Robert Scott writes......
>If it were possible to record 20 kHz and 20.5 kHz tones
> and mix them with no distortion at all, then there would be no
500 Hz
> resultant. ........
>If my theory is correct, the resultant tone should
> be even stronger when played on such a system with more
>distortion.
----- Original Message -----
From: Robert Scott <rscott@wwnet.net>
To: <pianotech@ptg.org>
Sent: Tuesday, December 03, 2002 3:39 PM
Subject: Re: audible resultant from two supersonic frequencies?
> Play C6 and F#6 together, as loudly as you can. If the theory
is >correct, the resultant between F#6 (1484 Hz) and C6 (1050.5
Hz) >will be 433.5 Hz, which is just 25 cents below A4.
Not to nit pick but in my book, (Braid White) the freq for F#6 is
1479.978 and C6 is 1046.502. The differnce between these two is
433.5 the same as yours. Maybe you were at 442?
C6--F#6 on the Kimball consolette isn't up to it, but my Yamaha
PSR-340 digital keyboard does it very good in the "GrandPno"
voice, you can easily hear (now that you mention it) a "resultant"
A4 and at C5--F#5 a resultant at A3 very clear.
On F#5--C6 I hear a resultant of Eb4. The arithmetical
difference between these two tones or 740 and 1046 is 306 which
is close to 311 hz which is Eb 4.
Another experiment might be to sound the F#--C while holding
down Eb4 and then see if Eb4 has been excited by the resultant.
I doubt it since it is 5 cycles per second away. Can't do this on
the digital, but hmmmmmm Yes it does seem to beat at 5 bps. ! !
when Eb4 is held while listening to F#5--C6.
>
> I tried this, not expecting to hear any resultant. I was quite
surprised
> to hear a faint A4, or something like it. But as I listened, I
noticed
> that the A4 resultant behaved strangely. As soon as the C6 and
F#6 began to decay, the A4 resultant suddenly dropped out. It
did not gradually decay.
This would have to be done on various pianos. On a digital piano
mentioned above I could not hear the resultant drop out. And it
sounded no matter what level (loud) I played.
> And if I play C6 and F#6 together softly, I hear no A4
resultant at >all. This leads me to believe that the A4
resultant is not being caused >by C6 and F#6 beating together in
the usual fashion. If that were >the case, the resultant would
decay exactly the same as the two >primary notes.
But it does on a digital piano and I think therefore it must on a
pipe organ. I think the evanescent quality of piano tone , (decay
in synth lingo although it should be "decay sustain decay" to
better describe piano tone) makes it hard to hear the resultant at
some volume levels. There is a lot in the Helmholtz-Ellis
translation (1865-75) that talks resultant tones being hard to
hear and other tones of summation and differentiation we don't
even consider about today.
> Instead, I think this is a case of non-linear mixing (as alluded
to
> by Sarah Fox several days ago).
I understand up to when the terms "linear" and "non-linear"
appear. What kind of ear is a linear? ;) So you see most of
us are lost when college level terms are bantered about, and a
few would appreciate when it is "dumbed down". As technicians we
are rather keen on precise definitions, or we would be
counting 880 oscillations of a vibrating wire giving what is
called
440 hz, or cps today.
> If there is a loose screw
> somewhere, or if two parts of the piano are just barely
touching, then you will have non-linear mixing. (In the extreme
case, these conditions cause noticeable buzzing.)
This was explained to me as a loose screw "resonating" at a
certain frequency. "Sympathetic vibration" was another term used
depending on who was trying to explain it.
> At high volume levels, there can even be non-linear
> mixing in our ears, when the bones that carry the sound from the
>eardrum reach the limit of their travel.
This is a statement that needs to be explained to us techs who may
or may not have a college education and have forgotten a lot.
What does, "when the bones that carry the sound from the ear drum
reach the limit of their travel" mean? First of all, do you
really know that the bones actually reach the limit of their
travel or do they just get closer without reaching the limit? If
they reached the limit of their travel couldn't that mean they
might be stressed or over taxed so that their limit of travel
would not be the same for the next sound that came along? Oh and
I almost forgot, what does "non-linear mixing in our ears" mean?.
>In fact it is difficult to guarantee a
> purely linear addition of two tones. Any distortion of the
sound >causes the component sounds to interact non-linearly, and
thus >produce real acoustic energy at the resultant frequency.
What is a linear addition of two tones? What is a distortion of
the sound?
If there were no distortion would there be no real acoustic energy
at the resultant frequency? How do component sounds interact
non-linearly? What is non-linear? What is linear?
---rm Linmeanear
ps
Of your comments below I want to ask, is it possible for a
computer to generate a tone of 20,400 ? If so with how much
precision? Is it then possible for a computer to generate a tone
of 20,000? Is it the loud speakers or earphones that do not
faithfully reproduce the tones? What then is the problem of
comparing one audio frequency of 20,500 to another audio frequency
of 20,000 cps.? ---rm
> For example, Don Mannino's
> suggestion:
>
> >....Back to your original question, Ric. I made a recording for
you, and
> >you can hear the resultant tone. I made a wav file of 20K and
20.5K
> >tones, one in each ear. Then I combined them into a single
mono file. If
> >you play it in good headphones and turn up the volume loud, you
can just
> >hear the 500hz tone as a pitch.
>
> The process of making wav files includes time-sampling and
quantization,
> both of which are somewhat non-linear. If there is a 500 Hz
resultant in
> Don's wav file, it is probably an artifact of the imperfections
of the
> recording process. If it were possible to record 20 kHz and
20.5 kHz tones
> and mix them with no distortion at all, then there would be no
500 Hz
> resultant. Since it is hard to find such a perfect recording
system, then
> the theory can perhaps be tested by finding a poorer recording
system - one
> with more distortion. If my theory is correct, the resultant
tone should
> be even stronger when played on such a system with more
distortion.
>
> -Bob Scott
> Ypsilanti, MI
>
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