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Hi,
Can you explain why you are mentioning this, that is, what purpose does it serve to know that?
Q of 3rd order filters
- Thread starter PeteHL
- Start date
Can you explain why you are mentioning this, that is, what purpose does it serve to know that?
YES - good question.
More than that - Pete, what is your goal? What do you want to realize and/or to understand?
I remember, you have started with a bandpass!
According to my view of Q, if the slope of the curves of two filters from 0 dB to -3 dB is the same, then thiis means that Q of the filters is identical.
My question is, how to determine the Q of a high or low pass filter? You seem to be taking the position that there is no such thing as Q generally of a filter. That I would say is nonsensical. When I started the thread, my question was, what is the Q of the example 3rd order filter that I provided.
IAll that I want to know is how to determine Q of a filter, simple or complex.
Certainly this thread lacks focus and I'm not sure how that came about.
Hi,
Ok i think i see your goal here. You have stated that you would like to be able to determine the Q of a filter from the slope near the -3d point.
What i think i can say is that SOMETIMES that may be possible.
The reason i say this is because we can set up a 2nd order filter with low Q and one with high Q and we will see the slope around the -3db point get more steep. That is pretty certain.
However, we can not say this will be true for EVERY filter. That's the problem.
We can see this by looking at a bandpass i think where the passband is relatively flat for a wide frequency range yet the slopes near the two -3db points may be steep or dip down very slowly with frequency.
An extreme example would be a digital bandpass filter with extremely steep slopes. With -3db points 10Hz and 100kHz and center frequency 50kHz, That means the Q is calculated to be approximately 2. The same filter with -3db points at 49kHz and 51kHz and same center frequency of 50kHz the Q comes out to be 25. Yet both filters can have the same slopes near the -3db points.
Now perhaps we could look at low and high pass filters too. But if we do find a difference, then whatever theory we find from this will mean we have to know what kind of filter it is first so we can figure out if the theory holds for that kind of filter or not.
What do you think of this?
I think it was mentioned several times now that to determine the Q you should really look at the poles, but if you have a plot of the response you can sometimes tell because you'll see the Q=Frequency/Bandwidth where the frequency is the center frequency where you can determine is roughly (F1+F2)/2 where F1 and F2 are the -3db points.
We can look for other relationships for the low pass and high pass.
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Pete - with all respect, I am really surprised about your answer. I cannot believe what you wrote.
In all of my answers (13, 16, 18, 54, 57, 61, 65, 67, 85, 91, 100,...) I have mentioned the importance of the Q-value and - several times (!!) - I gave you the DEFINITION of the Q-value (with numbers for the different lowpass functions).
And your only comment is that I would deny the existence of a Q-value.
Did you not read my answers? Or didn`t you understand the contents?
In this case, I cannot help you further (and I have no motivation) .... good luck.
Hi,
Maybe he is not saying "Q" alone but "a general Q for any filter", but i can see how this can get confusing for someone that is not used to looking at these kinds of attributes. That leads to some back and forth thinking patterns until it has time to sink in and start to make sense. Confusion is a nasty bedfellow.
Given some more time to think this through and maybe some more examples i think he will get it simply because he had the intelligence to ask in the first place.
What i think is that we can find the Q related to the slope but only for a CERTAIN particular filter we happen to be dealing with, and as soon as we switch to a different design we lose all bets for that "theory". It could also be that much of this kind of thinking will be confined to 2nd order filters only, or a particular kind of 3rd and up.
For example, if we start with a multiple feedback op amp 2nd order filter with Q=1, and we see a certain slope. Then we use the same filter design but make the Q=2, we see the slope increase (probably with more overshoot also). So we see some correlation but ONLY when dealing with that and only that particular filter.
It may or may not hold for another particular kind of filter.
We could probably see this idea better with a 3rd order filter though where we can see a lot more variation in the response type.
I am starting to think maybe we should delve into Bode plots, the original asymptotic type.
When speaking about "slope", I think it is important to DEFINE this term. Otherwise, misunderstandings cannot be avoided.
For my opinion, the term "slope" is reserved for the filter behaviour far above the passband end of a lowpass.
And, therefore, the slope depends on the filter order only (-20 dB/dec for 1st order, -40dB/dec for 2nd order...) and NOT on the Q-value.
The Qp-value (which is defined for 2nd-order only) determines the filter behaviour in the region of the pole frequency only (transition region):
* Max. flat for Butterworth functions
* With peaking effects (0.1dB, 0.5 dB, 1dB, 3 dB,..) for the various Chebyshev responses
* Rather broad transition region for Bessel-Thomson responses.
All these properties are independent on the particular circuit which is chosen for realization (passive or active).
Hello again,
Ok, well then you got some 'splainin' to do
1. First, didnt i specify the slope around the -3db point(s) ? (Also why did I specify that particular point?)
2. Second, what would be the point in increasing the Q of a second order low pass filter if the "roll off" was always the same for multiple values of Q?
3. This is a side issue not as important. Since you insist that the first order filter has no Q then how would you explain the fact that when we connect two isolated firsrt order LP stages in cascade we get EXACTLY point by point the same complete response for a single 2nd order low pass filter designed specifically for a Q of 0.5, do you honestly believe that there is no relationship between anything we might call a Q of all three of these filters?
In other words, you might be suggesting that we start with two filters with no Q at all and end up with a valid Q after connecting them in cascade even when the two first order ones are isolated from each other.
Again this is a side issue though, not directly on point of determining the Q of a general filter.
I am sure you can answer both of the first two easily, but here is a set of three plots that illustrate.
Note with three different values of Q as the Q gets higher the slope near the -3db point gets steeper.
Also note for frequencies much higher than the -3db cutoff frequency we see roughly the same response, as you noted previously.
When we want to determine a difference between things we dont look for characteristics that are the same we look for those that are different, and the more differences we can find the easier it gets to be able to define what the things are from those unique characteristics. If we have two buildings both 40 feet high one made of brick and the other made of wood we cant determine anything from the height alone but we can easily tell the difference between the two by finding out what they were made of. In other words, we dont go and say, "See, both buildings are the same height so we cant tell them apart."
Hi MrAl - here are my answers. (By the way - I like to answer clear questions)Hello again,
Ok, well then you got some 'splainin' to do
1. First, didnt i specify the slope around the -3db point(s) ? (Also why did I specify that particular point?)
2. Second, what would be the point in increasing the Q of a second order low pass filter if the "roll off" was always the same for multiple values of Q?
3. This is a side issue not as important. Since you insist that the first order filter has no Q then how would you explain the fact that when we connect two isolated firsrt order LP stages in cascade we get EXACTLY point by point the same complete response for a single 2nd order low pass filter designed specifically for a Q of 0.5, do you honestly believe that there is no relationship between anything we might call a Q of all three of these filters?
In other words, you might be suggesting that we start with two filters with no Q at all and end up with a valid Q after connecting them in cascade even when the two first order ones are isolated from each other.
Again this is a side issue though, not directly on point of determining the Q of a general filter.
I am sure you can answer both of the first two easily, but here is a set of three plots that illustrate.
Note with three different values of Q as the Q gets higher the slope near the -3db point gets steeper.
Also note for frequencies much higher than the -3db cutoff frequency we see roughly the same response, as you noted previously.
When we want to determine a difference between things we dont look for characteristics that are the same we look for those that are different, and the more differences we can find the easier it gets to be able to define what the things are from those unique characteristics. If we have two buildings both 40 feet high one made of brick and the other made of wood we cant determine anything from the height alone but we can easily tell the difference between the two by finding out what they were made of. In other words, we dont go and say, "See, both buildings are the same height so we cant tell them apart."
1.) As I have mentioned, it is common practice in filter theory to use the term "slope" for the attenuation characteristics (far enough) above the end of the passband - and to express it in dB/dec or dB/oct. Therefore, it is also common practice to display the frequency characteristics not in a linear manner but in dB on the vertical axis (and log spacing on the frequency axis). Only in this case, the different slopes (n*20dB/dec) are visible.
2.) This question touches the filter theory. And the answer is simple: In order to approach the IDEAL "brick wall" lowpass response (which never can be realized) several different approximations to the desired response have been proposed in the last century - and I am sure you have heard about the various names like Butterworth, Chebyshev, Bessel-Thomson, Cauer, Papoulis, Gauss, Mullik, ....
All these different functions have a different behaviour in the transition region between the passband and the stopband - and therefore, they all have different pole locations and, hence, different pole-Qs (Qp values) for each second-order pole pair.
Are you really asking why we have different approximations? I am sure you know the specific properties of a 2nd-order Butterworth-lowpass if compared with a 2nd-order Bessel-lowpass - and the different application areas (for example).
More than that, for realizing higher-order filters with a specific characteristic we need a variety of different pole distributions.
3.) Sorry, but I do not "insist" that a first order filter would "have" no Q-value. As I have mentioned already, the DEFINITION of the pole-Q applies to a second-order pole pair only!! And I am not responsible for the definition!!
Furthermore, I must admit, I cannot follow your example.
When cascading two isolated first-order filters we get a 2nd-order filter with a double real pole and a Qp=0.5. Of course, we get the same response for an active 2nd-order filter (one single stage) when we design it for Qp=0.5. So what?
Does this mean (in your view) that each of the 1st-order blocks also must have a Qp=0.5? Why?
Quote: "you might be suggesting that we start with two filters with no Q at all and end up with a valid Q after connecting them in cascade "
This sentence deserves a separate answer:
I repeat again that the pole-Q was defined and introduced in order to have a measure for the different pole locations associated with the different ways a second-order transfer function may look like!
In contrary - all first order functions look the same and there is absolutely no reason to define any additional parameter. For which purpose?
It would be a new term without any meaning - it is not needed to descrribe the 1st-order function. More than that it does not appear in the denominator of the 1st-order transfer function (1+sT). So - where do you see a need or a reason to define something which is without any sense and meaning? I really do not understand!
___________________________________________________________________
I hope, the following links work. They clearly show the different results for modifying the order of the filter and/or the Qp value of the filter.
https://www.google.com/imgres?imgurl=https://upload.wikimedia.org/wikipedia/commons/thumb/c/cd/Butterworth_Filter_Orders.svg/350px-Butterworth_Filter_Orders.svg.png&imgrefurl=https://en.wikipedia.org/wiki/Butterworth_filter&tbnid=fgjDDWYAo9Z50M&vet=12ahUKEwiz4eDmndTpAhVHiqQKHcxOBA0QMygDegUIARDVAQ..i&docid=q8858bgq6-_A-M&w=350&h=247&q=Butterworth transfer functions different order &client=firefox-b-d&ved=2ahUKEwiz4eDmndTpAhVHiqQKHcxOBA0QMygDegUIARDVAQ
https://www.google.com/url?sa=i&url=https://www.electronics-tutorials.ws/filter/sallen-key-filter.html&psig=AOvVaw1yYW7ErgnoSSZqCTK46vcZ&ust=1590674910342000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCNDJx_qb1OkCFQAAAAAdAAAAABAZ
With regards
LvW
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Pete - may I propose something to you?
Without in-depth knowledge of filter theory you should be careful with words like "nonsensical".
May I inform you that I am involved in design and development of active filters since more than 25 years?
More than that, I have published a book on filters and oscillators - however, in German but it was published also in the Russian language.
Well, I should have said nonsensical to me, because I have read other material about filters that treat Q as determining the characteristic of output voltage with respect to input voltage of filters.
For example, from "A Basic Introduction to Filters", the author writes,
"Second order filters are characterized by four basic properties: [including] filter Q. The Q of a second-order filter of a given type will determine the relative shape of the amplitude response. Q can be found from the denominator of the transfer function if the denominator is written in the form
D(s) = s^2 + (ws/ Q) + w^2
......Q relates to the 'sharpness' of the amplitude response curve. As Q increases, so does the sharpness of the response. "
In the above quotation, w = omega, and in the application note it is given as omega sub O.
So you see that the author directly refers to "filter Q", and he is connecting it to the amplitude response of the filter.
Regards,
Pete
Hi again,
Well you dont seem to understand why i am asking these questions and i am not blaming you so i will just try again.
1. Do you understand WHY i am asking this question of what happens around the -3db points (the slope)?
I dont care what happens at high frequencies that's NOT the point of what Pete and I are trying to get at.
2. You implied or even said that the responses are the same at some high frequency. You seem to be using that as an example of WHY 2nd order filter responses do not change depending on Q. So i asked if that was true then why would we ever want to change the Q.
In my example, if the -3db point slope never changed there would be no reason to go to a higher or lower Q.
Now if you say that the Q does matter then you cant say that we CAN NOT determine anything from the value of Q, as you are stating that the slope only depends on the ORDER of the filter (the slope at -3db does change even if the high frequency slope does not even with the SAME order filter just with different Q).
3. [Low pass filterers] If you put two 2nd order filters with Q=0.5 in cascade, you get a 4th order filter with Q=0.5. If you put 10 2nd order filters in cascade you get a 20th order filter with Q=0.5.
If you put two 3rd order filters in cascade both with Q=0.5 you get a 6th order filter with Q=0.5.
If you put N Mth order filters in cascade all with Q=0.5, you get a (N*M)th order filter with Q=0.5.
If you put N Mth order filters in cascade all with Q=Qx you get an (N*M)th order filter with Q=Qx.
Now let's back up just a little:
If you put a 2nd order filter with Q=Qx in cascade with a filter with Q=0 you get a filter with Q=Qx.
What does this tells us?
This tells us that it does not matter what the value of Q is for the filter placed LAST, if the Q is the same or zero, you always get the same Q.
But now let's put two in cascade both with Q=0, what do we get? We get Q=0:
First filter Q=0 order=M, second filter Q=0 order=M, following the rule that N Mth order filters with Q=Qx we get a filter with Q=Qx.
So now let us apply that to two first order filters with Q=0, and two with Q=0.5.
You stated that two first order filters with whatever Q result in a 2nd order filter with Q=0.5 and that was "OK" because it is now a 2nd order filter.
So now the two with Q=0 we get Q=0, but that cant be possible because you said it was OK to have a 2nd order filter with Q=0.5, so that must not be right, so the two first order filters can not have a Q=0 or else we would get a Q=0 of the 2nd order filter and we both know that is not right.
So it makes sense that if we get a Q=0.5 of the cascade arrangement that AT LEASE ONE of the filters must have had a NON ZERO Q, or at least a Q that can be said to EXIST.
Make sense now?
Yes - all this is exactly what I have also claimed in my former answers - but, of course, in other words .
And if you then look at the first order transfer function (compared with the 2nd-order),
H(s)=Ao/(1+s/wo) ,
there is no parameter Q and no necessity to "introduce" such a quantity.
By the way: It is very helpful to have a look on the various curves as can be found in the the two links I have given in my last post to MrAl. These curves show what is meant with the influence of Q on the "sharpness" of the amplitude response.
Several times you have used the term "slope" which always is used to describe the response for higher frequencies above the end of the passband. More than that, we were discussing 1st-order and second order functions which differ primarily in there slopes (-20dB/dec vs. -40dB/dec)
That is simply not true. I never have claimed that "the filter responses do not change depending on Q".
And I also never have said that "we cannot determine anything from the value of Q".
Both statements are nonsensical. Please, be correct when you are quoting me.
Sorry to say, but this is totally wrong.
Please, accept what I have stated several times: The Q-value for lowpass and highpass filters is defined for second-order only. The Q-value is called "pole-Q (Qp)" and desribes the position of the pole pair !!
Please - answer this question: How would you define the Q of a 4th-order lowpass filter?
Have you ever had a look into the tables (available in each filter design books) in which the pole parameters are listed for higher-order filters?
Example: A 2nd-order Bessel lowpass has a Q-value Qp=0.5773.
For a 4th-order Bessel lowpass you need two 2nd-order stages with Qp=0.5219 and Qp=0.8055.
To be honest - no, it does not make any sense at all.
I would be interesting to learn from you how a lowpass with Q=0 looks like.
An infinite value of Q is - in principle - possible. Inserting Q>>infinite into the second-order denominator, the midterm disappears - and we have an oscillator (poles on the imag. axis in accordance with the definition of Qp).
But did you ever check what happens when you insert Q=0 into the denominator?
The whole transfer function disappears (equal to zero).
So - I really do not know what you are trying to show ....
Regards
LvW
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Hi again,
Ok you disagreed with every point i made so either you dont understand me or i dont understand you, so let's keep it simpler so we can reply faster it takes too long to reply to so much stuff at once.
For now let's concentrate on the first order "Q" because that is part of what you disagree with.
You are saying that the first order lowpass filter has no Q or is it zero Q? I doubt you are saying zero Q but i am asking just to be sure.
Please try to keep your reply as short as possible TIA.
Actually your reply could be as short as either:
1. "no Q", or
2. "zero Q".
We can then go on, and i'd really like to understand your point of view.
The next thing we have to look at is the topic of "slope" in reference to a low pass filter response.
Hello again...in short:
* Denominator of a 2nd-order lowpass: D(s)=[1+s/wpQp+(s/wp)²].
There are two quantities which describe the response as a function of s=jw: wp and Qp.
* Denominator of a 1st-order lowpass(s)=(1+s/wp)=(1+s/wo)=(1+sT)
(pole frequency wp equal to 3dB cutoff wo; Time constant T=1/wo)
There is only one single quantity which describes the filter response: wp=wo=1/T.
* Is there any necessity to define a new parameter for describing the response?
If yes, please explain it to me.
I NEVER have claimed that there would be any lowpass circuit with a Q=0. That would be crazy!
As everybody can see - the 2nd-order transfer function would disappear and in the 1st-order function there is no Q.
So - I really do not understand your point.
I forgot to repeat my question: When speaking about the Q value for a 4th-order lowpass - how do you DEFINE this parameter?
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Keeping this simple and short...
Be patient my friend
So you said that Q can not be zero for a LP filter, that's great, now we are getting somewhere. So that must mean you are saying that there is no Q then?
So keeping this short to a one-liner:
This is a very very simple question, how do you describe the Q of a first order low pass filter?
I'll be more than happy to answer YOUR questions as soon as we get past this one. We have plenty of time i think to explore other questions no problem on this end. I will address each and every question posed at the proper time in order to avoid confusion and make it faster to reply too.
Hello again and thanks for the reply,
Be patient my friend
So you said that Q can not be zero for a LP filter, that's great, now we are getting somewhere. So that must mean you are saying that there is no Q then?
So keeping this short to a one-liner:
This is a very very simple question, how do you describe the Q of a first order low pass filter?
I'll be more than happy to answer YOUR questions as soon as we get past this one. We have plenty of time i think to explore other questions no problem on this end. I will address each and every question posed at the proper time in order to avoid confusion and make it faster to reply too.
MrAl - is this really your question? I am afraid, we are moving in a circle...
How often I have mentioned that there is no definition of a quantity "Q" for a first order equation?
And now you are asking me how I would "describe the Q of a first-order lowpass".
So you are asking me to describe something which does not exist...are you joking?
Do you realize that there is no Q at all in the equation for a 1st-order lowpass H(s)=[Ao/(1+s/wo)]?
PS: I have tried to answer all of your questions. Where is your answer to my simple question (How do you define the quantity you are speaking of - the Q for a 4th order filter?)
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My goal is very simple: for me to understand your point and for you to understand my point. If you cant up with a little questioning to make things perfectly clear then you are too temperamental.
So what if even i asked a question three times? Are you going to go crazy or jump out a window or what?
You might also think this is the only thing i have to worry about but i work with many other things in between readings and replies here so it makes it harder to keep track anyway, and so sometimes i will ask a simple question to make sure everyone is on the same page, not just me but any other readers (like the TS for example).
Questions are for the asking by definition.
I THINK you have answered mine, but you seem to be incapable of giving a direct answer without an associated mocking attempt.
The answer you gave, as indirect as it was, was that you state that there is no Q for a first order filter, that it does not exist.
Then ask yourself why i would think it did. But i'll answer your question first since it is an easy one.
You asked this question:
"Do you realize that there is no Q at all in the equation for a 1st-order lowpass H(s)=[Ao/(1+s/wo)]?"
There is no *variable* named "Q" (no symbol 'Q') in that first order filter equation, period. Anything else? (chuckle)
But i have to comment slightly here. You have given me an equation with no variable or symbol Q in it and asking me if there is a Q in it. That implies a few things.
First, that you are defining the Q of a filter based on what a first order filter equation has in it.
Second, that means that you cant understand how something can have a Q when it is not in the equation.
Third, you seem to be stuck with that definition for life.
Ok so we have established that you are defining the existence of a Q by what an equation has in it, if it has it then it has a Q, if it does not have it then it does not have a Q.
That is completely understandable, but rather old school in reference to this thread. That is because we are seeking a method to figure out the Q of different filters, without resorting to old fixed definitions, if it even is one. It doesnt matter to me or probably the TS if the Q is in the equation or not, because we are COMPARING filters and drawing conclusions from what we find out, not from something that might be written in a book somewhere (if it is even written in a book).
What this leads to is a more general definition of Q apparently, but to be honest i dont think it is actually new because it appears elsewhere also.
What you are suggesting is that the existence of a Q can arise out of two filters that "have no Q because it doesnt exist" so i can restate my previous statement about combining two filters based on what you have told me, the questions you have answered.
Now my statement according to what you told me would have to read like this:
"Take two first order LP filters (that dont have a Q because it doesnt exist for first order filters) and connect them in cascade and you get a 2nd order filter with a Q of 0.5."
Does that sound right to you, because it does not sound right to me. It seems more logical that the two first order filters must have a Q of something.
Corollary:
If we wanted to compare two filters with unknown topology let's say these three:
1. Q=0.5
2. Q=1.3
3. Q=5
We can see right off that #3 has higher Q then #2 and that has Q higher than #1.
Also, because the Q of #2 and #3 is above *0.5* (the star of the show) we know they must be of 2nd order or greater.
The next simple question is, what order is #1?
Now if you want to say that ti *must* be 2nd order because a first order can not have a Q because it does not exist that's fine, but i cant agree with that for the following reason (with this example only):
We could be referring to a first or second order filter with item #1 to keep the list shorter. If we did not do that, we''d have to list like this:
1. Q=0.5
2. Q=1.3
3. Q=5
4. Q does not exist
Granted that takes the ambiguity out of it.
Let me state this point another way...
If we are discussing what kind of filter to use for a given application, the dialog may go like this:
"Hey Dexter, what kind of filter should we use for this new product?"
"Well Henry, do you think we should use a filter with a Q of 4 or a Q of 7, or how about a filter that does not have a Q because it does not exist?"
"What? We just need a reasonable sharpness and we know that correlates to the Q of the filter, so how sharp is the filter you say has no Q because it does not exist?"
"Gee Dexter, i cant tell because it has no Q because it does not exist."
"Well then how do we know if it has a sharp enough response for our needs?"
"I guess we dont, unless of course you would like to assign a Q that correlates to the sharpness of the response, even though the Q does not appear in any formulas yet."
"Oh ok that sounds reasonable, then we could tell the difference when comparing filters."
"Ok as long as everyone agrees then we'll do that."
"Ok. By the way, my name is not Dexter it is Qdexter."
"Oh gee, sorry i was under the impression that the Q did not exist."
"Ha ha, very funny, so being Henry, i guess you are an inductor then?."
<Henry quietly walks out of the room>
Although the above is a little funny, it makes a point about comparing filters with regard to Q.
It is ok if you dont agree with this but then it is not my job to convince everyone on earth to accept a definition of any kind, electronical or other.
Note the use of a word that is NOT YET considered to be "real" because it does not YET exist in a dictionary.
Then why use it? Because due to the way language changes it is under consideration to finally be accepted as a new, real word, and the use of it puts more emphasis on something that pertains to electronics.
Note some people will argue that "electronic" covers all the bases, but adding the "al" adds more emphasis sort of like redundancy can.
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