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Re: Aircraft Doppler?



AM Modulation is used in aircraft communications so that a weak signal from 
another aircraft can be heard as a hetrodyne on a stronger signal allowing 
the control tower to quirrey the weaker signal by having the stronger stand 
by.
FM provides for capture by the strongest signal, and SSB does not give a 
hetrodyne.
 Control tower radios use circular polorized antennas to prevent loss of 
signal due to aircraft orientation. Aircraft use vertical antennas for 
communication, assuming level flight
Art,
KC6UQH
----- Original Message ----- 
From: "Roger Kolakowski" <rogerkola@aol.com>
To: <AMSAT-BB@amsat.org>; "jmfranke" <jmfranke@cox.net>
Sent: Saturday, November 19, 2005 2:12 PM
Subject: Re: [amsat-bb] Aircraft Doppler?


> 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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> ----
> Sent via amsat-bb@amsat.org. Opinions expressed are those of the author.
> Not an AMSAT member? Join now to support the amateur satellite program!
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----
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