From: John Woodgate
Subject: Re: New simple LC filter design program
Date: Thu, 26 Dec 2002 22:58:05 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Thu, 26 Dec 2002 23:15:17 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Peter O. Brackett
wrote (in )
about 'New simple LC filter design program', on Thu, 26 Dec 2002:
>> >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
Well, with that special definition of Ap, eps is indeed not zero for a
Butterworth filter. But then, how do you distinguish between a
Butterworth and a Chebyshev with 3 dB ripple? How do you reconcile that
special definition of Ap with the definition of Ap for a Chebyshev with
0.01 dB pass-band ripple?
I think it would be best to use a symbol other than Ap for the
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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