Am 12.10.2021 um 19:29 schrieb Michael Koch:
Am 12.10.2021 um 18:52 schrieb Paul B Mahol:
On Tue, Oct 12, 2021 at 10:55 AM Michael Koch
<astroelectro...@t-online.de>
wrote:
I have a question about the "fftfilt" filter. What's the default value
of the weight_U and weight_V options? I'm asking because I get an
unexpected result.
This command line creates my input image for testing:
ffmpeg -f lavfi -i color=black:s=300x50 -lavfi
drawgrid=c=white:y=-1:w=2:h=51,split[a][b];[b]crop=iw/2:x=0,scale=2*iw:ih:flags=neighbor,split[b][c];[c]crop=iw/2:x=0,scale=2*iw:ih:flags=neighbor,split[c][d];[d]crop=iw/2:x=0,scale=2*iw:ih:flags=neighbor,split[d][e];[e]crop=iw/2:x=0,scale=2*iw:ih:flags=neighbor,split[e][f];[f]crop=iw/2:x=0,scale=2*iw:ih:flags=neighbor[f];[a][b][c][d][e][f]vstack=6,split[h][v];[v]transpose[v];[v][h]hstack
-frames 1 -y test.png
This is the fftfilt lowpass example from the official documentation:
ffmpeg -i test.png -vf fftfilt=dc_Y=0:weight_Y='squish((Y+X)/100-1)' -y
out1.png
Problem: The output has a greenish tint.
Expressions by default for U and V are copied from Y if are unset.
filter works only in YUV or gray space thus in above combination one
gets
green tint.
I'm trying to make the filter's cutoff frequency independant of the
image size. But that's not so easy because I don't know the size of the
FFT array. It's calculated in vf_fftfilt.c lines 185 and 297.
This calculation is difficult (and slow) to replicate in an expression,
because either a loop or a logarithm is required.
Would it be possible to add two new variables so that the FFT array size
can be used in an expression?
ARRAY_H = 1 << rdft_hbits
ARRAY_V = 1 << rdft_vbits
It the array size is known, things would become much easier.
Thanks,
Michael
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