Solving symbolic Matlab equations in term of two variables (Vo/Vi)

syms Vx Vy Vi Vo Vg R1 R2 R3 C1 C2 C3 H k s  
eqn1 = (Vx - Vi) / R1 + (Vx - Vg) / (s * C3)^-1 + (Vx - Vo) / R2 == 0
eqn2 = (Vy - Vi) / (s * C1)^-1 + (Vy - Vg) / R3 + (Vy - Vo) / (s * C2)^-1 == 0
eqn3 = (Vi - Vx) / R1 + (Vi - Vy) / (s * C1)^-1 == 0
eqn4 = (Vo - Vx) / R2 + (Vo - Vy) / (s * C2)^-1 == 0
eqn5 = Vg == 0
eqn6 = H == Vo / Vi
S = solve([eqn1, eqn2, eqn3, eqn4, eqn5, eqn6],H)

I'm stuck been spinning my wheels for months..

I want to know H = (terms without Vx, Vy, Vi, Vo, Vg)

H is transfer function which is Voltage out / Voltage in

Honestly any help is appreciated. Thanks again.

[KSSV]

You must copy your code here. Attaching code as image is not useful.

[유진 정]

sorry rewrited

[Star Strider]

Post an image of the circuit.

I suspect those node-voltage equations are not set up correctly.

Answer 1

[Nathan Hardenberg]

From my experience trying to solve transfer-function with this approach did not work. If someone knows a better way please comment 😃 Not quite sure what your exact task is, but I had to get a transfer-function from a electrical filter (see diagram).

%       Z1 ____
% Vi o----|____|--.----o Va
%                 |
%                .┸.
%                | | Z2
%                |_|
%                 |
%                 ╧ GND

I'm not quite sure if your problem is similar, but it seems you already did some kind of laplace transform (Sorry I'm not into the topic that deep). I wrote my functions first with a symbolic variable p instead of ω. I'm guessing you also used but you seem to have a lot more voltages. Then I wrote my equation directly, without having to solve ( in my case). This still has the advantage of not having to work with long terms and it is possible to simplify the term.

Only after that, I substituted p. I used a "zero point development" or "biliniar transformation", which is probably not the way you want to tackle the problem.

Since it is very useful to have the transfer function in a specific form I wrote a function that does this afterwards.

If you want more info on the problem I worked with, I can eleborate. If not, I hope I could give you a few ideas.

And one important gerneral thing to note is that you have to tell the solve()-function what you want to solve for. That means that you have to also solve for the variables you do not want/know as well!

syms x y z a
eq1 = x^2+a*x+5 == y;
eq2 = y-a == 3;
eq3 = z+3 == y;

sol = solve([eq1; eq2; eq3], [x, y, z]);
sol.x

ans = 

sol.y

ans = 

sol.z

ans = 

In your case this would be the following, but in my tests it did not find a result this way either.

solve([eqn1 eqn2 eqn3 eqn4 eqn5 eqn6], [H Vx Vy Vi Vo Vg])