Neat stuff to be sure and I appreciate the math. So, as I understand it from my limited knowledge, is that once we know the frequencies up and down, then designing the ground station antenna will be easier. My sense is that a 3m dish, with the right slewing and control will be able to *hear* the orbiter at it's closest point to earth without experiencing too many dropouts.
CW with a BW of 500Hz appears to be the mode of choice -that seems reasonable to me, as does PSK- and the craft won't be in peril power-wise to do that.
I still don't understand, albeit from a novice viewpoint, why the craft and the radio system can't be tethered making it two separate units in a way. The craft can angle for it's purpose and so can the radio. Is that possible, given the state of craftsmanship or art we have?
Dave
----- Original Message ----- From: "i8cvs" domenico.i8cvs@tin.it To: "Bob Bruninga" bruninga@usna.edu; "'Joe'" nss@mwt.net; "'Edward Cole'" kl7uw@acsalaska.net Cc: "'AMSAT-BB'" amsat-bb@amsat.org; "'G0MRF David Bowman'" g0mrf@aol.com Sent: Saturday, July 05, 2008 11:20 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
----- Original Message ----- From: "Robert Bruninga" bruninga@usna.edu To: "'Joe'" nss@mwt.net; "'Edward Cole'" kl7uw@acsalaska.net Cc: "'AMSAT-BB'" amsat-bb@amsat.org; "'G0MRF David Bowman'" g0mrf@aol.com Sent: Saturday, July 05, 2008 4:09 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
The whole part that is confusing me on all this power budget stuff is the to me, the seemingly HIGH budget. I've done moon bounce. And many of these numbers seem to be not too far from Moonbounce numbers, and that is a horrid dead piece of rock reflector. that has a efficiency of a wet sponge. ...And it only reflects 6% of the energy it gets.
My guess is ... That 6% is an awful lot of power considering the 3.6 million square miles of surface doing the reflecting. Conversly, any amateur transmitter at the moon would have a much smaller receiving/transmitting antenna. Though lots more concentrated power.
So what you gain in changing from a 1/R^4 to a 1/R^2 path loss you lose a lot of it in the loss of signal receive aperture. Or something like that maybe.
Bob, WB4APR
but i would think anything there that is active circutry is a thousand times more efficient at sendinga signal back as compared to the moons surface. or what am I missing?
Joe
Hi Bob, WB4APR
Your guess is.......absolutely correct.
I also did 432 MHz EME from 1977 to 1980 and I will try to demonstrate in more datails to Joe that your analysis hit the centre of his question.
Hi Joe
Suppose to be in the center of a sphere with radius of 380.000 km that is the average distance from the earth to the moon.
The internal surface S of the above sphere computed in square meters is:
6 2 18S= 4 x 3,14 x ( 380 x 10 ) = 1,81 x 10 square meters
Suppose now to have in your hand an isotropic antenna radiating all around and uniformly the power P = 1 watt at 432 MHz
As soon the wave has reached the internal surface of the above sphere the full power of 1 watt will be collected on it so that the power density D collected in each square meter is:
1 -19D = ------------------------ = 5,52 x 10 watt / square meter 18 1,81 x 10
But in one point of the above sphere there is the disc of the moon which radius is 1735 km =1735000 meters and so the surface S1 of the lunar disc is: 2 12 S1 = 1735000 x 3,14= 9,45 x 10 square meters
The full power density P1 collected over the disc of the lunar surface will be D x S1 and so
-19 12P1= 5,52 x 10 x 9,45 x 10 = 0,0000052164 watt
Only the 7% of P1 at 432 MHz is reflected back by the lunar surface and very important the reflected power P2 is reirradiated and scattered back "isotropically" by the lunar disc and so the reflected power is
P2=(0,0000052164 / 100) x 7= 0,0000003651 watt
Now P2 make another trip of 380.000 km from the moon to the earth but actually the power P3 collected by each square meter over the earth surface will be only:
0,0000003651 -25P3= ----------------------- = 2,017 x 10 watt / square meter 18 1,81 x 10
Since we have in our hand an isotropic antenna at 432 MHz originally radiating 1 watt we want to know what actually is the power Pr received back from the moon into the same isotropic antenna.
The aperture area A of an isotropic antenna at 432 MHz i.e. at a wavelenght of 0,6944 meters is:
/ 2 2 /\ 0,6944A = -------- = ----------- = 0,0383 square meters 4 x 3,14 4 x 3,14
It follow that the power Pr received by the isotropic antenna on the earth is Pr = P3 x A and so
-25 -27Pr= 2,017 x 10 x 0,0383 = 7,725 x 10 watt
Consequently the round trip isotropic attenuation (Att) earth-moon-earth for 380.000 km at 432 MHz off the moon is P / Pr and so in dB
1(Att) = 10 log ------------------- = 261 dB 10 -27 7,725 x 10
The average of 432 MHz EME active stations are using the following:
Antenna gain = 30 dBi Power at the antenna feed = 1000 watt Overall RX noise figure NF= 0.6 dB = 43 kelvin BW for CW = 500 Hz Equivalent antenna temperature Ta when pointed at the cold-sky = 50 kelvin
With the above data NF, BW and Ta the noise floor of the receiving system KTB = -182 dBW or -152 dBm
Link budged calculation 432 MHz:
TX power at the feed.............................+30 dBW TX Antenna gain....................................+30 dBi -------------- Transmitted EIRP toward the moon.....+60 dBW = 1 Megawatt Round trip attenuation 380.000 km.. - 261 dB -------------- Received power Pr on isotropic antenna at the earth .............................-201 dBW RX antenna gain................................. +30 dB --------------
Available power at RX input............. - 171 dBW RX noise floor.....................................- 182 dBW --------------
CW signal received with a S/N ratio + 11 dB
To get a S/N ratio of 11 dB off the moon on CW it was necessary to 6 transmit + 60 dBW = 10 watt = 1 Megawatt toward the moon but calculating the round trip attenuation we remember that transmitting isotropically 1 watt from the earth the power collected by the lunar disc was
P1= 0,0000052164 watt 6 If now we multiply P1 by 10 we get the full power Pc collected by the lunar disc while transmitting on CW toward the moon and so: 6 Pc = 0,0000052164 x 10 = 5.21 watt ( an awful lot of power as Bob said)
Only the 7% of Pc at 432 MHz is reflected back by the lunar surface and very important the reflected power P is reirradiated and scattered back "isotropically" by the lunar disc and so the reflected power is
P = ( 5.21 / 100) x 7= 0, 3651 watt
If I want to receive a CW signal of 0,3651 watt transmitted isotropically from the moon and if I want to receive it with a S/N ratio of 11 dB it is evident that I need a 30 dBi antenna gain and a receiving system with a noise floor of - 182 dBW.........no way !
If instead I want to receive a SSB or CW signal transmitted in 2 meters by a satellite or from the moon with a power of 10 watt feed into a 10 dBi antenna gain and using a 2 meters ground station antenna with gain of only 13 dBi and a receiving system with a noise floor of - 178 dBW then everyting in SSB and CW becomes very easy as calculation shows.
2 meters downlink budged calculation:
Satellite power ................................... + 10 dBW Satellite antenna gain.......................... + 10 dBi -------------- Satellite EIRP..................................... + 20 dBW (100 W EIRP) 2 m isotr. attenuation 400.000 km.. -188 dB -------------- power density received on a ground isotropic 2 meters antenna..................-168 dBW
2 m ground station antenna gain.........+ 13 dBi
Power density at 2 m RX input...........- 155 dBW 2 m receiver noise floor......................- 178 dBW ---------------
Received CW signal S/N.................... + 23 dB
If we increase the BW to 2500 Hz for a SSB QSO than the noise floor of the receiving system increases by 10 log (2500/500) = 7 dB i.e. 10 it becames about -171 dB and the SSB signal will be received with a S/N ratio = 23-7 = 16 dB wich is a very strong SSB signal.
Be aware that the above figures are based on the assumption that the satellite antennas are pointig toward the earth wich is not the case with a moon orbiting satellite.
In addition we assume that the station in QSO with you has a 70 cm EIRP capability in order to get 10 watt from the 2m transponder only for you.
On the other side if a fixed 10 dBi 2 meters antenna is placed over the moon and it is oriented toward the earth could easily cover the inclination X libration window without any adjustement and only from the point of view of the downlink with 10 watt it can be easily used for a transponder on the moon.
If you make again the downlink budged calculation considering that the 2 meter transponder will develope only 2.5 watt for you then you will realize that the transponder will accomodate 3 more stations if each one is getting 2.5 watt as well. In this case your S/N ratio will be still +15.5 dB on CW and +8.5 dB in SSB and the same is true for the other 3 users.
I hope this helps
73" de
i8CVS Domenico
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