Difference between revisions of "Aufgaben:Exercise 4.6Z: Basics of Product Codes"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}} |
− | [[File:P_ID3002__KC_Z_4_6_v3.png|right|frame| | + | [[File:P_ID3002__KC_Z_4_6_v3.png|right|frame|Generator matrices of <br>the component codes]] |
− | + | We consider here a product code according to the description in section [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_product_code|"Basic structure of a Product Code"]]. The two component codes $\mathcal{C}_1$ and $\mathcal{C}_2$ are defined by the generator matrices $\mathbf{G}_1$ and $\mathbf{G}_2$ given on the right. | |
− | |||
− | |||
− | |||
− | === | + | <u>Hints:</u> |
+ | *This exercise belongs to the chapter [[Channel_Coding/The_Basics_of_Product_Codes|"Basics of a Product Code"]]. | ||
+ | |||
+ | *Reference is made to the section [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_Product_Code|"Basic structure of a product code"]]. | ||
+ | |||
+ | *The two component codes are also covered in the [[Aufgaben:Exercise_4.6:_Product_Code_Generation|$\text{Exercise 4.6}$]] . | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What statements does the generator matrix $\mathbf{G}_1$ allow about the code $\mathcal{C}_1$? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The code rate of $\mathcal{C}_1$ is $R_1 = 4/7$. |
− | + | + | + The code $\mathcal{C}_1$ is systematic. |
− | - $ | + | - $\mathcal{C}_1$ is a truncated Hamming code. |
− | + | + | + The minimum distance of this code is $d_1 = 3$. |
− | { | + | {What statements does the generator matrix $\mathbf{G}_2$ allow about the code $\mathcal{C}_2$? |
|type="[]"} | |type="[]"} | ||
− | - | + | - The code rate of $\mathcal{C}_2$ is $R_2 = 4/7$. |
− | + | + | + The code $\mathcal{C}_2$ is systematic. |
− | + $ | + | + $\mathcal{C}_2$ is a truncated Hamming code. |
− | + | + | + The minimum distance of this code is $d_2 = 3$. |
− | { | + | {Specify the parameters of the product code $\mathcal{C} = \mathcal{C}_1 × \mathcal{C}_2$ . |
|type="{}"} | |type="{}"} | ||
− | $k \ = \ ${ 12 3% } | + | $k \hspace{0.25cm} = \ ${ 12 3% } |
− | $n \ = \ ${ 42 3% } | + | $n \hspace{0.25cm} = \ ${ 42 3% } |
− | $d \ = \ ${ 9 3% } | + | $d \hspace{0.25cm} = \ ${ 9 3% } |
− | $R \ = \ ${ 0.286 3% } | + | $R \hspace{0.15cm} = \ ${ 0.286 3% } |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' Correct are the <u>statements 1, 2 and 4</u>: |
− | + | * The number of rows of the generator matrix $\mathbf{G}_1$ indicates the length of the information block ⇒ $k = 4$. | |
− | + | ||
− | '''( | + | * The code word length is equal to the number of columns ⇒ $n=4$ ⇒ Code rate $R = k/n = 4/7$. |
− | '''( | + | |
+ | * The code is systematic because the generator matrix $\mathbf{G}_1$ starts with a $4 × 4$ diagonal matrix. | ||
+ | |||
+ | *This is a "normal" Hamming code. | ||
+ | |||
+ | *For this, with the code word length $n$ and the number of check bits ⇒ $m = n - k$, the relation $n = 2^m - 1$ holds. | ||
+ | |||
+ | *In the present case, this is the (normal) Hamming code $\rm (7, \ 4, \ 3)$. | ||
+ | |||
+ | *The last parameter in this code label specifies the minimum distance ⇒ $d_{\rm min} = 3$. | ||
+ | |||
+ | |||
+ | '''(2)''' Correct are the <u>statements 2, 3 and 4</u>: | ||
+ | *This is a truncated Hamming code with parameter $n = 6, \ k = 3$ and $d_{\rm min} = 3$, also in systematic form. | ||
+ | |||
+ | *The code rate is $R = 1/2$. | ||
+ | |||
+ | |||
+ | '''(3)''' The basic structure of the product code is shown in the [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_product_code|"Basic structure of a Product Code"]] section. | ||
+ | * You can see the information block with $k = k_1 \cdot k_2 = 4 \cdot 3 \ \underline{= 12}$, | ||
+ | |||
+ | * The code word length is the total number of all bits: $n = n_1 \cdot n_2 = 7 \cdot 6 \ \underline{= 42}$. | ||
+ | |||
+ | *The code rate is thus given by $R = k/n = 12/42 = 2/7$. | ||
+ | |||
+ | *Or: $R = R_1 \cdot R_2 = 4/7 \cdot 1/2 \ \underline{= 2/7} \approx 0.289$. | ||
+ | |||
+ | * The free distance is $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Channel Coding: Exercises|^4.2 About the Product Codes^]] |
Latest revision as of 17:29, 6 December 2022
We consider here a product code according to the description in section "Basic structure of a Product Code". The two component codes $\mathcal{C}_1$ and $\mathcal{C}_2$ are defined by the generator matrices $\mathbf{G}_1$ and $\mathbf{G}_2$ given on the right.
Hints:
- This exercise belongs to the chapter "Basics of a Product Code".
- Reference is made to the section "Basic structure of a product code".
- The two component codes are also covered in the $\text{Exercise 4.6}$ .
Questions
Solution
(1) Correct are the statements 1, 2 and 4:
- The number of rows of the generator matrix $\mathbf{G}_1$ indicates the length of the information block ⇒ $k = 4$.
- The code word length is equal to the number of columns ⇒ $n=4$ ⇒ Code rate $R = k/n = 4/7$.
- The code is systematic because the generator matrix $\mathbf{G}_1$ starts with a $4 × 4$ diagonal matrix.
- This is a "normal" Hamming code.
- For this, with the code word length $n$ and the number of check bits ⇒ $m = n - k$, the relation $n = 2^m - 1$ holds.
- In the present case, this is the (normal) Hamming code $\rm (7, \ 4, \ 3)$.
- The last parameter in this code label specifies the minimum distance ⇒ $d_{\rm min} = 3$.
(2) Correct are the statements 2, 3 and 4:
- This is a truncated Hamming code with parameter $n = 6, \ k = 3$ and $d_{\rm min} = 3$, also in systematic form.
- The code rate is $R = 1/2$.
(3) The basic structure of the product code is shown in the "Basic structure of a Product Code" section.
- You can see the information block with $k = k_1 \cdot k_2 = 4 \cdot 3 \ \underline{= 12}$,
- The code word length is the total number of all bits: $n = n_1 \cdot n_2 = 7 \cdot 6 \ \underline{= 42}$.
- The code rate is thus given by $R = k/n = 12/42 = 2/7$.
- Or: $R = R_1 \cdot R_2 = 4/7 \cdot 1/2 \ \underline{= 2/7} \approx 0.289$.
- The free distance is $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$.