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
<mailto: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
<mailto: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
<mailto: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
--
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