I’m trying to solve a group of equations to find the transfer function of a circuit, nontheless I get the banner of "Unable to find explicit solution" and Empty sym: 0-by-1. If I remove the "ReturnConditions=true" then I do get a solution but I’m not sure if it is the correct one. its just that I tried solving a more easy circuit and the "ReturnConditions=true" was necesary in that case.

I planned on solving an even more complex one so I wanted to know if I was doing something wrong.
syms R0 R1 R2 R3 R4 R5 C1 C2 C3 Vin I1 I2 I3 I4 Vout s H
eq1= Vin==I1*R0+I1/(C2*s)-I2/(C2*s)+I1*R3-I3*R3+I1/(C1*s)-I4/(C1*s);
eq2=0==I2*R1+I2/(C3*s)-I4/(C3*s)+I2*R4-I3*R4+I2/(C2*s)-I1/(C2*s);
eq3=0==I3*R3-I1*R3+I3*R4-I2*R4+I3*R5-I4*R5;
eq4=0==I4*R2+I4/(C1*s)-I1/(C1*s)+I4*R5-I3*R5+I4/(C3*s)-I2/(C3*s);
eq5=Vout==I4*R2;
eq6=H==Vout/Vin;
result=solve([eq1,eq2,eq3,eq4,eq5,eq6],[Vin,I1,I2,I3,I4,Vout,H],ReturnConditions=true);
H=collect(result.H,s)I’m trying to solve a group of equations to find the transfer function of a circuit, nontheless I get the banner of "Unable to find explicit solution" and Empty sym: 0-by-1. If I remove the "ReturnConditions=true" then I do get a solution but I’m not sure if it is the correct one. its just that I tried solving a more easy circuit and the "ReturnConditions=true" was necesary in that case.

I planned on solving an even more complex one so I wanted to know if I was doing something wrong.
syms R0 R1 R2 R3 R4 R5 C1 C2 C3 Vin I1 I2 I3 I4 Vout s H
eq1= Vin==I1*R0+I1/(C2*s)-I2/(C2*s)+I1*R3-I3*R3+I1/(C1*s)-I4/(C1*s);
eq2=0==I2*R1+I2/(C3*s)-I4/(C3*s)+I2*R4-I3*R4+I2/(C2*s)-I1/(C2*s);
eq3=0==I3*R3-I1*R3+I3*R4-I2*R4+I3*R5-I4*R5;
eq4=0==I4*R2+I4/(C1*s)-I1/(C1*s)+I4*R5-I3*R5+I4/(C3*s)-I2/(C3*s);
eq5=Vout==I4*R2;
eq6=H==Vout/Vin;
result=solve([eq1,eq2,eq3,eq4,eq5,eq6],[Vin,I1,I2,I3,I4,Vout,H],ReturnConditions=true);
H=collect(result.H,s) I’m trying to solve a group of equations to find the transfer function of a circuit, nontheless I get the banner of "Unable to find explicit solution" and Empty sym: 0-by-1. If I remove the "ReturnConditions=true" then I do get a solution but I’m not sure if it is the correct one. its just that I tried solving a more easy circuit and the "ReturnConditions=true" was necesary in that case.

I planned on solving an even more complex one so I wanted to know if I was doing something wrong.
syms R0 R1 R2 R3 R4 R5 C1 C2 C3 Vin I1 I2 I3 I4 Vout s H
eq1= Vin==I1*R0+I1/(C2*s)-I2/(C2*s)+I1*R3-I3*R3+I1/(C1*s)-I4/(C1*s);
eq2=0==I2*R1+I2/(C3*s)-I4/(C3*s)+I2*R4-I3*R4+I2/(C2*s)-I1/(C2*s);
eq3=0==I3*R3-I1*R3+I3*R4-I2*R4+I3*R5-I4*R5;
eq4=0==I4*R2+I4/(C1*s)-I1/(C1*s)+I4*R5-I3*R5+I4/(C3*s)-I2/(C3*s);
eq5=Vout==I4*R2;
eq6=H==Vout/Vin;
result=solve([eq1,eq2,eq3,eq4,eq5,eq6],[Vin,I1,I2,I3,I4,Vout,H],ReturnConditions=true);
H=collect(result.H,s) transfer function, equation, solve MATLAB Answers — New Questions

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