OK, I think I see it.  My Total is your Filt, except Total is right=20
justified, and Filt is left justified.  Both contain 14 significant bits=20
(assuming N=3D16).

My Filt is the Total (or your Filt) right justified and truncated (or=20
rounded) to 10 bits, to give a filtered A/D between 0 and 1023.  This=20
step is optional, if one wanted the 14 bit number, one could skip the=20
conversion.

And your FF(NEW - FILT) is really (NEW - FILT) / FF.

Is this correct?

Kerry


Olin Lathrop wrote:
> Kerry Wentworth wrote:
>  =20
>>> Right.  You generally want to use Log2 fraction bits to avoid this.
>>>      =20
>> Where are your fractional bits?
>>    =20
>
> Where they ususally are, to the right of the binary point.
>
>  =20
>> In Filt?
>>    =20
>
> In all the values.  Everything is a fixed point number with the specified
> number of fraction bits.  To be clear, the algorithm I mentioned was:
>
>   FILT <-- FILT + FF(NEW - FILT)
>
> NEW and FILT are expressed as fixed point values.  This is usually easy o=
n a
> PIC when NEW is successive A/D readings by using the left justify A/D res=
ult
> mode.  For example, with the common 10 bit PIC A/D, you automatically get
> each reading in 10.6 fixed point format in left justify mode.
>
> The minimum number of fraction bits you need to not loose information
> is -Log2(FF).  For example, if FF =3D 1/16, then you need 4 fraction bits=
..
>
> In the other method that uses TOTAL, you still need the extra 4 bits but
> they are on the high end of the number.  For example, if filtering 10 bit
> values with FF =3D 1/16, then TOTAL needs to be 14 bits wide.  Note that =
TOTAL
> then has exactly the same representation as FILT.  It makes no difference
> whether you consider it in the scale of the original value with 4 fractio=
n
> bits, or in the scale of the original value times 16.  The bits inside th=
e
> processor are exactly the same.
>
> The only real difference between these methods is that yours requires the
> additional piece of persistant state, being the TOTAL accumulator.  You c=
an
> do some substitutions in your algorithm to reduce it to what I described
> above.
>
>  =20
>> Are you doing floating point math?
>>    =20
>
> You can, but this discussion was about fixed point.  There have been a fe=
w
> cases where I've done that where I needed precise control over the time
> constant or on the host computer with floating point hardware or where
> cycles don't matter anyway.
>
>
> ********************************************************************
> Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
> (978) 742-9014.  Gold level PIC consultants since 2000.
>  =20



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