# Check matrix H \boldsymbol{H} H

## Definition of zero space:

H \boldsymbol H The zero space or kernel of H is written as
N ( H ) ≜ { x ∈ R n : H x = 0 } \mathcal{N}(\boldsymbol{H}) \triangleq\left\{\boldsymbol{x} \in \mathbb{R}^{n}: \boldsymbol{H} \boldsymbol{x}=\mathbf{0}\right\} N(H)≜{x∈Rn:Hx=0}

So, the BCH code space is H \boldsymbol H Zero space of H.

# Decoding principle

The specific principle derivation will not be repeated. Please refer to reference [1], and only its core idea is summarized here.

## Decoding idea

### 1. Establishment of equations

According to adjoint S i S_i Si and wrong pattern e ( X ) e(X) Relationship of e(X)
S i = e ( α i ) S_i=\boldsymbol e(\alpha ^i) Si​=e(αi)

Get the wrong location j 1 , j 2 , ... , j v j_1,j_2,\ldots,j_v j1, j2,..., jv 2 t 2t 2t adjoint relational equations:

Solving the equations to get the error location can be used for error correction decoding. However, the equations are nonlinear and direct solution is not feasible, so the indirect solution method is adopted.

### 2 Newton identity

definition

By observing the above formula, we can see that using the complete square expansion, this construction can connect the adjoint component with the coefficients of the wrong position polynomial to obtain the Newton identity

The error position polynomial is obtained by solving the equation with Newton identity σ ( X ) \sigma(X) σ (10) And its coefficient σ ( X ) \sigma(X) σ (10) And the reciprocal of the root is the wrong location. The decoding process is as follows.

### 4 iterative solution of Newton identity (core of BM algorithm)

The Newton identity in decoding process (2) can be solved by Berlekamp Massey iterative algorithm (BM algorithm).

BM algorithm passed 2 t 2t The error position polynomial is solved by 2t iterations σ ( X ) \sigma(X) σ (10) , a bit like induction.

The first k k k times to get a minimum number of times σ ( k ) ( X ) \sigma^{(k)}(X) σ (k)(X), before meeting k k k equations. The first k + 1 k+1 k+1 times, based on σ ( k ) ( X ) \sigma^{(k)}(X) σ (k)(X) find the error position polynomial with the lowest degree next σ ( k + 1 ) ( X ) \sigma^{(k+1)}(X) σ (k+1)(X) to satisfy the previous k + 1 k+1 k+1 equations.

First, check σ ( k ) ( X ) \sigma^{(k)}(X) σ (k) Whether the coefficient of (x) meets the requirements of paragraph k + 1 k+1 k+1 equations, if satisfied, then
σ ( k + 1 ) ( X ) = σ ( k ) ( X ) \sigma^{(k+1)}(X)=\sigma^{(k)}(X) σ(k+1)(X)=σ(k)(X)

Otherwise, yes σ ( k ) ( X ) \sigma^{(k)}(X) σ (k)(X) obtained by adding correction items σ ( k + 1 ) ( X ) \sigma^{(k+1)}(X) σ (k+1)(X). The details are as follows:

calculation

# Restore BM iterative decoding process with an example

Taking (15,5,3)BCH code as an example, the code length is 15 and the error correction ability is good t = 3 t=3 t=3. The domain generating polynomial is p ( X ) = X 4 + X + 1 p(X)=X^4+X+1 p(X)=X4+X+1 (expressed as [1 0 1 1] by degree, expressed as 23 by octal number, hereinafter uniformly expressed as polynomial by octal), and the code generation polynomial is expressed as 2467 by octal number.

Generated domain G F ( 2 4 ) GF(2^4) GF(24) is shown in the table below

First initialize as follows

i i The initial value of i is - 1, i − l i = − 1 i-l_i=-1 i − li = − 1, start iteration:
① k = 1 k=1 When k=1, due to d 0 = 1 ≠ 0 d_0=1\ne0 d0 = 1  = 0, so the calculated correction term is d 0 d − 1 − 1 X 0 − ( − 1 ) σ ( − 1 ) ( X ) = X d_0d_{-1}^{-1}X^{0-(-1)}\sigma^{(-1)}(X)=X d0​d−1−1​X0−(−1) σ (− 1)(X)=X, so σ ( 1 ) ( X ) = 1 + X \sigma^{(1)}(X)=1+X σ (1)(X)=1+X, from which d 1 = S 1 + 1 + σ 1 ( 1 ) S 1 = S 2 + 1 ⋅ S 1 = 0 d_1=S_{1+1}+\sigma_1^{(1)}S_1=S_2+1\cdot S_1=0 d1​=S1+1​+ σ 1 (1) S1 = S2 + 1 ⋅ S1 = 0, so l 1 = 1 l_1=1 l1​=1， 1 − l 1 = 0 1-l_1=0 1−l1​=0， i = 0 i=0 i=0， i − l i = 0 i-l_i=0 i−li​=0.
Update table to

② k = 2 k=2 When k=2, due to d 1 = 0 d_1=0 d1 = 0, so σ ( 2 ) ( X ) = σ ( 1 ) ( X ) = 1 + X \sigma^{(2)}(X)=\sigma^{(1)}(X)=1+X σ (2)(X)= σ (1)(X)=1+X, from which d 2 = S 2 + 1 + σ 1 ( 2 ) S 2 = S 3 + 1 ⋅ S 2 = α 10 + 1 = α 5 d_2=S_{2+1}+\sigma_1^{(2)}S_2=S_3+1\cdot S_2=\alpha^{10}+1=\alpha^5 d2​=S2+1​+ σ 1(2)​S2​=S3​+1⋅S2​= α 10+1= α 5, so l 2 = 1 l_2=1 l2​=1， 2 − l 2 = 1 2-l_2=1 2−l2​=1， i = 0 i=0 i=0， i − l i = 0 i-l_i=0 i−li​=0.
Update table to

③ k = 3 k=3 When k=3, due to d 2 = α 5 ≠ 0 d_2=\alpha^5\ne0 d2​= α 5  = 0, so the calculated correction term is d 2 d 0 − 1 X 2 − 0 σ ( 0 ) ( X ) = α 5 X 2 d_2d_{0}^{-1}X^{2-0}\sigma^{(0)}(X)=\alpha^5X^2 d2​d0−1​X2−0 σ (0)(X)= α 5X2, so σ ( 3 ) ( X ) = σ ( 2 ) ( X ) + α 5 X 2 σ ( 0 ) ( X ) = 1 + X + α 5 X 2 \sigma^{(3)}(X)=\sigma^{(2)}(X)+\alpha^5X^2\sigma^{(0)}(X)=1+X+\alpha^5X^2 σ (3)(X)= σ (2)(X)+ α 5X2 σ (0)(X)=1+X+ α 5x2, resulting in d 3 = S 3 + 1 + σ 1 ( 3 ) S 3 + σ 2 ( 3 ) S 2 = S 4 + 1 ⋅ S 3 + α 5 S 2 = 1 + α 10 + α 5 = 0 d_3=S_{3+1}+\sigma_1^{(3)}S_3+\sigma_2^{(3)}S_2=S_4+1\cdot S_3+\alpha^5S_2=1+\alpha^{10}+\alpha^5=0 d3​=S3+1​+ σ 1(3)​S3​+ σ 2(3)​S2​=S4​+1⋅S3​+ α 5S2​=1+ α 10+ α 5 = 0, so l 3 = 2 l_3=2 l3​=2， 3 − l 3 = 1 3-l_3=1 3−l3​=1， i = 2 i=2 i=2， i − l i = 1 i-l_i=1 i−li​=1.
Update table to

④ And so on, and finally 2 t 2t The results of 2t iterations are as follows:

The result of the simplified BM algorithm is:

# Simulation performance and analysis

simulation result

## Main program code

File name: bch_codec.m

clear; close all; clc

%% (15,11,1) BCH code
% px=[1 0 0 1 1];  % 23 Primitive polynomial
% p = length(px)-1;
% [a,a_dec,log_a] = sub_gf_gen( px ); % Finite field
% gx=[1 0 0 1 1]; % 23
% t=1;

%% (15,7,2) BCH code
% gx=[1 1 1 0 1 0 0 0 1]; % 721
% t=2;

%% (31,26,1) BCH code
% px=[1 0 0 1 0 1];  % 45
% p = length(px)-1;
% [a,a_dec,log_a] = sub_gf_gen( px ); % Finite field
% gx=[1 0 0 1 0 1]; % 45
% t=1;
%% (31,21,2) BCH code
% gx=[1 1 1 0 1 1 0 1 0 0 1]; % 3551
% t=2;

%% (63,57,1) BCH code
px=[1 0 0 0 0 1 1];  % 103
p = length(px)-1;
[a,a_dec,log_a] = sub_gf_gen( px ); % Finite field
% gx=[1 0 0 0 0 1 1]; % 103
% t=1;

%% (63,51,2) BCH code
gx=[1 0 1 0 1 0 0 1 1 1 0 0 1]; % 12471
t=2;

m = length(px)-1;
n=2^m-1; % Code length
pg = length(gx) -1; % Generating polynomial degree
k=n-pg; % Information chief

codnum = 100000;
snr = 5:0.5:10;
snrlen = length(snr);
num_bef = zeros(1,snrlen);
num_aft = zeros(1,snrlen);
tic
parfor i = 1:snrlen
SNRi = snr(i);
for j = 1:codnum
%         rng('shuffle');
u = randi([0,1],1,k);
v = sub_encod_bch(u,gx,0);
vn = awgn(1-2*v,SNRi,'measured');
r = double(vn<0);
num_bef(i) = num_bef(i)+biterr(r,v);
[rdec,pos_err,flag]=sub_decod_bch(r,t,a,log_a,p);
num_aft(i) = num_aft(i)+biterr(rdec,v);
end
disp(['Signal to noise ratio ',num2str(SNRi),' complete']);
end
toc
ber_bef = num_bef/n/codnum;
ber_aft = num_aft/n/codnum;


## Drawing code

File name: plot_bch_dec.m

clear; clc; close all;

snr=5:0.5:10;
len = length(snr);
lw = 2;

%% Bit error rate before decoding
figure;semilogy(snr,ber_bef,'Marker','x','color',[0.49 0.18 0.56],'LineWidth',lw);hold on;

%% (15,11,1)BCH code
semilogy(snr,ber_aft,'Marker','s','color',[0 161/255 59/255],'LineWidth',2);hold on;

%% (15,7,2)BCH code
semilogy(snr,ber_aft,'o-b','LineWidth',lw);hold on;

%% (31,26,1)BCH code
semilogy(snr,ber_aft,'Marker','d','color',[0.85 0.33 0.10],'LineWidth',lw);hold on;

%% (31,21,2)BCH code
semilogy(snr,ber_aft,'^-r','LineWidth',lw);hold on;

%% (63,57,1)BCH code
semilogy(snr(4:end),ber_aft(4:end),'Marker','v','color',[0 0 0],'LineWidth',lw);hold on;

%% (63,51,2)BCH code
semilogy(snr,ber_aft,'Marker','p','color',[0 0.4470 0.7410],'LineWidth',lw);hold on;

xlabel('SNR (dB)');ylabel('decoding BER'); grid on;
legend('Before decoding','(15,11,1)','(15,7,2)','(31,26,1)','(31,21,2)','(63,57,1)','(63,51,2)');
set(gca,'FontSize',13,'Fontname', 'Microsoft YaHei ');


## Partial subroutine code

### Finite field multiplication subroutine

File name: sub_gf_mul.m

function [ z ] = sub_gf_mul( x, y, n )
% Galois field element multiplication, parameter meaning:
%% input:
% x y: Represents the power of two elements and the value range-inf or[0,n-1]
% n: 2^m-1

%% output:
% z: Represents the power of the product

if x>=0 && y>=0
z = mod(x+y,n);
else
z=-inf;
end


function [ z ] = sub_gf_add( x, y, a, log_a )
% Galois domain element addition, parameter meaning:
%% input:
% x y: Represents the power of two elements and the value range-inf or[0,n-1]
% n: 2^m-1

%% output:
% z: Represents the power of the sum of two elements. When the sum is 0, the power is specified as-inf，Convenient calculation

if x<0
z=y;
elseif y<0
z=x;
elseif x==y
z=-inf;
else
sum=mod(a(x+1,:)+a(y+1,:),2);
sum_deci = bi2de(sum, 'left-msb');
z = log_a(sum_deci);
end



file namefunction
bch_codec.mmain program
sub_gf_gen.mFinite field generation subroutine
sub_gf_mul.mFinite field multiplication subroutine
sub_poly_div.mPolynomial division subroutine
sub_encod_bch.mBCH coding subroutine
sub_decod_bch.mSimplified BM iterative decoding subroutine for BCH code
plot_bch_dec.mDrawing subroutine
ber_aft_bch_15_7_2.mat(15,7,2) code decoding error rate data file
ber_aft_bch_15_11_1.mat(15,11,1) code decoding error rate data file
ber_aft_bch_31_26_1.mat(31,26,1) code decoding error rate data file
ber_aft_bch_31_21_2.mat(31,21,2) code decoding error rate data file
ber_aft_bch_63_57_1.mat(63,57,1) code decoding error rate data file
ber_aft_bch_63_51_2.mat(63,51,2) code decoding error rate data file

## a) Local tyrant, please feel free

You can also subscribe Channel coding column , refer to the detailed explanation of each subroutine

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[BCH code 1] system BCH code coding principle and MATLAB implementation (without MATLAB library function)

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Pay attention to the public number [reverse communication], password: BCH, which can be obtained by simply supporting one wave

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# reference

[1]William E Ryan, Shu Lin. channel coding: classic and modern [M]. Electronic Industry Press, 2017

Posted by BinaryBird on Sun, 21 Nov 2021 02:28:23 -0800