[HPiper]

Can anyone explain to me what is going on when a square wave sags. I know it is supposed to go instantly to a set voltage, remain there for a time and then instantly go to the negative voltage producing a nice 'square' wave. In the real world though you hardly ever see that with headphones. They go to the voltage okay but then start to instantly decay until the voltage swings negative, producing a straight line at an angle until the voltage switches. But what is happening when the rate of decay takes an almost exponential rate producing a trailing wave that looks like an old horses back with a marked sag before the negative voltage swing. I would assume that is a bad thing as I don't see too many headphones that do that, but I was wondering what causes that? Is that related to the impedance of the phone, or the quickness of the driver or something else?

 

[castleofargh]

well you push a fixed direct voltage (instead of the alternating mess that is music). so the driver is pushed away from its resting position.
 membrane's speed depends on how much power is used to move it(obviously it will be fast nough to make high frequencies so pretty fast stuff), and also how much the membrane weights and resists(trying to go back to the resting position), how much air it's trying to move, the type of coil(do I forget some stuff?).
when it overshoots I would guess one of the reason could be because the membrane has momentum and doesn't succeed in decelerating fast enough.
why it slowly goes down could be that once the membrane has lost it's momentum, it tries to get back to its resting position by a mechanical pull. I would guess that induction plays a part too, as the more the voltage, the more the inductance, the more opposition to the voltage. but I don't know the scale of such an effect. maybe it doesn't actually matters at all?
 
I'm going on a limb here, but maybe with a lasting voltage, some coils get warmer so the impedance increases slightly, being a reason for another parameter of changes?
 
and I guess on closed drivers, each movement might create a few pressure problems inducing some short term misbehaving of the membrane.
 
the little vibrato could simply be that all the forces have a hard time getting to a balanced position. getting totally rid of all energy/momentum might not be that easy.
 
what I'm wondering also is how the amp behaves, it might be a challenge for it too. at least at low freqs.
 
 
tbh I don't find the square waves to be that meaningful, as most of what it reveals are a tendency to not hold a point, and a tendency to behave as a wave. so mostly it will all be ok anyway once playing music as music doesn't stay at a value, and music is a wave signal. in the end I don't know how much more intel you get compared to the impulse response.

 

[jcx]

even if the diaphragm holds its position the air leaks away - "tilt"/exponential decay of the "flat" of a square wave is a consequence of a high pass filter - most speakers/headphones don't make a DC air pressure change - because they don't seal to the head well - iem do have the possibility of being better with good seal to your ear canal

 

[esldude]

http://users.tpg.com.au/users/ldbutler/Waveforms.htm  Mostly limited low frequency capabilities.

 

[superjawes]

well you push a fixed direct voltage (instead of the alternating mess that is music). so the driver is pushed away from its resting position.
 membrane's speed depends on how much power is used to move it(obviously it will be fast nough to make high frequencies so pretty fast stuff), and also how much the membrane weights and resists(trying to go back to the resting position), how much air it's trying to move, the type of coil(do I forget some stuff?).
when it overshoots I would guess one of the reason could be because the membrane has momentum and doesn't succeed in decelerating fast enough.
why it slowly goes down could be that once the membrane has lost it's momentum, it tries to get back to its resting position by a mechanical pull. I would guess that induction plays a part too, as the more the voltage, the more the inductance, the more opposition to the voltage. but I don't know the scale of such an effect. maybe it doesn't actually matters at all?

I'm going on a limb here, but maybe with a lasting voltage, some coils get warmer so the impedance increases slightly, being a reason for another parameter of changes?

Note: my background is in electrical engineering, so if you want more information, you might ask a mechanical engineer.

What you're describing is pretty correct. Mechanical systems are often defined as mass-spring-dampener systems for easier analysis. EE's do the same thing with RLC, or Resistor-Inductor-Capacitor equivalents. Basically, you can use three elements in a second order diffrerential equation and do a pretty common analysis. You don't have to have the actual elements in your system to simplify to such a model. Things like air resistance will create dempening effects, but you can just include that using a measured response.

Anyway, the mass element plays into inertia and momentum, meaning that you need power to get it moving (more mass = more force to move), and once it's moving, it takes some effort to stop (more mass = more momentum = more stopping force). Springs exert force based on displacement, so if you stretch or compress a spring, you get a force trying to push it back to the resting position. Dampeners resist movement (velocity), and like springs, they resist no matter which way you go.

I don't remember the equations (sorry), but you can tell a few things about the system from a square response (which is basically a step response). The overshoot happens when you are under damped. If you are over damped, it will take a long time for the system to reach steady state--in this case, where the voltage is trying to put the diaphragm.

You'll have to ask someone who does this for a living about what is ideal...obviously you don't want to be too much over/under damped because the transient responses will alter the sound, either cutting the audio power (over damped) or creating harmonics (under damped). Square waves are useful because you can measure these responses, but you can also change the frequency to see if and how the response/dampening changes.

I would guess that induction plays a part too, as the more the voltage, the more the inductance, the more opposition to the voltage. but I don't know the scale of such an effect. maybe it doesn't actually matters at all?

Side note, inductors basically act as resistors to current change. The inductance is determined by the magnetic core, number of turns, and shape, and this value stays the same. However, the inductor will create a voltage response opposing a change in current passing through it. So a higher frequency current will create a more dramatic change than a lower frequency one. The applied voltage only affects the magnitude of the current (and it would be a linear effect on the response).

I do think you are onto something, WRT the square response. After the initial bump, the inductor will settle back to zero (volts across the inductor), and this probably will change diaphragm's response. After all, the driver does act as a separate system between your square wave generator and your diaphragm. If you are measuring diaphragm displacement, the input to your diaphragm is actually the output of the driver, which has its own response to the square wave input.

 

[castleofargh]

I would guess that induction plays a part too, as the more the voltage, the more the inductance, the more opposition to the voltage. but I don't know the scale of such an effect. maybe it doesn't actually matters at all?

Side note, inductors basically act as resistors to current change. The inductance is determined by the magnetic core, number of turns, and shape, and this value stays the same. However, the inductor will create a voltage response opposing a change in current passing through it. So a higher frequency current will create a more dramatic change than a lower frequency one. The applied voltage only affects the magnitude of the current (and it would be a linear effect on the response).

I used once induction, once inductance so my sentence doesn't make sense at all. hurray to me ^_^.

I do think you are onto something, WRT the square response. After the initial bump, the inductor will settle back to zero (volts across the inductor), and this probably will change diaphragm's response. After all, the driver does act as a separate system between your square wave generator and your diaphragm. If you are measuring diaphragm displacement, the input to your diaphragm is actually the output of the driver, which has its own response to the square wave input.

yes! because the beginning of the wave is an instant jump, if there had to be any meaningful effect, I though it should be maxed out at that point or right after, and maybe affect the driver itself or at least the magnetic field, as it's the only thing that really matters in the end. at least more than with a sinusoidal signal where changes are much slower accross time.

 

[chlyhne]

A square wave is basically a series of step responses. A step response is just an impulse responses that is integratedover time and if you take the fourier transform of the impulse response you get the frequency response. But that is only true for linear systems (e.g. resistor-inductor-capacitor + mass-spring-damper systems). That is good enough for headphones. They are for the most part well modelled using linear systems.
 
The step response is good for evaluation of headphones as it provides an intuitive intepretation of performance.
 
If the leading edge is very sharp and it does not overshoot or undershoot it shows that the treble is well balanced with the bass. A very sharp overshoot, which quickly returns indicates a more bright sounding headphone. If the ideal steady state level decays instead of staying constant it is because the pressure that has been built by the diaphram is leaking out from the cavity between the diaphram and the ear. This indicates a poor seal and if it is too servere it affects bass response. See for example the AKG 701 on innerfidelity, it has these characteristics. As a counter example see the LCD-X. If there is a large low frequency ripple where the response should be flat, it indicates a boomy and muddy bass combined with a comparative lack of mids.
 
BR