09-13-11: Paperw8
the equation for dB measurement is:
q_dB=10*log(q1/q2)
where q1 and q2 are measured quantities and q_dB is the dB measurement for those quantities. when q1=q2/2, q_dB is -3dB.
if you read closely the reference that you cited, you can see that what the writer is saying is clearly wrong. 6dB/octave is the falloff for a *second* order filter; the filter shown in the cited reference is a first order filter.
Paperw8, we had discussed the definition of db for electrical signals in another thread a while back, and as I indicated then, with all due respect you are simply wrong. Please do some further research, and I think you will see that:
db = 10log(P1/P2) = 20log(V1/V2)
where P1/P2 is the ratio of two powers, and V1/V2 is the ratio of two voltages. 6db is 6db, regardless of whether it is derived on the basis of power (where it represents a factor of 4) or voltage (where it represents a factor of 2).
I should qualify my statement about 6db/octave rolloff for a first order filter, though, by adding that at frequencies that are just a small amount higher than the 3db point (Fc) the rolloff will be somewhat less than 6db/octave. As frequency increases further the rate of rolloff will become progressively closer to 6db/octave. The actual equation is:
Vout/Vin = 1/(square root(1 + (f/Fc)^2))
During the first octave above Fc, a rolloff of about 4db will occur. As f increases further, the "1" in the denominator becomes progressively less significant, bringing the rolloff rate progressively closer to 6db/octave.
Regards,
-- Al