Hi,
I guess you *can* do the right shifting 3 times before addition, if you
save the shifted-out values also (it is the modulus by division with 8).
They can be also cumulated such way as the originals and at the end shift
right 3 times also right and the result is to be added to the sum of the
previous data. It requires only one more RAM cell.
I hope this helps.
Imre
P.S: Here also an example
Original        DIV by 8        Remainder
123             15              3
45               5              5
66               8              2
23               2              7
12               1              4
36               4              4
82              10              2
 1               0              1
----------------------------------
388             45              28      SUM
 48.5           45               3.5    AVG



On Mon, 26 Apr 1999, Matt Bennett wrote:

> I'm trying to figure out how to compute an average of 8 discrete
> measurements to reduce noise (on a 16C71).   The measurements appear to
> have a gaussian type of distribution, centered around some DC value, with
> perhaps a 2 bit sigma (noise amplitude).  I want that DC value.
>
> Average = (X1+X2+X3+X4+X5+X6+X7+X8)/8  But the summation of X1 ... X8 would
> be possibly greater than eight bits.
>
> Here's my situation: I recieve 50 bytes of data, (X1.1 .. X1.50) in the
> first burst, and store them (in the PICs limited memory).  I then recieve
> another 50 bytes (X2.1 .. X2.50).  Eventually I will recieve 8 bursts (up
> to X8.1 .. X8.50).  I want to average X1.1, X2.1, ... X8.1, and so on.  Is
> there any way to do this without losing precision?  There is time between
> XA.N and XA.N+1 to do some calculations.
>
> This can be done with just 2 samples (X1.1+X2.1)/2 (utilizing the carry
> generated with the add, and keeping the result in 8 bits)  I'm wondering if
> there is any way to extrapolate this method into a greater number of
> samples.
>
> Doing the division (right shift) before the add won't help- since that
> would serve to eliminate immediately the small variations I'm trying to
> compensate for.
>
> This message is not really as clear as I would like it to be, but I don't
> see any way to explain it better without a white board in front of me. I
> hope I've gotten the point across adequately.
>
> Matt Bennett
>
> --
> Matt Bennett
> mjb@arlut.utexas.edu
>
>