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7th July 2004, 02:05 AM  #1 
diyAudio Member
Join Date: Apr 2001
Location: Brisbane Australia

Linkwitz Transform
Why when using the Linkwitz transform circuits available on the web do I get negative resistances for certain speaker/eq. combinations. I have tried to raise the variable capacitance to bring the resistances within range but to no avail.
Any ideas? I thought maybe the circuit doesn't work for all speaker/eq. scenarios. Thanks Dan 
7th July 2004, 05:58 AM  #2 
diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.

Dan,
What numbers for Fp/Qp and Fz/Qz are you using?? Davey. Are you using a spreadsheet like this to compute the values? http://sound.westhost.com/linkxfrm.zip 
7th July 2004, 07:39 AM  #3  
diyAudio Member
Join Date: May 2002
Location: Switzerland

Quote:
In other words: The LTF circuit has its limitations regarding the "transforming range" (maybe someone else comes up wit abetter expression). For extreme alignments you will have to use a biquad filter. Regards Charles 

8th July 2004, 04:56 AM  #4 
diyAudio Member
Join Date: Apr 2001
Location: Brisbane Australia

Charles,
I don't find it necessary to end sentences with not one, but five exclamation marks. As all advice given and points made will be given equal merit no matter how they are punctuated. As far as I was aware the Linkwitz transform circuit is a biquad filter. Perhaps what you are inferring is that I require a different type of biquad filter. Can anyone suggest a filter topology that will take a high Q, low Fs driver/box combo and turn it into a very low Q, higher Fs system. Thanks Dan 
8th July 2004, 06:17 AM  #5 
diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.

Dan,
Yep, the LT circuit is a biquad. The LT circuit will do what you want. Although most folks use this to shift the cutoff frequency lower it can also be used the other way. Siegfrieds web site has the theory and a spreadsheet to compute component values. http://www.linkwitzlab.com/images/graphics/f0Q0fpQp.gif http://www.linkwitzlab.com/pzeql.xls Davey. 
8th July 2004, 07:34 AM  #6  
diyAudio Member
Join Date: May 2002
Location: Switzerland

Quote:
When I was talking about a biquad I meant the topology, i.e. the "real one" that is using two integrators. Its advantage is that it allows much more extreme Q and fc variations compared to the original LTF circuit. This restriction lead to the strange values Dan got when trying to calculate one. If the LTF suffices it is of course preferable to use that one because it uses only one OPAMP. If you search the forum for biquad you will get one circuit by a Dutch guy that he actually used. What he did was splitting LTF = (s^2*T1^2 + s*T1/Q1 + 1) / (s^2*T2^2 + s*T2/Q2 + 1) into LTF = (s^2*T1^2)/(s^2*T2^2 + s*T2/Q2 + 1) + (s*T1/Q1)/(s^2*T2^2 + s*T2/Q2 + 1) + 1/(s^2*T2^2 + s*T2/Q2 + 1) I.e. he is summing the (scaled) outputs of a highpass, a bandpass and a lowpass. You can do the same thing with a socalled universal filter (i.e. the one with the two integrators). Regards Charles 

8th July 2004, 04:10 PM  #7 
diyAudio Member
Join Date: Dec 2001
Location: Bremerton, WA.

Charles,
As you say, the typical biquad topology would look similar/identical to a statevariable filter. It is a kind of "extension" to that topology but with a pole/zero shifting capability that can move the zero(s) outside the passband. The LT topology also has this pole/zero shifting capability so I believe it would be classified as a "biquad" filter......although there would be limitations to the Q correction relating to the single opamp and passive components requrired. I don't think that would be an issue for those of us working in the audio world. I've been fiddling a bit with a couple LT spreadsheets that I have, but I've yet to find a combination that yields negative resistor values. Possibly Dan's original problem is related to a formula in the spreadsheet itself vice the actual theory. He didn't mention which spreadsheet he was using, but since he's requiring an opposite function than the one typically used for the LT topology maybe the author of that spreadsheet didn't account for this? I dunno. Cheers, Davey. 
8th July 2004, 11:00 PM  #8 
diyAudio Member
Join Date: Apr 2001
Location: Brisbane Australia

Thanks Charles and Davey for your input.
Charles: I am having trouble deciphering your formulae. Could you add some nomenclature to clear things up for me. Thanks. Davey: Sorry for not responding to your earlier questions, they were overlooked. I am using the True Audio spreadsheet modified by Elliot Sound to accept litres for Vas & Vb. I think what you are saying about the spreadsheet: "maybe the author of that spreadsheet didn't account for this" is correct and I will try to reverse engineer the cell formulae to work out what's going on. In answer to your other question the exact values I am working with is as follows: Fo = 32 Hz Qo = 1.32 Fp = 50 Hz Qp = 0.15 I realise this is out of the ordinary, but I don't see why it cannot be achieved. Thanks Dan 
9th July 2004, 09:21 AM  #9 
diyAudio Member
Join Date: May 2002
Location: Switzerland

The secret lies in a small detail: It is the poleshifting factor k (line 18, column D in the Linkwitz spreadsheet). It shall not be smaller than ZERO, otherwise you can't use the LTF circuit. In your case the value is 1.13 !
I have been searching the forum for the example that someone once posted butdidn't succeed. He used a highpass, a lowpass and a bandpass whose outputs he summed with predetermined gain. You can use an universal filter instead. An example can be found here: http://www.marchandelec.com/ftp/wm8man.pdf I made an error within my formulae above: Instead of T1 and T2 it should of course be f1 and f2. Where f2 = the desired pole frequency and f1 the given pole frequency is. The values of Q1 and Q2 are accordingly. Regards Charles 
9th July 2004, 09:27 AM  #10 
Did it Himself
diyAudio Member

Correct, Charles. The way I get around this is to make the LT go down much lower than needed, then use a highpass filter to get the response I want. Often ends up needing large cap values (which usually leads to electros), but the response is not critical as it's so far down so you can get away with it.

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