Optimisation problem in matlab

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wallflower
wallflower 2021-6-21
Hello guys,
So I have this scatte plot in the form of a measured impedance according to frequency Z(f), and I have an RL ladder circuit seen in the following figure
.
I need some help in writing an optimisation problem that will allow me to get the values of the parameters R_i L_i that will eventually give an equivalent impedance of the RL ladder similar to the one measured- Z(f) -.
Thanks in advance.
Wallflower
  2 个评论
Rik
Rik 2021-6-21
This looks like a homework assignment. You can find guidelines for posting homework on this forum here. If you have trouble with Matlab basics you may consider doing the Onramp tutorial (which is provided for free by Mathworks). If your main issue is with understanding the underlying concept, you may consider re-reading the material you teacher provided and ask them for further clarification.
wallflower
wallflower 2021-6-21
It is not a homework assignment. I am trying to reproduce a methology I've seen in a scientific paper.

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回答(1 个)

Sergey Kasyanov
Sergey Kasyanov 2021-6-21
编辑:Torsten 2022-7-12
Hello!
Try that:
% tune only next 2 lines
L = 5;% steps in ladder
Z_max = 1e2;% max Z for one R or Xl
% goal Z(w)
% w0 - array with measured angular frequences
% Z0 - array with measured impedances
% R0 - Rs_1.txt
% L0 - 2*Self_Energy_1.txt
w0 = (2.*pi.*[[1e-3,50,500],1e3:2e3:500e3])';
% Start inserted to make code work
R0 = rand(size(w0));
L0 = rand(size(w0));
% End inserted to make code work
Z0 = R0(:,1) + 1i*w0.*L0(:,1);
w0 = reshape(w0, 1, []);
Z0 = reshape(Z0, 1, []);
%% calculate z(w) for ladder
R = sym('R', [1,L+1]);
L = sym('L', [1,length(R)]);
w = sym('w');
par = @(a, b) 1/(1/a + 1/b);
h1 = @(a, i) R(i) + par(a, 1i*w*L(i));
Z = R(end);
for i = (length(R)-1):-1:1
Z = h1(Z, i);
end
% get lambda function for optimization
fun1 = matlabFunction(Z);
% do not look at that code
s = 'fun2 = @(vals, w) fun1(';
for i = 1:(length(R)*2 - 1)
s = [s, sprintf('vals(%i), ', i)];
end
s = [s, 'w);'];
eval(s);
% create optimization function
fun3 = @(vals) sum(abs(fun2(vals, w0) - Z0).^2);
% number of variables
N = length(symvar(Z)) - 1;
%the better way is to use more powerfull algorithms such as genetic algoritms which can stands
%against a huge amount of local min. There are a lot of algorithms that
%realized in matlab toolboxes. Try!
options = optimoptions('ga', 'Display', 'iter', 'PopulationSize', N*200, 'MaxGenerations', N*200);
res = ga(fun3, N, [], [], [], [], zeros(1, N), Z_max*[ones(1,floor(N/2))/2/pi, ones(1, ceil(N/2))], [], options);
Single objective optimization: 11 Variable(s) Options: CreationFcn: @gacreationuniform CrossoverFcn: @crossoverscattered SelectionFcn: @selectionstochunif MutationFcn: @mutationadaptfeasible Best Mean Stall Generation Func-count f(x) f(x) Generations 1 4400 2.953e+14 2.953e+14 0 2 6490 2.953e+14 2.953e+14 0 3 8580 2.953e+14 2.953e+14 0 4 10670 2.953e+14 2.953e+14 0 5 12760 2.953e+14 2.953e+14 0 6 14850 2.953e+14 2.953e+14 0 7 16940 2.953e+14 2.953e+14 0 8 19030 2.953e+14 2.953e+14 0 9 21120 2.953e+14 2.953e+14 0 10 23210 2.953e+14 2.953e+14 0 11 25300 2.953e+14 2.953e+14 0 12 27390 2.953e+14 2.953e+14 0 13 29480 2.953e+14 2.953e+14 0 14 31570 2.953e+14 2.953e+14 0 15 33660 2.953e+14 2.953e+14 0 16 35750 2.953e+14 2.953e+14 0 17 37840 2.953e+14 2.953e+14 0 18 39930 2.953e+14 2.953e+14 0 19 42020 2.953e+14 2.953e+14 0 20 44110 2.953e+14 2.953e+14 0 21 46200 2.953e+14 2.953e+14 0 22 48290 2.953e+14 2.953e+14 0 23 50380 2.953e+14 2.953e+14 0 24 52470 2.953e+14 2.953e+14 1 25 54560 2.953e+14 2.953e+14 0 26 56650 2.953e+14 2.953e+14 0 27 58740 2.953e+14 2.953e+14 0 28 60830 2.953e+14 2.953e+14 0 29 62920 2.953e+14 2.953e+14 0 30 65010 2.953e+14 2.953e+14 1 Best Mean Stall Generation Func-count f(x) f(x) Generations 31 67100 2.953e+14 2.953e+14 0 32 69190 2.953e+14 2.953e+14 1 33 71280 2.953e+14 2.953e+14 2 34 73370 2.953e+14 2.953e+14 0 35 75460 2.953e+14 2.953e+14 0 36 77550 2.953e+14 2.953e+14 0 37 79640 2.953e+14 2.953e+14 0 38 81730 2.953e+14 2.953e+14 0 39 83820 2.953e+14 2.953e+14 1 40 85910 2.953e+14 2.953e+14 2 41 88000 2.953e+14 2.953e+14 3 42 90090 2.953e+14 2.953e+14 0 43 92180 2.953e+14 2.953e+14 0 44 94270 2.953e+14 2.953e+14 0 45 96360 2.953e+14 2.953e+14 0 46 98450 2.953e+14 2.953e+14 1 47 100540 2.953e+14 2.953e+14 2 48 102630 2.953e+14 2.953e+14 3 49 104720 2.953e+14 2.953e+14 4 50 106810 2.953e+14 2.953e+14 0 51 108900 2.953e+14 2.953e+14 0 52 110990 2.953e+14 2.953e+14 0 53 113080 2.953e+14 2.953e+14 1 54 115170 2.953e+14 2.953e+14 0 55 117260 2.953e+14 2.953e+14 1 Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
% first floor(N/2) of res - L(1), L(2), ...
% last ceil(N/2) of res - R(1), R(2), ...
%plot results
figure;plot(w0, abs(Z0), w0, abs(fun2(res,w0)));
figure;plot(w0, angle(Z0), w0, angle(fun2(res,w0)));
  14 个评论
Anes MESSADI
Anes MESSADI 2022-7-12
there are no errors, it works, but the fitting is not good, I used only the Rs_1.txt and Self_Energy_1.txt just to test the code.
Thank you for your help
Sergey Kasyanov
Sergey Kasyanov 2022-7-13
Try to use that code with your w0 and Z0. Also you need function ladder_z in the file.
% tune only next 2 lines
L = 10;% steps in ladder
Z_max = 1e2;% max Z for one R or Xl
% w0 - array with measured angular frequences
% Z0 - array with measured impedances
% create data for testing
w0 = 2 * pi * logspace(-3, 2, 10);
z0 = [rand(1, L);rand(1, L)];
Z0 = [];
for i = 1:length(w0)
Z0(i) = ladder_z(z0(1, :) + z0(2, :) * 1i * w0(i));
end
% main code
h1 = @(vals, w) ladder_z(vals(1:fix(length(vals) / 2)) + 1i * w * vals(fix(length(vals) / 2) + 1:end));
h2 = @(vals) arrayfun(@(iw) h1(vals, iw), w0);
f = @(vals) sum(abs(h2(vals) - Z0).^2);
options = optimoptions('ga', 'Display', 'iter', 'PopulationSize', L*200, 'MaxGenerations', L*200, 'UseParallel', true);
res = ga(f, L * 2, [], [], [], [], zeros(1, 2 * L), Z_max*[ones(1, L), ones(1, L) / 2 / pi], [], options);
% check
figure;plot(w0, abs(Z0), w0, abs(h2(res)));

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