X-GM-THRID: 1234888028149353688 X-Gmail-Labels: rsgb lf Delivered-To: daveyxm@gmail.com Received: by 10.35.22.5 with SMTP id z5cs334217pyi; Fri, 27 Apr 2007 17:07:36 -0700 (PDT) Received: by 10.67.116.3 with SMTP id t3mr3350881ugm.1177718855822; Fri, 27 Apr 2007 17:07:35 -0700 (PDT) Return-Path: Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by mx.google.com with ESMTP id o24si7933972ugd.2007.04.27.17.07.30; Fri, 27 Apr 2007 17:07:35 -0700 (PDT) Received-SPF: neutral (google.com: 193.82.116.20 is neither permitted nor denied by best guess record for domain of owner-rsgb_lf_group@blacksheep.org) DomainKey-Status: bad (test mode) Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1HhaQ8-0000B6-Lg for rs_out_1@blacksheep.org; Sat, 28 Apr 2007 01:04:12 +0100 Received: from [83.244.159.144] (helo=relay3.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1HhaQ7-0000Aw-QO for rsgb_lf_group@blacksheep.org; Sat, 28 Apr 2007 01:04:11 +0100 Received: from smtp810.mail.ird.yahoo.com ([217.146.188.70]) by relay3.thorcom.net with smtp (Exim 4.63) (envelope-from ) id 1HhaQ5-0004VK-RB for rsgb_lf_group@blacksheep.org; Sat, 28 Apr 2007 01:04:11 +0100 Received: (qmail 49667 invoked from network); 28 Apr 2007 00:04:04 -0000 DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=s1024; d=btopenworld.com; h=Received:X-YMail-OSG:Received:Message-ID:From:To:References:Subject:Date:MIME-Version:Content-Type:X-Priority:X-MSMail-Priority:X-Mailer:X-MimeOLE; b=4yZu6sb2yZ/vIauo3UbSECH+txIo42ZDVifkqs7afznaJu5FjB8tH+6SbXHtm1FUlRWe+N1UDtPBFVftVcckF1G272RLT0EQx18rPz/IcCR5UVouiMSIcGUlZvcDNfGtQijoOtO3G3dMOMGFdCHftdhI3BOigzBx/oQEIWULSWM= ; Received: from unknown (HELO w4o8m9) (james.moritz@btopenworld.com@213.122.44.178 with login) by smtp810.mail.ird.yahoo.com with SMTP; 28 Apr 2007 00:04:02 -0000 X-YMail-OSG: Ms3JTNoVM1kB1r_GmoDUlEyBGQZGfvUTwLgS9iK6K0LDX559cli6sRzaTUCVEUkCoRJhiiXq9w-- Received: from 127.0.0.1 (AVG SMTP 7.5.463 [269.6.1/777]); Sat, 28 Apr 2007 01:04:34 +0100 Message-ID: <001901c78928$cd6e3f00$b22c7ad5@w4o8m9> From: "James Moritz" To: References: <001701c788d3$612266f0$6405a8c0@MJUSonyLaptop> Date: Sat, 28 Apr 2007 01:04:33 +0100 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2800.1106 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2800.1106 DomainKey-Status: good (testing) X-Spam-Score: 0.0 (/) X-Spam-Report: autolearn=disabled,HTML_MESSAGE=0.001 Subject: Re: LF: RE: Filter expert needed - Mystery of the lost real zero. Content-Type: multipart/alternative; boundary="----=_NextPart_000_0016_01C78931.2EE7E260" X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.0 required=5.0 tests=HTML_MESSAGE autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false Status: O X-Status: X-Keywords: X-UID: 2008 This is a multi-part message in MIME format. ------=_NextPart_000_0016_01C78931.2EE7E260 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Dear Andy, LF Group, Well, this should put an end to anyone's insomnia :-) For your particular circuit, the "missing" zero is at minus 40MHz. The = poles and zeroes in a transfer function have to be symmetrical about the = real axis of the complex frequency plane, so have to occur in conjugate = pairs if they are not on the real axis. In this case there are two zeros = at s =3D 0 +/- j*2pi*40Meg. But this does not mean that a bandpass = frequency response has to be symmetrical about the centre frequency - it = is quite OK for there to be a rejection notch on one side of the = passband and not on the other. It does mean the response will be the = same for positive and negative frequencies, i.e. symmetrical about zero = frequency. One classic way of designing a bandpass filter with a symmetrical = response is to start with a low-pass ladder filter (i.e. series = inductors, shunt capacitors) response with the desired bandwidth, = ripple, etc., then to perform "lowpass to bandpass transformation", = which results in each shunt C being replaced with a parallel LC with the = original value of C, and each series L being replaced by a series LC = with the original value of L. The other component in each LC has a value = such that each LC is resonant at the desired centre frequency. This = gives a response with the same bandwidth and shape as the original = low-pass, but with perfect geometrical symmetry about the centre = frequency, i.e. if centre frequency =3D fo, the response at k*fo is the = same as at fo/k . This process also works if you start with an elliptic = low-pass, in which case the series arm capacitors in the low-pass = elliptic become parallel LCs and shunt arm inductors become series LCs. = This will again give a geometrically symetrical response, with any zeros = (notches) above the passband mirrored by corresponding zeroes below the = passband. The trouble with this type of design is that for passbands = that are a small fraction of the centre frequency, the component values = vary over an impractically wide range (very small shunt inductors, very = large series inductors and so on). There are a whole range of tricks for = making the design more practical, as Mike pointed out. The initial circuit Andy has designed is a "coupled resonator" filter, = which is the classic way of designing a narrow-band bandpass filter. = This has the advantage that the starting point is a number of identical = LC resonators (or other types of resonator), and the required bandwidth = and response shape is obtained by changing the coupling capacitors (or = it could be inductors, irises etc.), and the load resistances. Again = there are lots of tricks for making the design more practical, but you = only have to worry about getting the right coupling. The trouble with = this is that it relies on a "narrow band approximation", where the = mathematical synthesis assumes (so I'm told...) that the coupling = impedance does not vary with frequency. This is a good approximation in = a narrow bandwidth, but gives rise to increasing deviation from the = assumed frequency response as you move further from the centre = frequency. Thus, the type of filter in Andy's design is inherently = asymmetrical, as can be seen from the frequency response plots - the = high frequency side has lower attenuation. The limitation of this type = of design comes when the passband becomes unacceptably lop-sided with = wide bandwidths. On the other hand, the transformed low-pass type can = have any combination of centre frequency and bandwidth, and will still = give a response which is perfectly symmetrical when plotted on a log = frequency axis. Andy's modification results in a circuit with a combination of = asymmetrical band-pass and notch filter responses - provided the = desirable notch reponse does not turn out to be horribly sensitive to = slight component value changes, or to depend on the precise interaction = of several components, this is OK. You will probably also find that = stop-band attenuation is reduced due to the by-passing effect of the = added capacitor - from the plots, a few dB attenuation have been lost at = the HF end. But this might be perfectly acceptable. Another possible = problem is the depenence on the unloaded Q of the components. In the = circuit shown on Andy's web page, there is no loss, so infinite Q. It = might be found that the notch is much less pronounced if some loss = resistance is added in series with each inductor. Cheers, Jim Moritz 73 de M0BMU ------=_NextPart_000_0016_01C78931.2EE7E260 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Dear Andy, LF Group,
 
Well, this should put an end to = anyone's insomnia=20 :-)
 
For your particular circuit, the = "missing"=20 zero is at minus 40MHz. The poles and = zeroes in a=20 transfer function have to be symmetrical about the real axis of the = complex=20 frequency plane, so have to occur in conjugate pairs if they are not on = the real=20 axis. In this case there are two zeros at s =3D 0 +/- j*2pi*40Meg. But = this does=20 not mean that a bandpass frequency response has to be = symmetrical=20 about the centre frequency - it is quite OK for there to be a rejection = notch on=20 one side of the passband and not on the other. It does mean the response = will be=20 the same for positive and negative frequencies, i.e. symmetrical about = zero=20 frequency.
 
One classic way of designing a = bandpass filter=20 with a symmetrical response is to start with a low-pass ladder filter = (i.e.=20 series inductors, shunt capacitors) response with the desired bandwidth, = ripple,=20 etc., then to perform "lowpass to bandpass transformation", which = results in=20 each shunt C being replaced with a parallel LC with the original = value of=20 C, and  each series L being replaced by a series LC with the = original value=20 of L. The other component in each LC has a value such that each LC = is=20 resonant at the desired centre frequency. This gives a response with the = same=20 bandwidth and shape as the original low-pass, but with perfect = geometrical=20 symmetry about the centre frequency, i.e. if centre frequency =3D fo, = the response=20 at k*fo is the same as at fo/k . This process also works if you = start with=20 an elliptic low-pass, in which case the series arm capacitors in the = low-pass=20 elliptic become parallel LCs and shunt arm inductors become series LCs. = This=20 will again give a geometrically symetrical response, with any zeros = (notches)=20 above the passband mirrored by corresponding zeroes below the passband. = The=20 trouble with this type of design is that for passbands that are a small = fraction=20 of the centre frequency, the component values vary over an impractically = wide=20 range (very small shunt inductors, very large series inductors and so = on). There=20 are a whole range of tricks for making the design more practical, as = Mike=20 pointed out.
 
The initial circuit Andy has designed = is a "coupled=20 resonator" filter, which is the classic way of designing a = narrow-band=20 bandpass filter. This has the advantage that the starting point is a = number of=20 identical LC resonators (or other types of resonator), and the required=20 bandwidth and response shape is obtained by changing the coupling = capacitors (or=20 it could be inductors, irises etc.), and the load resistances. Again = there are=20 lots of tricks for making the design more practical, but you only have = to worry=20 about getting the right coupling. The trouble with this is that it = relies on a=20 "narrow band approximation", where the mathematical synthesis assumes = (so I'm=20 told...) that the coupling impedance does not vary with frequency. = This is=20 a good approximation in a narrow bandwidth, but gives rise to increasing = deviation from the assumed frequency response as you move further from = the=20 centre frequency. Thus, the type of filter in Andy's design is = inherently=20 asymmetrical, as can be seen from the frequency response plots - the = high=20 frequency side has lower attenuation. The limitation of this type of = design=20 comes when the passband becomes unacceptably lop-sided with wide = bandwidths. On=20 the other hand, the transformed low-pass type can have any combination = of centre=20 frequency and bandwidth, and will still give a response = which is=20 perfectly symmetrical when plotted on a log frequency axis.
 
Andy's modification results in a = circuit with a=20 combination of asymmetrical band-pass and notch filter responses - = provided the=20 desirable notch reponse does not turn out to be horribly sensitive to = slight=20 component value changes, or to depend on the precise interaction of = several=20 components, this is OK. You will probably also find that stop-band=20 attenuation is reduced due to the by-passing effect of the added = capacitor -=20 from the plots, a few dB attenuation have been lost at the HF = end. But=20 this might be perfectly acceptable. Another possible problem is the = depenence on=20 the unloaded Q of the components. In the circuit shown on Andy's web = page, there=20 is no loss, so infinite Q. It might be found that the notch is much = less=20 pronounced if some loss resistance is added in series with each=20 inductor.
 
Cheers, Jim Moritz
73 de M0BMU
 
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