> - Keep last average of individual square values and
> corresponding square root;
> - Calculate the difference between new average and
> that last average of individual square values.
> - Divide the difference by 2 (do you remember what's
> the derivative of square function?)

Typo, sorry. It should be "Divide the difference by 2 multiplied by
old square root";

For example:
- old average of individual square values = 40000
- old square root 200
- new average of individual square values = 41000

41000-40000=1000
1000/(2*200)=2.5
202.5*202.5=41006.25

41006.25 is almost equal to 41000
Thus 202.5 should be taken as square root of 41000
New voltage 202.5 has been found in one cycle.
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