> Now define a maximum amplitude V > 0.
The problem being discussed is one in which the user cranks up the volume such that the maximum volume level is far exceeded, and the amplifier circuit, which cannot reproduce the higher level, instead clips the waveform at the supply voltage.
> The claim you made that sparked this conversation was this: "the resulting clipped waveform carries a lot more energy than an unclipped one."
And it does, as you can see from this diagram:
http://i.imgur.com/oE5NFZ9.png
The red trace is the 100% volume level for the system, and it is not clipped. The green, "clipped" trace is far above 100%. It's really very simple.
> But this claim seems to contradict my Claim 2.
Yes. Your claim 2 doesn't apply to a case in which the maximum output voltage of the amplifier cannot reproduce the input waveform. So instead the amplifier clips its input signal at its voltage limits, as the above diagram shows.
The interesting thing about this analysis is that two ways to analyze it produce the same result.
If we use a simple integration of the unclipped and clipped cases, we see a power increase of 2.46, as discussed earlier.
If instead we use Fourier analysis to examine the harmonics of a sine wave and a square wave, the square wave's higher harmonic energies sum to 2.46 of the original sine wave's energy level in the speaker (but pi/2 above the sinewave voltage levels):
Sine wave harmonic lines = 1f
Square wave harmonic lines = (4/pi) sum(sin(2 pi f t (2k-1) / (2k-1),k,1,oo) (http://en.wikipedia.org/wiki/Square_wave#Examining_the_squar...)
The outcomes obviously have to come out the same. The above shows how they do it.