# Algebraic way for finding a transfer function of a filter

I've the following filter and I tried to write the transfer function for it: simulate this circuit – Schematic created using CircuitLab

And I wrote for the current nodes the following equations:

1. $$\text{I}_1=\frac{\text{V}_\text{in}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_1}{\frac{1}{\text{s}\text{C}_1}}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=0\tag1$$
2. $$\text{I}_2=\frac{\text{V}_2-\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_\text{out}}{\frac{1}{\text{s}\text{C}_3}}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}=0\tag2$$
3. $$\text{I}_3=\frac{\text{V}_3-\text{V}_2}{\text{R}_3}+\frac{\text{V}_3}{\frac{1}{\text{s}\text{C}_2}}=0\tag3$$
4. $$\text{V}_+=\text{V}_-\space\implies\space\text{V}_3=\text{V}_+=\text{V}_-=\text{V}_\text{out}\tag4$$

Question: are my equations correct? And how can I find $\frac{\text{V}_\text{out}}{\text{V}_\text{in}}$ from this (if they are correct of coruse)?

Your first node equation should be:

$$\frac{\text{V}_\text{1}-\text{V}_{in}}{\text{R}_1}+\frac{\text{V}_1}{\frac{1}{\text{s}\text{C}_1}}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=0\tag1$$ Rest are correct: $$\frac{\text{V}_2-\text{V}_1}{\text{R}_2}+\frac{\text{V}_2-\text{V}_\text{out}}{\frac{1}{\text{s}\text{C}_3}}+\frac{\text{V}_2-\text{V}_3}{\text{R}_3}=0\tag2$$ $$\frac{\text{V}_3-\text{V}_2}{\text{R}_3}+\frac{\text{V}_3}{\frac{1}{\text{s}\text{C}_2}}=0\tag3$$ $$V_3 = V_{out}\tag4$$ We can simplify (3) using (4) $$\frac{\text{V}_{out}-\text{V}_2}{\text{R}_3}+\frac{\text{V}_{out}}{\frac{1}{\text{s}\text{C}_2}}=0$$ $$\implies V_2 = V_{out}(1+\frac {R_3}{1/sC_2})\tag5$$ You can use (5) to eliminate $V_2$ from (1) and (2) to end up in two equations with 3 independent variables of form: $$f(V_1,V_{in}, V_{out}) = 0 \tag6$$ $$g(V_1,V_{in}, V_{out}) = 0 \tag7$$ You can then find an expression for $V_1$ in terms of $V_{in}$ and $V_{out}$ from (6) as well as from (7). Equate both of them to get a final equation of form: $$h(V_{in},V_{out}) = 0$$ You can then sort out $V_{out}/V_{in}$ to derive the transfer function.

Basically what you have are 5 unknown variables and 4 equations, allowing you to find an expression in the form of $\frac{V_{out}}{V_{in}}$.

Be carefull with Kirchhoffs' Current Law: You state that $I_1 = 0$ which is not true unless you consider that as beign the current flowing in/out of node $I_1$ through a virtual wire. You can nevertheless correctly state that $$\sum_{i}I_{n,i} = 0$$ where $I_{n,i}$ is the current $I_i$ flowing into node $n$. With this mentioned you can insert one equation into another like you would in a normal algebraic equation.

• Ok, I know. I assume an ideal opamp. And this does not answer my question (because I'm asking IF my equations are correct?), but thank for the remark. Mar 18, 2018 at 17:50
• I know, because of eq. 4, that you are assuming an ideal opamp. Your equations seem correct, so using algebra you should be able to solve your problem? Mar 18, 2018 at 17:53
• I get a different result than they get in a different answer (electronics.stackexchange.com/questions/338350/… or electronics.stackexchange.com/questions/361665/…). Mar 18, 2018 at 18:08
• @Looper: your result in your previous thread is wrong. Post your result I will check it for you. Mar 18, 2018 at 18:57
• @Looper: This is my result (hope I didn't make any mistake). ibb.co/hGgeAH Mar 18, 2018 at 19:04