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Message-ID: <000701c79f80$2a56a460$c4227ad5@w4o8m9>
From: "James Moritz" <james.moritz@btopenworld.com>
To: <rsgb_lf_group@blacksheep.org>
References: <000701c79ef8$58d0b7a0$42e8fc3e@g3kev>
Date: Sat, 26 May 2007 11:25:21 +0100
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Subject: LF: Re: 500 khz ERP
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Dear Mal, LF Group,


> To justify my ERP argument and keep the maths simple, look at examples
> below.
>
> 1. 10W rf output to the antenna and an efficiency of the average antenna
1%
> the ERP would be 0.1W erp...

The maths is not quite as simple as this - to calculate ERP in this way, the
directivity of the antenna compared to the reference 'half-wave dipole in
free space' must also be included:

ERP = (TX power) x (Antenna Efficiency) x (Directivity)

Assuming the antenna is a vertical over a ground plane that is a small
fraction of a wavelength long (or a small loop over a ground plane), its
directivity can be assumed to be 2.6dB over a dipole, i.e. a factor of 1.8.
So the ERP in your example would be 0.18W ERP. It is important to bear this
in mind in correspondence with Ofcom!

Cheers, Jim Moritz
73 de M0BMU