More information about the above filter can be found here. From first look it seems like R1,C2,R3,R2,C1 forms a 2nd order low pass filter while R4,C3 makes a 1st order low pass filter. The 3rd order is acheived by cascading of 2 filters (one 2nd order filter & one 1st order filter). If the transfer function is found of the filter based on FDA than the overall transfer function of the 3rd order filter can be found easily.

Some important points to note of a FDA before proceeding forward for deriving the transfer function :

1. Voltage  \(V_2\) & \(V_4\) at Node 2 and Node 4 respectively are almost equal. This is because of the  finite output divided by very high amplification factor of the FDA.
2. The input impedance is very high (ideally infinite).
3. The components should be matched with high degree of precision. Higher the match, higher the CMRR. So in the diagram the components are already matched.
4. The \(V_{out+}\) is connected to \(V_{in-}\) to provide a negative feedback. Similarly for \(V_{out-}\) & \(V_{in+}\).
5. \(V_{out+} = V_5\) & \(V_{out-} = V_6\).

Nodal analysis @ 1

\[\frac{V_{in-} – V_1}{R_1} + \frac{V_{out+} – V_1}{R_2} + \frac{V_2 – V_1}{R_3} + (V_3 – V_1).sC_2 = 0\]

Nodal analysis @ 2

\[\frac{V_1 – V_2}{R_3} + (V_{out+} – V_2).sC_1 = 0\]

By symmetry,

Nodal analysis @ 3

\[\frac{V_{in+} – V_3}{R_1} + \frac{V_{out-} – V_3}{R_2} + \frac{V_4 – V_3}{R_3} + (V_1 – V_3).sC_2 = 0\]

Nodal analysis @ 4

\[\frac{V_3 – V_4}{R_3} + (V_{out-} – V_4).sC_1 = 0\]
Taking difference b/w Node equation of 2 and 4 & putting \(V_2 – V_4 = 0\), we get

\[V_3 – V_1 = (V_{out+} – V_{out-}).sR_3C_1\]

Taking difference b/w Node equation of 1 and 3, we get

\[\frac{V_{in+} – V_{in-}}{R_1} = \frac{V_{out+} – V_{out-}}{R_2} + (V_3 – V_1).(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+2sC_2)\]

Putting \(V_3 – V_1\) and after few adjustments we get,

\[\frac{V_{out+} – V_{out-}}{V_{in+} – V_{in-}} = \frac{\frac{1}{2R_1R_3C_1C_2}}{s^2 + \frac{s}{2(R_1||R_2||R_3)C_2} + \frac{1}{2R_2R_3C_1C_2}}\]

From aforementioned 2nd order equation we find,

\[Q = \frac{2(R_1||R_2||R_3)C_2}{\sqrt{2R_2R_3C_1C_2}} = \frac{\sqrt{2R_2R_3C_2C_1}}{R_2C_1 + (1+K)R_3C_1}\]

Where K is pass band gain,

\[K = \frac{R_2}{R_1}\]

\[\omega{}_n = \frac{1}{\sqrt{2R_2R_3C_1C_2}}\]

Now as per the instruction given in the application note, we have to find some practical values for the resistors and capacitors.

The desired cutoff frequency has to be 133KHz as per \(R_{max}\), when the time shift would be maximum and the corresponding beat frequency.