To: | "[email protected]" <[email protected]> |
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Subject: | Re: LF: AW: 8.97kHz - Near field and radiated signals? |
From: | Chris Trayner <[email protected]> |
Date: | Wed, 26 May 2010 09:57:25 +0100 |
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In-reply-to: | <BF4A524700075746A6467658DFC7102C5106464EEA@ICTS-S-EXC2-CA.luna.kuleuven.be> |
References: | <[email protected]> <38A51B74B884D74083D7950AD0DD85E82A1C00@File-Server-HST.hst.e-technik.tu-darmstadt.de> <[email protected]> <[email protected]>,<[email protected]> <BF4A524700075746A6467658DFC7102C5106464EEA@ICTS-S-EXC2-CA.luna.kuleuven.be> |
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Thread-topic: | LF: AW: 8.97kHz - Near field and radiated signals? |
Hi Rik, > the near fiels signal drops with the 3rd power of the distance. I think you'll find it's the H field that goes as the inverse 3rd power. The voltage in the Rx ant is proportional to this, and the power is proportional to the square of that. Thus the power, which is what dBs describe, goes as the inverse 6th power. > So 3km versus 0.4km = 422 times weaker = -26dB In other words you need to convert your 422 not as 10 log(422) but as 20 log(422), which gives -52dB. Sorry to be the bringer of bad news ;-( Roger also said > I am confused by the near field attenuation rates quoted. Some quote 18dB for > doubling distance This fits: double the distance is 1/8 H field, 1/64 power, 18dB. 73, Chris G4OKW ----------------------- Dr Chris Trayner School of Electronic & Electrical Engineering, The University of Leeds, Leeds LS2 9JT, United Kingdom Tel: +44 113 34 32053 Fax: +44 113 34 32032 |
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