Hamming Code: Matlab Tutorial

behindthesciences | July 5, 2016 | Coding, Imported | No Comments

In this post we are going to develop the Hamming Code error correction algorithm in Matlab, so the first thing you need to do is to download our free Hamming Matlab Code from Mathworks: https://uk.mathworks.com/matlabcentral/fileexchange/57312-hamming-code–matlab-tutorial

Ready? Let’s get started! 🙂

We are going to create a Dual Hamming Code (15,4) from the systematic code (15,11).

Recall that:

The first step is to create a parity check and generator Hamming matrix (15,11) , H and G respectively.

We now that the identity matrix dimensions are: r x r=4 x 4 and, the transposed parity matrix: r x k = 4 x 11. Therefore, the H matrix dimensions are 4 x 15, for example:

1 0 0 0 1 0 0 1 1 0 1 0 1 1 1

H= 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0

0 0 1 0 0 1 1 0 1 0 1 1 1 1 0

0 0 0 1 0 0 1 1 0 1 0 1 1 1 1

In the same way, we can conclude that the G matrix dimensions are 11 x 15, for instance:

1 1 0 0 1 0 0 0 0 0 0 0 0 0 0

0 1 1 0 0 1 0 0 0 0 0 0 0 0 0

0 0 1 1 0 0 1 0 0 0 0 0 0 0 0

1 1 0 1 0 0 0 1 0 0 0 0 0 0 0

1 0 1 0 0 0 0 0 1 0 0 0 0 0 0

G= 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0

1 1 1 0 0 0 0 0 0 0 1 0 0 0 0

0 1 1 1 0 0 0 0 0 0 0 1 0 0 0

1 1 1 1 0 0 0 0 0 0 0 0 1 0 0

1 0 1 1 0 0 0 0 0 0 0 0 0 1 0

1 0 0 1 0 0 0 0 0 0 0 0 0 0 1

The next step is to find the Hamming dual, (15,11); therefore, we need to work with the Hamming (15, 4). Now the size of the code word is still the same, but the size of the input message is going to be 4 (before it was 11) and theredundancy, 11 (before it was 4)

In addition, our new generator matrix, G, will be the Hamming matrix, H (15,11); in this way, we will form the following “input message-code word” table:

m (input)

Code word

0000	000000000000000
0001	000100110101111
0010	001001101011110
0011	001101011110001
0100	010011010111100
0101	010111100010011
0110	011010111100010
0111	011110001001101
1000	100010011010111
1001	100110101111000
1010	101011110001001
1011	101111000100110
1100	110001001101011
1101	110101111000100
1110	111000100110101
1111	111100010011010

We select a code word to transmit, for example, the corresponding with the input message [0101], which is [010111100010011].

Let’s suppose that after transmitting by a channel, there is a bit error in a random position; in this case, after computing the module, we obtain: (20809379,7)=3.

We use the property that allow us to obtain the syndrome vector from the error vector (size 1×15), by using H transposed (this is equivalent to do: received vector multiplied by H transposed) and, knowing that the H of the Dual Hamming (15,4) is the G obtained obtained for the (15,11).

Therefore, if our error vector is: e=[001000000000000] and G^T is:

1 0 0 1 1 0 1 0 1 1 1

1 1 0 1 0 1 1 1 1 0 0

0 1 1 0 1 0 1 1 1 1 0

0 0 1 1 0 1 0 1 1 1 1

1 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0

G^T= 0 0 1 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 0 1

Finally, the syndrome vector that we obtain and we need to check with the syndrome vector to error vector map is:

S=[01101011110]

*This vector is the third row of H^T, as it has been places in the error position.

Now you are ready to fully understand the provided Matlab code (see link above)! We hope you found this post useful and don’t hesitate to ask questions or require further posts for any of the sections that we have!