Why don't you calculate the integral:
                   T
                 /  2         1/2
     ( 1/(2T) * / x  (t) dt )
               /
            -T
You can approximate this integral with a sum. The T is a multiple of one
 period.There may be some problems with the root but you could solve this with a
lookup table. If you want, you can store a couple of squared x:s and "recycle"
them to create a moving rms process (You might need a lot of RAM to store
one period of data.)
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Conny Andersson / LiTH