Aufgaben:Exercise 5.3Z: Analysis of the BSC Model - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:09Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 16:		Line 16:
	:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$		:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$
	* the  '''error correlation function'''  $\rm (ECF)$		* the  '''error correlation function'''  $\rm (ECF)$
	:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\~~rmE~~}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \\left\{ \begin{array}{c} p \\p^2 \end{array} \right.\quad\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$		:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \\left\{ \begin{array}{c} p \\p^2 \end{array} \right.\quad\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$


Line 104:		Line 104:

	[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]		[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]
	[[de:5.~~3Z_Analyse_des_BSC~~-Modells]]		[[de:Aufgaben:Aufgabe 5.3Z: Analyse des BSC-Modells]]

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

2026-03-16T13:48:50Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 16:		Line 16:
	:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$		:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$
	* the  '''error correlation function'''  $\rm (ECF)$		* the  '''error correlation function'''  $\rm (ECF)$
	:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm		:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rmE}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \\left\{ \begin{array}{c} p \\p^2 \end{array} \right.\quad\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$
	E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \
	\left\{ \begin{array}{c} p \\
	p^2 \end{array} \right.\quad
	\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},
	\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm}.\\ \end{array}$$

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

2026-03-16T13:31:15Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:31, 16 March 2026
Line 70:		Line 70:

	*For the error correlation function and the error distance distribution are valid		*For the error correlation function and the error distance distribution are valid
	:$$\varphi_{e}(k) =		:$$\varphi_{e}(k) =\left\{ \begin{array}{c} p \\p^2 \end{array} \right.\quad\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
	\left\{ \begin{array}{c} p \\
	p^2 \end{array} \right.\quad
	\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},
	\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
	:$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$		:$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$

Line 96:		Line 92:

	'''(4)'''  According to the equation   ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$   we obtain:		'''(4)'''  According to the equation   ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$   we obtain:
	:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=		:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=0.1}\hspace{0.05cm},$$
	0.1}\hspace{0.05cm},$$		:$${\rm Pr}(a = 2) = 0.9 \cdot 0.1\hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
	:$${\rm Pr}(a = 2) = 0.9 \cdot 0.1		:$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm Pr}(a = 10)= 0.9^9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.0387}\hspace{0.05cm}.$$
	\hspace{0.15cm}\underline {= 0.09}\hspace{0.05cm},$$
	:$${\rm Pr}(a = {\rm E}[a]) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}
	{\rm Pr}(a = 10)= 0.9^9 \cdot 0.1 \hspace{0.15cm}\underline {= 0.0387}\hspace{0.05cm}.$$



	'''(5)'''  From the relation  $V_a(k) = (1–p)^{k–1}$  we obtain		'''(5)'''  From the relation  $V_a(k) = (1–p)^{k–1}$  we obtain
	:$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm}		:$$V_a(k = 2) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^1 \hspace{0.15cm}\underline {= 0.9 } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k =1) - V_a(k = 2)= 0.1\hspace{0.05cm},$$
	\Rightarrow \hspace{0.3cm}{\rm Pr}(a = 1) = V_a(k =		:$$V_a(k = 10)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^9\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}\hspace{0.15cm}\underline {=0.3487}.$$
	1) - V_a(k = 2)
	= 0.1\hspace{0.05cm},$$
	:$$V_a(k = 10)\hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.9^9
	\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}
	\hspace{0.15cm}\underline {=0.3487}.$$
	*To check in comparison with subtask  '''(4)''':		*To check in comparison with subtask  '''(4)''':
	:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =		:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$
	11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Maintenance script: Add German interlanguage link

2026-03-16T12:25:52Z

Add German interlanguage link

← Older revision		Revision as of 14:25, 16 March 2026
Line 122:		Line 122:

	[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]		[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]
			[[de:5.3Z_Analyse_des_BSC-Modells]]

Guenter at 13:52, 5 September 2022

2022-09-05T13:52:12Z

← Older revision		Revision as of 15:52, 5 September 2022
Line 67:		Line 67:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  In the BSC model, the mean error probability $p_{\rm M}$ is always equal to the characteristic probability $p$.		'''(1)'''  In the BSC model,  the mean error probability $p_{\rm M}$  is always equal to the characteristic probability  $p$.

	*For the error correlation function and the error distance distribution are valid		*For the error correlation function and the error distance distribution are valid
	:$$\varphi_{e}(k) =		:$$\varphi_{e}(k) =
Line 73:		Line 74:
	p^2 \end{array} \right.\quad		p^2 \end{array} \right.\quad
	\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},		\begin{array}{*{1}c} f{\rm or }\hspace{0.15cm}k = 0 \hspace{0.05cm},
	\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}		\\ f{\rm or }\hspace{0.15cm} k \ne 0 \hspace{0.05cm},\\ \end{array}$$
	~~\hspace{0.4cm}~~V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$		:$$V_a(k) = (1-p)^{k-1}\hspace{0.05cm}.$$

	*$p$  can be determined from all the given characteristics, except $V_a(k = 1)$. This EDD value is independent of  $p$  equal to  $(1–p)^0 = 1$.		*$p$  can be determined from all the given characteristics,  except $V_a(k = 1)$.  This EDD value is independent of  $p$   equal to   $(1–p)^0 = 1$.
	*Therefore, the <u>solutions 1, 2, 4 and 5</u> are correct.
			*Therefore, the  <u>solutions 1, 2, 4 and 5</u>  are correct.



	'''(2)'''  The relative error frequency of the given sequence is equal to $h_{\rm F} = 22/1000 \approx 0.022$.		'''(2)'''  The relative error frequency of the given sequence is equal to  $h_{\rm F} = 22/1000 \approx 0.022$.
	*It is quite obvious that the error sequence was generated by the model $M_2$  ⇒  $p_{\rm M} = 0.02$.		*It is quite obvious that the error sequence was generated by the model  $M_2$    ⇒   $p_{\rm M} = 0.02$.
	*Because of the short sequence, $h_{\rm F}$ does not match $p_{\rm M}$ exactly, but at least approximates  ⇒  <u>solution 2</u>.
			*Because of the short sequence,  $h_{\rm F}$  does not match  $p_{\rm M}$  exactly,  but at least approximates  ⇒  <u>solution 2</u>.



	'''(3)'''  The mean error distance – that is, the expected value of the random variable  $a$ – is equal to the inverse of the mean error probability ⇒ ${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}$.		'''(3)'''  The mean error distance – that is,  the expected value of the random variable  $a$  – is equal to the inverse of the mean error probability ⇒
			:$${\rm E}\big[a\big] = 1/0.1 \ \underline {= 10}.$$



	'''(4)'''  According to the equation  ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$  we obtain:		'''(4)'''  According to the equation   ${\rm Pr}(a = k) = (1–p)^{k–1} \cdot p$   we obtain:
	:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=		:$${\rm Pr}(a = 1) \hspace{0.15cm}\underline {=
	0.1}\hspace{0.05cm},$$		0.1}\hspace{0.05cm},$$
Line 109:		Line 113:
	\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}		\hspace{0.15cm}\underline {=0.3874}\hspace{0.05cm},\hspace{0.2cm}V_a(k = 11)= 0.9^{10}
	\hspace{0.15cm}\underline {=0.3487}.$$		\hspace{0.15cm}\underline {=0.3487}.$$
	To check in comparison with subtask (4):		*To check in comparison with subtask  '''(4)''':
	:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =		:$${\rm Pr}(a = 10) = V_a(k = 10) - V_a(k =
	11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$		11) = 0.3874 - 0.3487 {= 0.0387}\hspace{0.05cm}.$$

Guenter at 13:42, 5 September 2022

2022-09-05T13:42:13Z

← Older revision		Revision as of 15:42, 5 September 2022
Line 2:		Line 2:
	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)}}

	[[File:P_ID1832__Dig_Z_5_3.png\|right\|frame\|~~Given~~ error sequence]]		[[File:P_ID1832__Dig_Z_5_3.png\|right\|frame\|The given error sequence]]
	We consider two different BSC models with the following parameters:		We consider two different BSC models with the following parameters:
	* Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,		* Model $M_1 \text{:} \hspace{0.4cm} p = 0.01$,
Line 8:		Line 8:


	The graph shows an error sequence of length  $N = 1000$, but it is not known from which of the two models this sequence originates.		The graph shows an error sequence of length  $N = 1000$,  but it is not known from which of the two models this sequence originates.

	The two models are to be analyzed on the basis of		The two models are to be analyzed on the basis of
	* the error distance probabilities		* the  '''error distance probabilities'''
	:$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$		:$${\rm Pr}(a = k) = (1-p)^{k-1}\cdot p \hspace{0.05cm},$$
	* the error distance distribution		* the  '''error distance distribution'''  $\rm (EDD)$
	:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$		:$$V_a(k) = {\rm Pr}(a \ge k) = (1-p)^{k-1}\hspace{0.05cm},$$
	* the error correlation function		* the  '''error correlation function'''  $\rm (ECF)$
	:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm		:$$\varphi_{e}(k) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm
	E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \		E}\big[e_{\nu} \cdot e_{\nu + k}\big] \ \ = \ \
Line 25:		Line 25:


			Notes:
			* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)\| "Binary Symmetric Channel"]].

			*By counting,  we would see that the error sequence of length  $N = 1000$  contains exactly  $22$  "ones".


	~~''Notes:''~~
	* ~~The exercise belongs to the chapter~~  ~~[[Digital_Signal_Transmission/Binary_Symmetric_Channel_(BSC)\| "Binary Symmetric Channel (BSC)"]].~~
	*By counting, we would see that the error sequence of length  $N = 1000$  contains exactly  $22$  ones.


Line 49:		Line 46:
	\|type="()"}		\|type="()"}
	- Model $M_1$,		- Model $M_1$,
	+ ~~Model~~ $M_2$.		+ model $M_2$.

	{What is the mean error distance of model  $M_1$?		{What is the mean error distance of model  $M_1$?

Hwang at 13:45, 26 August 2022

2022-08-26T13:45:24Z

Show changes

Guenter at 17:23, 23 August 2022

2022-08-23T17:23:25Z

← Older revision		Revision as of 19:23, 23 August 2022
Line 120:		Line 120:


	[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel ~~(BSC)~~^]]		[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 5.3Z: Analyse des BSC-Modells to Aufgaben:Exercise 5.3Z: Analysis of the BSC Model

2022-08-23T16:34:00Z

Guenter moved page Aufgaben:Aufgabe 5.3Z: Analyse des BSC-Modells to Aufgaben:Exercise 5.3Z: Analysis of the BSC Model

← Older revision	Revision as of 18:34, 23 August 2022
(No difference)

Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:52:08Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

← Older revision		Revision as of 14:52, 23 March 2021
Line 120:		Line 120:


	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^5.2 Binary Symmetric Channel (BSC)^]]		[[Category:Digital Signal Transmission: Exercises\|^5.2 Binary Symmetric Channel (BSC)^]]