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| To: | [email protected] |
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| Subject: | LF: Estimating CFH's radiated power |
| From: | [email protected] |
| Date: | Sun, 21 Nov 2004 17:44:29 EST |
| Reply-to: | [email protected] |
| Sender: | [email protected] |
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Dear John and group, ok, I'll try... One kilowatt EMRP across 700 km of lossless flat earth would give 109.6 - 20 log (697) dBµV/m = 52.6 dBµV/m. The CCIR 1963 graph for seawater (4 S/m) says 45 dBµV/m at 150kHz, so we have lost ~8dB, mostly to the spherical diffraction. As a first approximation, we could take that and then simply add the extra ground loss for 200 km. Extracting this from the 1mS/m curve gives another 8dB, but I'd use 6dB as a realistic guess for 1.6mS/m conductivity. Thus you'd get 39 dBµV/m from a kW, give or take a couple of dB. Your measurement of 235 µV/m = 47.4 dBµV/m would thus indicate an EMRP of 8.4 dBkW = 7 kW, more or less in line with Alan's suggestion that it could be somewhat less than their 20kW maximum rating. Through ~ 4500 km of free space, CFH could theoretically produce 109.6 + 8.4 - 73 dBµV/m = 45 dBµV/m in Europe. At 20:00 today the plot beneath my grabber showed around 4 dBµV/m, which means 41dB path loss for reflections and D-layer absorption. An amateur at comparable distance using full 1W ERP (ie. 0.55W EMRP) would be 41 dB weaker than 7kW CFH. To copy him here in a bandwidth of 10.5mHz (QRSS 60), the SNR of CFH in my plot would have to climb from currently ~40dB to at least ~50dB. 73 and best wishes Markus, DF6NM In einer eMail vom 21.11.2004 21:12:14 Westeuropäische Normalzeit schreibt [email protected]: Markus, |
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