How can I solve 4 simultaneous equations ?

syms Rcon;
syms Ccon;
syms Rbulk;
syms Cbulk;
w1=6280;
w2=62800;
seta(2,2)=-88.8000;
seta(2,3)=-88.9600;
z(2,2)=213000;
z(2,3)=21800;
solveRcon=1;
solveCcon=1;
solveRbulk=1;
solveCbulk=1;
epns = [ tand(seta(2,2)) +  (Rcon + Rbulk)/(1/(w1*Ccon) + 1/(w1*Cbulk)) == 0 , tand(seta(2,3)) +  (Rcon + Rbulk)/(1/(w2*Ccon) + 1/(w2*Cbulk)) == 0, abs(z(2,2)) - 1/(sqrt((1/Rcon)^2 +(w1*Ccon)^2)) - 1/(sqrt((1/Rbulk)^2 +(w1*Cbulk)^2)) == 0, abs(z(2,3)) - 1/(sqrt((1/Rcon)^2 +(w2*Ccon)^2)) - 1/(sqrt((1/Rbulk)^2 +(w2*Cbulk)^2)) == 0]; 
vars = [ Ccon Rcon Cbulk Rbulk ];
[solCcon, solRcon, solCbulk, solRbulk] = solve(epns,vars)

i want to solve 4 simltaneous equation and get 4 syms values( Ccon, Rcon, Cbulk, Rbulk)

but i just get 0x1 sym in Ccon, Rcon, Cbulk, Rbulk

Answer 1

[Torsten]

You might want to test fsolve and lsqnonlin with different initial guesses for the parameters,

but I doubt you will find a solution.

w1=6280;
w2=62800;
seta(2,2)=-88.8000;
seta(2,3)=-88.9600;
z(2,2)=213000;
z(2,3)=21800;
epns = @(Ccon ,Rcon, Cbulk, Rbulk)[ tand(seta(2,2)) +  (Rcon + Rbulk)/(1/(w1*Ccon) + 1/(w1*Cbulk))  , tand(seta(2,3)) +  (Rcon + Rbulk)/(1/(w2*Ccon) + 1/(w2*Cbulk)) , abs(z(2,2)) - 1/(sqrt((1/Rcon)^2 +(w1*Ccon)^2)) - 1/(sqrt((1/Rbulk)^2 +(w1*Cbulk)^2)) , abs(z(2,3)) - 1/(sqrt((1/Rcon)^2 +(w2*Ccon)^2)) - 1/(sqrt((1/Rbulk)^2 +(w2*Cbulk)^2)) ]; 
sol0 = [1 1 1 1];
options = optimset('MaxFunEvals',100000,'MaxIter',100000);
%sol = fsolve(@(x)epns(x(1),x(2),x(3),x(4)),sol0,options)
sol = lsqnonlin(@(x)epns(x(1),x(2),x(3),x(4)),sol0,[],[],options)

Local minimum possible.

lsqnonlin stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

sol = 1×4

-0.0019 -0.0560 0.9207 -0.0560

epns(sol(1),sol(2),sol(3),sol(4))

ans = 1×4

1.0e+05 * -0.0005 -0.0004 2.1300 0.2180

[준형 박]

This result seems to be different from what I expected.

Can you explain why you said 'I doubt you will find a solution'?

[Torsten]

With my initial guesses for the unknowns, the solver gave a "solution" for which the error in the equations was in the order of 1e5 (see above). You were more successful ?

[준형 박]

I haven't calculated the unknown. But there are expected values ​​for the unknown.

All unknowns are positive,

and I thought that the difference between (Ccon, Cbulk) and (Rcon, Rbulk) would be less than 100 times.

[Torsten]

I haven't calculated the unknown.

?

So you didn't run the code and received a result ?

[준형 박]

When I ran the code, the same result as what you entered.

'I haven't calculated the unknown.' means I haven't tried calculating this in any other way than matlab.