Harold,

What you describe is an exponential decline (0.5, 0.25, 0.125, etc.)
and this is a first-order IIR low pass filter.

Sean


On Tue, Aug 10, 2010 at 12:04 PM, Harold Hallikainen
<harold@hallikainen.org> wrote:
>
>> On Tue, 10 Aug 2010 08:19:32 -0700 (PDT), "kris duff" said:
>>> Hello,
>>>
>>> My colleage told me about a smoothing/lpf formula which is :
>>>
>>> a(n) =3D {a(n-1) * param + ADC} / {param + 1}
>>>
>>> which
>>> ADC is the signal from adc
>>> a(n-1)=C2=A0 is the result of the last call to this formula
>>> the param is the part I don't understand.
>>>
>>> Is there anybody knows where this formula comes from ?
>>>
>>> it seems to work well for smoothing, but I want to tweek the param (
>>> which is certainly for the bandwith )
>
> I have used this with param=3D0 (just use current ADC) and with param=3D1
> (mean of old value plus current adc). The param determines how much of ol=
d
> values you're using. It's interesting to look at a step response to this
> LPF. If the ADC value goes from 1 to 0 in one sample, the output goes 1,
> 0.5, 0.25, 0.125, etc. Note that it's a linear decline while a single pol=
e
> RC would be an exponential decline.
>
> Harold
>
>
>
>
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