The simple answer is the output at pin 8 IC2 is a band-pass filter. The filter cut-off is determined by the current feeding into pins 1 and 16.

The OTA:

So the way the OTA's work is the input is a voltage, the voltage difference between pin 3 (V+) and pin 4 (V-). That voltage is multiplied by the trans-conductance gm which produce a current at pin 7, Iout = gm * (V+ - V-) The current feeds the impedance Z to ground on pin 9 which converts the output current into a voltage (on pin 7), Vout = gm * (V+ - V-) * Z. The chip has a buffer with input on pin 7 and output on pin 8. So the voltage on pin 8 is Vout = gm * (V+ - V-) * Z. The impedance to ground Z = 1/sC. So Vout = (gm / (sC)) * (V+ - V-). The transductance is a proportional to the current going into pin 1, see datasheet.

The dividers:

pin 3 is fed from a divider 10k + 1k fed from Vin

pin 4 is fed from two dividers 22k + 22k + 1k which sum and are fed from pin 8 and pin 9

pin 14 is fed from a divider 22k + 1k which is fed from pin 8

Suppose we call the output at pin 8 V1 and the output at pin 9 V2.

In order to simplify things, I'm going to say a divider of 10k + 1k is multiplying by a factor k (~ 1/10) and that the 22k + 1k divider is about half that so k/2. So,

V+ = k * Vin

V- = (k/2) * (V2 + V1)

The aim of the game is to solve the circuit to get the output terminal voltage V1 in terms of Vin.

From the way the OTA works,

V1 = (gm / (sC)) * (V+ - V-)

= (gm / (sC)) * ( k Vin - (k/2) (V2 + V1) ) ; substituting V+ and V- above

V2 = (gm / (sC)) * (k/2) V1

Next get rid of V2 from the V1 equation,

V1 = k (gm / (sC)) {Vin - (1/2) *[ (gm / (sC)) * (k/2) V1 + V1 ] }

Shuffle the V1's over to the left hand side,

(sC / (k gm)) * V1 = Vin - (1/2) [ (gm / (sC)) * (k/2) + 1] V1

{ (sC / (k gm)) + (1/2)*[ (gm / (sC)) * (k/2) + 1] } V1 = Vin

{ (sC / (k gm)) + (k gm / (4 sC)) + 1/2 } V1 = Vin

Multiply both sides by (4 sC) / (k gm)

{s^2 (2 C/(k gm))^2 + 1 + s (2 C) / (k gm)} V1 = Vin s (2C / (k gm))

Which gives the transfer function,

V1 / Vin = s (2 C / (k gm))

----------------------------------------------------------------

s^2 (2 C/(k gm))^2 + s (2 C) / (k gm) + 1

= ( s/w0)

--------------------------------------------

(s / w0)^2 + s / (Q w0) + 1

where w0 = k gm / (2C), Q = 1

This transfer function is a second order band-pass filter with center frequency w0 = 2 pi f0, and a Q of 1.

You can look-up in the data sheet how gm is a function of Iabc. If you work out the range of Iabc then you can work out the range of gm and then calculate the range of the center frequency.

It's a bit rough but that's the general way to approach it. I could have stuffed-up here and there above and I haven't tried to verify the f0 equation with a simulation.

It would be worthwhile trying to work through it yourself.

If you wanted to work out what each resistor does you would have to have many k's k1, k2 etc for each divider. Then you will end-up with a big messy equation similar to what I have done with many k's. Then you have to write the k's in terms of each of the R's in the divider and you get an even bigger mess.

You would need to do that if you wanted to try to change the Q for example, although you could do that just with the k's in the equation.