[chbot] EMA IIR cutoff calculation

Thu Jan 25 08:33:12 GMT 2024

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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