"Bill Everhart" wrote in message
...


I've been wondering: If a shortwave transmitter was put on Mars could
I pick it up - at night I mean?

Probably. The ionosphere would refract the received signals, but I think it
would still come in if it hit the ionosphere at a the correct angle. If the
signals were directly overhead, absorbtion would be the main problem.

There's a couple of small bands in SW allocated to radio astronomy. One is
around 13MHz and the other is around 26MHz.

Would I need an external antenna?

Only if the signal from Mars is weak when it gets to Earth. As long as this
is hypothetical, let's give the Martians a terawatt transmitter and a
steerable parabolic dish a mile across. In the real world, radio astronomy
needs good antennas.


BTW - I tune down.

Never up?

Frank Dresser

"Bill Everhart" wrote:
I've been wondering: If a shortwave transmitter was put on Mars could
I pick it up - at night I mean? Would I need an external antenna?

Interesting question. When Mars is at its closest point to Earth, it's still
about 35 million miles away. Only a tiny portion of the transmitted power
would arrive on Earth (the rest would "miss" us and go out into space in all
directions).

The "free space path loss" between Earth and Mars at 15 MHz would be 211 dB.
That's a HUGE loss. At UHF and microwave frequencies the path loss is even
greater, BUT the use of very high gain dish antennas both on Earth and Mars,
as well a low noise figure receivers, makes communication possible. At HF,
antenna gain of more than about 10 dB is hard to obtain. And the
atmosphereic noise at HF is a killer for weak signal reception.

Plus, you'd have to be listening at a time when your side of the Earth was
facing "their" side of Mars, and the E and F layers of the ionosphere were
NOT refracting signals.

BTW - I tune down.

Huh?

Art N2AH

Very little power is necessary in space. I had a QSO with an astronaut on
MIR with a 3 watt ht. With nothing in the way, it will go on virtually
forever.

"-=jd=-" wrote in message
...
I have no idea if it's an actual fact or not, but back in the mid 70's I
was third party to a conversation in which it was mentioned that the moon
missions communicated on something like 10 watts. I didn't believe it at
the time - not that I would have known any better... Perhaps someone in
here has the scoop on it?


-=jd=-
--
My Current Disposable Email:

(Remove YOUR HAT to reply directly)

In article ,
"Arthur Harris" wrote:

snip

The "free space path loss" between Earth and Mars at 15 MHz would be 211 dB.
That's a HUGE loss. At UHF and microwave frequencies the path loss is even
greater, BUT the use of very high gain dish antennas both on Earth and Mars,
as well a low noise figure receivers, makes communication possible. At HF,
antenna gain of more than about 10 dB is hard to obtain. And the
atmosphereic noise at HF is a killer for weak signal reception.

snip

How did you calculate this loss? I'm assuming you mean the 211 dB to be
an absorptive loss? Maybe you are considering the antenna on Mars to be
a point source off a 180 degree ground plane so the loss figure is power
distributed over a half sphere with the Earth - Mars distance?

--
Telamon
Ventura, California

"Telamon" wrote"
"Arthur Harris" wrote:

The "free space path loss" between Earth and Mars at 15 MHz would be 211
dB.
That's a HUGE loss. At UHF and microwave frequencies the path loss is
even
greater, BUT the use of very high gain dish antennas both on Earth and
Mars,
as well a low noise figure receivers, makes communication possible. At
HF,
antenna gain of more than about 10 dB is hard to obtain. And the
atmosphereic noise at HF is a killer for weak signal reception.

snip

How did you calculate this loss? I'm assuming you mean the 211 dB to be
an absorptive loss? Maybe you are considering the antenna on Mars to be
a point source off a 180 degree ground plane so the loss figure is power
distributed over a half sphere with the Earth - Mars distance?

The Free Space Path Loss Equation is:

Path Loss (dB) = 36.6 + 20 Log F + 20 Log D

Where F is freq in MHz and D is distance in miles. This assumes isotropic
antennas as both ends. The "path loss" represents the portion of the
transmitted signal that is NOT captured by the receiving antenna. It does
not include absorptive losses (which should be negligible in free space).

Some will argue that this is not a true dissipative loss, and that is
correct. But by knowing the "path loss" you can determine your needs as to
transmit power, receiver sensitivity, and antenna gain in order to assure
successful communication.

Art N2AH

"CW" wrote:
Very little power is necessary in space. I had a QSO with an astronaut on
MIR with a 3 watt ht. With nothing in the way, it will go on virtually
forever.

The signal will decrease by 6 dB every time you double the distance. MIR was
about 250 miles above Earth, and you could establish communicaion with
fairly low power when it was overhead. On the other hand, Mars is about 35
million miles away! You'd need a LOT more power and antenna gain to contact
Mars.

Art N2AH

"Arthur Harris" wrote in message
t...

"CW" wrote:
Very little power is necessary in space. I had a QSO with an astronaut
on
MIR with a 3 watt ht. With nothing in the way, it will go on virtually
forever.

The signal will decrease by 6 dB every time you double the distance. MIR
was
about 250 miles above Earth, and you could establish communicaion with
fairly low power when it was overhead. On the other hand, Mars is about 35
million miles away! You'd need a LOT more power and antenna gain to
contact
Mars.


I presume that the Mars rovers do not have extensive antenna arrays, nor
more than a few watts of power. Of course, NASA does have large receiving
arrays..

Arthur Harris wrote:


The Free Space Path Loss Equation is:

Path Loss (dB) = 36.6 + 20 Log F + 20 Log D

Where F is freq in MHz and D is distance in miles. This assumes isotropic
antennas as both ends. The "path loss" represents the portion of the
transmitted signal that is NOT captured by the receiving antenna. It does
not include absorptive losses (which should be negligible in free space).

Some will argue that this is not a true dissipative loss, and that is
correct. But by knowing the "path loss" you can determine your needs as to
transmit power, receiver sensitivity, and antenna gain in order to assure
successful communication.

The key here is a theorem that shows that the "capture area" of a
perfectly efficient isotropic antenna is (wavelength)^2/(4 Pi).

-jpd

for all of you who were speaking about what times you listen, here is what
happened to me when I was real young. I heard an advertisement on a shortwave
station for a program I wanted to hear, but at the time, I didn't understand 24
hour time since I had never heard of it before.

However, when I asked about it, either my uncle or my grandpa (maybe both)
explained it to me. and I then understood it.

But I STILL missed the program when I tuned in at the time said to hear it.

I didn't find out until later that the reason why I missed it is even though I
understood 24 hour time, I wasn't aware of different time zones in different
parts of the world.

The time given was in GMT and I was tuning in according to Eastern Time instead
of GMT.

The signal will decrease by 6 dB every time you double the distance. MIR
was
about 250 mi

but I believe that some people have dx'ed Mars in either the AM or FM broadcast
band.

That is,

Mars, Pennsylvania. ;)