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Re: Another way to calculate THD

Hi,

I made a spreadsheet of all the harmonics that shows what note
they produce, what relationship it is (ie. fifth, major 3rd, etc),
and how out of tune it is with the well tempered scale. Anybody
that wants it can request via email.

Ignoring masking effects for now that Cheever and Hiraga
take up in an absolute terms. This really complicates things as
I've learned before. I'm still trying to understand how to work
with this concept.

Anyway, one step at a time. First...

It looks like you might be able to take a normal THD calc:

sqrt(N2*N2 + N3*N3 + ...)

and apply a weighting factor to each harmonic. I know this has
been proposed many times before, but I'm looking at it as how
consonant or discordant the harmonics are in order to set the
weighting factor.

sqrt(W2*N2*N2 + W3*N3*N3 + ...)

Accounting for the 2nd, 3rd, 4th, 5th, 6th, 8th, 10th, 12th
harmonics being consonant and the 7th, 9th, 11th, 13th harmonics
being discordant, and the 7th, 11th, and 13th being out of tune
with the piano. These are somewhat based on Helmholtz Sensation
of Tone in order to adjust the extremes.

W2 = 0.1 (1 octave)
W3 = 0.3 (1 octave + fifth) (.1 + .2)
W4 = 0.2 (2 octaves) (.1 + .1)
W5 = 0.5 (2 octaves + major 3rd) (.1 + .1 + .3)
W6 = 0.4 (2 octaves + fifth) (.1 + .1 + .2)
W7 = 7.0 (2 octaves + out of tune dominant seventh)
W8 = 0.3 (3 octaves) (.1 + .1 + .1)
W9 = 9.0 (3 octaves + in tune second)
W10 = 0.6 (3 octaves + major 3rd) (.1 + .1 + .1 + .3)
W11 = 11.0 (3 octaves + way out of tune 4th)
W12 = 0.5 (3 octaves + fifth) (.1 + .1 + .1 + .2)
W13 = 13.0 (3 octaves + way out of tune 6th)

I wonder if the W7-W13 are strong enough?


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