Transfer Function for Sallen Key Lowpass Filter
To find the transfer function \(H(s) = \frac{V_o}{V_{in}}\) for this second order lowpass filter we replace the capacitors by with their impedances \(Z_C = \frac{1}{sC}\). Then, we write up node equations for \(V_a\) and \(V_b\).
$$\frac{V_a-V_{in}}{R_1}+\frac{V_a-V_b}{R_2}+(V_a-V_o)\cdot sC_2 = 0\tag{1}$$
$$\frac{V_b-V_a}{R_2}+V_b \cdot sC_1=0\tag{2}$$
We have a system of 2 equations with a total of 3 variables. To solve the system of equations, we need a constraint to reduce the number of variables.
Since the OP-amp is considered ideal, we have the following.
$$V_b = V_c = V_o \cdot \frac{R_4}{R_3+R_4} = \frac{V_o}{k}, \: \: \: k = 1+\frac{R_3}{R_4} $$
Notice here that \(k\) is the gain of the non-inverting amplifier and can be controlled by adjusting the values of \(R_3\) and \(R_4\). The equations from before now becomes
$$ V_a \bigg(\frac{1}{R_1}+\frac{1}{R_2}+sC_2 \bigg)-\frac{V_{in}}{R_1}-V_o \bigg(\frac{1}{kR_2}+sC_2 \bigg) = 0\tag{3}$$
$$\frac{V_o}{kR_2}-\frac{V_a}{R_2}+\frac{V_o\cdot sC_1}{k} = 0\tag{4}$$
Isolate \(V_a\) in (4).
$$V_a = \frac{V_o+V_osC_1R_2}{k}\tag{5} $$
Insert (5) in (3).
$$\bigg(\frac{V_o+V_osC_1R_2}{k}\bigg) \bigg(\frac{1}{R_1}+\frac{1}{R_2}+sC_2 \bigg)-V_o \bigg(\frac{1}{kR_2}+sC_2 \bigg) = \frac{V_{in}}{R_1} \tag{6}$$
Divide with \(V_o\) on both sides.
$$\bigg(\frac{1+sC_1R_2}{k}\bigg) \bigg(\frac{1}{R_1}+\frac{1}{R_2}+sC_2 \bigg)- \bigg(\frac{1}{kR_2}+sC_2 \bigg) = \frac{V_{in}}{V_o\cdot R_1} \tag{7}$$
Multiply with \(R_1\) on both sides.
$$\bigg(\frac{R_1+sC_1R_1R_2}{k}\bigg) \bigg(\frac{1}{R_1}+\frac{1}{R_2}+sC_2 \bigg)- \bigg(\frac{R_1}{kR_2}+sC_2R_1 \bigg) = \frac{V_{in}}{V_o} \tag{8}$$
Multiply the parantheses out.
$$\bigg(\frac{1+sC_1R_2}{k} \bigg)+\require{cancel}\cancel{\frac{R_1}{kR_2}}+\frac{sC_1R_1}{k}+\frac{sC_2R_1+s^2C_1C_2R_1R_2}{k}-\cancel{\frac{R_1}{kR_2}}-sC_2R_1 = \frac{V_{in}}{V_o}\tag{9}$$
$$\frac{s^2C_1C_2R_1R_2+sC_1R_1+sC_1R_2+sC_2R_1+1}{k}-sC_2R_1 = \frac{V_{in}}{V_o} \tag{10} $$
$$\frac{s^2C_1C_2R_1R_2+sC_1R_1+sC_1R_2+sC_2R_1-ksC_2R_1+1}{k} = \frac{V_{in}}{V_o} \tag{11} $$
$$H(s) = \frac{V_o}{V_{in}} = \bigg(\frac{V_{in}}{V_o} \bigg)^{-1} = \frac{k}{s^2C_1C_2R_1R_2+sC_1R_1+sC_1R_2+sC_2R_1-ksC_2R_1+1} \tag{12} $$
We have found the transfer function. But it is not in standard form. To bring it to that form, that term in front of \(s^2\) needs to be 1. We divide with \(C_1C_2R_1R_2\) in every term of the transfer function.
$$H(s) = \frac{V_o}{V_{in}} = \frac{\frac{k}{C_1C_2R_1R_2}}{s^2+s(\frac{1}{C_2R_2}+\frac{1}{C_2R_1}+\frac{1}{C_1R_2}-\frac{k}{C_1R_2})+\frac{1}{C_1C_2R_1R_2}} \tag{13}$$
And we are done.