# Re: Aircraft Doppler?

• Subject: Re: [amsat-bb] Aircraft Doppler?
• From: "Roger Kolakowski" <rogerkola@xxxxxxx>
• Date: Sat, 19 Nov 2005 17:12:21 -0500

```John...I needed you as a physics teacher in college!

One question I will throw out to everyone is about your final paragraph.

"One result 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."

Is this why aircraft still use AM modulation?

Thanks!

Roger
WA1KAT

----- Original Message -----
From: "Dave G." <dmg@bossig.com>
To: <AMSAT-BB@amsat.org>; "jmfranke" <jmfranke@cox.net>
Sent: Saturday, November 19, 2005 4:26 PM
Subject: Re: [amsat-bb] Aircraft Doppler?

> John.
>
> Beautiful exposition ---
> I appreciate it, even tho' I did not ask the question :-\
>
> Dave KK7SS
>
>
> On 19 Nov 2005 at 14:49, jmfranke wrote:
>
> > 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.
> ----
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```

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