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Old 31st August 2013, 09:33 AM   #21
AndrewT is offline AndrewT  Scotland
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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.
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Old 31st August 2013, 11:27 AM   #22
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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.
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Old 2nd September 2013, 04:28 PM   #23
DF96 is offline DF96  England
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Quote:
Originally Posted by mondogenerator
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.
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?

Quote:
Essentially he is mathematically correct but misses the point entirely, and hasnt contributed anything, except a perpetuation of a semantic arguement.
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.
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Old 2nd September 2013, 04:45 PM   #24
AndrewT is offline AndrewT  Scotland
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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?
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Old 2nd September 2013, 05:24 PM   #25
DF96 is offline DF96  England
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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.
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Old 2nd September 2013, 06:29 PM   #26
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Default LOL CHILL DF

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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Last edited by mondogenerator; 2nd September 2013 at 06:51 PM.
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Old 2nd September 2013, 08:11 PM   #27
DF96 is offline DF96  England
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Quote:
Originally Posted by DF96
Perhaps in the limit of vanishingly small Q a second-order can behave like two cascaded first-orders?
Just realised that an hour later I answered my own question:
Quote:
I believe the Q of two cascaded 1st order filters of the same corner frequency is 0.5.
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Old 3rd September 2013, 07:10 PM   #28
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Default 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)
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Old 3rd September 2013, 09:24 PM   #29
DF96 is offline DF96  England
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Quote:
2 RCs in series will add their Cs in the regular way (1/Ctot = 1/c+1/c) and the turnover point will change.
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.
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Old 5th September 2013, 11:14 PM   #30
RNMarsh is offline RNMarsh  United States
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Take a look at the filter decay time via an MLS waterfall plot.

Thx-RNMarsh
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