LTspice IV Computes That Over Unity AC Circuit Works!

[TinselKoala]

Power is not energy!!
A device isn't overunity unless it outputs more Energy than input.
I realize that is simplifying a bit. But it's still true.
Can you show, computed with your LTSpice simulation, the time integral of your watts waveforms?
Also, there can be a lot of energy gained or lost in those little wiggles at the top of your "square" wave, plus the rise and fall times of the "square" aren't all that steep. So I believe you are overestimating the energy in your output, even though your math, comparing some kind of average watts and assuming perfect squareness of your waves, might be right.

[TinselKoala]

I just took a closer look at the numbers on your graph. That wave is nowhere near "square"! My oscilloscope's calibration trace is a more accurate square wave than that, with less ripple on the top and 20 nanosecond rise and fall time. Your rise and fall times appear to be on the order of 50 microseconds or so!!  That is, if I'm reading your numbers right. 300 microseconds between vertical divisions on your scope, right?
That's not a square wave.

[TinselKoala]

This is a square wave.
Horizontal scale 50 microseconds/division. Vertical scale 10 milliVolts/division. Note the fast rise times and lack of ripple, compared to the "square" wave in the LTSpice simulation that you use in your power calculations under the assumption that it is perfectly square. On this scale your rise slope would probably be about 60 degrees, and I don't believe anybody would call it "square".
But I wouldn't even trust this square wave to be accurate in a real energy calculation.

But, using your numbers, again: I estimate that the deviation from true squareness in your output waveform brings its power down to something under 8 watts. But, unfortunately, 5.29 watts continuous input power, divided by the 64 percent duty cycle of the output waveform (an overestimate, by the way) is 8.265625 watts.

(A note on accuracy and significant digits: If you say that an answer to a computation is "8.265625" you are saying that it is NOT "8.265626" and NOT "8.265624" and so on. This accuracy is impossible if the input variables are not also known to at least the same accuracy. Since the input variables in any real circuit are impossible to know with this accuracy, at least without very special techniques and equipment, it is simply FALSE to state so many significant digits in an answer about the real world. Your answer cannot have more precision than the LEAST precise variable that goes into the computation.)

[TinselKoala]

"If you convert 5.29 Watts DC to a DC Signal Peak of 5.29W*1.414 = 7.48006W in a lossless  model (and this is based upon a sine wave peak rather than a square wave).  So the rms value of the sine wave peak will be 5.29 Watts, and our rms value of our square wave is the same as its average and peak which equals 8.23 Watts.  Understand?  See?  Thats over unity."

No, it's not. It's an error. You can confirm that it is an error by simply drawing the waveshapes (a constant DC at 5.29 watts, and your "square" wave, at sufficient resolution) on a piece of graph paper, cutting them out across the same time interval, and weighing them, or otherwise comparing their areas. (The area of the watts waveform is equal to the energy content). Your statement above does not even take into account duty cycle. It's easy to get high peak power levels from a low-power continuous input, as you show later when you do compute duty cycles. Your output waveform isn't square enough to justify your assumptions, the true duty cycle is less than 64 percent and the true power level is less than 8.29 watts, and once again, the appropriate measurement is ENERGY, which is power integrated over time, and your input energy is greater than your output energy. SO no overunity here.

For example, if we correct for your slow risetimes, the duty cycle might be, say, 62 percent instead of 64 percent. This means that the 5.29 watts constant input power could deliver as much as 8.532258064516129032258064516129 watts. No overunity, even if one accepts your own model's power calculations for that ripply signal,which I do not.

And see how ridiculous it looks when you just grab all the digits your calculator spits out? The proper answer from that computation is 8.5 watts.

Understand? See?

[TinselKoala]

Of course, you could just start yet another thread. But one wonders why you haven't addressed a single point I've made (and others have made) in this one.