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Re: linear fit of looped (and cutted) data set

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Subject	Re: linear fit of looped (and cutted) data set
From	Markus Grünwald <m_grueni@web.de>
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Am Sonntag, 27. November 2016 20:29:46 UTC+1 schrieb Ethan A Merritt:

> Looks like maybe you want to fit the first derivative of the
> trajectory rather than fitting the trajectory itself.

Good advice.

After little research and help of following blog https://stackoverflow.com/questions/15751226/how-can-i-plot-the-derivative-of-a-graph-in-gnuplot

if done following:
[code]
limitlo(x,lo) = (x > lo) ? x : 0/0
limithi(x,hi) = (x < hi) ? x : 0/0
limitlohi(x,lo,hi) = limitlo(limithi(x,hi),lo)
d2(x,y) = ($0 == 0) ? (x1 = x, y1 = y, 1/0) : (x2 = x1, x1 = x, y2 = y1, y1 = y, (y1-y2)/(x1-x2))

dx=.5; ylo=550; yhi=2500
fit [0:20] kwm "Data.txt" using ($2-dx):(d2($2,(limitlohi($3,ylo,yhi)))) via kwm
fm(x) = kwm*x + dwm
fit [0:20] [ylo:yhi] fm(x) "Data.txt" using 2:3 via dwm
[/code]

and now i got quite what i expected to.

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Thread

linear fit of looped (and cutted) data set Markus Grünwald <m_grueni@web.de> - 2016-11-27 07:31 -0800
  Re: linear fit of looped (and cutted) data set Ethan A Merritt <EAMerritt@gmail.com> - 2016-11-27 11:29 -0800
    Re: linear fit of looped (and cutted) data set Markus Grünwald <m_grueni@web.de> - 2016-11-27 14:56 -0800
      Re: linear fit of looped (and cutted) data set [solved] Markus Grünwald <m_grueni@web.de> - 2016-12-26 14:52 -0800
  Re: linear fit of looped (and cutted) data set Karl Ratzsch <mail.kfr@gmx.net> - 2016-11-27 22:11 +0100

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