Electronic – Handling Phase Shift with ANC

The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$ I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.

Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R. $$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$

From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.

Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.

A quote from my previous answer given yesterday regarding the phase shift oscillator with series resistors: -

Your transistor circuit with series resistors is self biasing - the DC flows through several resistors back to the base and, due to negative feedback, the collector dc quiescent voltage will be roughly mid-rail on low voltage supplies.

The transistor base must be biased for this to work correctly. R and RXXX are normally regarded as performing this function. Also, either R or RXXX can do the job of the final resistor in the phase shift network. The final resistor in the phase shift network can be tied to any supply rail because the phase shifting it performs is on an AC signal and either supply rail looks the same to AC.

With both resistors fitted, the bias to the base is better defined but maybe both resistors together attenuate the feedback signal too much and cause the circuit to fail to oscillate.

This can possibly be remedied by increasing the collector resistor (also called R to add to the confusion).