Capacitor Diodes Full-Bridge – How Does This Bridge Rectifier Work?

The transfer function makes sense only for linear systems (in summary, systems for which \$f(a+b)=f(a)+f(b)\$ and \$a\cdot f(x) = f(a \cdot x)\$). Yours is not a linear system, because it contains nonlinear elements (the diodes, due to the exponential relationship between voltage and current).

You can linearize the model at the neighborhood of a certain point (where f and df/dx are continuous). For a locally linearized system, you can find its transfer function (but, in your example, I don't think it will be very useful).

I say that df/dx has to be continuous, because you cannot linearize your diode bridge system around the point where the function that models it (which is f(x)=|x|) is zero (because there is no derivative there).

If you simulate nonlinear systems (like yours), your results (e.g. frequency response, attenuation, etc) will depend on the amplitude of the excitations, and those results, most of the times, will be of little use.

Updated: Just another intuitive view into this. A linear system can never "create new frequencies". The output spectrum of a linear system is equal to the input spectrum times some frequency response curve. It can attenuate, amplify and even eliminate frequency components present at the input, but it can never make new frequency components appear out of nowhere. If, at some frequency, the input has no spectral contents, it is sure that the output won't have, either.

Now think about your full-bridge (disregard the capacitor). The output is the absolute value of the input. If you apply sin(2·pi·\$f_0\$·t), you get |sin(2·pi·\$f_0\$·t)|. The input is a pure tone, of frequency \$f_0\$. All the spectral contents of the input is a delta (a single line) at f=\$f_0\$. However, the output does have spectral contents out of \$f_0\$. Why? Because |sin(2·pi·\$f_0\$·t)| contains sharp edges (at the points where the argument of the absolute value function changes sign). Those sharp edges mean spectral contents at (in theory) infinite frequencies (multiple of \$f_0\$). So:

Spectrum of input : \$f_0\$.
Spectrum of output: \$f_0\$, 2·\$f_0\$, 3·\$f_0\$, 4·\$f_0\$ ...

That system has created new frequencies out of nowhere. It cannot be linear. There is no transfer function that can make appear a (e.g.) "3·\$f_0\$" out of a zero, at that frequency, at the input. There is no finite number that, multiplied by 0, gives you for instance a 5.

The stray capacitance of pretty much any wire is going to be about the same order of magnitude as the value you're trying to measure. The best way to deal with this is to turn the measurement into a differential one. For example, if your moving plate is grounded, you could have two fixed plates, one on either side of it. The mechanical and electrical construction should be as symmetrical as possible, so that stray capacitances cancel out to the greatest extent possible.

Here's a circuit that's designed to measure small changes in a differential capacitor.

The 10V, 1MHz source causes the diodes to conduct in pairs — D1 and D4 conduct on the positive peaks, and D2 and D3 conduct on the negative peaks. Since D1 and D2 never conduct at the same time, the net current through them is directly porportional to the value of C1. Similarly, the current through D3 and D4 is proportional to the value of C2.

If C1 and C2 have the same value, the D1/D2 current equals the D3/D4 current, and the average voltage difference between points A and B is zero (although both are swinging up and down at 1 MHz). On the other hand, if the sensor is unbalanced, say C1 increases and C2 decreases, more current will flow in D1/D2 than in D3/D4, causing the average voltage at B to rise relative to A.

Note that the difference between A and B can't exceed two forward diode drops (about 1.5 V) in either direction, and in fact, the voltage between them will be related to the net current flow by the diode equation. For values less than 1 V, the voltage varies nearly linearly with the capacitance difference.