Exercise 4.6Z: Basics of Product Codes

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:

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

	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 $d_1 = 3$.

	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 $d_2 = 3$.

$k \hspace{0.25cm} = \ $

$n \hspace{0.25cm} = \ $

$d \hspace{0.25cm} = \ $

$R \hspace{0.15cm} = \ $

Solution

(1) Correct are 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 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 $d = d_1 \cdot d_2 = 3 \cdot 3 \ \underline{= 9}$.