Information Theory/Application to Digital Signal Transmission - Revision history

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Join 1 split formula blocks

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	:$$Y = \{\hspace{0.05cm}y_1\hspace{0.05cm}, \hspace{0.15cm} y_2\hspace{0.05cm},\hspace{0.15cm} \text{...}\hspace{0.1cm} ,\hspace{0.15cm} y_{\kappa}\hspace{0.05cm}, \hspace{0.15cm}\text{...}\hspace{0.1cm} , \hspace{0.15cm} Y_{\|Y\|}\hspace{0.05cm}\}\hspace{0.05cm}.$$		:$$Y = \{\hspace{0.05cm}y_1\hspace{0.05cm}, \hspace{0.15cm} y_2\hspace{0.05cm},\hspace{0.15cm} \text{...}\hspace{0.1cm} ,\hspace{0.15cm} y_{\kappa}\hspace{0.05cm}, \hspace{0.15cm}\text{...}\hspace{0.1cm} , \hspace{0.15cm} Y_{\|Y\|}\hspace{0.05cm}\}\hspace{0.05cm}.$$

	*Usually  $\|Y\| = \|X\|$is valid.  Also possible is  $\|Y\| > \|X\|$, for example with the  [[Channel_Coding/~~Kanalmodelle_und_Entscheiderstrukturen~~#~~Binary_Symmetric_Channel_.E2.80.93_BSC~~\|$\text{Binary Erasure Channel}$]]  (BEC):		*Usually  $\|Y\| = \|X\|$is valid.  Also possible is  $\|Y\| > \|X\|$, for example with the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Symmetric_Channel_–_BSC\|$\text{Binary Erasure Channel}$]]  (BEC):
	:$$X = \{0,\ 1\},\hspace{0.5cm} Y = \{0,\ 1,\ \ \text{E}\}\ ⇒ \ \|X\| = 2, \ \|Y\| = 3.$$		:$$X = \{0,\ 1\},\hspace{0.5cm} Y = \{0,\ 1,\ \ \text{E}\}\ ⇒ \ \|X\| = 2, \ \|Y\| = 3.$$

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	[[File:P_ID2780__Inf_T_3_3_S1b_neu.png\|right\|frame\|Channel model  "Binary Erasure Channel"  $\rm (BEC)$]]		[[File:P_ID2780__Inf_T_3_3_S1b_neu.png\|right\|frame\|Channel model  "Binary Erasure Channel"  $\rm (BEC)$]]
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 1}$:  In the book  "Channel Coding"  we also deal with the  [[Channel_Coding/~~Kanalmodelle_und_Entscheiderstrukturen~~#~~Binary_Erasure_Channel_.E2.80.93_BEC~~\|$\text{Binary Erasure Channel}$]]  $\rm (BEC)$, which is sketched in a somewhat modified form on the right.   The following prerequisites apply:		$\text{Example 1}$:  In the book  "Channel Coding"  we also deal with the  [[Channel_Coding/Channel_Models_and_Decision_Structures#Binary_Erasure_Channel_–_BEC\|$\text{Binary Erasure Channel}$]]  $\rm (BEC)$, which is sketched in a somewhat modified form on the right.   The following prerequisites apply:
	*Let the input alphabet be binary  ⇒   $X = \{0,\ 1 \}$   ⇒   $\vert X\vert = 2$  while three values are possible at the output   ⇒   $Y = \{0,\ 1,\ \text{E} \}$   ⇒   $\vert Y\vert = 3$.		*Let the input alphabet be binary  ⇒   $X = \{0,\ 1 \}$   ⇒   $\vert X\vert = 2$  while three values are possible at the output   ⇒   $Y = \{0,\ 1,\ \text{E} \}$   ⇒   $\vert Y\vert = 3$.
	*The symbol  $\text{E}$  indicates the case that the receiver cannot decide for one of the binary symbols  $0$  or  $1$  due to too much channel interference.  "E"  stands for erasure.		*The symbol  $\text{E}$  indicates the case that the receiver cannot decide for one of the binary symbols  $0$  or  $1$  due to too much channel interference.  "E"  stands for erasure.
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	[[File:EN_Inf_T_3_3_S2_vers2.png\|center\|frame\|Two information-theoretical models for digital signal transmission]]		[[File:EN_Inf_T_3_3_S2_vers2.png\|center\|frame\|Two information-theoretical models for digital signal transmission]]

	Let us now interpret the right graph starting from the general  [[Information_Theory/~~Anwendung_auf_die_Digitalsignalübertragung~~#Information-theoretical_model_of_digital_signal_transmission\|$\text{DMC model}$]]:		Let us now interpret the right graph starting from the general  [[Information_Theory/Application_to_Digital_Signal_Transmission#Information-theoretical_model_of_digital_signal_transmission\|$\text{DMC model}$]]:
	*The  »'''source entropy'''«  $H(X)$  denotes the average information content of the source symbol sequence.   With the symbol set size  $\|X\|$  applies:		*The  »'''source entropy'''«  $H(X)$  denotes the average information content of the source symbol sequence.   With the symbol set size  $\|X\|$  applies:

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	:$$\begin{align}\varepsilon_0 & = {\rm Pr}(Y\hspace{-0.1cm} = 1\hspace{0.05cm}\vert X \hspace{-0.1cm}= 0) = 0.01\hspace{0.05cm},\\\varepsilon_1 & = {\rm Pr}(Y\hspace{-0.1cm} = 0\hspace{0.05cm} \vert X \hspace{-0.1cm}= 1) = 0.2\hspace{0.05cm}\end{align}$$		:$$\begin{align}\varepsilon_0 & = {\rm Pr}(Y\hspace{-0.1cm} = 1\hspace{0.05cm}\vert X \hspace{-0.1cm}= 0) = 0.01\hspace{0.05cm},\\\varepsilon_1 & = {\rm Pr}(Y\hspace{-0.1cm} = 0\hspace{0.05cm} \vert X \hspace{-0.1cm}= 1) = 0.2\hspace{0.05cm}\end{align}$$

	:$$\Rightarrow \hspace{0.3cm} P_{\hspace{0.05cm}Y\hspace{-0.01cm} \vert \hspace{-0.01cm}X}(Y\hspace{-0.01cm} \vert \hspace{-0.01cm} X) =		:$$\Rightarrow \hspace{0.3cm} P_{\hspace{0.05cm}Y\hspace{-0.01cm} \vert \hspace{-0.01cm}X}(Y\hspace{-0.01cm} \vert \hspace{-0.01cm} X) =\begin{pmatrix}1 - \varepsilon_0 & \varepsilon_0\\\varepsilon_1 & 1 - \varepsilon_1\end{pmatrix} =\begin{pmatrix}0.99 & 0.01\\0.2 & 0.8\end{pmatrix} \hspace{0.05cm}.$$Furthermore, we assume source symbols that are not equally probable::$$P_X(X) = \big ( p_0,\ p_1 \big )=\big ( 0.1,\ 0.9 \big )\hspace{0.05cm}.$$With the  [[Information_Theory/Discrete_Memoryless_Sources#Binary_entropy_function\|$\text{binary entropy function}$]]  $H_{\rm bin}(p)$,  we thus obtain for the source entropy::$$H(X) = H_{\rm bin} (0.1) = 0.4690 \hspace{0.12cm}{\rm bit}\hspace{0.05cm}.$$For the probability mass function of the sink as well as for the sink entropy we thus obtain::$$P_Y(Y) = \big [ {\rm Pr}( Y\hspace{-0.1cm} = 0)\hspace{0.05cm}, \ {\rm Pr}( Y \hspace{-0.1cm}= 1) \big ] = \big ( p_0\hspace{0.05cm},\ p_1 \big ) \cdot\begin{pmatrix}1 - \varepsilon_0 & \varepsilon_0\\\varepsilon_1 & 1 - \varepsilon_1\end{pmatrix} $$:$$\begin{align}\Rightarrow \hspace{0.3cm} {\rm Pr}( Y \hspace{-0.1cm}= 0)& =p_0 \cdot (1 - \varepsilon_0) + p_1 \cdot \varepsilon_1 =0.1 \cdot 0.99 + 0.9 \cdot 0.2 = 0.279\hspace{0.05cm},\\{\rm Pr}( Y \hspace{-0.1cm}= 1) & = 1 - {\rm Pr}( Y \hspace{-0.1cm}= 0) = 0.721\end{align}$$:$$\Rightarrow \hspace{0.3cm}H(Y) = H_{\rm bin} (0.279) = 0.8541 \hspace{0.12cm}{\rm bit}\hspace{0.05cm}. $$The joint probabilities  $p_{\mu \kappa} = \text{Pr}\big[(X = μ) ∩ (Y = κ)\big]$  between source and sink are::$$\begin{align}p_{00} & = p_0 \cdot (1 - \varepsilon_0) = 0.099\hspace{0.05cm},\hspace{0.5cm}p_{01}= p_0 \cdot \varepsilon_0 = 0.001\hspace{0.05cm},\\p_{10} & = p_1 \cdot (1 - \varepsilon_1) = 0.180\hspace{0.05cm},\hspace{0.5cm}p_{11}= p_1 \cdot \varepsilon_1 = 0.720\hspace{0.05cm}.\end{align}$$From this one obtains forthe  »'''joint entropy'''«::$$H(XY) = p_{00}\hspace{-0.05cm} \cdot \hspace{-0.05cm}{\rm log}_2 \hspace{0.05cm} \frac{1}{p_{00} \rm } +p_{01} \hspace{-0.05cm} \cdot \hspace{-0.05cm}{\rm log}_2 \hspace{0.05cm} \frac{1}{p_{01} \rm } +p_{10}\hspace{-0.05cm} \cdot \hspace{-0.05cm} {\rm log}_2 \hspace{0.05cm} \frac{1}{p_{10} \rm } +p_{11} \hspace{-0.05cm} \cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.05cm} \frac{1}{p_{11}\rm } = 1.1268\,{\rm bit} \hspace{0.05cm},$$the  »'''mutual information'''«::$$I(X;Y) = H(X) + H(Y) - H(XY) = 0.4690 + 0.8541 - 1.1268 = 0.1963\hspace{0.12cm}{\rm bit}\hspace{0.05cm},$$
	\begin{pmatrix}
	1 - \varepsilon_0 & \varepsilon_0\\
	\varepsilon_1 & 1 - \varepsilon_1
	\end{pmatrix} =
	\begin{pmatrix}
	0.99 & 0.01\\
	0.2 & 0.8
	\end{pmatrix} \hspace{0.05cm}.$$Furthermore, we assume source symbols that are not equally probable::$$P_X(X) = \big ( p_0,\ p_1 \big )=
	\big ( 0.1,\ 0.9 \big )
	\hspace{0.05cm}.$$With the  [[Information_Theory/~~Gedächtnislose_Nachrichtenquellen~~#Binary_entropy_function\|$\text{binary entropy function}$]]  $H_{\rm bin}(p)$,  we thus obtain for the source entropy::$$H(X) = H_{\rm bin} (0.1) = 0.4690 \hspace{0.12cm}{\rm bit}
	\hspace{0.05cm}.$$For the probability mass function of the sink as well as for the sink entropy we thus obtain::$$P_Y(Y) = \big [ {\rm Pr}( Y\hspace{-0.1cm} = 0)\hspace{0.05cm}, \ {\rm Pr}( Y \hspace{-0.1cm}= 1) \big ] = \big ( p_0\hspace{0.05cm},\ p_1 \big ) \cdot
	\begin{pmatrix}
	1 - \varepsilon_0 & \varepsilon_0\\
	\varepsilon_1 & 1 - \varepsilon_1
	\end{pmatrix} $$:$$\begin{align*}\Rightarrow \hspace{0.3cm} {\rm Pr}( Y \hspace{-0.1cm}= 0)& =
	p_0 \cdot (1 - \varepsilon_0) + p_1 \cdot \varepsilon_1 =
	0.1 \cdot 0.99 + 0.9 \cdot 0.2 = 0.279\hspace{0.05cm},\\
	{\rm Pr}( Y \hspace{-0.1cm}= 1) & = 1 - {\rm Pr}( Y \hspace{-0.1cm}= 0) = 0.721\end{align*}$$:$$\Rightarrow \hspace{0.3cm}
	H(Y) = H_{\rm bin} (0.279) = 0.8541 \hspace{0.12cm}{\rm bit}
	\hspace{0.05cm}. $$The joint probabilities  $p_{\mu \kappa} = \text{Pr}\big[(X = μ) ∩ (Y = κ)\big]$  between source and sink are::$$\begin{align*}p_{00} & = p_0 \cdot (1 - \varepsilon_0) = 0.099\hspace{0.05cm},\hspace{0.5cm}p_{01}= p_0 \cdot \varepsilon_0 = 0.001\hspace{0.05cm},\\
	p_{10} & = p_1 \cdot (1 - \varepsilon_1) = 0.180\hspace{0.05cm},\hspace{0.5cm}p_{11}= p_1 \cdot \varepsilon_1 = 0.720\hspace{0.05cm}.\end{align}$$From this one obtains forthe  »'''joint entropy'''«::$$H(XY) = p_{00}\hspace{-0.05cm} \cdot \hspace{-0.05cm}{\rm log}_2 \hspace{0.05cm} \frac{1}{p_{00} \rm } +
	p_{01} \hspace{-0.05cm} \cdot \hspace{-0.05cm}{\rm log}_2 \hspace{0.05cm} \frac{1}{p_{01} \rm } +
	p_{10}\hspace{-0.05cm} \cdot \hspace{-0.05cm} {\rm log}_2 \hspace{0.05cm} \frac{1}{p_{10} \rm } +
	p_{11} \hspace{-0.05cm} \cdot \hspace{-0.05cm} {\rm log}_2\hspace{0.05cm} \frac{1}{p_{11}\rm } = 1.1268\,{\rm bit} \hspace{0.05cm},$$*the  »'''mutual information'''«::$$I(X;Y) = H(X) + H(Y) - H(XY) = 0.4690 + 0.8541 - 1.1268 = 0.1963\hspace{0.12cm}{\rm bit}
	\hspace{0.05cm},$$

	[[File:EN_Inf_T_3_3_S3b_vers2.png\|right\|frame\|Information theoretic model of the binary channel under consideration]]		[[File:EN_Inf_T_3_3_S3b_vers2.png\|right\|frame\|Information theoretic model of the binary channel under consideration]]
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	In the chapter  [[Information_Theory/~~AWGN_Channel_Capacity_for_Discrete_Input~~#AWGN_model_for_discrete-time_band-limited_signals\|"AWGN model for discrete-time band-limited signals"]]  it is explained in connection with the continuous  [[Channel_Coding/Channel_Models_and_Decision_Structures#AWGN_channel_at_binary_input\|$\text{AWGN channel model}$]]    what phenomenally great significance Shannon's theorem has for the entire field of information technology,  not only for those interested exclusively in theory,  but also for practitioners.		In the chapter  [[Information_Theory/AWGN_Channel_Capacity_for_Discrete-Valued_Input#AWGN_model_for_discrete-time_band-limited_signals\|"AWGN model for discrete-time band-limited signals"]]  it is explained in connection with the continuous  [[Channel_Coding/Channel_Models_and_Decision_Structures#AWGN_channel_at_binary_input\|$\text{AWGN channel model}$]]    what phenomenally great significance Shannon's theorem has for the entire field of information technology,  not only for those interested exclusively in theory,  but also for practitioners.


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	<br>		<br>
	[[File:P_ID2786__Inf_T_3_3_S3a.png\|right\|frame\|General model of the binary channel]]		[[File:P_ID2786__Inf_T_3_3_S3a.png\|right\|frame\|General model of the binary channel]]
	The mutual information of the general  (asymmetrical)  binary channel according to this sketch was calculated in  [[Information_Theory/~~Anwendung_auf_die_Digitalsignalübertragung~~#~~Transinformationsberechnung_f.C3.BCr_den_Bin.C3.A4rkanal~~\|$\text{Example 2}$]].  In this model, the input symbols  $0$  and  $1$  are distorted to different degrees:		The mutual information of the general  (asymmetrical)  binary channel according to this sketch was calculated in  [[Information_Theory/Application_to_Digital_Signal_Transmission#Calculation_of_the_mutual_information_for_the_binary_channel\|$\text{Example 2}$]].  In this model, the input symbols  $0$  and  $1$  are distorted to different degrees:

	:$$P_{\hspace{0.05cm}Y\hspace{-0.01cm}\|\hspace{-0.01cm}X}(Y\hspace{-0.01cm} \|\hspace{-0.01cm} X) =\begin{pmatrix}1 - \varepsilon_0 & \varepsilon_0\\\varepsilon_1 & 1 - \varepsilon_1\end{pmatrix} \hspace{0.05cm}.$$		:$$P_{\hspace{0.05cm}Y\hspace{-0.01cm}\|\hspace{-0.01cm}X}(Y\hspace{-0.01cm} \|\hspace{-0.01cm} X) =\begin{pmatrix}1 - \varepsilon_0 & \varepsilon_0\\\varepsilon_1 & 1 - \varepsilon_1\end{pmatrix} \hspace{0.05cm}.$$
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	==Properties of symmetrical channels ==		==Properties of symmetrical channels ==
	<br>		<br>
	The capacity calculation of the (general)  [[Information_Theory/~~Anwendung_auf_die_Digitalsignalübertragung~~#Information-theoretical_model_of_digital_signal_transmission\|$\text{discrete memoryless channel}$]]  $\rm (DMC)$  is often complex.  It simplifies decisively if symmetries of the channel are exploited.		The capacity calculation of the (general)  [[Information_Theory/Application_to_Digital_Signal_Transmission#Information-theoretical_model_of_digital_signal_transmission\|$\text{discrete memoryless channel}$]]  $\rm (DMC)$  is often complex.  It simplifies decisively if symmetries of the channel are exploited.

	[[File:EN_Inf_T_3_3_S6a_vers2.png\|right\|frame\|Examples of symmetrical channels]]		[[File:EN_Inf_T_3_3_S6a_vers2.png\|right\|frame\|Examples of symmetrical channels]]

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← Older revision		Revision as of 14:27, 16 March 2026
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	<references/>		<references/>
	{{Display}}		{{Display}}
			[[de:Informationstheorie/Anwendung_auf_die_Digitalsignalübertragung]]

Guenter at 14:10, 28 February 2023

2023-02-28T14:10:47Z

← Older revision		Revision as of 16:10, 28 February 2023
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	:$$ \text{Pr(bit error)} \equiv \text{Pr(block error)} \hspace{0.05cm}.$$		:$$ \text{Pr(bit error)} \equiv \text{Pr(block error)} \hspace{0.05cm}.$$

	[[File:~~EN_Inf_T_3_3_S7c_vers2~~.png\|frame\|Definition of different error probabilities ~~'''KORREKTUR: faults'''~~]]		[[File:EN_Inf_T_3_3_S7c.png\|frame\|Definition of different error probabilities]]
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 7:}$  The upper graph shows the first eight received blocks  $\underline{y}_j^{(n)}$  with  $n = 4$  bits each.  Here channel errors are shaded green.		$\text{Example 7:}$  The upper graph shows the first eight received blocks  $\underline{y}_j^{(n)}$  with  $n = 4$  bits each.  Here channel errors are shaded green.

Guenter at 14:09, 28 February 2023

2023-02-28T14:09:36Z

← Older revision		Revision as of 16:09, 28 February 2023
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	In order to interpret the channel coding theorem correctly, some basics of  »'''channel coding'''«.  This extremely important area of Communications Engineering is covered in our learning tutorial  $\rm LNTwww$  in a separate book called  [[Channel_Coding\|"Channel Coding"]].		In order to interpret the channel coding theorem correctly, some basics of  »'''channel coding'''«.  This extremely important area of Communications Engineering is covered in our learning tutorial  $\rm LNTwww$  in a separate book called  [[Channel_Coding\|"Channel Coding"]].

	[[File:~~EN_Inf_T_3_3_S7a_vers2~~.png\|center\|frame\|Model for binary–coded communication ~~'''KORREKTUR: syndrome'''~~]]		[[File:EN_Inf_T_3_3_S7a.png\|center\|frame\|Model for binary–coded communication]]

	The following description refers to the highly simplified model for  [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes\|$\text{binary block codes}$]]:		The following description refers to the highly simplified model for  [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes\|$\text{binary block codes}$]]:

Guenter at 14:08, 28 February 2023

2023-02-28T14:08:28Z

← Older revision		Revision as of 16:08, 28 February 2023
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	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	[[File:EN_Inf_T_3_3_S3b_vers2.png\|right\|frame\|Information theoretic model of the binary channel under consideration ~~'''KORREKTUR: ""'''~~]]		[[File:EN_Inf_T_3_3_S3b_vers2.png\|right\|frame\|Information theoretic model of the binary channel under consideration]]
	*the  »'''equivocation'''«:		*the  »'''equivocation'''«:

Guenter at 14:07, 28 February 2023

2023-02-28T14:07:09Z

← Older revision		Revision as of 16:07, 28 February 2023
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	\hspace{0.05cm}.\end{align*}$$		\hspace{0.05cm}.\end{align*}$$

	[[File:~~P_ID2788__Inf_T_3_3_S4a~~.png\|right\|frame\|Mutual information for the <br>"asymmetrical binary channel" ~~'''KORREKTUR: Alle Angaben...'''~~]]		[[File:EN_Inf_T_3_3_S4a.png\|right\|frame\|Mutual information for the <br>"asymmetrical binary channel"]]
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 3}$:		$\text{Example 3}$:

Guenter at 15:45, 19 February 2023

2023-02-19T15:45:19Z

← Older revision		Revision as of 17:45, 19 February 2023
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	Since the probability of the block errors can never be smaller than that of the bit errors,  the block error probability "zero" is also not possible for  $R > C$ .  <br>From the given bounds for the bit errors,		Since the probability of the block errors can never be smaller than that of the bit errors,  the block error probability "zero" is also not possible for  $R > C$ .  <br>From the given bounds for the bit errors,

	:$$ {1}/{k} \cdot {\rm Pr}({\rm block error}) \le {\rm Pr}({\rm bit error}) \le {\rm Pr}({\rm block error})\hspace{0.05cm},$$		:$$ {1}/{k} \cdot {\rm Pr}({\rm block\ error}) \le {\rm Pr}({\rm bit\ error}) \le {\rm Pr}({\rm block\ error})\hspace{0.05cm},$$

	a range for the block error probability can also be given:		a range for the block error probability can also be given:

	:$$ {\rm Pr}({\rm bit error}) \le {\rm Pr}({\rm block error}) \le k \cdot {\rm Pr}({\rm bit error})\hspace{0.05cm}.$$		:$$ {\rm Pr}({\rm bit\ error}) \le {\rm Pr}({\rm block\ error}) \le k \cdot {\rm Pr}({\rm bit\ error})\hspace{0.05cm}.$$

	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=

Guenter at 15:43, 19 February 2023

2023-02-19T15:43:00Z

← Older revision		Revision as of 17:43, 19 February 2023
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	[[File:EN_Inf_T_3_3_S7c_vers2.png\|frame\|Definition of different error probabilities '''KORREKTUR: faults''']]		[[File:EN_Inf_T_3_3_S7c_vers2.png\|frame\|Definition of different error probabilities '''KORREKTUR: faults''']]
	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 7:}$  The upper graph shows the first eight ~~reception '''KORREKTUR:~~ received~~?'''~~ blocks  $\underline{y}_j^{(n)}$  with  $n = 4$  bits each.  Here channel errors are shaded green.		$\text{Example 7:}$  The upper graph shows the first eight received blocks  $\underline{y}_j^{(n)}$  with  $n = 4$  bits each.  Here channel errors are shaded green.

	Below, the initial sequence  $\underline{v}$  after decoding is sketched, divided into blocks  $\underline{v}_j^{(k)}$  with  $k = 3$  bits each. Note:		Below, the initial sequence  $\underline{v}$  after decoding is sketched, divided into blocks  $\underline{v}_j^{(k)}$  with  $k = 3$  bits each. Note:
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	For this, some  (due to the short sequence)  only very vague information about the error probabilities:		For this, some  (due to the short sequence)  only very vague information about the error probabilities:
	*Half of the ~~reception '''KORREKTUR:~~ received~~'''~~ bits are shaded green.  From this follows:		*Half of the received bits are shaded green.  From this follows:
	:$$\text{Pr(channel error)} = 16/32 = 1/2.$$		:$$\text{Pr(channel error)} = 16/32 = 1/2.$$

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	*In contrast, with uncoded transmission would be:		*In contrast, with uncoded transmission would be:
	:$$\text{Pr(bit error)} = {~~\rm~~ Pr(channel error)} = 1/2.$$		:$$\text{Pr(bit error)} = \text{Pr(channel error)} = 1/2.$$

	*Half of the decoded blocks are outlined in blue. From this follows:		*Half of the decoded blocks are outlined in blue. From this follows:

Hwang at 15:49, 16 February 2023

2023-02-16T15:49:09Z

← Older revision		Revision as of 17:49, 16 February 2023
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	Since the probability of the block errors can never be smaller than that of the bit errors,  the block error probability ~~„zero”~~ is also not possible for  $R > C$ .  <br>From the given bounds for the bit errors,		Since the probability of the block errors can never be smaller than that of the bit errors,  the block error probability "zero" is also not possible for  $R > C$ .  <br>From the given bounds for the bit errors,

	:$$ {1}/{k} \cdot {\rm Pr}({\rm block error}) \le {\rm Pr}({\rm bit error}) \le {\rm Pr}({\rm block error})\hspace{0.05cm},$$		:$$ {1}/{k} \cdot {\rm Pr}({\rm block error}) \le {\rm Pr}({\rm bit error}) \le {\rm Pr}({\rm block error})\hspace{0.05cm},$$

← Older revision		Revision as of 16:09, 28 February 2023
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	In order to interpret the channel coding theorem correctly, some basics of  »'''channel coding'''«.  This extremely important area of Communications Engineering is covered in our learning tutorial  $\rm LNTwww$  in a separate book called  [[Channel_Coding\|"Channel Coding"]].		In order to interpret the channel coding theorem correctly, some basics of  »'''channel coding'''«.  This extremely important area of Communications Engineering is covered in our learning tutorial  $\rm LNTwww$  in a separate book called  [[Channel_Coding\|"Channel Coding"]].

	[[File:~~EN_Inf_T_3_3_S7a_vers2~~.png\|center\|frame\|Model for binary–coded communication ~~'''KORREKTUR: syndrome'''~~]]		[[File:EN_Inf_T_3_3_S7a.png\|center\|frame\|Model for binary–coded communication]]

	The following description refers to the highly simplified model for  [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes\|$\text{binary block codes}$]]:		The following description refers to the highly simplified model for  [[Channel_Coding/Allgemeine_Beschreibung_linearer_Blockcodes\|$\text{binary block codes}$]]: