From: RussellMc <apptechnz@gmail.com>
To: Microcontroller discussion list - Public. <piclist@mit.edu>
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Date: Sat, 15 Aug 2020 05:37:30 -0700
Subject: Re: [EE] averaging multiple samples to reject AC interference
Thread-Topic: [EE] averaging multiple samples to reject AC interference
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On Sat, 15 Aug 2020 at 20:56, Isaac Marino Bavaresco <
isaacbavaresco@gmail.com> wrote:

> Em 14 de agosto de 2020 22:04:36 BRT, RussellMc <apptechnz@gmail.com>
> escreveu:
>


> >> Latest/N + Sum x (N-1)/N
>
> Does N grow indefinitely?
> If yes, the new values will have less and less influence as time goes by.
> If not, there will be a steady increase in the final result.
>

Sorry - it didn't occur to me that that was ambiguous.
Note that this was an algorithm that I said was not as good as retaining K
samples and averaging them.


N is a selected factor that controls how much effect the latest sample has
compared to all prior samples.

Define "AVAL" as the averaged value
 Initially, set AVAL =3D 1st sample

Set N to say 4. This means that from now on new samples will account for
1/4 of the new value of AVAL and the old value will account for 3/4 of the
value of AVAL.

so AVAL =3D 1/4 x latest sample + 3/4 x AVAL

If N was 10 it would be  1/10x sample + 9/10 x AVAL.

This is not a marvellous method but has the advantage of not requiring
retaining any prior sample values and of the effects of past samples
diminishing relatively rapidly.

Example:
N=3D4
Sequence 5 5 5 5 5  20   20     20   20     20
Output     5 5 5 5 5 8.75 11.6  13.7 15.3  16.5 ...

For a single step function AVAL never reached the new value!

______________________________
Better:

If you retain say the last 4 and average them with the same input you get
5 5 5 5 5 8.75  12.5 16.25 20 20 ...

For a single outlier eg  5 5 5 5 5 20 5 5 5 5 5 5
You'd get 5 5 5 5 5 8.75 8.75 8.75 8.75 5 5 5

ie the outlier has an effect for N samples.


     R

>
>
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