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*Subject*: RE: [amsat-bb] Beginner birds*From*: Bob Bruninga <bruninga@xxxxxxxx>*Date*: Wed, 23 Jul 2003 20:37:32 -0400 (EDT)*In-Reply-To*: <B17EB7B34580D311BE38525405DF623201A9199F@atc-mail-db.atctraining.com.au>

> > ...why is the path loss lower in Mode A than in Mode B/J? The equation for path loss is: [ 4 * Pi * R ]2 [------------] Where R is the Range in meters. [ wavelength ] So space loss goes down as the square of wavelength. That is why 10m is 23 dB better, and 2m is 9 dB better than 70 cm and so on. This equation is fundamental to all satellite communications. It is EASY to figure out a link. The power received at a receiver is just the power transmitted plus antenna gains minus all losses. Of course, that relationship assumes the SAME kind of antenna as the baseline (lets say a dipole). It is obvious when you think about it because as Tony said, a 10m dipole collects the signal from 200 times more area then a 70 cm dipole. But conversly, it is EASIER to get high gain as you go up in frequency. In fact you get it all back if your antennas cover the same AREA. Thus a 10' TV dish will have the SAME link budget to a satellite using an OMNI antnenna no matter what frequency is used... (within reason Usually it needs to be 10 wavelengths across ore more))... Tony said it this way... > The capture area increases as the frequency is lowered. 2m has 9x the > capture area of 70cm, and 10m has over 20x the capture area of the > equivalent 2m antenna. This adds considerably to the link budget. Bob ---- Sent via amsat-bb@amsat.org. Opinions expressed are those of the author. Not an AMSAT member? Join now to support the amateur satellite program! To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

**Follow-Ups**:**Re: Beginner birds***From:*John Santillo

**References**:**RE: Beginner birds***From:*Tony Langdon

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