Convert spectrograms into best approximation of original audio.

takes:
	image filename
	two points indicating start and end of colour scale
	two values in -dB for the endpoints of the colours scale
		(or take range and normalise maximum to 0dB?)
	minx maxx miny and maxy of the spectrogram area
	length of the clip in seconds
	minimum Y frequency and hertz-per-pixel-step
then
	extract the colour scale and make some kind of reverse mapper to db
	reverse FFT each pixel column (with what window size?)
	reconstruct audio by fading between FFT outputs

e.g. Lowell, using a coordinate system with (0,0) at top left
	colour scale runs from -7 to -56dB at (29-41(say 32), 39-602)
	    and bands of 3 or 4 pixels have the same colour.
	graph is from (91,0) to (1013,622): entire image minus leftmost 91 pix)
	frequency scale is from 11Hz at y=622(=maxy) to 6808 at y=30.
	    (6808-11)/(622-30) = 11.48141891891891891891 Hz per pixel
	    ~= 44100/3841
	    Top of graph represents 6808+(11.48*30) = 7152.4
	    Each Hz value in the scale applies to the pixel row of the white
	    line under the digits. Here it starts at bin 1 omitting 0Hz.
	time is 0'32"-4'32" = 4x60 secs = 240 seconds in (1014-91=923) pixels
	    = 240/923=.2600216684 seconds per pixel (3.846 pix per sec)

e.g. Pot Au Feu early version
	colour scale runs from -9 to -67dB
	graph is from (91,0) to (1036,642): entire image minus leftmost 91 pix)
	frequency scale is marked from 46 at y=642 to 3515 at y=20
	    (3515-46)/(642-20) = 5.57717041800643086816 Hz/pixel
	    Top of graph represents 3515+(5.577*20) = 3626.54 Hz
	    BUT 46 is not a multiple of 5.577
		*8=44.61736334405144694528
		*9=50.19453376205787781344
	    so they do not necessarily start at the first bucket above 0Hz
	    nor is the first row necessarily a multiple of the difference
	    between bucket frequencies.
	    So we have to use a greater FFT size and either
	    interpolate between the buckets present in the input or
	    just fill in the frequencies in the FFT input that are
	    closest to the axis values of the input buckets (or the ones
	    either side of the exact mapped frequency.
	time is 2'51"-1'21"=1:30, 90 seconds in 1037-91=946 columns
	    90/(1037-91) = .09513742071881606765 seconds per pixel column
			 PPSEC = 10.511111111111111111 

Then for each pixel column, reverse map each pixel colour and un-dB it to obtain
the real values of an FFT taken every .26 seconds.

Reconstruct audio:
- set IFFT window size to twice .26002 seconds and fade between adjacent outputs
    using Hann window (which is cos^2 so the sum of adjacent windows == 1).
    e.g. Lowell column hop =
- Calculate phase values for each bin of each FFT such that a sine wave at
    the same frequency in two adjacent FFTs coincide in phase, then
    recreate real/imag from magnitude ready for inverse FFT. *Then* fade
    between outputs using a window function. The audio output resulting from
    the same bin in successive IFFTs should then sum in-phase.
    Phase value for a bin at f Hz at time t secs = t * f * 2_PI radians

The graph's time resolution is determined by the number of pixel columns per
second. For forward FFTs, we take an arbitrary-width window of samples from an
arbritrary section of the audio. If columns-per-second is less than the window
size, some samples do not count at all towards the output, so to give all
audio samples equal weigh towards the output, the FFT size should be twice
the distance in time between output pixel columns, or FFT_FREQ=PPSEC/2.

In the Lowell example,
IFFT window_length = 2 * column_interval = 2*.26002 = .52004 seconds.
At 44100 sampling rate, this makes the FFT size 22934 samples.
There are 623 spectro values from 11Hz to 7152.4Hz, one per 11Hz, while IFFT
values represent 0 to 22050Hz in 22934 samples, about one per Hz, giving one
input data point value every 11 FFT data points.

In the Pot Au Feu early version example,
IFFT window_length = 2 * column_interval = 2*.09513742 = .19027484 seconds.
At 44100 sampling rate, this makes the FFT size 8391.12 samples.
There are 643 spectro values from 46Hz to 3626.54Hz, one per 5.577Hz while IFFT
values represent 0 to 22050Hz in 8391 samples, about one per 2.63Hz, giving
roughly one input data point every other FFT data point.

Assume IFFT data points are sparser than our IFFT size, at least at 44100.

PROGRAM STRUCTURE
=================
Shell script crops out the db scale and the raw spectrogram and does
calculation of parameters for C routine.
C routine takes the raw spectrogram and raw frequency scale map as images
plus coordinates of start/end/length/minmax values in db and freq scales etc.
and writes a wav.

FURTHER DEVELOPMENT
===================
Better-quality output can probably be obtained using the Griffin-Lim algorithm
for the phase at each point.
Brief description:
https://www.quora.com/What-are-intuitive-explanations-of-the-Griffin-Lim-Algorithm
Python code:
https://github.com/bkvogel/griffin_lim
See also:
https://stackoverflow.com/questions/61132574/can-i-convert-spectrograms-generated-with-librosa-back-to-audio