Question

I have some periodic data, but the amount of data is not a multiple of
the period. How can I Fourier analyze this data? Example:

% Let’s create some data for testing:

data = Table[N[753+919*Sin[x/623-125]], {x,1,25000}] 

% I now receive this data, but have no idea that it came from the
formula above. I’m trying to reconstruct the formula just from ‘data’.

% Looking at the first few non-constant terms of the Fourier series:

ListPlot[Table[Abs[Fourier[data]][[x]], {x,2,20}], PlotJoined->True, 
 PlotRange->All] 

shows an expected spike at 6 (since the number of periods is really
25000/(623*2*Pi) or about 6.38663, though we don’t know this).

% Now, how do I get back 6.38663? One way is to “convolve” the data with
arbitrary multiples of Cos[x].

convolve[n_] := Sum[data[[x]]*Cos[n*x], {x,1,25000}] 

% And graph the “convolution” near n=6:

Plot[convolve[n],{n,5,7}, PlotRange->All] 

we see a spike roughly where expected.

% We try FindMaximum:

FindMaximum[convolve[n],{n,5,7}] 

but the result is useless and inaccurate:

FindMaximum::fmmp:  
   Machine precision is insufficient to achieve the requested accuracy or 
    precision. 

Out[119]= {98.9285, {n -> 5.17881}} 

because the function is very wiggly.

% By refining our interval (using visual analysis on the plots), we
finally find an interval where convolve[] doesn’t wiggle too much:

Plot[convolve[n],{n,6.2831,6.2833}, PlotRange->All] 

and FindMaximum works:

FindMaximum[convolve[n],{n,6.2831,6.2833}] // FortranForm 
List(1.984759605826571e7,List(Rule(n,6.2831853071787975))) 

% However, this process is ugly, requires human intervention, and
computing convolve[] is REALLY slow. Is there a better way to do this?

% Looking at the Fourier series of the data, can I somehow divine the
“true” number of periods is 6.38663? Of course, the actual result
would be 6.283185, since my data fits that better (because I’m only
sampling at a finite number of points).

Answer 1

Based on Mathematica help for the Fourier function / Applications / Frequency Identification:
Checked on version 7

n = 25000;
data = Table[N[753 + 919*Sin[x/623 - 125]], {x, 1, n}];
pdata = data - Total[data]/Length[data];
f = Abs[Fourier[pdata]];
pos = Ordering[-f, 1][[1]]; (*the position of the first Maximal value*)  
fr = Abs[Fourier[pdata Exp[2 Pi I (pos - 2) N[Range[0, n - 1]]/n], 
   FourierParameters -> {0, 2/n}]];
frpos = Ordering[-fr, 1][[1]];

N[(pos - 2 + 2 (frpos - 1)/n)]

returns 6.37072