| Butterworth, Bessel, Linkwitz-Riley... What's the difference, & what's your favorite? - Click HERE for Original Thread |
| Bricolo |
Hi all!
After the "what's your favorite crossover slope?", here's the "what's your favorite crossover type?" ;)
I was just wondering, what's the difference between all of them (if you know others, feel free to introduce them!), in terms of results/caracteristics, but also in the maths.
At first, I thought that the difference between them was the Q of the filter. But after reading some articles and threads, I have some doubts. Is this true?
Alex |
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| Bricolo |
| OH, BTW: what is a substractive crossover? :confused: |
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| Svante |
| In short: Butterworth is for odd-order filters (1, 3, 5 etc) and Linkwitz-Riley for even-order filters. Bessel filters have nothing to do in crossovers. |
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| Mr Evil |
Bessel (Q=0.58) has maximally flat delay. However you can't quite get a completely flat amplitude response. This is my favourite type.
Linkwitz-Riley (Q=0.49: critically damped) will have the best impulse response, if you like that sort of thing.
Butterworth (Q=0.71) is popular because it has the steepest slope you can get while retaining a flat amplitude response.
Yes, Q is the usual way of describing the difference between the different types of filter response.
A subtractive filter is one where you only have one filter section, then you derive the other by subtracting the filtered signal from a delayed version of the original. It has a few advantages, such as perfect impulse response, but a big disadvantage of shallow (6dB/octave) slope for the subtracted section. |
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| johnnyx |
I have a problem related to this; how do you change the Q of a passive crossover filter?
I am trying to "improve" a commercial speaker while replacing damaged woofers. The midrange frequency response looks like a two-hump camel (bactrian), so I thought to reduce the Q of the 2nd order filters, I change the ratio of L to C. So far I have found that everything increases the humps, rather than smooth them out. I want to slightly increase the high crossover frequency to better meet with the tweeter, but a reduction in inductance increases the hump.
So how does this talk of crossover slopes apply to passive networks?:xeye:
I wish I could just go active.:) |
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| Bricolo |
| quote: | Originally posted by Mr Evil
Bessel (Q=0.58) has maximally flat delay. However you can't quite get a completely flat amplitude response. This is my favourite type.
Linkwitz-Riley (Q=0.49: critically damped) will have the best impulse response, if you like that sort of thing.
Butterworth (Q=0.71) is popular because it has the steepest slope you can get while retaining a flat amplitude response.
Yes, Q is the usual way of describing the difference between the different types of filter response.
A subtractive filter is one where you only have one filter section, then you derive the other by subtracting the filtered signal from a delayed version of the original. It has a few advantages, such as perfect impulse response, but a big disadvantage of shallow (6dB/octave) slope for the subtracted section. |
That's odd, I read on another thread that classical XOs (not only 1st orders) are called substractive. Don't know why...
I started this thread because we started learning about filters, at school. L-R filters aren't mentionned (maybe they are specific to the audio use), but we saw tchebychev, butterworth and bessel.
One thing got my attention: the equations the teacher gave us were very different for all types. That's why I wondered if the difference was only the Q.
But that's maybe because the equations didn't explicitely show Q |
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| Nelson Pass |
| quote: | Originally posted by Bricolo
OH, BTW: what is a substractive crossover? :confused: |
A subtractive crossover is there you have a filter which provides
one output (either high or low pass usually) and the other
output is provided by a differential amplifier that looks at the
difference between the input and output of that filter. As a
result the output of the differential amplifier represents the
complement of the filter. If the filter is a low pass, then the
complement is a high pass, and so on. If you sum the two
outputs you get the original wave, with no phase or amplitude
distortion, in other words, square wave in, square wave out.
There's an article posted in the archives at www.passlabs.com
:cool: |
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| 5th element |
With regards to preference on these things I usually find that what dictates what xover you use (in passive xovers anyway) is what slopes/orders give you the flat or desired on axis response AND a good cancellation with reversed polarity.
Its not as simple as just saying yes I will use a 4th order LWR at 2500hz to mate this woofer and tweeter. Most likely just getting the two drivers acoustic slopes to meet 4th O LWR wont yeild a pleasent reverse null. Thats when you start to fiddle with things!:D
In an active crossover its just as simple as adding a delay circuit to the tweeter, but in passive crossovers this is not easily and cheaply done. |
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| Bricolo |
If Bessel is 0.58, LR 0.49 and Butterworth 0.71
What is Tchebichev? |
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| Evan Shultz |
actually, many things can determine the filter type. Chebyshev really is any filter that is underdamped, Butterworth is a critically damped filter, and Bessel is any over damped filter. however, there are typical Q values associated with each filter.
the implementation of the filter also determines type. Sallen-Key filters have the filter type determined by many things including gain.
LR filters are simple an even number of Butterworth filter cascaded togther. therefore, there cannot be LR3 or LR5 filters, only LR2, LR4, LR6, etc. and, any even number of Butterworth filter should be a LR, not just a Butterworth.
Rane has a great app note on LR filters and their advantages. they are excellent for audio applications in many cases.
the 'best' type of rolloff has more to do with enclosure and drivers than anything else. the filter must match and complement the driver's response to achieve whatever freq response the speaker needs. normally, underdamped filters are avoided in audio because they produce amplitude fluctuations in the passband which is sonically undesirable. overdamped filters have a slow rolloff which usually doesn't match well with speaker acoustic rolloff. BW/LR are a typical choice because they have flat amplitude response in the passband (sounds good) and sharp rolloffs. however, because the critically damped is such a finite value there are very few filters that are exactly BW or LR. acheiving the desired freq response for the speaker will push the filter type around a bit, but most filters hang around the BW/LR area, or else a smidge toward Bessel.
Also, something to keep in mind is that I believe that the damping changes with filter order. |
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| Mr Evil |
Chebyshev has Q=1. For most audio use I wouldn't recommend going outside a range of 0.49-0.71. Lower has too soft a cuttoff without offering any particular advantages. Higher will have lumpy frequency response. If you need the sharper cuttoff, better to use a higher order rather than higher Q.
I expect that Google would be able to find a site listing the various filter responses. |
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| JoshK |
If this isn't a confusing thread, I don't know what is. Everyone is saying something different than everyone else. Jeez.
My only knowledge of Chebeschev (sic) and Bessel come from my mathematics education, but not applied to crossovers and filters. Can someone define Q definitively for me? I guess I should not be so lazy and look it up. :rolleyes: |
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| 5th element |
Actually we are all saying pretty much the same thing.
What even said here
| quote: | | but most filters hang around the BW/LR area, or else a smidge toward Bessel |
is probably the best thing said here and what I do others probably do when designing a filter.
You dont tend to have exact filter types used in a loudspeaker. Its more what give you what you are after. Most speakers will use as even said from bessel to butterworth, many inbetween and most of the time maybe even none of the predetermined names given to set "Q's" for filters.
You may have a 0.65 Q high pass on the tweeter and a 0.57 Q lowpass on the woofer, all of these coming together to get you a flat response whilst maintaining good phase for symetrical and deep notch when the polarity is reversed. |
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| Bricolo |
Since someone mentionned Sallen Key filters,
How do we set the Q? Only with the 2 resistors making a divider bridge between -in, out, and ground?
By looking rapidly on the S-K schematic, I thought that those resistors were here to set the gain (and only the gain)
And, how can I set the Q with a follower? (not an opamp, a buffer, emitter follower...)
But that's maybe for another thread |
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| Bricolo |
this link has been posted several times on this forum:
http://beis.de/Elektronik/Filter/Ac....html#SallenKey
are the calculations correct?
I learnt at scool that Chebycheff filters had ripple in theyr passband (and not only a peak before the rolloff)
Spice and the values given by the online calculator don't result with a filter having ripples, even when 3dB chebycheff is chosen |
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| Mr Evil |
| quote: | Originally posted by Bricolo
...I learnt at scool that Chebycheff filters had ripple in theyr passband (and not only a peak before the rolloff)... |
Try a 4th order one, or higher. |
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| Bricolo |
thanks
that's obvious, a chebychef has as many max and mins as poles
2 poles, only one bump |
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| Mark25 |
| quote: | Originally posted by Bricolo
...........................At first, I thought that the difference between them was the Q of the filter. But after reading some articles and threads, I have some doubts. Is this true? Alex |
| quote: | Originally posted by 5th element
...................even none of the predetermined names given to set "Q's" for filters....................... |
But, the active MOX type X-over is an S-K type filter with a Q from 0.5 to 2.4:xeye: |
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| Svante |
| quote: | Originally posted by Mr Evil
Bessel (Q=0.58) has maximally flat delay. However you can't quite get a completely flat amplitude response. This is my favourite type.
Linkwitz-Riley (Q=0.49: critically damped) will have the best impulse response, if you like that sort of thing.
Butterworth (Q=0.71) is popular because it has the steepest slope you can get while retaining a flat amplitude response.
Yes, Q is the usual way of describing the difference between the different types of filter response.
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Are you taking about second order filter only? Otherwise, what is your definition of Q?
I don't agree that Q is the usual way of describing the difference, I even think it is inappropriate. |
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| Bricolo |
I've got some doubts, too
Sallen Key filters are sometibes described with this transfer function:
1/(1+j*1/Q*(w/wc)-(w/wc)²)
with a butterworth filter's equation, I managed to find the same thing (and Q is 0.707, that seems correct)
but with a bessel, whos equation is
3/(3+3*j*(w/wc)+(w/wc)²) , it doesn't seem possible to have something like the first equation
neither with a chebyshev whos equation is
1/(1+e²(2*(w/wc)+1)²)
I tried to solve those equation all day long; now it's time to relax and listen a few cds :) |
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| Mr Evil |
| quote: | Originally posted by Svante
Are you taking about second order filter only? Otherwise, what is your definition of Q?
I don't agree that Q is the usual way of describing the difference, I even think it is inappropriate. |
Q = resonant magnification factor, i.e. the ratio of magnitude at resonance to that at infinty or zero (for high-pass and low-pass respecively).
You're right, defining the filter responses by Q only works well at 2nd order. With higher orders, assuming they are constructed from multiple 2nd order sections, each section may have a different Q. Although the Q of each section is still fixed for each filter response, it's no longer the simplest way of describing them. |
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| Bricolo |
from what I've seen, for orders>2, you decompose them into multiples of 2nd and 1st orders
the total Q is Q1*Q2*Q3*....
and must be equal to the Q you want (0.71 for butterworth)
Someone found a way to write a bessel or chebychev like this?
1/(1+j*1/Q*(w/wc)-(w/wc)²) |
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| Svante |
Ok, let's go through this.
-A first order filter is described by a f0 only. f0 is the frequency at which the asymtotes for very low and very high frequencies cross.
-A second order filter is described by A f0 and a Q. The definition of f0 is the same as for the first order filter. The definition of Q can be the amplitude at f0. Other definitions exist, but they give the same numerical value for a given filter.
-A third order filter is a first and a second order filter in series. This filter has two f0:s and one Q. This Q is not the amplitude at either f0, but the Q of the 2nd order section.
- A fourth order filter is two 2nd order filters in series. This filter has two f0:s and two Q:s. Neither Q is the amplitude at either f0, but the Q of the 2nd order sections.
-etc, etc...
The bottom line here is that many filters (almost all crossover filters) can be decomposed into a cascade of 1st and 2nd order filters. Each of these sections have a f0, and each 2nd order section also has a Q value. Trying to describe filters of orders >2 with a single Q value is doomed to fail, and will only add confusion. In the light of this:
All first order filters of a given f0 have identical transfer functions.
Filters of any order >1 are commonly described by a f0 (which may be different from those of the subsections) and a "family name", eg butterworth, bessel etc. The f0 can be defined in different ways, two common ways are by the -3 dB point, or the frequency at which the high and low frequency asymptotes cross. If the filter does not fit a well defined family, it is preferrably described by the f0:s and Q:s of all the subsections. Again, a single Q value will not tell the whole story except for the 2nd order filter.
Butterworth filters have sections with identical f0:s, but different Q:s. These filters are 3 dB down at the asymtotic f0.
Linkwitz-Riley filters are two cascaded butterworth filters and are therefore always of even order. These filters are 6 dB down at the asymtotic f0.
Bessel filters filters have cascaded sections with different F0:s and Q:s. These filters are typically more than 3 dB down at the asymtotic f0.
Tjebychev filters have cascaded sections with different F0:s and Q:s. These filters are designed to have a "ripple" of x dB in the pass band, and the f0 is usually defined as when the response goes below x dB. This f0 is lower than the asymtotic f0.
For crossover filters, Butterworth filters produce a flat sum if and only if the filter is odd order, Linkwitz-Riley filters produce a flat sum for even orders.
Bessel and Tjebychev filters do not produce a flat sum, so they are not principally interesting for crossover filters.
PS. When I speak of cascaded filter sections, I think of active filter sections, where there are no interaction effects between from loading the previous section.
HTH |
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| Mark25 |
| Thank-you for the useful informative post Svante. |
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| phase_accurate |
| quote: | | Bessel and Tjebychev filters do not produce a flat sum, so they are not principally interesting for crossover filters. |
Just one addition:
2nd order Linkwitz is actually a 2nd order Bessel (with both branches having the same POLE frequency) with one driver wired out of phase.
Regards
Charles |
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| Ron E |
phase_Accurate wrote:
Just one addition:
2nd order Linkwitz is actually a 2nd order Bessel (with both branches having the same POLE frequency) with one driver wired out of phase.
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I don't think this is accurate. Bessel is a 2nd order TF with a Q of 0.577 and it is essentially a compromise between Butterworth and LR2. It has a compromise peak in amplitude to get a lower dip in power response.
Butterworth Q=0.707 is -3dB at Fc and has a 3dB peak in the vector sum when both sections are at the same frequency, but has flat power response.
LR2 is a Q of 0.5 and is -6dB at Fc and sums flat, but has a dip of 3dB in power response.
People still argue whether on axis frequency or power response is important. There are pros and cons, but on axis frequency response is more important unless the environment is very reverberant.
Getting hung up on alignments is not very useful, though. What we want is a flat sum and good off-axis behavior. Whatever gets you to that target on budget is what works. |
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| Bricolo |
I think I just have one thing to add to Svante's post
Svante, you say that defining a >2 order filter with only one Q is wrong, but I'm not sure about that.
Imagine a 4th order L-R. It's 2 Butterworth in series
a LR's Q is 0.5
a Butterworth's Q is 0.707
0.707*0.707=0.5
so I think you can also put a Q=1 and Q=0.5 in series, that will end in a LR4
But this is yet to confirm |
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| Svante |
| quote: | Originally posted by Ron E
phase_Accurate wrote:
Just one addition:
2nd order Linkwitz is actually a 2nd order Bessel (with both branches having the same POLE frequency) with one driver wired out of phase.
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I don't think this is accurate. Bessel is a 2nd order TF with a Q of 0.577 and it is essentially a compromise between Butterworth and LR2. It has a compromise peak in amplitude to get a lower dip in power response.
Butterworth Q=0.707 is -3dB at Fc and has a 3dB peak in the vector sum when both sections are at the same frequency, but has flat power response.
LR2 is a Q of 0.5 and is -6dB at Fc and sums flat, but has a dip of 3dB in power response.
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I was just about to say something similar. I think your Q's are right, Ron, but that you can say nothing about the power response in general. The power response will depend on the distance between the drivers, in relation to the wavelength at the crossover frequency. What you can say, though is that LR with swapped polarity on one driver is the only configuration that produces flat on-axis response (for 2nd order, given ideal drivers)
| quote: | Originally posted by Bricolo
I think I just have one thing to add to Svante's post
Svante, you say that defining a >2 order filter with only one Q is wrong, but I'm not sure about that.
Imagine a 4th order L-R. It's 2 Butterworth in series
a LR's Q is 0.5
a Butterworth's Q is 0.707
0.707*0.707=0.5
so I think you can also put a Q=1 and Q=0.5 in series, that will end in a LR4
But this is yet to confirm |
1+0.5 will not yield the same transfer function as 0.7+0.7. It is true that the amplitude at fc will be the same, but to the sides of fc, the responses will differ. The bottom line is, the definition of Q is not the level at fc, unless the filter is 2nd order. So 1+0.5 is not LR. |
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| Bricolo |
After looking at some exemple values; it seems you're right.
The total Q is still the same as the 2nd order equivalent, but there's a defined order for the intermediate Qs
LR is made of cascaded butterworth. But I've seen 2nd order LR, how are they made? |
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| richie00boy |
4th-order LR is made of 2 cascaded 2nd-order Butterworth.
2nd-order LR is made of 2 cascaded 1st-order Butterworth.
Butterworth is the only kind of 1st-order filter you can get.
Thanks for bringing up the Q=1 + Q=0.5 example, I had always thought that would result in the same as 2 cascaded Butterworth's. Some playing with sims is in order. |
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| Bricolo |
| quote: | Originally posted by richie00boy
Butterworth is the only kind of 1st-order filter you can get. |
I'm not 100% sure about that.
I have the equations for a Nth order Bessel, Butterworth and Chebyshev
If I put N=1, and fc at -3dB, all the equations are equal
In other words, I wouldn't say that butterworth are the only 1st order filters, but that all 1st order filters (Bu, Be, Ch) are equal |
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| Svante |
Right, there is only one kind of first order filter for a given -3dB frequency. You may call it Butterworth, Chebychev or Bessel but not Linkwitz-Riley. The only thing you can change is the cutoff frequency.
BTW, two cascaded 1st order filters with the same fc is a second order filter with Q=0.5. But please, stop thinking that Q is the amplitude at fc, it is not except for the 2nd order filter. Doing so is only confusing IHMO. |
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| Bricolo |
| I've never said that Q was the amplitude at fc ;) |
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| bser |
| OMG:xeye: Is there any english in this thread ;) I'm looking at building or buying a 2way crossover for the adire extremis I will be getting and I'd like to gather some info on crossovers however this info is way over my head. Would it be better for someone without much education of electrical components just to buy a complete crossover kit or is there a faq or noobie site I can visit to pick up basics for crossover building? |
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| DSP_Geek |
| quote: | Originally posted by bser
OMG:xeye: Is there any english in this thread ;) I'm looking at building or buying a 2way crossover for the adire extremis I will be getting and I'd like to gather some info on crossovers however this info is way over my head. Would it be better for someone without much education of electrical components just to buy a complete crossover kit or is there a faq or noobie site I can visit to pick up basics for crossover building? |
It can look a little overwhelming to start, eh? Try:
http://www.passivecrossovers.com
for a tutorial.
We were all n00bs once, so that's something to feel ok about. As for bolting on a crossover kit, that'd be like getting a set of rims on your car without knowing the hole pattern - it only fits by accident! You can either spent a fair bit of time coming up to speed on how to design crossovers, or you can find a kit designed around a specific set of speakers for which you build the cabinet. There's no shame in building a system like that - as a matter of fact some *really good* kits have been designed by well known top-notch engineers.
Francois. |
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| Mark25 |
| quote: | Originally posted by Svante
.................
Tjebychev filters have cascaded sections with different F0:s and Q:s..................... |
The above will hold true of any cascaded filter sections, to a certain degree, due to component tolerance, etc
I don't remember anybody claiming their ears prefered any X-over slope to 1st order, i think it's mainly LS drivers that like the higher order slopes. |
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| teodorom |
Good question, Bricolo !
It is precisely the question I'm asking since some time, and I find nowhere the answer !
The problem is that the transfer function of any filter is the ratio of any two polynomials in the complex variable s=iw. Then you start reasoning about zeros and poles.
As an example, the transfer function of a low-pass filter is given by
f(s)=w0^2/(s^2+s*w0/Q+w0^2)
My real problem, and it seems that someone scratched it, that Bessel, Chebychev, ..., polynomial where defined long before that filters came into interest to humanity.
So ... was first the egg or the chicken ?
In other terms: filters where first defined and then someone said <Wow ! this is archetyped by a Bessel polynomial !>.
Or someone said: <Let's try with Bessel polynomials and then let us see what happens !>.
Yes, I know the mantra about <maximally flat, om, equi-ripple, om, sums flat, om, ...> but it seems to me that "mathematical coincidence" should not be a coincidence. |
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| Svante |
| quote: | Originally posted by Mark25
The above will hold true of any cascaded filter sections, to a certain degree, due to component tolerance, etc
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Nope. Not for the Butterworth filter which have all F0 equal. Of course, if the component values are not exact, the F0s will differ, but on the other hand the filter will not be Butterworth any more ;) :cool: |
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| catapult |
| quote: | | Thanks for bringing up the Q=1 + Q=0.5 example, I had always thought that would result in the same as 2 cascaded Butterworth's. Some playing with sims is in order. |
Yep me too. Being the stubborn guy I am I had to build 'em in LspCAD and see for myself. ;)
In the pic, the top one is a real 100Hz LR4 and the bottom one is a .5 + 1 kludge. They are pretty close but there are small differences of a dB or so in the shoulder region below Fc.
I had always thought the Q values Linkwitz picked for his stacked filters were arbitrary but I guess we'd better stick to the Q values on his web page for building LR filters of orders up to 10. Thanks for the heads up, guys.
http://www.linkwitzlab.com/filters.htm#2 |
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| Bricolo |
| quote: | Originally posted by catapult
Yep me too. Being the stubborn guy I am I had to build 'em in LspCAD and see for myself. ;)
In the pic, the top one is a real 100Hz LR4 and the bottom one is a .5 + 1 kludge. They are pretty close but there are small differences of a dB or so in the shoulder region below Fc.
I had always thought the Q values Linkwitz picked for his stacked filters were arbitrary but I guess we'd better stick to the Q values on his web page for building LR filters of orders up to 10. Thanks for the heads up, guys.
http://www.linkwitzlab.com/filters.htm#2 |
could you also show us the phase of both, please?
maybe something more interesting is happening there |
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| catapult |
| quote: | | could you also show us the phase of both, please? |
Not with the demo copy of LspCAD I have. But simple lowpass filters like that are minimum phase so a Hilbert-Bode transform of the amplitude will give you the phase. They don't become non minimum phase until you mix in a highpass. |
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| tocs100 |
| Okay crossover/filter buffs, I want a low frequency contour for my electric guitar (in a small eq box). I want it to cascade like this: 1kHz -6dB & 250Hz -6dB & 60Hz -12dB. What's the fancy Q equation so I can convince my engineer-friend to build it? :smash: |
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| ctviggen |
| If you can take two cascaded first order filters and get a second order filter, then what's the difference between a quasi-second order filter and a second order filter? Further, isn't the "order" of a filter defined by the number of poles and not the slope (i.e., that the first order filter has a 6db slope is happenstance)? |
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| tocs100 |
| By "quasi 2nd order" do you mean "why a pole at 1kHz and another at 250Hz instead of just two at 250Hz?" I've tried a 2-pole at 250Hz and the cutoff sounds too abrupt. I want a g-e-n-t-l-e bass rolloff. (I'm a newbie, too, so don't laugh!):smash: |
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| forr |
Currently, I think there are two challenging active crossovers to the LR :
- the Le Cleac'h 18 dB/o or the Brooke 18 dB/o, -5 dB at Xover frequency (he called it a quasi LR)
- and the Hardman, of elliptic kind.
After having read this
http://www.diyaudio.com/forums/show...6655&highlight=
I built a two ways stereo Hardman at 1500 Hz. First listening tests were in agreement with what is said in the mentionned thread. I am very satisfied with the results. I have just finished to design a stereo three ways board.
~~~~~~~ Forr
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| tocs100 |
Thanks forr, that's neat stuff, but I'm not prototyping a crossover at the moment. What I need is a high pass tone control to reduce bass on my guitar. What values do I need to know to place a 1-pole filter at 1kHz, a 1-pole filter at 250Hz, and a 2-pole filter at 60Hz?
Or how about this as a toggle-switch option:
1-pole filter at 500kHz, and 3-pole filter at 30Hz? :whazzat: |
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