版权声明：转载请注释 https://blog.csdn.net/xinshuwei/article/details/84623397
function y = mls(n,flag)
%y = mls(n,{flag});
%
%Generates a Maximum Length Sequence of n bits by utilizing a
%linear feedback shift register with an XOR gate on the tap bits
%
%Function can accept bit lengths of between 2 and 24
%
%y is a vector of 1's & -1's that is (2^n)-1 in length.
%
%optional flag is:
%
% 1 for an initial sequence of all ones (repeatable)
% 0 for an initial sequence that is random (default)
%
%note: Because of the recursive nature of this process, it is not
%possible to completely vectorize this code (at least I don't know
%how to do it). As a result, longer bit lengths will take quite a
%long time to process, perhaps hours. If you figure out a way to
%vectorize the for loop, please let me know.
%
%reference:
% Davies, W.D.T. (June, July, August, 1966). Generation and
%properties of maximum-length sequences. Control, 302-4, 364-5,431-3.
%
%Spring 2001, Christopher Brown, [email protected]
switch n %assign taps which will yeild a maximum
case 2 %length sequence for a given bit length
taps=2; %I forget the reference I used, but theres
tap1=1; %a list of appropriate tap values in
tap2=2; %Vanderkooy, JAES, 42(4), 1994.
case 3
taps=2;
tap1=1;
tap2=3;
case 4
taps=2;
tap1=1;
tap2=4;
case 5
taps=2;
tap1=2;
tap2=5;
case 6
taps=2;
tap1=1;
tap2=6;
case 7
taps=2;
tap1=1;
tap2=7;
case 8
taps=4;
tap1=2;
tap2=3;
tap3=4;
tap4=8;
case 9
taps=2;
tap1=4;
tap2=9;
case 10
taps=2;
tap1=3;
tap2=10;
case 11
taps=2;
tap1=2;
tap2=11;
case 12
taps=4;
tap1=1;
tap2=4;
tap3=6;
tap4=12;
case 13
taps=4;
tap1=1;
tap2=3;
tap3=4;
tap4=13;
case 14
taps=4;
tap1=1;
tap2=3;
tap3=5;
tap4=14;
case 15
taps=2;
tap1=1;
tap2=15;
case 16
taps=4;
tap1=2;
tap2=3;
tap3=5;
tap4=16;
case 17
taps=2;
tap1=3;
tap2=17;
case 18
taps=2;
tap1=7;
tap2=18;
case 19
taps=4;
tap1=1;
tap2=2;
tap3=5;
tap4=19;
case 20
taps=2;
tap1=3;
tap2=20;
case 21
taps=2;
tap1=2;
tap2=21;
case 22
taps=2;
tap1=1;
tap2=22;
case 23
taps=2;
tap1=5;
tap2=23;
case 24
taps=4;
tap1=1;
tap2=3;
tap3=4;
tap4=24;
%case 25
% taps=2;
% tap1=3;
% tap2=25;
%case 26
% taps=4;
% tap1=1;
% tap2=7;
% tap3=8;
% tap4=26;
%case 27
% taps=4;
% tap1=1;
% tap2=7;
% tap3=8;
% tap4=27;
%case 28
% taps=2;
% tap1=3;
% tap2=28;
%case 29
% taps=2;
% tap1=2;
% tap2=29;
%case 30
% taps=4;
% tap1=1;
% tap2=15;
% tap3=16;
% tap4=30;
%case 31
% taps=2;
% tap1=3;
% tap2=31;
%case 32
% taps=4;
% tap1=1;
% tap2=27;
% tap3=28;
% tap4=32;
otherwise
disp(' ');
disp('input bits must be between 2 and 24');
return
end
if (nargin == 1)
flag = 0;
end
if flag == 1
abuff = ones(1,n);
else
% rand('state',sum(100*clock))
rng(sum(100*clock),'v5uniform')
while 1
abuff = round(rand(1,n));
%make sure not all bits are zero
if find(abuff==1)
break
end
end
end
for i = (2^n)-1:-1:1
xorbit = xor(abuff(tap1),abuff(tap2)); %feedback bit
if taps==4
xorbit2 = xor(abuff(tap3),abuff(tap4));%4 taps = 3 xor gates & 2 levels of logic
xorbit = xor(xorbit,xorbit2); %second logic level
end
abuff = [xorbit abuff(1:n-1)];
y(i) = (-2 .* xorbit) + 1; %yields one's and negative one's (0 -> 1; 1 -> -1)
end