solve() gives Empty sym: 0-by-1

syms Vx Vin Vs Io Vy Vo L C R Rp k s
eqn1 = ((Vx-Vin)/Rp) + ((Vx-Vs)/R) - (k*Io/2) + ((Vx-Vy)/((2*s*L)+(1/(s*C)))) == 0;
eqn2 = (Vy/Rp) + ((Vy-Vs)/R) - (k*Io/2) + ((Vy-Vx)/((2*s*L)+(1/(s*C)))) == 0;
eqn3 = (Vo*s*C) == ((Vx-Vy)/((2*s*L)+(1/(s*C))));
% sol = solve([eqn1, eqn2, eqn3, eqn4], [Vx, Vin, Vs, Io, Vy, Vz, Vo, L, C, R, Rp, k, s]);
sol = solve([eqn1, eqn2, eqn3], [Vo]);
sol
% sol.Vx



syms x y z c
eqn1 = 2*x + y + z == 2;
sol = solve([eqn1], [x]);
sol

I try to obtain transfer function of a circuit but I get empty sym. I have three equations with many more variables, so I expect symbolic solutions. Second part of the code is an example for me to see whether I can obtain such answer for different case, I get what I needed in this case. Outputs are below

sol =
 
Empty sym: 0-by-1
 
 
sol =
 
1 - z/2 - y/2

Can you help me?

Answer 1

[Ameer Hamza]

Why not just solve eqn3, it is enough to give expression of Vo

sol = solve(eqn3, Vo);

Result

>> sol
sol =
(Vx - Vy)/(2*C*L*s^2 + 1)

If you want to eliminate Vx and Vy from the solution then try this

sol = solve([eqn1, eqn2, eqn3], [Vo Vx Vy]);

Result

>> sol.Vo
ans =
(R*Vin)/(R + Rp + 2*C*L*R*s^2 + 2*C*L*Rp*s^2 + 2*C*R*Rp*s)

Also find the transfer function

>> sol.Vo/Vin
ans =
R/(R + Rp + 2*C*L*R*s^2 + 2*C*L*Rp*s^2 + 2*C*R*Rp*s)

[Serdar Ünal]

Thanks for the quick answer :)

[Ameer Hamza]

I am glad to be of help! :)