-----Original Message-----
From: Andy Talbot [mailto:[email protected]]
Sent: 25 February 2010 11:23
To: [email protected]
Subject: Re: LF: AW: 9 Dreamers

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 <[email protected]> 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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