Won Yang

It is working well on my PC. What error message(s) do you see? You know it may take some time and your patience.

I am sorry that I have no MATLAB codes on DAPSK/OQAM. However, if I could help you in composing the MATLAB codes on DAPSK/OQAM and there was any possibility that I can be a coauthor in your research paper, I would consider trying to compose the codes under your guidance
even if I am not so confident of my capability.
Thanks for being my reader.

I see no problem as far as the last line computing the BER is concerned. ber_QPSK_theory = prob_error(SNRbdB,'PSK',K,'BER')

gen_LDPC is what was made not by me:
function H=gen_ldpc(M,N)
%For examples and more details, please refer to the LDPC toolkit tutorial at
%http://arun-10.tripod.com/ldpc/ldpc.htm
bits_per_col=3;
for i=1:M
row_flag(i)=0; for j=1:N, H(i,j)=0; end
end
%add bits_per_col 1's to each column with the only constraint
% being that the 1's should be placed in distinct rows
for i=1:N
a=randperm(M);
for j=1:bits_per_col, H(a(j),i)=1; row_flag(a(j))=row_flag(a(j))+1; end
end
%row_flag
max_ones_per_row=ceil(N*bits_per_col/M);
%H

%add 1's to rows having no 1(a redundant row) or only one 1(that bit in the codeword becomes
%zero irrespective of the input)
for i=1:M
if row_flag(i)==1
j=unidrnd(N);
while H(i,j)==1, j=unidrnd(N); end
H(i,j)=1; row_flag(i)=row_flag(i)+1;
end
if row_flag(i)==0
for k=1:2
j=unidrnd(N);
while H(i,j)==1, j=unidrnd(N); end
H(i,j)=1; row_flag(i)=row_flag(i)+1;
end
end
end

%try to distribute the ones so that the number of ones per row is as uniform as possible
for i=1:M
j=1; a=randperm(N);
while row_flag(i)>max_ones_per_row;
if H(i,a(j))==1
H(i,a(j))=0; row_flag(i)=row_flag(i)-1; newrow=unidrnd(M);
k=0;
while row_flag(newrow)>=max_ones_per_row | H(newrow,a(j))==1
newrow=unidrnd(M); k=k+1;
if k>=M, break; end
end
if H(newrow,a(j))==0, H(newrow,a(j))=1; row_flag(newrow)=row_flag(newrow)+1;
else H(i,a(j))=1; row_flag(i)=row_flag(i)+1;
end
end%if loop
j=j+1;
end%while loop
end%for loop
%row_flag, H
%try to eliminate cycles of length 4 in the factor graph
for loop=1:10
ones_position(1)=0;
for r=1:M
ones_count=0;
for c=1:N
if H(r,c)==1
ones_count=ones_count+1; ones_position(ones_count)=c;
end
end
for i=1:r-1
common=0;
for j=1:ones_count
if H(i,ones_position(j))==1
common=common+1 ;
if common==1, thecol=ones_position(j); end
end
if common==2
common=common-1;
if (round(rand)==0), coltoberearranged=thecol; thecol=ones_position(j);
else coltoberearranged=ones_position(j);
end
H(i,coltoberearranged)=3;
%make this entry 3 so that we dont use of this entry again while getting rid of other cylces
newrow=unidrnd(M);
%while ((newrow==i)|(H(newrow,ones_position(j))==1))
iteration=0;
while H(newrow,coltoberearranged)~=0
newrow=unidrnd(M); iteration=iteration+1;
if iteration==5, break; end
end
if iteration==5
while H(newrow,coltoberearranged)==1, newrow=unidrnd(M); end
end
H(newrow,coltoberearranged)=1;
end
end
end
for i=r+1:M
common=0;
for j=1:ones_count
if H(i,ones_position(j))==1
common=common+1 ;
if common==1, thecol=ones_position(j); end
end
if common==2
common=common-1;
if (round(rand)==0), coltoberearranged=thecol; thecol=ones_position(j);
else coltoberearranged=ones_position(j);
end
H(i,coltoberearranged)=3;
%make this entry 3 so that we dont use of this entry again while getting rid of other cylces
newrow=unidrnd(M);
%while ((newrow==i)|(H(newrow,ones_position(j))==1))
iteration=0;
while H(newrow,coltoberearranged)~=0
newrow=unidrnd(M); iteration=iteration+1;
if iteration==5, break; end
end
if iteration==5
while H(newrow,coltoberearranged)==1, newrow=unidrnd(M); end
end
H(newrow,coltoberearranged)=1;
end
end
end
end
end;
%H;
for i=1:M
row_flag(i)=0;
for j=1:N
if H(i,j)==1, row_flag(i)=row_flag(i)+1; end
if eq(H(i,j),3), H(i,j)=0; end %replace the 3's with 0's
end
end

nice