Understanding rsenc (RS encoder) function

I was coding an RS encoder (n=7,k=5)

m = 3;
t = 1; % error correcting capacity
%%
n = 2^m-1;
k = n - 2*t;
[gx,~]  = rsgenpoly(n,k); % degree is (n-k)

the generator polynomial gx is [1 6 3] which goes well with my understanding that it represents polynomial = x^2 + (alpha)^4 * x + (alpha)^3 .

if the message is x, then its vector form is = [0 0 0 1 0];

encoded polynomial will be gx*x which is to my understanding [0 0 0 1 6 3 0]. i.e x^3 + (alpha)^4 * x^2 + (alpha)^3 * x.

But the following code gives

msg     = [0 0 0 1 0]; % degree is k-1
msg_gf  = gf(msg,m);
code_gf = rsenc(msg_gf, n, k,gx); % degree is n-1

code_gf = [0 0 0 1 0 1 1].

Why is rsenc giving this vector as output?

Answer 1

[Zinea]

Hi Sreekanth CB,

The discrepancy you are observing is due to a misunderstanding of how the Reed-Solomon encoding process works.

The following steps are involved in encoding a message using an RS encoder:

1. Generator Polynomial: The generator polynomial ( g(x) ) is used to multiply the message polynomial. For your case, the generator polynomial is given as ( g(x) = x^2 + \alpha^4 x + \alpha^3 ).
2. Message Polynomial: The message polynomial is represented by the vector [0 0 0 1 0], which corresponds to ( x^3 ).
3. Encoding: The encoding process involves multiplying the message polynomial by ( x^{n-k} ) and then dividing by the generator polynomial ( g(x) ). The remainder of this division is then added to the original message polynomial multiplied by ( x^{n-k} ).

Here is an explanation of the steps in the code:

1. Message Vector: The original message vector is [0 0 0 1 0], which corresponds to ( x^3 ).
2. Multiply by (x^{n-k}): Since ( n-k = 2 ), this means shifting the message polynomial by 2 degrees, resulting in ( x^5 ).
3. Polynomial Division: The shifted message polynomial is divided by the generator polynomial (g(x) ). The remainder of this division is the error correction code.
4. Encoded Message: The encoded message is the original message polynomial (shifted) plus the error correction code.

The ‘rsenc’ function performs these operations and returns the encoded codeword, which in your case is [0 0 0 1 0 1 1]. This vector represents the polynomial that includes both the original message and the error correction code. The output [0 0 0 1 0 1 1] indicates that after encoding, the polynomial is (x^3 + \alpha x + \alpha). The last two terms, 1 1, represent the error correction code added to the original message polynomial.

For more information on using 'rsenc', you may refer to the following link:

https://www.mathworks.com/help/comm/ref/rsenc.html#mw_0f82e6b2-1c17-4204-bf44-1250ba2e690f

Hope this clarifies the confusion!