what you're looking at is an active bandpass filter

The transfer function (Vout/ Vin) is messy.

K is 2

D is 1/3

Vo =

sC2R2KD Vi s^2C1C2R1R2D + s[C1R1D + C2R2(1-K)(1-D) +C2(R1+R2)D) + 1

where:

R1 = 11k3

R2 =7k87

C1= 0.047uF

C2= 0.047uF

get the co efficients down the bottom make into a quadratic then finding where Vo/Vi = infinte will give the two frequencies at each side of your bandpass

subing in values

Vo/Vi = s(2.466E-4) / s^2(6.548E-

+ s(1.77E-4 - 2.466E-4 + 3.003E-4) + 1

Vo/Vi = s(2.466E-4) / s^2(6.548E-

+ s(2.307E-4) + 1

and graphically i get one pole at 3908rad/sec => 622Hz and the other pretty close to it again (can't see graphically)

i hope this is right (haven't done the calc's in a long time)

so i think it's basically a bandpass filter that emphasises the region around 622Hz

grab a electronics textbook if you realy want to find out more.

Stu