Difference between revisions of "Aufgaben:Exercise 1.2: A Simple Binary Channel Code"

Latest revision as of 16:09, 17 August 2022

Clarification: Channel coding and decoding

The graph illustrates the channel coding $\mathcal{C}$ considered here:

There are four possible information blocks $\underline{u} = (u_{1},\ u_{2},\ \text{...}\hspace{0.05cm} ,\ u_{k})$.

Each information block $\underline{u}$ is uniquely assigned (recognizable by the same color) to the code word $\underline{x}= (x_{1},\ x_{2},\ \text{...}\hspace{0.05cm} ,\ x_{n})$ .

Due to decoding errors $(0 → 1, \ 1 → 0)$ there are more than 4, namely 16 different receive words $\underline{y} = (y_{1},\ y_{2},\ \text{...} \hspace{0.05cm} ,\ y_{n})$.

From subtask (4) we consider the following mapping:

$$\underline{u_0} = (0,\ 0) \leftrightarrow (0,\ 0,\ 0,\ 0) = \underline{x_0}\hspace{0.05cm},$$

$$\underline{u_1} = (0,\ 1) \leftrightarrow (0,\ 1,\ 0,\ 1) = \underline{x_1}\hspace{0.05cm},$$

$$\underline{u_2} = (1,\ 0) \leftrightarrow (1,\ 0,\ 1,\ 0) = \underline{x_2}\hspace{0.05cm},$$

$$\underline{u_3} = (1,\ 1) \leftrightarrow (1,\ 1,\ 1,\ 1) = \underline{x_3}\hspace{0.05cm}.$$

Hints:

This exercise belongs to the chapter "Objective of Channel Coding"

Reference is made in particular to the pages

Questions

$k \ = \ $

$n \ = \ $

$R \ = \ $

	Yes,
	No.

$ w_{\rm H} \ (\underline{x}_0) \ = \ $

$ w_{\rm H} \ (\underline{x}_1) \ = \ $

$ w_{\rm H} \ (\underline{x}_2) \ = \ $

$ w_{\rm H} \ (\underline{x}_3) \ = \ $

$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_1) \ = \ $

$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_3) \ = \ $

$ d_{\rm H} \ (\underline{x}_1,\ \underline{x}_2) \ = \ $

$ d_{\rm min} (\mathcal{C}) \ = \ $

Solution

(1) The code size here is given to $|\mathcal{C}| = 4$.

In general, $|\mathcal{C}|= 2^k$.
From this follows $\underline{ k = 2}$.

(2) Each code word $\underline{x}$ is uniquely associated with an information block $\underline{u}$.

By corruptions of a single of the total $n$ bits of a code word $\underline{x}$ the receive words $\underline{y}$ result.
From the number $(16 = 2^4$) of possible receive words follows $\underline{ n = 4}$.

(3) The code rate is by definition $R = k/n$. Using the above results, we obtain $\underline{R = 0.5}$.

(4) Correct is Yes: A systematic code is characterized by the fact that in each case the first $k$ bits of the code words are identical to the information block.

(5) The Hamming weight of a binary code is equal to the algebraic sum $x_1 + x_2 + \text{...} + x_n$ over all code word elements. Thus:

$$w_{\rm H}(\underline{x}_0) \hspace{0.15cm} \underline {= 0} \hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_1) \hspace{0.15cm} \underline {= 2} \hspace{0.05cm}, \hspace{0.3cm} w_{\rm H}(\underline{x}_2) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm}.$$

(6) The Hamming distance between two code words here can only take the values $2$ and $4$:

$$d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_1) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_2) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$

$$ d_{\rm H}(\underline{x}_1, \hspace{0.05cm}\underline{x}_2) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_1, \hspace{0.05cm}\underline{x}_3) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_2, \hspace{0.05cm}\underline{x}_3) = 2\hspace{0.05cm}.$$

(7) From the result of subtask (6) it follows $d_{\rm min}(\mathcal{C}) \hspace{0.15cm}\underline{= 2}$.

Generally, for this quantity:

$$d_{\rm min}(\mathcal{C}) = \min_{\substack{\underline{x},\hspace{0.05cm}\underline{x}' \hspace{0.05cm}\in \hspace{0.05cm} \mathcal{C} \ {\underline{x}} \hspace{0.05cm}\ne \hspace{0.05cm} \underline{x}'}}\hspace{0.1cm}d_{\rm H}(\underline{x}, \hspace{0.05cm}\underline{x}')\hspace{0.05cm}.$$

@@ Line 2: / Line 2: @@
 }}
-[[File:EN_KC_A_1_2.png|right|frame|Clarification of channel (de)coding]]
+[[File:EN_KC_A_1_2_bitte.png|right|frame|Clarification:&nbsp; Channel coding and decoding ]]
-The graph illustrates the channel coding considered here&nbsp; $\mathcal{C}$:
+The graph illustrates the channel coding&nbsp; $\mathcal{C}$&nbsp; considered here:
-*There are four possible information blocks&nbsp; $\underline{u} = (u_{1}, u_{2}, \text{...}\hspace{0.05cm} , u_{k})$.
+*There are four possible information blocks&nbsp; $\underline{u} = (u_{1},\ u_{2},\ \text{...}\hspace{0.05cm} ,\ u_{k})$.
-*Each information block&nbsp; $\underline{u}$&nbsp; is uniquely assigned (recognizable by the same color) to the codeword&nbsp; $\underline{x}= (x_{1}, x_{2}, \text{...}\hspace{0.05cm} , x_{n})$&nbsp;.
-*Due to decoding errors&nbsp; $(0 → 1, \ 1 → 0)$&nbsp; there are more than 4, namely 16 different receive words&nbsp; $\underline{y} = (y_{1}, y_{2}, \text{...} \hspace{0.05cm} , y_{n})$.
+*Each information block&nbsp; $\underline{u}$&nbsp; is uniquely assigned&nbsp; (recognizable by the same color)&nbsp; to the code word&nbsp; $\underline{x}= (x_{1},\ x_{2},\ \text{...}\hspace{0.05cm} ,\ x_{n})$&nbsp;.
-From subtask '''(4)''' we consider the following mapping:
+*Due to decoding errors&nbsp; $(0 → 1, \ 1 → 0)$&nbsp; there are more than 4,&nbsp; namely 16 different receive words&nbsp; $\underline{y} = (y_{1},\ y_{2},\ \text{...} \hspace{0.05cm} ,\ y_{n})$.
-:$$\underline{u_0} = (0, 0) \leftrightarrow (0, 0, 0, 0) = \underline{x_0}\hspace{0.05cm},$$
-:$$\underline{u_1} = (0, 1) \leftrightarrow (0, 1, 0, 1) = \underline{x_1}\hspace{0.05cm},$$
-:$$\underline{u_2} = (1, 0) \leftrightarrow (1, 0, 1, 0) = \underline{x_2}\hspace{0.05cm},$$
-:$$\underline{u_3} = (1, 1) \leftrightarrow (1, 1, 1, 1) = \underline{x_3}\hspace{0.05cm}.$$
+From subtask&nbsp; '''(4)'''&nbsp; we consider the following mapping:
+:$$\underline{u_0} = (0,\ 0) \leftrightarrow (0,\ 0,\ 0,\ 0) = \underline{x_0}\hspace{0.05cm},$$
+:$$\underline{u_1} = (0,\ 1) \leftrightarrow (0,\ 1,\ 0,\ 1) = \underline{x_1}\hspace{0.05cm},$$
+:$$\underline{u_2} = (1,\ 0) \leftrightarrow (1,\ 0,\ 1,\ 0) = \underline{x_2}\hspace{0.05cm},$$
+:$$\underline{u_3} = (1,\ 1) \leftrightarrow (1,\ 1,\ 1,\ 1) = \underline{x_3}\hspace{0.05cm}.$$
+Hints:
+*This exercise belongs to the chapter&nbsp; [[Channel_Coding/Objective_of_Channel_Coding|"Objective of Channel Coding"]]
-Hints:
-*This exercise belongs to the chapter&nbsp; [[Channel_Coding/Objective_of_Channel_Coding|Obective of Channel Coding]]
 *Reference is made in particular to the pages&nbsp;
-::[[Channel_Coding/Objective_of_Channel_Coding#Block_diagram_and_requirements|Block diagram and requirements]],
+:*[[Channel_Coding/Objective_of_Channel_Coding#Block_diagram_and_requirements|"Block diagram and requirements"]],
-::[[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|Important definitions for block coding]].
+:*[[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Important definitions for block coding"]].
@@ Line 37: / Line 37: @@
-{What is the codeword length $n$?
+{What is the code word length&nbsp; $n$?
 |type="{}"}
 $n \ = \ $ { 4 }
@@ Line 59: / Line 59: @@
 {Indicate the Hamming distances between the following code words.
 |type="{}"}
-$ d_{\rm H} \ (\underline{x}_0, \underline{x}_1) \ = \ $ { 2 }
+$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_1) \ = \ $ { 2 }
-$ d_{\rm H} \ (\underline{x}_0, \underline{x}_3) \ = \ $ { 4 }
+$ d_{\rm H} \ (\underline{x}_0,\ \underline{x}_3) \ = \ $ { 4 }
-$ d_{\rm H} \ (\underline{x}_1, \underline{x}_2) \ = \ $ { 4 }
+$ d_{\rm H} \ (\underline{x}_1,\ \underline{x}_2) \ = \ $ { 4 }
-{What is the minimum Hamming distance of the code under consideration&nbsp; $\mathcal{C}$?
+{What is the minimum Hamming distance of the code&nbsp; $\mathcal{C}$&nbsp; under consideration?
 |type="{}"}
 $ d_{\rm min} (\mathcal{C}) \ = \ $ { 2 }
@@ Line 72: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The code size here is given to $|\mathcal{C}| = 4$.
+'''(1)'''&nbsp; The code size here is given to&nbsp; $|\mathcal{C}| = 4$.
-*In general, $|\mathcal{C}|= 2^k$.
+*In general,&nbsp; $|\mathcal{C}|= 2^k$.
-*From this follows $\underline{ k = 2}$.
+*From this follows&nbsp; $\underline{ k = 2}$.
-'''(2)'''&nbsp; Each codeword $\underline{x}$ is uniquely associated with an information block $\underline{u}$.
-*By corruptions of single of the total $n$ bits of a codeword $\underline{x}$ the receive words $\underline{y}$ result.
-*From the number $(16 = 2^4$) of possible receive words follows $\underline{ n = 4}$.
+'''(2)'''&nbsp; Each code word&nbsp; $\underline{x}$&nbsp; is uniquely associated with an information block&nbsp; $\underline{u}$.
+*By corruptions of a single of the total&nbsp; $n$&nbsp; bits of a code word&nbsp; $\underline{x}$&nbsp; the receive words&nbsp; $\underline{y}$&nbsp; result.
+*From the number&nbsp; $(16 = 2^4$)&nbsp; of possible receive words follows&nbsp; $\underline{ n = 4}$.
-'''(3)'''&nbsp; The code rate is by definition $R = k/n$.
-*Using the above results, we obtain $\underline{R = 0.5}$.
+'''(3)'''&nbsp; The code rate is by definition&nbsp; $R = k/n$.&nbsp; Using the above results,&nbsp; we obtain $\underline{R = 0.5}$.
-'''(4)'''&nbsp; Correct <u>Yes</u>:
+'''(4)'''&nbsp; Correct is&nbsp; <u>Yes</u>:&nbsp; A systematic code is characterized by the fact that in each case the first&nbsp; $k$&nbsp; bits of the code words are identical to the information block.
-*A systematic code is characterized by the fact that in each case the first $k$ bits of the code words are identical to the information block.
-'''(5)'''&nbsp; The Hamming weight of a binary code is equal to the algebraic sum $x_1 + x_2 + \text{...} + x_n$ over all code word elements. Thus:
+'''(5)'''&nbsp; The Hamming weight of a binary code is equal to the algebraic sum&nbsp; $x_1 + x_2 + \text{...} + x_n$&nbsp; over all code word elements.&nbsp; Thus:
 :$$w_{\rm H}(\underline{x}_0) \hspace{0.15cm} \underline {= 0} \hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_1) \hspace{0.15cm} \underline {= 2} \hspace{0.05cm}, \hspace{0.3cm} w_{\rm H}(\underline{x}_2) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm}w_{\rm H}(\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm}.$$
-'''(6)'''&nbsp; The Hamming distance between two codewords here can only take the values $2$ and $4$:
+'''(6)'''&nbsp; The Hamming distance between two code words here can only take the values&nbsp; $2$&nbsp; and&nbsp; $4$:
 :$$d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_1) \hspace{0.15cm} \underline {= 2}\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_2) = 2\hspace{0.05cm}, \hspace{0.3cm} d_{\rm H}(\underline{x}_0, \hspace{0.05cm}\underline{x}_3) \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$
@@ Line 104: / Line 101: @@
-'''(7)'''&nbsp; From the result of subtask (6) it follows $d_{\rm min}(\mathcal{C}) \hspace{0.15cm}\underline{= 2}$.
+'''(7)'''&nbsp; From the result of subtask&nbsp; '''(6)'''&nbsp; it follows&nbsp; $d_{\rm min}(\mathcal{C}) \hspace{0.15cm}\underline{= 2}$.
-*Generally, for this quantity:
+*Generally,&nbsp; for this quantity:
 :$$d_{\rm min}(\mathcal{C}) = \min_{\substack{\underline{x},\hspace{0.05cm}\underline{x}' \hspace{0.05cm}\in \hspace{0.05cm} \mathcal{C} \ {\underline{x}} \hspace{0.05cm}\ne \hspace{0.05cm} \underline{x}'}}\hspace{0.1cm}d_{\rm H}(\underline{x}, \hspace{0.05cm}\underline{x}')\hspace{0.05cm}.$$