What I want to show here is that when you use a resonant circuit to fix a speaker resonance, the negative qualities do not remain after the correction has been made.
Like electronics, individual speaker drivers generally are minimum phase devices (this doesn't include effects like diffraction which include a delay component).
The first image is a plot of response vs frequency. I have split a flat response into a bandpassed and a bandstopped component, and then combined them. This combined response is flat like the original.
The second image is a plot of time vs level to show any ringing directly, on the same circuit test points. Using a square wave from the generator, the upper and lower circuit branches show resonance effects from their filters. The recombined result is again a perfect square wave.
Like electronics, individual speaker drivers generally are minimum phase devices (this doesn't include effects like diffraction which include a delay component).
The first image is a plot of response vs frequency. I have split a flat response into a bandpassed and a bandstopped component, and then combined them. This combined response is flat like the original.
The second image is a plot of time vs level to show any ringing directly, on the same circuit test points. Using a square wave from the generator, the upper and lower circuit branches show resonance effects from their filters. The recombined result is again a perfect square wave.
Last edited:
Awww gees ... here we go.
1) When this first came up we were talking about a speaker, not a pair of abstract filters fed into summing resistors. The Tweeter in question had it's own reactance that also became part of that ringing filter I pointed out.
2) Recombining the effects looks good but that's not was happening and in real life nobody would do that. It would be a waste of parts.
3) The circuit in question was NOT a simple bandpass filter.
Take a look at your own simulation, second thumbnail ...
In that circuit R1 would be analogous to the speaker on the output of a band stop filter... you coloured it Green ... now look at the green trace, does that look like a good square wave to you?
Similarly in a band pass filter R2 occupies the speaker's position and is charted in Red... does that look like a good square wave to you?
Finally the Blue trace after the two summing resistors would never happen in a real crossover circuit...
Seriously man ... don't play bafflegab with people who are just learning this stuff. Just be a MAN and admit you were wrong.
1) When this first came up we were talking about a speaker, not a pair of abstract filters fed into summing resistors. The Tweeter in question had it's own reactance that also became part of that ringing filter I pointed out.
2) Recombining the effects looks good but that's not was happening and in real life nobody would do that. It would be a waste of parts.
3) The circuit in question was NOT a simple bandpass filter.
Take a look at your own simulation, second thumbnail ...
In that circuit R1 would be analogous to the speaker on the output of a band stop filter... you coloured it Green ... now look at the green trace, does that look like a good square wave to you?
Similarly in a band pass filter R2 occupies the speaker's position and is charted in Red... does that look like a good square wave to you?
Finally the Blue trace after the two summing resistors would never happen in a real crossover circuit...
Seriously man ... don't play bafflegab with people who are just learning this stuff. Just be a MAN and admit you were wrong.
No they aren't. Resonance in both the circuits drawn by Allen involves circulating currents that go back and forth between the coil and capacitor. For the most part this is stored energy that stays in that part of the circuit. One of the more dangerous characteristics of sharp resonances is very high current loading on the filter's parts... many times normal.
Ringing is what happens after you take away the stimulus and relates to how long it takes the resonant circuit to dissipate that stored energy and settle down.
Resonance is when you hit the bell.
Ringing is what happens after you hit it.
If Allen wanted to explore that he should have used a very narrow pulse and examined the waveform at R1 and R2.
Exactly ... in fact it's not even looking at the problem. It's a nice sketch up trying to prove himself right from another thread. HERE Post #100 shows the actual ringing behaviour of an accidental resonance in a tweeter filter. It wasn't about speaker resonance... it was about an accidental LC resonance.
Unfortunately this kind of behaviour will mislead and misinform people who trust the guy and could lead to some very nasty problems in multiple projects.
Last edited:
Real world tests have shown that even if a resonant peak is notched out, its distrotion is still evident. With measurements.
dave
I explained this already.
At resonance ... while energy is still being input... while you are hitting the bell... you get circulating currents between the various parts of the resonant circuit. That is RESONANCE.
When you remove the stimulus ... when you stop hitting the bell ... what happens after, the time it takes to settle back down, is RINGING.
Yes they are related ... on is the result of the other... but they do not happen in the same time frames...
Ok dumbo here with a question.
Say I need to apply a very steep filter and electrically, the filter has a high Q (be it notch or steep slope). Does Q relate to resonance / ringing and the filter itself make the impulse of that driver "worse" at the knee / high point of Q on the filter?
I realise acoustically, the filter might result in a driver + filter = lower Q target.. I'm asking about electrical Q and whether very steep notch or very steep order filters increase ringing, make themselves apparent in impulse and therefore "worsen" the acoustic ressponse.
Say I need to apply a very steep filter and electrically, the filter has a high Q (be it notch or steep slope). Does Q relate to resonance / ringing and the filter itself make the impulse of that driver "worse" at the knee / high point of Q on the filter?
I realise acoustically, the filter might result in a driver + filter = lower Q target.. I'm asking about electrical Q and whether very steep notch or very steep order filters increase ringing, make themselves apparent in impulse and therefore "worsen" the acoustic ressponse.
The ringing from any high q filter produces both currents and voltages well above the input. That's what resonance is... an exaggerated reaction caused by stored energy... and yes it can be acoustically apparent as I demonstrated in the XSim critque thread only a couple of days ago.
Take a look in THIS thread posts 76 and 100.
In post 76 I worked up 4 versions of the tweeter circuit from Montana's project in thumbnail 1. In thumbnail 2 you can see that R3 is carrying far more current than it should and in thumbnail 3 you saw what happened when I took that resistor out of circuit... notice the power dissipation on the tweeter... over 100 watts, with 50 watts of input. Then in post 100 I showed through an LTSpice simulation that the same circuit, would produce a ringing effect at about 2khz that would be audible. Both posts are accompanied by detailed explanations.
Hoping I understand correctly. If an electrical filter is used to flatten a resonance peak, the time domain is improved too, not worsens. It doesn't matter how high Q is the electrical filter until it counteracts to the driver's peaking (in negative or positive magnitude) it improves the time domain too. That's easy to measure. At least that's what I experienced. Or I missed something?
Are we discussing electronics or semantics here?
Ringing ... after the stimulus is removed is the dissipation of stored energy which occurs at the resonant frequency. It's what you get after you hit the bell.
That's what you said ... and, not surprisingly, that's what I also said.
NEXT...
Gr8, that's what I thought. So if the time domain response doesn't improved after amplitude flattening, then the issue was not a minimum phase issue like the mentioned diffraction effect?
Agreed. That is one of thet hings i do. Somewhere north of 5k drivers so far.
dave