Aufgaben:Exercise 5.7Z: McCullough Model once more - Revision history

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	*In contrast, for the MC model we obtain:		*In contrast, for the MC model we obtain:
	:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rm		:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) + q(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G)}= \frac{0.5528}{0.5528 +0.3724}\hspace{0.15cm}\underline {= 0.5975}\hspace{0.05cm},\hspace{0.5cm}\alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 -\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$
	G\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + q(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)}= \frac{0.5528}{0.5528 +
	0.3724}\hspace{0.15cm}\underline {= 0.5975}\hspace{0.05cm},\hspace{0.5cm}
	\alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 -
	\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$

	*In subtask '''(3)''' of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.		*In subtask '''(3)''' of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.

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	[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|EDD and ECF of GE model and equivalent MC model]]		[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|EDD and ECF of GE model and equivalent MC model]]
	As in  [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration\|"Exercise 5.6"]],  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model\|"Exercise 5.6Z"]]  and  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]],  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters		As in  [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration\|"Exercise 5.6"]],  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model\|"Exercise 5.6Z"]]  and  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]],  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters
	:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,		:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}p(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.1, \hspace{0.2cm} p(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$
	\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}
	p(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \
	\hspace{-0.1cm} 0.1, \hspace{0.2cm} p(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

	From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely		From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely
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	The probabilities of the MC model were determined in  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]]  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:		The probabilities of the MC model were determined in  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]]  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:
	:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,		:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}q(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.5528, \hspace{0.2cm} q(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$
	\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}
	q(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \
	\hspace{-0.1cm} 0.5528, \hspace{0.2cm} q(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$

	The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.		The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.
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	*Further, using the data for Exercise 5.7:		*Further, using the data for Exercise 5.7:
	:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\rm		:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\rmB} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdotq_{\rm G}} \hspace{0.05cm}, \hspace{0.2cm}q({\rmG\hspace{0.05cm}\|\hspace{0.05cm} B })= \frac{\alpha_{\rmG}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)$$
	B} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \~~cdot~~		:$$\Rightarrow \hspace{0.3cm} S = q_{\rm G} \cdot q({\rmB\hspace{0.05cm}\|\hspace{0.05cm} G }) + q_{\rm B} \cdot\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G ) = q_{\rm G} \cdot q({\rmB\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rmG\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$
	q_{\rm G}} \hspace{0.05cm}, \hspace{0.2cm}q({\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B })= \frac{\alpha_{\rm
	G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G
	)$$
	:$$\Rightarrow \hspace{0.3cm} S = q_{\rm G} \cdot q({\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G }) + q_{\rm B} \cdot
	\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G ) = q_{\rm G} \cdot q({\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$

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	* From the above exercises, the following results can be further used:		* From the above exercises, the following results can be further used:
	:(a) The state probabilities of the GE model are		:(a) The state probabilities of the GE model are
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G}\hspace{0.05cm}.$$
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)}
	\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G
	}\hspace{0.05cm}.$$
	:(b) The mean error probability of the GE model is		:(b) The mean error probability of the GE model is
	:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}		:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$
	= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$
	:(c) The correlation duration of the GE model is calculated as		:(c) The correlation duration of the GE model is calculated as
	:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm}		:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm}B ) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )}-1\hspace{0.05cm}.$$
	B ) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )}-1
	\hspace{0.05cm}.$$


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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the state probabilities of the GE model was determined in Exercise 5.6Z:		'''(1)'''  For the state probabilities of the GE model was determined in Exercise 5.6Z:
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rmG\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G)} = 0.909\hspace{0.05cm},\hspace{0.5cm} w_{\rm B} = 1 - w_{\rm G}= 0.091\hspace{0.05cm}.$$
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G)} = 0.909
	\hspace{0.05cm},\hspace{0.5cm} w_{\rm B} = 1 - w_{\rm G
	}= 0.091\hspace{0.05cm}.$$

	*In contrast, for the MC model we obtain:		*In contrast, for the MC model we obtain:
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	*Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.		*Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.
	*By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by		*By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by
	:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rm		:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rmB}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$
	B}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =
	97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$

	*Of course, this value should be exactly the same as for the corresponding GE model.		*Of course, this value should be exactly the same as for the corresponding GE model.
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	'''(4)'''  In the GE model, the correlation duration is given as follows ($S$ stands for sum):		'''(4)'''  In the GE model, the correlation duration is given as follows ($S$ stands for sum):
	:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rm		:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rmPr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B ) + {\rm Pr}(\rmB\hspace{0.05cm}\|\hspace{0.05cm} G )\hspace{0.05cm}.$$
	Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B ) + {\rm Pr}(\rm
	B\hspace{0.05cm}\|\hspace{0.05cm} G )\hspace{0.05cm}.$$

	*Further, using the data for Exercise 5.7:		*Further, using the data for Exercise 5.7:
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	B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rm		B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$		G\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \~~cdot~~		:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$
	q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \~~cdot~~
	q({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$

	*So, the correct solution is <u>solution 2</u>. With the given parameter values, we obtain, for example:		*So, the correct solution is <u>solution 2</u>. With the given parameter values, we obtain, for example:
	:$$D_{\rm K} =\frac{1}{0.0061 \cdot 0.3724 + 0.1949 \~~cdot~~		:$$D_{\rm K} =\frac{1}{0.0061 \cdot 0.3724 + 0.1949 \cdot0.5528}-1=\frac{1}{0.11}-1 {\approx 8.09}\hspace{0.05cm}.$$
	0.5528}-1=\frac{1}{0.11}-1 {\approx 8.09}\hspace{0.05cm}.$$

	*The result is exactly the same value as in subtask '''(3)''' of Exercise 5.6.		*The result is exactly the same value as in subtask '''(3)''' of Exercise 5.6.

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	[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]
			[[de:5.7Z_Nochmals_MC-Modell]]

Hwang at 08:22, 21 September 2022

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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{~~Berechnen Sie die Wahrscheinlichkeiten~~  $\alpha_{\rm G}$  ~~und~~  $\alpha_{\rm B}$~~, dass sich das~~ MC~~–Modell im Zustand~~ "Good" ~~bzw. im Zustand~~ "Bad" ~~befindet~~.		{Calculate the probabilities  $\alpha_{\rm G}$  and  $\alpha_{\rm B}$ that the MC model is in the state "Good" and the state "Bad".
	\|type="{}"}		\|type="{}"}
	$\alpha_{\rm G} \hspace{0.05cm} = \ ${ 0.5975 3% }		$\alpha_{\rm G} \hspace{0.05cm} = \ ${ 0.5975 3% }
	$\alpha_{\rm B} \ = \ ${ 0.4025 3% }		$\alpha_{\rm B} \ = \ ${ 0.4025 3% }

	{~~Ermitteln Sie den mittleren Fehlerabstand des~~ MC~~–Modells~~.		{Determine the mean error distance of the MC model.
	\|type="{}"}		\|type="{}"}
	${\rm E}\big[a\big] \ = \ ${ 100.1 3% }		${\rm E}\big[a\big] \ = \ ${ 100.1 3% }

	{~~Wie groß ist der Fehlerkorrelationsfunktionswert für~~  $k = 0$?		{What is the error correlation function value for  $k = 0$?
	\|type="{}"}		\|type="{}"}
	$\varphi_e(k = 0) \ = \ ${ 0.01 3% }		$\varphi_e(k = 0) \ = \ ${ 0.01 3% }

	{~~Geben Sie die Fehlerkorrelationsdauer~~  $D_{\rm K}$  ~~als Funktion der~~ MC~~–Parameter~~  $q_{\rm G},  q_{\rm B},  q(\rm G\hspace{0.05cm}\|\hspace{0.05cm}B)$  ~~und~~  $q(\rm B\hspace{0.05cm}\|\hspace{0.05cm}G)$  ~~an.~~ <br>~~Welches Ergebnis ist richtig~~?		{Give the error correlation duration  $D_{\rm K}$  as a function of the MC parameters  $q_{\rm G},  q_{\rm B},  q(\rm G\hspace{0.05cm}\|\hspace{0.05cm}B)$  and  $q(\rm B\hspace{0.05cm}\|\hspace{0.05cm}G)$.  <br>Which result is correct?
	\|type="()"}		\|type="()"}
	- $D_{\rm K} = \big [q({\rm B\hspace{0.05cm}\|\hspace{0.05cm}G}) + q({\rm G\hspace{0.05cm}\|\hspace{0.05cm}B})\big]^{-1} \ -1$,		- $D_{\rm K} = \big [q({\rm B\hspace{0.05cm}\|\hspace{0.05cm}G}) + q({\rm G\hspace{0.05cm}\|\hspace{0.05cm}B})\big]^{-1} \ -1$,
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Für die Zustandswahrscheinlichkeiten des~~ GE~~–Modells wurde~~ in ~~Aufgabe~~ 5.6Z ~~ermittelt~~:		'''(1)'''  For the state probabilities of the GE model was determined in Exercise 5.6Z:
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm
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	}= 0.091\hspace{0.05cm}.$$		}= 0.091\hspace{0.05cm}.$$

	*~~Dagegen erhält man beim~~ MC~~–Modell~~:		*In contrast, for the MC model we obtain:
	:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rm		:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\rm
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	\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$		\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$

	*In ~~der Teilaufgabe~~ '''(3)''' ~~der Aufgabe~~ 5.7 ~~wurden diese Werte schon einmal ermittelt~~, ~~allerdings aus den Parametern des äquivalenten~~ Gilbert~~–~~Elliott~~–Modells~~.		*In subtask '''(3)''' of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.



	'''(2)'''  ~~Der mittlere Fehlerabstand im Kanalzustand~~ "GOOD" ~~ist gleich dem Kehrwert der dazugehörigen Fehlerwahrscheinlichkeit~~ $q_{\rm G}$.		'''(2)'''  The mean error distance in the channel state "GOOD" is equal to the reciprocal of the associated error probability $q_{\rm G}$.
	*~~Der mittlere Fehlerabstand im Zustand~~ "BAD" ~~ist dementsprechend~~ $1/q_{\rm B}$.		*Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.
	*~~Durch Gewichtung mit den beiden Zustandswahrscheinlichkeiten~~ $\alpha_{\rm G}$ ~~und~~ $\alpha_{\rm B}$ ~~ergibt sich der mittlere Fehlerabstand des~~ MC~~–Modells insgesamt zu~~		*By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by
	:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rm		:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rm
	B}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =		B}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =
	97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$		97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$

	*~~Dieser Wert sollte natürlich genau so groß wie beim entsprechenden~~ GE~~–Modell sein~~.		*Of course, this value should be exactly the same as for the corresponding GE model.
	*~~Die kleine Abweichung von~~ $0.1$ ~~ist auf Rundungsfehler zurückzuführen~~.		*The small deviation of $0.1$ is due to rounding errors.



	'''(3)'''  ~~Auch hier gilt der Zusammenhang~~ $\varphi_e(k = 0) = p_{\rm M}$.		'''(3)'''  Again, the relation $\varphi_e(k = 0) = p_{\rm M}$ holds.
	*~~Die mittlere Fehlerwahrscheinlichkeit ist aber gleich dem Kehrwert des mittleren Fehlerabstands~~ ${\rm E}[a]$.		*However, the mean error probability is equal to the reciprocal of the mean error distance ${\rm E}[a]$.
	*~~Daraus folgt~~ $\varphi_e(k = 0) \ \underline {= 0.01}$.		*It follows that $\varphi_e(k = 0) \ \underline {= 0.01}$.



	'''(4)'''  ~~Beim~~ GE~~–Modell ist die Korrelationsdauer wie folgt gegeben~~ ($S$ ~~steht für Summe~~):		'''(4)'''  In the GE model, the correlation duration is given as follows ($S$ stands for sum):
	:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rm		:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rm
	Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B ) + {\rm Pr}(\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}.$$		B\hspace{0.05cm}\|\hspace{0.05cm} G )\hspace{0.05cm}.$$

	*~~Weiter gilt mit den Angaben zur Aufgabe~~ 5.7:		*Further, using the data for Exercise 5.7:
	:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\rm		:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\rm
	B} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdot		B} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdot
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	q({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$		q({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$

	*~~Richtig ist also der~~ <u>~~Lösungsvorschlag~~ 2</u>. ~~Mit den gegebenen Parameterwerten erhält man zum Beispiel~~:		*So, the correct solution is <u>solution 2</u>. With the given parameter values, we obtain, for example:
	:$$D_{\rm K} =\frac{1}{0.0061 \cdot 0.3724 + 0.1949 \cdot		:$$D_{\rm K} =\frac{1}{0.0061 \cdot 0.3724 + 0.1949 \cdot
	0.5528}-1=\frac{1}{0.11}-1 {\approx 8.09}\hspace{0.05cm}.$$		0.5528}-1=\frac{1}{0.11}-1 {\approx 8.09}\hspace{0.05cm}.$$

	*~~Es ergibt sich exakt der gleiche Wert wie~~ in ~~der Teilaufgabe~~ '''(3)''' ~~von Aufgabe~~ 5.6.		*The result is exactly the same value as in subtask '''(3)''' of Exercise 5.6.
	{{ML-Fuß}}		{{ML-Fuß}}

Hwang at 06:57, 20 September 2022

2022-09-20T06:57:37Z

← Older revision		Revision as of 08:57, 20 September 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Burst_Error_Channels}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Burst_Error_Channels}}

	[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|~~FAV und FKF von GE–Modell und äquivalentem~~ MC~~-Modell~~]]		[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|EDD and ECF of GE model and equivalent MC model]]
	~~Wir betrachten wie auch~~ in  [[Aufgaben:5.6:~~_Fehlerkorrelationsdauer~~\|~~Aufgabe~~ 5.6]],  [[Aufgaben:5.~~6Z_GE~~-~~Modelleigenschaften~~\|~~Aufgabe~~ 5.6Z]]  ~~und~~  [[Aufgaben:5.~~7_McCullough~~-~~Parameter_aus_Gilbert-Elliott-Parameter~~\|~~Aufgabe~~ 5.7]]  ~~das Bündelfehler–Kanalmodell nach~~ Gilbert ~~und~~ Elliott (GE~~–Modell~~) ~~mit den Kenngrößen~~		As in  [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration\|"Exercise 5.6"]],  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model\|"Exercise 5.6Z"]]  and  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]],  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters
	:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,		:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,
	\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}		\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}
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	B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$		B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

	~~Aus diesen vier Wahrscheinlichkeiten lassen sich die entsprechenden Kenngrößen des Kanalmodells nach~~ McCullough (MC~~–Modell~~) ~~so ermitteln, dass beide Modelle die genau gleichen statistischen Eigenschaften besitzen~~, ~~nämlich~~		From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely
	* ~~exakt gleiche Fehlerabstandsverteilung~~ (~~FAV~~)  $V_a(k)$,		* exactly the same error distance distribution (EDD)  $V_a(k)$,
	* ~~exakt gleiche Fehlerkorrelationsfunktion~~ (~~FKF~~)  $\varphi_e(k)$.		* exactly the same error correlation function (ECF)  $\varphi_e(k)$.


	~~Die Wahrscheinlichkeiten des~~ MC~~–Modells wurden~~ in ~~der~~  [[Aufgaben:5.~~7_McCullough-Parameter_aus_Gilbert-Elliott~~-~~Parameter~~\|~~Aufgabe~~ 5.7]]  ~~wie folgt ermittelt~~ $($~~Bezeichnungen entsprechend der Grafik zur Aufgabe~~ 5.7, ~~alle mit~~  $q$  ~~anstelle von~~  $p)$:		The probabilities of the MC model were determined in  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]]  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:
	:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,		:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,
	\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}		\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}
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	B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$		B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$

	~~Die obere Grafik zeigt die aus~~  $~~N = 10^6~~$  ~~Folgenelementen simulativ ermittelten Funktionen~~  $~~V_a~~(k)$  ~~und~~  $~~\varphi_e(k)~~$  ~~für das~~ GE~~– und das~~ MC~~–Modell~~. ~~Hier ergeben sich noch leichte Abweichungen~~. ~~Im Grenzfall für~~  $N → ∞$  ~~stimmen dagegen Fehlerkorrelationsfunktion und Fehlerabstandsverteilung beider Modelle exakt überein~~.		The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.

	In ~~dieser Aufgabe sollen nun wichtige Beschreibungsgrößen des~~ GE~~-Modells wie~~		In this exercise, important descriptive variables of the GE model such as
	*~~Zustandswahrscheinlichkeiten~~,		*state probabilities,
	*~~mittlere Fehlerwahrscheinlichkeiten~~, ~~und~~		*mean error probabilities, and
	*~~Korrelationsdauer~~		*correlation duration


	~~direkt aus den~~ $q$~~–Parametern des~~ MC~~–Modells ermittelt werden~~.		should be determined directly from the $q$ parameters of the MC model.


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	''~~Hinweise~~:''		''Notes:''
	* ~~Die Aufgabe gehört zum Themengebiet des Kapitels~~  [[~~Digitalsignal%C3%BCbertragung~~/~~B%C3%BCndelfehlerkan%C3%A4le~~\| ~~Bündelfehlerkanäle~~]].		* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Burst_Error_Channels\| "Burst Error Channels"]].

	* ~~Aus den oben genannten Aufgaben können folgende Ergebnisse weiterverwendet werden~~:		* From the above exercises, the following results can be further used:
	:(a) ~~Die Zustandswahrscheinlichkeiten des~~ GE~~–Modells sind~~		:(a) The state probabilities of the GE model are
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm
	G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm		G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm
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	\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G		\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G
	}\hspace{0.05cm}.$$		}\hspace{0.05cm}.$$
	:(b) ~~Die mittlere Fehlerwahrscheinlichkeit des~~ GE~~–Modells beträgt~~		:(b) The mean error probability of the GE model is
	:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}		:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}
	= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$		= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$
	:(c) ~~Die Korrelationsdauer des~~ GE~~–Modells berechnet sich zu~~		:(c) The correlation duration of the GE model is calculated as
	:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm}		:$$D_{\rm K} =\frac{1}{{\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm}
	B ) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )}-1		B ) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )}-1
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	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Berechnen Sie die Wahrscheinlichkeiten  $\alpha_{\rm G}$  und  $\alpha_{\rm B}$, dass sich das MC–Modell im Zustand "Good" bzw. im Zustand "Bad" befindet.		{Berechnen Sie die Wahrscheinlichkeiten  $\alpha_{\rm G}$  und  $\alpha_{\rm B}$, dass sich das MC–Modell im Zustand "Good" bzw. im Zustand "Bad" befindet.

Hwang at 06:33, 20 September 2022

2022-09-20T06:33:02Z

← Older revision		Revision as of 08:33, 20 September 2022
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	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Bündelfehlerkanäle~~}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Burst_Error_Channels}}

	[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|FAV und FKF von GE–Modell und äquivalentem MC-Modell]]		[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|FAV und FKF von GE–Modell und äquivalentem MC-Modell]]

Guenter at 15:10, 9 September 2022

2022-09-09T15:10:11Z

← Older revision		Revision as of 17:10, 9 September 2022
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	[[Category:Digital Signal Transmission: Exercises\|^5.3 ~~Bündelfehlerkanäle~~^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 5.7Z: Nochmals McCullough-Modell to Aufgaben:Exercise 5.7Z: McCullough Model once more

2022-09-09T15:05:49Z

Guenter moved page Aufgaben:Aufgabe 5.7Z: Nochmals McCullough-Modell to Aufgaben:Exercise 5.7Z: McCullough Model once more

← Older revision	Revision as of 17:05, 9 September 2022
(No difference)

← Older revision		Revision as of 17:56, 16 March 2026
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	[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|EDD and ECF of GE model and equivalent MC model]]		[[File:P_ID1845__Dig_Z_5_7.png\|right\|frame\|EDD and ECF of GE model and equivalent MC model]]
	As in  [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration\|"Exercise 5.6"]],  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model\|"Exercise 5.6Z"]]  and  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]],  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters		As in  [[Aufgaben:Exercise_5.6:_Error_Correlation_Duration\|"Exercise 5.6"]],  [[Aufgaben:Exercise_5.6Z:_Gilbert-Elliott_Model\|"Exercise 5.6Z"]]  and  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]],  we consider the burst error channel model according to Gilbert and Elliott (GE model) with the parameters
	:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}p(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.1, \hspace{0.2cm} p(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$		:$$p_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.001,\hspace{0.2cm}p_{\rm B} = 0.1,\hspace{0.2cm}p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.1, \hspace{0.2cm} p(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.01\hspace{0.05cm}.$$

	From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely		From these four probabilities, the corresponding characteristics of the channel model according to McCullough (MC model) can be determined in such a way that both models have exactly the same statistical properties, namely
Line 12:		Line 12:

	The probabilities of the MC model were determined in  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]]  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:		The probabilities of the MC model were determined in  [[Aufgaben:Exercise_5.7:_McCullough_and_Gilbert-Elliott_Parameters\|"Exercise 5.7"]]  as follows $($labels according to the graph for Exercise 5.7, all with  $q$  instead of  $p)$:
	:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}q(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.5528, \hspace{0.2cm} q(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$		:$$q_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.0061,\hspace{0.2cm}q_{\rm B} = 0.1949,\hspace{0.2cm}q(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)\hspace{-0.1cm} \ = \\hspace{-0.1cm} 0.5528, \hspace{0.2cm} q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G) = 0.3724\hspace{0.05cm}.$$

	The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.		The upper graph shows the functions  $V_a(k)$  and  $\varphi_e(k)$  simulatively determined from  $N = 10^6$  sequence elements for the GE and MC models. There are still slight discrepancies here. In the limiting case for  $N → ∞$,  on the other hand, error correlation function and error distance distribution of both models agree exactly.
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	* From the above exercises, the following results can be further used:		* From the above exercises, the following results can be further used:
	:(a) The state probabilities of the GE model are		:(a) The state probabilities of the GE model are
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G}\hspace{0.05cm}.$$		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}\hspace{0.05cm},\hspace{0.2cm} w_{\rm B} = 1 - w_{\rm G}\hspace{0.05cm}.$$
	:(b) The mean error probability of the GE model is		:(b) The mean error probability of the GE model is
	:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$		:$$p_{\rm M} = w_{\rm G} \cdot p_{\rm G} + w_{\rm B} \cdot p_{\rm B}= \varphi_{e}(k = 0 )\hspace{0.05cm}.$$
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  For the state probabilities of the GE model was determined in Exercise 5.6Z:		'''(1)'''  For the state probabilities of the GE model was determined in Exercise 5.6Z:
	:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)} = 0.909\hspace{0.05cm},\hspace{0.5cm} w_{\rm B} = 1 - w_{\rm G}= 0.091\hspace{0.05cm}.$$		:$$w_{\rm G} = \frac{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{p(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + p(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)} = 0.909\hspace{0.05cm},\hspace{0.5cm} w_{\rm B} = 1 - w_{\rm G}= 0.091\hspace{0.05cm}.$$

	*In contrast, for the MC model we obtain:		*In contrast, for the MC model we obtain:
	:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B) + q(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G)}= \frac{0.5528}{0.5528 +0.3724}\hspace{0.15cm}\underline {= 0.5975}\hspace{0.05cm},\hspace{0.5cm}\alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 -\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$		:$$\alpha_{\rm G} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{q(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B)}{q(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B) + q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)}= \frac{0.5528}{0.5528 +0.3724}\hspace{0.15cm}\underline {= 0.5975}\hspace{0.05cm},\hspace{0.5cm}\alpha_{\rm B} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 -\alpha_{\rm G} \hspace{0.15cm}\underline {= 0.4025}\hspace{0.05cm}.$$

	*In subtask '''(3)''' of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.		*In subtask '''(3)''' of Exercise 5.7, these values have already been determined once, but from the parameters of the equivalent Gilbert-Elliott model.
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	*Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.		*Accordingly, the mean error distance in the state "BAD" is $1/q_{\rm B}$.
	*By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by		*By weighting with the two state probabilities $\alpha_{\rm G}$ and $\alpha_{\rm B}$, the mean error distance of the MC model as a whole is given by
	:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\~~rmB~~}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$		:$${\rm E}[a] =\frac{\alpha_{\rm G}}{q_{\rm G}} + \frac{\alpha_{\rm B}}{q_{\rm B}}=\frac{0.5975}{0.0061} + \frac{0.4025}{0.1949} =97.95 + 2.06\hspace{0.15cm}\underline { = 100.1}\hspace{0.05cm}.$$

	*Of course, this value should be exactly the same as for the corresponding GE model.		*Of course, this value should be exactly the same as for the corresponding GE model.
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	'''(4)'''  In the GE model, the correlation duration is given as follows ($S$ stands for sum):		'''(4)'''  In the GE model, the correlation duration is given as follows ($S$ stands for sum):
	:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\~~rmPr~~}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B ) + {\rm Pr}(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G )\hspace{0.05cm}.$$		:$$D_{\rm K} = {1}/{S}-1 \hspace{0.05cm},\hspace{0.2cm}S = {\rm Pr}(\rm G\hspace{0.05cm}\|\hspace{0.05cm} B ) + {\rm Pr}(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G )\hspace{0.05cm}.$$

	*Further, using the data for Exercise 5.7:		*Further, using the data for Exercise 5.7:
	:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\~~rmB~~} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdotq_{\rm G}} \hspace{0.05cm}, \hspace{0.2cm}q({\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B })= \frac{\alpha_{\~~rmG~~}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)$$		:$$q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) = \frac{\alpha_{\rm B} \cdot S}{\alpha_{\rm G} \cdot q_{\rm B} + \alpha_{\rm B} \cdotq_{\rm G}} \hspace{0.05cm}, \hspace{0.2cm}q({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })= \frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G)$$
	:$$\Rightarrow \hspace{0.3cm} S = q_{\rm G} \cdot q({\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G }) + q_{\rm B} \cdot\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G ) = q_{\rm G} \cdot q({\~~rmB~~\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\~~rmG~~\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} S = q_{\rm G} \cdot q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G }) + q_{\rm B} \cdot\frac{\alpha_{\rm G}}{\alpha_{\rm B}} \cdot q(\rm B\hspace{0.05cm}\|\hspace{0.05cm} G ) = q_{\rm G} \cdot q({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdot q({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B }) \hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}D_{\rm K} =\frac{1}{q_{\rm G} \cdotq({\rm B\hspace{0.05cm}\|\hspace{0.05cm} G })+ q_{\rm B} \cdotq({\rm G\hspace{0.05cm}\|\hspace{0.05cm} B })}-1 \hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]		[[Category:Digital Signal Transmission: Exercises\|^5.3 Burst Error Channels^]]
	[[de:5.~~7Z_Nochmals_MC~~-Modell]]		[[de:Aufgaben:Aufgabe 5.7Z: Nochmals McCullough-Modell]]