DF,
I am going to go back and re-read the various filter papers on my HDD.
There is obviously something that I have misunderstood.
At the moment, I see it as: all active filters can choose a Q value and that applies to all odd order and all even order filters, of the Bessel, Butterworth, Linkwitz type.
I don't use any of the ripple type filters and have not studied them.
I am going to go back and re-read the various filter papers on my HDD.
There is obviously something that I have misunderstood.
At the moment, I see it as: all active filters can choose a Q value and that applies to all odd order and all even order filters, of the Bessel, Butterworth, Linkwitz type.
I don't use any of the ripple type filters and have not studied them.
Andrew, to the best of my knowledge you are correct, eg a sallen key stage can be overdamped to get Q 0.7 Butterworth, or 0.5 Q Bessel 1st order. Df is correct that a simple RC has no Xi damping and has a simple TF but a quadratic TF can be made to perform identically. Essentially he is mathematically correct but misses the point entirely, and hasnt contributed anything, except a perpetuation of a semantic arguement. Once again Andrew thank you, you have (as usual) helped a great deal.
I'm not certain that a second-order filter can perform identically to a first-order filter. Perhaps in the limit of vanishingly small Q a second-order can behave like two cascaded first-orders?mondogenerator said:
Fortunately I would rather be mathematically correct and 'miss the point' than the converse. Assigning a fictitious value to a non-existent parameter might be a useful engineering trick, but it is best if people realise when they are doing this so they don't attach meaning to it.
What is the Q of two cascaded 1st order filters?
Or does this second order filter not have a Q?
What if we cascaded three first order passive filters each interposed with a buffer to ensure the near zero source impedance and near infinite load impedance, so that we have a third order passive filter? Would this too have no Q value?
Or does this second order filter not have a Q?
What if we cascaded three first order passive filters each interposed with a buffer to ensure the near zero source impedance and near infinite load impedance, so that we have a third order passive filter? Would this too have no Q value?
I believe the Q of two cascaded 1st order filters of the same corner frequency is 0.5. Different corners freqs give a different Q value. All second order filters have a Q.
A first order is 1+s - no other parameter so no Q.
Second order is 1+as+s^2. Q=1/a.
Third order is 1+as+bs^2+s^3. What is Q? It may be possible to define a Q-like parameter from some combination of a and b, but this will not fully characterise the filter as there are two parameters. As a minimum you would need Q1 and Q2.
A first order is 1+s - no other parameter so no Q.
Second order is 1+as+s^2. Q=1/a.
Third order is 1+as+bs^2+s^3. What is Q? It may be possible to define a Q-like parameter from some combination of a and b, but this will not fully characterise the filter as there are two parameters. As a minimum you would need Q1 and Q2.
LOL CHILL DF
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scratch that, just tried on filter modeller and im wrong Megalols 😀
phase maxima would still be 180 deg anyway. doh!
A single reactor cannot resonate, and such has no resonance magnification (Q)
Like a pendulum in a vacuum, which cannot exceed a max freq due to its mass/inertia, and unlike a 2nd order spring/damper suspension system which can have a resonant Q
EDIT:
scratch that, just tried on filter modeller and im wrong Megalols 😀
phase maxima would still be 180 deg anyway. doh!
A single reactor cannot resonate, and such has no resonance magnification (Q)
Like a pendulum in a vacuum, which cannot exceed a max freq due to its mass/inertia, and unlike a 2nd order spring/damper suspension system which can have a resonant Q
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Really? Really really?
Excuse me BUT.
2 RCs in series will add their Cs in the regular way (1/Ctot = 1/c+1/c) and the turnover point will change.*
*Disclaimer: I think that is correct, at least partially
Like you said, there is no Q in first order. You were right there.
There is no resonance with a single reactive component, only attenuation, since there is no negative reactance to counter the positive reactance. It is not a tuned circuit.
It will shift Cos Phi in a power engineering sense, and phase angle, but you need an inductor (or simulated inductance) to achieve resonance and Q magnification (either current resonance magnification in shunt reactors, or Voltage resonance magnification in series reactors)
I believe that the Q of 0.7 assigned to 1st order systems is due to the rate of attenuation, and half power -3dB / half voltage -6 dB point in the roll off (have to look this up)
Excuse me BUT.
2 RCs in series will add their Cs in the regular way (1/Ctot = 1/c+1/c) and the turnover point will change.*
*Disclaimer: I think that is correct, at least partially
Like you said, there is no Q in first order. You were right there.
There is no resonance with a single reactive component, only attenuation, since there is no negative reactance to counter the positive reactance. It is not a tuned circuit.
It will shift Cos Phi in a power engineering sense, and phase angle, but you need an inductor (or simulated inductance) to achieve resonance and Q magnification (either current resonance magnification in shunt reactors, or Voltage resonance magnification in series reactors)
I believe that the Q of 0.7 assigned to 1st order systems is due to the rate of attenuation, and half power -3dB / half voltage -6 dB point in the roll off (have to look this up)
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It is not that simple. Two options:
1. use a buffer amplifier between two first-order filters, then you can just multiply the responses;
2. directly connect the two CR, then you get interaction between them so you need to do some algebra to find the total response.
Filter Q has its affects -
Such as:
It does not sound flat even though it measures flat frequency response. When part of a cross-over, the decay time/energy at the cross-over region of a LP/HP filter will have its audible affects.
-RM
Such as:
It does not sound flat even though it measures flat frequency response. When part of a cross-over, the decay time/energy at the cross-over region of a LP/HP filter will have its audible affects.
-RM
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who cares about that? its only to illustrate a point I am making. the delay can be anything depending on your filter freq and Q.
-RM
Anything? Within bounds im sure,depending on phase rotation id imagine. Imnot sure I follow you entirely. I don't have the ability to model in the way you have demonstrated, any help here?
My LP for eg is 5th order, RC @ 500Hz, 3.8kHz @ Q=0.618, 2.8kHz @ Q = 1.618.
My LP for eg is 5th order, RC @ 500Hz, 3.8kHz @ Q=0.618, 2.8kHz @ Q = 1.618.
The plot was from an active filter... passive would do same (speaker cross-over etc.) Do you have access to MLSSA or MLS software?
Thx-RNMarsh
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