Difference between revisions of "Aufgaben:Exercise 5.6Z: Gilbert-Elliott Model"

Latest revision as of 03:54, 19 September 2022

Given Gilbert–Elliott model

We consider the burst error channel model according to "E.N. Gilbert" and E.O. Elliott (see sketch). For the transition probabilities let hold:

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

The error probability in the state "GOOD" is $p_{\rm G} = 0.1\%$. For the state "BAD" is $p_{\rm B} = 10\%$.

In the course of this exercise, further parameters are to be determined:

the average error probability $p_{\rm M}$,
the state probabilities $w_{\rm G} = \rm Pr(Z = G)$ and $w_{\rm B} = \rm Pr(Z = B)$,
the values of the correlation function, which for $k > 0$ is given analytically as follows:

$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot \big [1 - {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )- {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$

Notes:

The exercise belongs to the chapter "Burst Error Channels".
However, reference is also made to the chapter "Markov chains" in the book "Theory of Stochastic Signals" and in particular to the section "Error correlation function of the Gilbert-Elliott model" in the book "Channel Coding".

Questions

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

$\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}B) \hspace{0.2cm} = \ $

$w_{\rm G} \ = \ $

$w_{\rm B} \ = \ $

$p_{\rm M} \ = \ $

$\ \%$

$\varphi_e(k = 1) \hspace{0.35cm} = \ $

$\ \cdot 10^{-4}$

$\varphi_e(k = 2) \hspace{0.35cm} = \ $

$\ \cdot 10^{-4}$

$\varphi_e(k = 5) \hspace{0.35cm} = \ $

$\ \cdot 10^{-4}$

$\varphi_e(k = 50) \ = \ $

$\ \cdot 10^{-4}$

$\varphi_e(k = 0) \ = \ $

$\ \cdot 10^{-2}$

	sole change of $p_{\rm G}$,
	sole change of $p_{\rm B}$,
	sole change of $\rm Pr(G\hspace{0.05cm}\|\hspace{0.05cm}B)$,
	sole change of $\rm Pr(B\hspace{0.05cm}\|\hspace{0.05cm}G)$?

Solution

(1) $\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}G) = 1 \, –Pr(B\hspace{0.05cm}|\hspace{0.05cm}G) \ \underline {= 0.99}$ and $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}B) = 1 \, –Pr(G\hspace{0.05cm}|\hspace{0.05cm}B) \ \underline {= 0.9}$ are valid.

(2) The GE model is a stationary Markov chain.

For the probability that it is in the state "GOOD", taking into account the result of subtask (1):

$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) \cdot w_{\rm B}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) \cdot w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) \cdot w_{\rm B} \hspace{0.05cm}.$$

Further, $w_{\rm B} = 1 \, –w_{\rm G}$:

$${\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) \cdot w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)$$

$$\Rightarrow \hspace{0.3cm} w_{\rm G} = \frac{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)} = \frac{0.1}{0.1 + 0.01} \hspace{0.15cm}\underline {\approx 0.909} \hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G }\hspace{0.15cm}\underline {\approx 0.091}\hspace{0.05cm}.$$

(3) The mean error probability $p_{\rm M}$ is obtained from the error probabilities $p_{\rm G}$ and $p_{\rm B}$ weighted by $w_{\rm G}$ and $w_{\rm B}$:

$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B} = \frac{10}{11} \cdot 10^{-3} + \frac{1}{11} \cdot 10^{-1}= \frac{10+100}{11} \cdot 10^{-3}\hspace{0.15cm}\underline { = 1\%}\hspace{0.05cm}.$$

(4) According to the general equation on the specification section, for $k > 0$:

$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 + (p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot [1 - {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G )- {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009 \cdot 0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{\it k} \right )\hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 1 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1} \right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$

$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 2} \right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$

$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 5 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 5} \right )\hspace{0.15cm}\underline {= 5.523 \cdot 10^{-4}} \hspace{0.05cm},$$

$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 50 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 50} \right ) \hspace{0.15cm}\underline {= 1.024 \cdot 10^{-4}} \hspace{0.05cm}.$$

(5) For each channel model, because of $e_{\nu} ∈ \{0, 1\}$:

$$\varphi_{e}(k = 0 ) = {\rm E}[e_{\nu} ^2] = {\rm E}[e_{\nu} ] = p_{\rm M} \hspace{0.05cm}.$$

Using the result of subtask (3), $\varphi_e(k = 0) \ \underline {= 0.01}$ for the present case.

(6) According to subtask (3) holds:

$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \cdot p_{\rm B} \hspace{0.05cm}.$$

With given $p_{\rm B} = 0.1$, even for $p_{\rm G} = 0$ (no error in state "G"), the average error probability is $p_{\rm M} \approx 0.009$.
In contrast, with fixed $p_{\rm G} = 0.001$ the value $p_{\rm M} = 0.005$ is achievable:

$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1 = 4.5\%\hspace{0.05cm}.$$

Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:

$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)} = \frac{0.001 \cdot {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)+ 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}\hspace{0.05cm}.$$

With $\rm Pr(B|G) = 0.01$ and with $\rm Pr(G|B) = 0.1$, respectively, the following equations are obtained:

$${\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{0.001 \cdot {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B)+ 0.001 }{{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) + 0.01}\hspace{0.05cm},\hspace{0.5cm} {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) = 0.1\hspace{-0.15cm}:\hspace{0.2cm}{\it p}_{\rm M} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) }\hspace{0.05cm}.$$

From the last equation it can be seen that with no $\rm Pr(G|B)$ value the result $p_{\rm M} = 0.005$ is possible.
On the other hand, with a smaller $\rm Pr(B|G)$ the condition can be fulfilled:

$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) } \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) \le \frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$

Solutions 2 and 4 are therefore correct.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Bündelfehlerkanäle}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Burst_Error_Channels}}
-[[File:P_ID1843__Dig_Z_5_6.png|right|frame|GE–Modell]]
+[[File:P_ID1843__Dig_Z_5_6.png|right|frame|Given Gilbert&ndash;Elliott model]]
-Wir betrachten das <i>Bündelfehler&ndash;Kanalmodell</i> nach E.N. Gilbert und E.O. Elliott (siehe Skizze). Für die Übergangswahrscheinlichkeiten soll dabei gelten:
+We consider the <i>burst error channel model</i>&nbsp; according to &nbsp;[https://en.wikipedia.org/wiki/Edgar_Gilbert "E.N. Gilbert"]&nbsp; and E.O. Elliott (see sketch). For the transition probabilities let hold:
 :$${\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} B)=  0.1, \hspace{0.2cm} {\rm
 Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
-Die Fehlerwahrscheinlichkeit im Zustand &bdquo;GOOD&rdquo; betrage $p_{\rm G} = 0.1\%$ und für die im Zustand &bdquo;BAD&rdquo; gelte $p_{\rm B} = 10\%$. Im Verlaufe dieser Aufgabe sollen weitere Kenngrößen ermittelt werden:
+The error probability in the state "GOOD" is&nbsp; $p_{\rm G} = 0.1\%$. For the state "BAD" is&nbsp; $p_{\rm B} = 10\%$.
-* die mittlere Fehlerwahrscheinlichkeit $p_{\rm M}$,
-* die Zustandswahrscheinlichkeiten $w_{\rm G} = \rm Pr(Z = G)$ und $w_{\rm B} = \rm Pr(Z = B)$,
+In the course of this exercise, further parameters are to be determined:
-* die Werte der Korrelationsfunktion, die für $k > 0$ analytisch wie folgt gegeben ist:
+* the average error probability&nbsp; $p_{\rm M}$,
+* the state probabilities&nbsp; $w_{\rm G} = \rm Pr(Z = G)$&nbsp; and&nbsp; $w_{\rm B} = \rm Pr(Z = B)$,
+* the values of the correlation function, which for&nbsp; $k > 0$&nbsp; is given analytically as follows:
 :$$\varphi_{e}(k) = p_{\rm M}^2 + (p_{\rm B} -
 p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot
-  [1 - {\rm Pr}(\rm
+  \big [1 - {\rm Pr}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G )- {\rm Pr}(\rm
-G\hspace{0.05cm}|\hspace{0.05cm} B )]^{\it k} \hspace{0.05cm}.$$
+G\hspace{0.05cm}|\hspace{0.05cm} B )\big ]^{\it k} \hspace{0.05cm}.$$
-''Hinweis:''
-* Die Aufgabe bezieht sich auf das Kapitel [[Digitalsignal%C3%BCbertragung/B%C3%BCndelfehlerkan%C3%A4le| Bündelfehlerkanal]] des vorliegenden Buches sowie auf das Kapitel [[Stochastische_Signaltheorie/Markovketten| Markovketten]] im Buch &bdquo;Stochastische Signaltheorie&rdquo;.
+''Notes:''
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Burst_Error_Channels| "Burst Error Channels"]].
+*However, reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Markov_Chains| "Markov chains"]]&nbsp; in the book "Theory of Stochastic Signals" and in particular to the section&nbsp; [[Digital_Signal_Transmission/Burst_Error_Channels#Error_correlation_function_of_the_Gilbert-Elliott_model|"Error correlation function of the Gilbert-Elliott model"]]&nbsp; in the book "Channel Coding".
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lauten die folgenden Übergangswahrscheinlichkeiten?
+{What are the following transition probabilities?
 |type="{}"}
-$\rm Pr(G|G) \ = \ ${ 0.99 3% }
+$\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}G) \ = \ ${ 0.99 3% }
-$\rm Pr(B|B) \ = \ ${ 0.9 3% }
+$\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}B) \hspace{0.2cm} = \ ${ 0.9 3% }
-{Mit welchen Wahrscheinlichkeiten befindet sich das GE&ndash;Modell im Zustand &bdquo;GOOD&rdquo; ($w_{\rm G}$) bzw. im Zustand &bdquo;BAD&rdquo; ($w_{\rm B}$)?
+{With what probabilities is the GE model in the state "GOOD" &nbsp;$(w_{\rm G})$&nbsp; or in the state "BAD" &nbsp;$(w_{\rm B})$?
 |type="{}"}
 $w_{\rm G} \ = \ $ { 0.909 3% }
 $w_{\rm B} \ = \ $ { 0.091 3% }
-{Berechnen Sie die mittlere Fehlerwahrscheinlichkeit.
+{Calculate the mean error probability&nbsp; $p_{\rm M}$.
 |type="{}"}
-$p_{\rm M} \ = \ $ { 0.01 3% }
+$p_{\rm M} \ = \ $ { 1 3% } $\ \%$
-{Berechnen Sie die folgenden FKF&ndash;Werte:
+{Calculate the following ECF values:
 |type="{}"}
-$\varphi_e(k = 1) \ = \ ${ 8.209 3% } $\ \cdot 10^{&ndash;4}$
+$\varphi_e(k = 1) \hspace{0.35cm}  = \ ${ 8.209 3% } $\ \cdot 10^{-4}$
-$\varphi_e(k = 2) \ = \ ${ 7.416 3% } $\ \cdot 10^{&ndash;4}$
+$\varphi_e(k = 2) \hspace{0.35cm} = \ ${ 7.416 3% } $\ \cdot 10^{-4}$
-$\varphi_e(k = 5) \ = \ ${ 5.523 3% } $\ \cdot 10^{&ndash;4}$
+$\varphi_e(k = 5) \hspace{0.35cm} = \ ${ 5.523 3% } $\ \cdot 10^{-4}$
-$\varphi_e(k = 50) \ = \ ${ 1.024 3% } $\ \cdot 10^{&ndash;4}$
+$\varphi_e(k = 50) \ = \ ${ 1.024 3% } $\ \cdot 10^{-4}$
-{Wie groß ist der FKF&ndash;Wert $\varphi_e(k = 0)$?
+{What is the size of the ECF value&nbsp; $\varphi_e(k = 0)$?
 |type="{}"}
-$\varphi_e(k = 0) \ = \ ${ 1 3% } $\ \cdot 10^{&ndash;2}$
+$\varphi_e(k = 0) \ = \ ${ 1 3% } $\ \cdot 10^{-2}$
-{Lässt sich die mittlere Fehlerwahrscheinlichkeit $p_{\rm M} = 0.005$ erreichen durch
+{Can the mean error probability&nbsp; $p_{\rm M} = 0.005$&nbsp; be achieved by
 |type="[]"}
-- alleinige Änderung von $p_{\rm G}$,
+- sole change of&nbsp; $p_{\rm G}$,
-+ alleinige Änderung von $p_{\rm B}$,
++ sole change of&nbsp; $p_{\rm B}$,
-- alleinige Änderung von $\rm Pr(G|B)$,
+- sole change of&nbsp; $\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}B)$,
-+ alleinige Änderung von $\rm Pr(B|G)$?
++ sole change of&nbsp; $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}G)$?
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Es gilt $\rm Pr(G|G) = 1 \, &ndash;Pr(B|G) \ \underline {= 0.99}$ sowie $\rm Pr(B|B) = 1 \, &ndash;Pr(G|B) \ \underline {= 0.9}$.
+'''(1)'''&nbsp; $\rm Pr(G\hspace{0.05cm}|\hspace{0.05cm}G) = 1 \, &ndash;Pr(B\hspace{0.05cm}|\hspace{0.05cm}G) \ \underline {= 0.99}$ and $\rm Pr(B\hspace{0.05cm}|\hspace{0.05cm}B) = 1 \, &ndash;Pr(G\hspace{0.05cm}|\hspace{0.05cm}B) \ \underline {= 0.9}$ are valid.
-'''(2)'''&nbsp; Das GE&ndash;Modell ist eine stationäre Markovkette. Für die Wahrscheinlichkeit, dass sich diese im Zustand &bdquo;GOOD&rdquo; befindet, gilt unter Berücksichtigung des Ergebnisses der Teilaufgabe (1):
+'''(2)'''&nbsp; The GE model is a stationary Markov chain.
+*For the probability that it is in the state "GOOD", taking into account the result of subtask '''(1)''':
 :$$w_{\rm G} = {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} G) \cdot
 w_{\rm G} + {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) \cdot
@@ Line 72: / Line 83: @@
 \hspace{0.05cm}.$$
-Weiter gilt $w_{\rm B} = 1 \, &ndash;w_{\rm G}$:
+*Further, $w_{\rm B} = 1 \, &ndash;w_{\rm G}$:
 :$${\rm Pr}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G) \cdot w_{\rm G} + {\rm Pr}(\rm
@@ Line 85: / Line 96: @@
-'''(3)'''&nbsp; Die mittlere Fehlerwahrscheinlichkeit $p_{\rm M} ergibt sich aus den Fehlerwahrscheinlichkeiten $p_{\rm G}$ und $p_{\rm B}$, gewichtet mit $w_{\rm G}$  und $w_{\rm B}$:
+'''(3)'''&nbsp; The mean error probability $p_{\rm M}$ is obtained from the error probabilities $p_{\rm G}$ and $p_{\rm B}$ weighted by $w_{\rm G}$  and $w_{\rm B}$:
 :$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}
-= \frac{10}{11} \cdot 10^{-3} + \frac{1}{11} \cdot 10^{-1}\hspace{0.15cm}\underline { =
+= \frac{10}{11} \cdot 10^{-3} + \frac{1}{11} \cdot 10^{-1}= \frac{10+100}{11} \cdot 10^{-3}\hspace{0.15cm}\underline { =
-.01}\hspace{0.05cm}.$$
+\%}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Entsprechend der allgemeinen Gleichung auf dem Angabenblatt gilt für $k > 0$:
+'''(4)'''&nbsp; According to the general equation on the specification section, for $k > 0$:
 :$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} p_{\rm M}^2 +
 (p_{\rm B} - p_{\rm M}) \cdot (p_{\rm M} - p_{\rm G}) \cdot
   [1 - {\rm Pr}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G )- {\rm Pr}(\rm
-G\hspace{0.05cm}|\hspace{0.05cm} B )]^{\it k} =$$
+G\hspace{0.05cm}|\hspace{0.05cm} B )]^{\it k} = 10^{-4} + 0.09 \cdot 0.009
-:$$\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4} + 0.09 \cdot 0.009
 \cdot  0.89^{\it k} = 10^{-4} \cdot \left ( 1 + 8.1 \cdot
 .89^{\it k} \right )\hspace{0.05cm}.$$
@@ Line 103: / Line 114: @@
 \ \hspace{-0.1cm} 10^{-4} \cdot \left ( 1 + 8.1 \cdot 0.89^{ 1}
 \right ) \hspace{0.15cm}\underline {= 8.209 \cdot 10^{-4}} \hspace{0.05cm},$$
-:$$\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
+:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 2 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
 \cdot \left ( 1 + 8.1 \cdot 0.89^{ 2}
 \right )\hspace{0.15cm}\underline { = 7.416 \cdot 10^{-4}} \hspace{0.05cm},$$
-:$$\varphi_{e}(k = 5 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
+:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 5 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
 \cdot \left ( 1 + 8.1 \cdot 0.89^{ 5}
 \right )\hspace{0.15cm}\underline {= 5.523 \cdot 10^{-4}} \hspace{0.05cm},$$
-:$$\varphi_{e}(k = 50 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
+:$$\Rightarrow \hspace{0.3cm}\varphi_{e}(k = 50 ) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 10^{-4}
 \cdot \left ( 1 + 8.1 \cdot 0.89^{ 50} \right ) \hspace{0.15cm}\underline {= 1.024 \cdot
 ^{-4}} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; Für jedes Kanalmodell gilt wegen $e_{\nu} &#8712; \{0, 1\}$:
+'''(5)'''&nbsp; For each channel model, because of $e_{\nu} &#8712; \{0, 1\}$:
 :$$\varphi_{e}(k = 0 ) = {\rm E}[e_{\nu} ^2] = {\rm E}[e_{\nu} ] =
 p_{\rm M} \hspace{0.05cm}.$$
-Mit dem Ergebnis der Teilaufgabe (3) ergibt sich für den vorliegenden Fall $\varphi_e(k = 0) \ \underline {= 0.01}$.
+Using the result of subtask '''(3)''', $\varphi_e(k = 0) \ \underline {= 0.01}$ for the present case.
-'''(6)'''&nbsp; Entsprechend der Teilaufgabe (3) gilt
+'''(6)'''&nbsp; According to subtask '''(3)''' holds:
 :$$p_{\rm M} = {10}/{11} \cdot p_{\rm G} + {1}/{11} \cdot
 p_{\rm B} \hspace{0.05cm}.$$
-Bei vorgegebenem $p_{\rm B} = 0.1$ ergibt sich selbst für $p_{\rm G} = 0$ (kein Fehler im Zustand &bdquo;G&rdquo;) die mittlere Fehlerwahrscheinlichkeit zu $p_{\rm M} \approx 0.009$. Dagegen ist mit festem $p_{\rm G} = 0.001$ der Wert $p_{\rm M} = 0.005$
+*With given $p_{\rm B} = 0.1$, even for $p_{\rm G} = 0$ (no error in state "G"), the average error probability is $p_{\rm M} \approx 0.009$.
-erreichbar:
+*In contrast, with fixed $p_{\rm G} = 0.001$ the value $p_{\rm M} = 0.005$ is achievable:
 :$$0.005 = {10}/{11} \cdot 10^{-3} + {1}/{11} \cdot p_{\rm B}
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_{\rm B} \le 0.055 - 0.1
 = 4.5\%\hspace{0.05cm}.$$
-Weiterhin kann die mittlere Fehlerwahrscheinlichkeit (mit vorgegebenem $p_{\rm G}$ und $p_{\rm B}$) auch wie folgt dargestellt werden:
+*Furthermore, the mean error probability (with given $p_{\rm G}$ and $p_{\rm B}$) can also be represented as follows:
 :$$p_{\rm M} = \frac{p_{\rm G} \cdot {\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} B)+ p_{\rm B} \cdot {\rm Pr}(\rm
@@ Line 142: / Line 154: @@
 B\hspace{0.05cm}|\hspace{0.05cm} G)}\hspace{0.05cm}.$$
-Mit $\rm Pr(B|G) = 0.01$ bzw. mit $\rm Pr(G|B) = 0.1$ erhält man folgende Gleichungen:
+*With $\rm Pr(B|G) = 0.01$ and with $\rm Pr(G|B) = 0.1$, respectively, the following equations are obtained:
 :$${\rm Pr}(\rm B\hspace{0.05cm}|\hspace{0.05cm} G) =
 .01\hspace{-0.15cm}:\hspace{0.2cm} {\it p}_{\rm M} \hspace{-0.1cm} \ = \
@@ Line 148: / Line 160: @@
 G\hspace{0.05cm}|\hspace{0.05cm} B)+ 0.001 }{{\rm Pr}(\rm
 G\hspace{0.05cm}|\hspace{0.05cm} B) +
-.01}\hspace{0.05cm},$$
+.01}\hspace{0.05cm},\hspace{0.5cm}
-:$$
 {\rm Pr}(\rm G\hspace{0.05cm}|\hspace{0.05cm} B) =
 .1\hspace{-0.15cm}:\hspace{0.2cm}{\it p}_{\rm M} \hspace{-0.1cm} \ = \
@@ Line 156: / Line 167: @@
 G\hspace{0.05cm}|\hspace{0.05cm} B) }\hspace{0.05cm}.$$
-Aus der oberen Gleichung ist zu erkennen, dass mit keinem $\rm Pr(G|B)$&ndash;Wert das Ergebnis $p_{\rm M} = 0.005$ möglich ist. Dagegen lässt sich durch ein kleineres $\rm Pr(B|G)$ die Bedingung erfüllen:
+*From the last equation it can be seen that with no $\rm Pr(G|B)$ value the result $p_{\rm M} = 0.005$ is possible.
+*On the other hand, with a smaller $\rm Pr(B|G)$ the condition can be fulfilled:
 :$$0.005 = \frac{0.0001 + 0.1 \cdot {\rm Pr}(\rm
 B\hspace{0.05cm}|\hspace{0.05cm} G)}{0.1 +{\rm Pr}(\rm
@@ Line 163: / Line 175: @@
 \frac{0.0004}{0.095} \approx 0.0042\hspace{0.05cm}.$$
-Richtig sind somit die <u>Lösungsvorschläge 2 und 4</u>.
+*<u>Solutions 2 and 4</u> are therefore correct.
 {{ML-Fuß}}
-[[Category:Aufgaben zu Digitalsignalübertragung|^5.3 Bündelfehlerkanäle^]]
+[[Category:Digital Signal Transmission: Exercises|^5.3 Burst Error Channels^]]