# How to derive the transfer function for this Texas Instruments Application note narrowband pass filter

I've been trying to derive the transfer function for this filter but I'm absolutely stumped, and none of the functions I've come up with seem to reflect its actual response. (It is taken from here): • Have you tried nodal analysis?
– Chu
Oct 15 '19 at 22:04

It's a multiple feedback band pass filter like this: - R3 in the above picture is R3 + R4 in your picture. Cin in your picture is not modelled but it's probably irrelevant if high enough in value.

• even though the caps in the original post are between the postive and negative op amp inputs?? Oct 15 '19 at 12:55
• @ScottSeidman check again! Oct 15 '19 at 13:04
• That hurts my head Oct 15 '19 at 13:06

The calculation (by hand) is, of course, possible. I recommend the following steps:

1.) Perform - from the beginning - all calculations with transconductances only (Y instead of Z).....that means: Rename each path between two nodes. For example: 1/R1=Y1 and 1/sC1=Y2 and 1/R2=Y3 .... (So the following steps are simpler to write down....at the end you can go back to the actual parts).

2.) Now you have a circuit with five transconductances Y1...Y5.

3.) Allocate node voltages - for example: The voltage (to ground) of the common node of C1 and C2 is V1.

4.) Derive the corresponding node equations in the form (Va-Vb)Yi=(Vb-Vc)Yii+....

5.) From these equations you can easily isolate the ratio Vout/Vin.

6.) Now you can insert again the real parts into the equation..Y3=1/R2....

Comment: Cin and Cout are very large capacitances (just to block DC) and do not appear in the transfer function. During calculation, the DC voltage at the pos. opamp input can be set to ac ground.