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.
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.
*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.
Svante said:
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...
BEAR
(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.
john k... said:
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...
_-_-bear
>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 !
>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 !
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.bear said:
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
... 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.
_-_-bear
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.
_-_-bear
I'm with Svante and co on this one. Having played with LTs for almost 10 years now I cannot fault them. They compensate for amplitude, phase and damping. Perfectly. It's all about the manipulation of poles and zeroes and the cancelling effect this has. An LT is not just a simple amplitude correction.
bear said:
(JPK) Sounds like a reasonable comment IF you looked at the driver alone. But the driver alone is not the SYSTEM. The SYSTEM is everything that appears between the input and the acoustic output. Look at the response to the input. Ask yourself what it is? It's the convolution of the input with the impuse response of the pre Eq circuit convoluted the driver's impulse response. If the convolution of the EQ 's impulse with the driver's impulse yields a Q =0.527 response, then the response will have a Q= 0.527 behavior to any input. The only requirement is that the system is linear and time invariant, standard stuff.
(JPK)No that is not all we have done for F3. Yes, the stored energy in the driver won't change but the EQ will introduce a behavior in the applied signal (its onw storred energy if you like) which will control the driver motion through the Eq. That is the basic principle that applies to response shaping in loudspeaker to start with; a basic principle of linear system. It if didn't hold we couldn't build loudspeaker passive cross overs etc.
(JPK) I don't want to argure that because it has nothing to do with it.
The waterfall plot will reveal the correct behavior. It's not that the bell is hit less hard, it is that the bell is hit with a signal (the Eqed input) that controls the ringing. We aren't talking natural response of the driver here, we are talking about the forced response to an input that controls the way the driver moves. As long as we are talking about a system where the input is the signal applied to the input of the EQ circuit and the output is the acoustic output of the driver then eqing the amplitude is exactly the same thing as building a lower Q system. As long as the system is linear the amplitude and time domain responses are related by the Fourier transformation.
If you want to say that if you displace the cone physically, release it and watch its motion decay, then yes there will be a difference. But that is the response of a different system to a different input and not what is happening when an Eqed electical input is stimulating the driver.
While you guys are still arguing about that one, I've got another question.
How does current drive account for BL^2 variations? That's what Mills/Hawksford's paper says in the abstract.
http://www.essex.ac.uk/ese/research...J12 Distortion reduction MC current drive.pdf
How does current drive account for BL^2 variations? That's what Mills/Hawksford's paper says in the abstract.
http://www.essex.ac.uk/ese/research...J12 Distortion reduction MC current drive.pdf
(JPK) First, there is no argument as far as in can see. The Mills Hawksford paper is in agreement with what I have stated. It's just the physics of linear systems. Re BL nonlinearity, as I recall, the paper really isn't addressing BL nonlinearity passively. What is stated is that BL nonlinearity leads to nonlinear electrical damping and since current drive eliminates electrical damping one source of nonlinear distortion is removed. However, this doesn’t solve the problem of nonlinear distortion arising form BL nonlinearity and it relation to the drive or forward current. One other point is that current drive will eliminate the nonlinear compression effects associated with VC heating the increases in Re. The only problem here is that the VC heating in a current driven system may lead to thermal instability and potentially VC burn out. This is because the heat generated in the VC goes like I^2xRe. VC heating still occurs in a current driven system and Re increases. So as Re increases the heat generated increases in proportion to the increase in Re. This means further increases in Re and greater (faster) heating. Unless the system can reject heat to the surroundings at a high enough rate there exists the possibility the VC temp will increase without limit until burn out. In a voltage driven case, the heat generated goes like V^2/Re. This as Re increases due to heating the heat generation rate decrease and thus tends to be self limiting.
Going back to BL nonlinearity for a moment, it, along with other sources of nonlinear distortion, can be reduced with motional feedback. However, motional feedback is not in any way unique to current drive.
Going back to BL nonlinearity for a moment, it, along with other sources of nonlinear distortion, can be reduced with motional feedback. However, motional feedback is not in any way unique to current drive.
I stand corrected.
I was thinking about the response of the system after then input signal is "turned off." In this case, my thinking was much like that of hitting a bell... sure, it seems, if the bell rings at a certain frequency then anytime that frequency comes up the EQ simply hits the bell "less hard." Seems like it still rings if the signal is abrubtly cut off... yes?
Well, that is true only if the signal is physically cut. If the input to the EQ is cut, the EQ still responds to that signal. Anti-resonance is precisely what a LT does. I knew this deep down, and had for some reason convinced myself it only applied to semi-steady-state conditions. In reality, it applies equally to all transient signals. Charles is correct in that in minimum-phase systems it doesn't matter which part of the system alters the Q, it only matters what the final Q actually is.
As for the bell analogy... a LT acts like another hammer - one that hits the bell precisely at the peak of its ringing with perfect force to cancel the vibration completely.
Now, this brings me to a question I asked long ago in KWY's transconductance thread: are there any practical reasons why you would not want to use LT to control damping in a current driven system, as opposed to mechanical damping in the driver?
I was thinking about the response of the system after then input signal is "turned off." In this case, my thinking was much like that of hitting a bell... sure, it seems, if the bell rings at a certain frequency then anytime that frequency comes up the EQ simply hits the bell "less hard." Seems like it still rings if the signal is abrubtly cut off... yes?
Well, that is true only if the signal is physically cut. If the input to the EQ is cut, the EQ still responds to that signal. Anti-resonance is precisely what a LT does. I knew this deep down, and had for some reason convinced myself it only applied to semi-steady-state conditions. In reality, it applies equally to all transient signals. Charles is correct in that in minimum-phase systems it doesn't matter which part of the system alters the Q, it only matters what the final Q actually is.
As for the bell analogy... a LT acts like another hammer - one that hits the bell precisely at the peak of its ringing with perfect force to cancel the vibration completely.
Now, this brings me to a question I asked long ago in KWY's transconductance thread: are there any practical reasons why you would not want to use LT to control damping in a current driven system, as opposed to mechanical damping in the driver?
Serow said:
(JPK) well I would agree with this except for the VC heating effect which, in theory won't alter the response of a current driven system. However, all these effect are present in current and voltage driven system with or w/o Eq applied to alter F3, Q or both. The only way around thenm is motional feedback, and if motionaly feedback is emploied I can't see any advantage to a current driven systm at all. In such a case a correctly designed system make the amp supply the required voltage and current to meet the targeted response.
There is at least one: Current-drive eliminates the pole that is caused by the driver's Le and makes thus the feedback a little easier. One could of course eliminate the pole also by a zero in the forward path but two things come for free with current-drive:
1.) you don't have to measure Le exactly
2.) the nonlinear displacement-dependant behaviour of Le is also automatically accounted for.
Regards
Charles
RHosch said:
Ok. maybe I need to read about the LT some more... perhaps I missed something? Hmmm... citations, URLs that make the case?
The LT can't be another hammer, except in a situation where we are looking at a semi or actually steady state excitation.
The input "transform" no matter what it is set to do has no way to keep track of the response of the speaker without closing the feedback loop, in which case I would agree that the response of the system can be altered within the bounds of the ability of the motor to control the cone (and even there, not quite perfectly)...
The closest non closed loop method might be these home DSP based "speaker/room correction" boxes where you mic the speaker or the room and the box gives you a perfect response in return. Does it? Isn't this far more precise than the relatively crude LT filters? Does the waterfall response curve really change?
Show me any filter circuit that can correct the impulse response of a speaker so that it is just that little spike and nothing more, and I will show you some fairy tale books. No input filter can change the stored energy in a given driver/speaker system.
I agree that quite a bit of correction can be applied to a signal before it hits the speaker, but that correction's effectiveness is proportional to how *little* that change really is, and how linear the transducer already is.
Let's take an easier example to consider - like a speaker with a nice resonance peak at ~5kHz. There is no filter that will completely correct that resonance peak. It might work on axis but it won't work off axis, etc. Is this a different case?
In addition, if this LT method was so effective and sucessful, would we not see manufacturers slamming random drivers into essentially random boxes and applying an LT, resulting in perfect response?? Why isn't this happening if it works as suggested here?
Imho, it's got a far better shot at working if you take a too low Qts system and increase the Q. Then you are ok all other factors being reasonable in the design. The other way around, not likely. This is like the EBS bass idea, etc...
Is there a filter that we can apply to the input of a problematic amplifier (define it as you wish) that will correct the output completely and eliminate the problem(s)? If so we have a very easy solution to power amp issues, no?
I guess my point is that thus far "predistortion" schemes used to correct output issues for almost all of these sorts of problems range in sucess from pretty darn good approximations to impossible to suceed approximations - the best case coming where the problems to be overcome are really not too great in the first place...
_-_-bear
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