# Q of 3rd order filters

#### MrAl

Joined Jun 17, 2014
11,583
Example 16-4 on p.310 and Fig. 16-30 on p.311. Those pages are of the *.pdf.
Ok First, i dont see any page numbers. However, i found fig 16-30 the third order HP filter.
So what is it that you dont understand about it or a low pass version?

Either way it will be a cascade of two filters sections, where one is first order and the other is second order, and the total response is the convolution of the two filter functions.

So what are you having a hard time with?

Oh one interesting fact i think is that if you design a LP filter with a certain -3db frequency and you exchange parts to make a HP filter i dont think the -3db frequency is the same as with the LP. I think the -6db frequency is though, but not the -3db frequency.
So if a LP filter has 3db down frequency of w=10 the high pass would have maybe w=15 or something, but if the LP filter had a 6db down frequency of w=10 then the 6db down frequency for the HP would be the same w=10. This is something that might not be expected.
We could look at this in more detail.

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#### Papabravo

Joined Feb 24, 2006
21,264
Ok First, i dont see any page numbers. However, i found fig 16-30 the third order HP filter.
So what is it that you dont understand about it or a low pass version?

Either way it will be a cascade of two filters sections, where one is first order and the other is second order, and the total response is the convolution of the two filter functions.

So what are you having a hard time with?

Oh one interesting fact i think is that if you design a LP filter with a certain -3db frequency and you exchange parts to make a HP filter i dont think the -3db frequency is the same as with the LP. I think the -6db frequency is though, but not the -3db frequency.
So if a LP filter has 3db down frequency of w=10 the high pass would have maybe w=15 or something, but if the LP filter had a 6db down frequency of w=10 then the 6db down frequency for the HP would be the same w=10. This is something that might not be expected.
We could look at this in more detail.
The page numbers are a feature of Acrobat Reader. They are not printed on the pages themselves.

#### MrAl

Joined Jun 17, 2014
11,583
The page numbers are a feature of Acrobat Reader. They are not printed on the pages themselves.
Yeah thanks i didnt see it because you have to click an icon to enable the left pane with the pages in it.

#### Papabravo

Joined Feb 24, 2006
21,264
Yeah thanks i didnt see it because you have to click an icon to enable the left pane with the pages in it.
There is a page number indicator on the top bar. If you click in the box you can go to a particular page.

#### PeteHL

Joined Dec 17, 2014
476
Ok First, i dont see any page numbers. However, i found fig 16-30 the third order HP filter.
So what is it that you dont understand about it or a low pass version?

Either way it will be a cascade of two filters sections, where one is first order and the other is second order, and the total response is the convolution of the two filter functions.

So what are you having a hard time with?

Oh one interesting fact i think is that if you design a LP filter with a certain -3db frequency and you exchange parts to make a HP filter i dont think the -3db frequency is the same as with the LP. I think the -6db frequency is though, but not the -3db frequency.
So if a LP filter has 3db down frequency of w=10 the high pass would have maybe w=15 or something, but if the LP filter had a 6db down frequency of w=10 then the 6db down frequency for the HP would be the same w=10. This is something that might not be expected.
We could look at this in more detail.
When I simulated response of a high-pass 3rd order Bessel filter following the method of Mancini's book, I found that roll-off at 1 octave below the cut-off frequency equaled -12 dB or the same as that of a 2nd order filter in general. Given that I thought that I had miscalculated the component values. Then Crutschow explained.

Also I did not find an example calculation of a 3rd order Bessel low-pass filter in Mancini's book. Papabravo then pointed out that his book states that both HP and LP filters with the same cut-off frequency have the same component values.

Regards,
Pete

#### LvW

Joined Jun 13, 2013
1,766
Also I did not find an example calculation of a 3rd order Bessel low-pass filter in Mancini's book. Papabravo then pointed out that his book states that both HP and LP filters with the same cut-off frequency have the same component values.
1.) Pete, when you need a 3rd-order lowpass - why not using the transfer function and filter tables - and design it by your own?
2.) It is NOT true that HP and LP filters have the same cut-off frequency (for equal but interchanged parts). It is the POLE FREQUENCY that remain unchanged but not the cut-off !! That is an important difference.

#### MrAl

Joined Jun 17, 2014
11,583
When I simulated response of a high-pass 3rd order Bessel filter following the method of Mancini's book, I found that roll-off at 1 octave below the cut-off frequency equaled -12 dB or the same as that of a 2nd order filter in general. Given that I thought that I had miscalculated the component values. Then Crutschow explained.

Also I did not find an example calculation of a 3rd order Bessel low-pass filter in Mancini's book. Papabravo then pointed out that his book states that both HP and LP filters with the same cut-off frequency have the same component values.

Regards,
Pete
As i pointed out the LP and HP do not have the same cut off frequencies just because they have the same component values (interchanged R and C). The cut off frequency is always stated in terms of the 3db down frequency and the LP and HP do not show that characteristic. They do however show the same 6db down frequencies. I through i made this very clear but it is easy to get mixed up on the 3db vs 6db down frequency characteristics.
LvW also pointed this out in a different manner check out his post too.

If you want we can probably find the relationship between the true 3db down frequencies for LP and HP i didnt get that far yet. The key point in the 2nd order case is the numerator in the HP amplitude response has a dependency on 'w' while the low pass does not. So really we have the same denominator for both, but the numerators are different and so this changes the 3db down frequencies for the LP vs HP.

Here is a plot of a representative LP and HP with values of R and C swapped. Note the point where they are the same is at 1/2 amplitude which is the same as 6db down from the passband. If they crossed at the 3db down point they would both be around 0.71 in amplitude with the same value of 'w'.

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#### Papabravo

Joined Feb 24, 2006
21,264
In the Mancini work they use A₀ and A for the respective gain constants. It is certainly true that they can be different. The other complication is that Bessel filters can be designed in more than one way. You can use either a unit frequency normalization or a unit group delay normalization as your starting point. This leads to different component values for different lowwpass filters let alone highpass and lowpass. When I said swap the R's and C's, I was speaking about the topology of the realization. I didn't mean to imply that you could use the same values. The cookbook derivation of the expanded transfer function should have made that obvious, but maybe it would be hard to see if you were not looking closely.

#### MrAl

Joined Jun 17, 2014
11,583
No problem.

It looks like the frequency for LP and HP are related by the factor:
K=1+sqrt(2)

so if we multiply the 3db down frequency of the LP by K we get the 3db down frequency for the HP.
Similarly if we divide the 3db down frequency of the HP by K we get the 3db down frequency of the LP.
This worked for the two representative filters i tried but to be sure it works with all 2nd order filters i should do more experiments. The 3rd order filters could very well be different too i have not looked at them yet.

LATER:
Ok it varies a lot for different component values. That only works in certain cases, not worth mentioning really.

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#### Papabravo

Joined Feb 24, 2006
21,264
No problem.

It looks like the frequency for LP and HP are related by the factor:
K=1+sqrt(2)

so if we multiply the 3db down frequency of the LP by K we get the 3db down frequency for the HP.
Similarly if we divide the 3db down frequency of the HP by K we get the 3db down frequency of the LP.
This worked for the two representative filters i tried but to be sure it works with all 2nd order filters i should do more experiments. The 3rd order filters could very well be different too i have not looked at them yet.

LATER:
Ok it varies a lot for different component values. That only works in certain cases, not worth mentioning really.
I think what you need to do is swap the topology and redo your calculations. That way you can verify them and have some confidence in the results.

#### PeteHL

Joined Dec 17, 2014
476
Here is an interesting equation that I found in the section on low-pass filters in Opamps for Everyone in the middle of reader page 301-

Q = 1 / (3 - Ao)

This applies to 2nd order filters such as those that are the second stage in the two filters that are shown in my post #3 of this thread.

For the low-pass filter in my post #3,

Ao = (R9 + R10) / R9
Ao = 2

From this it follows that Q of that filter equals 1. If I want the filter to have Q < 1, I can simply make R10 < 10 k Ohm.

This is my understanding after reading that page carefully. Do you agree?

Regards,
Pete

#### Papabravo

Joined Feb 24, 2006
21,264
It will take a minute or two to look at your last question. Here I have an LTspice simulation for a Bessel Lowpass filter that use two different transfer functions. One is normalized for unit Group Delay (the whole point of the Bessel filter) and the other is normalized for unit frequency. You can see the performace tradeoff to get what we were after

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#### Papabravo

Joined Feb 24, 2006
21,264
Here is an interesting equation that I found in the section on low-pass filters in Opamps for Everyone in the middle of reader page 301-

Q = 1 / (3 - Ao)

This applies to 2nd order filters such as those that are the second stage in the two filters that are shown in my post #3 of this thread.

For the low-pass filter in my post #3,

Ao = (R9 + R10) / R9
Ao = 2

From this it follows that Q of that filter equals 1. If I want the filter to have Q < 1, I can simply make R10 < 10 k Ohm.

This is my understanding after reading that page carefully. Do you agree?

Regards,
Pete
Let me try to explain. To have a filter of a given type, you have to take the Q that you need for the filter to be of that type. If you change the Q you get a different type of filter. What you can do independently of filter type is change the gain and the corner frequency or the Group Delay, but you cannot, repeat cannot, optimize everything at the same time. But HEY, you're young and you've probably got lots of time on your hands so try and see if it works for you. Not to go all "Forest Gump" on you but filters are kind of like a seesaw: you change something here like Q, and all kinds of other things change over there that you won't expect and might not be aware of. Also remember the prime directive of "Arm Control Negotiators": Trust, but Verify.

Thar rule also applies only Sallen-Key 2nd Order filters with EQUAL Resistors and EQUAL Capacitors. Don't try to use it where it does not apply.

#### LvW

Joined Jun 13, 2013
1,766
Here is an interesting equation that I found in the section on low-pass filters in Opamps for Everyone in the middle of reader page 301-
Q = 1 / (3 - Ao)
This applies to 2nd order filters such as those that are the second stage in the two filters that are shown in my post #3 of this thread.
For the low-pass filter in my post #3,
Ao = (R9 + R10) / R9
Ao = 2
From this it follows that Q of that filter equals 1. If I want the filter to have Q < 1, I can simply make R10 < 10 k Ohm.
This is my understanding after reading that page carefully. Do you agree?
Regards,
Pete
Hi Pete,
I have the impression that you are trying to get to a circuit - without much effort of your own for dimensioning.
But this is not a good start - and I am not sure if you would succeed (and you will learn nothing).
So here is my recommendation for the following steps:
1.)
Determination of the filter function - based on the requirements for pass band and stop band (or timely) behaviour.
It is not clear to me yet whether you want to realize lowpass, highpass or bandpass.
2.) Choice of a circuit structure (several alternatives)
3.) Use of filter tables to define the corresponding low-pass parameters "Pole frequency" and "Pole quality" for the selected filter function.
Note that filter tables contain lowpass data only (and for high-pass and bandpass you need transformation formulas)
4.) If necessary, use the low-high-pass or low-band-pass transformation
5.) Use of available formulas for dimensioning the filter structure. These formulas are given in the filter literature for the most important circuits.

* This is the logical and straight-forward procedure for dimensioning a filter circuit - everything else is a "blind" combination of formulas and/or a trial and error process (without gaining any knowledge).

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#### PeteHL

Joined Dec 17, 2014
476
@LvW That is true that I'm trying to solve a difficulty with my circuit without doing the studying to have a thorough understanding of filters. Whether or not I can be successful at that without having an engineer's level of knowledge certainly is questionable. If not, then I will have to do some studying. Already I have a fair amount of greater knowledge than I did before working on my present circuit.

If seems like maybe there are some other people who need to maybe refresh their knowledge, given that nobody was able to answer the question that I posed in my first post.
-Pete

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#### Papabravo

Joined Feb 24, 2006
21,264
@LvW That is true that I'm trying to solve a difficulty with my circuit without doing the studying to have a thorough understanding of filters. Whether or not I can be successful at that without having an engineer's level of knowledge certainly is questionable. If not, then I will have to do some studying. Already I have a fair amount of greater knowledge than I did before working on my present circuit.

If seems like maybe there are some other people who need to maybe refresh their knowledge, given that nobody was able to answer the question that I posed in my first post.
-Pete
I guess we can't all live up to your expectations.

#### LvW

Joined Jun 13, 2013
1,766
If seems like maybe there are some other people who need to maybe refresh their knowledge, given that nobody was able to answer the question that I posed in my first post.
-Pete
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.

#### MrAl

Joined Jun 17, 2014
11,583
A Bessel filter with lowest possible Q is a resistor voltage divider

But really the lowest possible Q would be a filter with no peakedness at all.
But is that is what you really want i'll take another look at that angle today.

Yes maybe restate the goal here we might have lost track of that.

#### MrAl

Joined Jun 17, 2014
11,583
Here are amplitude plots of a 2nd order LP filter with Q factors of 0.6, 0.7, and 0.8 all designed with a w0 of 1.

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. The defunct case viewpoint comes from the comparison to the ideal low pass which is perfectly flat in the passband and cuts off sharply to zero at the cutoff point.

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. When i think back to when i worked in the industry the subject of Q hardly ever came up, it was always about applicability. That is, does the filter response fit the requirement. What i found in the past was that there are a lot of different types of filters and the most important thing is does the response do what the application needs. One filter i remember in particular was one that had to emulate an inductor capacitor filter with resistive load where the resistance had to vary in order to see the effect of the changing load. In that filter Q was of no interest only the degree to which the filter matched the original LC filter and changing load.

So the most important thing is does the response meet the requirement of the application.

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. If we go too low on the Q we end up with a very droopy response. Could that be useful anyway? I guess it could, but then why not just use an integrator which is a kind of low pass but there is no clear passband.

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#### Papabravo

Joined Feb 24, 2006
21,264
A triple pole at -1+j0 would be the lowest Q 3rd order transfer function I can think of.