Hmmm... could be this: if the FFT uses the *other* definition than the one in the wiki, which is to put a 1/sqrt[N] in front of both the transform and its inverse, then a bandwidth-limited signal (say a pure sin) would have an amplitude twice as large for the 4096 as for the 1024 (4/sqrt[4]=2). However, if you FFT the *same* white noise signal, once with 1024 and once with 4096 samples, there will be a factor of 4 in the amplitude of the 1024 FFT because each frequency "bin" is four times as large (that's a kind of aliasing) - so in the end the 1024 FFT will have twice the amplitude of the 4096.
Anyway there are clearly lots of factors of 2 floating around - if that's not the right explanation, hopefully it puts you on the right track. -- opaqueice ------------------------------------------------------------------------ opaqueice's Profile: http://forums.slimdevices.com/member.php?userid=4234 View this thread: http://forums.slimdevices.com/showthread.php?t=34206 _______________________________________________ audiophiles mailing list [email protected] http://lists.slimdevices.com/lists/listinfo/audiophiles
