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Re: Yes

Posted by SomeJoe (A) on April 24, 2007 at 18:08:58

In Reply to: Yes posted by Mark Kelly on April 23, 2007 at 18:10:22:

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

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