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Old 22nd January 2003, 05:38 AM   #1
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Default Microphone turns triangle wave into square wave?

Neglecting RIAA equalization, when a magnetic phono cartridge is required to produce a square wave signal, it traces a triangular shaped groove on the record. The linearly sloping part of the triangle waveform produces a side-to-side movement of the stylus that moves at a constant velocity one way then a constant velocity back the other way. When it is moving across at a constant rate it is generating a constant voltage and this is shown by the flat-topped part of the square wave output. When the stylus abruptly changes direction at the corners of the triangle wave this is manifested as the sudden positive to negative transition in the square wave output. OK. Nothing new here.

Now consider the situation where a moving coil microphone is subject to an acoustic waveform that varies in a triangular manner, i.e. the instantaneous air pressure ramps up and down linearly. The result is that the microphone diaphragm moves in a triangular motion in and out, in fact in the same manner as the phono cartridge just mentioned. This means that the microphone generates a square waveform that doesn’t really represent the triangular sound it is presented with. It has in fact *differentiated* the original waveform. If instead of a moving coil microphone we now use a condenser microphone, would not the triangular movement of the diaphragm produce a triangular output waveform instead of a square one?

It goes further. With a loudspeaker some people argue that it should be current fed rather than voltage fed, the reasoning being that the magnetic force exerted by the voice coil depends entirely on the current. The trouble is, the current controls *only* the force, not the position of the voice coil at any given moment. On the other hand, with the usual voltage drive, the voltage controls nothing but the velocity of the voice coil through the air gap, and indirectly it controls the current.

So what, you say? Well, a loudspeaker is just like an oversized moving coil microphone. One is a motor, the other is a generator. If we apply our triangular sound which became a square wave to the speaker, the speaker voice coil will move forward at a rate determined by the magnitude of the flat topped voltage of the square wave, then it will reverse direction and move back at a similar rate. IOW it moves in a triangular fashion. As a result the air does too and we are back to our original waveform that the microphone heard. What though if we use an electrostatic speaker? Even though it is driven with the same waveform, is it not going to move its membrane in a square wave fashion unlike the triangular movement of a conventional electromagnetic speaker? Will things only work properly if we have a condenser microphone and an electrostatic speaker, or a moving coil microphone and a moving coil speaker?? Is the sound really affected?
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Old 22nd January 2003, 08:22 AM   #2
halojoy is offline halojoy  Sweden
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Default So we should use voltage driven voice coils.

And I, and others, we were leaning towards
that Current driven Voice coils
should be better controlled.

Now we might have to reconsider
To get a better waveform from our
loudspeakers - we should let the amplifiers
STAY voltage controlled.

Thanks for sharing your findings and your thoughts
Circlotron

Hope some others will join this interesting question.

Kwei Yung Tan - come on!

/halo - waits for some verbal fights - between gentlemen
- who defend themselves, using some "Noble Art of Verbality"
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Old 22nd January 2003, 12:19 PM   #3
DRC is offline DRC  United Kingdom
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Circlotron,

Agreed about the cartridge bit. If the groove is described as (for mono) vertical displacment, S(t) then the output voltage is the derivative, dS/dt

Quote:
the speaker voice coil will move forward at a rate determined by the magnitude of the flat topped voltage of the square wave
I think you are wrong on this bit . The force on the speaker diaphram is (more or less) directly related to the current through the voice coil (in turn related to the applied voltage - ignoring the effect of inductance) - as this is working against a spring (mechanical and usually air too) the motion of the diaphram will be a 'square wave' just like the applied signal, at a sufficiently low frequency. Additionally the forces / mass involved will limit the acceleration of the diaphram rounding off the transitions. A speaker does not integrate !

Dave
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Old 22nd January 2003, 12:40 PM   #4
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Since the radiation resistance of a speaker cone is rising with frequency, the cone will definitely NOT be moving in square-wave fashion when the driver is fed with a square-wave.

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Charles
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Old 22nd January 2003, 05:05 PM   #5
DRC is offline DRC  United Kingdom
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Quote:
Since the radiation resistance of a speaker cone is rising with frequency, the cone will definitely NOT be moving in square-wave fashion when the driver is fed with a square-wave.
It is true that the speaker diaphram cannot reproduce a square wave exactly as this would imply infinite velocity and acceleration. Perhaps I did not make this sufficiently clear in my reply - I actually meant to say that the original authors statement that the speaker diaphram would continue to move, with a steady DC voltage applied, was incorrect. You can easily connect a battery to the speaker to test this. Of course, if you wind the frequency up high enough it will indeed start to look like a 'triangle like' due the limitations on the diaphram acceleration.

BTW, what do you mean by "radiation resistance", I have not heard this term before.

Dave
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Old 23rd January 2003, 07:41 AM   #6
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To keep it simple:
Every source of energy needs a load connected to it in order to deliver energy. For an electric source this can be a resistor or whatever. For a speaker cone this is the mechanical load presented by the athmosphere, which is called radiation resistance. It is proportional to frequency and radiating area (and dependant on other things as well).
The power delivered to the environment is sound pressure times velocity or pressure squared divided by radiation resistance or velocity squared times the radiation resistance.

The movement of a speaker cone is mainly compliance controlled below fs (which explains the finite cone travel for the battery example) and above fs it is mass-controlled.

This makes cone travel for a closed box speaker behave like a second-order lowpass. I.e. if a speaker is driven by a square-wave it's cone motion will never be rectangular !

The behaviour of soundpressure of being velocity AND radiation-resistance dependant makes the final acoustic response behave like a second order highpass.

Theoretically an IDEAL single fullrange-driver should therefore be capable of reproducing a square-wave as it was passed through a very wide bandpass filter !!!

Regards

Charles
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Old 23rd January 2003, 07:52 AM   #7
halojoy is offline halojoy  Sweden
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Isn't great luck that squarewaves are rather rare in music.
And that even most power-amps can not produce perfect sq-waves,
as they do not posses an infinite bandwidth.
This is also a luckily thing, as mixing the soundmaterial
with Radio signals, hardly can be good for TIM or IM-distortion.

An amplifier capabable of producing Sinuswaves with very low distortion
and Driving Lodspeakers who produces soundwaves from such signal
with low distortion Should be what I am aiming for.

---------------------------------------------------------------

But as we are living in the digital age!
some of you are probably preparing for to
listen to them "1" and "0"

VIA true digital speakers feed by unfiltered Digital Amplifiers
----------------------------

halo - oldfashion enough to listen to music - OFTEN of some acoustic nature
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Old 23rd January 2003, 08:02 AM   #8
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Hi halojoy

Though this example is a little extreme, I do of course agree that it wouldn't be good if the harmonics of audio could interfere with RF !!!!

But I don't agree that the perfect reproduction of sinusoids is accurate enough (I personally don't listen to pure sinusoids).
A square wave finally emerging from the speakers doesn't have to be of the same perfect shape as the one coming out of a signal generator. But it should resemble a squarewave with reasonable accuracy.

Regards

Charles
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Old 23rd January 2003, 10:18 AM   #9
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Quote:
Originally posted by DRC
The force on the speaker diaphram is (more or less) directly related to the current through the voice coil (in turn related to the applied voltage - ignoring the effect of inductance)
I suppose I am thinking about an ideal speaker that has a *zero* resistance voice coil. Then I think, if you apply say 1 volt, the voice coil will move at a steady rate proportional to 1 volt (to it's mechanical limit) all the while generating a *constant counter emf* of 1 volt. If you try and stop the cone motion it will slow down causing the voice coil's counter emf to drop and therefore the voice coil current will get very high forcing the voice coil etc to continue at basically the same rate as it did before. Just like in a dc electric motor. Resistance only complicates things. With an ideal speaker as described, you could push the cone in, short cct the voice coil connections and let the cone go and it would stay in, it wouldn't return to the rest positon. Real-life ones with dc resistance dont behave quite this way!

Thinking more about the speaker-to-air, and also the air-to-mic interface, I think there might be a differentiation thing there too, especially if the speaker is in an open area, not like inside a car or especially part of a pair of headphones. E.g. in open air (but mounted in a box of course) apply low frequency square wave to the speaker, I imagine this would produce a triangular movement; while the speaker is moving outward at a constant rate it holds the air pressure higher, when it is moving back in at a constant rate it holds the air pressure lower. Voila! We have our square pressure wave back again.

This does contradict what said in the first post but it is what I now think. Electrostatics OTOH when fed with a square wave voltage I expect would try and jump to one extreme, then jump to the other. If the differentiation phenomena I just mentioned does really happen then an electrostatic would produce a pulsed acoustic waveform that jumps up, then decays to zero, then jumps down and decays to zero. To all the REAL speaker experts, sorry for making you do this -->
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Old 23rd January 2003, 10:56 AM   #10
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Hi Circlotron

If you look at the real life dynamic driver at a specific frequency - then force is proportional to current and current is proportional to voltage and therefore force is proportional to voltage as well. What I haven't considered with this simplification is the electromagnetic interaction of course, i.e. it is of limited accuracy. But this simple analogy should explain why an ESL should basically behave like a dynamic driver re square wave response:
With ESLs the driving force on the diaphragm is also proportional to the drive voltage.
Therefore both types of drivers should behave the same way.

Regards

Charles
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