Won Yang

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

Thank you.

png is a function saved in an M-file named 'png'. If you had downloaded, it would be automatically run whenever the main program "DS_SS" is run. If you don't have the function, drag and paste the following:
 
function c=png(gm,Kc,x)
%generates an m-sequence from a feedback-shift register if Kc=0
% or from a high-speed feedback shift generator if KC>0
% gm: a generator tap coefficient vector
% or a set of connected tap position numbers
%Copyleft: Won Y. Yang, wyyang53@hanmail.net, CAU for academic use only
if sum(gm>1)>0, g(gm)=1; % say, if gm=[1 2 4 5]
 else g=gm; % if gm=[1 1 0 1 1]
end
m=length(g); % Number of feedback shift-registers
if nargin==2
  if length(Kc)>1, x=Kc; Kc=0; % If x0 is given as 2nd input argument
   else x=[zeros(1,m-1) 1]; % Initial state of the shift register
  end
elseif nargin<2, Kc=0; x=[zeros(1,m-1) 1];
end
x=x(end:-1:1); N=2^m-1;
if Kc==0 % Feedback shift register
  g=fliplr(g); for i=1:N, c(i)=x(1); x=[x(2:m) rem(g*x',2)]; end
 elseif Kc<0
  g=fliplr(g); c=x;
  for i=m+1:N, x=[x(2:m) rem(g*x',2)]; c(i)=x(m); end
 else % High-speed feedback shift generator
  c=x;
  for i=m+1:N, c(i)=x(1); x=[rem(x(2:m)+g(1:m-1)*x(1),2) x(1)]; end
end