Yes, you did, and when I asked how you determined this, you changed the subject. So, I'll ask again, since you brought it up: how did you determine that the frequency response changes were not the cause of your claimed audibility differences? (post 537)
Of course there is no filter in my example. The measurement is just a 8" woofer in air. The frequency response is extended but is NOT rising at all. It actually slopes down @ 12 dB per octave above 3 KHz. Being a treated paper cone it is rather smooth with no evidence of break-ups showing up. The CSD is clean, decay gets shorter and shorter going up in frequency and above all there are no nasty peaks and dips while decaying.
In fact the woofer is matched to its tweeter to work without crossover. The woofer has a phase plug, so in the "worst case" one can talk of mechanical cross-over but anything can be analyzed as a filter....
About the high Q there is no problem. The peak at resonance is more a theoretical problem. In reality the entity of the peak is quite small and completely nullified by the room modes in listening mode. The bass is fast, clean, detailed, all the best adjectives one can think of....
I think too much importance is attributed to T&S parameters which in reality don't have and one should use a real current amp to draw final conclusions.
Ok Joe, have it your way. You are my agenda. You got me. Boy is my face red. Ha, ha...
You want to be judged fairly. Ok, send me the impedance data files for the drivers so I can import them into my simulations and present the results.
But in the mean time, to be fair, you just posted this plot:
You state green is output Z = 0.01, with and w/o comp. I'm satisfied that it would also represent the response with output Z = 5 when the Z comp is present to reasonable accuracy. Red is Z out = 5, w/o Z comp. So we can all take red and green as representative of your triode amp driving the speaker with and w/o Z comp.
Are you still going to claim that the response differences shown between red and green are inaudible? Your claim is of increased clarity with the Z comp networks regardless of the amp. How can you make that claim,; how can you make that assessment; how can you make that subjective evaluation, when FR differences of that magnitude are present? Surely you aren't going to sweep FR differences of that magnitude under the rug and claim inaudibility, are you?
So, as far as I can see, you can listen to the speaker with your solid state amp, with output Z = 0.01, with and w/o Z comp and make the conclusion of increase clarity with the Z comp. You can make any claim you like, as far as I am concerned, as to why. I don't care. I, personally, would still point to the amp's characteristics into the different load as it is the most obvious. I could even go as far as saying the increased current draw from the amp places the operating point at a different point on the load line. Or increased current draw increases the junction temperature places the transistors in a more linear region. There are many reasons the amps characteristic would change under a different load. There aren't any reasons the speaker would sound different, other than the input signal is different. And again, the input signal is a function of the amplifier and it operating point. But, BUT, BUT, what ever the reason for increased clarity is that particular comparison, you can not carry it forward that increased clarity results for all amps, regardless of output Z, when, as you yourself have shown, the FR changes significantly. You simply can not make the same subjective or even objective evaluation when more than one parameter changes.
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Yes.
However, by making the loudspeaker load appear as a pure resistance load, secondary effects become less audible, and ime for the better.
To be expected.
So as per the above quotes, it's all about compensating the drivers/crossover correctly in order to make the loudspeaker appear as a 'purely' resistive load.
So that means, a load that absorbs all input/impetus energy and dissipates/transduces all of the supplied energy and does not store or return energy to the source....IOW a 'perfect resistor'.
This 'perfect' resistive load enables amplifiers to perform 'as advertised', and reduces amplifier load dependency effects which include THD, IMD and other effects aside from loudspeaker in room FR variation according to amplifier source impedance.
In my experience, correct driver compensation changes subjective listening for the better, way better.
Highs extend, lows go lower, overall cleaner, livelier and more realistic sound in the room, and with infinite depth portrayal.
The caveat is that the compensating networks need to be connected pretty much directly across the driver terminals.....hint use network connection leads same length as the driver flex VC leads.
Also the network capacitors are critical...bypass PP with the biggest value polystyrenes you can afford/find.
With compensation done sufficiently correctly, depth info becomes 3D out of two loudspeakers.
With compensation done sufficiently correctly, one downside is that acoustic polarity of sources becomes clearly evident.
For natural sound recordings the sense of 3D depth when running correct acoustic polarity is stunning/surprising.
With multi miked/close miked recordings individual sound sources in the mix can be identified as being in wrong polarity wrt the other sound sources in the mix.....common is vocals inverted wrt instruments, and instruments inverted wrt each other.
Joe, I would not say that your insistence on/advocating of impedance compensation is heralding.
Kudos for exploring this in detail and publishing.
Such compensation has been dismissed by typical loudspeaker manufacturers because of impractical high cost, and argued against by many on theory grounds....all true and correct.
IIRC KEF published a paper on why their networks were 'proven' to be correct by their 'sota at the time' simulation computer.
The theory is all good, but the proof is in the eating of course.
Dan.
Jan, this time you owe me an apology.
Do you not read my post carefully? Did I not say "offset" !!!
Good grief!
You must be talking about active systems because placing the comp network across the driver's terminals will flatten the drive's impedance, not the input impedance of a speaker with passive crossover.
Hi Jan
I accept that and wish I had not over-reacted. I also wish I could delete that post. But it has become a sign post just how poisoned the discussion has become. It does not sit well with me. So I thank you for your apology.
Hi Dan
Can I get back to you, likely in the morning, we are two hours later than you here in Sydney, and I need to find something on my computer to post. Been a long day....
Well with ll the colors and bold and very large fonts, it's easy to miss what isn't. ;-)
Jan
By the way Joe, regarding my post about the observed difference in your measured FR with the comp network in place for amps with near zero and 270 ohms output Z, here is the quote from your initial post:
http://www.diyaudio.com/forums/mult...-usher-s520-current-compatible-crossover.html.
So, I take back my apology. Hey, it's your data.
John I took the time to re-read again that first post, and something caught my eye.
It is stated that because a current source 'sees it's own very large output impedance' it will always deliver current to the load (xover & drivers) that is in-phase, whatever that load is, implying (I believe) that therefor the voltage and current on the load are also necessarily in phase.
I am not now at my desk but I will sim it tomorrow, because I have the gut feeling that this is not true. It is the load that determines what the relationship is between it's current and voltage and not whatever drives it.
What is the considered opinion here on this? Or did I misinterpret the statement?
This is the statement I am uncomfortable with:
'Now I want to make a statement that I believe is an essential issue. The voltage source can produce current - no doubt about that - the problem is that it can also produce current of any phase angle. But a genuine current source can only produce a phase angle of zero since it sees it own output impedance as a series resistor of (ideally) infinite value. The voltage produced across the final load will always track current and voltage together without contamination.'
Jan
It is stated that because a current source 'sees it's own very large output impedance' it will always deliver current to the load (xover & drivers) that is in-phase, whatever that load is, implying (I believe) that therefor the voltage and current on the load are also necessarily in phase.
I am not now at my desk but I will sim it tomorrow, because I have the gut feeling that this is not true. It is the load that determines what the relationship is between it's current and voltage and not whatever drives it.
What is the considered opinion here on this? Or did I misinterpret the statement?
This is the statement I am uncomfortable with:
'Now I want to make a statement that I believe is an essential issue. The voltage source can produce current - no doubt about that - the problem is that it can also produce current of any phase angle. But a genuine current source can only produce a phase angle of zero since it sees it own output impedance as a series resistor of (ideally) infinite value. The voltage produced across the final load will always track current and voltage together without contamination.'
Jan
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Let's look at something simple for a load: a capacitor. What happens if you hook a current source to it and apply a unit sine of angular frequency w? V = IZ, I = sin(wt), Z = -i/wC, so V =-isin(wt)/wC, which certainly looks 90° out of phase...
Yes I think the basic error is the apparent thinking that somehow a current source send out not just current magnitude but also a phase angle to the load.
But the phase angle of the voltage that that current develops across the load depends on the load and not the current source - it just 'sends out' current.
Just like (mirror images again!) a voltage source 'sends out' a voltage but the phase angle of the current drawn is determined by the load.
Your cap example is a clear one, but just to make sure I'll sim it tomorrow.
Jan
But the phase angle of the voltage that that current develops across the load depends on the load and not the current source - it just 'sends out' current.
Just like (mirror images again!) a voltage source 'sends out' a voltage but the phase angle of the current drawn is determined by the load.
Your cap example is a clear one, but just to make sure I'll sim it tomorrow.
Jan
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