Message-ID: <001601c4c196$8b7ce5d0$0702a8c0@parissn2>
From: "Stewart Nelson" <sn@scgroup.com>
To: rsgb_lf_group@blacksheep.org
References: <4188AA2B.21796.2CA200@localhost>
Date: Wed, 3 Nov 2004 12:16:18 +0100
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Subject: Re: LF: Rugby Loran station
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>> Not true.  For example, if you have a 100 watt CW transmitter
>> with the key held down, the carrier power is 100 watts.
>> Now, what happens when you send a string of dots (50% duty)?
>> Your total power out is indeed 50 watts, but only 25 watts
>> is in the carrier; the other half is in the keying sidebands
>> (if ideal envelope shaping).  In general, the carrier power
>> is multiplied by the square of the duty cycle.
> 
> But doesn't the lost power depend on the Morse speed? 
> Are you saying that for a dot lasting an hour, 50% of the 
> total key down power is taken up by the few milliseconds at 
> key up/down?

No.  If you send one-hour dots every two hours, and had a
"key click" filter with a time constant of an hour, then
a 100-watt Tx at 136 kHz would put out 25 W on 136000 Hz,
12.5 W on ~ 136000.000139 Hz, and 12.5 W on ~ 135999.999861 Hz.
If you use a filter with a time constant of a few milliseconds,
the carrier power is still only 25 W.  However, there are now
also sidebands at ~ 136000.000417 Hz, 136000.000694 Hz, etc.
In either of the above cases, nearly all the sideband energy
would fall within your receiver's bandwidth, so you could
indeed claim 50 W average useful output from the Tx.

However, a LORAN blanker would be gating the signal on and
off much faster than the desired signal's message modulation.
The sidebands created would be offset from the carrier by
> 10 Hz, and would not even appear on the Argo screen.

--Stewart