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Date: Thu, 25 Feb 2010 11:22:47 +0000
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From: Andy Talbot <andy.g4jnt@googlemail.com>
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>From a curve in an old 1981 copy of "Reference Data for Radio Engineers" ,
at 10kHz (lowest value shown) atmospheric noise is given as being in the in
the range 155 to 175 dB kToB.   This will have risen over the intervening
years due to the proliferation of industrial / domestic electrical
junk-noise, so if just we take the upper figure it shouldn't be outrageous
to start with.

In a reference 1Hz bandwidth kTB = -174dBm, into a theoretical 0dBi
antenna.  Therefore, you could expect to see about 0dBm, or about 1mW of
noise from such a beast

To get the V/m value from this, first calculate the effective area of a 9kHz
isotropic antenna :
At 9kHz, lambda = 33333m,   G = 4.pi.A/lambda^2,   G = 1,   so  A =
88*10^6  m^2

1mW of noise per Hz bandwidth, received in this aperture means noise density
Nd = is  1.1E-11 W/m^2
Field strength E volts/metre, == SQRT(Nd. 377)
= 65uV/m in 1 Hz bandwidth,    or 65uV/ (ROOT Hz)

So your 2.5uV/m is now 28dB below the external noise in a 1Hz bandwidth

Now, if you go to a 1mHz bandwidth and use QRSS .001, you will just about
detect the signal.   A callsign is, say, 50 dot symbols long, so expect to
take about 14 hours to send your callsign at teh power levels quoted.

Now, if you increase your Tx power by 30dB, communication in 1Hz bandwidth
become feasible, always assuming atmospheric noise levels aren't even higher
these days than those curves state.   Does anyone have access to a later set
of atmospheric noise measurements ?

So I'm afraid 9kHz communication at power levels we could generate
domestically is really not all that easy.    When I applied for the
unsucessful NoV a few yerars ago, I had intended from the start using a
round loop, and coherent signalling in probably 0.01Hz bandwidth.   That
would allow soundcard users to receive directly in more realistic timescales
- probably at no more than a few km

Andy
www.g4jnt.com


On 25 February 2010 17:53, Alexander S. Yurkov <fitec@omskcity.com> wrote:

> Hellow, Stefan.
>
> > If that calculation is reasonably correct, what distance could be
> > reached with 1,4mW @ 8,9 kHz
>
> With such a condition you'll get about 2.5 uV/m at 100 km. Seems it can be
> recievid. If there is no atmospheric and industrial noise it should be
> very strong signal. But all depends on noise on 9 kHz. I have no ideas
> about noise on 9 kHz. I neglet ionosphere in estimations. But at D=100 km
> it should be approximately right.
>
>

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<div>From a curve in an old 1981 copy of &quot;Reference Data for Radio En=
gineers&quot; , at 10kHz (lowest value shown) atmospheric noise is given=
 as being in the in the range 155 to 175 dB kToB.=A0=A0 This will have ris=
en over the intervening years due to the proliferation of industrial /=A0d=
omestic electrical junk-noise, so if just we take the upper figure it shou=
ldn&#39;t be outrageous to start with.</div>

<div>=A0</div>
<div>In a reference 1Hz bandwidth kTB =3D -174dBm, into a theoretical 0dBi=
 antenna.=A0=A0Therefore, you could expect to see about 0dBm, or about 1mW=
=A0of noise=A0from such a beast</div>
<div>=A0</div>
<div>To get the V/m value from this, first calculate the effective area of=
 a 9kHz isotropic antenna=A0:</div>
<div>At 9kHz, lambda =3D 33333m,=A0=A0 G =3D 4.pi.A/lambda^2,=A0=A0 G =3D=
=A01,=A0 =A0so=A0 A =3D=A0 88*10^6=A0 m^2</div>
<div>=A0</div>
<div>1mW of noise per Hz bandwidth, received in=A0this aperture=A0means no=
ise density Nd =3D is=A0 1.1E-11 W/m^2</div>
<div>Field strength E volts/metre, =3D=3D SQRT(Nd. 377)</div>
<div>=3D 65uV/m in 1 Hz bandwidth,=A0=A0=A0 or 65uV/=A0(ROOT Hz)</div>
<div>=A0</div>
<div>So your 2.5uV/m is now 28dB below the external noise in a 1Hz bandwid=
th</div>
<div>=A0</div>
<div>Now, if you go to a 1mHz bandwidth and use QRSS .001, you will just=
 about detect the signal.=A0=A0 A callsign is, say, 50 dot symbols long,=
 so expect to take about 14 hours to send your callsign at teh power level=
s quoted.</div>

<div>=A0</div>
<div>Now, if you increase your Tx power by 30dB, communication in 1Hz band=
width become feasible, always assuming atmospheric noise levels aren&#39;t=
 even higher these days than those curves state.=A0=A0 Does anyone have ac=
cess to a later set of atmospheric noise measurements ?=A0=A0=A0 </div>

<div>=A0</div>
<div>So I&#39;m afraid 9kHz communication=A0at power levels we could gener=
ate domestically is really not all that easy.=A0=A0=A0 When I applied for=
 the unsucessful NoV a few yerars ago, I had intended from the start using=
 a round loop, and coherent signalling in probably 0.01Hz bandwidth.=A0=A0=
 That would allow soundcard users to receive directly in more realistic ti=
mescales - probably at no more than a few km</div>

<div>=A0</div>
<div>Andy<br><a href=3D"http://www.g4jnt.com">www.g4jnt.com</a><br><br><br=
></div>
<div class=3D"gmail_quote">On 25 February 2010 17:53, Alexander S. Yurkov=
 <span dir=3D"ltr">&lt;<a href=3D"mailto:fitec@omskcity.com">fitec@omskcit=
y.com</a>&gt;</span> wrote:<br>
<blockquote style=3D"BORDER-LEFT: #ccc 1px solid; MARGIN: 0px 0px 0px 0.8e=
x; PADDING-LEFT: 1ex" class=3D"gmail_quote">Hellow, Stefan.<br>
<div class=3D"im"><br>&gt; If that calculation is reasonably correct, what=
 distance could be<br>&gt; reached with 1,4mW @ 8,9 kHz<br><br></div>With=
 such a condition you&#39;ll get about 2.5 uV/m at 100 km. Seems it can be=
<br>
recievid. If there is no atmospheric and industrial noise it should be<br>=
very strong signal. But all depends on noise on 9 kHz. I have no ideas<br>=
about noise on 9 kHz. I neglet ionosphere in estimations. But at D=3D100=
 km<br>
it should be approximately right.<br><br></blockquote></div>

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