Or put another way, if the filter is 10 bins wide (117 Hz = 10*4096/48000)
the noise power in that 117 Hz is the sum of the noise lower in those ten
bins and thus should be ten times or 10 dB bigger than the noise floor.
This is what the meter reads: all the power, noise + signal, in the ten bins
On Wed, Jun 29, 2011 at 5:48 AM, Robert McGwier rwmcgw...@gmail.com wrote:
...Great discussion!
It certainly has been a great discussion, very informative. Thanks to
Jean-Marc for bringing this up, and to Graham and Bob for their
explanations.
Tony KT0NY
Forgive the typo, 117 = 10 * 48000/4096 Hz.
On Wed, Jun 29, 2011 at 6:48 AM, Robert McGwier rwmcgw...@gmail.com wrote:
Or put another way, if the filter is 10 bins wide (117 Hz = 10*4096/48000)
the noise power in that 117 Hz is the sum of the noise lower in those ten
bins and thus should be
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