gold_2.zip_gold_gold序列的另一种实现

function [g,G,l,OG] = gold (pu,pv,shift,varargin) % GOLD Gold codes. % [g,G,l,OG] = GOLD(PU,PV,SHIFT) generates a binary Gold code g and Gold % code sequence G from two binary linear feedback shift registers. l % gives the length of the code. OG is an orthogonal Gold code % sequence. % The shift registers are described by the preferred polynomials PU and % PV. PU and PV must be vectors of the same length. SHIFT is an % integer scalar which specifies the offset of the singular Gold code % from the initial time. % % GOLD(..,RU,RV) generates Gold codes given RU and RV, the initial % states of the U and V shift registers. RU and RV must be vectors of % the same length as the preferred polynomials. % % Note that a {0,1}-valued binary sequence corresponds to a % cos(pi*{0,1}) = {1,-1}-transmitted sequence. % % The following table provides a short list of preferred pairs % (a trailing binary 1 is always assumed): % n N Preferred Polynomial 1 Preferred Polynomial 2 % -- ---- ---------------------- ---------------------- % 5 31 [ 5 2 ] = [1 0 0 1 0] [ 5 4 3 2 ] = [1 1 1 1 0] % 6 63 [ 6 1 ] = [1 0 0 0 0 1] [ 6 5 2 1 ] = [1 1 0 0 1 1] % 7 127 [ 7 3 ] = ... [ 7 3 2 1 ] = ... % 9 511 [ 9 4 ] = ... [ 9 6 4 3 ] = ... % 10 1023 [ 10 3 ] = ... [ 10 8 3 2 ] = ... % 11 2047 [ 11 2 ] = ... [ 11 8 5 2 ] = ... % See http://www.cs.umbc.edu/~lomonaco/f97/442/Peterson_Table.html for a % table of irreducible polynomials over GF(2). % % Given a polynomial representation in octal, the polynomial in binary % is de2bi(oct2dec(x),'left-msb'). % For example, octal [23] is [1 0 0 1 1] in binary. Since a trailing % binary 1 is always assumed, the preferred polynomial is [1 0 0 1]. % Joe Henning - Fall 2006 g = 0; G = 0; l = 0; OG = 0; % error(nargchk(3,5,nargin)); narginchk(3,5); [ru,rv,n,msg] = parseinput(pu,pv,shift,varargin{:}); error(msg); l = 2^n - 1; if mod(n,2) == 1 % odd t = 1 + 2^((n+1)/2); else % even t = 1 + 2^((n+2)/2); end %fprintf('Cross-correlation can take the values: %.3f, %.3f, %.3f.\n',-t,-1,t-2); u = zeros(1,l); for k = 1:l temp = mod( fliplr(pu)*ru.', 2 ); u(k) = ru(n); ru = [temp ru(1:n-1)]; end v = zeros(1,l); for k = 1:l temp = mod( fliplr(pv)*rv.', 2 ); v(k) = rv(n); rv = [temp rv(1:n-1)]; end shifted_v = circshift(v,[0 -shift]);%向做移动shift位 g = mod( u + shifted_v, 2 ); G = [u; v]; %G = []; for k = 0:l-1 shifted_v = circshift(v,[0 -k]); G = [G; mod( u + shifted_v, 2 )]; end OG = [G zeros(size(G,1),1)]; %-------------------------------------------------------------------------- function [ru,rv,n,msg] = parseinput(pu,pv,shift,varargin) % Parse the input and determine optional parameters: % % Outputs: % ru - initial values of register u % rv - initial values of register v % n - number of register bits % msg - possible error message % Set some defaults: ru = 0; rv = 0; n = 0; msg = ''; [mu,nu] = size(pu); if ( (mu ~= 1 & nu ~= 1) | (mu == 1 & nu == 1) ) msg = 'pu must be a vector.'; return end [mv,nv] = size(pv); if ( (mv ~= 1 & nv ~= 1) | (mv == 1 & nv == 1) ) msg = 'pv must be a vector.'; return; end if ( length(pu) ~= length(pv) ) msg = 'pu and pv must be of the same length.'; return end n = length(pu); if ~isnumeric(shift) msg = 'shift must be an integer.'; return end % Assume no register values until proven otherwise found_ru = 0; found_rv = 0; ru = ones(1,n); rv = ones(1,n); switch nargin case 4 % is ru if isnumeric(varargin{1}) if length(varargin{1}) == n found_ru = 1; ru = varargin{1}; end end case 5 % is ru and rv if isnumeric(varargin{1}) if length(varargin{1}) == n found_ru = 1; ru = varargin{1}; end end if isnumeric(varargin{2}) if length(varargin{2}) == n found_rv = 1; rv = varargin{2}; end end end if ~found_ru fprintf('Using default ru.\n'); end if ~found_rv fprintf('Using default rv.\n'); end