[sdiy] db to semitones, octaves.

[karl dalen]

>Dear Karl,
Yes please! :-)

> 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.

Well.........host.......hrrmmmm, well im a bit
puzzled my self to! :-)

>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.

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.

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.)

>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.

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