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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'"
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
            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
              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

(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
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
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:
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.
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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