Using symbolic solver, I seem to not be able to get a system of equations that give me a desired answer. Does a solution not exist?

syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
syms Rp Rx Rn  positive
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;
[Rn, Rp, Rx] = solve([equ1, equ2,equ3],[Rn , Rp, Rx]);

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

simplify(Rn)

ans = 

simplify(Rp)

ans = 

simplify(Rx)

ans = 

Rn=vpa(subs(Rn,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))

Rn = 

0.000044531163809619033020191303948558

Rp=vpa(subs(Rp,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))

Rp = 

0.000044531163809288526802731288836632

Rx=vpa(subs(Rx,{V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx},{270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013}))

Rx = 

So I am trying to find values for Rn,Rx and Rp. I have three equations and three unknowns. When I use solve to find symbolic equation, I get equations for each variable. But when I sub real values in, I get an answer that is wrong or doesnt make sense. Adding assume doesnt seem to change anything and the return conditions is the same as the equations. The real values of Rn, Rp and Rx should be 2e6,1.5e6 and 700k. If you were to put these values in the original equations, you can verify that these values are correct. How do I get a system of equations that will give me the desired answer?

Answer 1

[Walter Roberson]

The real values of Rn, Rp and Rx should be 2e6,1.5e5 and 700k.

Those are inconsistent with the equations.

Rn = 2e6;
Rp = 1.5e5;
Rx = 700e3;
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;

eqns = [equ1; equ2; equ3]

eqns = 

sol = subs(eqns, {V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx}, {270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013})

sol = 

vpa(lhs(sol) - rhs(sol))

ans = 

%above should be quite close to zero if the equations are consistent.

[Taiji]

So sorry about that. I meant 2e6,1.5e6 and 700k. Really sorry about that

[Walter Roberson]

Those are closer to matching, but still not so great

Rn = 2e6;
Rp = 1.5e6;
Rx = 700e3;
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;

eqns = [equ1; equ2; equ3]

eqns = 

sol = subs(eqns, {V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx}, {270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013})

sol = 

vpa(lhs(sol) - rhs(sol))

ans = 

%above should be quite close to zero if the equations are consistent.

[Walter Roberson]

If you use those values, then

Rn = 2e6;
Rp = 1.5e6;
Rx = 700e3;
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;

eqns = [equ1; equ2; equ3]

eqns = 

sol1 = subs(eqns, {Vn,Vxp,Vxn,Vpx,Vnx}, {147.38892,125.43132,122.61108,268.58987,271.41013})

sol1 = 

vpasolve(sol1)

ans = struct with fields:
    V1: 269.99980003972474906460769785699
    V2: 270.0001692516440955324278388692
    Vp: 144.56867609419185247248136333716

the equation holds only if V1 V2 Vp are slightly different than your suggested replacement values of 270, 270, 144.56868

[Taiji]

The thing is, I dont want exact solutions to the 5th decimal places. These values are from real circuits with real values. In reality resistances cannot be negative. The measured values will have a very small percent of error which is expected. What I am doing is taking these equations and making sure I can extrapolate it to other values of voltages and resistances. But it doesnt seem to work. It might also be because the currents( Ison, Isop etc) are not limited to a range and their values seem to be wrong as well. But even when I use assume(Ison < 10), nothing changes.

[Walter Roberson]

Using the hypthosis that perhaps the 68220 is not exact:

Q = @(v) sym(v);
Rn = Q(2e6);
Rp = Q(1.5e6);
Rx = Q(700e3);
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
%C = Q(68220);
syms C positive
Ison=((Vp/((Rp*C)/(Rp+C)))-(Vxp/((Rx*C)/(Rx+C))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*C)/(Rn+C)))-(Vxn/((Rx*C)/(Rx+C))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*C)/(Rn+C)))-(Vpx/((Rp*C)/(Rp+C))));
equ3= Vpx==V1+Rx*Isox;

eqns = [equ1; equ2; equ3]

eqns = 

sol = subs(eqns, {V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx}, {270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013})

sol = 

temp = lhs(sol) - rhs(sol)

temp = 

solve(temp(1))

ans = 

vpa(ans)

ans = 

68220.011297423462838554054936255

solve(temp(2))

ans = 

vpa(ans)

ans = 

68219.996154709413705379656029139

solve(temp(3))

ans = 

vpa(ans)

ans = 

68220.471392826108213922497359148

The disagreement between the values shows the hyptothesis to be wrong -- there is no single replacement value for the constant that solves all of the systems simultaneously.

This leaves open the possibility that there are slightly different constants for each of the three equations.

[Walter Roberson]

Under the hypothesis that perhaps Rn Rp Rx are incorrect:

Q = @(v) sym(v);
syms Rn Rp Rx
%Rn = Q(2e6);
%Rp = Q(1.5e6);
%Rx = Q(700e3);
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
C = Q(68220);
Ison=((Vp/((Rp*C)/(Rp+C)))-(Vxp/((Rx*C)/(Rx+C))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*C)/(Rn+C)))-(Vxn/((Rx*C)/(Rx+C))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*C)/(Rn+C)))-(Vpx/((Rp*C)/(Rp+C))));
equ3= Vpx==V1+Rx*Isox;

eqns = [equ1; equ2; equ3];
sol = subs(eqns, {V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx}, {270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013});
temp = lhs(sol) - rhs(sol)

temp = 

sol_Rn = solve(temp(1), Rn)

sol_Rn = 

temp2 = simplify(subs(temp(2:end), Rn, sol_Rn))

temp2 = 

sol_Rp = solve(temp2(1), Rp)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

sol_Rp = 

temp3 = simplify(subs(temp2(2:end), Rp, sol_Rp))

temp3 = 

sol_Rx = solve(temp3, Rx)

sol_Rx = 

back_Rx = sol_Rx

back_Rx = 

back_Rp = subs(sol_Rp, Rx, back_Rx)

back_Rp = 

back_Rn = subs(subs(sol_Rn, Rx, back_Rx), Rp, back_Rp)

back_Rn = 

subs(temp, [Rn, Rp, Rx], [back_Rn, back_Rp, back_Rx])

ans = 

Notice that Rx is negative.

[Walter Roberson]

We must conclude one of a few possibilities:

• that the equations are correct, but that resistance is indeed negative in the system; OR
• that the equations are wrong; OR
• that 270, 270, 144.56868 is inaccurate coefficients and the inaccuracies are significant

I would put it to you that the correct coefficients are V1 V2 Vp from https://www.mathworks.com/matlabcentral/answers/2148774-using-symbolic-solver-i-seem-to-not-be-able-to-get-a-system-of-equations-that-give-me-a-desired-ans#comment_3249499

Your coefficients might be accurate to 5 significant digits, but that isn't good enough.

Answer 2

[Torsten]

It's the same situation as in your other problem:

https://uk.mathworks.com/matlabcentral/answers/2138791-how-can-i-find-a-unique-solution-within-tolerance-using-solve?s_tid=prof_contriblnk

Your system of equations has a unique solution, but this solution has not all components positive.

Your estimated values of Rn, Rp and Rx as 2e6,1.5e6 and 700k are not bad, but they cannot be true solutions because your system only has one unique solution:

Rn = 2e6;
Rp = 1.5e6;
Rx = 700e3;
syms V1 V2 Vp Vn Vxp Vxn Vpx Vnx
Ison=((Vp/((Rp*68220)/(Rp+68220)))-(Vxp/((Rx*68220)/(Rx+68220))));
equ1=Vp==(V1+V2)-Ison*Rn;
Isop=((Vn/((Rn*68220)/(Rn+68220)))-(Vxn/((Rx*68220)/(Rx+68220))));
equ2=Vn==(V1+V2)-Isop*Rp;
Isox=((Vnx/((Rn*68220)/(Rn+68220)))-(Vpx/((Rp*68220)/(Rp+68220))));
equ3= Vpx==V1+Rx*Isox;
eqns = [equ1; equ2; equ3]

eqns = 

sol = subs(eqns, {V1,V2,Vp,Vn,Vxp,Vxn,Vpx,Vnx}, {270,270,144.56868,147.38892,125.43132,122.61108,268.58987,271.41013})

sol = 

vpa(lhs(sol) - rhs(sol))

ans = 

[Taiji]

The thing is, I dont want exact solutions to the 5th decimal places. These values are from real circuits with real values. In reality resistances cannot be negative. The measured values will have a very small percent of error which is expected. What I am doing is taking these equations and making sure I can extrapolate it to other values of voltages and resistances. But it doesnt seem to work. It might also be because the currents( Ison, Isop etc) are not limited to a range and their values seem to be wrong as well. But even when I use assume(Ison < 10), nothing changes.

[Torsten]

As long as your system has a unique solution, "solve" will report that solution. There is no way to tell MATLAB that you want the "second-best" values for the unknowns, but in a range of your choice.

[Taiji]

Does this mean the system is not defined properly? If I was to add more equations to the system, would I get a more accurate answer?