[sdiy] db to semitones, octaves.

[Magnus Danielson]

From: karl dalen <dalenkarl at yahoo.se>
Subject: Re: [sdiy] db to semitones, octaves.
Date: Sat, 29 May 2004 15:27:49 +0200 (CEST)
Message-ID: <20040529132749.6342.qmail at web50102.mail.yahoo.com>

Dear Karl,

> >Dear Karl,
> Yes please! :-)

> >Let's say that I am quite puzzled about what you are
> >really after.
> 
> Well.........host.......hrrmmmm, well im a bit puzzled my self to! :-)

(host <=> cough)

> If a sawtooth passes a BP then the width of the pass-
> band will decide wich harmonics will be adiouible,
> if the passband narrowes fewer harmonics will pass
> untill we have a sine depending on the Cf of the Bp
> filter. Based on the width of the Bp filters, the two
> Cf points we can tell the amount of semitones between
> these two Cf points att full db value.(whaterver db
> reference are used, lets say 0.775V reference). Now
> if we also know the db of the slopes starting at the
> -3db point of the slopes we also would be able to tell
> the amplitude of the harmonics outside theses two Cf
> points.

Well yes... but you need to know which slope you are having etc. It is not
hard to predict the strength of the individual overtones when passed throug a
BP filter (or any other responce).

> Think of a variable width spectrum analysator who
> uses a imagenary super BP filter for analyse.
> (dont get hocked at this, most specs dont use a
> sweepable BP as a analyser, it was just a analogy
> to it all.)

I know specs fairly well and happends to own one too (1mHz - 500 MHz,
3Hz to 3MHz filters).

> >Consider that >the slopes differes between
> >different grades of filters, so which one is the
> >right one? None!
> 
> >My point is that seminotes in dB form is meaningless.
> >Whatever definition you
> >make, you can break it as far as I can see. Also, I
> >see no use for it, we
> >already have adequate logarithmic scales.
> 
> >I still must ask: Why? What new thing would it solve?
> 
> It would not solve any "new" thing im just looking
> for a way to fit in semitones into a db value,sortoff.

Do this then:

1) Convert your tone (semitone and octave) into frequency.
2) Predict the strength of that overtone in that waveform.
3) Calculate the filter responce |H(jw)| for that frequency (w=2*pi*f).
4) Multiply the filterresponce with the preditect input strength and you got
   the output strength of that tone.

If you want to know how much a seminote rise increases or decreases a
particular overtone as you raises it from a particular tone, then just do the
above exercise for both of them and the subtract the logarithmic values with
each other. The point is that whatever value you get there it depends greatly
on the frequency you are starting from and the change is the derivate of the
amplitude-responce and is non-linear, so a linear match may only work for
particular parts of the curve. Also, the result is only valid for that
particular amplitude-responce curve. For another BP filter which may have much
steeper responce you will get quite different results.

Cheers,
Magnus