Aufgaben:Exercise 1.09: Extended Hamming Code: Difference between revisions
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[[File:|right|]] | [[File:EN_KC_A_1_9_neu.png|right|frame|$\text{HC (7, 4)}$ (yellow background) and $\text{(8, 4)}$ extension (green background).]] | ||
Two codes are to be compared, whose code tables are given on the right. | |||
*The first four bits of each code word $\underline{x}$ are equal to the respective information word $\underline{u}$ (black font). | |||
*Then follow $m = n- k$ parity bit (red font). | |||
=== | |||
The systematic $\text{(7, 4)}$ Hamming code has already been discussed in [[Aufgaben:Exercise_1.6:_(7,_4)_Hamming_Code|"Exercise 1.6"]] and [[Aufgaben:Exercise_1.6:_(7,_4)_Hamming_Code|"Exercise 1.7"]]. The parity-check matrix and generator matrix of this code are given as follows: | |||
:$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$ | |||
:$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$ | |||
In the further course of the exercise this (yellow highlighted) code is called $\mathcal{C}_{1}$. | |||
The right column in the above table specifies a block code with parameters $n = 8$ and $k = 4$, usually referred to in the literature as the "Extended Hamming Code". We refer to this code (highlighted in green) in the following $\mathcal{C}_{2}$ and denote its parity-check matrix by ${ \boldsymbol{\rm H}}_{2}$ and the corresponding generator matrix by ${ \boldsymbol{\rm G}}_{2}$. | |||
The questions for this exercise are related to | |||
*the [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"code rate"]], | |||
*the [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"minimum distance"]] between two code words, | |||
*the [[Channel_Coding/General_Description_of_Linear_Block_Codes#Code_definition_by_the_parity-check_matrix|"parity-check matrix"]] and the [[Channel_Coding/General_Description_of_Linear_Block_Codes#Code_definition_by_the_generator_matrix|"generator matrix"]] of the extended $\text{(8, 4)}$ Hamming code. | |||
Hints: | |||
*This exercise belongs to the chapter [[Channel_Coding/General_Description_of_Linear_Block_Codes|"General Description of Linear Block Codes"]]. | |||
*Note in the solution that $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are each [[Channel_Coding/General_Description_of_Linear_Block_Codes#Systematic_Codes|"systematic codes"]]. | |||
*The following [[Aufgaben:Exercise_1.09Z:_Extension_and/or_Puncturing|"Exercise 1.9Z"]] deals with the extension of codes in somewhat more general terms. | |||
===Questions=== | |||
<quiz display=simple> | <quiz display=simple> | ||
{ | |||
{Specify the code rates of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. | |||
|type="{}"} | |||
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $ { 0.571 3% } | |||
$\mathcal{C}_{2}\text{:}\hspace{0.4cm}R \ = \ $ { 0.5 3% } | |||
{Give the minimum distances of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. | |||
|type="{}"} | |||
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 3 } | |||
$\mathcal{C}_{2}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 4 } | |||
{What is the format of the parity-check matrix $\boldsymbol{\rm H}_{2}$ of $\mathcal{C}_{2}$? | |||
|type="{}"} | |||
$\text{Number of columns} \ = \ $ { 8 } | |||
$\text{Number of rows} \ = \ $ { 4 } | |||
{Derive the equation for the code bit $x_ {8} (= p_{4})$ from the code table. Which specification is correct? | |||
|type="()"} | |||
- $x_{8} = 0.$ | |||
- $x_{8} = x_{1}⊕x_{2}⊕x_{4}⊕x_{5}.$ | |||
+ $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$ | |||
{Which statements are true for ${ \boldsymbol{\rm H}}_{2}$? Hint: Correct are three out of four answers. | |||
|type="[]"} | |type="[]"} | ||
- | |||
+ | + Row 1 reads: $1 1 0 1 1 0 0 0$. | ||
+ Row 2 reads: $0 1 1 1 0 1 0 0$. | |||
- Row 3 reads: $0 0 0 0 1 1 1 1$. | |||
+ Row 4 reads: $1 1 1 1 1 1 1 1$. | |||
{Which transformation is allowed for the last row of ${ \boldsymbol{\rm H}}_{2}$ ? | |||
|type="()"} | |||
- $1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$ | |||
+ $1 1 1 1 1 1 1 1 → 1 1 1 0 0 0 0 1,$ | |||
- $1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$ | |||
{Give the corresponding generator matrix ${ \boldsymbol{\rm G}}_{2}$. Which statements are true? | |||
|type="[]"} | |||
- ${ \boldsymbol{\rm G}}_{2}$ has same format as matrix ${ \boldsymbol{\rm G}}_{1}$ of the $\text{(7, 4)}$ code. | |||
+ ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$. | |||
+ ${ \boldsymbol{\rm G}}_{2}$ in the considered example has the same format as ${ \boldsymbol{\rm H}}_{2}$ . | |||
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</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''1 | '''(1)''' The corresponding equation for the code rate in both cases is $R = k/n\text{:}$ | ||
'''2 | *$\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$ | ||
'''3 | |||
'''4 | *$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$ | ||
'''5 | |||
'''6 | |||
'''7 | |||
'''(2)''' The minimum distance of the $(7, 4, 3)$ Hamming code $\mathcal{C}_{1}$ is $d_{\rm min} \underline{= 3}$, which can be read from the naming alone. | |||
*From the table in the information section, it can be seen that for the extended Hamming code $d_{\rm min} \underline{= 4}$ holds. | |||
* $\mathcal{C}_{2}$ is therefore also called in the literature a $\rm (8, 4, 4)$ block code. | |||
'''(3)''' The parity-check matrix ${ \boldsymbol{\rm H}}$ generally consists of $n$ columns and $m = n - k$ rows, where $m$ indicates the number of parity-check equations. | |||
*For the $(7, 4, 3)$ Hamming code, ${ \boldsymbol{\rm H}}$ is a $3 × 7$ matrix. | |||
*For the extended Hamming code ⇒ $\mathcal{C}_{2}$, on the other hand: $\underline{n = 8}$ (column number) and $\underline{m = 4}$ (row number). | |||
'''(4)''' From the code table in the information section you can see that only <u>answer 3</u> is correct. | |||
*The parity bit $p_{4}$ is to be determined in such a way that the modulo 2 sum over all bits of the code word results in the value $0$. | |||
'''(5)''' It should first be noted that the specification of the parity-check matrix is never unambiguous, if only because the order of the parity-check equations is interchangeable. | |||
*However, considering that only one of the given rows is wrong, ${ \boldsymbol{\rm H}}_{2}$ is uniquely determined: | |||
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | |||
*Correct are therefore the <u>statements 1, 2 and 4</u>. The rows of this parity-check matrix represent the four parity-check equations in this order: | |||
:$$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$ | |||
:$$x_2 \oplus x_3 \oplus x_4 \oplus x_6 = 0 \hspace{0.05cm},$$ | |||
:$$ x_1 \oplus x_3 \oplus x_4 \oplus x_7 = 0 \hspace{0.05cm},$$ | |||
:$$ x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0 \hspace{0.05cm}.$$ | |||
'''(6)''' Correct is <u>answer 2</u>: | |||
*This result is obtained by replacing the last row with the modulo 2 sum over all four rows, which is allowed. | |||
*Proposition 1 does not represent a parity-check equation. | |||
*Proposal 3 represents the parity-check equation $x_{3}⊕x_{5} = 0$, which also does not correspond to the facts. | |||
According to the correct solution suggestion 2, on the other hand, the parity-check equation becomes | |||
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$ | |||
:is replaced by the following new parity-check equation: | |||
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$ | |||
*The modified parity-check matrix is now: | |||
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$ | |||
'''(7)''' After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes: | |||
:$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$ | |||
*Thus, the generator matrix is: | |||
:$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$ | |||
So the <u>statements 2 and 3</u> are correct: | |||
* ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ $($see specification sheet$)$ with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$, but unlike ${ \boldsymbol{\rm G}}_{1}$ now has 8 columns. | |||
*In the present case $n = 8, k = 4 \ ⇒ \ m = 4$ both ${ \boldsymbol{\rm G}}_{2}$ and ${ \boldsymbol{\rm H}}_{2}$ are $4×8$ matrices respectively. | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
[[Category: | [[Category:Channel Coding: Exercises|^1.4 Linear Block Code Description | ||
^]] | ^]] | ||
[[de:Aufgaben:Aufgabe 1.09: Erweiterter Hamming–Code]] | |||
Latest revision as of 17:58, 16 March 2026

Two codes are to be compared, whose code tables are given on the right.
- The first four bits of each code word $\underline{x}$ are equal to the respective information word $\underline{u}$ (black font).
- Then follow $m = n- k$ parity bit (red font).
The systematic $\text{(7, 4)}$ Hamming code has already been discussed in "Exercise 1.6" and "Exercise 1.7". The parity-check matrix and generator matrix of this code are given as follows:
- $${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$
- $${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$
In the further course of the exercise this (yellow highlighted) code is called $\mathcal{C}_{1}$.
The right column in the above table specifies a block code with parameters $n = 8$ and $k = 4$, usually referred to in the literature as the "Extended Hamming Code". We refer to this code (highlighted in green) in the following $\mathcal{C}_{2}$ and denote its parity-check matrix by ${ \boldsymbol{\rm H}}_{2}$ and the corresponding generator matrix by ${ \boldsymbol{\rm G}}_{2}$.
The questions for this exercise are related to
- the "code rate",
- the "minimum distance" between two code words,
- the "parity-check matrix" and the "generator matrix" of the extended $\text{(8, 4)}$ Hamming code.
Hints:
- This exercise belongs to the chapter "General Description of Linear Block Codes".
- Note in the solution that $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ are each "systematic codes".
- The following "Exercise 1.9Z" deals with the extension of codes in somewhat more general terms.
Questions
Solution
- $\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$
- $\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$
(2) The minimum distance of the $(7, 4, 3)$ Hamming code $\mathcal{C}_{1}$ is $d_{\rm min} \underline{= 3}$, which can be read from the naming alone.
- From the table in the information section, it can be seen that for the extended Hamming code $d_{\rm min} \underline{= 4}$ holds.
- $\mathcal{C}_{2}$ is therefore also called in the literature a $\rm (8, 4, 4)$ block code.
(3) The parity-check matrix ${ \boldsymbol{\rm H}}$ generally consists of $n$ columns and $m = n - k$ rows, where $m$ indicates the number of parity-check equations.
- For the $(7, 4, 3)$ Hamming code, ${ \boldsymbol{\rm H}}$ is a $3 × 7$ matrix.
- For the extended Hamming code ⇒ $\mathcal{C}_{2}$, on the other hand: $\underline{n = 8}$ (column number) and $\underline{m = 4}$ (row number).
(4) From the code table in the information section you can see that only answer 3 is correct.
- The parity bit $p_{4}$ is to be determined in such a way that the modulo 2 sum over all bits of the code word results in the value $0$.
(5) It should first be noted that the specification of the parity-check matrix is never unambiguous, if only because the order of the parity-check equations is interchangeable.
- However, considering that only one of the given rows is wrong, ${ \boldsymbol{\rm H}}_{2}$ is uniquely determined:
- $${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
- Correct are therefore the statements 1, 2 and 4. The rows of this parity-check matrix represent the four parity-check equations in this order:
- $$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$
- $$x_2 \oplus x_3 \oplus x_4 \oplus x_6 = 0 \hspace{0.05cm},$$
- $$ x_1 \oplus x_3 \oplus x_4 \oplus x_7 = 0 \hspace{0.05cm},$$
- $$ x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0 \hspace{0.05cm}.$$
(6) Correct is answer 2:
- This result is obtained by replacing the last row with the modulo 2 sum over all four rows, which is allowed.
- Proposition 1 does not represent a parity-check equation.
- Proposal 3 represents the parity-check equation $x_{3}⊕x_{5} = 0$, which also does not correspond to the facts.
According to the correct solution suggestion 2, on the other hand, the parity-check equation becomes
- $$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$
- is replaced by the following new parity-check equation:
- $$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$
- The modified parity-check matrix is now:
- $${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
(7) After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes:
- $${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$
- Thus, the generator matrix is:
- $${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
So the statements 2 and 3 are correct:
- ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ $($see specification sheet$)$ with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$, but unlike ${ \boldsymbol{\rm G}}_{1}$ now has 8 columns.
- In the present case $n = 8, k = 4 \ ⇒ \ m = 4$ both ${ \boldsymbol{\rm G}}_{2}$ and ${ \boldsymbol{\rm H}}_{2}$ are $4×8$ matrices respectively.