I agree! That's why I wanted base this conversation on the impulse response!!!
Anyone willing to help me understand it in the terms I laid out?
Let's start with:
All else equal, raise BL.
1) Does the peak and trough occur at the same time?
2) The peak and trough are higher, correct?
Anyone care to tell me what hapens when we change only the inductance?
And of course, can I raise the drive level and achieve the same response in either case?
I know, I'm stubborn....
herm
Also, the Linkwitz transform works on the system Q, not the driver Q.
An LT transforms a high pass second order response in another high pass second order response. In its piston range (all the following considers this region exclusively), a driver in an infinite baffle or a closed box has a second order high pass behaviour.
To transform only the Q of a second order filter keeping the same resonance frequency, a notch is needed. For example, a Q=1.0 can be lowered to 0.5 using a "bell" equalisation (as it is called in digital processors) set at the same frequency, with Q=1 and gain=-6 dB.
forr, I don't own a dual coil driver. I wish I could try this. Can you tell me what happens?
Of course, the resonance won't change. Let's compare the Qs and the reference efficiencies with some numeric examples. Let suppose :
both coils driven, Q = 0.5
one coil driven, Q=1.0
The BL is halved and the reference efficiency is halved, and mid band SPL is -6 dB.
With a " bell" equalisation as above, a Q=0.5 can be achieved.
Neglecting Joules's effect, doubling the applied voltage doubles the current and adds 6 dB SPL : the frequency response is then the same as when both coils are driven.
This is because the forces applied to the cone are the same in both cases :
two coils F = B*L*2*i
one coil F = B*L*i*2
There is an even more subtle effect.
Let's do not apply the bell equalization and voltage increase to the one coil driven case : the Q is 1.0, level at resonance is then at 0 dB compared to its mid band SPL.
In the both coils driven case, the Q is 0.5, level at resonance is -6 dB compared to its mid band SPL (which is 6 dB higher than in the one coil driven case).
So at resonance, the SPLs of both cases are the same, but the electrical power is halved for the one voice coil driven case, its efficiency around resonance is higher than when both coils are driven.
In no case, you can say that higher BL start and stop cones better than lower BL. There are good reasons to use high BL, but this last one is a fallacy and I do not know neither any serious paper agreeing with it nor any facts demonstrating it.
An LT transforms a high pass second order response in another high pass second order response. In its piston range (all the following considers this region exclusively), a driver in an infinite baffle or a closed box has a second order high pass behaviour.
To transform only the Q of a second order filter keeping the same resonance frequency, a notch is needed. For example, a Q=1.0 can be lowered to 0.5 using a "bell" equalisation (as it is called in digital processors) set at the same frequency, with Q=1 and gain=-6 dB.
forr, I don't own a dual coil driver. I wish I could try this. Can you tell me what happens?
Of course, the resonance won't change. Let's compare the Qs and the reference efficiencies with some numeric examples. Let suppose :
both coils driven, Q = 0.5
one coil driven, Q=1.0
The BL is halved and the reference efficiency is halved, and mid band SPL is -6 dB.
With a " bell" equalisation as above, a Q=0.5 can be achieved.
Neglecting Joules's effect, doubling the applied voltage doubles the current and adds 6 dB SPL : the frequency response is then the same as when both coils are driven.
This is because the forces applied to the cone are the same in both cases :
two coils F = B*L*2*i
one coil F = B*L*i*2
There is an even more subtle effect.
Let's do not apply the bell equalization and voltage increase to the one coil driven case : the Q is 1.0, level at resonance is then at 0 dB compared to its mid band SPL.
In the both coils driven case, the Q is 0.5, level at resonance is -6 dB compared to its mid band SPL (which is 6 dB higher than in the one coil driven case).
So at resonance, the SPLs of both cases are the same, but the electrical power is halved for the one voice coil driven case, its efficiency around resonance is higher than when both coils are driven.
In no case, you can say that higher BL start and stop cones better than lower BL. There are good reasons to use high BL, but this last one is a fallacy and I do not know neither any serious paper agreeing with it nor any facts demonstrating it.
Not according to absolutely everything I’ve been exposed to on the subject dating back to 1956, which as I later learned was worked out back in the early 1920s when Bell Labs/W.E. began experimenting with adding super tweeters to their full range horn systems. Indeed, contrary to popular belief that proper basic speaker design began with the advent of T/S filter theory cab design, it was actually a very mature science by the time I read Cohen’s book for the layman/hobbyist.
You’re right though that a 0.5 Qt ‘rings’ once at the ‘let go’ point, but it doesn’t change the fact that it takes a 1st order Butterworth filter [6 dB/octave] slope for a transient perfect response which in T/S parlance = 0.5 Q[ts, tb, tc, tp].
An over-damped system isn't able to accurately track the signal in any measurable way, stopping short of its 'let go' point plus it decays more rapidly, the price one pays for a too powerful motor for the application, so it may sound ‘tight’, ‘fast’, ‘dynamic’, but it will be tonally unbalanced [harmonically distorted].
There’s no ‘pulse of infinite amplitude and zero width’ in live or recorded music, so there’s a delayed start to go along with the delay at the ‘let go’ point [‘ringing’].
Thanks for the thought, but the way things have been going for over a decade now I figure it will be yet another year before I can get back to being an active speaker DIYer.
GM
True, its ability to ‘articulate’ usually refers to its harmonic distortion due to being over-damped and/or the Q and quantity of its break-up modes BW. Drivers such as Lowthers are nothing if not ‘dynamic’ due to being very large tweeters by design. This doesn’t mean they are accurate reproducers though except over a narrow BW, just that it gives one more tuning options to tailor a wider BW one to meet the needs of the app.
GM
Maybe, raising the drive level lowers dynamic headroom and raises the noise floor, so large changes in amplitude of both average and peak SPL such as occurs in symphonies may be even more distorted than if left alone, so it will depend on how compressed and/or BW limited whatever you listen to is.
Anyway, at a glance it seems your Qs and much more have all been answered in various posts, links.
GM
Thanks for the replies guys.
So far I think we all agree on the following:
All else equal,
1) A low BL driver will respond identically to a high BL driver if you increase the input power.
2) A High Mms driver will respond identically to the low Mms driver if you increase the input power.
3) the time that the peaks and troughs occur are identical.
...within the limits of headroom, power compression, of course...
(I think Forr, ScottMoose, and GM are all in agreement with me anyway)
SO:
A light cone, high BL driver plays LOUDER for the same input than a heavy cone, low BL driver.
The cone moves faster (reached a higher peak in the same amount of time).
But it started and reversed it's direction in the same amount of time.
This only means LOUDER.
Comments?
-herm
P.S. After we flame this one a bit longer I'm moving on to inductance.
P.S.S. onform, please don't let the percieved "tension" scare you off. this is a very confusing topic
that we battle every 3 months or so. Every time we go through it we all learn more.
So far I think we all agree on the following:
All else equal,
1) A low BL driver will respond identically to a high BL driver if you increase the input power.
2) A High Mms driver will respond identically to the low Mms driver if you increase the input power.
3) the time that the peaks and troughs occur are identical.
...within the limits of headroom, power compression, of course...
(I think Forr, ScottMoose, and GM are all in agreement with me anyway)
SO:
A light cone, high BL driver plays LOUDER for the same input than a heavy cone, low BL driver.
The cone moves faster (reached a higher peak in the same amount of time).
But it started and reversed it's direction in the same amount of time.
This only means LOUDER.
Comments?
-herm
P.S. After we flame this one a bit longer I'm moving on to inductance.
P.S.S. onform, please don't let the percieved "tension" scare you off. this is a very confusing topic
that we battle every 3 months or so. Every time we go through it we all learn more.
GM,
I'm still thinking what you said. Damping is an after-event. Everything that happens after is not part of the signal. All the systems are imperfect because none of them (Q<0.5, Q=0.5, Q>0.5) can stop at the very instant the signal stops. But an overdamped system would stop the earliest, staying truest to the music.
With regard to harmonic distortion/structure, there might be a difference, whether it sounds better or worse is up for debate (I guesst). I'll digest some more before coming back.
herm, I don't agree with anything that you've said in you posts. Those are your thoughts and your thoughts alone.
forr,
I'll read through yours before replying.
The force that displaces the cone depends on the signal level, but the amount of force generated depends on how powerful a magnet you are using.
Dynamics
Thinking of compression drivers with even lighter cones and bigger magnets, the details and lifelike dynamics and natural sound that they can produce. Lowthers and ANs are bascially the direct radiator equivalents of compression drivers. Put a horn in front of them and they will produce similar dynamics. Just some thoughts.
I'm still thinking what you said. Damping is an after-event. Everything that happens after is not part of the signal. All the systems are imperfect because none of them (Q<0.5, Q=0.5, Q>0.5) can stop at the very instant the signal stops. But an overdamped system would stop the earliest, staying truest to the music.
With regard to harmonic distortion/structure, there might be a difference, whether it sounds better or worse is up for debate (I guesst). I'll digest some more before coming back.
herm, I don't agree with anything that you've said in you posts. Those are your thoughts and your thoughts alone.
forr,
I'll read through yours before replying.
The force that displaces the cone depends on the signal level, but the amount of force generated depends on how powerful a magnet you are using.
Dynamics
Thinking of compression drivers with even lighter cones and bigger magnets, the details and lifelike dynamics and natural sound that they can produce. Lowthers and ANs are bascially the direct radiator equivalents of compression drivers. Put a horn in front of them and they will produce similar dynamics. Just some thoughts.
Ra7,
I'm still thinking what you said. Damping is an after-event. Everything that happens after is not part of the signal. All the systems are imperfect because none of them (Q<0.5, Q=0.5, Q>0.5) can stop at the very instant the signal stops. But an overdamped system would stop the earliest, staying truest to the music.
See Krekowski's way of thinking : Box-Q
Abstract
Finally I would like to touch on the ideal that proliferates many discussions: Systems with Q = 0.5 have the best transient response because they are critically damped. I don't recall ever reading that in any engineering text book.
I'm still thinking what you said. Damping is an after-event. Everything that happens after is not part of the signal. All the systems are imperfect because none of them (Q<0.5, Q=0.5, Q>0.5) can stop at the very instant the signal stops. But an overdamped system would stop the earliest, staying truest to the music.
See Krekowski's way of thinking : Box-Q
Abstract
Finally I would like to touch on the ideal that proliferates many discussions: Systems with Q = 0.5 have the best transient response because they are critically damped. I don't recall ever reading that in any engineering text book.
high BL/Mms ratio gives larger ds/dt where compliance is constant, or at least linear....thus a higher rate of change of displacment, accn/decn, which all ELSE being equal will track the signal more accurately than a lower BL/Mms driver.
I wouldve thought most people here with an engineering background would agree with that.
The problem with a simple analogy such as this, IMHO is that, compliance is another variable which needs to be addressed, and the ratio of BL/Mms does not in itself denote a 'quick' or 'slow' driver, just as Qts does not. Just as friction hampers accn, so the resistance of the compliance to movement acts like friction on the 'force' and accn of the VC.
in an outward excursion the accn is reduced due to a degree of surround 'tension' resistance, and in the returning cones' inward excursion, the reverse is true, until again the 'tension' returns. To me then it would still seem that the highest BL/Mms ratio one can find would be desirable over a lower one, everything ELSE all being equal....
I personally think VC design/cone design/magnetics/basket design....ALL impacts upon this. since there ARE no identical drivers...
like some wise man said:
damping is AFTER the signal has totally decayed, and at that point, overdamping of spurious movement is an oxymoron.
I wouldve thought most people here with an engineering background would agree with that.
The problem with a simple analogy such as this, IMHO is that, compliance is another variable which needs to be addressed, and the ratio of BL/Mms does not in itself denote a 'quick' or 'slow' driver, just as Qts does not. Just as friction hampers accn, so the resistance of the compliance to movement acts like friction on the 'force' and accn of the VC.
in an outward excursion the accn is reduced due to a degree of surround 'tension' resistance, and in the returning cones' inward excursion, the reverse is true, until again the 'tension' returns. To me then it would still seem that the highest BL/Mms ratio one can find would be desirable over a lower one, everything ELSE all being equal....
I personally think VC design/cone design/magnetics/basket design....ALL impacts upon this. since there ARE no identical drivers...
like some wise man said:
damping is AFTER the signal has totally decayed, and at that point, overdamping of spurious movement is an oxymoron.
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A gradient on the attack/decay envelope of a square wave reproduced by a driver surely has many causes: suspension 'drag' , reactance of VC and probably others i dont know of. I would postulate that a high BL/Mms ratio is relevant, but only when referenced to Lvc as in my sig. although its a qualitative measure rather than quantative. the most interesting thing is that Qts, BL and Lvc are normally stated as constants for ease of calcs, when clearly they arent depending on Xlin
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high BL/Mms ratio gives larger ds/dt where compliance is constant, or at least linear....thus a higher rate of change of displacment, accn/decn, which all ELSE being equal will track the signal more accurately than a lower BL/Mms driver.
I wouldve thought most people here with an engineering background would agree with that.
I would have thought the current wouldn't be forgotten in the equation.
Yes, for same currents, a higher Bl/Mms ratio means faster acceleration/deceleration.
Which translates in the fact that more travel of the cone is done in the same duration.
Which means more air moved and higher delivered acoustic pressure,
not the ability to reproduce the signal more accurately.
I wouldve thought most people here with an engineering background would agree with that.
I would have thought the current wouldn't be forgotten in the equation.
Yes, for same currents, a higher Bl/Mms ratio means faster acceleration/deceleration.
Which translates in the fact that more travel of the cone is done in the same duration.
Which means more air moved and higher delivered acoustic pressure,
not the ability to reproduce the signal more accurately.
It wasnt forgotten....hence Lvc...
although perhaps 'Lvc' should be replaced with Levc or even XLvc
overdamping is a 'drag' factor during reproduction of a signal, but since we already have many drag factors already.......
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OK, there's impedance compensation to render Lvc moot and all else equal between two drivers, i.e. same Fs, Sd, Vas, Qes, speed of sound and air pressure:
Re = 7 ohms, BL/Mms = 5
Re = 14 ohms, BL/Mms = ~7.07
Please explain how the higher Re driver can track a signal more accurately.
GM
@all, you might look at my last post over there, too : http://www.diyaudio.com/forums/full...making-baffle-less-speaker-2.html#post2651089
After studying the first paper in the references I think any discussion about what's really going on should be settled, I can't think of a better paper summarizing it all, the first three equations cover every aspect of a driver's linear behaviour (some extrapolating skills are required, though).
Qts = 0.5 is for sure critical (aperiodic) damping for the cone movement (which is 2nd order lowpass). SPL is the double integral of it, a highpass, and no concept of "transient perfect" or "aperiodically damped" exists for highpass, per se. When this Q is truly physical/mechanical and not just dialed in by EQ, then damping of the cone movement to any external exitation (other drivers, box and room modes, etc) is also aperiodic, as is recovery from overdrive and from DC offset.
Also please go ahead and simulate (with LTspice this is as easy as it gets) things like excursion and SPL at the pole freq (and around it) vs. different Q, things get pretty clear then (normalize levels with Q when at pole freq to get that qualitative difference the box Q makes, not swamped by level differences.
- Klaus
After studying the first paper in the references I think any discussion about what's really going on should be settled, I can't think of a better paper summarizing it all, the first three equations cover every aspect of a driver's linear behaviour (some extrapolating skills are required, though).
Qts = 0.5 is for sure critical (aperiodic) damping for the cone movement (which is 2nd order lowpass). SPL is the double integral of it, a highpass, and no concept of "transient perfect" or "aperiodically damped" exists for highpass, per se. When this Q is truly physical/mechanical and not just dialed in by EQ, then damping of the cone movement to any external exitation (other drivers, box and room modes, etc) is also aperiodic, as is recovery from overdrive and from DC offset.
Also please go ahead and simulate (with LTspice this is as easy as it gets) things like excursion and SPL at the pole freq (and around it) vs. different Q, things get pretty clear then (normalize levels with Q when at pole freq to get that qualitative difference the box Q makes, not swamped by level differences.
- Klaus
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