Exercise 4.11: Analysis of Parity-check Matrices
In the adjacent graphic, a product code is indicated at the top, which is characterized by the following parity-check equations:
- p1 = u1⊕u2,p2=u3⊕u4,p3 = u1⊕u3,p4=u2⊕u4.
Below are given the check matrices H1, H2 and H3 . To be checked is which of the matrices has the given product code according to the equation x_=u_⋅HT correctly describing the given product code, assuming the following definitions:
- the code word x_=(u1,u2,u3,u4,p1,p2,p3,p4),
- the code word x_=(u1,p1,u2,p2,u3,p3,u4,p4).
All H–matrices contain fewer ones than zeros. This is a characteristic of the so-called low–density parity–check codes (short: 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:
- A (n, k)–block code is systematic if the first k bits of the code word x_ contains the information word u_ . With the code word definition x_=(u1,u2,u3,u4,p1,p2,p3,p4) the parity-check matrix H must then end with a k×k diagonal matrix.
- A regular code (with respect to LDPC application) exists, if the Hamming–weight of all rows ⇒ wZ and the Hamming weight of all columns ⇒ wS are equal in each case. Otherwise, one speaks of an irregular LDPC code.
- The parity-check matrix H of a conventional linear (n, k) block code consists of exactly m=n−k rows and n columns. For LDPC codes, on the other hand, 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 H a lower bound for the code rate R can be given:
- R≥1−E[wS]E[wZ]mitE[wS]=1n⋅n∑i=1wS(i)undE[wZ]=1m⋅m∑j=1wZ(j).
- This equation applies equally to regular and irregular LDPC–codes, where the regular codes E[wS]=wS and E[wZ]=wZ can be considered.
Hints:
- This exercise belongs to the chapter "Basics of the Low–density Parity–check".
- Reference is made in particular to the page "Some characteristics of LDPC codes".
Questions
Solution
- With the code word definition x_=(u1,u2,u3,u4,p1,p2,p3,p4), the parity-check matrix H1 denotes the following check equations:
- u1⊕u2⊕p1=0,u3⊕u4⊕p2=0,u1⊕u3⊕p3=0,u2⊕u4⊕p4=0.
- This corresponds exactly to the assumptions made above. The same result is obtained for H2 and the code word definition x_=(u1,p1,u2,p2,u3,p3,u4,p4).
With the same code word definition x_=(u1,p1,u2,p2,u3,p3,u4,p4) the other parity-check matrices do not yield a meaningful set of equations:
- According to parity-check matrix H1:
- u1⊕p1⊕u3=0,u2⊕p2⊕p3=0,u1⊕u2⊕u4=0,p1⊕p2⊕p4=0;
- corresponding to parity-check matrix H3:
- u1⊕p2⊕u3⊕p4=0,u1⊕p2⊕p3⊕u4=0,p1⊕u2⊕u3⊕u4=0,p1⊕u2⊕p3⊕p4=0.
(2) Correct are the proposed solutions 1 and 4:
- The code is systematic because H1 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 (wZ=3), but not the second. Rather, there is (equally often) one one or two ones per column ⇒ E[wS]=1.5.
- For an irregular code, the lower bound for the code rate is:
- R≥1−E[wS]E[wZ]=1−1.53=1/2.
- 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 H3–rows result from linear combinations of H1–rows:
- The first H3–row is the sum of row 1 and row 4.
- The second H3–row is the sum of row 2 and row 3.
- The third H3–row is the sum of row 1 and row 3.
- The fourth H3–row is the sum of row 2 and row 4.
By linear combinations, the four linearly independent equations with respect to H1 now become four linearly independent equations with respect to H3
⇒ Therefore, the proposed solutions 2 and 3 are correct.
(4) Here, solutions 2, 3, and 4 are correct:
- If the code described by H3 were systematic, H3 should end with a 4×4–diagonal matrix. This is not the case here.
- The Hamming weights of all rows are equal (wZ=4) and also all columns each have the same Hamming weight (wS=2) ⇒ the code is regular.
- This gives R≥1−2/4=1/2 for the code rate. But since the four rows of H3 also describe four independent equations, the equal sign ⇒ R=1/2 also holds.