Message-ID: <01C4C197.D3FECC50.actalbot@southsurf.com>
From: Andy <actalbot@southsurf.com>
To: "'rsgb_lf_group@blacksheep.org'" <rsgb_lf_group@blacksheep.org>
Date: Wed, 3 Nov 2004 11:25:30 -0000
Importance: high
Organization: UKNWN(UK)
MIME-Version: 1.0
Subject: RE: LF: Rugby Loran station
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For a dot lasting an hour, followed by a space lasting an hour and then 
repeated,  the total power  is 50 Watts, with 25W in the carrier, and the rest 
in the entire set of sidebands stretching out to infinite frequency each side 
(negative frequencies included which just wrap round) and the sidebands spaced 
apart by 0.0001389Hz.  The amplitude roll-off depends on the pulse shaping, but 
the total energy is contained only in an infinite bandwidth.

So if YOU filter out this signal in your Rx, the received power could vary with 
CW speed as you lose some sideband energy.

Andy G4JNT


On Wednesday, November 03, 2004 9:52 AM, Mike Dennison 
[SMTP:mike.dennison@ntlworld.com] wrote:
> On 2 Nov 2004 at 20:18, Stewart Nelson wrote:
>
> > 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?
>
> Mike, G3XDV
> ==========