help needed for Butterworth crossover response
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From: gangadhar.m on 18 Jul 2007 03:18 I got the answer. I have to use Polarity inversion also as specified in the document availabe at the link www.lake.com.au/documentation/Lake%20Contour%20Classical%20Crossover%20Notes.pdf The document says that "When using 24 dB and 48 dB L-R crossovers, there is no need to reverse the polarity of adjacent crossover channels. However, the 12 dB and 36 dB L-R crossovers will require polarity reversals. In order to provide the desired constant magnitude summation, the polarity of an output channel must be inverted." Thank you all Gangadhar > >Dear Jerry, > >As i mentioned earlier i get the same kind of response in case of Linkwitz >Riley filters also..... In net, i find papers where they have flat all pass >response for Linkwitz Riley crossovers(LR-2, LR-3 and LR-4 filters) > >but for me, all pass response is observed only for the LR-4 crossover and >big dig at cutoff for the other LR Crossovers. Similar cases are observed >for Butterworth crossovers also. > >Since we are having butterworth crossovers available, i assume that all >butterworth crossovers shoud be able to exhibit 3dB increase at the cutoff >irrsepective of their order. How can I achieve that? > >Please help me if my understanding is not correct. > >Regards >Gangadhar >>gangadhar.m wrote: >>> Hi, >>> >>> ... Can you please tell >>> me how should I proceed now to achieve 3 dB increase at the cut off >>> frequency for all butterworth crossovers(2nd,3rd and 4th orders) >>> irrsepective of change in the order. ... >> >>Butterworth is Butterworth. If you change it in any way, it's not >>Butterworth any more. Try Linkwitz-Reilly. >> >>Jerry >>-- >>Engineering is the art of making what you want from things you can get. >>¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ >> >
From: Jerry Avins on 18 Jul 2007 12:22 gangadhar.m wrote: > Dear Jerry, > > As i mentioned earlier i get the same kind of response in case of Linkwitz > Riley filters also..... In net, i find papers where they have flat all pass > response for Linkwitz Riley crossovers(LR-2, LR-3 and LR-4 filters) > > but for me, all pass response is observed only for the LR-4 crossover and > big dig at cutoff for the other LR Crossovers. Similar cases are observed > for Butterworth crossovers also. > > Since we are having butterworth crossovers available, i assume that all > butterworth crossovers shoud be able to exhibit 3dB increase at the cutoff > irrsepective of their order. How can I achieve that? > > Please help me if my understanding is not correct. A Butterworth low-pass filter is one whose frequency-response derivatives are all zero at w = 0. Band-pass and high-pass Butterworth filters can be derived from it by the usual transformations or derived directly by setting all derivatives to zero at w = infinity (high-pass) or at the center frequency (band-pass). A consequence is that the transfer function of a Butterworth low-pass prototype of order n is H(s) = 1/sqrt(1 + (w/w_c)^2n) That's it, unalterably. If that response isn't suitable, then Butterworth isn't suitable. This response can only be approximated by a digital filter, and the approximation is usually crude by the standards of those who listen with their calculators instead of with their ears. There is another degree of freedom with crossovers. It is not written in stone that the corner frequencies of the high-pass and low-pass sections must be equal. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
From: Vladimir Vassilevsky on 18 Jul 2007 17:14 Jerry Avins wrote: > A Butterworth low-pass filter is one whose frequency-response > derivatives are all zero at w = 0. Band-pass and high-pass Butterworth > filters can be derived from it by the usual transformations or derived > directly by setting all derivatives to zero at w = infinity (high-pass) > or at the center frequency (band-pass). A consequence is that the > transfer function of a Butterworth low-pass prototype of order n is > > H(s) = 1/sqrt(1 + (w/w_c)^2n) > > That's it, unalterably. If that response isn't suitable, then > Butterworth isn't suitable. This response can only be approximated by a > digital filter, and the approximation is usually crude by the standards > of those who listen with their calculators instead of with their ears. Actually, if a filter is specified by the frequency response, going digital improves it. Warping provides for the better roloff then in the analog case. > There is another degree of freedom with crossovers. There are many degrees of freedom if you have nothing more important to do. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: Jerry Avins on 18 Jul 2007 19:02 Vladimir Vassilevsky wrote: ... > Actually, if a filter is specified by the frequency response, going > digital improves it. Warping provides for the better roloff then in the > analog case. Warping sharpens the band edge. Whether that's an improvement depends on circumstance. A filter with sharper-than-Butterworth transition isn't a Butterworth filter. ... Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
From: Robert Orban on 25 Jul 2007 19:17
In article <nNydnXeVrvc0IwDbnZ2dnUVZ_gCdnZ2d(a)giganews.com>, gangadhar.m(a)jasmin-infotech.com says... > > >I got the answer. I have to use Polarity inversion also as specified in the >document availabe at the link >www.lake.com.au/documentation/Lake%20Contour%20Classical%20Crossover% 20Notes.pdf > > >The document says that > >"When using 24 dB and 48 dB L-R crossovers, there is no need to reverse >the polarity of adjacent crossover channels. However, the 12 dB and 36 dB >L-R crossovers will require polarity reversals. In order to provide the >desired constant magnitude summation, the polarity of an output channel >must be inverted." To add a bit of info: Odd-order Butterworth highpass and lowpass pairs all have the allpass summation property regardless of the polarity of the summation. However, the order of the resulting allpass is different depending on polarity. If you want to have the summation be an allpass with the minimum number of poles, you need to invert polarity for 3rd-order (18 dB/oct), 7th order, 11th order... For example, the allpass sum of a 3rd-order Butterworth has two poles if the summation is in uninverted polarity and one pole if the sum has one input with inverted polarity (view with monospaced font): 3 2 1 s s - s + 1 _____________________ + _____________________ = ____________ 3 2 3 2 2 s + 2+s + 2+s + 1 s + 2+s + 2+s + 1 s + s + 1 3 1 s 1 - s _____________________ - _____________________ = _______ 3 2 3 2 s + 1 s + 2+s + 2+s + 1 s + 2+s + 2+s + 1 Polarity for 5th order, 9th order... should not be inverted, Finally, the allpass summation property is preserved if the Butterworth filter is transformed into an IIR digital filter via the binlear transform. |