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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.

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Gordon
Gordon 2022 年 6 月 13 日
編集済み: Torsten 2022 年 6 月 13 日
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 件のコメント
Gordon
Gordon 2022 年 6 月 13 日
Note: I actually gave the input as -1 as an initial guess. The actual solution via vpasolve() is -.6....

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回答 (1 件)

Torsten
Torsten 2022 年 6 月 13 日
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 件のコメント
Torsten
Torsten 2022 年 6 月 13 日
@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 2022 年 6 月 13 日
編集済み: Torsten 2022 年 6 月 13 日
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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