[sdiy] Ladder filters and gain drop, that old chestnut
Andrew Simper
andy at cytomic.com
Mon Aug 31 07:14:38 CEST 2015
Also my point is that dB to me seems like an arbitrary unit, it makes
more sense to me to see on a plot a drop of 1 unit per octave
(incidentally which is another power of 2) for a 1 pole filter, or a
rise of 1 unit per octave for a 1 pole high pass filter, 1 pole = 1
doubling / halving per octave. It just makes things easier to count in
my head instead of having to multiply and divide by 6 all the time.
Cheers,
Andy
On 31 August 2015 at 13:04, Andrew Simper <andy at cytomic.com> wrote:
> On 31 August 2015 at 12:05, <mskala at ansuz.sooke.bc.ca> wrote:
>> On Mon, 31 Aug 2015, Andrew Simper wrote:
>>> And a point to note is the half power point is not -3 dB exactly, as
>>> dB isn't exactly related to powers of 2:
>>>
>>> 20*log(10, sqrt(1/2)) ~= -3.0102
>>>
>>> where ~= means approximately equal to.
>>
>> Are your filters built with components accurate enough, and
>> temperature-stable enough, for a difference of 0.01dB in the response
>> curves to be meaningful?
>> --
>> Matthew Skala
>> mskala at ansuz.sooke.bc.ca People before principles.
>> http://ansuz.sooke.bc.ca/
>
> When working out the equations it is much easier to use exact values
> rather than dB, the solutions come out much neater!
>
> If you are interested in actual filter shapes you are best off working
> out the exact equations and then you can do monte-carlo analysis by
> randomising the component values and working out a realistic
> difference from the ideal values, and this can help choose filter
> topology and component tolerances that fit with your requirements.
>
> Cheers,
>
> Andy
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