Neil,
it will depend on your input and output impedances.
You required an output Z << R1 (Or add to R1 for and effective value) and
an input impedance >> R2 (or it form s a voltage divider, lowers the
effective inpedance of R2 & raises the cutoff  frequency)

Then the simplist method is to make R2 about 10x or more times the value of
R1 and then you can treat it as 2 simple low pass filters.

For the first stage, the voltage drop will be to ~70% (3db) when the
impedance of the Cap = that of the resistor i.e. R=1/(2*pi*f*C) or C = 1/(2
x pi x F x R)
(You might expect 6dB but the impedance is complex)

Since the second set of parts have much higher impedance, it will hardly
effect the response of the first part so the same equation holds.
E.g for a 1KHz cutoff

R1 = 1K then C1 = 1/(2 x pi x 1000 x 1000) = 0.16uF
 if R2 = 10k then C2 = 16nF

The susequent overall slope will be -12dB/ Octave = -40dB/decade with a
-6dB point at 1kHz.
Note, however the -3dB cutoff frequency has moved, as the loss at 1kHz is
now ~6dB. You can adjust the values to compensate if required.
Hope this assists, I know it's not an exact answer to the question - I
normally model these things on  Pspice to tune up convienient values & to
include source & load impedances..

Richard P



For some reason, I can't find (thru google) the formula for calculating the

best values of R's and C's in this circuit for a given cutoff frequency...

   >----`\/\/\,----o----`\/\/\,-----o------>
           R1      |       R2       |
                  ---              ---
               C1 ---           C2 ---
                   |                |
   >---------------o----------------o------>

Any of you have the formula?

Thanks,
-Neil.

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