Can't fit nonlinear filter equation by lsqcurvefit()

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Md Nure Alam Dipu
Md Nure Alam Dipu 2021-7-19
编辑： Walter Roberson 2021-7-20
I want to fit a passive RLC bandstop filter equation to the following data:
f = [5500 6000 6500 7000 7500 8000 8500 9000];
y = [0.998 0.997 0.994 0.988 0.947 0.255 0.956 0.987];
Where f is the frequency and y is the value of transfer function. Here is my code:
fun = @(p, x) abs(j.*(x.*p(2)-1./(x.*p(3)))./(p(1) + j.*(x.*p(2)-1./(x.*p(3))))); % transfer function of RLC bandstop
% x, p(1), p(2), p(3) correspond to w, R, L, C respectively
x = f(1):f(end); % for plotting purpose
x0 = [1 1e-3 1e-6] % starting points
lb = [1 1e-3 1e-6];ub = [100E3 100 1000E-6];
p_fit = lsqcurvefit(fun, x0, f, y, lb, ub) % p_fit is the vector of fitted parameters
semilogx(f, y, 'ro'); hold on;
semilogx(x, fun(p_fit, x), '--m'); grid on; % semilog plot of the transfer function
hold off;
I am not familiar with lsqcurvefit() parameter's starting points. Please, suggest any modification or any other method which could fit for any frequency range.
I have attached the image of actual plot and the transfer function.

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Walter Roberson
Walter Roberson 2021-7-20
编辑：Walter Roberson 2021-7-20
Ran in:
format long g
f = [5500 6000 6500 7000 7500 8000 8500 9000];
y = [0.998 0.997 0.994 0.988 0.947 0.255 0.956 0.987];
fun = @(p, x) abs(1i.*(x.*p(2)-1./(x.*p(3)))./(p(1) + 1i.*(x.*p(2)-1./(x.*p(3))))); % transfer function of RLC bandstop
% x, p(1), p(2), p(3) correspond to w, R, L, C respectively
x = f(1):f(end); % for plotting purpose
x0 = [1 1e-3 1e-6] % starting points
x0 = 1×3
1 0.001 1e-06
lb = [1 1e-3 1e-6];ub = [100E3 100 1000E-6];
ub = [1 0.02 2e-5];
x0 = [1 0.0038733109633344873895993294337111 4e-6]
x0 = 1×3
1 0.00387331096333449 4e-06
[p_fit, fval_fit] = lsqcurvefit(fun, x0, f, y, lb, ub) % p_fit is the vector of fitted parameters
Solver stopped prematurely. lsqcurvefit stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 2.000000e+02.
p_fit = 1×3
1 0.00342918710933376 4.51291192153049e-06
fval_fit =
0.000466077848839323
semilogx(f, y, 'ro', 'displayname', 'original'); hold on;
semilogx(x, fun(p_fit, x), '--m', 'displayname', 'lsq'); grid on; % semilog plot of the transfer function
E = @(p) sum((fun(p,f)-y).^2);
[Pfc, fval_fc] = fmincon(E, x0, [], [], [], [], lb, ub)
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
Pfc = 1×3
1 0.00387331096333449 4e-06
fval_fc =
0.000804295477866486
[Pfs, fval_fs] = fminsearch(E, x0)
Pfs = 1×3
1.20098480142405 0.00390552292365765 3.96072393734393e-06
fval_fs =
0.00041975861436785
semilogx(x, fun(Pfs,x), '--g', 'displayname', 'fminsearch');
semilogx(x, fun(Pfc,x), ':b', 'displayname', 'fmincon');
hold off
legend show
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Walter Roberson
Walter Roberson 2021-7-20
编辑：Walter Roberson 2021-7-20
Ran in:
format long g
f = [5500 6000 6500 7000 7500 8000 8500 9000];
y = [0.998 0.997 0.994 0.988 0.947 0.255 0.956 0.987];
fun = @(p, x) abs(1i.*(x.*p(2)-1./(x.*p(3)))./(p(1) + 1i.*(x.*p(2)-1./(x.*p(3))))); % transfer function of RLC bandstop
% x, p(1), p(2), p(3) correspond to w, R, L, C respectively
x = f(1):f(end); % for plotting purpose
x0 = [1 1e-3 1e-6] % starting points
x0 = 1×3
1 0.001 1e-06
lb = [1 1e-3 1e-6];ub = [100E3 100 1000E-6];
ub = [1 0.02 2e-5];
%x0 = [1 0.0038733109633344873895993294337111 4e-6]
E = @(P) sum((fun(P,f)-y).^2);
syms P [1 3]
E2 = simplify(E([1, P(2), 4e-6]))
E2 = 
dE2 = diff(E2,P(2))
dE2 = 
fplot(E2, [0.003 0.004])
fplot(E2, [0.0038 0.004])
E2f = matlabFunction(E2);
[bestP2, fval_E2] = fminsearch(E2f,50)
bestP2 =
0.00393867492675781
fval_E2 =
0.000695343018803931
x0 = [1, bestP2, 4E-6]
x0 = 1×3
1 0.00393867492675781 4e-06
[p_fit, fval_fit] = lsqcurvefit(fun, x0, f, y, lb, ub) % p_fit is the vector of fitted parameters
Solver stopped prematurely. lsqcurvefit stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 2.000000e+02.
p_fit = 1×3
1 0.00336360777147123 4.69157914821343e-06
fval_fit =
9.96139569469118e-05
semilogx(f, y, 'ro', 'displayname', 'original');
hold on;
semilogx(x, fun(p_fit, x), '--m', 'displayname', 'lsq'); grid on; % semilog plot of the transfer function
[p_fit2, fval_fit2] = lsqcurvefit(fun, x0, f, y, lb, ub)
Solver stopped prematurely. lsqcurvefit stopped because it exceeded the function evaluation limit, options.MaxFunctionEvaluations = 2.000000e+02.
p_fit2 = 1×3
1 0.00336360777147123 4.69157914821343e-06
fval_fit2 =
9.96139569469118e-05
[Pfc, fval_fc] = fmincon(E, x0, [], [], [], [], lb, ub)
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
Pfc = 1×3
1 0.00393867492675781 4e-06
fval_fc =
0.000695343018803941
[Pfs, fval_fs] = fminsearch(E, x0)
Pfs = 1×3
1.26924675790965 0.0039541197292072 3.9938320146631e-06
fval_fs =
2.13462497232316e-06
semilogx(x, fun(Pfs,x), '--g', 'displayname', 'fminsearch');
semilogx(x, fun(Pfc,x), ':b', 'displayname', 'fmincon');
hold off
legend show
The result of [bestP2, fval_E2] = fminsearch(E2f,50) shows that you do not need to start the search from anywhere close by in order to hit one of the two best regions. The one found, the second of the two dips in the w, gives an even better fit (using fminsearch) than using the first of the w dips.
Walter Roberson
Walter Roberson 2021-7-20
编辑：Walter Roberson 2021-7-20
Ran in:
The following was not able to find a good solution... but
format long g
f = [80 190 216 224 234 243 254 314]*1e3
f = 1×8
80000 190000 216000 224000 234000 243000 254000 314000
y = [999 984 897 679 205 801 938 993]*1e-3
y = 1×8
0.999 0.984 0.897 0.679 0.205 0.801 0.938 0.993
fun = @(p, x) abs(1i.*(x.*p(2)-1./(x.*p(3)))./(p(1) + 1i.*(x.*p(2)-1./(x.*p(3))))); % transfer function of RLC bandstop
% x, p(1), p(2), p(3) correspond to w, R, L, C respectively
x = f(1):f(end); % for plotting purpose
lb = [1 1e-3 1e-6]; %ub = [100E3 100 1000E-6];
ub = [1 0.02 2e-5];
%x0 = [1 0.0038733109633344873895993294337111 4e-6]
E = @(P) sum((fun(P,f)-y).^2);
syms P [1 3]
E2 = simplify(E([1, P(2), 4e-6]))
E2 = 
dE2 = diff(E2,P(2))
dE2 = 
%fplot(E2, [0.003 0.004])
%fplot(E2, [0.0038 0.004])
E2f = matlabFunction(E2);
[bestP2, fval_E2] = fminsearch(E2f,50)
bestP2 =
-1.26659870147705e-06
fval_E2 =
0.480361871668374
x0 = [1, bestP2, 4E-6]
x0 = 1×3
1 -1.26659870147705e-06 4e-06
Notice that search came out with a negative P(2), which is out of range.
[p_fit, fval_fit] = lsqcurvefit(fun, x0, f, y, lb, ub) % p_fit is the vector of fitted parameters
Initial point is a local minimum. Optimization completed because the size of the gradient at the initial point is less than the value of the optimality tolerance.
p_fit = 1×3
1 0.0105 4e-06
fval_fit =
0.789425744655355
Those search coordinates are in range, but the result is not great
semilogx(f, y, 'ro', 'displayname', 'original');
hold on;
semilogx(x, fun(p_fit, x), '--m', 'displayname', 'lsq'); grid on; % semilog plot of the transfer function
[Pfc, fval_fc] = fmincon(E, x0, [], [], [], [], lb, ub)
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
Pfc = 1×3
1 0.00100169385945684 1.10465327170057e-06
fval_fc =
0.789396955822885
Again within range, but the results are not great
[Pfs, fval_fs] = fminsearch(E, x0)
Pfs = 1×3
0.983635708916737 -2.43204590273628e-06 6.21518622374251e-06
fval_fs =
0.467740743880735
and that one is out of range again
%semilogx(x, fun(Pfs,x), '--g', 'displayname', 'fminsearch');
semilogx(x, fun(Pfc,x), ':b', 'displayname', 'fmincon');
hold off
legend show
I did some grid searching, brute force. The locations I found varied a lot depending how fine a grid I used. The best I found so far was
63535.7188297698 0.710929969491426 0.001
but there was a lot of variability -- such as near 1000 0.01 1e-3 was only marginally more.
In short: there is no good fit.

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