Hi Lawrence,

Keep in mind that for a constant window width, an average really isn't all
that different than a sum.  For example, take the series

            sum     average
2, 3, 4      9         3
3, 4, 5      12        4
4, 5, 6      15        5

If, say, you wanted to exclude windows with average > 4, and you knew the
window width was constant at 3, you could simply exclude windows with sum
> 12.  That would save you the computational cost of division.

Cheers,
Craig

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ craign@uic.edu ~~~~~~~~~~~~~~~~~~~~~~~~~
Craig Niederberger MD FACS          | Chief, Division of Andrology
University of Illinois at Chicago   | Director Urologic Basic Research
Department of Urology M/C 958       | 312-996-2779   Fax: 312-996-1291
840 South Wood St, Chicago IL 60612 | http://godot.urol.uic.edu

On Tue, 23 Sep 1997, Lawrence Lile wrote:

> Anybody got a good averaging algorithm?
>
> I've got a stream of numbers from 0 to 15 coming out of a PIC '620
> comparator representing temperature of an appliance.  When I turn on
> the flourescent lamp on my bench, the program picks up a spike, and
> the relay in this thing fires momentarily, because it reads the
> noise.  "NO PROBLEM" I sez, and sat down to write an averaging
> algoritm.
>
> This ought to be so easy.  The first "Algorithm" anybody learns in
> kindy-garden Math 1 is how to average numbers.  I've been checking
> out verious ways of averaging a stream of data, though, and don't
> have one that satisfies on a PIC with Two byte arithmetic and no
> floating point math.
>
> Try this.  Make a spreadsheet with 300 or 400 random integers between
> 0 and 16.  Take the arithmetic average of all of 'em, probably
> around 8.   Then try to simulate a PIC program processing each number
> in succession that would get near the average, without floating point
> math.  I tried a half dozen ways this morning.
>
> One trick I worked out is that the average of N numbers, given the
> NTH piece of data is being processed now, is (N-1)*(old average)/N
> plus (Nth piece of data)/N.  If you pick 256 data points, given the
> old average AVG and the new data point DATA,
>
> NEWAVG = {(AVG*255)+DATA}/256
>
> That oughta work!  Not so fast.  Using only integer values,no
> floating point math,  the average climbs slowly toward zero and stays
> there.  PICs can probably do floating point math, but there's gotta
> be an easier way!
>
>
> Best Regards,
>
> Lawrence Lile
>