[chbot] EMA IIR cutoff calculation

Thu Jan 25 21:49:58 GMT 2024

Years ago I was looking at a first order low pass filter that someone
found on the net. It didn't seem to be doing what it should so I tested
ity in excel. It wasn't right there either. Then I used the speadsheet
to find an expression that was a simple simulation of and RC filter.
The result was simple, and looked much like one mentioned earlier
here. 
I have lost that sheet, but I suggest using a spreadsheet to test your
formula.
Martin

On Thu, 2024-01-25 at 21:33 +1300, Robin Gilks wrote:
> Hi Richard
> Don't think I can get a 'scope into the depths of my code!!
> Think what I'm looking for is actually
> Fcutoff =-ln(1-alpha) * Fsample / (2 * pi)
> 
> I was missing the '1' in the  (1-alpha)
> 
> 
> 
> On Thu, Jan 25, 2024 at 8:44 PM Richard Jones
> <richard.jones.1952 at gmail.com> wrote:
> > One way you could observe the frequency response of the filter is
> > to generate sine wave samples and feed them into the filter. Sweep
> > the sine wave at say 1Hz per ms. Observe the output by feeding the
> > results into a file or into a d->a convertor. If you wish to
> > observe the results on an oscilloscope the time from the beginning
> > of the sweep to 3dB attenuation gives you the cutoff frequency.
> > Adjust the filter parameters if the response is not as you desire.
> > Hope that helps or am I missing something?
> > 
> > Richard Jones
> > 
> > On Thu, Jan 25, 2024 at 4:51 PM Robin Gilks <gb7ipd at gmail.com>
> > wrote:
> > > Thats one of the pages I've visited and still none the wiser. I
> > > followed a link to another page and was immediately hit by:
> > > The impulse invariant discrete formulation of the ideal low pass
> > > filter is obtained from the solution to the first order linear
> > > differential equation of the scalar (one dimensional)
> > > differential equation corresponding to the Laplace transform of
> > > the filter .......wah wah wah wah
> > > Just the sort of stuff I'm too old and not bothered enough to
> > > spend hours trying to catch up on from when I was at college over
> > > 50 years ago.
> > > 
> > > I think what I'm looking for is 
> > > Fcutoff =-ln(alpha) * Fsample / (2 * pi)
> > > 
> > > but I can't find any confirmation of that amongst all the
> > > smartarse answers on StackExchange etc
> > > 
> > > Looks like I'll just have to carry on guessing
> > > 
> > > 
> > > On Thu, Jan 25, 2024 at 9:29 AM Charles Manning
> > > <cdhmanning at gmail.com> wrote:
> > > > Here's a relatively simple article on first order IIRs
> > > > 
> > > > https://dsp.stackexchange.com/questions/34969/cutoff-frequency-of-a-first-order-recursive-filter
> > > > 
> > > > On Wed, Jan 24, 2024 at 11:47 PM Robin Gilks <gb7ipd at gmail.com>
> > > > wrote:
> > > > > 
> > > > > 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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