Aufgaben:Exercise 5.2: Error Correlation Function - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:53:47Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:53, 16 March 2026
Line 59:		Line 59:

	'''(3)'''  According to the definition equation and  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]],  the following result is obtained:		'''(3)'''  According to the definition equation and  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]],  the following result is obtained:
	:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\~~rmE~~}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot(e_{\nu + 1}=1)]={\rm Pr}(e_{\nu + 1}=1\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot {\~~rmPr~~}(e_{\nu} = 1)\hspace{0.05cm}.$$		:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot(e_{\nu + 1}=1)]={\rm Pr}(e_{\nu + 1}=1\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot {\rm Pr}(e_{\nu} = 1)\hspace{0.05cm}.$$

	*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:		*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:
Line 80:		Line 80:

	[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]		[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]
	[[de:Aufgaben:5.~~2_Fehlerkorrelationsfunktion~~]]		[[de:Aufgaben:Aufgabe 5.2: Fehlerkorrelationsfunktion]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:48Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 59:		Line 59:

	'''(3)'''  According to the definition equation and  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]],  the following result is obtained:		'''(3)'''  According to the definition equation and  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]],  the following result is obtained:
	:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm		:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rmE}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot(e_{\nu + 1}=1)]={\rm Pr}(e_{\nu + 1}=1\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot {\rmPr}(e_{\nu} = 1)\hspace{0.05cm}.$$
	E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot
	(e_{\nu + 1}=1)]={\rm Pr}(e_{\nu + 1}=1
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot {\rm
	Pr}(e_{\nu} = 1)
	\hspace{0.05cm}.$$

	*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:		*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:31:14Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:31, 16 March 2026
Line 5:		Line 5:
	For the characterization of digital channel models one uses among other things		For the characterization of digital channel models one uses among other things
	* the  "error correlation function"  $\rm (ECF)$		* the  "error correlation function"  $\rm (ECF)$
	:$$\varphi_{e}(k) = {\rm E}\big[e_{\nu} \cdot e_{\nu +		:$$\varphi_{e}(k) = {\rm E}\big[e_{\nu} \cdot e_{\nu +k}\big]\hspace{0.05cm}, \hspace{0.2cm} k \ge 0\hspace{0.05cm},$$
	k}\big]\hspace{0.05cm}, \hspace{0.2cm} k \ge 0\hspace{0.05cm},$$

	* the  "error distance probabilities"		* the  "error distance probabilities"
	:$${\rm Pr}( a =k) \hspace{0.05cm}, \hspace{0.2cm} k \ge		:$${\rm Pr}( a =k) \hspace{0.05cm}, \hspace{0.2cm} k \ge1\hspace{0.05cm}.$$
	1\hspace{0.05cm}.$$

	Here denote:		Here denote:
Line 53:		Line 51:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The mean error probability is equal to the ECF value for  $k = 0$.  Namely,  because of  $e_{\nu} ∈ \{0, 1\}$:		'''(1)'''  The mean error probability is equal to the ECF value for  $k = 0$.  Namely,  because of  $e_{\nu} ∈ \{0, 1\}$:
	:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=		:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}\hspace{0.05cm}.$$
	p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}
	\hspace{0.05cm}.$$


Line 71:		Line 67:

	*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:		*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:
	:$$\varphi_{e}(k = 1) = {\rm Pr}(a = 1) \cdot p_{\rm M} = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}		:$$\varphi_{e}(k = 1) = {\rm Pr}(a = 1) \cdot p_{\rm M} = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


	'''(4)'''  The ECF value  $\varphi_e(k = 2)$  can be interpreted  (approximately)  as follows:		'''(4)'''  The ECF value  $\varphi_e(k = 2)$  can be interpreted  (approximately)  as follows:
	:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1		:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} $$
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} $$		:$$\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) = \frac{\varphi_{e}(k= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) = \frac{\varphi_{e}(k
	= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$

	*This probability is composed of  "at time  $\nu+1$  an error occurs"  and  "at time  $\nu+1$  there is no error":		*This probability is composed of  "at time  $\nu+1$  an error occurs"  and  "at time  $\nu+1$  there is no error":
	:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =		:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$
	1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$

	*In the calculation,  it was assumed that the individual error distances are statistically independent of each other.		*In the calculation,  it was assumed that the individual error distances are statistically independent of each other.

Maintenance script: Add German interlanguage link

2026-03-16T12:34:03Z

Add German interlanguage link

← Older revision		Revision as of 14:34, 16 March 2026
Line 95:		Line 95:

	[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]		[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]
			[[de:Aufgaben:5.2_Fehlerkorrelationsfunktion]]

Hwang at 02:39, 18 September 2022

2022-09-18T02:39:49Z

← Older revision		Revision as of 04:39, 18 September 2022
Line 82:		Line 82:
	= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$		= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$

	*This probability is composed of  "at time  $\nu+1$  an error occurs"  and  "At time  $\nu+1$  there is no error":		*This probability is composed of  "at time  $\nu+1$  an error occurs"  and  "at time  $\nu+1$  there is no error":
	:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =		:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =
	1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}		1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}

Hwang at 02:38, 18 September 2022

2022-09-18T02:38:25Z

← Older revision		Revision as of 04:38, 18 September 2022
Line 41:		Line 41:
	${\rm E}\big[a\big] \ = \ ${ 10 3% }		${\rm E}\big[a\big] \ = \ ${ 10 3% }

	{Calculate the value of the error correlation function  \rm (ECF)$  for  $k = 1$.		{Calculate the value of the error correlation function  $\rm (ECF)$  for  $k = 1$.
	\|type="{}"}		\|type="{}"}
	$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }		$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }

Guenter at 15:47, 3 September 2022

2022-09-03T15:47:04Z

Guenter at 15:32, 3 September 2022

2022-09-03T15:32:45Z

← Older revision		Revision as of 17:32, 3 September 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models}}

	[[File:P_ID1854__Dig_A_5_2_version1.png\|right\|frame\|~~Probabilities of~~ error ~~distances and ECF~~]]		[[File:P_ID1854__Dig_A_5_2_version1.png\|right\|frame\|Error distance probability & error correlation function]]
	For the characterization of digital channel models one uses among other things		For the characterization of digital channel models one uses among other things
	* the error correlation function (ECF)		* the  "error correlation function"  $\rm (ECF)$
	:$$\varphi_{e}(k) = {\rm E}\big[e_{\nu} \cdot e_{\nu +		:$$\varphi_{e}(k) = {\rm E}\big[e_{\nu} \cdot e_{\nu +
	k}\big]\hspace{0.05cm}, \hspace{0.2cm} k \ge 0\hspace{0.05cm},$$		k}\big]\hspace{0.05cm}, \hspace{0.2cm} k \ge 0\hspace{0.05cm},$$

	* the error distance probabilities		* the  "error distance probabilities"
	:$${\rm Pr}( a =k) \hspace{0.05cm}, \hspace{0.2cm} k \ge		:$${\rm Pr}( a =k) \hspace{0.05cm}, \hspace{0.2cm} k \ge
	1\hspace{0.05cm}.$$		1\hspace{0.05cm}.$$

	Here denote:		Here denote:
	* $〈e_{\rm \nu}〉$  is the error sequence with  $e_{\rm \nu} ∈ \{0, 1\}$.		* $〈e_{\rm \nu}〉$  is the error sequence with  $e_{\rm \nu} ∈ \{0, 1\}$.
	* $a$  indicates the error distance.
			* $a$  indicates the error distance with  $a_{\rm \nu} ∈ \{0, 1, 2, \text{...} \}$.


	Two directly consecutive ~~bit~~ errors are thus characterized by the error distance  $a = 1$.		Two directly consecutive symbol errors are thus characterized by the error distance  $a = 1$.

	The table shows exemplary values of the error distance probabilities  ${\rm Pr}(a = k)$  as well as the error correlation function  $\varphi_e(k)$.		The table shows exemplary values of the error distance probabilities  ${\rm Pr}(a = k)$  as well as the error correlation function  $\varphi_e(k)$.
Line 26:		Line 27:


			Note:  The exercise covers the subject matter of the chapter  [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models\|"Parameters of Digital Channel Models"]].

	''Note:''
	* The exercise covers the subject matter of the chapter  [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models\|"Parameters of Digital Channel Models"]].


Line 35:		Line 33:
	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{Which value results for the ~~average~~ error probability?		{Which value results for the mean error probability?
	\|type="{}"}		\|type="{}"}
	$p_{\rm M} \ = \ ${ 0.1 3% }		$p_{\rm M} \ = \ ${ 0.1 3% }
Line 43:		Line 41:
	${\rm E}\big[a\big] \ = \ ${ 10 3% }		${\rm E}\big[a\big] \ = \ ${ 10 3% }

	{Calculate the value of the error correlation function (ECF) for  $k = 1$.		{Calculate the value of the error correlation function  \rm (ECF)$  for  $k = 1$.
	\|type="{}"}		\|type="{}"}
	$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }		$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }

Hwang at 10:54, 26 August 2022

2022-08-26T10:54:33Z

← Older revision		Revision as of 12:54, 26 August 2022
Line 91:		Line 91:
	In the calculation, it was assumed that the individual error distances are statistically independent of each other.		In the calculation, it was assumed that the individual error distances are statistically independent of each other.
	*However, this assumption is valid only for a special class of channel models called "renewing".		*However, this assumption is valid only for a special class of channel models called "renewing".
	*The ~~bundle fault~~ model considered here does not satisfy this condition.		*The burst error model considered here does not satisfy this condition.
	*The actual probability  ${\rm Pr}(a = 2) = 0.1675$  therefore deviates slightly from the value calculated here  $(0.1715)$.		*The actual probability  ${\rm Pr}(a = 2) = 0.1675$  therefore deviates slightly from the value calculated here  $(0.1715)$.

Hwang at 10:35, 26 August 2022

2022-08-26T10:35:46Z

← Older revision		Revision as of 12:35, 26 August 2022
Line 17:		Line 17:


	~~Zwei direkt aufeinanderfolgende Bitfehler werden somit durch den Fehlerabstand~~  $a = 1$  ~~gekennzeichnet.~~		Two directly consecutive bit errors are thus characterized by the error distance  $a = 1$.

	~~Die Tabelle zeigt beispielhafte Werte der Fehlerabstandswahrscheinlichkeiten~~  ${\rm Pr}(a = k)$  ~~sowie der Fehlerkorrelationsfunktion~~  $\varphi_e(k)$.		The table shows exemplary values of the error distance probabilities  ${\rm Pr}(a = k)$  as well as the error correlation function  $\varphi_e(k)$.
	*~~Einige Angaben fehlen~~ in ~~der Tabelle~~.		*Some data are missing in the table.
	*~~Diese Werte sollen aus den gegebenen Werten berechnet werden~~.		*These values are to be calculated from the given values.


Line 28:		Line 28:


	''~~Hinweis~~:''		''Note:''
	* ~~Die Aufgabe behandelt den Lehrstoff des Kapitels~~  [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models\|"Parameters of Digital Channel Models"]].		* The exercise covers the subject matter of the chapter  [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models\|"Parameters of Digital Channel Models"]].



	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{~~Welcher Wert ergibt sich für die mittlere Fehlerwahrscheinlichkeit~~?		{Which value results for the average error probability?
	\|type="{}"}		\|type="{}"}
	$p_{\rm M} \ = \ ${ 0.1 3% }		$p_{\rm M} \ = \ ${ 0.1 3% }

	{~~Welcher Wert ergibt sich für den mittleren Fehlerabstand~~?		{Which value results for the mean error distance?
	\|type="{}"}		\|type="{}"}
	${\rm E}\big[a\big] \ = \ ${ 10 3% }		${\rm E}\big[a\big] \ = \ ${ 10 3% }

	{~~Berechnen Sie den Wert der Fehlerkorrelationsfunktion~~ (~~FKF~~) ~~für~~  $k = 1$.		{Calculate the value of the error correlation function (ECF) for  $k = 1$.
	\|type="{}"}		\|type="{}"}
	$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }		$\varphi_e(k = 1) \ = \ ${ 0.0309 3% }

	{~~Welche Näherung gilt für die Wahrscheinlichkeit des Fehlerabstands~~  $a = 2$?		{What is the approximation for the probability of the error distance  $a = 2$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(a = 2) \ = \ ${ 0.1715 3% }		${\rm Pr}(a = 2) \ = \ ${ 0.1715 3% }
	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Die mittlere Fehlerwahrscheinlichkeit ist gleich dem FKF–Wert für~~ $k = 0$. ~~Wegen~~ $e_{\nu} ∈ \{0, 1\}$ ~~gilt nämlich~~:		'''(1)'''  The mean probability of error is equal to the ECF value for $k = 0$. Namely, because of $e_{\nu} ∈ \{0, 1\}$:
	:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=		:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=
	p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}		p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}
Line 60:		Line 60:


	'''(2)'''  ~~Der mittlere Fehlerabstand ist gleich dem Kehrwert der mittleren Fehlerwahrscheinlichkeit~~. ~~Das heißt~~:		'''(2)'''  The mean error distance is equal to the reciprocal of the mean error probability. That is:
	:$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$		:$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$


	'''(3)'''  ~~Nach der Definitionsgleichung und dem~~ [[Theory_of_Stochastic_Signals/~~Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit~~#~~Bedingte_Wahrscheinlichkeit~~\| ~~Satz von~~ Bayes]] ~~erhält man folgendes Ergebnis~~:		'''(3)'''  According to the definition equation and [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]], the following result is obtained:
	:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm		:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm
	E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot		E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot
Line 72:		Line 72:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*~~Die erste Wahrscheinlichkeit ist gleich~~ ${\rm Pr}(a = 1)$ ~~und die zweite Wahrscheinlichkeit ist gleich~~ $p_{\rm M}$:		*The first probability is equal to ${\rm Pr}(a = 1)$ and the second probability is equal to $p_{\rm M}$:
	:$$\varphi_{e}(k = 1) = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}		:$$\varphi_{e}(k = 1) = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(4)'''  ~~Der FKF–Wert~~ $\varphi_e(k = 2)$ ~~kann~~ (~~näherungsweise~~) ~~folgendermaßen interpretiert werden~~:		'''(4)'''  The ECF value $\varphi_e(k = 2)$ can be interpreted (approximately) as follows:
	:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1		:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} \hspace{0.3cm}		\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} \hspace{0.3cm}
Line 84:		Line 84:
	= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$		= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$

	~~Diese Wahrscheinlichkeit setzt sich zusammen aus~~  "~~Zum Zeitpunkt~~ $\nu+1$ ~~tritt ein Fehler auf~~"  ~~sowie~~  "~~Zum Zeitpunkt~~ $\nu+1$ ~~gibt es keinen Fehler~~":		This probability is composed of  "At time $\nu+1$ an error occurs"  and  "At time $\nu+1$ there is no error":
	:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =		:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =
	1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}		1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$

	~~Bei der Rechnung wurde davon ausgegangen~~, ~~dass die einzelnen Fehlerabstände statistisch voneinander unabhängig sind~~.		In the calculation, it was assumed that the individual error distances are statistically independent of each other.
	*~~Diese Annahme gilt allerdings nur für eine besondere Klasse von Kanalmodellen~~, ~~die man als~~ "~~erneuernd~~" ~~bezeichnet~~.		*However, this assumption is valid only for a special class of channel models called "renewing".
	*~~Das hier betrachtete Bündelfehlermodell erfüllt diese Bedingung nicht~~.		*The bundle fault model considered here does not satisfy this condition.
	*~~Die tatsächliche Wahrscheinlichkeit~~  ${\rm Pr}(a = 2) = 0.1675$  ~~weicht deshalb vom hier berechneten Wert~~  $(0.1715)$  ~~geringfügig ab.~~		*The actual probability  ${\rm Pr}(a = 2) = 0.1675$  therefore deviates slightly from the value calculated here  $(0.1715)$.

	{{ML-Fuß}}		{{ML-Fuß}}

	[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]		[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]

← Older revision		Revision as of 17:47, 3 September 2022
Line 52:		Line 52:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The mean probability ~~of error~~ is equal to the ECF value for $k = 0$. Namely, because of $e_{\nu} ∈ \{0, 1\}$:		'''(1)'''  The mean error probability is equal to the ECF value for  $k = 0$.  Namely,  because of  $e_{\nu} ∈ \{0, 1\}$:
	:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=		:$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=
	p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}		p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}
Line 58:		Line 58:


	'''(2)'''  The mean error distance is equal to the reciprocal of the mean error probability. That is:		'''(2)'''  The mean  $($or:  "average"$)$  error distance is equal to the reciprocal of the mean error probability.  That is:
	:$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$		:$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$


	'''(3)'''  According to the definition equation and [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]], the following result is obtained:		'''(3)'''  According to the definition equation and  [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability\| "Bayes' theorem"]],  the following result is obtained:
	:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm		:$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm
	E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot		E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot
Line 70:		Line 70:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The first probability is equal to ${\rm Pr}(a = 1)$ and the second probability is equal to $p_{\rm M}$:		*The first probability is equal to  ${\rm Pr}(a = 1)$  and the second probability is equal to  $p_{\rm M}$:
	:$$\varphi_{e}(k = 1) = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}		:$$\varphi_{e}(k = 1) = {\rm Pr}(a = 1) \cdot p_{\rm M} = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(4)'''  The ECF value $\varphi_e(k = 2)$ can be interpreted (approximately) as follows:		'''(4)'''  The ECF value  $\varphi_e(k = 2)$  can be interpreted  (approximately)  as follows:
	:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1		:$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} ~~\hspace{0.3cm}~~		\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} $$
	\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1		:$$\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1
	\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) = \frac{\varphi_{e}(k		\hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} = 1) = \frac{\varphi_{e}(k
	= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$		= 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$

	This probability is composed of  "At time $\nu+1$ an error occurs"  and  "At time $\nu+1$ there is no error":		*This probability is composed of  "at time  $\nu+1$  an error occurs"  and  "At time  $\nu+1$  there is no error":
	:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =		:$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}\|\hspace{0.05cm} e_{\nu} =
	1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}		1) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$

	In the calculation, it was assumed that the individual error distances are statistically independent of each other.		*In the calculation,  it was assumed that the individual error distances are statistically independent of each other.
	*However, this assumption is valid only for a special class of channel models called "renewing".		#However,  this assumption is valid only for a special class of channel models called  "renewing".
	*The burst error model considered here does not satisfy this condition.		#The burst error model considered here does not satisfy this condition.
	*The actual probability  ${\rm Pr}(a = 2) = 0.1675$  therefore deviates slightly from the value calculated here  $(0.1715)$.		#The actual probability  ${\rm Pr}(a = 2) = 0.1675$  therefore deviates slightly from the value calculated here  $(0.1715)$.

	{{ML-Fuß}}		{{ML-Fuß}}

	[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]		[[Category:Digital Signal Transmission: Exercises\|^5.1 Digital Channel Models^]]