2011/11/6 Paul Davis :
Thanks for the hint, Paul, and to Fons of course.
An interresting paper, found it here:
Still, I wonder:
Why not compute the time of any sample - relative - to a random start
time (using sample frequency) ?
( The start time would be the time, a recording or playback was started. )
At f = 48000 Hz, I would expect a soundcard to deliver exactly 48000 *
10 = 48E4 samples in 10 seconds.
Is this assumtion wrong ?
If this applied, one could easily calculate the time-position of
Sample S = 96000
based on a random start time (offset) T0 = 10 seconds:
(We know the answer : 10 sec + 2 sec = 12 sec )
1 sec / 48000 Hz * 96000 + 10 sec = 12 sec
(position of sample nr. 96000)
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