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.
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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
le 13 Juin 2022
Réponses (1)
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
Gordon
le 13 Juin 2022
Evaluating your equation for lambda = 0 gives Inf.
Thus use a reasonable starting value or interval for lambda in the call to fzero.
Or use
expr = cp(k) - 1e7*(ta_kf.Data(k))/(rho*pi*N^5*wr_kf.Data(k)^2)*2*lambda^3;
Gordon
le 13 Juin 2022
Torsten
le 13 Juin 2022
As stated before the function should not evaluate to inifinte at the initial guess because I already found an alternate solution using vpasolve() at the same initial guess that is finite and real.
You specify that the solution for lambda lies in the interval [-1 0]. Consequently, fzero evaluates your function at lambda = -1 and lambda = 0, and for lambda = 0 you get a division by zero.
I don't see that you prescribe the solution from vpasolve as initial guess for fzero anywhere.
Gordon
le 13 Juin 2022
Torsten
le 13 Juin 2022
Then you have to include executable code. In the code supplied, data are missing.
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().
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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