We propose to design a channel code, called C$\_1,2,$ by combining two linear block codes $C_{-}1$

and $C_{2}2.$

The parameters of $C_1$ are $k_1,n_1,$ and $d_{m}in1,$ whereas the parameters of $C_2$ are $k_2,$

$n_2,$ and $d_{?}$min $2,$ where $k_1$ and $k_2$ denote the message lengths, $n_1$ and $n_2$ designate the

codeword lengths, and $d_{m}in1$ and $d_{m}in2$ are the minimum Hamming distances between

codewords.

The encoding algorithm is explained below.

Each message composed of $k_1{k}_2$ information bits is stored in a two$-$dimensional $(2-$D$)$

array of size $k1k2$ bits. This array thus has $k_1$ columns and $k_2$ rows.

At the first step of encoding, each row of this array is encoded into a codeword of length $n_1$

bits using C$\_1.$ The row encoding results in a $2-$D array composed of $n_1$ columns and $k_{-}2$

rows.

At the second and final step of encoding, each of the $n_1$ columns is encoded into a

codeword of length $n_2$ using C$\_2.$ The column encoding results in a $2-$D array composed of

$n_1$ columns and $n_2$ rows. This array is the codeword associated with the $k_1{k}_2-$bit array

present at the encoder output.

The minimum Hamming distance, dmin, between codewords for $C1,2$ is the product of

$d_{m}in1$ and $d_{m}in2.$ In other words, we can write $d_{m}in=d_{m}in1*d_{m}in2.$ How would

you demonstrate this result?

How would you decode the code C$\_1,2$ in practice? To illustrate your explanations, feel

free to draw the decoder structure that you propose.