what's the relation between A , B and M,G for this Nonlinear system of equation ?
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YOUSSEF El MOUSSATI
le 11 Fév 2024
Commenté : YOUSSEF El MOUSSATI
le 11 Fév 2024
A*sin(3*Phi)-B*sin(Phi) = G ;
A*cos(3*Phi)-B*cos(Phi) = M ;
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Torsten
le 11 Fév 2024
A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2
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Torsten
le 11 Fév 2024
Modifié(e) : Torsten
le 11 Fév 2024
If each of the equations could be solved uniquely for Phi, you could get what you want.
Assume that the first equation would uniquely yield Phi = f(A,B,G) and the second equation would uniquely yield Phi = g(A,B,M), then f(A,B,G) - g(A,B,M) = 0 would be your relation. But unfortunately, the equations cannot be solved uniquely for Phi.
syms A B G M Phi
eqn = A*sin(3*Phi)-B*sin(Phi) == G ;
sol1 = solve(eqn,Phi,'ReturnConditions',1,'MaxDegree',3)
sol1.Phi
sol1.conditions
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John D'Errico
le 11 Fév 2024
Modifié(e) : John D'Errico
le 11 Fév 2024
This is not even remotely a question about MATLAB. As such, it should arguably not even be on Answers. But I have a minute to respond, so I will choose to do so.
Trivial! What is the relation? Admittedly, the relation itself is a slightly complex thing, composed of two equations. The relations are:
A*sin(3*Phi)-B*sin(Phi) = G
A*cos(3*Phi)-B*cos(Phi) = M
which is exactly what you wrote.
It is not a nonlinear system of equations though. Not at all! Phi there is simply a parameter, not one of the parameters involved. That makes your problem fully a LINEAR system of equations. As such, if you want to view it in that form, then we could write:
M = [sin(3*Phi), -sin(Phi); ..
cos(3*Phi), -cos(Phi)]
So M is a matrix function of the parameter Phi. Biven the matrix M, then we could write:
M*[A;B] = [G;M]
There is no simpler relation between those variables. And it is NOT at all nonlinear. Purely linear.
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