Difference between revisions of "Aufgaben:Exercise 4.6Z: Basics of Product Codes"

From LNTwww
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
{{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}
 
{{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}
  
[[File:P_ID3002__KC_Z_4_6_v3.png|right|frame|Generator matrices of the component codes]]
+
[[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&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_product_code|"Basic structure of a Product Code"]]. The two component codes&nbsp; $\mathcal{C}_1$&nbsp; and&nbsp; $\mathcal{C}_2$&nbsp; are defined by the generator matrices&nbsp; $\mathbf{G}_1$&nbsp; and&nbsp; $\mathbf{G}_2$&nbsp; given on the right.
+
We consider here a product code according to the description in section&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_product_code|"Basic structure of a Product Code"]].&nbsp; The two component codes&nbsp; $\mathcal{C}_1$&nbsp; and&nbsp; $\mathcal{C}_2$&nbsp; are defined by the generator matrices&nbsp; $\mathbf{G}_1$&nbsp; and&nbsp; $\mathbf{G}_2$&nbsp; given on the right.
  
  
  
  
 +
<u>Hints:</u>
 +
*This exercise belongs to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes|"Basics of a Product Code"]].
  
 +
*Reference is made  to the section&nbsp; [[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&nbsp; [[Aufgaben:Exercise_4.6:_Product_Code_Generation|$\text{Exercise 4.6}$]]&nbsp;.
 
 
Hints:
 
*This exercise belongs to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes|"Basics of a Product Code"]].
 
*Reference is made in particular to the section&nbsp; [[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&nbsp; [[Aufgaben:Aufgabe_4.6:_Produktcode–Generierung|"Exercise 4.6"]]&nbsp;.
 
  
  
Line 46: Line 44:
 
===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Correct are <u>statements 1, 2 and 4</u>:
+
'''(1)'''&nbsp; Correct are the&nbsp; <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 &nbsp; &#8658; &nbsp; $k = 4$.  
+
* The number of rows of the generator matrix&nbsp; $\mathbf{G}_1$&nbsp; indicates the length of the information block &nbsp; &#8658; &nbsp; $k = 4$.
 +
 
* The code word length is equal to the number of columns &nbsp; &#8658; &nbsp; $n=4$ &nbsp; &#8658; &nbsp; Code rate $R = k/n = 4/7$.
 
* The code word length is equal to the number of columns &nbsp; &#8658; &nbsp; $n=4$ &nbsp; &#8658; &nbsp; Code rate $R = k/n = 4/7$.
* The code is systematic because the generator matrix $\mathbf{G}_1$ starts with a $4 &times 4$ diagonal matrix.
+
 
*This is a "normal" Hamming code.  
+
* The code is systematic because the generator matrix&nbsp; $\mathbf{G}_1$&nbsp; starts with a&nbsp; $4 &times 4$&nbsp; diagonal matrix.
*For this, with the code word length $n$ and the number of check bits &nbsp; &#8658; &nbsp; $m = n - k$, the relation $n = 2^m - 1$ holds.
+
 
*In the present case, this is the (normal) Hamming code $\rm (7, \ 4, \ 3)$.  
+
*This is a&nbsp; "normal"&nbsp; Hamming code.
 +
 +
*For this,&nbsp; with the code word length&nbsp; $n$&nbsp; and the number of check bits &nbsp; &#8658; &nbsp; $m = n - k$,&nbsp; the relation&nbsp; $n = 2^m - 1$&nbsp; holds.
 +
 
 +
*In the present case,&nbsp; this is the&nbsp; (normal)&nbsp; Hamming code $\rm (7, \ 4, \ 3)$.
 +
 
*The last parameter in this code label specifies the minimum distance &nbsp; &#8658; &nbsp; $d_{\rm min} = 3$.
 
*The last parameter in this code label specifies the minimum distance &nbsp; &#8658; &nbsp; $d_{\rm min} = 3$.
  
  
'''(2)'''&nbsp; Correct <u>statements 2, 3 and 4</u>:  
+
'''(2)'''&nbsp; Correct are the&nbsp; <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.
+
*This is a truncated Hamming code with parameter&nbsp; $n = 6, \ k = 3$&nbsp; and&nbsp; $d_{\rm min} = 3$,&nbsp; also in systematic form.  
*The code rate is $R = 1/2$.
 
  
 +
*The code rate is&nbsp; $R = 1/2$.
  
'''(3)'''&nbsp; 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 $k = k_1 \cdot k_2 = 4 \cdot 3 \ \underline{= 12}$,  
+
'''(3)'''&nbsp; The basic structure of the product code is shown in the&nbsp; [[Channel_Coding/The_Basics_of_Product_Codes#Basic_structure_of_a_product_code|"Basic structure of a Product Code"]]&nbsp; section.  
 +
* You can see the information block with&nbsp; $k = k_1 \cdot k_2 = 4 \cdot 3 \ \underline{= 12}$,
 +
 
* The code word length is the total number of all bits: $n = n_1 \cdot n_2 = 7 \cdot 6 \ \underline{= 42}$.
 
* The code 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 $R = k/n = 12/42 = 2/7$.  
+
 
 +
*The code rate is thus given by&nbsp; $R = k/n = 12/42 = 2/7$.
 +
 
*Or: &nbsp; $R = R_1 \cdot R_2 = 4/7 \cdot 1/2 \ \underline{= 2/7} \approx 0.289$.
 
*Or: &nbsp; $R = R_1 \cdot R_2 = 4/7 \cdot 1/2 \ \underline{= 2/7} \approx 0.289$.
* The free distance is $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$.
+
 
 +
* The free distance is&nbsp; $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Latest revision as of 17:29, 6 December 2022

Generator matrices of
the component codes

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:



Questions

1

What statements does the generator matrix  $\mathbf{G}_1$  allow about the code  $\mathcal{C}_1$?

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  $d_1 = 3$.

2

What statements does the generator matrix  $\mathbf{G}_2$  allow about the code  $\mathcal{C}_2$?

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  $d_2 = 3$.

3

Specify the parameters of the product code  $\mathcal{C} = \mathcal{C}_1 × \mathcal{C}_2$ .

$k \hspace{0.25cm} = \ $

$n \hspace{0.25cm} = \ $

$d \hspace{0.25cm} = \ $

$R \hspace{0.15cm} = \ $


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  $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$.