I sent an off-group message explaining this. The reasoning is quite simple.
The audio signal must supply enough voltage to just cut off the DC power
supply at maximum negative peaks. This would imply that at positive peaks
it will double (for a brief instant) the power supply voltage. Double the
voltage will cause (by E squared / R power formula) the *peak* power to be 4
times the unmodulated carrier. Now, consider the audio voltage. It's
*peak* voltage equals the DC voltage to the final. Hmmmm ... the RMS
voltage of that audio signal is only .707 (square root of two divided by
two) of its' peak voltage. Feeding into the same impedance, it will develop
only half of the power (E squared over R or .707 times .707); therefore, you
only need 1/2 of the unmodulated carrier power in audio power to modulate at
100%. This leads to the numbers in my original response.


73 from Rochester, NY
Jim


"Jim Hampton" wrote in message
...
A 1 KW carrier will be modulated 100% with 500 watts of high level
modulation. You end up with 1,000 watts of carrier, 250 watts upper
sideband, and 250 watts of lower sideband. Most inefficient.

73 from Rochester, NY
Jim


"Marco Ferra" wrote in message
om...
Salut

If an audio signal (cosine) with 4 KHz of frequency and 10 V of
amplitude modulates a carrier of 100 MHz, with a percentual modulation
of 85%, does the carrier have 11.76 V of amplitude?

Supposing that the transmitter power is of 1 kW how can I know the
power of the carrier and sidebands? How can I know this values in
Watt, dBW and dBm?

If I just hadn't fluke the exame...

Thank everybody in advance, Marco.


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