Aufgaben:Exercise 5.2: Error Correlation Function - Revision history

Hwang at 02:39, 18 September 2022

2022-09-18T02:39:49Z

Hwang at 02:38, 18 September 2022

2022-09-18T02:38:25Z

Guenter at 15:47, 3 September 2022

2022-09-03T15:47:04Z

Guenter at 15:32, 3 September 2022

2022-09-03T15:32:45Z

Hwang at 10:54, 26 August 2022

2022-08-26T10:54:33Z

Hwang at 10:35, 26 August 2022

2022-08-26T10:35:46Z

Hwang at 07:59, 19 August 2022

2022-08-19T07:59:35Z

Guenter at 15:32, 15 August 2022

2022-08-15T15:32:47Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.2: Fehlerkorrelationsfunktion to Aufgaben:Exercise 5.2: Error Correlation Function

2022-08-15T15:31:07Z

Guenter moved page Aufgabe 5.2: Fehlerkorrelationsfunktion to Exercise 5.2: Error Correlation Function

Javier: Text replacement - "„" to """

2021-05-28T14:45:58Z

Text replacement - "„" to """

@@ Line 82: / Line 82: @@
 = 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$
-*This probability is composed of &nbsp;"at time&nbsp; $\nu+1$&nbsp; an error occurs"&nbsp; and &nbsp;"At time&nbsp; $\nu+1$&nbsp; there is no error":
+*This probability is composed of &nbsp;"at time&nbsp; $\nu+1$&nbsp; an error occurs"&nbsp; and &nbsp;"at time&nbsp; $\nu+1$&nbsp; there is no error":
 :$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}|\hspace{0.05cm} e_{\nu} =
 ) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}

@@ Line 41: / Line 41: @@
 ${\rm E}\big[a\big] \ = \ ${ 10 3% }
-{Calculate the value of the error correlation function&nbsp; \rm (ECF)$&nbsp; for&nbsp; $k = 1$.
+{Calculate the value of the error correlation function&nbsp; $\rm (ECF)$&nbsp; for&nbsp; $k = 1$.
 |type="{}"}
 $\varphi_e(k = 1) \ = \ ${ 0.0309 3% }

@@ Line 52: / Line 52: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The mean probability of error is equal to the ECF value for $k = 0$. Namely, because of $e_{\nu} &#8712; \{0, 1\}$:
+'''(1)'''&nbsp; The mean error probability is equal to the ECF value for&nbsp; $k = 0$.&nbsp; Namely,&nbsp; because of&nbsp; $e_{\nu} &#8712; \{0, 1\}$:
 :$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=
 p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}
@@ Line 58: / Line 58: @@
-'''(2)'''&nbsp; The mean error distance is equal to the reciprocal of the mean error probability. That is:
+'''(2)'''&nbsp; The mean&nbsp; $($or:&nbsp; "average"$)$&nbsp; error distance is equal to the reciprocal of the mean error probability.&nbsp; That is:
 :$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$
-'''(3)'''&nbsp; According to the definition equation and [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability| "Bayes' theorem"]], the following result is obtained:
+'''(3)'''&nbsp; According to the definition equation and&nbsp; [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability| "Bayes' theorem"]],&nbsp; the following result is obtained:
 :$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm
 E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot
@@ Line 70: / Line 70: @@
   \hspace{0.05cm}.$$
-*The first probability is equal to ${\rm Pr}(a = 1)$ and the second probability is equal to $p_{\rm M}$:
+*The first probability is equal to&nbsp; ${\rm Pr}(a = 1)$&nbsp; and the second probability is equal to&nbsp; $p_{\rm M}$:
-:$$\varphi_{e}(k = 1) = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}
+:$$\varphi_{e}(k = 1) = {\rm Pr}(a = 1) \cdot p_{\rm M} = 0.3091 \cdot 0.1\hspace{0.15cm}\underline { = 0.0309}
   \hspace{0.05cm}.$$
-'''(4)'''&nbsp; The ECF value $\varphi_e(k = 2)$ can be interpreted (approximately) as follows:
+'''(4)'''&nbsp; The ECF value&nbsp; $\varphi_e(k = 2)$&nbsp; can be interpreted&nbsp; (approximately)&nbsp; as follows:
 :$$\varphi_{e}(k = 2) ={\rm Pr}(e_{\nu + 2}=1
-\hspace{0.05cm}|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} \hspace{0.3cm}
+\hspace{0.05cm}|\hspace{0.05cm} e_{\nu} = 1) \cdot p_{\rm M} $$
-\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1
+:$$\Rightarrow \hspace{0.3cm}{\rm Pr}(e_{\nu + 2}=1
 \hspace{0.05cm}|\hspace{0.05cm} e_{\nu} = 1) = \frac{\varphi_{e}(k
 = 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$
-This probability is composed of &nbsp;"At time $\nu+1$ an error occurs"&nbsp; and &nbsp;"At time $\nu+1$ there is no error":
+*This probability is composed of &nbsp;"at time&nbsp; $\nu+1$&nbsp; an error occurs"&nbsp; and &nbsp;"At time&nbsp; $\nu+1$&nbsp; there is no error":
 :$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}|\hspace{0.05cm} e_{\nu} =
 ) = {\rm Pr}( a =1) \cdot {\rm Pr}( a =1) + {\rm Pr}( a =2)\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}{\rm Pr}( a =2)= 0.267 - 0.3091^2 \hspace{0.15cm}\underline {= 0.1715}\hspace{0.05cm}.$$
-In the calculation, it was assumed that the individual error distances are statistically independent of each other.
+*In the calculation,&nbsp; 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,&nbsp; this assumption is valid only for a special class of channel models called&nbsp; "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&nbsp; ${\rm Pr}(a = 2) = 0.1675$&nbsp; therefore deviates slightly from the value calculated here&nbsp; $(0.1715)$.&nbsp;
+#The actual probability&nbsp; ${\rm Pr}(a = 2) = 0.1675$&nbsp; therefore deviates slightly from the value calculated here&nbsp; $(0.1715)$.&nbsp;
 {{ML-Fuß}}
 [[Category:Digital Signal Transmission: Exercises|^5.1 Digital Channel Models^]]

@@ Line 2: / Line 2: @@
 {{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
-* the error correlation function (ECF)
+* the&nbsp; "error correlation function"&nbsp; $\rm (ECF)$
 :$$\varphi_{e}(k) = {\rm E}\big[e_{\nu} \cdot e_{\nu +
 k}\big]\hspace{0.05cm}, \hspace{0.2cm} k \ge 0\hspace{0.05cm},$$
-* the error distance probabilities
+* the&nbsp; "error distance probabilities"
 :$${\rm Pr}( a =k)  \hspace{0.05cm}, \hspace{0.2cm} k \ge
 \hspace{0.05cm}.$$
 Here denote:
 * $&#9001;e_{\rm \nu}&#9002;$&nbsp; is the error sequence with&nbsp; $e_{\rm \nu} &#8712; \{0, 1\}$.
 * $a$&nbsp; indicates the error distance.
-Two directly consecutive bit errors are thus characterized by the error distance&nbsp; $a = 1$.&nbsp;
+Two directly consecutive symbol errors are thus characterized by the error distance&nbsp; $a = 1$.&nbsp;
 The table shows exemplary values of the error distance probabilities&nbsp; ${\rm Pr}(a = k)$&nbsp; as well as the error correlation function&nbsp; $\varphi_e(k)$.
@@ Line 26: / Line 27: @@
+Note:&nbsp; The exercise covers the subject matter of the chapter&nbsp; [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models|"Parameters of Digital Channel Models"]].
@@ Line 35: / Line 33: @@
 ===Questions===
 <quiz display=simple>
-{Which value results for the average error probability?
+{Which value results for the mean error probability?
 |type="{}"}
 $p_{\rm M} \ = \ ${ 0.1 3% }
@@ Line 43: / Line 41: @@
 ${\rm E}\big[a\big] \ = \ ${ 10 3% }
-{Calculate the value of the error correlation function (ECF) for&nbsp; $k = 1$.
+{Calculate the value of the error correlation function&nbsp; \rm (ECF)$&nbsp; for&nbsp; $k = 1$.
 |type="{}"}
 $\varphi_e(k = 1) \ = \ ${ 0.0309 3% }

@@ Line 91: / Line 91: @@
 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".
-*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&nbsp; ${\rm Pr}(a = 2) = 0.1675$&nbsp; therefore deviates slightly from the value calculated here&nbsp; $(0.1715)$.&nbsp;

@@ Line 17: / Line 17: @@
-Zwei direkt aufeinanderfolgende Bitfehler werden somit durch den Fehlerabstand&nbsp; $a = 1$&nbsp; gekennzeichnet.
+Two directly consecutive bit errors are thus characterized by the error distance&nbsp; $a = 1$.&nbsp;
-Die Tabelle zeigt beispielhafte Werte der Fehlerabstandswahrscheinlichkeiten&nbsp; ${\rm Pr}(a = k)$&nbsp; sowie der Fehlerkorrelationsfunktion&nbsp; $\varphi_e(k)$.
+The table shows exemplary values of the error distance probabilities&nbsp; ${\rm Pr}(a = k)$&nbsp; as well as the error correlation function&nbsp; $\varphi_e(k)$.
-*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&nbsp; [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models|"Parameters of Digital Channel Models"]].
+* The exercise covers the subject matter of the chapter&nbsp; [[Digital_Signal_Transmission/Parameters_of_Digital_Channel_Models|"Parameters of Digital Channel Models"]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welcher Wert ergibt sich für die mittlere Fehlerwahrscheinlichkeit?
+{Which value results for the average error probability?
 |type="{}"}
 $p_{\rm M} \ = \ ${ 0.1 3% }
-{Welcher Wert ergibt sich für den mittleren Fehlerabstand?
+{Which value results for the mean error distance?
 |type="{}"}
 ${\rm E}\big[a\big] \ = \ ${ 10 3% }
-{Berechnen Sie den Wert der Fehlerkorrelationsfunktion (FKF) für&nbsp; $k = 1$.
+{Calculate the value of the error correlation function (ECF) for&nbsp; $k = 1$.
 |type="{}"}
 $\varphi_e(k = 1) \ = \ ${ 0.0309 3% }
-{Welche Näherung gilt für die Wahrscheinlichkeit des Fehlerabstands&nbsp; $a = 2$?
+{What is the approximation for the probability of the error distance&nbsp; $a = 2$?
 |type="{}"}
 ${\rm Pr}(a = 2) \ = \ ${ 0.1715 3% }
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die mittlere Fehlerwahrscheinlichkeit ist gleich dem FKF&ndash;Wert für $k = 0$. Wegen $e_{\nu} &#8712; \{0, 1\}$ gilt nämlich:
+'''(1)'''&nbsp; The mean probability of error is equal to the ECF value for $k = 0$. Namely, because of $e_{\nu} &#8712; \{0, 1\}$:
 :$$\varphi_{e}(k = 0) = {\rm E}[e_{\nu}^2 ]= {\rm E}[e_{\nu} ]=
 p_{\rm M} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm M}\hspace{0.15cm}\underline { = 0.1}
@@ Line 60: / Line 60: @@
-'''(2)'''&nbsp; Der mittlere Fehlerabstand ist gleich dem Kehrwert der mittleren Fehlerwahrscheinlichkeit. Das heißt:
+'''(2)'''&nbsp; The mean error distance is equal to the reciprocal of the mean error probability. That is:
 :$${\rm E}\big[a\big] = 1/p_{\rm M} \ \underline {= 10}.$$
-'''(3)'''&nbsp; Nach der Definitionsgleichung und dem [[Theory_of_Stochastic_Signals/Statistische_Abh%C3%A4ngigkeit_und_Unabh%C3%A4ngigkeit#Bedingte_Wahrscheinlichkeit| Satz von Bayes]] erhält man folgendes Ergebnis:
+'''(3)'''&nbsp; According to the definition equation and [[Theory_of_Stochastic_Signals/Statistical_Dependence_and_Independence#Conditional_Probability| "Bayes' theorem"]], the following result is obtained:
 :$$\varphi_{e}(k = 1) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm
 E}[e_{\nu} \cdot e_{\nu + 1}] = {\rm E}[(e_{\nu} = 1) \cdot
@@ Line 72: / Line 72: @@
   \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}
   \hspace{0.05cm}.$$
-'''(4)'''&nbsp; Der FKF&ndash;Wert $\varphi_e(k = 2)$ kann (näherungsweise) folgendermaßen interpretiert werden:
+'''(4)'''&nbsp; The ECF value $\varphi_e(k = 2)$ can be interpreted (approximately) as follows:
 :$$\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}
@@ Line 84: / Line 84: @@
 = 2)}{p_{\rm M}} = \frac{0.0267}{0.1} = 0.267\hspace{0.05cm}.$$
-Diese Wahrscheinlichkeit setzt sich zusammen aus &nbsp;"Zum Zeitpunkt $\nu+1$ tritt ein Fehler auf"&nbsp; sowie &nbsp;"Zum Zeitpunkt $\nu+1$ gibt es keinen Fehler":
+This probability is composed of &nbsp;"At time $\nu+1$ an error occurs"&nbsp; and &nbsp;"At time $\nu+1$ there is no error":
 :$${\rm Pr}(e_{\nu + 2}=1 \hspace{0.05cm}|\hspace{0.05cm} e_{\nu} =
 ) = {\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}.$$
-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&nbsp; ${\rm Pr}(a = 2) = 0.1675$&nbsp; weicht deshalb vom hier berechneten Wert&nbsp; $(0.1715)$&nbsp; geringfügig ab.
+*The actual probability&nbsp; ${\rm Pr}(a = 2) = 0.1675$&nbsp; therefore deviates slightly from the value calculated here&nbsp; $(0.1715)$.&nbsp;
 {{ML-Fuß}}
 [[Category:Digital Signal Transmission: Exercises|^5.1 Digital Channel Models^]]

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