Q of 3rd order filters

LvW

Joined Jun 13, 2013
1,247
Yes but then again i amended that spec for this kind of viewpoint and stated that the first order could be said to have a virtual Q of 0.25 not 0.5, but then i amended it a second time (perhaps not as clear) as having a Q of zero because in the limit the response becomes IDENTICAL from w=0 to w=infinity. I'll double check this though.
Which responses became "identical"? It is not clear to me....

I guess the main theme i am seeing from all these tests and thoughts is that although saying the Q should be 0.5 may not be the best idea, it can be said that the first order filter has an equivalent "Lowest possible Q" maybe without assigning a number to it, or Q=0.
Please, can you tell me why we should even try to define a Q value for a 1st-order function?
Is there any sense or advantage or even a necessity?

Please keep in mind that the pole position idea came from the distant past. From that point of view, this is the future. The future always holds changes sometimes big changes sometimes small changes. When we seek to understand things we cant always rely on a mechanical application of theory sometimes we have to amend or it may be good to amend. Along the way i have found many things overlooked in theory even though most if it works just fine.
I understand what you mean - and, in general, I agree with you.
However, in this specific case, I think we can and should stick to the existing definitions. As mentioned before, I see no advantage or even necessity for a redefinition of the quality factor for lowpass functions.
 

Tesla23

Joined May 10, 2009
442
Both 4th-order filters are "uncommon" because each of them consists of a series connection of two identical 2nd-order stages (your wording: "squaring of a 2nd-order function").
I think, in practice, nobody would use such a 4th-order filter - however, here it can serve as part of such a hypothetical and interesting allpass realisation.
These filters are central to the Linkwitz-Riley active crossover which, apparently, is very useful / popular in the audio community. The all-pass response is part of it's appeal. So it appears that many folk do, in fact, use such a filter.

Here is the link the OP posted in post 106 https://www.linkwitzlab.com/filters.htm, or simply google 'linkwitz riley crossover'.

UPDATE 1: I have simulated the sum of two such 4th-order functions.....not really an allpass. In the pole frequency area there is a peak of app. 6 dB.
The problem is as follows: For s=jw=j1 the last expression (from wolfram alpha) gives "-1". However, the 4th-order original gives 2/j*SQRTT(2) when I introduce s=jw=j*1.
It's clearly all-pass, Wolfram confirms that, you can't dismiss it with one miscalculation.

\[ \frac{s^4+\omega_0^4}{(s^2+\sqrt{2}\omega_0s+\omega_0^2)^2}\vert_{s=j\omega_0}=\frac{2\omega_0^4}{(j\sqrt{2}\omega_0^2)^2}=-1 \]


UPDATE 2: Tesla, as far as algebra is concerned, I agree with you. But we must not forget the system point of view. What does it mean to convert a 4th-order function to a 2nd-order function? What about the several poles and zeroes ? A 4th-order highpass has 4 poles in the origin. Did they disappear (because an allpass has no poles in the origin) ? I am afraid, that a pure mathematical view does not tell us the truth in this case...
Do you know what I mean?
What are you talking about?? We move/cancel poles and zeroes all the time when we add and subtract system functions. A high pass filter has poles at the origin - yes, and when we sum this output with a low pass filter that passes signal at the origin, these move. Poles and zeroes aren't like energy - they can be created and cancelled! Sure, in any implementation the cancellation won't be perfect, and there will be ripples in the response, but that's understandable and predictable. We're talking about poles and zeroes in the audio range - electronic circuits behave pretty well here.
 

LvW

Joined Jun 13, 2013
1,247
Hi Tesla

At first, you are right......In my update 1 there was a computational error (which was corrected in the mean time)
Secondly, thank you for reminding me that there is an application for 4th-order lowpass/highpass sections as a cross-over device...

What are you talking about?? We move/cancel poles and zeroes all the time when we add and subtract system functions.
OK - I can tell you. I like to understand from the system point of view (and not only from the math side) what`s going on by adding two functions and thereby reducing the order from n=4 to n=2.
You know that in most cases ("normally") a pole-zero cancellation takes place when we multiply two functions (two stages in series). In these cases, it is straightforward to see what happens in the numerator resp. denominator of the overall transfer function. Thats all I want....
 

MrAl

Joined Jun 17, 2014
8,231
["LvW, post: 1519501, member: 209076"]
Which responses became "identical"? It is not clear to me....
>First order and 2nd order with Q approaching zero.


Please, can you tell me why we should even try to define a Q value for a 1st-order function?
Is there any sense or advantage or even a necessity?
>So we have a way to decide what filter we want to use.
We know that Q=4 has more peaking than Q=2, and Q=1 lower, and Q=0.5 lower yet (none) and Q=0.01 not only has no peaking it is very close to a first order filter.
So the main idea is when we are in the process of deciding what kind of filter to use. If we want low Q it may help to realize that a (low pass in these examples) first order filter has very low Q if not zero. If we did not think of a first order has having a zero Q then we might instead use a 2nd order filter which is more complicated and if the Q is very low a first order filter would work just as well (if we know that).
I even gave a real life problem that could come up in real life that showed how recognizing a Q value (or virtual Q if you will) would immediately lead to a solution without having to consider "peaking" for one filter and "Q" for the others. WE'd only have to refer to "Q" alone as in the example.

I understand what you mean - and, in general, I agree with you.
However, in this specific case, I think we can and should stick to the existing definitions. As mentioned before, I see no advantage or even necessity for a redefinition of the quality factor for lowpass functions.
[End quotes]

>So the 'advantage' is so that we ONLY have to refer to Q and NOT also peaking.

Note: i might have to edit this because i used ">" to show my replies and it may not work right.
 

LvW

Joined Jun 13, 2013
1,247
These filters are central to the Linkwitz-Riley active crossover which, apparently, is very useful / popular in the audio community. The all-pass response is part of it's appeal.
Hello Tesla - finally, I am convinced now.
I like to mention that I am involved in filter analyses and design since more than 25 years - and, up to now, I was sure to know the most important things about active and passive filters (approximations, transformations, circuit alternatives with pros and cons, cascade and direct design, active-L and FDNR techniques, delay properties,...).
However, I must admit that I never have heard about the combination of two 4th-order filters for realizing a 2nd-order allpass (as described by the Linkwitz-Riley paper). Thank you for your examples and providing the link - I have filled a knowledge gap.
 
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LvW

Joined Jun 13, 2013
1,247
So the main idea is when we are in the process of deciding what kind of filter to use. If we want low Q it may help to realize that a (low pass in these examples) first order filter has very low Q if not zero. If we did not think of a first order has having a zero Q then we might instead use a 2nd order filter which is more complicated and if the Q is very low a first order filter would work just as well (if we know that).
I even gave a real life problem that could come up in real life that showed how recognizing a Q value (or virtual Q if you will) would immediately lead to a solution without having to consider "peaking" for one filter and "Q" for the others. WE'd only have to refer to "Q" alone as in the example.
MrAl - hello again.
As far as the question "deciding what kind of filter to use" is concerned, I think the classical and most realistic approach is as follows (without any thinking about Q values):
* We start with the filter requirements (damping scheme).
* From this, we derive the necessary (mimnimum) filter order and the corresponding approximation function (depending on the allowed ripple within the pass band and/or the correponding phase/delay properties)
* Depending on this order, we decide if a cascade approach (series connection of filter stages of max. 2nd-order) or a "direct" realizaton seems to be appropriate (leapfrog, L-active, FDNR, ...)
* In case of a cascade approach we select either S&K or MFB stages (in most cases) and use filter tables to derive the corresponding pole data (pole-Q and normalized pole freqency wp) for each stage.
That means, we do not "want" specific Q-values. We simply have to accept the tabulated values.
* These pole data are used to calculate the parts values for each stage.
 

MrAl

Joined Jun 17, 2014
8,231
MrAl - hello again.
As far as the question "deciding what kind of filter to use" is concerned, I think the classical and most realistic approach is as follows (without any thinking about Q values):
* We start with the filter requirements (damping scheme).
* From this, we derive the necessary (mimnimum) filter order and the corresponding approximation function (depending on the allowed ripple within the pass band and/or the correponding phase/delay properties)
* Depending on this order, we decide if a cascade approach (series connection of filter stages of max. 2nd-order) or a "direct" realizaton seems to be appropriate (leapfrog, L-active, FDNR, ...)
* In case of a cascade approach we select either S&K or MFB stages (in most cases) and use filter tables to derive the corresponding pole data (pole-Q and normalized pole freqency wp) for each stage.
That means, we do not "want" specific Q-values. We simply have to accept the tabulated values.
* These pole data are used to calculate the parts values for each stage.
Hi again, and thanks for the reply,

Oh yes i realized there are other methods available no problem with that. I guess i thought that thinking about Q (as is often done with some filter applications) would be one of those methods.

Back in the day i cant remember needing to know the "Q" of any filter when we had to design something to fit some need in the field i was working in. It was all about the absolute response. The main question was always something like, "Does the response do what we need it to do" and never "Is the Q high enough".
But the Texas Instruments example did in fact rely entirely on the Q values of the individual stages. How important to me is that really? Probably not that much, because i would always check the theoretical response possibly tamed by the limitations of all the parts that go into it.
Things have changed in the past 30 years as i am sure you know. We have computers now that can solve and plot responses easily. Back then my main tool was a TI58 or TI59 calculator with red LED display :)
 

Papabravo

Joined Feb 24, 2006
16,075
Hi again, and thanks for the reply,

Oh yes i realized there are other methods available no problem with that. I guess i thought that thinking about Q (as is often done with some filter applications) would be one of those methods.

Back in the day i cant remember needing to know the "Q" of any filter when we had to design something to fit some need in the field i was working in. It was all about the absolute response. The main question was always something like, "Does the response do what we need it to do" and never "Is the Q high enough".
But the Texas Instruments example did in fact rely entirely on the Q values of the individual stages. How important to me is that really? Probably not that much, because i would always check the theoretical response possibly tamed by the limitations of all the parts that go into it.
Things have changed in the past 30 years as i am sure you know. We have computers now that can solve and plot responses easily. Back then my main tool was a TI58 or TI59 calculator with red LED display :)
If you design filters by placing the poles and zeros, the Q's will take care of themselves. They will be unique for a given filter order and response type.
 

MrAl

Joined Jun 17, 2014
8,231
If you design filters by placing the poles and zeros, the Q's will take care of themselves. They will be unique for a given filter order and response type.
Hi,

Yes that is a good point. I guess we got off into talking about Q so much because it was part of the main subject of this thread and i know the TS was thinking along those lines, and i sympathize with that because we read about Q all the time.

But i think Q is more important when it comes to a passive filter made with real life passive components because then we are talking about losses in the filter, losses that are important, and the Q of the filter reflects that. The Q of individual components also, and it provides a means for comparison. Of course today we go by ESR so maybe Q is becoming an outdated concept anyway. In my days in the field we always paid very close attention to ESR because the efficiency of a device was of very high importance. If the device didnt meet some standard of efficiency nobody would want to buy that product.

So maybe we should give Q a rest unless someone can come up with a good reason to keep paying so much attention to it.
 

Tesla23

Joined May 10, 2009
442
Hello Tesla - finally, I am convinced now.
I like to mention that I am involved in filter analyses and design since more than 25 years - and, up to now, I was sure to know the most important things about active and passive filters (approximations, transformations, circuit alternatives with pros and cons, cascade and direct design, active-L and FDNR techniques, delay properties,...).
However, I must admit that I never have heard about the combination of two 4th-order filters for realizing a 2nd-order allpass (as described by the Linkwitz-Riley paper). Thank you for your examples and providing the link - I have filled a knowledge gap.
I thought the way the two squared filters when summed provided all the roots of \( -\omega_0^4 \) in the numerator, and the LHP ones cancel with one of the denominators, is really neat. I learnt something too - I never knew these networks existed either!
 

LvW

Joined Jun 13, 2013
1,247
But i think Q is more important when it comes to a passive filter made with real life passive components because then we are talking about losses in the filter, losses that are important, and the Q of the filter reflects that. The Q of individual components also, and it provides a means for comparison. Of course today we go by ESR so maybe Q is becoming an outdated concept anyway. In my days in the field we always paid very close attention to ESR because the efficiency of a device was of very high importance. If the device didnt meet some standard of efficiency nobody would want to buy that product.

So maybe we should give Q a rest unless someone can come up with a good reason to keep paying so much attention to it.
I think, we should not (must not) mix two different quality factors.
Up to now, we spoke about the "pole-Q (Qp)" - that is a quality factor describing the location of a pole pair in the complex s-plane.
The quality factor Q of a passive component is something else - it concerns the parasitic losses (resistive portions) of a reactive part (L or C).
 
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MrAl

Joined Jun 17, 2014
8,231
I think, we should not (must not) mix two different quality factors.
Up to now, we spoke about the "pole-Q (Qp)" - that is a quality factor describing the location of a pole pair in the complex s-plane.
The quality factor Q of a passive component is something else - it concerns the parasitic losses (resistive portions) of a reactive part (L or C).
Hi,

Yes i thought i made that clear too when i said:

But i think Q is more important when it comes to a passive filter made with real life passive components because then we are talking about losses in the filter
:)
 

LvW

Joined Jun 13, 2013
1,247
... because then we are talking about losses in the filter, losses that are important, and the Q of the filter reflects that.....
Yes.....it is interesting to note that for an LC bandpass - without any additional resistive part and with an ideal capacitor C - the pole-Q (Qp) will be identical to the quality factor Qi of the inductor L as well as the Q of the resonant curve Qr=fo/B (3dB-bandwidth B): Qp=Qi=Qr.
 

Papabravo

Joined Feb 24, 2006
16,075
Yes.....it is interesting to note that for an LC bandpass - without any additional resistive part and with an ideal capacitor C - the pole-Q (Qp) will be identical to the quality factor Qi of the inductor L as well as the Q of the resonant curve Qr=fo/B (3dB-bandwidth B): Qp=Qi=Qr.
It is your position that the Q of a given 2nd order section is not determined by the pole locations?
 

LvW

Joined Jun 13, 2013
1,247
It is your position that the Q of a given 2nd order section is not determined by the pole locations?
Sorry, I do not understand the question.....did I say something wrong?
Of course, the quality factor Qp is determined (and DEFINED) by the pole location.
Thats what i have repeatadly stated in this thread...
 

Papabravo

Joined Feb 24, 2006
16,075
Sorry, I do not understand the question.....did I say something wrong?
Of course, the quality factor Qp is determined (and DEFINED) by the pole location.
Thats what i have repeatadly stated in this thread...
The Q of a bandpass section is not a function of the Q of the inductor alone.
 

LvW

Joined Jun 13, 2013
1,247
The Q of a bandpass section is not a function of the Q of the inductor alone.
Perhaps You have overlooked the restriction I have mentioned: ",,without any additional resistive part and with an ideal capacitor C".
However, I admit that this is a rather theoretical case and - more or less - without a realistic background.
But I think - from the system point of view - it is interesting to point to the similarities and the relations between the several quality factors we have defined. In any case, the bandwidth related quality factor Qr=fo/B for a second-order bandpass is always identical to the corresponding pole Q (Qp=Qr).

Remark: I am still waiting for your reply to my post#154 (Qp for 2nd-order functions only).
What are your arguments to assign Q=0.5 to a 1st-order circuit (your post#144) ?
 
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