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From: "Peter O. Brackett"
Subject: Re: New simple LC filter design program
Date: Thu, 26 Dec 2002 10:39:23 -0500
Organization: MindSpring Enterprises
X-Server-Date: 26 Dec 2002 15:39:21 GMT
X-Newsreader: Microsoft Outlook Express 5.00.2919.6600
> >Where: eps = sqrt(10^(Ap/10) - 1.0) is the passband ripple factor with
> >in dB.
> What is 'eps' for the Butterworth filters you say these equations apply
> to? It appears to be the square root of zero.
> Regards, John Woodgate, OOO - Own Opinions Only.
Hmmm... No, I don't believe that eps evaluates to zero. Of course the way
your news reader
renders the formula I typed may be different from the way mine does, but in
my case the
expression for epsilon does not evaluate to zero unless Ap is zero. For the
maximally flat characteristic function case the passband ripple is
conventionally taken to be
the value of the passband attenuation in dB at the band edge, namely Ap in
eps = sqrt(10^(Ap/10) - 1.0) Where Ap in dB is the passband attenuation at
Aside: Since the Butterworth or maximally flat characteristic function
places all of the filter's
N reflection zeros [attenuation zeros] at zero frequency, there is really
only *one* ripple
and that is the smooth monotonic roll-up of attenuation from zero to
infinity which attains the
value of Ap [dB] exactly at the band edge.
Incidently in case you have problems with my equation rendering I try to
guidelines given at the interesting and useful URL:
Indialantic By-the-Sea, FL.
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