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Old 29th January 2005, 02:00 AM   #1
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Default Loss of electrical damping for bass drivers powered by current drive

Could it be fixed with something as simple as a Linkwitz transform?
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Old 29th January 2005, 04:27 AM   #2
RHosch is offline RHosch  United States
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You can fix the amplitude response, i.e. remove the resonant peak under sine input conditions, but under transient conditions you will not be able to prevent the mechanical ringing without some form of damping in the system. Voltage drive uses electrical damping... current drive practically requires mechanical damping.
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Old 29th January 2005, 08:43 PM   #3
Svante is offline Svante  Sweden
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Wrong. It is possiblie* to perfectly compensate for a resonant system, by putting an anti-resonance in the amplifier, and as far as I understand it, that is what the linkwitz transform is all about. Both frequency and transient response will be well-behaved.

*There are some issues regarding aging and drift of fs that can become hard to deal with in practice, but in principle, the compensation circuit fixes also the transient response.
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Old 29th January 2005, 11:03 PM   #4
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Quote:
Originally posted by Svante
Wrong. It is possiblie* to perfectly compensate for a resonant system, by putting an anti-resonance in the amplifier, and as far as I understand it, that is what the linkwitz transform is all about. Both frequency and transient response will be well-behaved.

*There are some issues regarding aging and drift of fs that can become hard to deal with in practice, but in principle, the compensation circuit fixes also the transient response.

No you can't.

As noted you can compensate for frequency response just fine.
But damping factor isn't the same as frequency response.
The outcome is similar under static conditions - a change in amplitude.

Under dynamic conditions it's less straightforward. DF is a function of the output impedance, and the load does not have a resistive characteristic, it's reactive. The effect of a high DF/low output Z connected to a reactive load, especially with an amp with negative feedback is not quite predictable with a dynamic music signal.

Merely exciting a resonant system less (frequency response compensation) does not actually change it's behavior. Increasing the damping of a resonant system does change its behavior (high DF vs. low).

The result of freq response correction, imho, in many cases could be subjectively similar, IF the speaker is already low in non-linearities.

But the bottom line is that a high DF amp with a series resistor still doesn't sound the same as a low DF amp on the same load.

Ymmv of course...


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Old 30th January 2005, 12:38 AM   #5
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(JPK) The problem here is that you can not think of the driver alone. You must analyze the system which includes the amplifier, voltage or current source. It's the forced response that is of interest, not the natural response of the driver alone. For example ask yourself what happens if a driver with Qms = 10 and Qes = .527 (Qts= .5 ) is driven by a low impedance voltage source and a step input is applied. How does the driver response? Now ask yourself what happens if the same driver is connected to a current source with infinite output impedance so Qts = Qms = 10. but an equalization network is in place so that the steady state frequency response of the system is that of Q=0.5 response. Now hit this system with a step input. What do you suppose the system response will be? Obviously the question is rhetorical.
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Old 1st February 2005, 02:38 AM   #6
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Quote:
Originally posted by john k...
(JPK) The problem here is that you can not think of the driver alone. You must analyze the system which includes the amplifier, voltage or current source. It's the forced response that is of interest, not the natural response of the driver alone. For example ask yourself what happens if a driver with Qms = 10 and Qes = .527 (Qts= .5 ) is driven by a low impedance voltage source and a step input is applied. How does the driver response? Now ask yourself what happens if the same driver is connected to a current source with infinite output impedance so Qts = Qms = 10. but an equalization network is in place so that the steady state frequency response of the system is that of Q=0.5 response. Now hit this system with a step input. What do you suppose the system response will be? Obviously the question is rhetorical.
Ok, so we pre equalize the amplitude response of the incoming signal so the frequency response *looks* the same as if the driver had a Qt of 0.527.

At the resonant frequency of the driver in the enclosure the thing takes off like a scared rabbit. Excited it rings. Qt = 10.

All we have done for the F3 frequency is reduce the excitation. The system is still a Qt=10. You can reduce the excitation and get an overall freq response that appears similar to the low Q response, but no matter how you cut it, the response to say 3-4 cycles of F3 energy is still going to be that of a Qt=10 speaker, just lower in amplitude.

The stored energy and ringing will not change.

IF you want to argue that with the Qt=10 the F3 will be lower, and so it is possible to roll the driver off, operating *above* the now outrageous resonance, maybe then you can create a system with effectively higher damping.

The result of amplitude equalization may be *similar to* but it is not the same as a lower Q system. Perhaps checking the waterfall plot... not the step response might reveal more?

The inherent "ringy-ness" of the bell doesn't change because you hit it less hard...

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Old 1st February 2005, 02:52 AM   #7
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>But the bottom line is that a high DF amp
>with a series resistor still doesn't sound
>the same as a low DF amp on the same load.

Yep! I've been experimenting with using an
output xfmr.
But backwards to the tubetype implementation
I.E.
8 Ohm primary ........ 32 Ohm secondary
Hooked to 8 Ohm driver ....
Mr. Fat Sound !
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Old 1st February 2005, 03:53 AM   #8
Mr Evil is offline Mr Evil  United Kingdom
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Quote:
Originally posted by bear
...The inherent "ringy-ness" of the bell doesn't change because you hit it less hard....
Maybe not, but surely you can add anti-ringing to the signal to cancel it out? For instance, simplistically, if you know that inputing signal X will result in ringing at 50Hz for three cycles, then just add three cycles of 50Hz 180 degrees out of phase and the end result is that it looks like there was no ringing to start with.
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Old 1st February 2005, 06:39 AM   #9
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Maybe not, but surely you can add anti-ringing to the signal to cancel it out?
Thats not a bad description of what happens.
A 2nd order highpass like a closed-box speaker is a minimum-phase device. Its temporal behaviour is therefore determined by its behaviour in the frequency domain. A series of minimum-phase networks are still a minimum-phase network in total. So if we have a 2nd order highpass of a given frequency response it will have a given transient response independant of how we approached this function.
One cannot compare it with hitting a bell less hard - it is like hitting it in a different way !

The only disadvantage of relying on the driver's own damping and controlling overall behaviour by an LTF is the higher susceptibility to parameter drift (an ordinary driver's Qms is more likely to drift than it's Qes) than using electrical damping.

Regards

Charles
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Old 1st February 2005, 11:57 AM   #10
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... sure the theory says that they are identical. In reality they are similar. How similar is the question.

I suggest you listen to it and see what you think.

Also, measure the waterfall and see what you see.

No matter how you cut it the speaker in the box doesn't know about what the equalizer is doing to the rest of the spectrum, around F3 it thinks it's a high Q system - presented with signal in that area it still acts like a high Q system, right? Lower amplitude, yes, but the control and damping doesn't change with amplitude, the real Q is unchanged.

It's being "hit" the same way, just lower in amplitude in this example.

Clearly, given relatively small differences in Qt the equalization method will be more sucessful, with large differences like the posited example, it seems unlikely to have a result that is identical when measured or listened to.

Regardless we can agree that if the output is identical after EQ then it is identical.

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