Re: NASA's American Student Moon Orbiter...
In a message dated 04/07/2008 01:16:33 GMT Standard Time, [email protected] writes:
Hi Ed, KL7UW
If we put AO40 at a distance of 400.000 km instead of 60.000 km from the earth the increase of isotropic attenuation at 2400 MHz is about 16 dB etc etc etc.........
Hi Ed / Dom
On the other hand, if you were to reduce path loss by using 70cm as the uplink band and 2m as the downlink the numbers begin to look quite possible.
Also, if the satellite is orbiting the moon, then it's quite likely that the attitude will be such that the experimental end of the satellite is pointing at the moons surface. This probably also means that the communication antennas are not pointing at the earth, so high gain will not be possible. Maybe 3 or 4dB is the limit.
So how about 10W of 2m on the satellite and a passband that's say 5kHz wide? Not good for SSB, but passable for CW or reasonable speed coherent BPSK
Regards
David
At 05:44 PM 7/3/2008, [email protected] wrote:
In a message dated 04/07/2008 01:16:33 GMT Standard Time, [email protected] writes:
Hi Ed, KL7UW
If we put AO40 at a distance of 400.000 km instead of 60.000 km from the earth the increase of isotropic attenuation at 2400 MHz is about 16 dB etc etc etc.........
Hi Ed / Dom
On the other hand, if you were to reduce path loss by using 70cm as the uplink band and 2m as the downlink the numbers begin to look quite possible.
Also, if the satellite is orbiting the moon, then it's quite likely that the attitude will be such that the experimental end of the satellite is pointing at the moons surface. This probably also means that the communication antennas are not pointing at the earth, so high gain will not be possible. Maybe 3 or 4dB is the limit.
So how about 10W of 2m on the satellite and a passband that's say 5kHz wide? Not good for SSB, but passable for CW or reasonable speed coherent BPSK
Regards
David
David,
I think you meant to say 5-Hz vs 5-KHz bandwidth. That is one of the best ways to improve the link equation and CW or WSJT will work well.
When you lower isotropic path loss by lowering frequency, keeping the same antenna gain means much bigger antennas. The footprint on a Moon orbiter would be probably too small to get enough gain on 2m or 70cm.
But this exercise of using AO-40 as a benchmark has its limits. One should just do the complete pathlink analysis to come up with good numbers. The one factor that is always there is a big jump in pathloss due to 400,000 km vs. earth orbit.
Ed I have a pathlink excell calculator on my website for MRO that can be modified to work for a lunar orbiter. http://www.kl7uw.com/raseti.htm
***************************************************** 73, Ed - KL7UW BP40iq, 6m - 3cm 144-EME: FT-847, mgf-1801, 4x-xp20, 185w http://www.kl7uw.com AK VHF-Up Group NA Rep. for DUBUS: [email protected] *****************************************************
----- Original Message ----- From: [email protected] To: [email protected] Sent: Friday, July 04, 2008 3:44 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
In a message dated 04/07/2008 01:16:33 GMT Standard Time, [email protected] writes:
Hi Ed, KL7UW
If we put AO40 at a distance of 400.000 km instead of 60.000 km from the earth the increase of isotropic attenuation at 2400 MHz is about 16 dB etc etc etc.........
Hi Ed / Dom
On the other hand, if you were to reduce path loss by using 70cm as the uplink band and 2m as the downlink the numbers begin to look quite possible.
Hi David, G0MRF
Decreasing the frequency the absolute value of the isotropic attenuation decreases but the difference in path loss between 400.000 km and 60.000 km is the same 16.5 dB at any frequency so that to compensate for the above attenuation using lower frequencies you need bigger antennas both on the satellite and at the ground station.
Also, if the satellite is orbiting the moon, then it's quite likely that the attitude will be such that the experimental end of the satellite is pointing at the moons surface. This probably also means that the communication antennas are not pointing at the earth, so high gain will not be possible. Maybe 3 or 4dB is the limit.
This is why it does not make sense to put a transponder orbiting around the moon just for the simple reason that it's very much more simple and cheap to put it into a HEO earth orbit.
So how about 10W of 2m on the satellite and a passband that's say 5kHz wide? Not good for SSB, but passable for CW or reasonable speed coherent BPSK
Regards
David
Only considering the 2 meters downlink suppose to put AO40 at 400.000 km with the antennas pointing at the earth with low squint angle let say less than 10 degrees. The gain of the AO40 2 meters antennas was 10 dBi and we put your 10 watt on it.
Suppose that your 2 meter antenna has a gain of 13 dBi and the overall noise figure of your receiving system is NF= 0,7 dB = 51 kelvin so that the noise floor into a CW passband of 500 Hz with the antenna looking at the moon (200 kelvin) is about -178 dBW
Suppose that the station in QSO with you has a 70 cm EIRP capability to get the full 2 meters 10 watt from the transponder only for you and we can calculate it later on.
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 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.
73" de
i8CVS Domenico
Probably a dumb question born of ignorance, but why isn't there more than 1 antenna on a sat, so that it's always pointing towards earth? Is Ion propulsion for stabilization using solar power too far away to be practical stabilization system?
----- Original Message ----- From: "i8cvs" [email protected] To: "G0MRF David Bowman" [email protected]; "AMSAT-BB" [email protected] Sent: Friday, July 04, 2008 7:42 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
----- Original Message ----- From: [email protected] To: [email protected] Sent: Friday, July 04, 2008 3:44 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
In a message dated 04/07/2008 01:16:33 GMT Standard Time, [email protected] writes:
Hi Ed, KL7UW
If we put AO40 at a distance of 400.000 km instead of 60.000 km from the earth the increase of isotropic attenuation at 2400 MHz is about 16 dB etc etc etc.........
Hi Ed / Dom
On the other hand, if you were to reduce path loss by using 70cm as the uplink band and 2m as the downlink the numbers begin to look quite possible.
Hi David, G0MRF
Decreasing the frequency the absolute value of the isotropic attenuation decreases but the difference in path loss between 400.000 km and 60.000 km is the same 16.5 dB at any frequency so that to compensate for the above attenuation using lower frequencies you need bigger antennas both on the satellite and at the ground station.
Also, if the satellite is orbiting the moon, then it's quite likely that the attitude will be such that the experimental end of the satellite is pointing at the moons surface. This probably also means that the communication antennas are not pointing at the earth, so high gain will not be possible. Maybe 3 or 4dB is the limit.
This is why it does not make sense to put a transponder orbiting around the moon just for the simple reason that it's very much more simple and cheap to put it into a HEO earth orbit.
So how about 10W of 2m on the satellite and a passband that's say 5kHz wide? Not good for SSB, but passable for CW or reasonable speed coherent BPSK
Regards
David
Only considering the 2 meters downlink suppose to put AO40 at 400.000 km with the antennas pointing at the earth with low squint angle let say less than 10 degrees. The gain of the AO40 2 meters antennas was 10 dBi and we put your 10 watt on it.
Suppose that your 2 meter antenna has a gain of 13 dBi and the overall noise figure of your receiving system is NF= 0,7 dB = 51 kelvin so that the noise floor into a CW passband of 500 Hz with the antenna looking at the moon (200 kelvin) is about -178 dBW
Suppose that the station in QSO with you has a 70 cm EIRP capability to get the full 2 meters 10 watt from the transponder only for you and we can calculate it later on.
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 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.
73" de
i8CVS Domenico
Sent via [email protected]. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
Hi Dave,
The following article will show you how a "three axis stabilization wheels" works on a satellite in order to keep the antennas constantly pointed toward the earth.
Unfortunately due to the explosion many systems failed aboard of AO40 and the above wheels never were running.
Only the magnetorquing system was used on AO40 to keep the Z axis spinning and get the best compromise between the solar panels illumination and the antenna pointing to the earth.
http://www.amsat.org/amsat/sats/phase3d/wheels/index.html
73" de
i8CVS Domenico
----- Original Message ----- From: "D. Mynatt" [email protected] To: "i8cvs" [email protected]; "G0MRF David Bowman" [email protected]; "AMSAT-BB" [email protected] Sent: Friday, July 04, 2008 6:03 PM Subject: Re: [amsat-bb] Re: NASA's American Student Moon Orbiter...
Probably a dumb question born of ignorance, but why isn't there more than
1
antenna on a sat, so that it's always pointing towards earth? Is Ion propulsion for stabilization using solar power too far away to be
practical
stabilization system?
----- Original Message ----- From: "i8cvs" [email protected] To: "G0MRF David Bowman" [email protected]; "AMSAT-BB" [email protected] Sent: Friday, July 04, 2008 7:42 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
----- Original Message ----- From: [email protected] To: [email protected] Sent: Friday, July 04, 2008 3:44 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
In a message dated 04/07/2008 01:16:33 GMT Standard Time, [email protected] writes:
Hi Ed, KL7UW
If we put AO40 at a distance of 400.000 km instead of 60.000 km from the earth the increase of isotropic attenuation at 2400 MHz is about 16 dB etc etc etc.........
Hi Ed / Dom
On the other hand, if you were to reduce path loss by using 70cm as the uplink band and 2m as the downlink the numbers begin to look quite possible.
Hi David, G0MRF
Decreasing the frequency the absolute value of the isotropic attenuation decreases but the difference in path loss between 400.000 km and 60.000
km
is the same 16.5 dB at any frequency so that to compensate for the above attenuation using lower frequencies you need bigger antennas both on the satellite and at the ground station.
Also, if the satellite is orbiting the moon, then it's quite likely
that
the attitude will be such that the experimental end of the satellite is pointing at the moons surface. This probably also means that the communication antennas are not pointing at the earth, so high gain will not be possible. Maybe 3 or 4dB is the limit.
This is why it does not make sense to put a transponder orbiting around the moon just for the simple reason that it's very much more simple and cheap to put it into a HEO earth orbit.
So how about 10W of 2m on the satellite and a passband that's say 5kHz wide? Not good for SSB, but passable for CW or reasonable speed coherent BPSK
Regards
David
Only considering the 2 meters downlink suppose to put AO40 at 400.000 km with the antennas pointing at the earth with low squint angle let say less than 10 degrees. The gain of the AO40 2 meters antennas was 10 dBi and we put your 10 watt on it.
Suppose that your 2 meter antenna has a gain of 13 dBi and the overall noise figure of your receiving system is NF= 0,7 dB = 51 kelvin so that the noise floor into a CW passband of 500 Hz with the antenna looking at the moon (200 kelvin) is about -178 dBW
Suppose that the station in QSO with you has a 70 cm EIRP capability to get the full 2 meters 10 watt from the transponder only for you and we can calculate it later on.
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 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.
73" de
i8CVS Domenico
Sent via [email protected]. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite
program!
Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
At 05:42 AM 7/4/2008, i8cvs wrote:
----snip----
Only considering the 2 meters downlink suppose to put AO40 at 400.000 km with the antennas pointing at the earth with low squint angle let say less than 10 degrees. The gain of the AO40 2 meters antennas was 10 dBi and we put your 10 watt on it.
Suppose that your 2 meter antenna has a gain of 13 dBi and the overall noise figure of your receiving system is NF= 0,7 dB = 51 kelvin so that the noise floor into a CW passband of 500 Hz with the antenna looking at the moon (200 kelvin) is about -178 dBW
Suppose that the station in QSO with you has a 70 cm EIRP capability to get the full 2 meters 10 watt from the transponder only for you and we can calculate it later on.
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 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.
73" de
i8CVS Domenico
Good example of path link analysis, keeping it simple!
But the trick is limiting input to four stations with a linear transponder and they all running an equal uplink. Reality is this doesn't happen so the shared portion of downlink power may and most likely will be less with reduced S/N. My experience with AO-40 was that to have a reasonably good SSB contact you needed at least S/N of 10-dB. In fact that resulted in a fairly weak signal which was difficult to copy. 20-dB S/N made for arm-chair reception.
Not discussed were the 70cm uplink requirements. I suppose one could run high power to achieve that. My AO-40 experience was running up to 60w at a 16.5 dBdc antenna (18.6 dBic). Most of the time I was good with about 5-10w if the satellite was lightly loaded. But with high numbers of stations trying to operate I needed the full EIRP = 72x60 = 4320w or in dB: 18.6 + 47.8 = 66.4 dBW
My AO-40 mode-US station consisted of a FT-847+60w linear at the antenna (M2-436CP42UG) for uplink. The 2.4 GHz downlink was a 33-inch dish with helix feed+MKU-232A2 preamp+Drake converter+FT-847 (on 123-MHz).
I'm not going to go into those calculations.
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. I assume it reflects radio waves not too far efficiently than it reflects sunlight. And it only reflects 6% of the energy it gets.
I haven't found any numbers of the moos efficiency at reflecting radio signals.
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
Edward Cole wrote:
At 05:42 AM 7/4/2008, i8cvs wrote:
----snip----
Only considering the 2 meters downlink suppose to put AO40 at 400.000 km with the antennas pointing at the earth with low squint angle let say less than 10 degrees. The gain of the AO40 2 meters antennas was 10 dBi and we put your 10 watt on it.
Suppose that your 2 meter antenna has a gain of 13 dBi and the overall noise figure of your receiving system is NF= 0,7 dB = 51 kelvin so that the noise floor into a CW passband of 500 Hz with the antenna looking at the moon (200 kelvin) is about -178 dBW
Suppose that the station in QSO with you has a 70 cm EIRP capability to get the full 2 meters 10 watt from the transponder only for you and we can calculate it later on.
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 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.
73" de
i8CVS Domenico
Good example of path link analysis, keeping it simple!
But the trick is limiting input to four stations with a linear transponder and they all running an equal uplink. Reality is this doesn't happen so the shared portion of downlink power may and most likely will be less with reduced S/N. My experience with AO-40 was that to have a reasonably good SSB contact you needed at least S/N of 10-dB. In fact that resulted in a fairly weak signal which was difficult to copy. 20-dB S/N made for arm-chair reception.
Not discussed were the 70cm uplink requirements. I suppose one could run high power to achieve that. My AO-40 experience was running up to 60w at a 16.5 dBdc antenna (18.6 dBic). Most of the time I was good with about 5-10w if the satellite was lightly loaded. But with high numbers of stations trying to operate I needed the full EIRP = 72x60 = 4320w or in dB: 18.6 + 47.8 = 66.4 dBW
My AO-40 mode-US station consisted of a FT-847+60w linear at the antenna (M2-436CP42UG) for uplink. The 2.4 GHz downlink was a 33-inch dish with helix feed+MKU-232A2 preamp+Drake converter+FT-847 (on 123-MHz).
I'm not going to go into those calculations.
Sent via [email protected]. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
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.
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 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.
73" de
i8CVS Domenico
Good example of path link analysis, keeping it simple!
But the trick is limiting input to four stations with a
linear
transponder and they all running an equal uplink. Reality is
this
doesn't happen so the shared portion of downlink power may
and most
likely will be less with reduced S/N. My experience with
AO-40 was
that to have a reasonably good SSB contact you needed at
least S/N of
10-dB. In fact that resulted in a fairly weak signal which
was
difficult to copy. 20-dB S/N made for arm-chair reception.
Not discussed were the 70cm uplink requirements. I suppose
one could
run high power to achieve that. My AO-40 experience was
running up
to 60w at a 16.5 dBdc antenna (18.6 dBic). Most of the time
I was
good with about 5-10w if the satellite was lightly loaded.
But with
high numbers of stations trying to operate I needed the full
EIRP =
72x60 = 4320w or in dB: 18.6 + 47.8 = 66.4 dBW
My AO-40 mode-US station consisted of a FT-847+60w linear at
the
antenna (M2-436CP42UG) for uplink. The 2.4 GHz downlink was
a
33-inch dish with helix feed+MKU-232A2 preamp+Drake
converter+FT-847
(on 123-MHz).
I'm not going to go into those calculations.
Sent via [email protected]. Opinions expressed are those of
the author.
Not an AMSAT-NA member? Join now to support the amateur
satellite program!
Subscription settings:
http://amsat.org/mailman/listinfo/amsat-bb
Sent via [email protected]. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings:
----- Original Message ----- From: "Robert Bruninga" [email protected] To: "'Joe'" [email protected]; "'Edward Cole'" [email protected] Cc: "'AMSAT-BB'" [email protected]; "'G0MRF David Bowman'" [email protected] 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
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" [email protected] To: "Bob Bruninga" [email protected]; "'Joe'" [email protected]; "'Edward Cole'" [email protected] Cc: "'AMSAT-BB'" [email protected]; "'G0MRF David Bowman'" [email protected] Sent: Saturday, July 05, 2008 11:20 AM Subject: [amsat-bb] Re: NASA's American Student Moon Orbiter...
----- Original Message ----- From: "Robert Bruninga" [email protected] To: "'Joe'" [email protected]; "'Edward Cole'" [email protected] Cc: "'AMSAT-BB'" [email protected]; "'G0MRF David Bowman'" [email protected] 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
Sent via [email protected]. Opinions expressed are those of the author. Not an AMSAT-NA member? Join now to support the amateur satellite program! Subscription settings: http://amsat.org/mailman/listinfo/amsat-bb
Hello All,
For those who would like to try and experience a signal from a moon orbiter, the Kaguya lunar probe is transmitting a carrier around 2263.6MHz. I use a medium size BBQ grill on a camera tripod, a noisy MMDS downconverter that has filtering against this frequency (it's tuned for 2300-2400MHz) and a PCR1000 as IF on 312.5MHz (6kHz filter on USB). It shouldn't work, but it does, and I get a 6-10dB spike above the noise using DL4YHF's spectrum lab software. It is audible.
The orbital period is around 2 hours and it can be behind the moon for up to 50 minutes. Doppler is about 45kHz so it can be seen as different from the birdies from the MMDS.
73 de David VK5DG
participants (7)
-
D. Mynatt -
David Giles -
Edward Cole -
G0MRF＠aol.com -
i8cvs -
Joe -
Robert Bruninga