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Conversion to logical from symfun is not possible

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Alex Muniz
Alex Muniz 2023 年 1 月 10 日
回答済み: Walter Roberson 2023 年 1 月 10 日
How to fix this error: Conversion to logical from symfun is not possible ?
Code:
syms s eqn1 eqn2 eqn3;
kp = sym('kp','real');
ki = sym('ki','real');
kd = sym('kd','real');
w = sym('w','real');
i0 = 1;
i1 = 1;
i2 = 1;
eqn1 = (subs(V_jw_even,w,0))*i0 > 0;
eqn2 = (subs(V_jw_even,w,w_roots))*i1 > 0;
eqn3 = (subs(V_jw_even,w,inf))*i2 > 0;
% [ V_jw_even] is a symbolic polynomial. V_jw_even = (- kd - 1)*w^4 + (ki - 9*kd + 17)*w^2 + 9*ki
% [w_roots] is a real and positive root . w_roots ~= 1.4638
% Define o passo de busca para x e y
step = 0.5;
solution_found = false;
% Enquanto a solução ainda não foi encontrada
for ki = x_min:step:x_max
for kd = y_min:step:y_max
% Verifica se as inequações são satisfeitas para os valores atuais de x e y
solution_found = eqn1 && eqn2 && eqn3;
if solution_found
fprintf('Solucao encontrada: x = %.2f, y = %.2f\n', ki, kd);
solution_found = true;
end
end
end

採用された回答

Walter Roberson
Walter Roberson 2023 年 1 月 10 日
There are two possibilities here:
  1. eqn1, eqn2, eqn3 do not involve any unresolved symbolic variables before the loop, so you can double() them before the loops start and pre-calculate eqn1 && eqn2 && eqn3 -- the condition would either be true for all loop iterations or for none at all; or
  2. at least one of eqn1, eqn2, eqn3 involve any unresolved symbolic variables, In this case, eqn1 && eqn2 && eqn3 cannot be resolved, and looping changing ki and kd is not going to change that
The second situation would be different if you were to subs() to evaluate the conditions.
I want you to consider for a moment the code segment
a = 2
b = a^2 + 1
a = 7
What is the value of b after that code? Does b get automatically recalculated accounting for the new version of a and so become 7^2+1, keeping some kind of internal history of how it was calculated and getting recalculated when the variables involved changed? If so then how would you possibly be able to handle b = b + 1 ? Or... does the current value of a get used to calculate b and then b promptly forgets how it was calculated, so that when a changes, b does not change?
The situation is the same with
syms a
b = a^2 + 1
a = 7
At the time the b = a^2 + 1 is calculated, a is a reference to a symbolic variable and b gets assigned a symbolic result. When you then a=7 then b still uses that reference to a as a symbolic variable and does not get recalculated with the change in a

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