[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 03:34:14 +0200 (CEST)
Message-ID: <20040529013414.85256.qmail at web50110.mail.yahoo.com>

Dear Karl,

> Some people say for instance:
> The fist bandpass filter are spaced 1 octave
> above the 2 bandpass filter, both wich have
> about ca 20db slopes.(they asume Cf to Cf i suppose).
> 
> So if we have a BP filter wich we dont know exatly
> the slope db but want to figure out how many
> semitones fitts in that range the BP covers
> by the means of centerfrequency (easy to see on
> a scope.)how do we do that?
> 
> Ohh, did i answer my own question perhaps?!

?

Let's say that I am quite puzzled about what you are really after.

The slope of a filter is ~6.02 dB per Octave and per zero on that side (in
bandpass filters you place zeroes on both sides of the passband, so you must
examine the slopes of each side individually). You need at least as many poles
as you have zeroes in total (usually all zeroes is used for slope-shaping, so
then the amount of poles and zeros is equalent). But just because the slopes
is per octave (which is for conveience, we could equally well be talking about
20 dB per Decade) there is nothing about the passband width which naturally
becomes frequency-related in dB form. 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?

Cheers,
Magnus