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请大家帮我看一下下面的程序,这是小数据量法求最大Lyapunov指数的程序,我想问问里面那个delt_t是什么意思?怎么取?
function lambda_1=largest_lyapunov_exponent(data,N,m,tau,P)
%the function is used to calcultate largest lyapunov exponent with the
%mended algorithm,which put forward by lv jing hu.
%data:the time series
%N:the length of data
%m:enbedding dimention
%tau:time delay
%P:the mean period of the time series,calculated with FFT
%lambda_1:return the largest lyapunov exponent
%skyhawk
data=[
];
N=length(data);
P=38;
m=5;
tau=3;
delt_t=1;
Y=reconstitution(data,N,m,tau );%reconstitute state space
M=N-(m-1)*tau;%M is the number of embedded points in m-dimensional space
for j=1:M
d_max=1e+100;
for jj=1:M %寻找相空间中每个点的最近距离点,并记下
d_s=0; %该点下标
if abs(j-jj)>P %限制短暂分离
for i=1:m
d_s=d_s+(Y(i,j)-Y(i,jj))*(Y(i,j)-Y(i,jj));
d_min=d_max;
if d_s<d_min
d_min=d_s;
idx_j=jj;
end
end
end
end
% index(j)=idx_j;
max_i=min((M-j),(M-idx_j));%计算点j的最大演化时间步长i
for k=1:max_i %计算点j与其最近邻点在i个离散步后的距离
d_j_i=0;
for kk=1:m
d_j_i=d_j_i+(Y(kk,j+k)-Y(kk,idx_j+k))*(Y(kk,j+k)-Y(kk,idx_j+k));
d(k,j)=d_j_i;
end
end
end
%对每个演化时间步长i,求所有的j的lnd(i,j)平均
[l_i,l_j]=size(d);
for i=1:l_i
q=0;
y_s=0;
for j=1:l_j
if d(i,j)~=0
q=q+1;
y_s=y_s+log(d(i,j));
end
end
y(i)=y_s/(q*delt_t);
end
x=1:length(y);
y
pp=polyfit(x,y,1);
lambda_1=pp(1);
yp=polyval(pp,x);
plot(x,y,'-o',x,yp,'--') |
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