change with excursion?
Seem to me, the more force on the cone, the farther the excursion of the bending modes and the more it should ring? Just like hitting a bell with little or high force.
Also, how does cone breakup cause peaks? I understand the different parts of the cone will be moving in different phase, causing acoustic interference, but how can the sound from the cone be louder when it is breaking up than when the driver is isophase? When the driver is pistonic, sound emanating from one side of the cone is in phase with sound emanating from the other side (at least directly on axis...depending on driver size), so how can sound be louder when the cone is not isophase?
Sorry if this has already been asked!
Thanks everyone.
Seem to me, the more force on the cone, the farther the excursion of the bending modes and the more it should ring? Just like hitting a bell with little or high force.
Also, how does cone breakup cause peaks? I understand the different parts of the cone will be moving in different phase, causing acoustic interference, but how can the sound from the cone be louder when it is breaking up than when the driver is isophase? When the driver is pistonic, sound emanating from one side of the cone is in phase with sound emanating from the other side (at least directly on axis...depending on driver size), so how can sound be louder when the cone is not isophase?
Sorry if this has already been asked!
Thanks everyone.
A bell does not change its sound from soft to moderate excitation, provided that for soft excitation one uses a lighter stick than for moderate excitation. Stick speed hence spectrum remain the same, but kinetic energy hence loudness is smaller.
At high levels, the bell starts to produce intermodulation tones, because it does not vibrate linearily anymore -- it is stressed.
At ring tones there may be higher level, because at these frequencies mechanical damping is less, so kinetic efficiency is higher.
At high levels, the bell starts to produce intermodulation tones, because it does not vibrate linearily anymore -- it is stressed.
At ring tones there may be higher level, because at these frequencies mechanical damping is less, so kinetic efficiency is higher.
Because at breakup the cone resonates, so the amplitude is larger than at other frequencies.
Hello monkish54, you are not the only one, thinking this way. Back in 1954, Edgar M. Villchur authored the AR-1 loudspeaker, promoting the idea of the "acoustic suspension" acting as dominant spring in the system, compared to the internal spyder and cone surround actions. He wrote that the "acoustic suspension" has the advantage of spreading the reaction force on the whole cone surface, instead of localizing the reaction force on the spider edge and on the cone surround edge. He wrote that the "acoustic suspension" could somewhat decrease the cone breakup susceptibility.
You will find people taking attitude about this, contesting Edgar M. Villchur intuition and conclusion. Unfortunately, most people rely on a mindset and math equations basing on the "small signal" hypothesis. Unfortunately, this appears to be insufficient. The reality is more complicated, especially when you take hysteresis into account, that's exhibited by the spider and by the cone surround suspension.
Personally, I support Edgar M. Villchur intuition and conclusion.
Your question about some speakers exhibiting a high frequency resonance (say + 10 dB) at a certain frequency just before the beginning of the extinction, is challenging. In your question you assume this is caused by cone breakup. Why such assumption ?
IMO, such high frequency resonance is caused by the spring formed by the cone Young's modulus, that's acting as an elastic bridge joining the moving coil (source of movement) to the main cone surface (destination of movement), aggravated by the fact that the destination of movement is not properly terminated, thus reflecting most of the kinetic energy back to the moving coil. I'm not calling this a "cone breakup". I'm calling this a "high frequency resonance before extinction". As soon as the system can be described as a 2nd order system (or 3rd order, or 4th order, ...) comes the possibility of such resonance.
Many say that the art of loudspeaker design, is to combine and mitigate all material properties, for ending up with a smooth lowpass transfer function exhibiting no peak, or possibly, exhibiting the beginning of some transfer function that can be completed by a simple lowpass filtering, for ending up with a double-Butterworth transfer function - in case of targeting a Linkwitz-Riley multiway system. Take the JBL 2206 or JBL 2226 as examples. Their on-axis curves look marvelous indeed.
Now look the off-axis curves. They are radically different. When you allow a loudspeaker to deliver frequencies until its natural cutoff frequency, you get directivity pattern problems. Which means that the curve that you need to base on, is not the on-axis curve, but the off-axis curve. Hoping that both curves stay in overlap until a certain frequency, that's going to be your crossover frequency. When there is no decent overlap, or when the overlap is constrained to very low frequencies, consider such loudspeaker as unusable.
IMO, most of the time, cone breakup starts at a much higher frequency. Starting from a certain frequency, you will see valleys and hills. Valleys are at the frequencies where the cone breaks up in such a way that half the cone surface is moving "in phase" and half of the cone surface is moving "anti-phase". Hills are at the frequencies where no anti-phase islands do show.
Hills don't exceed the nominal level (no gain), unless the loudspeaker is equipped with a cone that's breaking-up at a frequency that's lower than the natural extinction frequency. You may consider this a terribly poor loudspeaker, but unfortunately, that's what you will get each time you need an inexpensive subwoofer. Not a big issue as anyway, you will lowpass filter it at 120 Hz or so.
Regards,
Steph
You will find people taking attitude about this, contesting Edgar M. Villchur intuition and conclusion. Unfortunately, most people rely on a mindset and math equations basing on the "small signal" hypothesis. Unfortunately, this appears to be insufficient. The reality is more complicated, especially when you take hysteresis into account, that's exhibited by the spider and by the cone surround suspension.
Personally, I support Edgar M. Villchur intuition and conclusion.
Your question about some speakers exhibiting a high frequency resonance (say + 10 dB) at a certain frequency just before the beginning of the extinction, is challenging. In your question you assume this is caused by cone breakup. Why such assumption ?
IMO, such high frequency resonance is caused by the spring formed by the cone Young's modulus, that's acting as an elastic bridge joining the moving coil (source of movement) to the main cone surface (destination of movement), aggravated by the fact that the destination of movement is not properly terminated, thus reflecting most of the kinetic energy back to the moving coil. I'm not calling this a "cone breakup". I'm calling this a "high frequency resonance before extinction". As soon as the system can be described as a 2nd order system (or 3rd order, or 4th order, ...) comes the possibility of such resonance.
Many say that the art of loudspeaker design, is to combine and mitigate all material properties, for ending up with a smooth lowpass transfer function exhibiting no peak, or possibly, exhibiting the beginning of some transfer function that can be completed by a simple lowpass filtering, for ending up with a double-Butterworth transfer function - in case of targeting a Linkwitz-Riley multiway system. Take the JBL 2206 or JBL 2226 as examples. Their on-axis curves look marvelous indeed.
Now look the off-axis curves. They are radically different. When you allow a loudspeaker to deliver frequencies until its natural cutoff frequency, you get directivity pattern problems. Which means that the curve that you need to base on, is not the on-axis curve, but the off-axis curve. Hoping that both curves stay in overlap until a certain frequency, that's going to be your crossover frequency. When there is no decent overlap, or when the overlap is constrained to very low frequencies, consider such loudspeaker as unusable.
IMO, most of the time, cone breakup starts at a much higher frequency. Starting from a certain frequency, you will see valleys and hills. Valleys are at the frequencies where the cone breaks up in such a way that half the cone surface is moving "in phase" and half of the cone surface is moving "anti-phase". Hills are at the frequencies where no anti-phase islands do show.
Hills don't exceed the nominal level (no gain), unless the loudspeaker is equipped with a cone that's breaking-up at a frequency that's lower than the natural extinction frequency. You may consider this a terribly poor loudspeaker, but unfortunately, that's what you will get each time you need an inexpensive subwoofer. Not a big issue as anyway, you will lowpass filter it at 120 Hz or so.
Regards,
Steph
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Cone breakup does not change significantly with level because the speed of sound in the material does not change with amplitude.
I accept this premise. The question then becomes, with increased excursion, and thus SPL, why does the ratio of non-resonance spl to resonant spl stay the same?
That is, if I have a resonance with a nice Seas Excel cone at 7khz that is 15db above the non-resonating response (lower in frequency), why does the resonance always stay 15db above the non-resonance SPL?
I go to Klippel's site for actual speaker info, https://www.klippel.de/know-how/literature/papers.html
try: https://www.klippel.de/fileadmin/kl...ture/Papers/KLIPPEL_Cone_Vibration_Poster.pdf
for a few points - note the air load vs vacuum plot shows some frequency shift but highly similar shape/Q cone resonances ~= acoustic suspension air spring "support"/air load not suppressing cone resonances?
try: https://www.klippel.de/fileadmin/kl...ture/Papers/KLIPPEL_Cone_Vibration_Poster.pdf
for a few points - note the air load vs vacuum plot shows some frequency shift but highly similar shape/Q cone resonances ~= acoustic suspension air spring "support"/air load not suppressing cone resonances?
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Why should it not? A ratio independent of level is characteristic of a ________ system/process.
Brief Answer
Material Characteristics,
Diaphragm Geometry and
Radiator Size.
In this setting two sound velocities are responsible for determining radiation characteristics of a driver: One reflects the speed of sound traveling in the diaphragm material, and the other reflects the speed of sound traveling in the surrounding air.
They jointly determine the frequencies at which wavelengths become comparable to radiator dimensions. The resonances that arise in these domains combine to either reinforce or cancel driver response at these frequencies.
The severity of these anomalies is determined by the mechanical properties of the diaphragm materials and geometry used.
Regards,
WHG
The frequency response of a loudspeaker cone is affected by three main factors:
Material Characteristics,
Diaphragm Geometry and
Radiator Size.
In this setting two sound velocities are responsible for determining radiation characteristics of a driver: One reflects the speed of sound traveling in the diaphragm material, and the other reflects the speed of sound traveling in the surrounding air.
They jointly determine the frequencies at which wavelengths become comparable to radiator dimensions. The resonances that arise in these domains combine to either reinforce or cancel driver response at these frequencies.
The severity of these anomalies is determined by the mechanical properties of the diaphragm materials and geometry used.
Regards,
WHG
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