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In Reply to: Is there a formula for calculating the f-1dB pole? posted by SomeJoe on April 23, 2007 at 08:28:26:
You have a series element and a shunt element, one of which is reactive and the other resistive. For the reactive element Z = w.L or 1/w.C , for the resistive element Z = R.dB = 20 log 10 SQRT (Zshunt^2 / (Zseries^2 + Zshunt^2) or
Zshunt^2 + Zseries^2 = Zshunt^2 / 10^(dB/10)
You know one and want to find the other, so
Zseries = Zshunt.SQRT(10^(-dB/10) - 1)
Zseries = Zshunt / SQRT(10^(-dB/10) - 1)
Please note that you must express dB correctly for this to work eg it must always be negative so the expression 10^(-dB/10) is greater than 1.
Follow Ups:
Thank you Mark,So if i understand well, if i want to calculate the f-1dB frequency of a RC filter, where C is the series element of the potential divider:
Zseries = Zshunt.SQRT(10^(-dB/10) - 1)
Replacing Zseries:
1/w.C = Zshunt.SQRT(10^(-dB/10) - 1)or:
(1) 1/(2.pi.f.C) = Zshunt.SQRT(10^(-dB/10) - 1)
(2) 2.pi.f.C.Zshunt.SQRT(10^(-dB/10) - 1) = 1
(3) f = 1/(2.pi.C.Zshunt.SQRT(10^(-dB/10) - 1)for dB = -1, C=.1uF, R=1meg this gives me f = 3.13Hz, where the classic 1/2.pi.R.C formula gives a -3dB point at 1.59Hz. The values seem to make sense, but i must admit math is not my strengh... nor is english :O)
Am i on the right track?
Joris
Sounds right.This leads to a simplification: for a 1st order filter there is about an octave between the -3dB and -1 db points.
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