Thank you for your responses.
In a sense I feel you're only restating my puzzlement in different terms. I can see that, yes, if one removes the average voltage of the signal, then you would get the effect I am observing. But what is the mechanism by which the capacitor removes the average voltage? How does it know what the average is?
Don't think of it in terms of removing anything. Think of the capacitor in terms of it's frequency dependent impedance. At DC (0 Hz), the impedance of the capacitor is infinite...i.e, it looks like an open circuit.
So what happens if you have 5VDC on the left side, and 2VDC on the right side of the capacitor? The same thing that happens if you have an open circuit. When you measure the right side to ground, you'll see 2VDC, when you measure the left side to ground, you'll measure 5VDC, and when you measure across, you'll measure 3VDC....just like an open circuit.
As the frequency increases, the impedance of the capacitor goes down. So put a 100Hz, 1V peak to peak signal in on the left side. Toss that on the scope and you'll see a 1V peak to peak, 100Hz signal with a 5V DC offset. What do you see on the right side? Well, it depends on the value of the capacitor. Possibly, you'll measure a .5V peak to peak signal with a 2VDC offset. At 100MHz, perhaps you'll measure a .999V peak to peak signal at a 2V DC offset (I didn't calculate anything...I'm just making up numbers for illustration).
Thinking about it in terms of magically blocking DC while letting through AC will get you into all sorts of trouble. The right way to think about it is in terms of it's impedance.
So to answer your specific question, the DC isn't being taken away. When you first turn on your system, the DC goes into charging up the capacitor. If you have a scope probe on the input side when the system first turns on, you'll see the DC offset rise until you get to steady state. That's the cap charging up.
Now you have some 9V peak to peak signal. Let's say you want to amplify that with an opamp, but you don't want to use a dual supply. You can't just AC couple it and feed it in because it will clip when it hits the negative 0V rail. You need to bias the opamp so that it's operating at, say, 10V. No problem. Between the cap and the opamp input, simply put some large resistor (1M, 470K...something like that) to some 10V bias voltage. When you turn the system on, after a little time everything will reach steady state, and you'll end up with your 9V peak to peak signal riding on top of a 10V DC offset.
Now let's use it to filter a power supply. Put a cap from positive to ground. When you turn the system on, you'll see the voltage rise on the + side until it hits steady state. Remember, it's a DC voltage you want, but when you first turn it on, it looks like an AC signal...something approximating the first half of a square wave.
But after a bit, the cap charges up and you reach stead state. Then there is some noise on the line, some AC signal riding on top of that. You select your cap to be large enough so that the impedance at whatever noise frequencies you're expecting is very low. Ideally, you want DC to look like an open circuit to ground, and any noise to look like a dead short to ground. In real life, neither of these is ever achievable, but you just have to get good enough that the noise is attenuated sufficiently.
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