AVR-Chat

Heard that this is a common technique.

Mean of N samples is (1/N) sum(Xi) where Xi is one of the samples X0, X1,
... XN-1

Now if you are doing a total mean rather than a windowed mean (most recent
N samples), then the new mean value is

Mean = (N*MeanPrevious + Xnew)/(N+1)

Not sure that there is a way to do this for standard deviation because if
the mean changes, then all the difference values change. That is the  hard
 part.

Jim Wagner
Oregon Research Electronics


> I'm playing with an idea to make sense of analog sensors.
> This is borrowed from a technique for background supression in video
> cameras.
> Having trouble figuring an economical way to do the means and standard
> deviations.
>
> Take a single analog input, sampled at some appropriate rate.
>
> Allocate N (call it five for now..) slots for mean and standard
> deviation, and "strength".
>
> When a new sample arrives, for each of five means, test wether the
> sample falls within two standard deviations of a mean.
> If so, then "strengthen" that mean, and recalculate it's standard
> deviation.
> If not, then kill the "weakest" mean, and start a new one with this
> samples value as the mean, and some appropriate guess at standard
> deviation.
>
> I've got code that buffers 50 samples on two inputs, and calculates
> mean and standard deviation, and "alarms" when either of the input
> falls outside 2sd, but the thing is computationally intensive, and I'm
> told that the video guys do NOT keep N frames of data (which makes
> sense, as it would get enormous FAST). Rather they have some method of
> figuring a mean and standard deviation without history data...
>
> Does that make any sense?
> How could I implement a function like that?
>
>
> --
>
> "The very powerful and the very stupid have one thing in common. Instead
> of
> altering their views to fit the facts, they alter the facts to fit their
> views... which can be very uncomfortable if you happen to be one of the
> facts that needs altering." Doctor Who, Face of Evil
>