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Undescribed Horrific Abuse, One Victim & Survivor of Many
gmkarl at gmail.com
Thu Nov 10 11:02:50 PST 2022
ummmmm
so max_period is turned into k in the micro_ functions., and passed to
the exponent.
looks like it's the angle advancement per sample for each sinusoid, so
the frequency is proportional to its inverse.
meanwhile, sample_idcs i believe are frequencies, as a portion of the
total data length
np.fft.fftfreq(4) returns [0,0.25,-.5,-.25]
0 is DC
0.25 then would be period = 4
-0.5 is then the nyquist frequency, period=2
so it looks like fftfreq expects frequency to be 1/period
draft saved at 1:53
then if we have k in the exponent as 2j * pi * k * n, where n is the index
what is the period in samples? well, a period goes by when the
exponent is 2j * pi
so that's when k * n == 1 . then, the whole thing is N long ...
so when k == 1/N, that's an fftfreq .. of ... N?
k == 1/N means the period is N. the fftfreq is 1/period, so yes.
this means k == fftfreq ... I think?
so a possible difference here is the difference in np.linspace(min_k /
N, 1, N - 1)
from arange(N) * (N * 2 - 1) / (N - 1)
if a bisect/troubleshoot the issue by simply trying that arange
directly, it may still fail for another reason, sadly.
the other reason could be missing data in the transform. i know the
current transform can reconstruct a wave; i don't know if the other
will.
maybe i can test it alone.
draft saved a little late at 2:00p
oh i don't have the freqs yet, i have sample indices
i'm using numpy's default frequencies
these default frequencies, are periods equal to 2, 4, and infinity.
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