Bourdeau, pipolin, pourriez vous m'expliquer que quoi s'agit ? :-)
31 réponses
kowalski
Back to ITU
We have now have the 576 active lines at 52 =CE=BCs per line length, in
which the picture
information is contained, which we need for our digital final product.
=E2=97=8F 576 lines a 52=CE=BCs
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 =CE=BCs "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64=CE=BCs / 864 =3D 74.074ns
So, in a line of 64 =CE=BCs there are 864 samples with a =E2=80=9Clength=E2=
=80=9D of 74.074
ns each.
=E2=97=8F Period duration for a single pixel =3D 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 =3D 53.3333=CE=BCs
The "window" suggested by the ITU is thus larger, than the 52 =CE=BCs
existing video
information.
=E2=97=8F - 720 pixel =3D 53.3333 =CE=BCs
We now divide the "genuine" active 52 =CE=BCs by the calculated period
duration/pixel:
52=CE=BCs / 74.074ns =3D 702.0007 =3D ~702 pixel
There are only 702 active pixels!
