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#1 |
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diyAudio Member
Join Date: Aug 2005
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I have been slowly correcting all the issues with my first crossover. One finding is that one reason the tweeter needs a larger than calculated time delay network, is because of its 12db lowpass behavior affecting the phase of my LR4 2000 hz crossover point.
I would like to apply a linkwitz transform to move the 1200 hz natural rolloff up to 2000 hz, and then cascade it with a second order butterworth to form LR4. This would allow me to keep the tweeter better in phase throughout the whole crossover region, and to reduce the time delay of its allpass network. Does anyone know why I shouldn't do it? I don't ever see this done, and I'm not sure if its because few use active crossovers, or there is some reason. Before anyone says just try it, I'm using an analog active crossover, so changes require some effort. I could prototype it on my DEQ2496, but that only works with two channels (tweeter and mid), and would only give me one half of the stereo image. |
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#2 |
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diyAudio Member
Join Date: Dec 2004
Location: Next door
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Hi Mazurek,
Many people think of the Linkwitz transform as a technique to lower cut-off frequency and the fact that it can be used to increase it, with benefit, seems to be forgotten. I even forgot it myself till the day I check it in the Linkwitz writings. Increasing the cut-off frequency (which probably means the resonance frequency) of a driver using a Linkwitz transform should be less error prone than lowering it. The reason is that the resonance is now in the stop band. ---I don't ever see this done--- I think Siegfried has done it in one of its first published design. I may once try the idea myself. |
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#3 |
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diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.
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Yes, this option was always possible with the original LT concept. I think the first practical use I remember seeing of it was from SL himself in the Speaker Builder article 20 years ago. I can't recall any since then however.
http://www.linkwitzlab.com/filters.htm#9 (look at the bottom of section 9.) http://www.linkwitzlab.com/sb80-3wy.htm http://www.linkwitzlab.com/images/graphics/f0Q0fpQp.gif http://www.linkwitzlab.com/images/graphics/sb186-48.gif You can use one of the many pole-zero LT spreadsheets available to calculate the proper values and give it a try. Cheers, Davey. |
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#4 |
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diyAudio Member
Join Date: Aug 2005
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Interesting citations, thanks for the facts everyone. I had not seen linkwitz's early designs that use this technique.
I'll be integrating that into my next crossover then, it may be a month or two before I report back. |
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#5 |
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diyAudio Member
Join Date: May 2005
Location: Californication
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Quote from Likwitz about the late Dueland's work. Hope this helps
Note that all mathematical functions are meant to describe the acoustic response of a 3-way loudspeaker. To the extend that woofer, midrange or tweeter drivers have a non-flat frequency response they must be either equalized to be flat, or their response must be made part of the filter function. In particular, the midrange driver's natural low frequency roll-off could be equalized with a "Linkwitz Transform" to obtain the desired LR2 highpass behavior of c) above. The tweeter can be treated similarly for two of the poles of d). The woofer's natural highpass behavior causes a phase lead which is probably far from zero at the first crossover frequency and thus affects the proper addition of woofer and midrange outputs. This can be corrected by placing a 1st order allpass in the midrange channel which simulates the highpass phase shift of the woofer. The Duelund 3-way crossover could provide a useful option for building a loudspeaker
__________________
like four million tons of hydrogen exploding on the sun like the whisper of the termites building castles in the dust |
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#6 |
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diyAudio Member
Join Date: Apr 2002
Location: Singapore
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I had thought of this before, I see some potential issues though:
- ferrofluid dampened tweeters may not show a clean "classic" rolloff behavior, looking at some FR graphs - not all tweeters come with fully published specs necessary for the LT calculation I have one related question: in an LR4, you cascade two 2-pole Butterworth filters (Q=0.7) with f0 equal to the final f0 of the desired LR4. If you wanted to use a driver's rolloff's first pole (call it f1) as a filter section of a Q=0.7 filter, which corresponding electrical filter frequency "f2" do you use to make the combined, now 2-pole section, an acoustic Q=0.7? And what will the final 2-pole filter's f0 be? What I am looking for is a formula of the style f0=f(f1, f2, Q) where I would know f1 and Q (=0.7) and just derive the required f2. |
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#7 |
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diyAudio Member
Join Date: Dec 2004
Location: Next door
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Hi MBK,
I may have not understand your question well. I hope this may help. --- an LR4, you cascade two 2-pole Butterworth filters (Q=0.7)--- Not necessarily. You final is Q = Q1 x Q2 = 0.7 x 0.7 = 0.49. It's the aim which is important. You can get the same result using other values for Q1 and Q2. For example Q1 =1 and Q2 = 0.49. |
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#8 |
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diyAudio Member
Join Date: Apr 2002
Location: Singapore
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Hi Forr,
thanks, ok I agree, I had thought of the "classic" combo but it's the product of the Q's that matters really. My question was basically, which pole frequencies belong to a specific Q and f0 - in this case f1 and f2 for a Q=0.7. Filter handbooks usually just show the horrific looking full transfer functions, for deterrence, and move then on swiftly to the C and R values needed assuming a specific topology, without stopping at the partial filter frequencies they represent. But if you have *one* acoustic cutoff frequency given as the f3 of a driver, how do you calculate the *other* frequency needed to realize a specific Q and combined f0? |
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#9 |
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diyAudio Member
Join Date: Dec 2004
Location: Next door
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MBK
--- if you have *one* acoustic cutoff frequency given as the f3 of a driver, how do you calculate the *other* frequency needed to realize a specific Q and combined f0?--- An LR 24 dB/o isa combination of two cascaded 12 dB/o filters. For the acoustic filter you have a resonance frequency fs (and its Qs) of the speaker, eventually Linkwitz transformed. The second filter may be a Sallen and Key filter tuned to the same frequency as fs and with a Q adjusted so that the product of both Q's give 0.49. MacCauley used this technique in a design described in Electronics World. |
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#10 |
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diyAudio Member
Join Date: Aug 2005
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I agree that the ferro fluid damping may be confusing. For about one octave, the seas tdfc looks like it rolls of at 6db/oct to me. I will measure its response with a simulated linkwitz transform, and post it. My guess is that the linkwitz transform correction magnitude is so much greater than the error magnitude that it will be fine.
I think there are ways to infer the f3 of the tweeter, also it wouldn't be too hard to do a sensitivity analysis. I know we're not supposed to use winISD for these things, but it is a good way of visuallizing cascaded transfer functions (I don't have anything better). |
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