Note that the one with a Q of 0.7 is flat in the passband, and the one with Q of 0.8 is slightly peaked, and the one with Q of 0.6 is rolling off too soon which suggests that the general character of the low pass is being lost.
So in one case it starts to look like a sort of bandpass, and in the other case it starts to look like a defunct low pass.

This leads me to the conclusion that the idea of the Q of a filter is not always the best way to view the quality of a filter and that would apply to any kind not just low pass.

For the academic view, for a second order filter i think the lowest Q is 1/sqrt(2) before we lose the character of the low pass itself where we have a clearly defined passband and a clearly noticeable roll off.

With all respect, I think some comments are really necessary :

* No, the character of a lowpass is not lost - even for Q=0.5 (which belongs to a simple RC 1st-order lowpass) we speak, of course, of a "lowpass" with a 20dB/dec roll-off

* ...looks like a "sort of bandpass" ? A 2nd-order Chebyshev lowpass (ripple w=0.5 dB) has a Q=0.864 and Q=0.956 (ripple w=1dB)

* All filter tables which are used for designing filters for 2nd and higher order filters are based (and give) Q-values for the different approximations (Butterworth, Chebyshev, Bessel, Cauer,...). Note that all the various well-known lowpass functions differ in the Q-values only!

* The quality factor Q=1/sqrt(2)=0.7071 belongs to a 2nd-order Butterworth filter - it is, of course, not the lowest possible Q value (1st-order RC-lowpass with Q=0.5)

 

Pete - you are not fair.
What was the question in your first post?
As far as I can see, you were asking for a 3-rd-order BESSEL filter with "the lowest possible Q".
But thats nonsense!!
As I have mentioned in one of my my answers, a 3rd-order BESSEL filter has three poles - one real pole (Q=0.5) and a complex pole pair (Q=0.691).
Either you want a BESSEL response (with fixed Q-values) or you want any other response (not tabulated) with a lower Q.
I gave you one example (passive filter with buffer stages) for three real poles.

In my post starting the thread, what I ask first is if it might be possible to determine the Q of 3rd order low and high -pass filters from the resistance and capacitance of the filters. You can see the example filters in my post #3. Maybe it would have helped if I had pointed out that in each filter each of the two resistors have the same resistance, and each of the capacitors have the same capacitance. But at the time I wasn't aware that doing so greatly simplifies determining Q of the filter.

Then secondly I was wondering about a 3rd order Bessel filter that I had configured following Mancini's book that had a 2nd order roll-off at an octave after the cut-off frequency in the stop band. Because of that roll-off, I thought that perhaps I had made a mistake in solving component values.

Almost all of the time I get great help at this forum, so I basically have nothing bad to say about the members here.
-Pete

Edit: One question- For the two stage 3rd order filters shown in my post #3 in this thread, is Q of each filter equal to Q of the second stage, or does the first stage affect overall Q?

 

Well here is something to think about.
The amplitude of a 1st order LP with wo=1 and Q=0.5 is:
A1=1/(sqrt(w^2+1)

The amplitude of a 2nd order LP with wo=1 and Q=0.5 is:
A2=A1^2

The amplitude of a 3rd order LP with wo=1 and unknown Q is:
A3=A1^3

So what would you say the Q of this 3rd order filter was?

The 2nd order is two first order LP in cascade, the 3rd order is three first order LP in cascade.
The rate of decrease increases going from 1st to 2nd order, and the rate of decrease increases more when going from 2nd to 3rd order. So the rate of decrease increase for each additional stage, but note when we put two 1st orders in cascade the Q did not change as the 2nd order filter was constructed.
The wo was never changed for any filter.

 

With all respect, I think some comments are really necessary :

* No, the character of a lowpass is not lost - even for Q=0.5 (which belongs to a simple RC 1st-order lowpass) we speak, of course, of a "lowpass" with a 20dB/dec roll-off

* ...looks like a "sort of bandpass" ? A 2nd-order Chebyshev lowpass (ripple w=0.5 dB) has a Q=0.864 and Q=0.956 (ripple w=1dB)

* All filter tables which are used for designing filters for 2nd and higher order filters are based (and give) Q-values for the different approximations (Butterworth, Chebyshev, Bessel, Cauer,...). Note that all the various well-known lowpass functions differ in the Q-values only!

* The quality factor Q=1/sqrt(2)=0.7071 belongs to a 2nd-order Butterworth filter - it is, of course, not the lowest possible Q value (1st-order RC-lowpass with Q=0.5)

Well i guess there are different views on this. The first order has a droopy response in the pass band.
Maybe i should have just said that it gets farther from the idealized LP filter while with slightly higher Q it becomes closer to the idealized LP. With a still higher Q it starts to become bandpass-like.

It doesnt really matter to me what the names of the filters are or who they were named after except for historical purposes or when the filters are being compared based on known responses. Here we are looking at the overall quality of the filters so we are really reexamining them in the hopes of making some making more sense out of all this in a numerical sense.

 

In my post starting the thread, what I ask first is if it might be possible to determine the Q of 3rd order low and high -pass filters from the resistance and capacitance of the filters. You can see the example filters in my post #3.

Edit: One question- For the two stage 3rd order filters shown in my post #3 in this thread, is Q of each filter equal to Q of the second stage, or does the first stage affect overall Q?

Hi Pete.....it seems that I have misunderstood your question in post#1. The reason is as follows: The normal procedure for designing a filter is to determine the required characteristif FIRST (cut-off frequency and Q) and the - in a SECOND step to select the correspondiung parts values. However, you did the inverse: You designed a filter and then you were asking for the quality factor.

OK - here is an answer to your question:
(1) For a so-called equal-component Sallen-Key lowpass (your circuit) the transfer function of the second stage (2nd-order) is
H(s)=A/[1+sRC(3-A)+s²R²C²] with A=closed loop gain (A=2 in your case).
From this, it is easy to derive the corresponding Q-value:
Q=1/(3-A)=1/(3-2)=1.
(2) The Q-value of the first passive stage (1st-order lowpass) is Q=0.5

(3) For the corresponding high-pass we have the same expressions.

(4) Answer to your (last) question: As I have mentioned already - there is not something like an "overall Q" ! The first stage has a certain selectivity expressed by Q=0.5 and the selctivity of the second stage is characterized with Q=1.
Of course, both stages together have a certain response and a certain selectivity - but his cannot be expressed with an "overall Q". We need other data to describe such a filter:
* amount of ripple withinthe passband,
* end of passband (cut-off),
* attenuation characteristic (roll-off) in db/dec (order of the filter).

 

Here is another interesting thing.

Looking at the roll off for four filters all with the same designed cutoff frequency wc:
1st order
2sd order
3rd order
4th order

all with a Q=0.5 we get the following results for the roll offs in that order:
(10*(log(4*w^2+1)-log(w^2+1)))/log(10)
(20*(log(4*w^2+1)-log(w^2+1)))/log(10)
(30*(log(4*w^2+1)-log(w^2+1)))/log(10)
(40*(log(4*w^2+1)-log(w^2+1)))/log(10)

Note first that the difference between the four here are that we get the next result by multiplying the first result by the enumeration of the results (1,2,3,4). So the roll off for the second result is 2 times the first, and the entry for the third result is 3 times the first, etc. So we see the roll off does follow the basic rule for the filter order.
The same happens with a second order filter with a Q=2 and a fourth order with Q=2, the roll off of the 4th order LP is twice the roll off of the 2nd order LP. The 4th order filter is created by cascading two 2nd order filters both with a Q=2.

So i tend to think that when two filters with the same Q and same cutoff frequency are connected in cascade
the Q stays the same but the roll off follows the enumeration rule so for this case the resulting filter would roll off at twice the 2nd order filter.

 

So i tend to think that when two filters with the same Q and same cutoff frequency are connected in cascade
the Q stays the same but the roll off follows the enumeration rule so for this case the resulting filter would roll off at twice the 2nd order filter.

The Q for first-order and second-order lowpass filters are clearly defined: Q=1/(2cos(α). The angle α is the angle defined by the position of the poles in the left half of the s-plane.
Therefore my question: While saying "the Q stays the same" - which definition have you in mind ?
I think, for a 3rd-order filter there is no new definion of a resulting pole.
A third-order lowpass has three poles and two pole-Q values.

 

The Q for first-order and second-order lowpass filters are clearly defined: Q=1/(2cos(α). The angle α is the angle defined by the position of the poles in the left half of the s-plane.
Therefore my question: While saying "the Q stays the same" - which definition have you in mind ?
I think, for a 3rd-order filter there is no new definion of a resulting pole.
A third-order lowpass has three poles and two pole-Q values.

The filters i had shown all have the same wc for all sections. If they didnt then the second 2nd order LP of the 4th order filter might have a different Q.

 

@MrAl thanks for addressing my question from my post # 62. Can't say that I see where your enumeration equations came from, but I see equal phase of output voltage with respect to frequency of each stage as being the important ramification of each filter having the same Q and cut-off frequency. Then it becomes a matter of each filter attenuating according to frequency in the same manner as the other filters. If this is not the case, then good luck explaining what happens.

 

@MrAl thanks for addressing my question from my post # 62. Can't say that I see where your enumeration equations came from, but I see equal phase of output voltage with respect to frequency of each stage as being the important ramification of each filter having the same Q and cut-off frequency. Then it becomes a matter of each filter attenuating according to frequency in the same manner as the other filters. If this is not the case, then good luck explaining what happens.

Oh the equations came from cascading the same first order filter over and over again.
The first order is of course an RC LP filter with output unity gain buffer, the second order is two of those sections, the third three, the fourth four of those sections cascaded. It's very simple.
An important point is that all sections are tuned for the very same cutoff frequency.

Here is some of the reasoning that went behind the declaration that the Q of some of those combinations is 0.5 ...
The rationale for declaring the second order to have a Q of 0.5 is because when we cascade two first order filters both with a Q of 0.5 the transfer function comes out exactly the same as a second order filter designed with one op amp that has a Q of 0.5, so the Q of the two cascade combo must also have a Q of 0.5 there is no doubt there.
The rationale for declaring the fourth order to have a Q of 0.5 is because if we cascade two single op amp second order filters we get the same transfer function as four first order filters that all have a Q of 0.5 so that seems logical too.
I would imagine that if we combined a single op amp second order with a first order we would get the same response as three first orders in cascade. Thus we might declare the 3rd order to have a Q of 0.5 also. This comparison is less direct but i think it holds.

Note again all of these filters are designed with the same cutoff frequency, and also unity gain but i dont think the gain itself affects the Q because the gain amplifies all frequencies the same so the relative local amplitudes will always have the same ratio relationships.

Also note that declaring the "transfer function" to be exactly the same in each comparison is actually an over specification. We dont have to show that exact relationship all we have to show is that their "Amplitudes" are the same. But declaring the transfer functions to be the same means the phases must be the same too so it would be impossible to argue that the Q of the final construction is not 0.5 .

 

MrAl's post #66-

Doesn't Q of a low-pass filter correspond to roll-off, at least the -3 dB frequency? In other words, if I find by simulation the cut-off frequency of a low-pass filter, whatever the circuit of the filter is, doesn't that tell me what the overall (sorry!) Q of that filter is?

From a table of cut-off frequency (-3 dB down) versus Q, where Fc is the resonant frequency,

Q ------------------------- F3/ Fc
0.5------------------------ 0.65
0.6 ----------------------- 0.82
0.7------------------------ 1.00

and so on. If this is so, based on Mr Al's enumeration, then it seems that a 4th order low-pass filter that is made up of four 1st order low-pass filters in cascade does not have Q= 0.5 (output with respect to input of the entire filter).

Regards,
Pete

 

MrAl's post #66-

Doesn't Q of a low-pass filter correspond to roll-off, at least the -3 dB frequency? In other words, if I find by simulation the cut-off frequency of a low-pass filter, whatever the circuit of the filter is, doesn't that tell me what the overall (sorry!) Q of that filter is?

From a table of cut-off frequency (-3 dB down) versus Q, where Fc is the cut-off frequency,

Q F3/ Fc
0.5 1.55
0.6 1.22
0.7 1.00

and so on. If this is so, based on Mr Al's enumeration, then it seems that a 4th order low-pass filter that is made up of four 1st order low-pass filters in cascade does not have Q= 0.5 (output with respect to input of the entire filter).

Regards,
Pete

No it does not. Cutoff frequency, ω₀, and Q are independent parameters of any filter section. The rolloff is determined by the highest order term in the denominator of the transfer function. For a third order filter, the rolloff should be 18 dB/octave or 60 db/decade. The Q of a filter is related to the locations of the poles of the transfer function. A pole on the negative real axis (at least one is required for a 3rd order filter) has a fixed Q of 0.5. This is the minimum Q that any filter section can have. Poles that are not on the negative real axis, occur in complex conjugate pairs and the Q of a second order section is related to the distance of the pole from the jω-axis. For example, the pole of a 2nd order section with a Q of 10 will make an angle of just over 87 degrees with the negative real axis. In other words, it is very close to the jω-axis (less than 3° from). I would expect such a 2nd order section to exhibit extreme peaking in the neighborhood of ω₀ (the corner frequency of the 2nd order section).

You can establish conditions on the location(s) of the poles of a filter to achieve various beneficial effects, such as

1. Maximal flatness in the passband
2. Maximal rolloff in the transition band
3. Constant group delay in the passband

I'm not persuaded there are any benefits to cascading 1st order sections.

 

Here is the fairly droopy frequency response of both your simulated filters connected together to make a bandpass filter. Notice that the gain is almost +12dB.

 

MrAl's post #66-

Doesn't Q of a low-pass filter correspond to roll-off, at least the -3 dB frequency? In other words, if I find by simulation the cut-off frequency of a low-pass filter, whatever the circuit of the filter is, doesn't that tell me what the overall (sorry!) Q of that filter is?

From a table of cut-off frequency (-3 dB down) versus Q, where Fc is the cut-off frequency,

Q ------------------------- F3/ Fc
0.5------------------------ 0.65
0.6 ----------------------- 0.82
0.7------------------------ 1.00

and so on. If this is so, based on Mr Al's enumeration, then it seems that a 4th order low-pass filter that is made up of four 1st order low-pass filters in cascade does not have Q= 0.5 (output with respect to input of the entire filter).

Regards,
Pete

Hi,

What is that F3/Fc supposed to be?

Try this...
1. Design a 2nd order filter with cutoff frequency Fc1 using ONE op amp (multi feedback, whatever) with a Q of 0.5 (the damping factor will have to be set to a certain value to get this and you can look at transfer functions with Q in them if you like).
2. Design a 2nd order filter using two first order filters both having the same cutoff frequency as before Fc1.
3. Plot the frequency response of both and compare the two.

Now if the #1 filter has the exact same response as the #2 filter above we would have to say that they both have the same Q which in these two cases was 0.5.

So i am not sure how you are attempting to calculate the Q in your filters. Perhaps you can explain exactly what you are doing and be sure to mention what F3/Fc is.

 

@MrAl
My apologies, Fc is misleading; it should perhaps be named Fr and is the resonant frequency of the filter (not cut-off).

What I was suggesting in my post #71 was that you could determine the Q of a filter by determining the ratio of the -3 dB frequency of the filter to its resonant frequency.

F3 = the frequency at which the filter attenuates by -3 dB
Fr = the resonant frequency of the filter

 

No it does not. Cutoff frequency, ω₀, and Q are independent parameters of any filter section. The rolloff is determined by the highest order term in the denominator of the transfer function. For a third order filter, the rolloff should be 18 dB/octave or 60 db/decade. The Q of a filter is related to the locations of the poles of the transfer function. A pole on the negative real axis (at least one is required for a 3rd order filter) has a fixed Q of 0.5. This is the minimum Q that any filter section can have. Poles that are not on the negative real axis, occur in complex conjugate pairs and the Q of a second order section is related to the distance of the pole from the jω-axis. For example, the pole of a 2nd order section with a Q of 10 will make an angle of just over 87 degrees with the negative real axis. In other words, it is very close to the jω-axis (less than 3° from). I would expect such a 2nd order section to exhibit extreme peaking in the neighborhood of ω₀ (the corner frequency of the 2nd order section).

@MrAl
My apologies, Fc is misleading; it should perhaps be named Fr and is the resonant frequency of the filter (not cut-off).

What I was suggesting in my post #71 was that you could determine the Q of a filter by determining the ratio of the -3 dB frequency of the filter to its resonant frequency.

F3 = the frequency at which the filter attenuates by -3 dB
Fr = the resonant frequency of the filter

Lowpass and highpass filters do not have resonant frequencies. They have corner frequencies and half power frequencies. The corner frequency marks the beginning or the end of the transition band. A bandpass filter will have a center frequency that is usually the geometric mean of the two half power frequencies. Note that a bandpass filter is NOT required to have a sharp peak. It may have a passband that one or more decades wide.

In a lowpass filter with a high Q you can get peaking in the vicinity of the corner frequency by placing the poles very close to the jω-axis. This is not resonance, because there is no inductor involved to create one, it just happens because of the pole location.

 

@MrAl
My apologies, Fc is misleading; it should perhaps be named Fr and is the resonant frequency of the filter (not cut-off).

What I was suggesting in my post #71 was that you could determine the Q of a filter by determining the ratio of the -3 dB frequency of the filter to its resonant frequency.

F3 = the frequency at which the filter attenuates by -3 dB
Fr = the resonant frequency of the filter

Hi,

See post #76. What are you calling the "resonate" frequency for a first order filter?

 

<snip>
In a lowpass filter with a high Q you can get peaking in the vicinity of the corner frequency by placing the poles very close to the jω-axis. This is not resonance, because there is no inductor involved to create one, it just happens because of the pole location.

Not to be nit picky as you did a good job of explaining i think, but sometimes a peak response from a bandpass filter is also called resonance. That's a second form of resonance and there are two more one i'd have to look it up. The other form is usually referred to as physical resonance which is when capacitance and inductance cancel out and we are left with resistance only. In the peaking form, the resonant frequency is just the peak if there is one. I guess we could call that electrical resonance. But there is one more i'd have to look up to remember. So that's three forms in total but that would be application specific calling it a resonance or not.

I agree though there is no resonance for a first order filter with Q of 0.5 although there may be if the response is peaked with Q>0.5 though.

 

Not to be nit picky as you did a good job of explaining i think, but sometimes a peak response from a bandpass filter is also called resonance. That's a second form of resonance and there are two more one i'd have to look it up. The other form is usually referred to as physical resonance which is when capacitance and inductance cancel out and we are left with resistance only. In the peaking form, the resonant frequency is just the peak if there is one. I guess we could call that electrical resonance. But there is one more i'd have to look up to remember. So that's three forms in total but that would be application specific calling it a resonance or not.

I agree though there is no resonance for a first order filter with Q of 0.5 although there may be if the response is peaked with Q>0.5 though.

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?

 

You cannot make a first order filter with any value of Q except 0.5.

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