# Any Field Strength Guys Here?

• posted

I am trying to find the field strength of a 1 Watt (30 dBm) signal with an isotropic antenna at a distance of 15 kilometers.

I need to compare this to a GPS signal of -125 dBm.

Anyone know how to find this?

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Well, note that GPS signals are below ambient noise floor on Earth's surface. So, the solution must include a large dose of correlation gain. Like from a lock-in amplifier et al.

Joe Gwinn

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Calculate the surface area of a sphere of 15 km radius and divide your transmitting power onto it.

500 W, including antenna gain. The radius of its sphere is about 20000 km.

The calculations give power per square meter. If you need to have the more conventional measurement in volts/m, just calculate the voltage to get the power into 377 ohms.

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If the radiation is 'isotropic' in the ground plane, it might have a factor of three more power intensity than that calculation recognizes, because it's a vertical dipole. It isn't much gain, but that's the implication I'd draw from 'isotropic'.

• posted

says "It’s transmit power is 44.8 Watt at 1575.43 MHz and the antenna gain is 12 dBi."

Thus about 500W as an isotropic radiator as seen from inside the antenna beam.

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whit3rd snipped-for-privacy@gmail.com wrote in news: snipped-for-privacy@googlegroups.com:

The GPS signal is the lowest power signal in use today.

-125dB down That is a very small signal femtowatts... likely less. It sits right above baseline noise. Amazing that we can discern it at all.

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Traditionally, "isotropic" is used to refer to antennas which are truly isotropic, radiating equal power in all directions. This is a useful fiction - no such antenna actually exists - but it's the "least common denominator" in antenna patterns because it has no directional bias at all.

A vertical dipole would (in principle) have 2.15 dB of gain (in its preferred direction) over an isotropic reference.

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467uV/m

piglet

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Fantastic. That is a lot more than I expected. Very interesting. Can you show your calculations?

If my calculations are correct, that is (467e-6)^2 / 377 = 5.78e-10 W/m = -62.381 dBm, or 63 db above the GPS signal.

Perfect. Thanks.

• posted

The one I know is E = (7*sqrt(P))/d which gives higher results so tends to get used for exposure compliance calcs

E in V/m; P in W; d in m

1W at 15km = 467uV/m

see:

But there is another formula often seen which is ballpark same:

E = sqrt(30*P)/d

1W at 15km = 365uV/m

Real life may be wildly different.

Isn't this stuff Ham Radio 101 that you needed to get your call?

piglet

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This is very valuable information. Thank you for taking the time, and for the links.

I got my first license in 1958. There was not much to the exam, mostly regulations and a code test at 5 wpm. A kid could have passed it. Come to think of, I was a kid then:)

Thanks,

Mike

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Many licenses are a joke. I got my first class license radiotelephone when I was 22 and had never seen a TV transmitter, but was licensed to work on them.

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My FCC /Restricted/ Radiotelephone Operator Permit allows me to operate aircraft radios and instruments. It doesn't license me to work on them. Danke,

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