"Self tuning piano" questions

[Isaac sur Noos]

Don,

While i did not get really the inside of the ^process, I understood
that you are able to measure pitch on a very short moment .

Is the measure occurring as long as the system is on (and corrections
applied with heat all along ?)

Is the measure/pitch regulation occurring always at the same time, or
is it the fact that the initial tuning is recorded by the same process
that gives enough accuracy - the question is about the lack of
stability of pitch, if the measure occur only once, it should be
precisely at the same moment to be accurate.

But I suppose it is a real time system is not it ?

Why are the pickup not catching overtones (are they partials ?)

Very interesting ...

Best Regards.


------------------------------------
Isaac OLEG
accordeur - reparateur - concert
oleg-i@noos.fr
19 rue Jules Ferry
94400 VITRY sur SEINE
tel: 033 01 47 18 06 98
fax: 33 01 47 18 06 90
mobile: 033 06 60 42 58 77
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> -----Message d'origine-----
> De : pianotech-bounces@ptg.org
> [mailto:pianotech-bounces@ptg.org]De la
> part de Don A. Gilmore
> Envoyé : jeudi 25 décembre 2003 06:30
> À : robert.scott@tunelab-world.com; Pianotech
> Objet : Re: "Self tuning piano" questions
>
> I can indeed get that accuracy and I can do it in a few
> milliseconds,
> believe it or not.  You have made the assumption that
> frequency counting is
> the only solution to determining frequency (as many in the
> past have).  To
> get 1/6000th of a cent takes a fast processor like we
> intend to use, but I
> can (and did) do it with three simple chips from Radio
> Shack and still get
> an accuracy of 1/200th of a cent and do it in 36 ms!
>
> The signal from the pickups is super-clean (virtually no
> overtones).  I
> convert this wave to a square wave using a simple chip
> called a Schmitt
> trigger ($0.16 at Radio Shack).  Then, rather than count
> waves for a long
> time to get a frequency, I determine the "period" (the time
> for one complete
> vibration), which is just the reciprocal of the frequency
> and is just as
> useful.
>
> How do I get it so accurate?  Well, if I'm willing to
> splurge another $0.86
> I can get a little 10 MHz crystal oscillator.  This puts
> out 10 million
> square-wave pulses each second and is ultra-accurate.  Then
> for another five
> bucks (this is getting expensive!) I buy me a programmable
> counter chip.
> The counter chip has several independent counters that can
> count pulses this
> fast.  I use two of them.
>
> The first one is a one-shot counter and I feed it the
> square wave from the
> piano string and tell it to count to "one".  The way the
> counter works (I
> can get into more detail if you like), this produces an
> output pulse equal
> in duration to exactly one vibration of the string.  I feed
> this signal to
> the "gate" of the second counter.  All the gate does is
> tell the second
> timer when to start and stop counting.  So if I feed it my
> one-vibration
> pulse it will turn the counter "on" for one vibration and
> then back "off".
> What is it counting?  The 10 MHz oscillator!  Then I read
> the number out of
> it to see what I got.  What I get is how many times a 10
> MHz oscillator
> pulses during one vibration of the string.  Do you see
> where we're going
> here?
>
> Let's use A-440 for an example.  An oscillator pulsing 10
> million times a
> second would pulse
>
> 10,000,000 / 440 = 22,727 times
>
> during one period of A-440.  If I wanted a string to
> vibrate at exactly 440
> Hz, I would have to tune it until my counter read 22,727.
> And remember, the
> time it takes to do this measurement is just one period, or
>
> 1 / 440 = 2.3 milliseconds.
>
> Obviously the largest, slowest waves would be from the Big
> Daddy A0 string
> at 27.5 Hz.  Then the counter would read 10,000,000 / 27.5
> = 363,636 pulses.
>
> The resolution of the counter is, obviously, one count (it
> can't count
> fractions of a count, only integers).  The frequency at one
> cent above A0 is
>
> f = 27.5 x 2^(1/1200) = 27.5159 Hz
>
> This results in a count of 10,000,000 / 27.5159 = 363,426.
> This differs
> from the other count by
>
> 363,636 - 363,426 = 210 pulses
>
> That's 210 pulses difference in reading to detune the note
> by just one cent.
> And it took
>
> 1 / 27.5 = 36.4 milliseconds
>
> For about $6.00.
>
> Actually I have found that an accuracy of better than a
> tenth of a cent or
> so is futile since the string naturally wavers more than
> this even when held
> at a constant volume.  It's also far better than any but
> the finest ear
> could detect.
>
> Don A. Gilmore
> Mechanical Engineer
> Kansas City
>
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