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fsolve giving error that solution is not finite and real. When I test using vpasolve() I get a solution, then I input the same solution and get same error of not finite/real.

1 vue (au cours des 30 derniers jours)
c1 = 0.5176;
c2 = 116;
c3 = .4;
c4 = 0.4;
c5 = 5;
c6 =21;
c7 = .08;
c8 = .035;
syms lambda
for k = 1:600
theta = pitch.Data(k);
%cp(k) = c1*(c2/lambda - c3*theta -c5)*exp(c6/lambda);
cp(k) = c1*(c6*lambda + (-c4 - c3*(2.5 + theta) + c2*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3)))/exp(c5*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3))));
eqn(k) = cp(k)/(2*lambda^3) == 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2);
tsr(k) = vpasolve(eqn(k),lambda);%
tsr_check(k) = fzero(@(lambda)cp(k),[-1 0]);
end
  1 commentaire
Gordon
Gordon le 13 Juin 2022
Note: I actually gave the input as -1 as an initial guess. The actual solution via vpasolve() is -.6....

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Réponses (1)

Torsten
Torsten le 13 Juin 2022
c1 = 0.5176;
c2 = 116;
c3 = .4;
c4 = 0.4;
c5 = 5;
c6 =21;
c7 = .08;
c8 = .035;
syms lambda
for k = 1:600
theta = pitch.Data(k);
%cp(k) = c1*(c2/lambda - c3*theta -c5)*exp(c6/lambda);
cp(k) = c1*(c6*lambda + (-c4 - c3*(2.5 + theta) + c2*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3)))/exp(c5*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3))));
eqn(k) = cp(k)/(2*lambda^3) == 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2);
tsr(k) = vpasolve(eqn(k),lambda);%
expr = cp(k)/(2*lambda^3) - 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2);
fun = matlabFunction(expr,'Vars',lambda);
tsr_check(k) = fzero(fun,[-1 0]);
end
  8 commentaires
Torsten
Torsten le 13 Juin 2022
@Gordon comment moved here:
fsolve() is not a good enough solver in this situation because of the rate of change of the data. Therefore, vpasolve() needed to be used. The reason for trying to implement fsolve is because simulink does not allow vpasolve() a solution therefore is to use code.extrinsic() to implement function including vpasolve().
Torsten
Torsten le 13 Juin 2022
Modifié(e) : Torsten le 13 Juin 2022
If the solution for index k is "near" to the solution of index k-1, it is usually a good idea to take the solution of step k-1 as initial guess for the solution of index k. Something like
c1 = 0.5176;
c2 = 116;
c3 = .4;
c4 = 0.4;
c5 = 5;
c6 =21;
c7 = .08;
c8 = .035;
tsr_guess = 1.0;
syms lambda
for k = 1:600
theta = pitch.Data(k);
%cp(k) = c1*(c2/lambda - c3*theta -c5)*exp(c6/lambda);
cp(k) = c1*(c6*lambda + (-c4 - c3*(2.5 + theta) + c2*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3)))/exp(c5*(1/(lambda + ...
c7*(2.5 + theta)) - c8/(1 + (2.5 + theta)^3))));
eqn(k) = cp(k)/(2*lambda^3) == 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2);
tsr(k) = vpasolve(eqn(k),lambda);%
expr = cp(k)/(2*lambda^3) - 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2);
fun = matlabFunction(expr,'Vars',lambda);
tsr_check(k) = fsolve(fun,tsr_guess);
tsr_guess = tsr_check(k);
end

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