Table of Contents

Note that for a gaussian signal, differencing subsequent elements will create a distribution with a noise sqrt(2) larger than the input signal.

**in=**- Input file name. No default.
**xcol=**- Column(s) to use [1]
**nmax=**- max size if a pipe [100000]
**cumul=t|f**- Cumulative, instead of differences. Default: false
**orig=t|f**- Also show the original column (if selected it will
be shown
*after*the trend column [f] **first=t|f**- Add the first row? [Default:
f]

set n = 10000 seq $n | tabmath - - ’rang(0,1)’ all seed=-1 | tabhist - 1 -2 2 nmax=$n # sigma=1 seq $n | tabmath - - ’rang(0,1)’ all seed=-1 | tabsmooth - | tabhist - 1 -2 2 nmax=$n # sigma=0.612 (sqrt(3/8)) seq $n | tabmath - - ’rang(0,1)’ all seed=-1 | tabtrend - | tabhist - 1 -2 2 nmax=$n # sigma=1.414 (sqrt(2)) seq $n | tabmath - - ’rang(0,1)’ all seed=-1 | tabsmooth - | tabtrend - | tabhist - 1 -2 2 nmax=$n # sigma=0.5 seq $n | tabmath - - ’rang(0,1)’ all seed=-1 | tabtrend - | tabsmooth - | tabhist - 1 -2 2 nmax=$n # sigma=0.5Note that smoothing and trending are related, so their resulting dispersions in the last two examples do not multiply.

This also means if you don’t know if a signal had been applied a Hanning smoothing, compute the sigma before and after a difference operator. If that ratio is sqrt(2) , it means it was not, it was a pure un-correllated gaussian. However, if the ratio is sqrt(2/3) = 0.816, it was a Hanning smoothed signal to begin with.

26-Nov-2012finally documentedPJT 13-oct-2014better examplesPJT 15-jun-2016added cumul=PJT 26-may-2020added orig=PJT