hi i need help understanding the filter arrangement of this circuit

 

C2 and R1 form a High pass filter.
Fc = 0.16/RC = 0.034Hz
Next C1 and R7 form a low pass filter Fc = 0.16/RC = 10.6MHz

Next we have a op-amp non-inverting amp. The C3 and C4 and R2 and R4 form a bandpass type response filter.
With maximum gain equal to

Av = 1 + (C3*R3)/(R2*C3 + R4*C4) = 2.4 [V/V]

At frequency equal to

\(Fo = \frac{1}{2 *\pi \sqrt{C3*C4*R2*R4}} = 4.9KHz\)

And at the end we have another High pass filter (C6, R6).
Fc = 0.16/RC = 0.16Hz And that is the end of this simplified analysis.

Is it those information you were looking for?

 

Hi Jony,

Next C1 and R7 form a low pass filter Fc = 0.16/RC = 10.6MHz

I am curious to know why C1 and R7 are separated.
Why don't we put C1 in parallel with R8?
And could you explain the role of R8 here?

 

I am curious to know why C1 and R7 are separated.
Why don't we put C1 in parallel with R8?

You can place C1 in parallel with R8 if you like. This change don't have any effect on the circuit operation.
Because C2>>C1 we can treat C2 as a short circuit.
So C1 time constant is equal to:
t = C1 * Rth and Fc = 1/(2 * pi *t)

Rth = R7||R8||R1 ≈ R7

And could you explain the role of R8 here?

R8 provide a path to ground for C2 left plate.
So when we disconnect input source (signal). We don't leave input (C2) floating. So we use R8 as a pull-down resistor.

R6 at the output is also pull-down resistor.

 

For other parts of the circuit, I see that thread help me a lot.
http://forum.allaboutcircuits.com/showthread.php?t=67075
Thanks Jony!