G})
@@ Line 62: / Line 62: @@
 \hspace{0.2cm}\alpha_{\rm B} = 1-\alpha_{\rm G}\hspace{0.05cm}.$$
-$w_{\rm G}$ und $w_{\rm B}$ sind die Zustandswahrscheinlichkeiten für &bdquo;GOOD&rdquo; und &bdquo;BAD&rdquo; des GE&ndash;Modells. In der  [[Aufgaben:5.6Z_GE-Modelleigenschaften| Aufgabe 5.6Z]] wurden diese wie folgt berechnet:
+$w_{\rm G}$&nbsp; and &nbsp;$w_{\rm B}$&nbsp; are the state probabilities for "GOOD" and "BAD" of the GE model. In&nbsp;  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model| "Exercise 5.6Z"]],&nbsp; these were calculated as follows:
 :$$w_{\rm G} = {10}/{11}\hspace{0.05cm}, \hspace{0.2cm}w_{\rm B}
 = {1}/{11}\hspace{0.05cm}.$$
-Die entsprechenden Zustandswahrscheinlichkeiten des MC&ndash;Modells sind $\alpha_{\rm G}$ und $\alpha_{\rm B}$.
+The corresponding state probabilities of the MC model are&nbsp; $\alpha_{\rm G}$&nbsp; and &nbsp;$\alpha_{\rm B}$.
+''Notes:''
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Burst_Error_Channels| "Burst Error Channels"]].
+* In the following&nbsp; [[Aufgaben:Exercise_5.7Z:_McCullough_Model_once_more|"Exercise 5.7Z"]]&nbsp; the most important descriptive quantities are calculated directly from the MC parameters:
+** Error correlation function,
+** correlation duration,
+** mean error probability and
+** error distance distribution
-''Hinweise:''
-* Die Aufgabe gehört zum Kapitel [[Digitalsignal%C3%BCbertragung/B%C3%BCndelfehlerkan%C3%A4le| Bündelfehlerkanäle]].
-* In der nachfolgenden [[5.7Z_Nochmals_MC-Modell|Aufgabe 5.7Z]] werden die wichtigsten Beschreibungsgrößen wie
-** Fehlerkorrelationsfunktion,
-** Korrelationsdauer,
-** mittlere Fehlerwahrscheinlichkeit und
-** Fehlerabstandsverteilung
-: direkt aus den MC&ndash;Parametern berechnet.
-* Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Berechnen Sie die folgenden Hilfsgrößen:
+{Calculate the following auxiliary variables:
 |type="{}"}
-$u_{\rm GG}\ = \ $ { 0.98901 3% }
+$u_{\rm GG}\hspace{0.05cm} = \ $ { 0.98901 3% }
 $u_{\rm BG}\ = \ $ { 0.09 3% }
 $u_{\rm GB}\ = \ $ { 0.00999 3% }
@@ Line 93: / Line 96: @@
 $\beta_{\rm B}\hspace{0.32cm} = \ $ { 0.8051 3% }
-{Wie lauten die beiden Fehlerwahrscheinlichkeiten des MC&ndash;Modells?
+{What are the two error probabilities of the MC model?
 |type="{}"}
-$q_{\rm G} \ = \ $ { 0.061 3% }
+$q_{\rm G} \hspace{0.08cm} = \ $ { 0.0061 3% }
 $q_{\rm B} \ = \ $ { 0.1949 3% }
-{Berechnen Sie die weiteren Hilfsgrößen:
+{Calculate the other auxiliary quantities:
 |type="{}"}
 $x_{\rm G} \ = \ $ { 0.5432 3% }
@@ Line 105: / Line 108: @@
 $\alpha_{\rm B} \ = \ $ { 0.4025 3% }
-{Berechnen Sie die Übergangswahrscheinlichkeiten des MC&ndash;Modells:
+{Calculate the transition probabilities of the MC model:
 |type="{}"}
 $q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)\ = \ $ { 0.3724 3% }
@@ Line 111: / Line 114: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Für die $u$&ndash;Hilfsgrößen gilt:
+'''(1)'''&nbsp; For the $u$ auxiliary variables:
 :$$u_{\rm GG} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} G ) \cdot (1-{\it p}_{\rm G}) = 0.99
@@ Line 128: / Line 131: @@
 \cdot (1 -0.1) \hspace{0.15cm}\underline {\approx 0.81}\hspace{0.05cm}.$$
-Daraus folgt für die $\beta$&ndash;Hilfsgrößen:
+*It follows for the $\beta$ auxiliary quantities:
-:$$\beta_{\rm G} \hspace{-0.1cm} \ = \
+:$$\beta_{\rm G} =\frac{u_{\rm GG} + u_{\rm BB} + \sqrt{(u_{\rm GG} -
-\hspace{-0.1cm}\frac{u_{\rm GG} + u_{\rm BB} + \sqrt{(u_{\rm GG} -
 u_{\rm BB})^2 + 4 \cdot u_{\rm GB}\cdot u_{\rm BG}}}{2}
-\hspace{0.05cm},$$
+=\frac{0.98901 + 0.81 +
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{0.98901 + 0.81 +
 \sqrt{(0.98901 - 0.81)^2 + 4 \cdot 0.00999\cdot 0.09}}{2}
-\hspace{0.05cm},$$
+$$
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{1.79901 + \sqrt{0.03204
+:$$\Rightarrow \hspace{0.3cm}\beta_{\rm G}= \frac{1.79901 + \sqrt{0.03204
 + 0.003596}}{2} = \frac{1.79901 + 0.18877}{2} \hspace{0.15cm}\underline {= 0.9939}
 \hspace{0.05cm},$$
 :$$\beta_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm
 GG} + u_{\rm BB} - \sqrt{(u_{\rm GG} - u_{\rm BB})^2 + 4 \cdot
-u_{\rm GB}\cdot u_{\rm BG}}}{2}\hspace{0.05cm},$$
+u_{\rm GB}\cdot u_{\rm BG}}}{2}\hspace{0.05cm}, = \ \hspace{-0.1cm}\text{...} =  \frac{1.79901 - 0.18877}{2}
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm}... =  \frac{1.79901 - 0.18877}{2}
 \hspace{0.15cm}\underline {= 0.8051} \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Mit dem Ergebnis der Teilaufgabe (1) erhält man:
+'''(2)'''&nbsp; Using the result of subtask '''(1)''', we obtain:
 :$$q_{\rm G} = 1-\beta_{\rm G} =
 - 0.9939 \hspace{0.15cm}\underline {= 0.0061}\hspace{0.05cm}, \hspace{0.3cm}q_{\rm B} = 1-\beta_{\rm B}= 1 - 0.8051 \hspace{0.15cm}\underline {=
@@ Line 152: / Line 152: @@
-'''(3)'''&nbsp; Entsprechend dem Angabenblatt ist hier anzusetzen
+'''(3)'''&nbsp; According to the data section, the following is to be applied here
 :$$x_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{u_{\rm
 BG}}{\beta_{\rm G}-u_{\rm BB}}=
@@ Line 161: / Line 162: @@
 :$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{(w_{\rm
 G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}\cdot x_{\rm G})(
-x_{\rm B}-1)}{p_{\rm M} \cdot( x_{\rm B}-x_{\rm G})} = $$
+x_{\rm B}-1)}{p_{\rm M} \cdot( x_{\rm B}-x_{\rm G})} =  \frac{(0.9091 \cdot 0.001 +
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{(0.9091 \cdot 0.001 +
 .0909 \cdot 0.1\cdot 0.5432)( -20.388-1)}{0.01 \cdot( -20.388
 -0.5432)} \hspace{0.15cm}\underline {= 0.5975} \hspace{0.05cm},$$
@@ Line 170: / Line 170: @@
-'''(4)'''&nbsp; Entsprechend den vorgegebenen Gleichungen gilt:
+'''(4)'''&nbsp; According to the given equations:
 :$$q(\rm B\hspace{0.05cm}|\hspace{0.05cm} G ) \hspace{-0.1cm} \ = \
 \hspace{-0.1cm}\frac{\alpha_{\rm B} \cdot[{\rm Pr}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G ) + {\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} B )]}{\alpha_{\rm G} \cdot q_{\rm
-B} + \alpha_{\rm B} \cdot q_{\rm G}} \hspace{0.05cm},$$
+B} + \alpha_{\rm B} \cdot q_{\rm G}} =\frac{0.4025 \cdot[0.1 +
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm}\frac{0.4025 \cdot[0.1 +
 .01]}{0.5975 \cdot 0.1949 + 0.4025 \cdot 0.0061}\hspace{0.15cm}\underline {= 0.3724}
 \hspace{0.05cm},$$
@@ Line 187: / Line 186: @@
-[[Category:Aufgaben zu Digitalsignalübertragung|^5.3 Bündelfehlerkanäle^]]
+[[Category:Digital Signal Transmission: Exercises|^5.3 Burst Error Channels^]]