# Q of 3rd order filters

#### PeteHL

Joined Dec 17, 2014
442
Attached to this post is a LTspice simulation of a high-pass (Fc = 400 Hz) and a low-pass (Fc = 800 Hz) 3rd-order active filters with op amps. My question is whether or not it is possible to calculate Q of the filters based on the values of resistance and capacitance of the filters.

Based on what is advised in the book Op Amps for Everyone by Mancini, as I am interested in getting as low of a Q as possible, I simulated a Bessel 3rd order unity gain high-pass filter and found that with respect to frequency one octave below the cut-off frequency, attenuation is only -12 dB when it should be -18 dB given that it is supposed to be 3rd order. I would prefer the design from Mancini as I know in advance that it has the lowest possible Q, but it also needs to truly be a 3rd order filter.

If anyone knows the answer to this I will be appreciative.

Pete

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#### Papabravo

Joined Feb 24, 2006
20,377
The attenuation for a given order changes for each filter type. In other words, attenuation away from the corner is not the same for different types

3rd order Butterworth ≠ 3rd order Bessel ≠ 3rd order Chebyshev.
For a given active filter you can derive the performance from the component values

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#### PeteHL

Joined Dec 17, 2014
442
Supposing that some of you here don't want to download attachments, here is a screen shot of the attachment in my first post of this thread. -Pete

#### PeteHL

Joined Dec 17, 2014
442
The attenuation for a given order changes for each filter type. In other words, attenuation away from the corner is not the same for different types

3rd order Butterworth ≠ 3rd order Bessel ≠ 3rd order Chebyshev.
For a give active filter you can derive the performance from the component values
-12 dB at one octave below the cut-off frequency is the same as 2nd order.

#### crutschow

Joined Mar 14, 2008
32,848
with respect to frequency one octave below the cut-off frequency, attenuation is only -12 dB when it should be -18 dB
At one octave below cutoff you are still seeing the affect of the Q at the cut-off frequency.
You need to look at the roll-off below the one octave point.
That should be -18dB/octave.

And the lower the Q the softer the roll-off at the corner frequency.

#### PeteHL

Joined Dec 17, 2014
442
At one octave below cutoff you are still seeing the affect of the Q at the cut-off frequency.
You need to look at the roll-off below the one octave point.
That should be -18dB/octave.

And the lower the Q the softer the roll-off at the corner frequency.
Thanks, Crutschow, you stated it perfectly.

#### crutschow

Joined Mar 14, 2008
32,848
If interested, here's a tool to build a single op amp, 3-pole filter.
Edit: It's basically the same as the two op amp circuit except the input single-pole filter buffer op amp is eliminated.
That makes the math rather complex, so a dedicated calculator is almost a necessity.

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#### Audioguru again

Joined Oct 21, 2019
6,151
A filter with a low Q has a very droopy frequency response. A Q that produces a Butterworth frequency response has a sharp cutoff corner.
Your lowpass filter (I didn't look at the highpass) has some gain which is usually not used in a complex filter like this.

#### crutschow

Joined Mar 14, 2008
32,848
Your lowpass filter (I didn't look at the highpass) has some gain which is usually not used in a complex filter like this.
It's not common but it can have gain and still provide a proper 2-pole response as long as the component values are adjusted accordingly.

#### Audioguru again

Joined Oct 21, 2019
6,151
I have made many second-order Sallen-Key Butterworth filters that have equal component values and a gain of 1.6 times.

Joined Mar 10, 2018
4,057
If interested, here's a tool to build a single op amp, 3-pole filter.
Edit: It's basically the same as the two op amp circuit except the input single-pole filter buffer op amp is eliminated.
That makes the math rather complex, so a dedicated calculator is almost a necessity.
Just an FYI, some of the heavy lifting, the transfer function provided. Maybe this would help
getting at the Q and BW (approximations, polynomial factored) -

https://www.electronics-tutorials.ws/filter/filter_8.html

Regards, Dana.

Joined Mar 10, 2018
4,057
As an aside when working with circuits doing LaPlace consider learning Signal Flow
Graphs. The T(s) is easily computed from the graph, and you can physically "see"
the affects components have on the circuit and its nodes. Quite powerful. There
are utilities out there where you can draw the graph and an algebraic calculator will
gen the T(s) for you.

Regards, Dana.

#### LvW

Joined Jun 13, 2013
1,667
Attached to this post is a LTspice simulation of a high-pass (Fc = 400 Hz) and a low-pass (Fc = 800 Hz) 3rd-order active filters with op amps. My question is whether or not it is possible to calculate Q of the filters based on the values of resistance and capacitance of the filters.
You are speaking about a quantity "Q". What do you mean with that?

* For a bandpass we can define a quality factor Q that gives an information about the realtive bandwidth.
* For a second-order lowpass (or highpass) the "Q" is defined as a "pole Q" - and gives an information about the pole location
* For a first order low- or highpass the pole-Q is always lower than 0.5 (also defined by the location of the pole on the real axis)
* Hence, a third order low- or highpass has (a) one real pole (Q<0.5) and (b) two other poles (real or complex) - either with two different Q-values (real poles) or with one pole-Q>0.5 (complex).
But the 3rd-order filter has NOT one single Q-value. So - what is your goal?

Question: Can you define the quantity "Q" you are speaking of?

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#### PeteHL

Joined Dec 17, 2014
442
You are speaking about a quantity "Q". What do you mean with that?

* For a bandpass we can define a quality factor Q that gives an information about the realtive bandwidth.
* For a second-order lowpass (or highpass) the "Q" is defined as a "pole Q" - and gives an information about the pole location
* For a first order low- or highpass the pole-Q is always lower than 0.5 (also defined by the location of the pole on the real axis)
* Hence, a third order low- or highpass has (a) one real pole (Q<0.5) and (b) two other poles (real or complex) - either with two different Q-values (real poles) or with one pole-Q>0.5 (complex).
But the 3rd-order filter has NOT one single Q-value. So - what is your goal?

Question: Can you define the quantity "Q" you are speaking of?
By Q I mean damping of the filter. The higher the Q the less the damping and then a ringing response to a transient signal becomes more likely. A low Q filter has more damping than a high one.

For example in the book (which I have as a *.pdf), Opamps for Everyone, there is a listing of Bessel, Butterworth and Tschebysheff high-pass and low-pass filters indicating Q of the filter with respect to order.

Poles is a concept that I would like to understand but currently don't. In the past I have made a stab at understanding that, but could not quickly grasp it. I'm not (heaven forbid!) an analog engineer, so I feel okay admitting that I don't understand.

-Pete

#### PeteHL

Joined Dec 17, 2014
442
A filter with a low Q has a very droopy frequency response. A Q that produces a Butterworth frequency response has a sharp cutoff corner.
Your lowpass filter (I didn't look at the highpass) has some gain which is usually not used in a complex filter like this.
"Very droopy frequency response"- think good transient response. Those filters are from Rod Elliott, ESP Project 123. They are impressive in having a strictly 3rd Order response and resistance and capacitance is the same for both stages.

#### LvW

Joined Jun 13, 2013
1,667
By Q I mean damping of the filter. The higher the Q the less the damping and then a ringing response to a transient signal becomes more likely. A low Q filter has more damping than a high one.

For example in the book (which I have as a *.pdf), Opamps for Everyone, there is a listing of Bessel, Butterworth and Tschebysheff high-pass and low-pass filters indicating Q of the filter with respect to order.

Poles is a concept that I would like to understand but currently don't. In the past I have made a stab at understanding that, but could not quickly grasp it. I'm not (heaven forbid!) an analog engineer, so I feel okay admitting that I don't understand.

-Pete
Hi Pete - no, your understanding is not correct.
Damping of frequency components outside of the passband is determined by the filter ORDER only!!
This has nothing to do with Q-values. Hence, it is false to say "A low Q filter has more damping than a high one".
Look again into the relevant knowledge sources (books or papers, including Mancinis "opamps for everyone") - and you will se that for a third-order filter there will be not one single Q value.
For example, a BUTTERWORTH filter of 3rd order has one complex pole pair with a Q-value of "1" and a real pole. (It makes not much sense to define a pole-Q for a first order block).
Note that "poles" are not an alternative "concept" - the pole location is the basis for defining Q-values for low- and high-pass circuits.
Therefore, the correct name is "pole-quality factor" or only "pole-Q".

Perhaps your misunderstandings result from the fact that the term "damping" is used in the time domain as well as in the frequency domain.

* Time domain: The response of a filter upon a voltage step at the input depends on the location of the various poles and, hence, from the various Q-values of the several stages (if more than one stage). And this step response can be underdamped (with ringing effects, overshoot) or overdamped (without ringing/overshoot).

* Frequency domain: Here the term "damping" has another meaning - the damping factor is given in dB and tells you how much the output is attenuated (damped) at some certain frequencies outside the passband. And this property has nothing to do with Q-values.
It is determind by the filter order only!!

Final question: Why do you speak about 3rd-order filters? Where does this requirement comes from?

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#### PeteHL

Joined Dec 17, 2014
442
Hi Pete - no, your understanding is not correct.
Damping of frequency components outside of the passband is determined by the filter ORDER only!!
This has nothing to do with Q-values. Hence, it is false to say "A low Q filter has more damping than a high one".
Look again into the relevant knowledge sources (books or papers, including Mancinis "opamps for everyone") - and you will se that for a third-order filter there will be not one single Q value.
For example, a BUTTERWORTH filter of 3rd order has one complex pole pair with a Q-value of "1" and a real pole. (It makes not much sense to define a pole-Q for a first order block).
Note that "poles" are not an alternative "concept" - the pole location is the basis for defining Q-values for low- and high-pass circuits.
Therefore, the correct name is "pole-quality factor" or only "pole-Q".

Perhaps your misunderstandings result from the fact that the term "damping" is used in the time domain as well as in the frequency domain.

* Time domain: The response of a filter upon a voltage step at the input depends on the location of the various poles and, hence, from the various Q-values of the several stages (if more than one stage). And this step response can be underdamped (with ringing effects, overshoot) or overdamped (without ringing/overshoot).

* Frequency domain: Here the term "damping" has another meaning - the damping factor is given in dB and tells you how much the output is attenuated (damped) at some certain frequencies outside the passband. And this property has nothing to do with Q-values.
It is determind by the filter order only!!

Final question: Why do you speak about 3rd-order filters? Where does this requirement comes from?
What I mean by damping is the extent of resonance magnification, and I'm referring to how the filter behaves in the passband, not outside that.

Linkwitz in describing how to go about calculating the component values of for example a second order low or high pass LR filter stipulates that Q0 (Q sub-zero) of that filter equals 0.5. That value of Q affects the resulting component values.

For my application, roll-off of a second order filter is too gradual while roll-off of a fourth order filter is too abrupt.

#### LvW

Joined Jun 13, 2013
1,667
What I mean by damping is the extent of resonance magnification, and I'm referring to how the filter behaves in the passband, not outside that.

Linkwitz in describing how to go about calculating the component values of for example a second order low or high pass LR filter stipulates that Q0 (Q sub-zero) of that filter equals 0.5. That value of Q affects the resulting component values.

For my application, roll-off of a second order filter is too gradual while roll-off of a fourth order filter is too abrupt.
OK - from your last contribution I derive that you need two 3rd- order filters (lopass and highpass) having "critical damping", which means: Step response without any overshoot - correct? Therefore, the BESSEL-THOMSON approximation is not "good enough" because it exhibits a small overshoot.
In this case, you need a triple real pole filter.
The most simple and most direct realization is to use three identical simple passive RC lowpass (resp. CR highpass) stages (pole Q=0.5) - decoupled from each other with unity-gain buffers. The resulting cut-off formula is
w,c=(1/RC)*SQRT[2*exp(1/3)-1].

Comment:
I think, the problems/misunderstandings came from the fact that you in your first post did mention attenuation values of 12dB and 18 dB .... and it was not clear if you your damping requirements are in the time domain or in the frequency domain.

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#### PeteHL

Joined Dec 17, 2014
442
@LvW Thanks for the advice. For the past three days I've been without internet access and this explains my delay in responding to you.

As I'm not an engineer, with my latest project I've been doing "seat of the pants" engineering. This is in audio reproduction, so based on what I was hearing, and the following quote, I figured that I needed a 3rd order filter roll-off and/or reduced Q of the filter.

Here is some text from an introduction to filters by Kerry Lacanette, National Semiconductor:
"As a rule of thumb, filters with sharper cutoff characteristics or higher Q will have more pronounced ringing."

-Pete

#### crutschow

Joined Mar 14, 2008
32,848
Below is the LTspice simulation of a single opamp 3-pole Bessel LP filter with a -3dB point of 800Hz using the tool I referenced in post #7 (although I had to set the characteristic frequency to 440Hz to get the -3dB 800Hz rolloff).

Notice that the pulse transient response has very little overshoot (likely inaudible).

You can use the tool to similarly design your 400Hz high-pass filter.