what's the relation between A , B and M,G for this Nonlinear system of equation ?

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A*sin(3*Phi)-B*sin(Phi) = G ;
A*cos(3*Phi)-B*cos(Phi) = M ;

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Torsten
Torsten 2024-2-11
A^2 + B^2 - 2*A*B*cos(2*Phi) = G^2 + M^2
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Torsten
Torsten 2024-2-11
编辑:Torsten 2024-2-11
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 = struct with fields:
Phi: [6×1 sym] parameters: k conditions: [6×1 sym]
sol1.Phi
ans = 
sol1.conditions
ans = 

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更多回答(1 个)

John D'Errico
John D'Errico 2024-2-11
编辑:John D'Errico 2024-2-11
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.
  1 个评论
YOUSSEF El MOUSSATI
Ok ,thanks for your answer , yes this system is linear but the parameters A,B and C,D are Nonlinear A=f(x^3,x^2,x); B=g(x^3,x^2,x); C=h(x^3,x^2,x); D=l(x^3,x^2,x);

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