In electrical engineering class, they teach the idea of doing a DC analysis (to find the operating point) and then doing a small-signal AC analysis (to find the AC behavior).
The fact that we can do the DC analysis separately from the AC analysis is a direct consequence of the glorious principle that linearity implies superposition. The AC signal is superposed on the DC operating levels.
We can introduce one bit of terminology: The swing. If at a certain time a certain node changes from V_{0} to V_{0}+ΔV, we say the swing was ΔV.
Tangential remark: The swing is gauge-independent.
We can use this to analyze the problem set forth in reference 1. Hint: this is a linear circuit.
This problem is easy enough that you should be able to solve it in your head, if you look at it the right way. When solving problems in your head, it pays to make simplifying assumptions, to minimize the number of things that you need to keep track of.
Since the circuit is linear, we can WLoG (Without Loss of Generality) set V_{r} equal to 1. To say the same thing another way, we choose to measure voltages in units such that V_{r} equals 1. Not one volt, just 1.
For homework, you can re-do the calculation step by step, putting a factor of V_{r} in each place where it belongs.
In round 1 step 1, when S_{1} is closed for the first time, the swing on node V_{1} is 1.
At this same time, the swing on on node V_{2} is some fraction of this. Hint: Voltage divider. Capacitive voltage divider. Denote this fraction by β.
In round 1 step 3, when S_{2} is closed for the first time, the swing on node V_{2} is −β. At the same time the swing on node V_{1} is also −β. It has to be, because C_{1} is open-circuited at the top, so no current can flow through it, so the voltage across C_{1} cannot change.
In round 2 step 1, when S_{1} is closed for the second time, the swing on node V_{1} is β. What else could it be? In step 3 of the previous round it went down by that much, so now it has to go up by that much. At the same time, the swing on node V_{2} is a factor of β less than this, namely β^{2}. Again, this is just a capacitive voltage divider.
In round 2 step 3, when S_{2} is closed for the second time, the swing on node V_{2} is −β^{2}. The swing on node V_{1} is the same.
And so forth, around and around.
At the end of the Nth round, the voltage on node V_{1} is (1−β^{N}) for all N≥0. This agrees with what we already know for N=0, N=1, and N=2. You can also easily verify that it is correct in the large-N limit, by showing that V_{1}=1, V_{2}=0 is a fixed point.
Remark: I would like this problem better if it showed some token resistance R in series with each switch. With out that, the real-world behavior is pathological when you close a switch across a charged capacitor. The R doesn’t change the result given above, if at each step you allow the system to settle for a time long compared to RC.Other than that, this circuit has some features in common with a wide class of real-world circuits. See reference 2 and reference 3 for an important example of a circuit with capacitors on top of capacitors. Voltage multipliers of this sort have many applications in industry and commerce, as well as in the physics research lab.
It’s worth knowing how to analyze such things.