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
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