Difference between revisions of "Aufgaben:Exercise 4.11: Analysis of Parity-check Matrices"

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<u>Hints:</u>
 
<u>Hints:</u>
* This exercise belongs to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes|"Basics of the Low&ndash;density Parity&ndash;check"]].
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* This exercise belongs to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes|"Basics of the Low&ndash;density Parity&ndash;check codes"]].
  
 
* Reference is made in particular to the page&nbsp; [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes#Some_characteristics_of_LDPC_codes|"Some characteristics of LDPC codes"]].
 
* Reference is made in particular to the page&nbsp; [[Channel_Coding/The_Basics_of_Low-Density_Parity_Check_Codes#Some_characteristics_of_LDPC_codes|"Some characteristics of LDPC codes"]].

Revision as of 18:41, 16 December 2022

Product code described by the parity-check matrix  $\mathbf{H}$

In the adjacent graphic,  a product code is indicated at the top,  which is characterized by the following parity-check equations:

$$p_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_1 \oplus u_2\hspace{0.05cm},$$
$$p_2 = u_3 \oplus u_4\hspace{0.05cm},$$
$$p_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_1 \oplus u_3\hspace{0.05cm},$$
$$p_4 = u_2 \oplus u_4\hspace{0.05cm}.$$

Below are given the check matrices  $\mathbf{H}_1, \ \mathbf{H}_2$  and  $\mathbf{H}_3$.  To be checked:  Which of the matrices correctly describe the given product code according to the equation  $\underline{x} = \underline{u} \cdot \mathbf{H}^{\rm T}$  assuming the following definitions:

  • the code word  $\underline{x} = (u_1, \, u_2, \, u_3, \, u_4, \, p_1, \, p_2, \, p_3, \, p_4)$,
  • the code word  $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$.


All  $\mathbf{H}$  matrices contain fewer  "ones"  than  "zeros".  This is a characteristic of the so-called  "low–density parity–check codes"  $($short:  $\rm LDPC$ codes$)$.  In the case of LDPC codes relevant to practice,  however,  the share of  "ones"  is still significantly lower than in these examples.

Furthermore, note for the exercise:

  • An  $(n, \ k)$  $\underline{x} = \underline{u} \cdot \mathbf{H}^{\rm T}$  block code is systematic if the first  $k$  bits of the code word  $\underline{x}$  contains the information word  $\underline{u}$.  With the code word  $\underline{x} = (u_1, \, u_2, \, u_3, \, u_4, \, p_1, \, p_2, \, p_3, \, p_4)$  the parity-check matrix  $\mathbf{H}$  must then end with a  $k × k$ diagonal matrix.
  • A  "regular code"  $($with respect to the LDPC application$)$  exists,  if the Hamming weight of all rows   ⇒   $w_{\rm R}$ and the Hamming weight of all columns   ⇒   $w_{\rm C}$ are equal in each case.  Otherwise,  one speaks of an  "irregular LDPC code".
  • The parity-check matrix  $\mathbf{H}$  of a conventional linear  $(n, \ k)$  block code consists of exactly  $m = n - k$  rows and  $n$  columns.  For LDPC codes,  the requirement is   $m ≥ n - k$.  The equal sign is true if the  $m$  parity-check equations are statistically independent.
  • From the parity-check matrix  $\mathbf{H}$  a lower bound for the code rate  $R$  can be given:
$$R \ge 1 - \frac{{\rm E}[w_{\rm C}]}{{\rm E}[w_{\rm R}]} \hspace{0.5cm}{\rm with}\hspace{0.5cm} {\rm E}[w_{\rm C}] =\frac{1}{n} \cdot \sum_{i = 1}^{n}w_{\rm C}(i) \hspace{0.5cm}{\rm and}\hspace{0.5cm} {\rm E}[w_{\rm R}] =\frac{1}{m} \cdot \sum_{j = 1}^{ m}w_{\rm R}(j) \hspace{0.05cm}.$$
  • This equation applies equally to regular and irregular LDPC codes,  where for regular codes holds  ${\rm E}[w_{\rm C}] = w_{\rm C}$  and  ${\rm E}[w_{\rm R}] = w_{\rm R}$.



Hints:


Questions

1

Which parity-check matrix describes the given product code according to the sketch above?

$\mathbf{H}_1$  given  $\underline{x} = (u_1, \, u_2, \, u_3, \, u_4, \, p_1, \, p_2, \, p_3, \, p_4)$.
$\mathbf{H}_1$  given  $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$.
$\mathbf{H}_2$  given  $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$.
$\mathbf{H}_3$  given  $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$.

2

For the remaining subtasks we shall always assume  $\underline{x} = (u_1, \, u_2, \, u_3, \, u_4, \, p_1, \, p_2, \, p_3, \, p_4)$.
Which statements are valid for the parity-check matrix  $\mathbf{H}_1$?

The code is systematic.
The code is regular.
For the code rate holds  $R > 1/2$.
For the code rate holds  $R = 1/2$.

3

What statements hold for the parity-check matrix  $\mathbf{H}_3$?

No relation between  $\mathbf{H}_1$  and  $\mathbf{H}_3$  is discernible.
The  $\mathbf{H}_3$  rows are linear combinations of two  $\mathbf{H}_1$  rows each.
The four parity-check equations according to  $\mathbf{H}_3$  are linearly independent.

4

Which statements apply to the code denoted by  $\mathbf{H}_3$ ?

The code is systematic.
The code is regular.
For the code rate holds  $R ≥ 1/2$.
For the code rate holds  $R = 1/2$.


Solution

(1)  Correct are the solutions 1 and 3:

  • With the code word definition $\underline{x} = (u_1, \, u_2, \, u_3, \, u_4, \, p_1, \, p_2, \, p_3, \, p_4)$, the parity-check matrix $\mathbf{H}_1$ denotes the following check equations:
$$u_1 \oplus u_2 \oplus p_1 = 0\hspace{0.05cm},\hspace{0.3cm} u_3 \oplus u_4 \oplus p_2 = 0\hspace{0.05cm},\hspace{0.3cm} u_1 \oplus u_3 \oplus p_3 = 0\hspace{0.05cm},\hspace{0.3cm} u_2 \oplus u_4 \oplus p_4 = 0\hspace{0.05cm}.$$
  • This corresponds exactly to the assumptions made above. The same result is obtained for $\mathbf{H}_2$ and the code word definition $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$.


With the same code word definition $\underline{x} = (u_1, \, p_1, \, u_2, \, p_2, \, u_3, \, p_3, \, u_4, \, p_4)$ the other parity-check matrices do not yield a meaningful set of equations:

  • According to parity-check matrix $\mathbf{H}_1$:
$$u_1 \oplus p_1 \oplus u_3 = 0\hspace{0.05cm},\hspace{0.3cm} u_2 \oplus p_2 \oplus p_3 = 0\hspace{0.05cm},\hspace{0.3cm} u_1 \oplus u_2 \oplus u_4 = 0\hspace{0.05cm},\hspace{0.3cm} p_1 \oplus p_2 \oplus p_4 = 0\hspace{0.05cm};$$
  • corresponding to parity-check matrix $\mathbf{H}_3$:
$$u_1 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_2 \hspace{-0.12cm}\oplus\hspace{-0.06cm} u_3 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_4 \hspace{-0.05cm} = \hspace{-0.05cm} 0\hspace{0.05cm},\hspace{0.15cm} u_1 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_2 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_3 \hspace{-0.12cm}\oplus\hspace{-0.06cm}u_4 \hspace{-0.05cm} = \hspace{-0.05cm} 0\hspace{0.05cm},\hspace{0.15cm} p_1 \hspace{-0.12cm}\oplus\hspace{-0.06cm} u_2 \hspace{-0.12cm}\oplus\hspace{-0.06cm} u_3 \hspace{-0.12cm}\oplus\hspace{-0.06cm}u_4 \hspace{-0.05cm} = \hspace{-0.05cm} 0\hspace{0.05cm},\hspace{0.15cm} p_1 \hspace{-0.12cm}\oplus\hspace{-0.06cm} u_2 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_3 \hspace{-0.12cm}\oplus\hspace{-0.06cm} p_4 \hspace{-0.05cm} = \hspace{-0.05cm} 0\hspace{0.05cm}.$$


(2)  Correct are the proposed solutions 1 and 4:

  • The code is systematic because $\mathbf{H}_1$ ends with a $4 × 4$ diagonal matrix.
  • For a regular (LDPC) code, there should be an equal number of ones in each row and in each column.
  • The first condition is satisfied $(w_{\rm Z} = 3)$, but not the second. Rather, there is (equally often) one one or two ones per column  ⇒  ${\rm E}[w_{\rm S}] = 1.5$.
  • For an irregular code, the lower bound for the code rate is:
$$R \ge 1 - \frac{{\rm E}[w_{\rm S}]}{{\rm E}[w_{\rm Z}]} = 1 - \frac{1.5}{3} = 1/2 \hspace{0.05cm}.$$
  • Because of the given code structure ($k = 4$ information bits, $m = 4$ check bits  ⇒  $n = 8$ code bits) the code rate can also be given in the conventional form: $R = k/n$   ⇒   Correct is solution 4 in contrast to answer 3.


(3)  The $\mathbf{H}_3$–rows result from linear combinations of $\mathbf{H}_1$–rows:

  • The first $\mathbf{H}_3$–row is the sum of row 1 and row 4.
  • The second $\mathbf{H}_3$–row is the sum of row 2 and row 3.
  • The third $\mathbf{H}_3$–row is the sum of row 1 and row 3.
  • The fourth $\mathbf{H}_3$–row is the sum of row 2 and row 4.


By linear combinations, the four linearly independent equations with respect to $\mathbf{H}_1$ now become four linearly independent equations with respect to $\mathbf{H}_3$  
⇒   Therefore, the proposed solutions 2 and 3 are correct.


(4)  Here, solutions 2, 3, and 4 are correct:

  • If the code described by $\mathbf{H}_3$ were systematic, $\mathbf{H}_3$ should end with a $4 × 4$–diagonal matrix. This is not the case here.
  • The Hamming weights of all rows are equal $(w_{\rm Z} = 4)$ and also all columns each have the same Hamming weight $(w_{\rm S} = 2)$   ⇒   the code is regular.
  • This gives $R ≥ 1 - 2/4 = 1/2$ for the code rate. But since the four rows of $\mathbf{H}_3$ also describe four independent equations, the equal sign   ⇒   $R = 1/2$ also holds.