Jean Marc:

I think it would be impossible (or at least difficult) to rescale the noise
without also
rescaling the signal levels in the same bandwidth.  Even though you can tell
them apart, the
signal processing algorithms can not, so it would introduce errors in the
reported levels of the incoming signals.  Currently, the panadaptor
accurately
reports the signal levels, and the noise floor that a receiver would see if
it was
set to the same bandwidth.

We normally use 0.5 kHz (500 Hz) for the 'normalizing filter width for noise
measurement."
It is the normal choice for reporting amateur radio sensitivity
specifications, and has been
used on all worldwide noise level charts I have seen from the ITU.

Best Regards,
--- Graham / KE9H

==

On Fri, Jul 1, 2011 at 1:28 AM, F1HDI <f1...@orange.fr> wrote:

> **
> Hello,
> Thanks for the explanations, now I have another silly question :
> Why not normalizing the noise floor value to either 1Hz (consistant with
> the -174dBm/Hz) or 2,5kHz which in general is the 'normalizing filter width
> for noise measurement'.
> I would vote for the 1Hz norm , I understand that it will expand the dBm
> axis of the panadapter/panafall by approx 13dB but , at least comparison
> could be made on the noise floor in between f1,5k, f3000 and f5000.
> 2,5kHz would give a noise floor greater than usual one with cw,ssb filters
> so will be useless on a graphical display.
>
> What would you bargain ?.
> Kind regards
> Jean-marc
>
> Le 29/06/2011 15:53, Robert McGwier a écrit :
>
> 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 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
>> of the main part of the filter.  This does not account for the filter edges
>> perfectly but the difference is minuscule because of the tremendous shape
>> factor on the filter.
>>
>> Great discussion!
>> Bob
>> N4HY
>>   On Jun 28, 2011 7:55 PM, "Graham Haddock" <gra...@flexradio.com> wrote:
>> > Hello Jean-Marc:
>> >
>> > Not a silly question at all. In fact, a great question.
>> >
>> > The panadaptor is fixed at 4096 bins. Each bin is essentially a receiver
>> > with bandwidth
>> > equal to the sample rate divided by 4096.
>> >
>> > Example: for a FLEX-1500, which has sample rate of 48,000 samples per
>> > second,
>> > the bin bandwidth is 48000/4096 = 11.7 Hz
>> >
>> > The noise level is lower on the panadaptor, because of the lower (bin)
>> > bandwidth.
>> > The difference in dB relative to the "S-Meter" window can be calculated
>> as
>> > the log of the difference in bandwidths.
>> >
>> > Example: The difference in noise level seen on the S-Meter, receiving
>> > through
>> > a CW filter of 500 Hz, and the panadaptor on a FLEX-1500 would be:
>> >
>> > dB = 10*LOG(Bandwidth-of-Receiver / Bandwidth-of-Panadaptor-Bin)
>> > = 10 * LOG (500/11.7) = 16.3 dB
>> >
>> > So the panadaptor noise floor will appear to be 16 dB lower than the S
>> Meter
>> > value with a 500 Hz filter selected.
>> >
>> > This only applies to noise. The level of a narrow signal (one that will
>> fit
>> > entirely inside
>> > of your filter) is independent of the filter bandwidth, so the level
>> shown
>> > is the true level
>> > and no correction difference applies.
>> >
>> > --- Graham / KE9H
>> > FlexRadio Systems
>> >
>> > ==
>> >
>> >
>> > On Tue, Jun 28, 2011 at 11:14 AM, F1HDI <f1...@orange.fr> wrote:
>> >
>> >> Hello,
>> >>
>> >> In Powersdr, how is computed the values of the noise floor across the
>> >> spectrum in spectrum, panafall modes ?.
>> >> To be more specific, those values relate to which bandwitdh ?.
>> >>
>> >>
>> >> Kind regards
>> >> Jean-marc F1HDI
>> >>
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>
>
>
> --
> Bob McGwier
> ARS: N4HY
>
>
>
> --
> 73's from F1HDI *http://www.f1hdi.org*
>
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