I estimate my desired vector with the given code but can't when.....
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I want to estimate a desired vector u=[1 2 30 70] defined on the dafault line 2 inside "main.m". For this I use a function fobj.m and an algorithm WHO.m. Both of them are attached here along with the main.m. The extra two m-files namely exchange and initialization are the support files for WHO.m.
When I run the main.m, I get the result in the command window like my desired vector u along with the fitness each time I run the main.m. But when I add some Noise of 0dB, and then I run the main.m, then I get a result which is very far from the desired vector u. What to do to get the result nearly identical to the desired vector u like before?
Note:- When I want to add Noise, I comment my numbered lines which are as follows: Line 2 inside main.m, line 4 inside fobj.m and uncomment these numbered lines: line 3 inside main.m, line 5 inside fobj.m, line 6 inside fobj.m.
Now when I run the main again, I get a result in the command window which is vey far from my desired vector. What to do to reduce the effect of added Noise so that I get nearly the identical result like u before?
clear;clc
u=[1 2 30 50]; % My desired vector
Noise=0;%-------------------------------------------------->(1)
N=30; % Number of search agents
Max_iter=500; % Maximum number of iterations
lb=zeros(1,4);ub=[5 5 90 90];dim=length(u);
%[Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,@(b)fobj(b,u));%-------->(2)
[Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,@(b)fobj(b,u,Noise));%-->(3)
estimatedResult=gBest
fmin=gBestScore
function [Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,fobj)
if size(ub,1)==1
ub=ones(1,dim).*ub;
lb=ones(1,dim).*lb;
end
PS=0.2; % Stallions Percentage
PC=0.13; % Crossover Percentage
NStallion=ceil(PS*N); % number Stallion
Nfoal=N-NStallion;
Convergence_curve = zeros(1,Max_iter);
gBest=zeros(1,dim);
gBestScore=inf;
%create initial population
empty.pos=[];
empty.cost=[];
group=repmat(empty,Nfoal,1);
for i=1:Nfoal
group(i).pos=lb+rand(1,dim).*(ub-lb);
group(i).cost=fobj(group(i).pos);
end
Stallion=repmat(empty,NStallion,1);
for i=1:NStallion
Stallion(i).pos=lb+rand(1,dim).*(ub-lb);
Stallion(i).cost=fobj(Stallion(i).pos);
end
ngroup=length(group);
a=randperm(ngroup);
group=group(a);
i=0;
k=1;
for j=1:ngroup
i=i+1;
Stallion(i).group(k)=group(j);
if i==NStallion
i=0;
k=k+1;
end
end
Stallion=exchange(Stallion);
[value,index]=min([Stallion.cost]);
WH=Stallion(index); % global
gBest=WH.pos;
gBestScore=WH.cost;
Convergence_curve(1)=WH.cost;
l=2; % Loop counter
while l<Max_iter+1
TDR=1-l*((1)/Max_iter);
for i=1:NStallion
ngroup=length(Stallion(i).group);
[~,index]=sort([Stallion(i).group.cost]);
Stallion(i).group=Stallion(i).group(index);
for j=1:ngroup
if rand>PC
z=rand(1,dim)<TDR;
r1=rand;
r2=rand(1,dim);
idx=(z==0);
r3=r1.*idx+r2.*~idx;
rr=-2+4*r3;
Stallion(i).group(j).pos= 2*r3.*cos(2*pi*rr).*(Stallion(i).pos-Stallion(i).group(j).pos)+(Stallion(i).pos);
else
A=randperm(NStallion);
A(A==i)=[];
a=A(1);
c=A(2);
% B=randperm(ngroup);
% BB=randperm(ngroup);
% b1=B(1);b2=BB(1);
x1=Stallion(c).group(end).pos;
x2=Stallion(a).group(end).pos;
y1=(x1+x2)/2; % Crossover
Stallion(i).group(j).pos=y1;
end
Stallion(i).group(j).pos=min(Stallion(i).group(j).pos,ub);
Stallion(i).group(j).pos=max(Stallion(i).group(j).pos,lb);
Stallion(i).group(j).cost=fobj(Stallion(i).group(j).pos);
end
% end
%
% for i=1:NStallion
R=rand;
% z=rand(1,dim)<TDR;
% r1=rand;
% r2=rand(1,dim);
% idx=(z==0);
% r3=r1.*idx+r2.*~idx;
% rr=-2+4*r3;
if R<0.5
k= 2*r3.*cos(2*pi*rr).*(WH.pos-(Stallion(i).pos))+WH.pos;
else
k= 2*r3.*cos(2*pi*rr).*(WH.pos-(Stallion(i).pos))-WH.pos;
end
k=min(k,ub);
k=max(k,lb);
fk=fobj(k);
if fk<Stallion(i).cost
Stallion(i).pos =k;
Stallion(i).cost=fk;
end
end
Stallion=exchange(Stallion);
[value,index]=min([Stallion.cost]);
if value<WH.cost
WH=Stallion(index);
end
gBest=WH.pos;
gBestScore=WH.cost;
Convergence_curve(l)=WH.cost;
l = l + 1;
end
end
function [X]=initialization(N,dim,up,down)
if size(up,1)==1
X=rand(N,dim).*(up-down)+down;
end
if size(up,1)>1
for i=1:dim
high=up(i);low=down(i);
X(:,i)=rand(1,N).*(high-low)+low;
end
end
end
function Stallion=exchange(Stallion)
nStallion=length(Stallion);
for i=1:nStallion
[value,index]=min([Stallion(i).group.cost]);
if value<Stallion(i).cost
bestgroup=Stallion(i).group(index);
Stallion(i).group(index).pos=Stallion(i).pos;
Stallion(i).group(index).cost=Stallion(i).cost;
Stallion(i).pos=bestgroup.pos;
Stallion(i).cost=bestgroup.cost;
end
end
end
%function e=fobj(b,u)%------------------------->(4)
function e=fobj(b,u,Noise)%----------------->(5)
[R,C]=size(b);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
xo1 = sum(1*exp(1i*((k).*(-pi/2).*sind(u(P+i))+((k).^2.*pi./(16*u(i))).*cosd(u(P+i)).^2)),2);
xo=awgn(xo1,Noise);% Add Noise %----------------->(6)
norm(xo1-xo)
xe = sum(1*exp(1i*((k).*(-pi/2).*sind(b(P+i))+((k).^2.*pi./(16*b(i))).*cosd(b(P+i)).^2)),2);
%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
e=norm(xo-xe).^2/(M);
end
Accepted Answer
As you can see, the difference between xo and the noise-added xo is quite large. Thus you can expect that the difference in the solutions will be quite large as well.
Further, your problem makes no sense for me because you change the added noise in each call of fobj. Thus you solve a problem with a stochastically varying objective function. This is stochastic optimization, not deterministic optimization. Your solution method is not suited to deal with that situation.
Maybe the b you get is best suited to approximate the mean of the stochastic xo values you try to fit. Test it.
23 Comments
Sadiq
on 26 Dec 2023
awgn(xo,0) doesn't return xo.
Did you read the documentation for awgn
?
This might also be of interest:
rng("default")
x = 5;
x + randn()
rng("default")
awgn(x,0)
Sadiq
on 27 Dec 2023
Sadiq
on 27 Dec 2023
I ran your given piece of code and it gives the same result as you have shown but I didn't understand what do you mean by this i.e., what do you want to convey to me?
I want to tell you two things:
- : awgn(x,0) is the same as x + randn and
- : if you add the command "rng("default")" before calling awgn, you will always get the same noise vector.
for i = 1:5
rng("default")
awgn(1,0)
end
If you want to always use the same noise vector, don't generate it in fobj, but generate it once before calling WHO.m and pass it as an additional parameter to fobj.
But if you add a (constant) noise vector n to xo, the estimated values for b to best approximate xo + n will no longer be the same as those for xo. So b = u will no longer hold.
Sadiq
on 27 Dec 2023
As I said, if you add constant noise to xo, the result of your computations will no longer be b = u (or some permuted version of u), but something different.
As you can see from the code below, gBestScore for the determined vector b = gBest is smaller than the one you get for b = u.
Maybe if you computed b for many different noise vectors xo and the solution for b would be unique in each of these cases (what it isn't in your case), the mean of the solutions for b would come out to be u.
clear;clc
u=[1 2 30 50]; % My desired vector
Noise=0;%-------------------------------------------------->(1)
[R,C]=size(u);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
xo = sum(1*exp(1i*((k).*(-pi/2).*sind(u(P+i))+((k).^2.*pi./(16*u(i))).*cosd(u(P+i)).^2)),2);
xo=awgn(xo,Noise);% Add Noise %----------------->(6)
N=30; % Number of search agents
Max_iter=500; % Maximum number of iterations
lb=zeros(1,4);ub=[5 5 90 90];dim=length(u);
%[Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,@(b)fobj(b,u));%-------->(2)
[Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,@(b)fobj(b,u,xo));%-->(3)
estimatedResult=gBest
fmin=gBestScore
fobj(u,u,xo)
fobj(estimatedResult,u,xo)
function [Convergence_curve,gBest,gBestScore]=WHO(N,Max_iter,lb,ub,dim,fobj)
if size(ub,1)==1
ub=ones(1,dim).*ub;
lb=ones(1,dim).*lb;
end
PS=0.2; % Stallions Percentage
PC=0.13; % Crossover Percentage
NStallion=ceil(PS*N); % number Stallion
Nfoal=N-NStallion;
Convergence_curve = zeros(1,Max_iter);
gBest=zeros(1,dim);
gBestScore=inf;
%create initial population
empty.pos=[];
empty.cost=[];
group=repmat(empty,Nfoal,1);
for i=1:Nfoal
group(i).pos=lb+rand(1,dim).*(ub-lb);
group(i).cost=fobj(group(i).pos);
end
Stallion=repmat(empty,NStallion,1);
for i=1:NStallion
Stallion(i).pos=lb+rand(1,dim).*(ub-lb);
Stallion(i).cost=fobj(Stallion(i).pos);
end
ngroup=length(group);
a=randperm(ngroup);
group=group(a);
i=0;
k=1;
for j=1:ngroup
i=i+1;
Stallion(i).group(k)=group(j);
if i==NStallion
i=0;
k=k+1;
end
end
Stallion=exchange(Stallion);
[value,index]=min([Stallion.cost]);
WH=Stallion(index); % global
gBest=WH.pos;
gBestScore=WH.cost;
Convergence_curve(1)=WH.cost;
l=2; % Loop counter
while l<Max_iter+1
TDR=1-l*((1)/Max_iter);
for i=1:NStallion
ngroup=length(Stallion(i).group);
[~,index]=sort([Stallion(i).group.cost]);
Stallion(i).group=Stallion(i).group(index);
for j=1:ngroup
if rand>PC
z=rand(1,dim)<TDR;
r1=rand;
r2=rand(1,dim);
idx=(z==0);
r3=r1.*idx+r2.*~idx;
rr=-2+4*r3;
Stallion(i).group(j).pos= 2*r3.*cos(2*pi*rr).*(Stallion(i).pos-Stallion(i).group(j).pos)+(Stallion(i).pos);
else
A=randperm(NStallion);
A(A==i)=[];
a=A(1);
c=A(2);
% B=randperm(ngroup);
% BB=randperm(ngroup);
% b1=B(1);b2=BB(1);
x1=Stallion(c).group(end).pos;
x2=Stallion(a).group(end).pos;
y1=(x1+x2)/2; % Crossover
Stallion(i).group(j).pos=y1;
end
Stallion(i).group(j).pos=min(Stallion(i).group(j).pos,ub);
Stallion(i).group(j).pos=max(Stallion(i).group(j).pos,lb);
Stallion(i).group(j).cost=fobj(Stallion(i).group(j).pos);
end
% end
%
% for i=1:NStallion
R=rand;
% z=rand(1,dim)<TDR;
% r1=rand;
% r2=rand(1,dim);
% idx=(z==0);
% r3=r1.*idx+r2.*~idx;
% rr=-2+4*r3;
if R<0.5
k= 2*r3.*cos(2*pi*rr).*(WH.pos-(Stallion(i).pos))+WH.pos;
else
k= 2*r3.*cos(2*pi*rr).*(WH.pos-(Stallion(i).pos))-WH.pos;
end
k=min(k,ub);
k=max(k,lb);
fk=fobj(k);
if fk<Stallion(i).cost
Stallion(i).pos =k;
Stallion(i).cost=fk;
end
end
Stallion=exchange(Stallion);
[value,index]=min([Stallion.cost]);
if value<WH.cost
WH=Stallion(index);
end
gBest=WH.pos;
gBestScore=WH.cost;
Convergence_curve(l)=WH.cost;
l = l + 1;
end
end
function [X]=initialization(N,dim,up,down)
if size(up,1)==1
X=rand(N,dim).*(up-down)+down;
end
if size(up,1)>1
for i=1:dim
high=up(i);low=down(i);
X(:,i)=rand(1,N).*(high-low)+low;
end
end
end
function Stallion=exchange(Stallion)
nStallion=length(Stallion);
for i=1:nStallion
[value,index]=min([Stallion(i).group.cost]);
if value<Stallion(i).cost
bestgroup=Stallion(i).group(index);
Stallion(i).group(index).pos=Stallion(i).pos;
Stallion(i).group(index).cost=Stallion(i).cost;
Stallion(i).pos=bestgroup.pos;
Stallion(i).cost=bestgroup.cost;
end
end
end
%function e=fobj(b,u)%------------------------->(4)
function e=fobj(b,u,xo)%----------------->(5)
[R,C]=size(b);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
%xo1 = sum(1*exp(1i*((k).*(-pi/2).*sind(u(P+i))+((k).^2.*pi./(16*u(i))).*cosd(u(P+i)).^2)),2);
%xo=awgn(xo1,Noise);% Add Noise %----------------->(6)
xe = sum(1*exp(1i*((k).*(-pi/2).*sind(b(P+i))+((k).^2.*pi./(16*b(i))).*cosd(b(P+i)).^2)),2);
%%%%%%%%%%%%%%%%%%
% MSE
%%%%%%%%%%%%%%%%%%
e=norm(xo-xe).^2/(M);
end
Sadiq
on 28 Dec 2023
Torsten
on 28 Dec 2023
Maybe it's time to explain what you are trying to do in the end by stating your optimization problem in a mathematical way.
Sadiq
on 29 Dec 2023
Thanks a lot dear Torsten for your kind response.
In my work I will choose a vector of any diemsnion. May be 1x4, 1x6 or 1x8. Then I have a mathematical equation for that which is as below: In this equation xl is the received signal on the lth element, "Si" is kept as 1, "theta" is the angle of a target and "ri" is the distance of that target from the element. "Eta" is the noise. For this we take all these parameters in a vector u and assume their values and write those assumed values inside a desired vector u in the format u=[R1 R2 Theta1 Theta2]. So if we assume two targets and lets say that distance of the 1st target is 1 and the distance of the 2nd target is 2 kilometers and the angle of the 1st target is 30 and that of the 2nd target is 70 degrees, then we assume our desired vector like this: u=[1 2 30 70]; Then we put these values in the below equation and find out the received signal xl which we call as our observed signal xo. Then we make our function of this problem which is based on mean sqaure error. Then we will run an algorithm WHO that will generate random solutions and will put those solutions one by one in our problem function and will note corresponding fitness of that solution vector. The solution vector whose fitness has minimm value will be the best solution and will be nearly equal to our desired vector u. This is all what I am going to do. Then most of the time we add noise also to our observed signal xo. When we don't add the noise then we get very very good estimated vector but when we add noise, say for example, of 0 level, then we normally don't get good solution vector in estimation. For that I have posted this question.
You helped me a lot and you have helped me before also and I have accepted your answer which I felt were correct. But this time I am not satisfied though you tried your bet to guide me but since my object is not met that is why I am changing my method. That 1st I will run the algorithm WHO and will find the estimated vetor and then i will give that estimated vector to the local search optimizer SQP i.e., Sequential Quadratic Programming which will refine my estimated solution vector and will make it like my desired vector u. But I don;t know how to invoke the SQP and how to give it my estimated solution vector and how to run that sqp 100 times to get 100 fitness values and 100 different estimated refined solutions and how to choose the best among those refined solutions and the corresponding fitness? In this regard, I want your help.
I guess in the real application, u is unknown and you only have the noisy signals xl from which you want to estimate the u-vector ? And at the moment, you only add noise to the signals with a known u-vector to test your solution method ?
The way to proceed is as follows:
You are given parameters u and "measurement data" xl. From the given parameter vector, you compute the sum expression xo as "simulated data". You pass the error vectors real(xl-xo) and imag(xl-xo) to the solver. Then the parameters vector u will be determined best to fit your measurement data.
Use "lsqcurvefit" for this purpose and pass the real and imaginary parts of the above sum which will be compared with the real and imaginary parts of your measurement vector.
Sadiq
on 29 Dec 2023
I think you misunderstand the whole procedure.
You have a function that - given its parameters - your measured data should follow in the ideal case. This is the sum-term of your formula (without the noise term). Call this xsim which is a function of the parameter vector: xsim = xsim(u)
But measurements contain noise. So to determine the parameter vector u, you only have available data xmeasured (the sum term of your formula with the noise term).
Now you want to determine the parameter vector p such that xsim(p) gets as close as possible to xmeasured.
Use "lsqcurvefit" to accomplish this - it's the usual parameter fitting tool.
Sadiq
on 30 Dec 2023
Given u, you will have to generate a lot of data for xl with different values for the noise eta_l. The (artificial) ydata for lsqcurvefit are then
ydata = [real(xl),imag(xl)]
The xdata don't exist in your case, but that doesn't matter. Just pass an arbitrary array to lsqcurvefit that has the same length as "ydata".
In the function of "lsqcurvefit", you are given a parameter vector that you can interpret as a suggestion for u. With this parameter vector, you compute xl without the noise contribution eta_l. Name it xo. From xo, you compute the real and imaginary part, concatenate them as you did for ydata and return it to the optimizer.
This is the process to recover u from noisy test data xl.
If this is not what you want, I cannot help you any longer because in this case, I cannot understand your problem.
Sadiq
on 31 Dec 2023
Thanks dear Torsen for your consistent help. I tried to do as you suggested. The following is my work in the light of your discussion and the above site,I quoted.
clear all;clc
u=[1 2 30 50];
[R,C]=size(u);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
xl = sum(1*exp(1i*((k).*(-pi/2).*sind(u(P+i))+((k).^2.*pi./(16*u(i))).*cosd(u(P+i)).^2)),2);%------(1)
xo=awgn(xl,0);
x1=awgn(xl,1);
x2=awgn(xl,2);
x3=awgn(xl,3);
x4=awgn(xl,4);
%-------------------------
% Real data
xlr=[real(xo) real(x1) real(x2) real(x3) real(x4)];
xlr=xlr(:);
% Imaginary data
xli=[imag(xo) imag(x1) imag(x2) imag(x3) imag(x4)];
xli=xli(:);
%ydata = [real(xlr),imag(xl)]
ydata=[xlr,xli];
xdata=rand(40,2);% Artificial data
fun = @(x,xdata)x(1)*exp(x(2)*xdata);
x0 = [100,-1];
x = lsqcurvefit(fun,x0,xdata,ydata)
% Plot the data
times = linspace(xdata(1),xdata(end));
plot(xdata,ydata,'ko',times,fun(x,times),'b-')
legend('Data','Fitted exponential')
title('Data and Fitted Curve')
And the plot is in the attachment. But I don't see any fitting in this or may be I don't understand it.
clear all;clc
u=[1 2 30 50];
[R,C]=size(u);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
xl = sum(1*exp(1i*((k).*(-pi/2).*sind(u(P+i))+((k).^2.*pi./(16*u(i))).*cosd(u(P+i)).^2)),2);%------(1)
xo=awgn(xl,0);
x1=awgn(xl,1);
x2=awgn(xl,2);
x3=awgn(xl,3);
x4=awgn(xl,4);
%-------------------------
% Real data
xlr=[real(xo) real(x1) real(x2) real(x3) real(x4)];
xlr=xlr(:);
% Imaginary data
xli=[imag(xo) imag(x1) imag(x2) imag(x3) imag(x4)];
xli=xli(:);
ydata=[xlr,xli];
xdata=rand(40,2);% Arbitrary data (same size as ydata)
x0 = [u(1)+0.1;u(2)-0.1;u(3)+5;u(4)-7]; % Change the expected result u a bit
x = lsqcurvefit(@fun,x0,xdata,ydata)
x = x.';
xlfit = sum(1*exp(1i*((k).*(-pi/2).*sind(x(P+i))+((k).^2.*pi./(16*x(i))).*cosd(x(P+i)).^2)),2);
% Plot the data
figure(1)
hold on
plot(1:numel(xo),real(xo),'o')
plot(1:numel(x1),real(x1),'o')
plot(1:numel(x2),real(x2),'o')
plot(1:numel(x3),real(x3),'o')
plot(1:numel(x4),real(x4),'o')
plot(1:numel(xl),real(xl),'*')
plot(1:numel(xlfit),real(xlfit),'+')
hold off
figure(2)
hold on
plot(1:numel(xo),imag(xo),'o')
plot(1:numel(x1),imag(x1),'o')
plot(1:numel(x2),imag(x2),'o')
plot(1:numel(x3),imag(x3),'o')
plot(1:numel(x4),imag(x4),'o')
plot(1:numel(xl),imag(xl),'*')
plot(1:numel(xlfit),imag(xlfit),'+')
hold off
function res = fun(x,xdata)
x = x.';
[R,C]=size(x);
P=C/2;
M=2*C;
k = (-(M/2-1):M/2).';
i = (1:P);
xl = sum(1*exp(1i*((k).*(-pi/2).*sind(x(P+i))+((k).^2.*pi./(16*x(i))).*cosd(x(P+i)).^2)),2);
res = [repmat(real(xl),5,1),repmat(imag(xl),5,1)];
end
Sadiq
on 1 Jan 2024
2- Why have you used the following statement?
x0 = [u(1)+0.1;u(2)-0.1;u(3)+5;u(4)-7]; % Change the expected result u a bit
If instead of the above, I give any random xo, I mean like the following:
x0=rand(1,4);
will it be ok?
x0 is arbitrary. If "lsqcurvefit" converges to a reaonable solution, every vector is ok.
3- The output of "lsqcurvefit()" is x which is a 1 x 4 vector, is this the estimated result?
x is the estimated u.
4- Why have you put this vector x back in my problem equation? Do you want to calculate the corresponding fit data which is the same as above ydata=[xlr,xli];?
Which line of code are you referring to ? If you mean in the script part after the call to lsqcurvefit, it's to plot ydata against the y-values obtained with the x from the solution.
5- Why have you plotted the noisy data separately like the following:
plot(1:numel(xo),real(xo),'o')
plot(1:numel(x1),real(x1),'o')
plot(1:numel(x2),real(x2),'o')
plot(1:numel(x3),real(x3),'o')
plot(1:numel(x4),real(x4),'o')
plot(1:numel(xl),real(xl),'*')
Could you not put it in a single variable and then plot that?
You mean
plot(1:numel(xo),[real(xo);real(x1);real(x2);real(x3);real(x4);real(xl)])
? Yes, I could have done this. But why don't you try it yourself ?
6- Further, why have you used the x1 data twice? i.e., I mean this "plot(1:numel(xl),real(xl),'*') "
It's an "l", not a "1". It's the y-data you get to approximate xo,x1,x2,x3 and x4 simultaneously with the x from the solution of lsqcurvefit. Thus real(xl) and imag(xl) should be approximately "in the middle" of the noisy real and imaginary parts of xo,x1,x2,x3 and x4.
7- The same question is for the imiginary plot data also?
Yes, I could have done this. But why don't you try it yourself ?
8- Can't we combine the real and imiganiary data in a single variable and plot a single graph of the original data and then plot the fit data over it?
How can we combine them ? By plotting abs(x0),...,abs(x4), abs(xl) instead of the real and imaginary parts separately ? Yes, it's possible. Try it.
9-If "x" is the estimated vector by lsqcurvefit(), then where is the fitness value?
10-Also how will we find the corresponding execution time?
tic before, toc after the call to lsqcurvefit.
But I don't know why I have to explain your own problem to you - assuming I understood what you are trying to do (what I'm still not certain of). Especially the dependence of your target function on l in the sum expression for x_l in your graphics is confusing and is different from what you coded.
Sadiq
on 2 Jan 2024
"lsqcurvefit" is a suitable solver for your problem, and the results look fine to me.
The plot of the fitted ydata against the data you prescribed after using "awgn" looks ok to me: The fitted y-data (those with a "+") are approximately in the middle of the ones you wanted to fit (those with a "o"). Look at the spreading of your xo,x1,x2,x3 and x4: you cannot expect x to be identical to u if you add as much noise as you did.
"lsqcurvefit" is not "ga". The variable I named "fitness_value" is the norm of the error between the data you wanted to fit and the fit result. The error norm for x is smaller than the error norm for your prescribed u - thus "lsqcurvefit" produced a better solution. This can be seen if you add the lines
norm(ydata-fun(x,xdata))
norm(ydata-fun(u.',xdata))
after the call to "lsqcurvefit".
If you want to try another solver or have further questions, open a new question in the forum.
Sadiq
on 2 Jan 2024
Torsten
on 2 Jan 2024
Though my objective is not met,
As I tried to explain: it's impossible to meet your objective because u as a solution gives a worse fit than the x returned by "lsqcurvefit".
If you want u as solution, you must use many more of your noisy data xi such that the condition mean(noise) = 0 is exactly met.
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