So let's say you have a circuit, which generates a carrier wave at some frequency (let's say 27MHz) and it's connected to a 50 ohm dummy load (which I gather is equivalent to an antenna for circuit analysis purposes). And it is powered by a regulated 12V power supply.
So imagine the carrier wave is 12 volts peak-peak, which is 4.242 volts RMS. As per the formula \$P = (V_{rms})^2/R\$, this gives a power output of about 0.36W. Even disregarding average power, 12V into 50\$\Omega\$ is 2.88W. And the peak of the waveform is actually 6V, and at 50 ohms that's only 0.72W.
How then do circuits like these output of 5W or more with a 12V (give or take a few volts) power supply?
http://www.rason.org/Projects/transmit/transmit.pdf (This one reports that when built the output was actually over 7W)
http://www.radanpro.com/Radan2400/Transmitter/5-Watt%20Transmitter%20by%20SM0VPO.htm
If you wanted an average 5W out of a 50 ohm load, you'd need a peak-peak voltage of almost 45V. For 100W, you'd need a signal that's 200V peak to peak! Somehow I doubt that people are powering their radios with such high voltages.
What I'm not understanding is how one gets more power out of a circuit with a fixed load and a fixed supply voltage. Even if your amplifier can deliver 100A, I=V/R; With a 12V supply, Ohm's law says that even at the peak, it's only going to be delivering 0.12A, with the load dissipating 0.72W.
I think one could somehow use a step-up transformer to increase the voltage to the necessary level, trading current on the primary for voltage on the secondary, but neither of the circuits above do this. Aside from that, all the impedance matching networks in the world aren't going to get you more voltage across that load.
All of what I explained may well be wrong, and that's why I explained it. Please help me sort out my conceptual misunderstandings :)