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

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유진 정 2023 年 5 月 26 日

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.
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Star Strider 2023 年 5 月 26 日
Post an image of the circuit.
I suspect those node-voltage equations are not set up correctly.

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### 回答 (1 件)

Nathan Hardenberg 2023 年 5 月 26 日
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])

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