# efficiency of speaker driver below resonance frequency

Discussion in 'Physics' started by praondevou, Oct 9, 2013.

1. ### praondevou Thread Starter AAC Fanatic!

Jul 9, 2011
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Any speaker specialists here?

I worked my way through the Thiele/Small analysis and I understand more or less the impedance curves and what mechanical driver parameters affect it.

I'm still not sure that I understand why below resonance frequency there is this negative magnitude roll-off. They talk about bad impedance matching of driver to air the lower the frequency gets.

Can somebody enlighten me? I need to visualize it, formulas don't help much.

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3. ### #12 Expert

Nov 30, 2010
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I am going to have to assume some things.

Just assuming I know what you mean by, "below resonance frequency" "magnitude roll off", you must be referring to, below the resonance of the speaker system, including the enclosure.

The enclosure can be defined as a way to attach the speaker cone to the air while avoiding the back side pressure wave from quickly coming into contact with the front side pressure wave (cancellation). This can be done by using an infinite baffle, like cutting a hole in the wall of the building and allowing all the back side energy to escape into the infinite reaches of the outdoor world. It can be done by building a closed box such that the compression of the internal air is a spring that is tuned to, or lower than, the natural frequency of the mechanical springs of the cone mounting material and the mass of the cone. It can be done by redirecting the back side energy so it arrives at the front of the enclosure at the right time (ported enclosure or folded horn). Using a ducted port provides 2 air springs you can tune at different frequencies.

When a resonance frequency is the point of interest, the speaker cone tends to move at the correct speed with minimum of power applied (compared to some other frequency). It is like pushing a pendulum at the correct frequency to maintain its natural frequency. The pendulum is already moving at its natural frequency and you are merely replacing a little bit of energy lost in air resistance and suspension resistance. There is an electrical resonance frequency and one or more mechanical resonance frequencies. These show up as peaks in efficiency.

Your lowest efficiency peak is mechanical. If the enclosure is tuned to the mechanical peak of the speaker, all the bass notes will seem to resemble the mechanical resonant frequency of the speaker. The goal of most bass enclosures is to provide one or more mechanical efficiency peaks that are lower than the speaker resonance. This is for the purpose of achieving acoustic power delivery at lower frequencies than the natural frequency of the speaker.

By now, you are thinking in terms that I am thinking in. By now, you are visualizing the speaker cone mass and its suspension parts as a pendulum of sorts. You can see that the speaker alone has a resonant frequency and the enclosure full of springy air has resonant frequencies based on air mass and air compressibility. Now you see that all of the mechanical resonant frequencies will use less energy for the same cone excursion as a Hz higher or lower than the resonance point. Therefore, at any particular energy input, the cone excursion will diminish between or below the mechanical resonance frequencies. When lowering the drive frequency toward zero, the cone excursion must diminish simply because the driving frequency is continuously getting farther away from the lowest mechanical resonance frequency.

In addition, the total excursion of the cone must lose efficiency (diminish) as the voice coil gets farther out of center from the magnetic field center. In the subsonic range of frequencies, the cone would have to move several inches to deliver the same acoustic power it delivered at 100 Hz. There is no such thing as a voice coil that is a foot long, and that would be required for the cone to move +/- 6 inches. You are also reaching a mechanical limit to the perimeter surround and spider support for the cone. They are resisting movement more and more as they are pulled away from their rest position. Eventually, they become fully distended and become nearly rigid.

Did this get you a model you can picture in your head?

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4. ### praondevou Thread Starter AAC Fanatic!

Jul 9, 2011
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He , thanks for that lengthy explanation.

There is one more immediate question I have.

From researching the internet I came to the conclusion that the phase shift "input voltage to cone displacement" will be always 90° at the speaker driver resonance frequency. Is that correct? Is there a way to circumvent this?
My guess is no. I'm not concerned about phase or impedance at higher frequencies (where the voice coil inductance comes into play). What I wanted was efficiently drive a speaker/enclosure with near zero phase shift "input voltage to cone movement" up to 200Hz but this seems to be a contradiction. Maybe it would work if the cone/voice coil had no mass.

That's also something I do not really understand. Why is this not true for high frequencies (even away from a tweeters resonance). Why do we have to move much more air at low frequencies? Sounds like a stupid question but I can't visualize it.

5. ### nsaspook AAC Fanatic!

Aug 27, 2009
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A tire pressure analogy. A bicycle tire(200hz) at 35psi and a monster truck tire(20hz) at 35psi, to maintain the same pressure at any point on the larger tire requires more air to be moved inside.

Air has mass, elastic and inertial properties that causes it to react to the compression forces in a way that is time and volume related. The volume of air needed to create a 20hz wave at a set sound level is much larger than what's needed to create one at 200hz because the 20hz wave is physically larger and requires more energy for it to move (greater mass/properties of air).

Last edited: Oct 10, 2013
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6. ### studiot AAC Fanatic!

Nov 9, 2007
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I don't have time for discussion now, but since you are online, praondevou, here is the 1 dimensional equation of motion of a speaker cone.

$F\sin \omega t = m\frac{{{d^2}x}}{{d{t^2}}} + r\frac{{dx}}{{dt}} + sx$

Where m is mass, r is the resistive dissipation and s the elastic stiffness.

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7. ### #12 Expert

Nov 30, 2010
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One of my rules is, "Start with the obvious. By the time you get done with that, your brain will have had time to analyze the other parts", so here I go:

The math for (effective) driver size and acoustic power go like this:
A = cone area
E = cone excursion
F = lowest frequency desired
P = acoustic power delivered
K is a constant

A1E1 = A2E2
E1F1^2 = E2F2^2
E1P2 = E2P1
KE = KP
P2A1^2 = P1A2^2

Verbally:
1) For a certain frequency and power delivered, half the effective driver area requires twice the excursion.
2) For a certain cone area and power delivered, half the frequency requires 4 times the excursion.
3) For a certain cone area and frequency, twice the excursion will produce twice the acoustic power delivered.
4) For a certain frequency and the same excursion, twice as much acoustic power delivered requires 4 times the cone area.

"Rules of good practice" include limiting the cone excursion to +/- 1/2 inch and 1 acoustic watt is a good reference point.

Now, if a 12 inch (effective area) speaker is delivering 80Hz at one acoustic watt by moving forward and backward by 1/2 inch, what has to change to get the same loudness at 40Hz? We're out of excursion, so it must be area. I will go through excursion to get to area.

E1F1^2 = E2F2^2
1 x 80^2 = E2 40^2
6400 = E2 x 1600
E2 = 4E1

A1E1 = A2E2
A1 x (1) = A2 x (4)
A1/4 = A2
A2 = 4A1

OK. We have established that half the frequency requires 4 times the driver area, but the question was, "why?".

I have spent 3.6 hours checking 3 reference books and the answer is, "I don't know". It must be about the power in a compression wave traveling through a square foot of area at a certain distance from the driver for a time duration, but I can't find the "why". It has to be simple physics, but I'm just not getting it today. Perhaps you could post this particular question in "Chat" and get somebody else to contribute.

____________________________________________________

The voltage to cone phase at electrical resonance (Qes) seems valid at frequencies where the inductive component is important because the phase lag makes sense when you think about the low amount of energy required to maintain excursion at resonance and the fact that a pure inductor dissipates no energy, but I think this is wrong at low frequencies where we are talking about mechanical resonance (Qms). Modern amplifiers have damping factors in the hundreds, and that means the position of the cone is very much controlled by the driver. If the cone tries to get out of phase, the back EMF will be swallowed up by the output drivers. Can you point me to a reference?
___________________________________________________________

"I don't know" and, "I disagree" are pretty weak answers, but I'm trying.

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8. ### praondevou Thread Starter AAC Fanatic!

Jul 9, 2011
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Ok, I "think" I got it.

Example: A speaker is driven with a 100Hz signal, the wave is measured with a microphone at 1m distance. Then the speaker is driven with 200Hz. Assuming the cone peak displacement is the same for both signals, the measured sound pressure at 1m distance will be higher for 200Hz? (microphone etc has a flat frequency response)

9. ### #12 Expert

Nov 30, 2010
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Good to see I have help on this one. Now, if that \$%^& verification page would stay out of the way!

10. ### #12 Expert

Nov 30, 2010
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True. Think: the energy required to move the mass of the cone twice as fast must be more, therefore, more energy will be delivered as sound.

11. ### praondevou Thread Starter AAC Fanatic!

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This was from a klippel.de. I was asking them what the phase shift would be and here is what he answered: "
Typical behavior of a loudspeaker is as shown below (0° below fs, -90° at fs, -180° above fs):

I am trying to find this info on their website but have not had success yet. He is talking about the phase shift from input voltage to the corresponding cone displacement. It seems logical that there must be a delay because of the inertia of the cone and air to be pushed away.

They seemed to be specialists in this field but of course they won't give detailed free advice per email.

12. ### #12 Expert

Nov 30, 2010
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I found KLIPPEL but I have to go to work now. Can you describe how to find the page you were looking at?

13. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes, exactly. The energy of any oscillator increases with frequency. (Plank's Law)

So the energy of a sound wave increases with frequency.

So the energy you need to input to obtain a sound wave of a given energy increases with frequency, all other things being equal.

But they are not equal in normal sound reproduction. The frequency spectum of normal sound (speech, music etc) is heavily biased towards the lower registers. This more than outweighs the increase required by increasing frequency.

So in sound reproduction you need more oomph in the bass amp and bass speakers.

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14. ### nsaspook AAC Fanatic!

Aug 27, 2009
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An important factor with low frequency sound reproduction is the Acoustic impedance of the speaker medium when coupled to the air medium.

The effect at the low end with a speaker cone ratio much smaller than the wavelength is similar to a RF mismatch at the antenna. You can have large displacements of the cone but very little energy coupled to air at a distance because the energy is reactive (out of phase and stored locally in near air particles or the speaker cone)

Our ear is a classic example of an acoustic impedance matching horn. The acoustic impedance of our ear drum is well matched at normal sound frequencies to that of the ear canal and it (ear canal) acts like a matching network from free air to the drum. When the ear canal is sealed like when wearing earbuds the sound pressure from even the tiny cone inside them reproduces low bass because the transducers on both ends are transferring real power but if you move the earbud just slightly from your ear the sound now has an acoustic impedance discontinuity to cross and bass response drops sharply.

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15. ### praondevou Thread Starter AAC Fanatic!

Jul 9, 2011
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Now we are getting to my main question. How exactly is that impedance matching of cone to air working?

Then you say "stored locally in near air particles" . How near? Within a wavelength or half a wavelength or even more close to the cone? What you are saying is that the energy is not being radiated, right?

@#12 The graph I posted is from an e-mail they sent me, I asked them what the phase shift of a speaker driver was. But then, as I said, they won't spend hours (like you ) trying to explain something to me in detail.

16. ### #12 Expert

Nov 30, 2010
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I am not seeing that graph. Post #11? I only see a quote, not a link, not a graph.

17. ### praondevou Thread Starter AAC Fanatic!

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Here again their response: "The Relationship between voltage signal and pressure/displacement signal is best described with the linear loudspeaker model. Using the analogies defined by Thiele/Small, displacement/velocity/pressure can be calculated using these analogies. Please have a look at the following document on how to measure these measures. http://www.klippel.de/fileadmin/kli..._Accurate_Linear_Parameter_Measurement_01.pdf

Typical behavior of a loudspeaker is as shown below (0° below fs, -90° at fs, -180° above fs)"

Their website is www.klippel.de. But i did not find a document where this transfer function is being described. A lot about impedances though...

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18. ### #12 Expert

Nov 30, 2010
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This certainly reveals that I did my study 28 years ago!

The pdf shows Fs to be 58Hz to 70Hz and the graph shows the phase angle to be 30 degrees to 35 degreees at those frequencies, therefore, the statement, "Typical behavior of a loudspeaker is as shown below (0° below fs, -90° at fs, -180° above fs)" can not be valid.

Last edited: Oct 11, 2013
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19. ### nsaspook AAC Fanatic!

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Sure, we can have air particle velocity changes that don't result in air pressure changes that propagate and result in evanescent standing waves (the air particles are just pumped around the speaker by spilling over the edge of the speaker cone).

The near field is where sound power (air particle velocity changes) generates sound pressure that we can hear. Near field would be <<λ/2pi and the very near field (where air is close to a uncompressible fluid and pressure is low compared to velocity) <<L/2pi where L is the size of the sound source.

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20. ### studiot AAC Fanatic!

Nov 9, 2007
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Sounds good, but what does it mean?