# Re: R: L-Band Sanity - was "being heard over LEILA siren"

```Hi Dom,

Nice to hear from you, I haven't been on AO-40 much
lately, I hope you have had more time to operate than me!

Ok, I will explain:

The "path loss" is a convenient mathematical convention that
allows us to make system comparisons using standard antenna types,
such as a dipole or an "isotropic radiator." The path loss isn't
a "dissipative" loss like that of a resistor.

I can show how this works with an example:

Lets say the ground station can send 1.5Kw EIRP on 1269 MHz and
150 watts EIRP on 435 MHz. Expressed in dBm0 (where 0 = 1 milliwatt)

+62 dBm0 on 1269 MHz
+52 dBm0 on 435 MHz.

The path loss to a satellite at range 60,000 Km is about:

-190 dB on 1269 MHz
-180 dB on 435 MHz.

A "standard" dipole antenna at any frequency has 2.14 dBi gain
so, the power delivered to a dipole antenna on the satellite from
the ground station would be about:

+62 -190 +2.14 = -126 dBm0 at 1269 MHz
+52 -180 +2.14 = -126 dBm0 at 435 MHz

Notice that the power delivered to the satellite receiving
antenna is the same. If the antennas are not the same at the
satellite, just use the gain of the respective receiving
antenna to calculate the received power.

As Scott pointed out, the AO-40 antennas have about 5 dB
more gain at 1269 vs 435 MHz so if you include this,
the L-band uplink on AO-40 would really be 5 dB greater than
the U-band for 1.5 Kw (L-band) versus 150 w (U-band.)

Now, you are quite correct that the power density is not
a function of frequency so if you know the effective apperture
sizes of the satellite antennas, you can calculate the power
delivered to the satellite without using the "path loss" convention.

Usually, however, the satellite antenna gains are what is given
so it is more convenient to use the path loss convention to correct
for the different apperture sizes of the reference antennas.

Best 73,

Tony AA2TX

---

At 10:27 AM 8/23/03 +0200, i8cvs wrote:

>Hi Tony AA2TX,
>
>If you radiate 1.5 kW  EIRP at 1269 MHz the power density in W/square meters
>at 60,000 km is the same if you radiate the same 1.5 kW EIRP at 435 MHz
>
>The path loss at 1269 MHz is about 10 dB greater (9.299 dB) only because
>the path attenuation is referred to a receiving isotropic antenna wich
>capture area A is:
>
>              /  2
>            / \
>A= -------------           [square meters]
>          4 x Pi
>
>At 1269 MHz the isotropic capture area is A1 =0.00444  square meters
>At   435 MHz the isotropic capture area is A2 =0.0378   square meters
>
>And so 0.00444/0.0378 = 0.1175 or  -9.299 dB
>
>Radiating 1.5 kW EIRP to collect  the full power concentrated in
>1 square meter in 1269 MHz at any distance you need
>1/0.00444 = 224.84 isotropic antennas in phase.
>
>Radiating 1.5 kW EIRP to collect the full power concentrated in
>1 square meter at 435 MHz at any distance you need
>1/0.0378 =26.42 isotropic antennas in phase.
>
>Again 26.42/224.84= 0.1175 or  -9.29 dB
>
>To collect the full power concentrated in 1 square meter at 435 MHz
>you need an antenna with isotropic gain G=26.42  or 14.21 dBi
>
>To collect the full power concentrated in 1 square meter at 1269 MHz
>you need an antenna with isotropic gain G= 224.84 or  23.51 dB
>
>If the gain of your receiving antennas is 14.21 dBi at 435 MHz and
>23.51 dBi at 1269 MHz than you collect the same amount of power
>provided the EIRP is the same at 435 MHz and 1269 MHz
>
>And this is the case on AO40 (about)
>
>In conclusion your assumption that radiating 1.5 kW EIRP at 1269 MHz is
>like running  150 watt EIRP at 435 MHz is incorrect otherwise my question
>is:
>
>Where the -10 dB of losted power is dissipated ? ?
>
>73" de
>
>i8CVS Domenico

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