equation for dB filter frequencies if you know cut off

The filter's frequency response is

Vout/Vin = 1/sqrt(1 + (f/f_c)^2). (1)

Well below f_c, that's close to

Vout/Vin ~= 1 - 0.5*(f/f_c)^2, (2)

which of course is close to 1. Now (*)

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

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.

Yes, there is an equation, which is easily solved. I get,

|H| = 0.99947 at 20kHz

It's not a complicated equation, but if you insist, an approximation would be: |H| ~= 1 - (f/f0)^2 / 2 when f Hello,

This is basic signal processing. You really should learn some of the theory in a cohesive fashion.

A single-order low-pass filter, if it's continuous-time and linear, has a transfer function

where s is the Laplace frequency operator, and w_0 (which should be omega_0) is the 3dB cutoff frequency in radians per second.

To find the response at any frequency replace s with j * w, where j = sqrt (-1) and w (omega) is the radian frequency.

At 20kHz, w = 2 * pi * 20kHz.

So

which has magnitude 1/sqrt(0.0327^2 + 1), or around 0.9995

Low pass filter? BOTE (with no envelop) is about -12db?

Sorry?

Cheers

Phil Hobbs

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.

1 octave = 1224hz = -6db

3 octaves = 4896 hz = -18db

5 octaves = 19584 hz = -30db

"panfilero"

.... Phil

my mistake - I was seeing a low pass filter with the corner frequency of