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? -- MRM
Any Field Strength Guys Here?
Started by ●December 3, 2022
Reply by ●December 3, 20222022-12-03
On Sat, 3 Dec 2022 17:49:17 -0000 (UTC), Mike Monett VE3BTI <spamme@not.com> wrote:>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?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
Reply by ●December 3, 20222022-12-03
On 3.12.2022 19.49, Mike Monett VE3BTI wrote:> 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?Calculate the surface area of a sphere of 15 km radius and divide your transmitting power onto it. To compare, the radiating power of a GPS satellite is about 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. -- -TV
Reply by ●December 3, 20222022-12-03
On Saturday, December 3, 2022 at 12:49:25 PM UTC-5, Mike Monett VE3BTI wrote:> 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?50 dBuV/m ( this is called field strength) -65 dBm/m^2 (this is called power density, power per unit area perpendicular to propagation vector) Convert to numerical mW Multiply by lambda ( in meters) squared over 4 pi to derive power actually intercepted by receive antenna. If it's not isotropic receive antenna, multiply that number by the antenna gain. This is called baby-techno: https://www.phys.hawaii.edu/~anita/new/papers/militaryHandbook/pwr-dens.pdf> > > > > > -- > MRM
Reply by ●December 3, 20222022-12-03
On Saturday, December 3, 2022 at 2:58:09 PM UTC-5, Tauno Voipio wrote:> On 3.12.2022 19.49, Mike Monett VE3BTI wrote: > > 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? > Calculate the surface area of a sphere of 15 km radius > and divide your transmitting power onto it. > > To compare, the radiating power of a GPS satellite is about > 500 W, including antenna gain. The radius of its sphere is > about 20000 km.Where are you getting that 500W bullshit? This source says 45W. https://www.nxp.com/docs/en/brochure/75016740.pdf This says the same thing: https://www.waterboards.ca.gov/water_issues/programs/swamp/docs/cwt/guidance/6120.pdf A whole handbook here: https://www.gps.gov/technical/icwg/IS-GPS-200D.pdf> > 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. > > -- > > -TV
Reply by ●December 3, 20222022-12-03
On Saturday, December 3, 2022 at 11:58:09 AM UTC-8, Tauno Voipio wrote:> On 3.12.2022 19.49, Mike Monett VE3BTI wrote: > > 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? > Calculate the surface area of a sphere of 15 km radius > and divide your transmitting power onto it.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'.
Reply by ●December 3, 20222022-12-03
whit3rd <whit3rd@gmail.com> wrote in news:d851b90a-4faa-493f-a1ca-fc0d1afaec5en@googlegroups.com:> On Saturday, December 3, 2022 at 11:58:09 AM UTC-8, Tauno Voipio > wrote: >> On 3.12.2022 19.49, Mike Monett VE3BTI wrote: >> > 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? >> Calculate the surface area of a sphere of 15 km radius >> and divide your transmitting power onto it. > > 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'. >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.
Reply by ●December 3, 20222022-12-03
In article <d851b90a-4faa-493f-a1ca-fc0d1afaec5en@googlegroups.com>, whit3rd <whit3rd@gmail.com> wrote:>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'.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.
Reply by ●December 3, 20222022-12-03
On Saturday, December 3, 2022 at 6:00:51 PM UTC-5, Jasen Betts wrote:> On 2022-12-03, Fred Bloggs <bloggs.fred...@gmail.com> wrote: > > On Saturday, December 3, 2022 at 2:58:09 PM UTC-5, Tauno Voipio wrote: > >> On 3.12.2022 19.49, Mike Monett VE3BTI wrote: > >> > 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? > >> Calculate the surface area of a sphere of 15 km radius > >> and divide your transmitting power onto it. > >> > >> To compare, the radiating power of a GPS satellite is about > >> 500 W, including antenna gain. The radius of its sphere is > >> about 20000 km. > > > > Where are you getting that 500W bullshit? This source says 45W. > > https://www.nxp.com/docs/en/brochure/75016740.pdf > 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.I see it now. Looks like they upgraded the L-band antennas a few years ago to broaden the pattern 3dB knees to cover the entire hemisphere of the Earth with a single interception, with a line of sight elevation no less than 5o. https://www.gpsworld.com/gps-iir-iir-m-satellite-antenna-patterns-released-for-worldwide/ In that 3-D illustration they list the distance to satellite as 26,553 km, but that's from Earth center. They list Earth radius as 6,378 km, so that leaves 26,553-6378=20,175 of propagation distance to Earth's surface. NPX article says 20,200, so take that as a match. Then the angle subtended by the Earth as seen from the satellite is 2x6378/26,553x180/pi=27.5o, confirming they're illuminating an entire hemispheric surface. That 44.8W at 12dBi makes for xmit power of 10Log(44.8 x 1000) + 12dBi=58.5 dBm, over the Earth hemisphere with range loss 10Log((20200 x 1000)^2x4pi) (range squared)= 157 dB, making for an incident power of 58.5-157=-98.5 dBm. Then plugging into the receive antenna with 4 dBi gain and taking lambda to be 0.19m ( at 1575 MHz nominal) gets -21dB receive antenna aperture and directivity gain, Gain x Lamda squared/4pi. Total input power to the receiver is then or -98.5 dBm -21= -120dBm as stated in NPX article, which is a match. Add in another 6dB atmospheric loss for -125dBm, not that even 6dB is all that significant. This was a calculation for a receiver looking straight up, 90o elevation, and the power levels won't be that much different at the Earth edges- distance is not that significant but beam pattern being down 3dB is, a little. Fact check on NPX write-up: TRUE https://www.nxp.com/docs/en/brochure/75016740.pdf Thanks for your input.> > > > > > -- > Jasen.
Reply by ●December 3, 20222022-12-03
On Saturday, December 3, 2022 at 6:00:51 PM UTC-5, Jasen Betts wrote:> On 2022-12-03, Fred Bloggs <bloggs.fred...@gmail.com> wrote: > > On Saturday, December 3, 2022 at 2:58:09 PM UTC-5, Tauno Voipio wrote: > >> On 3.12.2022 19.49, Mike Monett VE3BTI wrote: > >> > 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? > >> Calculate the surface area of a sphere of 15 km radius > >> and divide your transmitting power onto it. > >> > >> To compare, the radiating power of a GPS satellite is about > >> 500 W, including antenna gain. The radius of its sphere is > >> about 20000 km. > > > > Where are you getting that 500W bullshit? This source says 45W. > > https://www.nxp.com/docs/en/brochure/75016740.pdf > 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.Getting back to that 1W isotropic radiator at 15km, incident on the -21dB gain antenna ( 4dBi gain and L-Band wavelength) gives an receiver input power of -65 dBm -21 dB= -86 dBm.> > > > > > -- > Jasen.
