Let's do some quick math.

The change in frequency (delta F in Hz) is equal to (-) (Vs-Vr)/wavelength (a 
close 1st order approximation when the speed of the wave is much greater than 
the speed of the observer)

Vs is the velocity of source.  For AO-51, that is about 8000m/s
Vr is the velocity of the receiver, which would be 0 in this case since I am 
not taking into account the velocity contributed by Earth's rotation.

The wavelength of a 145 MHz signal is 2.05m
The wavelength of a 435 MHz signal is 0.69m

The delta F for 145 MHz is -3902 Hz
The delta F for 435 MHz is -11,594 Hz

Pretty close to a 3/1 ratio.


-Tim


-----Original Message-----
From: flexradio-boun...@flex-radio.biz 
[mailto:flexradio-boun...@flex-radio.biz] On Behalf Of Ray - K9DUR
Sent: Wednesday, August 04, 2010 10:09 PM
To: 'Brian Lloyd'; 'Ronald G. Parsons'
Cc: flexradio@flex-radio.biz
Subject: Re: [Flexradio] Another Satellite Question

Brian,

I believe that you will find that the Doppler shift is a proportional to the
frequency.  Therefore, the shift at 432MHZ would be approximately 3 times
the shift at 144MHz.  

A simple thought experiment demonstrates this.  Assume you have a 144MHz
frequency source & it is moving away from the receiver fast enough to cause
a 5kHz negative shift.  If you decrease the frequency of the source and the
shift remains 5kHz, consider what the result would be if you decrease the
source frequency all the way down to 5kHz -- the resulting received
frequency would be 0Hz.  That is obviously not correct.  The frequency shift
is dependent on BOTH the relative speed and the frequency.
 
73, Ray, K9DUR
http://k9dur.info




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