Doppler question - Moon-net

4 May 2013

Sri I made, at least a mistake, I send it again.

Hi all,
Very interesting answers.  I come from Sats world, where the Doppler
is a serious issue and it is so quick that it is impossible to work
JT65.

I will speak thinking in one carrier.  Sats need to receive in only
one fix frequency (it is an active RX with a wideband), if the Sat
receives a carrier in this frequency then it will be repeated.  The
signal TX from the Sat will be in one fix frequency.  So people who
work Sats must correct its TX frequency (according with actual
Doppler) to reach the sat in its fix frequency, in the same way
because of the Sats will repeat this signal in a fix frequency, people
must correct its RX frequency (according with actual Doppler).

This way of dealing with Doppler is very interesting because NOBODY
KNOWS the Doppler of the other stations and nobody needs to know it,
we only need to know our own Doppler.
If we would use this agreement to EME communications we will always
speak about the frequency we hit the Moon, in other words the
frequency which the Moon hear us and repeat our signal.

The actual EME software shows us the Doppler as the sum of the Doppler
effects to reach the Moon and to receive the bounce signal, so Moon
will receive us at ½ Doppler announced, the other ½ effect will appear
after de bounce.

I will try to set a couple of examples.

1.- We will agree to hit the Moon in 1296.070. This is the frequency I
will announce in the log.  In our software the dates will be:

*Self Doppler
+3KHz
*DX Doppler
-2KHz

The sign that the software shows will aplly to RX frequency and the
opposite to the TX frequency, and always ½ magnitude on each one.
Both of us must deal with Doppler, each one must do their task to hit
the moon in the agreed frequency and to receive the bounce according
our Doppler.  Then we must set on our transceivers:

*Me
TX:  1296.070 -1.5KHz = 1296.0685MHz
RX: 1296.070 +1.5KHz= 1296.0715MHz

*DX
TX:  1296.070 +1.0KHz = 1299.071MHz
RX: 1296.070 -1.0KHz= 1296.069MHz

Each station who sees the post will act only according its own
Doppler. Don´t mind where is the DX station.

2.-  In this case I see a signal in my receiver in 1296.070.  What
kind of pair of frequencies I must set (I know nothing about the DX
station I have still decode nothing), I only know me own Doppler, and
it is +2KHz.  I must hit the moon in the same frequency that my
partner:

TX:  1296.070 -1.0KHz = 1296.069MHz
RX: 1296.070 = 1296.070MHz

WE WILL NOT NEED TO KNOW NO MORE THAN OUR DATES.

I haven´t spoken about what frequencies will be the echoes (both
echoes), to make the understanding easy.

As a drawback we should change our habits, we should always use the
RIT as follow:

TX: The frequency we choose plus the opposite sign of Doppler and ½
Doppler magnitude.
RIT: the sign of our Doppler and ½ Doppler magnitude.

Although at first glance it seems not to be interesting in 144MHz but
we will always hit the target in whatever band, even in 144MHz.

Somebody can tell that they need a computer, to do JT65 of course, if
you do CW surely have a smart phone with the proper software to know
your own Doppler or you can print it before.

If I made some mistakes let me know.

It is my 2 cents.

Juan Antonio
EA4CYQ