Q of 3rd order filters

MrAl

Joined Jun 17, 2014
11,687
You cannot make a first order filter with any value of Q except 0.5. I know this because a Q of greater than 0.5 requires a complex pole and there is no such thing as a first order complex pole that you can realize with real components. Can we at least put a nail in that one?
Hi,

Yes i am not sure why i worded it that way. It would really be a second order filter with Q of 0.5 vs one with higher Q.
 

MrAl

Joined Jun 17, 2014
11,687
Given that a first order filter does not resonate, wouldn't it be correct to say that it is Q-less? If it does have a quality factor, then this implies that there exists some resonance frequency of the filter and this is impossible.

So this is one advantage of a fourth order filter composed of four first order filters in cascade, relative to one consisting of two second order filters in cascade- resonance is eliminated.

A second order active filter with an op amp does resonate in a way in that there is a frequency at which gain of the filter is at a maximum. I was wondering about that, given that as you say only resistors and capacitors are used in configuring an op amp as an active filters and not any coils.

Regards,
Pete
Hi,

That's an interesting idea, however, again, if you make a second order filter with Q=0.5 and you connect two first order filters in cascade, you get the same response. That is with wc the same for all the filters. If you connect two second order filters with Q=0.5 in cascade you get the same as four first order filters iwth Q=0.5. So we could ask the question how could we connect two first order filters with"no Q" in cascade and get a filter with a Q of 0.5.
Also, if you look at other definitions it makes more sense to say it has Q=0.5.
 

Papabravo

Joined Feb 24, 2006
21,301
Given that a first order filter does not resonate, wouldn't it be correct to say that it is Q-less? If it does have a quality factor, then this implies that there exists some resonance frequency of the filter and this is impossible.

So this is one advantage of a fourth order filter composed of four first order filters in cascade, relative to one consisting of two second order filters in cascade- resonance is eliminated.

A second order active filter with an op amp does resonate in a way in that there is a frequency at which gain of the filter is at a maximum. I was wondering about that, given that as you say only resistors and capacitors are used in configuring an op amp as an active filters and not any coils.

Regards,
Pete
No, not at all. It is true that once upon a time a connection between Q and LC resonant circuits was discovered, It turns out that its application was much broader than that. It is simply a way to describe pole placement of second and higher order systems regardless of their ability to resonate or not.

I can make a maximally flat 4th order Butterworth filter that has no obvious signs of resonance. I can make of 4th order Chebyshev filter with a defined amount of ripple in the passband in exchange for steeper rolloff in the transition band, and lastly I can make a 4th order Bessel filter with a nearly constant group delay in the passband. None of these properties have anything resembling resonance in the magnitude response.

The TS wanted filters with minimum Q because "resonance was to be avoided. This is a nonsensical goal". Mathematics is the answer when it comes to getting the response you want and anecdotal approaches will lead to suboptimal results.

The maximum gain of an ideal 2nd order lowpass filter is a maximum at DC for example. It is monotonically decreasing from there to infinity. The maximum gain of an ideal 2nd order highpass filter is a maximum at infinite frequency. Looking at the datasheet for a real opamp you will notice that the open loop gain starts falling off at a fairly low frequency. This is due to a pole on the negative real axis fairly close to the origin. The gain continues to decrease until you reach a gain of 1 at some high frequency and the rollof gets steeper again. This defines the unity gain bandwidth of the opamp. It also suggest a second pole near the unity gain bandwidth frequency.

I can make filters of any order greater than 1 that will not resonate
 
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MrAl

Joined Jun 17, 2014
11,687
Hello again,

Here is a mathematical derivation of the Q of a first order filter.

Starting with the 2nd order transfer function with damping factor d:
wn^2/(s^2+s*2*d*wn+wn^2)

and same transfer function with Q instead of d:
wn^2/(s^2+s*wn/Q+wn^2)

therefore Q is:
Q=1/(2*d)

Damping factor d is:
d=a/sqrt(a^2+b^2), where a is real part b is imaginary part of pole,

therefore Q is:
Q=sqrt(a^2+b^2)/(2*a)

with b=0 because there is no imaginary part in the first order filter, we get:
Q=sqrt(a^2)/(2*a)
Q=|a|/(2*a)

and with 'a' negative we get:
Q=-1/2

but Q is always expressed as positive so the Q is 1/2 for a first order filter because there is no imaginary part.

Another way of looking at it is:
Q=1/(2*cos(atan2(b,a)))

and this leads to the same value.
 

LvW

Joined Jun 13, 2013
1,772
MrAl.....perhaps not too important (and with all respect), but at the end of your derivation there is a small error.
The quality factor Q is defined for second-order lowpass (resp. highpass) filters only - and the Q-value contains information about the pole location and its nearness to the imaginary axis of the s-plane. Therefore the name: Pole-Q (Qp)
Hence, for a 1st-order lowpass there is no definition for a Q value.

Using your notation (a, b) we only can say the following: Starting with
Q=sqrt(a^2+b^2)/(2*a)
we can make b (the imaginary part) smaller and smaller ...and the pole pair comes closer to the real axis...until we have two real poles for b=0. This is the case for a second-order lowpass with Q=0.5 (Example: Two decoupled passive RC sections).
 

Papabravo

Joined Feb 24, 2006
21,301
For a 1st order filter, the Q of 0.5 is derived from 1/2Q is equal to the cosine of 0° = 1. This is the angle made by a vector from the origin to the pole location on the negative real axis. so if 1/2Q = 1, the Q must be 0.5

No hocus-pocus or resonance.
 

MrAl

Joined Jun 17, 2014
11,687
MrAl.....perhaps not too important (and with all respect), but at the end of your derivation there is a small error.
The quality factor Q is defined for second-order lowpass (resp. highpass) filters only - and the Q-value contains information about the pole location and its nearness to the imaginary axis of the s-plane. Therefore the name: Pole-Q (Qp)
Hence, for a 1st-order lowpass there is no definition for a Q value.

Using your notation (a, b) we only can say the following: Starting with
Q=sqrt(a^2+b^2)/(2*a)
we can make b (the imaginary part) smaller and smaller ...and the pole pair comes closer to the real axis...until we have two real poles for b=0. This is the case for a second-order lowpass with Q=0.5 (Example: Two decoupled passive RC sections).
Hi,

Well i have to disagree and i will point out why but you can elaborate more if you want too.

First, if we have two poles that coincide such that the Q of both poles comes out to 2 then the Q is still 2. If we have 4 poles that coincide such that he Q of each pole is 2 then the Q of the entire circuit is still 2.

Second, much of the literature on this also says that the Q of the first order is 1/2.

Third keep in mind that there are even definitions for a single real inductor, and also a Q for a single real capacitor. The inductor is 'real' in that it has a non zero series ESR, and ditto for the capacitor.

Fourth, the angle of any real pole is zero degrees and 1/(2*cos(Angle)) with Angle=0 is 1/2.

Fifth, coincidentally, we dont have to take a limit as b goes to zero we can just set b=0. This may not be as important that just dawned on me.

Also, can you show an analog filter that has a Q greater than zero but less than 1/2 ?
I think i can show a circuit that has a Q of zero so dont worry about that case, but maybe we can discuss that too ater.
 
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Thread Starter

PeteHL

Joined Dec 17, 2014
478
The TS wanted filters with minimum Q because "resonance was to be avoided.
My reason for wanting a filter with the minimum possible Q was to eliminate ringing based on the statement,

"As a rule of thumb, filters with sharper cutoff characteristics or higher Q will have more pronounced ringing." (National Semiconductor Application Note 779)

Most of it I'm not following, but I really appreciate the lengths that you are going to in explaining.

-Pete
 

Thread Starter

PeteHL

Joined Dec 17, 2014
478
If you connect two second order filters with Q=0.5 in cascade you get the same as four first order filters iwth Q=0.5.
I tried that in a simulation with LTspice and you are correct. So I see your point and thank you.

Regards,
Pete
 

Papabravo

Joined Feb 24, 2006
21,301
My reason for wanting a filter with the minimum possible Q was to eliminate ringing based on the statement,

"As a rule of thumb, filters with sharper cutoff characteristics or higher Q will have more pronounced ringing." (National Semiconductor Application Note 779)

Most of it I'm not following, but I really appreciate the lengths that you are going to in explaining.

-Pete
The way you get ANY filter with the absolute least amount of disturbance is to ask for "maxim flatness in the passband". This will guarantee, that whatever else happens -- there will be no ringing whatsoever. To make a higher order filter according to the maximum flatness requirement you will arrange the sections in the following order:
  1. 1st order section if there is one with Q=0.5
  2. 2nd order section with the lowest Q, poles as far way from the jω-axis as possible
  3. One or more 2nd order sections with increasing Q
  4. Last 2nd order section, with the highest Q
You won't need to apply all four steps for some filters. When constructing your maximum flatness filter, each stage will have a unique value of Q that you will shoot for. Component variations may prevent you from hitting it exactly but those deviation won't cause any real problems with tight tolerance components that are available these days. For example if the target Q is 0.618 and you implement 0.615 you may have trouble telling the difference.

BTW In case you did not know about it LTspice, has a collection of 2nd order sections in the "Special Functions" block where you specify corner frequency and Q without having to search for actual components to implement the function.
 

LvW

Joined Jun 13, 2013
1,772
For a 1st order filter, the Q of 0.5 is derived from 1/2Q is equal to the cosine of 0° = 1. This is the angle made by a vector from the origin to the pole location on the negative real axis. so if 1/2Q = 1, the Q must be 0.5
No hocus-pocus or resonance.
Hi - this discussion is perhaps not too important for the design of filter stages - however, rather intersting from the systems point of view (at least in my opinion).
Let me first ask: What is the purpose of defining a quantity like Q for lowpass filters?
There is only one purpose: To desrcibe the different frequency response alternatives for second-order transfer functions (different approximations like Butterworth, Chebysheff,...).
And the defnition uses the pole location of the conjugate-complex pole pair in the s-plane.
Therefore, we speak about the pole-Q (symbol Qp).
The value of Qp is defined as Qp=wp/2σ=SQRT(σ²+wn²)/2σ.
Definition: Pole frequency wp is the magnitude (length) of the pointer from the origin to the pole position. Note that σ and wn are the real resp. imaginary part of the pole.

Now we can clearly distinguish between the different approximations - and for designing purposes we have filter tables with Qp-values (of course, for second-order stages) and associated pole frequencies wp - normalized to the specific cut-off frequencies.

As a special case, we can have a double real pole on the negative-real axis with wn=0 and σ=wp which is identical with Qp=0.5. (For two different real poles the definition gives Qp<0.5).

Question: Does it makes sense to speak about Q=0.5 for a first-order filter?
Do we need such a parameter for design puposes ? The answer is: NO.
All first order filters have the same response - aktive or RC-passiv or RL-passive and the only design parameter is the cut-off frequency.
More than that, looking into the tables of pole parameters for the diffent approximations (contained in filter books), for all uneven functions, we find wp and Qp for the various 2nd-order stages - and for the required first order stage just the normalized cut-off frequency (no necessity for something like a Qp value)

More than that, in this case we would have two different filter responses - one 1st-order and one 2nd-order - with the same Qp value. Does this makes sense? For which purpose? For my opinion, such a definition would be superfluous and would not fit into the logical system of describing filter parametes.
However, as mentioned at the beginning. If somebody likes to speak about Q=0.5 for 1st-order filters - no problem. But this contains no information at all.
 
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LvW

Joined Jun 13, 2013
1,772
Hi,
Well i have to disagree and i will point out why but you can elaborate more if you want too.
Hi again - sorry, but I must disagree as well because there are some errors in your derivation. Let me explain in detail:

MrAl: First, if we have two poles that coincide such that the Q of both poles comes out to 2 then the Q is still 2. If we have 4 poles that coincide such that he Q of each pole is 2 then the Q of the entire circuit is still 2.

What do you mean with "coincide" ? A 2nd-order function ALWAYS has a conjugate-complex pole pair (with a certain Qp-value) and a 4th-order function has two pole pairs - each with an associated Qp-value. There are no poles which "coincide".

MrAl: Second, much of the literature on this also says that the Q of the first order is 1/2.

Please see my response to Papabravo`s contribution.

MrAl: Third keep in mind that there are even definitions for a single real inductor, and also a Q for a single real capacitor. The inductor is 'real' in that it has a non zero series ESR, and ditto for the capacitor.

The definition of Q-values for passive parts is completely different as there are no poles.

MrAl: Fifth, coincidentally, we dont have to take a limit as b goes to zero we can just set b=0. This may not be as important that just dawned on me.

As I have mentioned already, there is an error in your derivation (post 84).
You gave a 2nd-order transfer function which contains the quantity wn. However, you did not define it.
In fact, the denumerator of this 2nd-order function must contain the pole-Q (Qp) and the pole frequency wp (and NOT the frequency wn). Now - you may argue that it does not matter if this parameter is named wn or wp.
However, this is very important because - finally - you set wn=0 (which means that, in reality, you have set the pole frequency equal to zero which is identical with a pole at the origin).
Hence, this operation is not a transition to a first order function.
However, using the correct expression for the quality factor: Qp=wp/2σ=SQRT(σ²+wn²)/2σ , we now can set wn=0 and we have
Qp=σ/2σ=0.5 which applies to a double pole on the real axis (as I have mentioned already in my post 85)

MrAl: Also, can you show an analog filter that has a Q greater than zero but less than 1/2 ?

Yes - no problem. This is the case for a second-order function with two different real poles .
In principle, we are using the same definition as for a conjugate-complexe pole pair Qp=wp/2σ.
Here, wp is the geometric mean of both real poles.

Regards
LvW
 
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MrAl

Joined Jun 17, 2014
11,687
Hi again - sorry, but I must disagree as well because there are some errors in your derivation. Let me explain in detail:

MrAl: First, if we have two poles that coincide such that the Q of both poles comes out to 2 then the Q is still 2. If we have 4 poles that coincide such that he Q of each pole is 2 then the Q of the entire circuit is still 2.

What do you mean with "coincide" ? A 2nd-order function ALWAYS has a conjugate-complex pole pair (with a certain Qp-value) and a 4th-order function has two pole pairs - each with an associated Qp-value. There are no poles which "coincide".

MrAl: Second, much of the literature on this also says that the Q of the first order is 1/2.

Please see my response to Papabravo`s contribution.

MrAl: Third keep in mind that there are even definitions for a single real inductor, and also a Q for a single real capacitor. The inductor is 'real' in that it has a non zero series ESR, and ditto for the capacitor.

The definition of Q-values for passive parts is completely different as there are no poles.

MrAl: Fifth, coincidentally, we dont have to take a limit as b goes to zero we can just set b=0. This may not be as important that just dawned on me.

As I have mentioned already, there is an error in your derivation (post 84).
You gave a 2nd-order transfer function which contains the quantity wn. However, you did not define it.
In fact, the denumerator of this 2nd-order function must contain the pole-Q (Qp) and the pole frequency wp (and NOT the frequency wn). Now - you may argue that it does not matter if this parameter is named wn or wp.
However, this is very important because - finally - you set wn=0 (which means that, in reality, you have set the pole frequency equal to zero which is identical with a pole at the origin).
Hence, this operation is not a transition to a first order function.
However, using the correct expression for the quality factor: Qp=wp/2σ=SQRT(σ²+wn²)/2σ , we now can set wn=0 and we have
Qp=σ/2σ=0.5 which applies to a double pole on the real axis (as I have mentioned already in my post 85)

MrAl: Also, can you show an analog filter that has a Q greater than zero but less than 1/2 ?

Yes - no problem. This is the case for a second-order function with two different real poles .
In principle, we are using the same definition as for a conjugate-complexe pole pair Qp=wp/2σ.
Here, wp is the geometric mean of both real poles.

Regards
LvW
Ok so you disagreed with four out of five points, but you dont disagree with this one:
@MrAl: "Fourth, the angle of any real pole is zero degrees and 1/(2*cos(Angle)) with Angle=0 is 1/2. "

Is that right?

Also, you said:
"It makes no sense to define a first order filter as having a Q of 1/2".
But so what if it happens to have a Q of 1/2 even if other filters have that too?
I think what you are saying here is that for a single filter it may not do any good. But when we combine it with any other filter it makes perfect sense, so why scrap the definition altogether.
What you would be suggesting then is that we can not say it has a Q when it is used in a single stage, but in a multiple stage construction then it does. So sometimes it does and sometimes it doesnt, for the very same sub circuit.
See that makes less sense to me. If we can have a rule where we can deduce the Q of a multi stage circuit from the Q of the sub circuits, then each sub circuit must have a Q of some value or else we could not be able to use that rule anymore.

A "coincidence of poles" is when two or more are the same.
 
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LvW

Joined Jun 13, 2013
1,772
Ok so you disagreed with four out of five points, but you dont disagree with this one:
@MrAl: "Fourth, the angle of any real pole is zero degrees and 1/(2*cos(Angle)) with Angle=0 is 1/2. "
Is that right?
Yes, I agree that a vector from the origin to a single pole on the neg.-real axis creates an angle of 0 deg. That is pure geometry and does not mean anything else.

Also, you said:
"It makes no sense to define a first order filter as having a Q of 1/2".
But so what if it happens to have a Q of 1/2 even if other filters have that too?
I think what you are saying here is that for a single filter it may not do any good. But when we combine it with any other filter it makes perfect sense, so why scrap the definition altogether.
What you would be suggesting then is that we can not say it has a Q when it is used in a single stage, but in a multiple stage construction then it does. So sometimes it does and sometimes it doesnt, for the very same circuit.
See that makes less sense to me.
A "coincidence of poles" is when two or more are the same.
To say a single pole filter would have a Qp=0.5 does not bear any information. Thats all I can say. Everybody may define such a quantity for his own purpose. However, as mentioned already, flter tables in relevant textbooks do not contain any Q-values for single pole stages. Additional arguments are mentioned in my response to Papabravo.

Concerning the last line: Why do you mention this theoretical case? What will you say or proove? There is no known lowpass approximation where "two or more "poles are the same. One single exception: Two identical real poles (Qp=0.5)
 

Thread Starter

PeteHL

Joined Dec 17, 2014
478
From the general point of view of Q as the quality factor, then we have to say that it makes perfect sense to assign a Q to the first order filter. The first order filter rather poorly discriminates between signal frequencies in the pass- and stop- bands, but it does discriminate. It has a large grey area between passing and stopping that filters with higher Q reduce.
 

MrAl

Joined Jun 17, 2014
11,687
Yes, I agree that a vector from the origin to a single pole on the neg.-real axis creates an angle of 0 deg. That is pure geometry and does not mean anything else.


To say a single pole filter would have a Qp=0.5 does not bear any information. Thats all I can say. Everybody may define such a quantity for his own purpose. However, as mentioned already, flter tables in relevant textbooks do not contain any Q-values for single pole stages. Additional arguments are mentioned in my response to Papabravo.

Concerning the last line: Why do you mention this theoretical case? What will you say or proove? There is no known lowpass approximation where "two or more "poles are the same. One single exception: Two identical real poles (Qp=0.5)
In the "limit" when a complex pole become real.
But it's up to you how you want to think of a first order filter but dont get alarmed when you see this elsewhere also.

It is geometry perhaps, but math applied to the physical is always an interpretation. It may or may not have practical significance and the practicality may depend on who is using it.
Maybe it is being used by "Ancient Aliens" :) (see other thread) ha ha
 

LvW

Joined Jun 13, 2013
1,772
From the general point of view of Q as the quality factor, then we have to say that it makes perfect sense to assign a Q to the first order filter. The first order filter rather poorly discriminates between signal frequencies in the pass- and stop- bands, but it does discriminate. It has a large grey area between passing and stopping that filters with higher Q reduce.
PeteHL - you state that it would make "perfect sense" to assign a Q to a 1st-order filter.
Up to now - this is an assertion only . Can you explain why and where and in which situation it will "make sense"?
In my answer to Papabravo I have tried to EXPLAIN why I don`t think it would make sense.
Can you also deliver an explanation?

Please be aware that in the denominator of each second order function there is (in the middle) an s-term that contains the pole-Q (Qp) or its inverse ζ=1/2Qp.
Such a term does NOT exist, of course, for a 1st-order function.
May I finally ask you: Where is a need to define a property of a function (a parameter) that bears (a) no information at all and (b) does not appear in the function and (c) is never used in any calculation or filter design ?
It is totally superfluous. But you can use it for your own pupose if it helps you during designing filters.
 
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Thread Starter

PeteHL

Joined Dec 17, 2014
478
PeteHL - you state that it would make "perfect sense" to assign a Q to a 1st-order filter.
Up to now - this is an assertion only . Can you explain why and where and in which situation it will "make sense"?
It's only possible for an ideal filter to have infinite Q. If Q of a "filter" is zero, then it is in reality not a filter, as filters are intended to be selective over a certain range of the input frequency. Thus a first order filter must have some Q less than infinity and greater than zero.

In other words, all filters function imperfectly. Those filters that are more selective are assigned a higher Q value. The Q value lets you know how much grey area there is between the pass-band and stop-band of the filter. The less gray area, then the higher the Q of the filter.
 
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MrAl

Joined Jun 17, 2014
11,687
It's only possible for an ideal filter to have infinite Q. If Q of a "filter" is zero, then it is in reality not a filter, as filters are intended to be selective over a certain range of the input frequency. Thus a first order filter must have some Q less than infinity and greater than zero.

In other words, all filters function imperfectly. Those filters that are more selective are assigned a higher Q value. The Q value lets you know how much grey area there is between the pass-band and stop-band of the filter. The less gray area, then the higher the Q of the filter.
What about the Allpass filter, that might have Q=0 it passes all frequencies.
 

LvW

Joined Jun 13, 2013
1,772
In other words, all filters function imperfectly. Those filters that are more selective are assigned a higher Q value. The Q value lets you know how much grey area there is between the pass-band and stop-band of the filter. The less gray area, then the higher the Q of the filter.
Sorry....it is the order of the filter function (and NOT the pole-Q) which is responsible for the width of the transition area between passband and stopband. The so-called "quality factor" only describes the magnitude response very near to the pole frequency....that means: If there is some ripple resp. peaking within the passband. More than that, the Q-factor influences the phase response...remember the well-known BESSEL-Thomson approximation which has a maximum flat group delay (Q=0.5773 for 2nd order).

Remark: By the way, even if I would follow your logic it would make no sense to assign a Q-value to a first order lowpass. In this case, we would have the same quality factor for a 1st-order lowpass (-20dB/dec) and a 2nd-order lowpass with a double pole on the real axis (-40 dB/dec). And, as you know, both have different attenuation characteristics between passband and stopband.
 
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