Olin Lathrop wrote:
> Sean Breheny wrote:
>  =20
>> For example, if we have a filter factor of 16, with Olin's example,
>> and using integer math, we have:
>>
>> FILT<--FILT + (NEW-FILT)/16
>>
>> If the absolute value of the difference NEW-FILT is less than 16,
>> nothing ever happens.
>>    =20
>
> Right.  You generally want to use Log2 fraction bits to avoid this.
>  =20
Where are your fractional bits?  In Filt?  Are you doing floating point=20
math?  If you are, converting to integer math without loss of resolution=20
IS a 'free lunch'.
> However, the other version doesn't give you a free lunch.  It looks like
> TOTAL will float around the value times the same factor that you're divid=
ing
> by in the equation above.  That just means you're adding the bits on the
> left instead of the right.
>  =20
Yes, Total will float around that number.
Do you mean adding bits on the right instead of the left?  The=20
fractional bits in mine are on the right side of the equation.  Am I=20
getting confused?
> So you have to leave overflow bits on the left or add the same number of
> fraction bits on the right.
>
>
> ********************************************************************
> Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
> (978) 742-9014.  Gold level PIC consultants since 2000.
>  =20

Kerry, fresh off the turnip truck



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