blueintheface wrote on Mon, 18 September 2006 11:38 
I always thought that as a square wave is progressively (lowpass) filtered, it approcimates a sine wave of the same peak value . . . ?

An understandable error, but an error nonetheless.
This would be easier to show graphically, but I'll try to do it in words and numbers.
A square wave is a sum of a fundamental, then odd harmonics scaled by the harmonic multiplier.
So you have a waveform that is
Fundamental + 1/3 * 3rd harmonic + 1/5 * 5th harmonic + 1/7 * 7th harmonic
and so on.
Now for simplicities sake lets assume a perfect brick wall lowpass filter, so anything above the cutoff is removed completely, and everything below it passes unhindered.
And rather than sweeping it down, let's sweep it up.
So we start with the filter set so that only the fundamental passes through, and then fundamental has a peak value of 1. So our waveform is a sine wave, peak 1.
Now we sweep the filter up so it lets through the third harmonic, what is the peak of the waveform now?
Well let's start by looking at the point in the waveform which was the peak before, i.e. one quarter of the way through the cycle, that is where the fundamental peaks, but it is also the point where the third harmonic peaks in the opposite direction. So the contribution of the fundamental is +1, but the contribution of the harmonic is 1/3, giving a level at that point of +2/3, or 0.667
In fact with just the third harmonic being allowed through as well as the fundamental the peak points in the waveform occur at different points and have a value of 0.833.
Basically adding the harmonic lifts some parts of the waveform up, and pushes others down, making it flatter on top, but also reducing the peak amplitude. As you add more and more harmonics it gets flatter and flatter, but each time the peak also gets slightly less.