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# Re: Aircraft Doppler?

• Subject: Re: [amsat-bb] Aircraft Doppler?
• From: "jmfranke" <jmfranke@xxxxxxx>
• Date: Sat, 19 Nov 2005 14:49:05 -0500

```The maximum Doppler shift would be when the aircraft was moving direct
towards or away from the tranceiver.  When the aircraft is approaching, the
shift would be to a higher frequency, when moving away, the shift would be
to a lower frequency.  One cycle of Doppler occurrs for every half
wavelength of change of distance.  This is because the transmitter to
receiver path changes one wavelength when the aircraft moves a half
wavelength.  For a direct approach or departure, the Doppler shift is
independent of distance.  For other angles, just multiply the direct
approach valus by the Cosine of the approach or departure angle.

The Doppler shift "D" is found by starting with calculating the wavelength
L:

L(in feet) = 984,000,000/F(in Hz)

D = V/(L/2)

But V is in mph and it must be converted to feet/sec.  An easy convesion to
remember is 60mph = 88 ft/sec

Or, D(in Hz) = F(in MHz) times V(in mph) times 88 divided by ((984/2) times
60)

Running the number for a police traffic radar using 10.525 GHz and a speed
of 1mph:

D = (10525*88)/((984/2)*60) = 926200/29520 = 31.375 Hz/mph which is correct

Now, returning to your problem;
F = 144, let V = 1mph

D = (144*88)/((984/2)*60) = 12672/29520 = 0.429 Hz/mph

So:
for 200mph, the maximum shift is 200* 0.429 = 85.8 Hz
for 300 mph, the maximum shift is 300* 0.429 = 128.7 Hz
for 500 mph, the maximum shift is 500* 0.429 = 214.5 Hz

One resuly of the calculation is that it shows that aircraft talking to each
other do not need to worry about the Doppler shift moving the signals more
than the channel bandwidth.

Hope this helps,

John  WA4WDL

----- Original Message -----
From: "KT2Q" <dxdx@optonline.net>
To: <AMSAT-BB@amsat.org>
Sent: Saturday, November 19, 2005 12:55 PM
Subject: [amsat-bb] Aircraft Doppler?

> All:
>
> What is the maximum aircraft doppler shift expected on 2 meters from an
> aicraft moving at 200, 300 and 500 mph with the aircraft posistion at a
> distance of 5, 10 and 20 miles?
>
> An example of the calculation would be appreciated.
>
> Thanks,
>
> Tony KT2Q
> ----
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