Miller compensates for the interferometer drift or slope over each page, ie 20 turns. Advantage calculation are on single page. see p214
The purpose of this test is to find the effect of slope on the sin[ ] component in an interval of 2 Pi.
See ApexMeansTest.nb
Fourier analysis assumes that a signal repeats on some fixed cycle. If a signal does not do this, then FA cannot be used. At least without closing the cycle or the like.
This program shows that when the “c” parameter is removed from the model expression, then the “a” parameter (on Sin[]) increases wildly. But when the “c” is present, the “a” parameter matches the “a1” in the data(xy) generation code. This means that the linear term “c x” completely prevents the linear slope from affecting the true Sin[ ] component.
also to find out how to structure data for FindFit “data”
Clear[x,y,x1]; imax=20; x2=Table[i/2.,{i,imax}]; Print["x ",x2]; a1=1.01; b1=25; c1=10; xy= Table[{x1/2.,a1 Sin[2 Pi .1 x1/2.]+ b1 + c1 x1/2.},{x1,imax}]; Print["xy ",xy]; ff=FindFit[xy,a Sin[2 Pi .1 x] + b + c x,{a,b,c},x]; Print["angles ",Table[2 Pi .1 x1/2.,{x1,imax}]]; Print["ff ",ff]; x {0.5,1.,1.5,2.,2.5,3.,3.5,4.,4.5,5.,5.5,6.,6.5,7.,7.5,8.,8.5,9.,9.5,10.} xy {{0.5,30.3121},{1.,35.5937},{1.5,40.8171},{2.,45.9606},{2.5,51.01},{3.,55.9606},{3.5,60.8171},{4.,65.5937},{4.5,70.3121},{5.,75.},{5.5,79.6879},{6.,84.4063},{6.5,89.1829},{7.,94.0394},{7.5,98.99},{8.,104.039},{8.5,109.183},{9.,114.406},{9.5,119.688},{10.,125.}} angles {0.314159,0.628319,0.942478,1.25664,1.5708,1.88496,2.19911,2.51327,2.82743,3.14159,3.45575,3.76991,4.08407,4.39823,4.71239,5.02655,5.34071,5.65487,5.96903,6.28319} ff {a->1.01,b->25.,c->10.}
I could do the same over whole day: 24h * 60t/h = 1440 turns. The daily compensation could change could be overall zero but be cyclic over day. This would cause a daily cycle in the interferometric sin component, albeit 1/1440 as large each day.
Or even an entire set of days in name-list.
Effect of slope on sin[ ] in interval 2 Pi.