I have the following third-order filter circuit:
simulate this circuit – Schematic created using CircuitLab
And I know that the transfer function looks like:
$$\mathscr{H}\left(\text{s}\right)=\frac{1}{\alpha_1\cdot\text{s}^3+\alpha_2\cdot\text{s}^2+\alpha_3\cdot\text{s}+1}\tag1$$
Where:
$$\alpha_1=\text{C}_1\cdot\text{C}_2\cdot\text{C}_3\cdot\text{R}_1\cdot\frac{\text{R}_2\cdot\text{R}_3}{\text{R}_2+\text{R}_3}\cdot\left(\text{R}_1+\text{R}_2\right)\tag2$$
- $$\alpha_2=\text{C}_1\cdot\text{C}_2\cdot\text{R}_1\cdot\left(\text{R}_1+\text{R}_2\right)+\text{C}_2\cdot\text{C}_3\cdot\text{R}_3\cdot\left(\text{R}_1+\text{R}_2\right)\tag3$$
- $$\alpha_3=\text{C}_1\cdot\text{R}_1+\text{C}_2\cdot\left(\text{R}_1+\text{R}_2+\text{R}_3\right)\tag4$$
But how can I use current node analysis to find this transfer function (assuming an ideal op-amp)?
My work:
I wrote the current node equations:
- $$\text{I}_1=\text{I}_{\text{R}_1}+\text{I}_{\text{C}_1}+\text{I}_{\text{R}_2}\tag5$$
- $$\text{I}_2=\text{I}_{\text{R}_2}+\text{I}_{\text{R}_3}+\text{I}_{\text{C}_3}\tag6$$
- $$\text{I}_3=\text{I}_{\text{R}_3}+\text{I}_{\text{C}_2}\tag7$$
- $$\text{I}_4=\text{I}_{\text{C}_3}\tag8$$
- $$\text{I}_5=\text{I}_4=\text{I}_{\text{C}_3}\tag9$$
- And of course I know that ideal op amp equation: $$\text{V}_+=\text{V}_-\tag{10}$$
But now I do not know how to continue using the voltages at those nodes.