Algebraic way for finding a transfer function of a filter

Question

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)?

Answer 1

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.

Answer 2

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.