Undescribed Horrific Abuse, One Victim & Survivor of Many gmkarl at gmail.com
Thu Nov 10 06:31:38 PST 2022

so if recording_bufsize is the length of the data passed in,
then recording_bufsize // 2 + 1 is the output of ftt.rfft . it removes
one side of the data.
then we have rfft_bufsize = recording_bufsize // 2

so at rfft index 0, this represents a period of infinity
at rfft index 1, this represents a period of recording_bufsize
at rfft index 2, i suspect this would be a period of recording_bufsize / 2
and at the last rfft index, i.e. recording_bufsize // 2 + 1 - 1, this
would be a period of 2

so there's a quotient relationship between recording_bufsize, and the
expression recording_bufsize // 2, that yields 2 . one is the
numerator, and one is the denominator

recording_bufsize / (recording_bufsize / 2) = 2

that works for the others too
recording_bufsize / 0 = infinity
recording_bufsize / 1 = recording_bufsize
recording_bufsize / 2 = recording_bufisze / 2

so I think you divide the original buffer size, by the rfft index, and
this simple relation gets the exact period, still not sure.
and the reason half is missing in that calculation ... is not stable
in me but not super important here ... i think it's because the lowest
frequency is the nyquist, which is 2.

so, i guess i'll checking: recording_bufsize / rfft_index = equivalent
period in indices

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