Q of 3rd order filters

LvW

Joined Jun 13, 2013
980
What about the Allpass filter, that might have Q=0 it passes all frequencies.
Hi again....no, that is not correct.
We should not forget that the quality factor Q is in reality defined with respect to the POLE POSITION and, therefore, should be called "POLE-Q Qp".
An allpass filter has poles AND zeros which are symmertrical to the pole distribution.
Hence, each pole pair has a certain Qp value (like lowpass filters). More than that, also the zeroes have associated quality factors Qz. In both cases, the Q-values are a measure for the nearness to the imaginary axis.
Here is the classical Allpass function (2nd-order):
H(s)=N(s)/D(s) with
N(s)=[1-s/wzQz+s²/wz²]
D(s)=[1+s/wpQp+s²/wp²]
 

sparky 1

Joined Nov 3, 2018
138
The various electric fields are not easily unified.
In compact fusion elements in an implosion field travel opposite confinement paths to fuse then into transmutation. The excess energy travels outward and is absorbed by a spiral coil. In constructive interference the waves can mix in various medias. We find the left overs tend to go to the ground state as noise, In space we can see the stars forming probably from fragments. In silicon the small electromagnetic fields contain small parasitics mostly seen as a hovering menace to the ultra precision being sought. In the media around a crystal radio employing high Q coils some of the better designs using surface area and parallel coils are said to work better. The subjects are difficult to combine. There is a lot of ideas in cyber space about all this so it would naturally be a concern.

Any charge separating mechanism has a corresponding shaped field the best known is the dipole, there can be stages. The relationship with Q can be quantified at each stage. It is the charge separating mechanism that is responsible for establishing the accumulation (the diode for visual purposes)
Or a wimshurst generator has a charge separating mechanism.

Take for example a tornado, there are real forces outside but it is the charge separating mechanism Terra Watts. In a radio antenna the forces are from a radio transmitter.
The expression would declare the field let's say a single dipole, the Quality factor could be found.
However the expression in a mixer/amplifier the boundary conditions and the media type would be a different case not a special case like fusion.
 
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MrAl

Joined Jun 17, 2014
7,511
Hi again....no, that is not correct.
We should not forget that the quality factor Q is in reality defined with respect to the POLE POSITION and, therefore, should be called "POLE-Q Qp".
An allpass filter has poles AND zeros which are symmertrical to the pole distribution.
Hence, each pole pair has a certain Qp value (like lowpass filters). More than that, also the zeroes have associated quality factors Qz. In both cases, the Q-values are a measure for the nearness to the imaginary axis.
Here is the classical Allpass function (2nd-order):
H(s)=N(s)/D(s) with
N(s)=[1-s/wzQz+s²/wz²]
D(s)=[1+s/wpQp+s²/wp²]
Well then how do you yourself define a straight wire with little resistance?

It is interesting that the phase changes with the Allpass filter even though the frequency amplitude response remains flat.
 

LvW

Joined Jun 13, 2013
980
Well then how do you yourself define a straight wire with little resistance?

It is interesting that the phase changes with the Allpass filter even though the frequency amplitude response remains flat.
Yes - the amplitude response remains flat - and different Q-values have influence on the phase response (resp. the group delay) only. That is the main applicaton of allpass filters - to correct unwanted phase excursions of forgoing stages.
 

sparky 1

Joined Nov 3, 2018
138
Possibly the history of all pass filters with telephone transmission might illustrate attributes to make the explanation more coherent with filters and amplifiers. I am not an expert. I want to to place some emphasis that filter type does not change the electrical principles. The practical examples briefly touch on how it works reminding them that a more in depth study will be needed to understand how various filters have an effect on the output of an amplifier. A demonstration might show the all pass in an op amp circuit.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
337
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.
While I'm not certain if this shows conclusively that the order and Q of a filter are independent, today I did a simulation (attached) with LTspice that shows two low-pass active filters that have the identical cut-off frequency (-3 dB, 1 kHz) and yet the top one has a 1st order roll-off, while the second one has a second order roll-off.
The component values of the second order filter I calculated following equations that Linkwitz has made available at his web site. One of the equations is for Q0 (Q sub-zero). Do you know if giving it this particular name has a special meaning? Below is a link to the article giving the equations.

Regards,
Pete
https://www.linkwitzlab.com/filters.htm

Edit May 23: In the attached simulation, if C2 is made equal to 15nF, then I believe that this makes Q of the two filters more nearly equal as then the curves of the two filters are very nearly equal to 0 dB at the same frequency.
 

Attachments

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LvW

Joined Jun 13, 2013
980
While I'm not certain if this shows conclusively that the order and Q of a filter are independent,
...............................
One of the equations is for Q0 (Q sub-zero). Do you know if giving it this particular name has a special meaning?
Yes, the order of a filter and the associated Qp-values are completely independent. Simple example: For a 2nd-order lowpass, there are several classical filter functions defined (Butterworth, Chebyshev, Cauer, Bessel-Thomson,m....) - each with different Q-values.
In the Linkwitz article, the Qo parameter is defined in the shown 2nd-order function - it i the classical pole-Q (Qp), defined by the pole location.

EDIT/UPDATE: Pete - here comes another argument against the attempt to assign a Q-value of 0.5 to a 1st-order lowpass: For ALL 2nd-order lowpass functions, the value of Qp always gives you the magnitude of the function at the frequency w=wp: A(w=wp)=Ao*Qp.
Example: For a double real pole we have Q=0.5 and the magnitude is A(w=wp)=0.5Ao.

However, for a 1st-order lowpass, we have A(w=wp==0.7071.
 
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MrAl

Joined Jun 17, 2014
7,511
Yes and the example of connecting one, two, three, or four first order filters in cascade creates filters first order, second order, third order, and fourth order, all with the same Q.
 

Thread Starter

PeteHL

Joined Dec 17, 2014
337
Yes, the order of a filter and the associated Qp-values are completely independent. Simple example: For a 2nd-order lowpass, there are several classical filter functions defined (Butterworth, Chebyshev, Cauer, Bessel-Thomson,m....) - each with different Q-values.
In the Linkwitz article, the Qo parameter is defined in the shown 2nd-order function - it i the classical pole-Q (Qp), defined by the pole location.

EDIT/UPDATE: Pete - here comes another argument against the attempt to assign a Q-value of 0.5 to a 1st-order lowpass: For ALL 2nd-order lowpass functions, the value of Qp always gives you the magnitude of the function at the frequency w=wp: A(w=wp)=Ao*Qp.
Example: For a double real pole we have Q=0.5 and the magnitude is A(w=wp)=0.5Ao.

However, for a 1st-order lowpass, we have A(w=wp==0.7071.
What do A and Ao stand for? The equation relating A and Ao of the 1st order low-pass filter is not clear to me.

-Pete

I
 

Thread Starter

PeteHL

Joined Dec 17, 2014
337
Yes and the example of connecting one, two, three, or four first order filters in cascade creates filters first order, second order, third order, and fourth order, all with the same Q.
With LTspice, I compared the AC small signal response of a first order filter, and a second order filter that is two first order filters in cascade. If the cut-off frequency of the second order filter is made 1.5 times that of the first order filter, then the two traces are very nearly coincident all the way to 0 dB. So I think that what you are saying is true.
 

MrAl

Joined Jun 17, 2014
7,511
With LTspice, I compared the AC small signal response of a first order filter, and a second order filter that is two first order filters in cascade. If the cut-off frequency of the second order filter is made 1.5 times that of the first order filter, then the two traces are very nearly coincident all the way to 0 dB. So I think that what you are saying is true.
Well why didnt you use the same frequency for BOTH?
What was the reason you used 1.5 times the frequency for the 2nd filter?
 

Thread Starter

PeteHL

Joined Dec 17, 2014
337
Well why didnt you use the same frequency for BOTH?
What was the reason you used 1.5 times the frequency for the 2nd filter?
If you put two identical first order filters in cascade to form a 2nd order filter, then naturally you would expect that, for a given frequency, attenuation of the 2nd order filter is twice that effected by one of the 1st order filters. If both the 2nd order filter and a 1st order filter have the same cut-off frequency, then the 2nd order filter is -3 dB at a frequency that is about 2/3rds that of the - 3 dB frequency of the first order filter.

Thus making the cut-off frequency of the 2nd order filter 1.5 times that of the first order filter should result in overlapping traces at -3 dB. Doing this results in the traces of the outputs of the 1st and 2nd order filters coincident from 0 dB to -3 dB. Given my perception of what Q is, this tells me that the filters have equal Q.
 

LvW

Joined Jun 13, 2013
980
What do A and Ao stand for? The equation relating A and Ao of the 1st order low-pass filter is not clear to me.
Pete - in general, I agree with you that symbols should be defined. However, we are speaking here about lowpass filters - and I have mentioned the term "...magnitude is A(w=wp)...". Hence, I think "A" was defined. From this it should not be a problem to deduce that Ao is the magnitude for w=0 (DC).
 

LvW

Joined Jun 13, 2013
980
With LTspice, I compared the AC small signal response of a first order filter, and a second order filter that is two first order filters in cascade. If the cut-off frequency of the second order filter is made 1.5 times that of the first order filter, then the two traces are very nearly coincident all the way to 0 dB. So I think that what you are saying is true.
Pete...one important information is missing in your test setup description: Di you use an isolating buffer amplifier between the two 1st-order stages? If not, the 2nd stage will load the first one - and you must not simply multiply both 1st-order equations to get the 2nd-oder transfer function.
 

MrAl

Joined Jun 17, 2014
7,511
Pete - in general, I agree with you that symbols should be defined. However, we are speaking here about lowpass filters - and I have mentioned the term "...magnitude is A(w=wp)...". Hence, I think "A" was defined. From this it should not be a problem to deduce that Ao is the magnitude for w=0 (DC).
A sub zero sometimes means the amplitude at the center frequency or cutoff frequency.
 

MrAl

Joined Jun 17, 2014
7,511
If you put two identical first order filters in cascade to form a 2nd order filter, then naturally you would expect that, for a given frequency, attenuation of the 2nd order filter is twice that effected by one of the 1st order filters. If both the 2nd order filter and a 1st order filter have the same cut-off frequency, then the 2nd order filter is -3 dB at a frequency that is about 2/3rds that of the - 3 dB frequency of the first order filter.

Thus making the cut-off frequency of the 2nd order filter 1.5 times that of the first order filter should result in overlapping traces at -3 dB. Doing this results in the traces of the outputs of the 1st and 2nd order filters coincident from 0 dB to -3 dB. Given my perception of what Q is, this tells me that the filters have equal Q.
Hi again,

Are you using isolated stages or completely passive filters?

For a single stages:
Amplitude=w1/sqrt(w1^2+w^2)

and for two isolated stages in cascade (or two stages where the 2nd stages has much higher impedance than the first):
Amplitue=(w1*w2)/(sqrt(w1^2+w^2)*sqrt(w2^2+w^2))

where w1 and w2 are the two cutoff frequencies respectively.

The ratio of the first to the second is:
r12=sqrt(w2^2+w^2)/w2

and with w2=a*w1 we have:
r12=sqrt(a^2*w1^2+w^2)/(a*w1)

and now making a=1.5 and w=w1 we end up with:
r12=1.201850425154663

so this isnt exactly equal to 1 but it is close to 1.2 if that is what you meant.

Just setting w=w1 we end up with:
r12=sqrt(a^2+1)/a

and with this we find that if we make a=7.05 we get close to 1 with an error of about 1 percent.
We only get a perfect 1 if we let a tend toward infinity, however a=7 is probably close enough.
a=1.5 isnt too bad i guess, off by 20 percent, and the higher we go with 'a' the closer we get to exactly 1.

This can change significantly if we allow the 2nd stage to load the first stage, but we might have to know the impedances to be able to come to any conclusions about 'a'.
With an impedance ratio (R2&C2 vs R1&C1) of 10:1 we get about 1.6 for w1=1 and about 1.9 for w1=10, so the match changes with frequency now. As the impedance ratio goes up toward infinity we get get the same results as with isolated stages.
 
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Thread Starter

PeteHL

Joined Dec 17, 2014
337
Pete...one important information is missing in your test setup description: Di you use an isolating buffer amplifier between the two 1st-order stages? If not, the 2nd stage will load the first one - and you must not simply multiply both 1st-order equations to get the 2nd-oder transfer function.
The first order low-pass filters are with an op amp, that is, a passive resistor-capacitor filter with the junction of the resistor and capacitor connected to the non-inverting input terminal of the op amp configured as a voltage follower. So yes, there is buffering between the stages.

At a given frequency, assuming that first and second first order filters in cascade forming a 2nd order filter are identical, then it follows that the attenuation effected by the second stage is fully added to that of the first stage.
 

MrAl

Joined Jun 17, 2014
7,511
The first order low-pass filters are with an op amp, that is, a passive resistor-capacitor filter with the junction of the resistor and capacitor connected to the non-inverting input terminal of the op amp configured as a voltage follower. So yes, there is buffering between the stages.

At a given frequency, assuming that first and second first order filters in cascade forming a 2nd order filter are identical, then it follows that the attenuation effected by the second stage is fully added to that of the first stage.
Yes the magnitudes can be multiplied to get the total output magnitude.
|VoutTotal|=|Vin|*|Vout1|*|Vout2|
 

LvW

Joined Jun 13, 2013
980
The first order low-pass filters are with an op amp, that is, a passive resistor-capacitor filter with the junction of the resistor and capacitor connected to the non-inverting input terminal of the op amp configured as a voltage follower. So yes, there is buffering between the stages.
At a given frequency, assuming that first and second first order filters in cascade forming a 2nd order filter are identical, then it follows that the attenuation effected by the second stage is fully added to that of the first stage.
Ok.....is there still any problem ? Do you have any further open questions?
 

Thread Starter

PeteHL

Joined Dec 17, 2014
337
Hi again,
Are you using isolated stages or completely passive filters?
so this isnt exactly equal to 1 but it is close to 1.2 if that is what you meant.
The stages are isolated by an op amp configured as a voltage follower. Probably I would have to study your math some to understand what you mean. But my conclusion of a ratio of 1.5 causing the traces of amplitude vs. frequency to coincide to the -3 dB frequency is based on simulation with LTspice. If the ratio of the cut-off frequency of the 2nd order filter is made 1.2 times that of the first order filter, then in the region from 0 dB to -3 dB, the traces don't even come close to coincidence.

-Pete
 
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