Aufgaben:Exercise 1.09: Extended Hamming Code: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Channel_Coding/General_Description_of_Linear_Block_Codes | ||
}} | }} | ||
[[File: | [[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: | |||
:$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$ | :$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$ | ||
:$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$ | ||
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="{}"} | |type="{}"} | ||
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $ { 0.571 3% } | $\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $ { 0.571 3% } | ||
| Line 50: | Line 49: | ||
{ | {Give the minimum distances of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. | ||
|type="{}"} | |type="{}"} | ||
$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 3 } | $\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $ { 3 } | ||
| Line 56: | Line 55: | ||
{ | {What is the format of the parity-check matrix $\boldsymbol{\rm H}_{2}$ of $\mathcal{C}_{2}$? | ||
|type="{}"} | |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="()"} | |type="()"} | ||
| Line 68: | Line 67: | ||
+ $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$ | + $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$ | ||
{ | {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="()"} | |type="()"} | ||
- $1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$ | - $1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$ | ||
| Line 82: | Line 81: | ||
- $1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$ | - $1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$ | ||
{ | {Give the corresponding generator matrix ${ \boldsymbol{\rm G}}_{2}$. Which statements are true? | ||
|type="[]"} | |type="[]"} | ||
- ${ \boldsymbol{\rm G}}_{2}$ | - ${ \boldsymbol{\rm G}}_{2}$ has same format as matrix ${ \boldsymbol{\rm G}}_{1}$ of the $\text{(7, 4)}$ code. | ||
+ ${ \boldsymbol{\rm G}}_{2}$ | + ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$. | ||
+ ${ \boldsymbol{\rm G}}_{2}$ | + ${ \boldsymbol{\rm G}}_{2}$ in the considered example has the same format as ${ \boldsymbol{\rm H}}_{2}$ . | ||
| Line 95: | Line 94: | ||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' The corresponding equation for the code rate in both cases is $R = k/n\text{:}$ | ||
*$\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$ | *$\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$ | ||
*$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$ | *$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$ | ||
'''(2)''' | '''(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)''' | '''(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)''' | '''(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: | ||
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | ||
* | *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: | ||
:$$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$ | :$$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$ | ||
| Line 134: | Line 137: | ||
'''(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 | |||
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$ | :$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$ | ||
:is replaced by the following new parity-check equation: | |||
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$ | :$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$ | ||
*The modified parity-check matrix is now: | |||
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$ | ||
'''(7)''' After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes: | |||
'''(7)''' | |||
:$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$ | ||
*Thus, the generator matrix is: | |||
:$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$ | ||
So the <u>statements 2 and 3</u> are correct: | |||
* ${ \boldsymbol{\rm G}}_{2}$ | * ${ \boldsymbol{\rm G}}_{2}$ 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.