Ultra sound transducer for servo feedback subwoofer? - Page 2 - diyAudio
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Old 7th February 2003, 04:21 AM   #11
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I worked on a bipole feedback sub last year for a school project. Was all in the analog domain using an AD accelorometer. Noise was a huge problem and really limited the amount of feedback we could add. Also mounting the accelorometer took some engineering to damp high freq vibrations in the cheap plastic cone. Low pass filters in the feedback loop mess with the phase. Still, after much tweaking, we had a sub that was +- 1 db from 16-200 hz using 2 10" drivers in a 1.5 cubic foot box, at decent SPLs. Wasn't too reliable though; the accelorometer fell off during our demo for the prof! Hearing the sub try to correct for the loose banging of the chip on the cone was not pretty.
- John
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Old 7th February 2003, 04:24 AM   #12
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Quote:
Originally posted by MRehorst
I believe the "problem" with piezo devices is that they provide a voltage output that is proportional to acceleration when what you want is an output that is proportional to position. The piezo signal needs to be integrated to get the velocity, and integrated again to get the position info.

MR

Just saw your post after I put mine up. The problem with integrating the signal (low pass filter) is that phase delays are created. This is why it's good to calculate the feedback signal in the digital domain.

Also, is voltage proportional to position in a speaker?
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Old 7th February 2003, 07:18 AM   #13
e96mlo is offline e96mlo  Sweden
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Quote:
Originally posted by MRehorst
I believe the "problem" with piezo devices is that they provide a voltage output that is proportional to acceleration when what you want is an output that is proportional to position. The piezo signal needs to be integrated to get the velocity, and integrated again to get the position info.
Yes, but double integration of a sine leads to an inverted sine, hence no integrators are needed in the feedback for that reason. Maybe for other reasons, though, as cogs mentioned.

Quote:
Originally posted by cogs
Also, is voltage proportional to position in a speaker?
I've though about that to. I think that the speakers position is proportional to the current. The reason is that the magnetic field in the speaker coil is proportional to current and in the end it's the two magnetic fields that are determining the position.

However, since feedback is applied it doesn't matter if we monitor voltage or current. Since we know that the voltage from the amp is a correct copy of the music signal that is the only thing we can rely on. The current is created when our correct voltage "hits" the non-linear speaker load.

/Marcus
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Old 12th February 2003, 06:16 PM   #14
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Default acceleration sensor

Hi coqs,

im planing to experiment with an accerometer to build
a servo controlled subwoofer. Is the work You did
for Your school available. If yes could You plese post it
or mail it to me.

Greetings

Zelter
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Old 12th February 2003, 06:33 PM   #15
JohnG is offline JohnG  United States
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Default I like the capacitor idea!

Here is a way that may make it simpler to sense:

Mount a metal tube to the magnet, on the same axis as the rest of the speaker. Mount a wire or smaller diameter tube to the dust cap, so that it can travel in and out of the tube. Leave enough space so that the wire does not touch the tube wall. As the diaphragm moves in and out, the wire moves in and out of the tube and changes capacitance.

This is potential more linear than puttinga plate on the cone and an exteranl plate. In that case, the capacitance will not change linearly with the spacing, unless the plate area is much larger than the plate spacing. The non-linearity is due to changing of effective plate area becuase of the effect of fringing fields.

There will be fringing fields with the wire and the tube also, but they will act more like a constant capacitance in parallel with the variable capacitance, if a reasoanble section of the wire is always out of the tube.

John
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Old 12th February 2003, 06:40 PM   #16
e96mlo is offline e96mlo  Sweden
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Interesting idea!

Have you tried it?

/Marcus
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Old 12th February 2003, 07:04 PM   #17
JohnG is offline JohnG  United States
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No, I haven't tried it. My interest was piqued by MRehorst's post, and I just thought of a possibly imporved way to implement the capacitor.

John
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Old 12th February 2003, 09:19 PM   #18
SY is offline SY  United States
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Quote:
Yes, but double integration of a sine leads to an inverted sine, hence no integrators are needed in the feedback for that reason. Maybe for other reasons, though...
Yes, like the f^2 coefficient.
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Old 12th February 2003, 10:09 PM   #19
e96mlo is offline e96mlo  Sweden
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You lost me there. f^2? Could you please explain more?

/Marcus
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Old 12th February 2003, 10:23 PM   #20
SY is offline SY  United States
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I'll use "w" as omega. i.e., 2 pi f. For a constant amplitude signal A*sin(wt), the acceleration is -A(w^2)sin(wt), because of the nature of the double differentiation- chain rule, twice applied. So an accelerometer's output will scale to displacement but for the square of frequency.

It would be a lot easier with an equation editor!
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