Difference between revisions of "Aufgaben:Exercise 1.09: Extended Hamming Code"
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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}_{2}\text{:}\hspace{0.4cm}R \ = \ $ { 0.5 3% } |
− | { | + | {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}_{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="{}"} | |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="()"} |
- $x_{8} = 0.$ | - $x_{8} = 0.$ | ||
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+ $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$ | + $x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$ | ||
− | { | + | {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="()"} |
- $1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$ | - $1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$ | ||
+ $1 1 1 1 1 1 1 1 → 1 1 1 0 0 0 0 1,$ | + $1 1 1 1 1 1 1 1 → 1 1 1 0 0 0 0 1,$ | ||
- $1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$ | - $1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$ | ||
− | { | + | {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 ${ \boldsymbol{\rm H}}_{2}$ . |
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' The corresponding equation for the code rate in both cases is $R = k/n\text{:}$ |
− | *$ | + | *$\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$ |
− | *$ | + | |
+ | *$\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(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. |
− | '''(5)''' | + | *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: | ||
:$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | ||
− | + | *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 | |
:$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$ | :$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$ | ||
− | + | :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)''' | + | '''(7)''' After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes: |
:$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$ | ||
− | + | *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: | |
+ | * ${ \boldsymbol{\rm G}}_{2}$ 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 |
^]] | ^]] |
Latest revision as of 17:01, 23 January 2023
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.