Azzera filtri
Azzera filtri

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 visualizzazione (ultimi 30 giorni)
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

Risposte (1)

Torsten
Torsten il 13 Giu 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 Commenti
Torsten
Torsten il 13 Giu 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 il 13 Giu 2022
Modificato: Torsten il 13 Giu 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

Accedi per commentare.

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Prodotti


Release

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by