[chbot] EMA IIR cutoff calculation

[Robin Gilks]

I'm already using (and have been for years) IIR filters like this but I've
always used seat-of-the-pants trial and error to determine the ALPHA value.
I'd like to be a bit more precise.
BTW: I'm using a micro with hardware floating point !

On Wed, Jan 24, 2024 at 8:17 PM Charles Manning <cdhmanning at gmail.com>
wrote:

> Have a play around on https://fiiir.com/. This has sections for both
> FIR and IIR.
>
> These are the simplest filters you can build, but I would recommend
> using fixed point if you are running it on a low power micro. You can
> avoid multiplies by doing things like
>
> out = out + (in - out)/4;  // (which will turn into +, - and a shift).
>
>
>
>
> On Wed, Jan 24, 2024 at 4:59 PM Robin Gilks <gb7ipd at gmail.com> wrote:
> >
> > Greetings all
> >
> > I'm after the simplest answer to the question: what is the cutoff
> frequency of an Exponential Moving Average Infinite Impulse Response filter.
> > This code snippet is called at 8KHz with a sample from the ADC. I know
> there is a natural log of ALPHA somewhere in the equation but trying to
> find the simple solution on the 'net buries me in Bode plots and sumof
> equations. I specifically don't want a mathematical treatise on EMAs, just
> a simple formula I can plug into my code 😀
> >
> >
> > #define ALPHA     0.3
> > static float EMA_S;                     // Exponential Moving Average -
> Signal
> >
> >    EMA_S = (ALPHA*sample) + ((1-ALPHA)*EMA_S);
> >
> >
> > Cheers
> > --
> > Robin Gilks
> >
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