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Re: NASA's American Student Moon Orbiter...



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                  18
> S= 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                                        -19
> D =  ------------------------   = 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                     12
> P1= 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                        -25
> P3= ----------------------- = 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
>          /\              0,6944
> A = -------- =  ----------- = 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                                         -27
> Pr= 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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