How to find largest lyapunov for Mackey Glass data?

Hello.

I want to find largest Lyapunov for Mackey Glass data. How can I do that?

Answer 1

[Zahra kamkar]

Hello

I have another question. I used a Lyapunov Exponent Matlab code to find the largest Lyapunov result. The code seems to be correct but I don't know what should be the value of Maximum Lag in this code!

here is the code:

% close all;
clear all;

y= xlsread('MackeyGlass.xlsx');

warning off
tic

%___________________________Defining lags for y____________________________
%%Here is the my problem
maxlag=3;
y=y(:);
[~,nyc]=size(y);
yL=lagmatrix(y,1:maxlag);
y=y(maxlag+1:end,1);
yL=yL(maxlag+1:end,:);
[ryL cyL]=size(yL);
%_________________ Regressors Up to degree 3_______________________________

X1=yL;
num1=0;
X2ij=[];
for i=1:cyL
     for j=i:cyL
      X2ij=[X2ij yL(:,i).*yL(:,j)];
      Indexij(num1+1,1)=i;
      Indexij(num1+1,2)=j;
      num1=num1+1;
     end 
end

num2=0;
X3ijk=[];
for i=1:cyL
     for j=i:cyL
         for k=j:cyL
           X3ijk=[ X3ijk yL(:,i).*yL(:,j).*yL(:,k)];
           Indexijk(num2+1,1)=i;
           Indexijk(num2+1,2)=j;
           Indexijk(num2+1,3)=k;
           num2=num2+1;

           end 
       end
  end

X=[ones(ryL,1) X1 X2ij  X3ijk];

beta =inv (X'*X)*X'*y;
betaX1=beta(2:cyL+1);
betaX2ij=beta(cyL+2:cyL+1+num1);
betaX3ijk=beta(cyL+2+num1:cyL+1+num1+num2);

for i=1:cyL
    Inij1=find(Indexij(:,1)==i);
    Inij2=find(Indexij(:,2)==i);
    Inijk1=find(Indexijk(:,1)==i);
    Inijk2=find(Indexijk(:,2)==i);
    Inijk3=find(Indexijk(:,3)==i);
    Df1(:,i)=betaX1(i,1)*ones(ryL,1);

      Df210=zeros(ryL,1);
      for h21=1:length(Inij1)

          Df210=betaX2ij(Inij1(h21),1)*yL(:,Indexij(Inij1(h21),2))+Df210;

      end
      Df21(:,i)=Df210;

      Df220=zeros(ryL,1);
       for h22=1:length(Inij2)

          Df220=betaX2ij(Inij2(h22),1)*yL(:,Indexij(Inij2(h22),1))+Df220;

       end
       Df22(:,i)=Df220;

       Df310=zeros(ryL,1);
       for h31=1:length(Inijk1)

           Df310=betaX3ijk(Inijk1(h31),1)*yL(:,Indexijk(Inijk1(h31),2))...
               .*yL(:,Indexijk(Inijk1(h31),3))+Df310;

       end
        Df31(:,i)=Df310;

       Df320=zeros(ryL,1);
       for h32=1:length(Inijk2)

           Df320=betaX3ijk(Inijk2(h32),1)*yL(:,Indexijk(Inijk2(h32),1))...
               .*yL(:,Indexijk(Inijk2(h32),3))+Df320; 

       end
       Df32(:,i)=Df320;

       Df330=zeros(ryL,1);
       for h33=1:length(Inijk3)

           Df330=betaX3ijk(Inijk3(h33),1)*yL(:,Indexijk(Inijk3(h33),1))...
               .*yL(:,Indexijk(Inijk3(h33),2))+Df330;

       end
        Df33(:,i)=Df330;
  end

Df=Df1+Df21+Df22+Df31+Df32+Df33;

%QR decomposition

M=floor(ryL/50);
Lyap=[Df(1,:);eye(maxlag-1,maxlag)]-[Df(1,:);eye(maxlag-1,maxlag)];
for bl=1:50
LAMBDA=[Df(1,:);eye(maxlag-1,maxlag)]-[Df(1,:);eye(maxlag-1,maxlag)];

Q0=eye(maxlag);
for i=(bl-1)*M+1:bl*M;
    [Q,R]=qr([Df(i,:);eye(maxlag-1,maxlag)]*Q0);
    LAMBDA1=logm(R);
    LAMBDA=LAMBDA+real(LAMBDA1);
    Q0=Q;
end
Lyap=LAMBDA+Lyap;

end
Lyap=Lyap/(M*50);

t=toc
%_________________________________END______________________________________

[Zahra kamkar]

I know the lag is the value of delay in this code but I can't understand why the result of this code becomes N*N when I choose N as the lag's value????!!!

[Star Strider]

You probably have to experiment to see what lag gives you the best result. The references I provided could provide a clue.

In the first reference, they write: ‘In this paper, tau is chosen as the lag time at which the autocorrelation function of the time series falls to exp(-l) = 0.37.’ (They define tau as the time delay.)

[Zahra kamkar]

I've read this article but tau refers to the time series generation. In my code, I have generated the Mackey Glass Data as a chaotic time series, now I decide to use Lyapunov.

[Star Strider]

It is still a heuristic exercise. Use the delay that gives you the result you want. I haven’t analysed chaotic time series in a while, but I’ve not discovered any particular method for determining the best delay without simply trying different ones to see which worked best according to whatever criteria you set. This Scholarpedia article on the Lyapounov exponent may provide you with some guidance.

[alireza ghavami]

Hi dear ,

i just followed your conversation with ms.Zahra,

since Im preparing for my M.Sc thesis, is it possible to help me too?

i need to find the largest Lyapunov exponent for my EEG signal as well as other non-linear features.

could you please help me with the code?

Answer 2

[alireza ghavami]

Hi dear Zahra ,

i just found your problem by searching in google since im trying to find the way to calculate or find the largest Lyapunov exponent in my EEG signal,

could you please help me where can i find the solution?

i tried in youtube, aparat and also anywhere i thought i could find but no result!...

i consider it has long time ago you made your question but if its possible for you to help me, i will be grateful.

here's my gmail address: [e-mail]

thanks alot