# How can I solve 4 simultaneous equations ?

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준형 박 on 29 Sep 2022
Edited: 준형 박 on 29 Sep 2022
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

Torsten on 29 Sep 2022
Edited: Torsten on 29 Sep 2022
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 on 29 Sep 2022
Edited: 준형 박 on 29 Sep 2022
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.

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