In article <XvydnQv1cPdCKZDcRVn-qg@comcast.com>, Peter Nachtwey <peter@deltacompsys.com> wrote:>ftp://ftp.deltacompsys.com/public/mcd/Bessel%20High%20Pass.htm >A high pass Bessel filter. It doesn't look much different from the >Butterworth filter. Beware, the corner frequencies are really higher than >the value used in the calculations. The filters aren't hard to make if you >have Mathcad.Thanks! I don't have Mathcad. The problem I had with the Butterworth (modified for critical damping) is that it was critically-damped only for 1 pass, but stacking together multiple filters to get better stopband attenuation seemed to lose the critical damping characteristic. When I get home today I'll see how the Bessel stacks up (so to speak). -A
Wanted: critically-damped high-pass IIR filter
Started by ●July 31, 2004
Reply by ●August 2, 20042004-08-02
Reply by ●August 2, 20042004-08-02
"axlq" <axlq@spamcop.net> wrote in message news:cejuqq$uia$1@blue.rahul.net...> In article <410d2a5c$0$2845$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >> Jerry suggested looking at Bessel filters, but after researching all > >> afternoon, (unbelievable how hard it is to find good information > >> about this online!) I finally discovered from > >> http://users.ece.gatech.edu/~mleach/ece4435/filtrpot.pdf that a > >> Bessel "can only be realized as a low-pass filter. This is because > >> a time delay cannot be realized with, for example, a high-pass > >> filter." > > > >Oops! I'm sorry for the lapse. > > Maybe not a lapse... http://www.rane.com/note147.html shows a Bessel > high-pass filter, but I have no idea how it was made. I found > several references to highpass Bessels by googling for the string > (with quotes) "high pass bessel" and also "highpass bessel" but so > far I can't see how it's constructed.You can certainly transform a LP Bessel filter into a HP one with the standard methods. It may not retain some of the desirable characteristics of a Bessel filter such as linear phase in the pass band, but it can be done just the same. My suggestion, do you already know how to construct HP filters with varying damping or "Q"? If so, just experiment with different values of Q with your cascaded filters in order to get the response you are after. If it rings more than you like, decrease Q.
Reply by ●August 2, 20042004-08-02
axlq wrote:> In article <XvydnQv1cPdCKZDcRVn-qg@comcast.com>, > Peter Nachtwey <peter@deltacompsys.com> wrote: > >>ftp://ftp.deltacompsys.com/public/mcd/Bessel%20High%20Pass.htm >>A high pass Bessel filter. It doesn't look much different from the >>Butterworth filter. Beware, the corner frequencies are really higher than >>the value used in the calculations. The filters aren't hard to make if you >>have Mathcad. > > > Thanks! I don't have Mathcad. > > The problem I had with the Butterworth (modified for critical > damping) is that it was critically-damped only for 1 pass, > but stacking together multiple filters to get better stopband > attenuation seemed to lose the critical damping characteristic. > When I get home today I'll see how the Bessel stacks up (so to > speak). > > -AYou don't design a 4th-order filter by stacking two 2nd-order filters. Check this: A pair of cascaded 2-pole Butterworth filters doesn't have a 4-pole butterworth response. Neither of the two 2-pole filters which when cascaded give a 4-pole Butterworth response are themselves Butterworth filters. Moreover, they are different. In the foregoing, replace "Butterworth" with "Bessel", Chebychev, Cauer, or whatever. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 2, 20042004-08-02
Jerry Avins <jya@ieee.org> writes:> > You don't design a 4th-order filter by stacking two 2nd-order filters.Why not? If you consider the "order" the number of dB/octave the composite filter rolls off in the stopband divided by 12, then this is exactly what a Bode plot will tell you. If the two 2nd-order filters have different cutoff frequencies, then of course you'll have to measure the rolloff past the highest one. Or have I completely forgotten Electronic Systems 101? -- % Randy Yates % "Rollin' and riding and slippin' and %% Fuquay-Varina, NC % sliding, it's magic." %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Living' Thing', *A New World Record*, ELO http://home.earthlink.net/~yatescr
Reply by ●August 2, 20042004-08-02
Jerry Avins <jya@ieee.org> writes:> > You don't design a 4th-order filter by stacking two 2nd-order filters.I should also add that I'm fairly certain I've seen Nth order digital filters split into N/2 biquad (2nd-order) sections (N = 2n). -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://home.earthlink.net/~yatescr
Reply by ●August 3, 20042004-08-03
Randy Yates wrote:> Jerry Avins <jya@ieee.org> writes: > >>You don't design a 4th-order filter by stacking two 2nd-order filters. > > > Why not? If you consider the "order" the number of dB/octave the composite > filter rolls off in the stopband divided by 12, then this is exactly what > a Bode plot will tell you. If the two 2nd-order filters have different > cutoff frequencies, then of course you'll have to measure the rolloff > past the highest one. > > Or have I completely forgotten Electronic Systems 101?Maybe. For simplicity, cascade two low-pass sections, 1/(1+st)^2 and examine the sharpness of the corner. Compare to 1/(1+ast)*(1+st/a). The first example is the equivalent of a=1. The corner is as sharp as it can get without peaking when a=sqrt(2). Disclaimer: that's all from memory. I didn't work it our or even graph it, so I know that my reputation for (ahem) infallibility (choke kaff) is on the line. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●August 3, 20042004-08-03
In article <4qnlt0ax.fsf@ieee.org>, Randy Yates <yates@ieee.org> wrote:>Jerry Avins <jya@ieee.org> writes: >> >> You don't design a 4th-order filter by stacking two 2nd-order filters. > >I should also add that I'm fairly certain I've seen Nth order digital >filters split into N/2 biquad (2nd-order) sections (N = 2n).Yes, here is was confused me: http://kwon3d.com/theory/filtering/fil.html Near the bottom there are some recursive formulas for generating filter coefficients for a Butterworth filter of order 2N. Following those formulas there's this statement: "For example, a 6th-order low-pass filter has 3 (6 divided by 2) elementary 2nd-order filters. Passing the data through these 3 2nd-order filters consecutively is the same to passing them through the 6th-order filter once." So, I believed it until someone else demonstrated to me by sending graphics in private email that cascading two 2nd-order Butterworths isn't the same as a 4th-order Butterworth. I must ask: besides computation time, is there an advantage to using a single 4th-order or higher-order filter, rather than cascading some 2nd-order filters to acheive approximately the same result? -A
Reply by ●August 3, 20042004-08-03
axlq@spamcop.net (axlq) writes:> In article <4qnlt0ax.fsf@ieee.org>, Randy Yates <yates@ieee.org> wrote: >>Jerry Avins <jya@ieee.org> writes: >>> >>> You don't design a 4th-order filter by stacking two 2nd-order filters. >> >>I should also add that I'm fairly certain I've seen Nth order digital >>filters split into N/2 biquad (2nd-order) sections (N = 2n). > > Yes, here is was confused me: > > http://kwon3d.com/theory/filtering/fil.html > > Near the bottom there are some recursive formulas for generating > filter coefficients for a Butterworth filter of order 2N. > > Following those formulas there's this statement: "For example, > a 6th-order low-pass filter has 3 (6 divided by 2) elementary > 2nd-order filters. Passing the data through these 3 2nd-order > filters consecutively is the same to passing them through the > 6th-order filter once." > > So, I believed it until someone else demonstrated to me by sending > graphics in private email that cascading two 2nd-order Butterworths > isn't the same as a 4th-order Butterworth. > > I must ask: besides computation time, is there an advantage to using > a single 4th-order or higher-order filter, rather than cascading > some 2nd-order filters to acheive approximately the same result?For digital filters, the answer is Yes. The nastiness of numerical problems (overflow and circulated roundoff errors) is easier managed in s series of simple second-order filters. -- % Randy Yates % "And all that I can do %% Fuquay-Varina, NC % is say I'm sorry, %%% 919-577-9882 % that's the way it goes..." %%%% <yates@ieee.org> % Getting To The Point', *Balance of Power*, ELO http://home.earthlink.net/~yatescr
Reply by ●August 3, 20042004-08-03
Randy Yates <yates@ieee.org> writes:> axlq@spamcop.net (axlq) writes: >> I must ask: besides computation time, is there an advantage to using >> a single 4th-order or higher-order filter, rather than cascading >> some 2nd-order filters to acheive approximately the same result? > > For digital filters, the answer is Yes. The nastiness of numerical > problems (overflow and circulated roundoff errors) is easier managed > in s series of simple second-order filters.Sorry! I answered the opposite question of the one you were asking! No, not that I know of (to answer your question). And also, now that I think about this for a second more, it MUST be the case, at least theoretically, that cascaded filters give the same response as one big filter. This is basic linear system theory. The sections are linear systems, and the frequency response of the composite filter is the product of the invidividual filters. If there is some problem in cascading, it must be a practical one. Can you expound? -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr
Reply by ●August 3, 20042004-08-03
Jerry Avins <jya@ieee.org> writes:> Randy Yates wrote: > >> Jerry Avins <jya@ieee.org> writes: >> >>>You don't design a 4th-order filter by stacking two 2nd-order filters. >> Why not? If you consider the "order" the number of dB/octave the >> composite >> filter rolls off in the stopband divided by 12, then this is exactly what >> a Bode plot will tell you. If the two 2nd-order filters have different >> cutoff frequencies, then of course you'll have to measure the rolloff >> past the highest one. Or have I completely forgotten Electronic >> Systems 101? > > Maybe. For simplicity, cascade two low-pass sections, > 1/(1+st)^2 and examine the sharpness of the corner. Compare to > 1/(1+ast)*(1+st/a). The first example is the equivalent of a=1. The > corner is as sharp as it can get without peaking when a=sqrt(2).First of all, what is t? These are s-domain equations, right? What is a "t" doing in them (if it is time)? Second, of course these two filters won't be the same. They're different (if a != 1). If a = 1, then they should be identical. And if a != 1, if you cascaded 1/(1 + ast) with 1/(1 + st/a) you should get identical results and one filter with response 1/[(1 + ast)(1 + st/a)]. There are second-order systems that can't be factored into real, first-order systems. Is that what you're getting at? -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr
