Theorical question about Fs

andreaemme

Member
2008-11-12 6:24 pm
Hello, really I wasn't able to find an answer to this question: why the response of any loudspeaker goes down below its Fs?
I understand that resonance of a driver is a critical point because moving at that frequency is easier for coil and impedance has its highest peak but what does it change in response between higher and lower frequencies? Why highers can be reproduced and lowers not?
Thank you!
 

Sangram

Moderator
Paid Member
2002-09-25 11:01 am
India
You can extend response below Fs in a number of ways. Most ported boxes anyway have declining response below Fb and the system -10dB point is usually a little below the driver Fs after accounting for room/boundary gain.

However for low Q drivers in ported boxes, more extension becomes problematic because of the high amounts of energy required from the drive system and the strain on the suspension. Low Q drivers by necessity have high loss suspensions, so the excursion demands may become excessive and you may hit the end stops.

For higher Q drivers you may use the box + electrical means to usefully extend the response below Fs. I've used a single capacitor technique very fruitfully, though the gain is only half an octave or so. But for full range drivers in a smallish sealed box, it does provide a very useful extension at the low end. Power limits do get compromised further though, but if you watch the excursion and respect thermal limits it is possible to get excellent results from this technique.
 
I think the question is why there is roll-off below resonance. You can indeed lower the cut-off frequency to some extent by various tricks, the most common one being a vented box, but essentially you get roll-off below a cut-off frequency that is intimitely related to the resonance frequency.

Anyway, I can only partly answer this. The sound pressure produced by a conventional loudspeaker in an infinitely large closed box (I take an infinitely large closed box because it is the simplest case) is related to the acceleration of the cone of the loudspeaker.

Above resonance, the mass of the cone and airload has more influence on the cone movement than the compliance of the suspension. Hence, the acceleration is essentially independent of frequency and you get a flattish response.

Below resonance, the compliance of the suspension dominates and the cone excursion rather than its acceleration becomes frequency-independent. A constant excursion corresponds to an acceleration that is proportional to the square of the frequency. Hence the second-order roll-off.

The bit I can't explain is why the sound pressure depends on the cone acceleration in the first place. I hope some other member can fill in this gap.
 

andreaemme

Member
2008-11-12 6:24 pm
Above resonance, the mass of the cone and airload has more influence on the cone movement than the compliance of the suspension. Hence, the acceleration is essentially independent of frequency and you get a flattish response.

Great explanation! That is exactly the sense of my question. All I can think is that for the Newton's law F=m*a there can't be any force without any acceleration and viceversa; probably, said in rawest words, if coil doesn't "feel" a mass to move accelerating it (and feel instead only a compliance to fight), in it doesn't flow current and it doesn't product force.
On this base in last minutes I tried a more refined search and found an article that probably meets our needs this article by TNT group (there's also in italian version), but I haven't for today time to read it deeply.
The same mass problem, at this point, should explain also the roff-off of the response at higher frequency, because of the decreasing mass involved by the membrane movement whit increasing frequency to reproduce (obviously I'm talking about a woofer).
Thanks a lot, for every other contribution too.
 

TBTL

Member
2013-10-08 12:26 pm
Low Q drivers by necessity have high loss suspensions, so the excursion demands may become excessive and you may hit the end stops.
Not true. For most drivers the Q factor is primarily determined by electrical damping, not mechanical. Secondly, the excursion required for a certain sound pressure level is not related to suspension loss.
All I can think is that for the Newton's law F=m*a there can't be any force without any acceleration and viceversa; probably, said in rawest words, if coil doesn't "feel" a mass to move accelerating it (and feel instead only a compliance to fight), in it doesn't flow current and it doesn't product force.
If I understand you correctly you say that the coil does not exert a force on the cone at frequencies below the resonance frequency. That is not the case; current flows and the coil does exert a force on the cone. That force is however (to some extent) counteracted by the force of the suspension (acting on the cone). Note that the force in Newtons law is the net force, it includes both these forces.
 
Last edited:
Low Q drivers by necessity have high loss suspensions, so the excursion demands may become excessive and you may hit the end stops.

On the first part, if you replace 'low Q' with 'low Vas' then it would be right because they need very high mass and therefore much, much stiffer and robust suspension. 'Best' example are the car-subwoofer. But there are actually drivers which have a quite soft suspension AND a low Qts. That ofcourse leads to high Vas.

For the 2nd part, the 'end stops' aren't a 'hard hit' anymore for most modern drivers since the suspension becomes more progressive and save the driver from too much excursion, at least if it's not a permanent mechanical overload or way over the limit of the driver - ofcourse you can destroy anything with too much force/power though.
 
If I understand you correctly you say that the coil does not exert a force on the cone at frequencies below the resonance frequency. That is not the case; current flows and the coil does exert a force on the cone. That force is however (to some extent) counteracted by the force of the suspension (acting on the cone). Note that the force in Newtons law is the net force, it includes both these forces.

Thanks TBTL, of course, but what is to make difference between higher and lower than Fs frequencies? Assuming that with small signals counteracting by suspension compliance is constant and independent by frequency, necessarily below Fs there must be a decreasing in force exerted by the coil on the membrane. Since effectively the impedance graph below Fs shows that current still flows, if energy is still spent but force decreases, where does it go energy not used?
May it come into play an auto-braking effect by the coil due to electrical damping factor, that below Fs is not counteracted by effects of cone mass and up Fs is instead? If yes, is auto-braking effect due to cone velocity? Should this explain the necessity for low Qts (and so Qes) drivers to work in little space to contain cone velocity and mitigate the auto-braking effects?
If yes, roll-off below Fs is due to a velocity question rather than an acceleration one?
Excuse my confusion, it's just a way to investigate about.
 
Last edited:
Explanation in terms of velocity and acoustic impedance

The velocity of the cone rises with 6 dB/oct below Fs and falls with 6 dB/oct above Fs (band-pass filter). The radiated sound pressure rises with 6 dB/oct in relation to velocity (differentiator) due to the acoustic impedance. The combined response is a rise of 12 dB/oct below Fs and flat above Fs.
 
The velocity of the cone rises with 6 dB/oct below Fs and falls with 6 dB/oct above Fs (band-pass filter). The radiated sound pressure rises with 6 dB/oct in relation to velocity (differentiator) due to the acoustic impedance. The combined response is a rise of 12 dB/oct below Fs and flat above Fs.

Excuse me, what does it mean velocity rises with 6 dB/oct? Isn't velocity measured in m/s? So sound pressure level is proportional to air velocity and not to air displacement?
What dos exactly is acoustic impedance, does it increase with increasing frequency or viceversa?
Thanks..
 

Ron E

Member
2002-06-27 10:41 pm
USA, MN
Thanks TBTL, of course, but what is to make difference between higher and lower than Fs frequencies? Assuming that with small signals counteracting by suspension compliance is constant and independent by frequency, necessarily below Fs there must be a decreasing in force exerted by the coil on the membrane. Since effectively the impedance graph below Fs shows that current still flows, if energy is still spent but force decreases, where does it go energy not used?
May it come into play an auto-braking effect by the coil due to electrical damping factor, that below Fs is not counteracted by effects of cone mass and up Fs is instead? If yes, is auto-braking effect due to cone velocity? Should this explain the necessity for low Qts (and so Qes) drivers to work in little space to contain cone velocity and mitigate the auto-braking effects?
If yes, roll-off below Fs is due to a velocity question rather than an acceleration one?

.....what does it mean velocity rises with 6 dB/oct? Isn't velocity measured in m/s? So sound pressure level is proportional to air velocity and not to air displacement?
What dos exactly is acoustic impedance, does it increase with increasing frequency or viceversa?

Coil force does not vary with frequency, it is the same Bl*i for all frequencies. As has been said several times, sound pressure is proportional to acceleration.

For a speaker in a sealed box or on a large baffle, cone displacement is approximately constant below Fs. The box/suspension dominates the motion. In that region, the constant force from the coil moves the cone a constant amount x = F/spring constant. Velocity is then proportional to Frequency * displacement below resonance. The amount it moves back and forth times the frequency it moves back and forth.

If cone displacement x
X(t)=x*sin(wt)
then Velocity u
u(t)=x*w*cos(wt)
and acceleration a
a(t)=x*w^2(-sin(wt))

Above resonance the mass dominates, and you have a mass that it takes more and more effort to move, the faster you move it. Since you only have a fixed force input, the acceleration stays constant, which means displacement drops. F=m*a (ignoring the time varying part) is also F=m*x*w^2. So to have constant acceleration a=x*w^2, the displacement must drop as a/w^2.

Auto braking, back EMF, are the same thing. They play a part at resonance but are not important to the behavior above and below resonance.
The peak at resonance is the velocity maximum, the place where it is easiest to get the cone moving, where the stored energy in the suspension and cone mass are equal. Near resonance the motion can be considered to be damping-dominant. The amount of total damping (so Qts, not Qes or Qms alone) affects the amount of motion (sound output) at resonance. The ratio of total to mechanical affects the impedance rise.

Peak impedance at resonance = Zmax=Re(1+Qms/Qes)
 
Last edited:

Ron E

Member
2002-06-27 10:41 pm
USA, MN
Sound pressure is proportional to frequency times volume velocity, which is Pressure=k*w*Cone area*velocity
Note that ignoring the time varying part,
x(max) = x
u(max) = x*w
a(max) = x*w^2 = u(max)*w
w=2*pi*frequency

The first equation in the post again
Pressure=k*Cone area*velocity * w
Saying that pressure is proportional to velocity * frequency is the same as saying it is proportional to acceleration.