don't get what solve should return as solution

Hi! I am new to symbolical equations in MATLAB. I have written the dynamics of a mobile inverted pendulum. The solution of the problem should be in this form: [ x y theta 0 0 0 0] but I get strange values. I have 3 solution per variable...I have attached the problem's parameters if someone were so nice to try the code.

%% Rappresentazione ISU
syms x y theta alpha dalpha v dtheta real
stato=[x y theta alpha dalpha v dtheta]';
syms dx dy dtheta dalpha ddalpha dv ddtheta real
d_stato=[dx dy dtheta dalpha ddalpha dv ddtheta]';
syms tau_r tau_l real
u=[tau_r tau_l]';


f1=v*cos(theta);
f2=v*sin(theta);
f3=dtheta;
f4=dalpha;
f5=(sin(2*alpha)*(b3*b1*(dtheta)^2-b5^2*(dalpha)^2)+(2*b5*b3*g*sin(alpha)))/(2*(b2*b3-b5^2*(cos(alpha)^2)));
f6=(sin(2*alpha)*(-b1*b5*(dtheta)^2*cos(alpha)-b5^2*g)+(2*b2*b5*(dalpha)^2*sin(alpha)))/(2*(b2*b3-b5^2*(cos(alpha)^2)));
f7=(-b1*dalpha*dtheta*sin(2*alpha))/(b4+b1*sin(alpha)^2);

f=[f1 f2 f3 f4 f5 f6 f7]';

T1=b2*b3-b5^2*(cos(alpha))^2;
T2=b4+b1*(sin(alpha))^2;
g5=b3*R+b5*cos(alpha);
g6=b2+(R*b5*cos(alpha));

g1=[0 0 0 0 -g5/(R*T1) g6/(R*T1) b/(R*T2)]';
g2=[0 0 0 0 -g5/(R*T1) g6/(R*T1) -b/(R*T2)]';
g=[g1 g2];

d_stato=f+g*u;


%% Risoluzione 
assume(alpha<pi/2 & alpha>-pi/2)
eqns = [d_stato(1)==0 , d_stato(2)==0, d_stato(3)==0, d_stato(4)==0, d_stato(5)==0, d_stato(6)==0, d_stato(7)==0];
S= solve(eqns,[stato; tau_r; tau_l]','ReturnConditions',true);

Answer 1

[Walter Roberson]

%% Parametri

M_b=3.9;%Kg massa del pendolo
M_w=0.375;%Kg massa singola ruota
R=0.1025;%m raggio ruota
c_z=0.1;%m (0,0, c_z) centro di massa del pendolo rispetto al frame body pendolo
b=0.1620;%m
g=9.8;%N gravit
I_x=0.02;%Kg*m^2
I_y=0.015;%Kg*m^2
I_z=0.01;%Kg*m^2
I_wd=0.002;%Kg*m^2 momento di inerzia ruota attorno diametro
I_wa=0.001;%Kg*m^2 momento di inerzia ruota attorno asse di roll

%
b1=(M_b*c_z^2)+I_x-I_z;
b2=(M_b*c_z^2)+I_y;
b3=((M_b*R^2)+(2*(I_wa+M_w*R^2)))/R^2;
b4=R^2*(I_z+2*I_wd+2*b^2*M_w)+(2*b^2*I_wa);
b5=M_b*c_z;


%% Rappresentazione ISU
syms x y theta alpha dalpha v dtheta real
stato=[x y theta alpha dalpha v dtheta]';
syms dx dy dtheta dalpha ddalpha dv ddtheta real
d_stato=[dx dy dtheta dalpha ddalpha dv ddtheta]';
syms tau_r tau_l real
u=[tau_r tau_l]';


f1=v*cos(theta);
f2=v*sin(theta);
f3=dtheta;
f4=dalpha;
f5=(sin(2*alpha)*(b3*b1*(dtheta)^2-b5^2*(dalpha)^2)+(2*b5*b3*g*sin(alpha)))/(2*(b2*b3-b5^2*(cos(alpha)^2)));
f6=(sin(2*alpha)*(-b1*b5*(dtheta)^2*cos(alpha)-b5^2*g)+(2*b2*b5*(dalpha)^2*sin(alpha)))/(2*(b2*b3-b5^2*(cos(alpha)^2)));
f7=(-b1*dalpha*dtheta*sin(2*alpha))/(b4+b1*sin(alpha)^2);

f=[f1 f2 f3 f4 f5 f6 f7]';

T1=b2*b3-b5^2*(cos(alpha))^2;
T2=b4+b1*(sin(alpha))^2;
g5=b3*R+b5*cos(alpha);
g6=b2+(R*b5*cos(alpha));

g1=[0 0 0 0 -g5/(R*T1) g6/(R*T1) b/(R*T2)]';
g2=[0 0 0 0 -g5/(R*T1) g6/(R*T1) -b/(R*T2)]';
g=[g1 g2];

d_stato=f+g*u;


%% Risoluzione 
assume(alpha<pi/2 & alpha>-pi/2)
eqns = [d_stato(1)==0 , d_stato(2)==0, d_stato(3)==0, d_stato(4)==0, d_stato(5)==0, d_stato(6)==0, d_stato(7)==0];
S= solve(eqns,[stato; tau_r; tau_l]','ReturnConditions',true);
S

S = struct with fields:
             x: [3×1 sym]
             y: [3×1 sym]
         theta: [3×1 sym]
         alpha: [3×1 sym]
        dalpha: [3×1 sym]
             v: [3×1 sym]
        dtheta: [3×1 sym]
         tau_r: [3×1 sym]
         tau_l: [3×1 sym]
    parameters: [1×4 sym]
    conditions: [3×1 sym]

S.parameters

ans = 

S.x

ans = 

S.y

ans = 

S.theta

ans = 

S.conditions

ans = 

What this is telling you is that there are three families of solutions.

The first family says that x can be any arbitrary real number, and y is an arbitray real number (possibly different than x), and theta is any integer multiple of pi. In other words if theta is an integer multiple of pi, then a number of your terms vanish and so any arbitrary real x and y work.

The third family is nearly the same as the first, but theta is an odd half-integer multiple of pi; again that causes a lot of terms to vanish so it does not matter which real x and y you use.

The second family says that x and y and theta can all be arbitrary real values; possibly some of the other variables are tightly constrained, but you did not ask to look at those.

[Valeria Leto]

yeah, but the look equivalent to me