Got a schematic?
Don't total up individual parts. You need to calculate bass roll-off from R-C at each step through the circuit. Let's do a pretend example.
Let's say you have a 12AX7 gain stage with a 100kΩ plate load, a 0.022µF coupling cap and a 1MΩ resistor after the cap from the next-stage's grid to ground. Our 12AX7 stage has a 1.5kΩ cathode resistor and a 25µF cathode bypass cap. The power supply node for this stage is a 16µF filter cap, which has an upstream 10kΩ decoupling resistor running to the filter cap from the next-higher supply voltage node. Oh wait... this is a simplified version of the Vibrato channel's 2nd gain stage in the
AB763 Deluxe Reverb.
1. Calculate bass roll-off "going forward" of the audio circuit.
First place to look would be the cathode circuit of the gain stage. Simplify the calculation to assume the cathode resistor is the total resistance to be considered (the actual effective resistance is somewhat lower).
Use 1/(2*π*R*C) to find the -3dB point ("C" is in farads, so you'll need to add decimal places).
F
-3dB = 1/(2*π*R*C) = 1/(2*π*1500Ω*0.000025F) = ~4Hz
Do the same R-C calculation for the coupling cap. However, "R" is all resistance between each of the cap's leads. That's a 1MΩ to ground on one leg, and a 100kΩ to a.c. ground, so the 2 are "in series" as far as the cap is concerned. (100kΩ is an approximation; the output impedance of the 12AX7 is the real number to use, which includes the tube's apparent internal resistance in parallel with the 100kΩ, for a lower total. But this moves the roll-off up a little, so our approximation is safe.)
F
-3dB = 1/(2*π*R*C) = 1/(2*π*1.1MΩ*0.000000022F) = ~7Hz
So for audio, the coupling cap roll-off dominates somewhat (which is typical unless you fiddling bypass caps to trim bass).
2. Calculate the bass roll-off "going backward" through the power supply.
Let's keep things simple and consider just this stage's filter cap, and the decoupling resistor going backwards to the next-higher supply voltage. So 16µF and 10kΩ
F
-3dB = 1/(2*π*R*C) = 1/(2*π*10kΩ*0.000016F) = ~1Hz.
So the power supply R-C sets a bass roll-off 1/7th that of the dominant bass roll-off of the audio path in our example. I would have guessed 10:1 would be a good ratio to meet PRR's suggested "significantly lower" recommendation. If you make C bigger or R bigger in the power supply, the roll-off moves lower.
It's a good idea for a preamp to have supply node voltages staggered such that the R is measured at least in kΩ's or tens-of-kΩ's. That way, C doesn't have to be very big, and you have reasonably-good assurance without checking that the supply's bass roll-off is well below the audio path roll-offs.