If you know the -3dB point of a single pole low pass filter, is there an equation you can use to find your dB at other frequencies relative to the cut off?
For example I know that my -3dB cut off is 612kHz, but I'm curious to know how much I've lost at 20kHz... it's a non-linear problem, is there quick way to approximate it?
dB = 20*log(Pout/Pin) = 20/ln(10)*ln(Pout/Pin), (3)
and 20/ln(10) ~= 8.68, and for x near 1,
ln(x) ~= x-1. (4)
so combining (2) and (3)
dB ~= 8.68 ln(Vout/Vin) (5)
plugging (5) into (4),
dB ~= 8.68 (Vout/Vin - 1) (6)
and finally using (2),
dB ~= -4.34 (f/f_c)^2 .
Thus at 20 kHz you're down by
4.34 * (20/612)^2 ~= 0.005 dB.
This formula is pretty good--you note that it's only off by 1.34 dB at f = fc. At 20 kHz it's good to 6 decimals, i.e. 0.000001 dB, which is a lot better than your 612 kHz number.
Cheers
Phil Hobbs
(*) Strictly speaking, decibels are always a power ratio, but RC filters are usually followed by a buffer or other high impedance load, so it's common to use squared voltage in place of power.
A low pass RC filter will cut the output in half every time the frequency doubles. Which is a 6db per octave rolloff or 20db per decade.
The rate of rolloff is very close to 6db starting at 3Fc on up. Because it is a straight line, you can estimate the loss at any frequency by counting the number of octaves or fractions of an octave, then multiply by 6db.
Below 3 or 4 FC the curve is not a straight line, so you might get to within only a db.
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