[sdiy] Formant filters, yay!

mskala at ansuz.sooke.bc.ca mskala at ansuz.sooke.bc.ca
Thu Mar 9 17:48:40 CET 2017


On Thu, 9 Mar 2017, Elain Klopke wrote:
> For example, 
> Tenor 'o'        f1   f2   f3   f4   f5 
> frequency (Hz)   400  800 2600 2800 3000 
> Amp (dB)         0   -10  -12  -12  -26 
> bw (Hz)          40   80   100  120  120
>
> I am, of course, assuming that the signal goes to the five filters in
> parallel and then the outputs get mixed back together into, say, an op amp.
> f1 would be going through a 100k resistor and then the rest would be
> progressively higher values of resistors. Yes?

That's one way to do it.  You want the voltage gains to be the square
roots of the power gains shown, and voltage gains are in turn inverse to
the resistor values.  So if you're using 100kOhm for amplitude 0dB, then
for amplitude -10dB you can calculate antilog(-(-10)/20) = 3.162,
resistance = 316.2kOhm.  Substitute other decibel numbers for the -10 in
that equation as appropriate.  The overall minus sign accounts for the
inverse relationship between resistance and gain, and the 20 accounts for
the square root operation and the conversion from decibels to "bels"
(powers of ten).

However, using such large resistances is going to introduce some thermal
noise.  It would be preferable, if you aren't constrained by needing very
high input impedance for interfacing or something, to start with some
smaller resistance, like 10kOhm or even less, for 0dB and scale everything
(including the feedback resistor) accordingly.

-- 
Matthew Skala
mskala at ansuz.sooke.bc.ca                 People before principles.
http://ansuz.sooke.bc.ca/


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