So, looking at that reactance chart...
.01uF intersects 100K Ohms at about 150Hz.
What is that telling me? ...
Capacitive Reactance: 1 / (2 x π x f x C)
= 1 / (2 x π x 150Hz x 0.00000001 Farads)
= 1 / (π x 0.000003 Farads) = 1 / 0.0000094248 = 106kΩ
0.01µF = 0.00000001 Farads; the formula for reactance uses units of Farads (or "Henries" when you remove the 1/x term)
Capacitance, Resistance, Frequency.
Pick any 2: The intersection of those values tells you the value of the 3rd item.
Or you can continue to do the math demonstrated above.
So, looking at that reactance chart...
.01uF intersects 100K Ohms at about 150Hz.
What is that telling me? ...
edit: Oh, I guess it's the resonance frequency. Maybe?
Get a reactance chart and work it out. The cap is "open" at DC and "short" at infinite frequency. The reactance chart tells you where it passes 10k and 2.7k, the corner frequencies.
Corner Frequency (or "Cutoff Frequency" or "-3dB Frequency")That article doesn't do a great job of just telling you the essential points, but
this one does sorta: "corner frequency" is also the "half-power frequency" because the circuit-resistance equals the circuit-reactance (both in Ohms), so half-power is dropped across each element.
PRR told you what to do:
bypass cap is "220nF" which is "0.22µF" ---> Goes a little past the 0.2µF line
One of 2 resistors is 10kΩ ---> find the point where the "0.22µF line" intersects the 10kΩ horizontal line
From the intersection above ----> find the nearest vertical line ---> Corner Frequency
Rinse & repeat for 2.7kΩ.
Takeaway: The bypass made of series resistor & cap implements a stepped-frequency-response in that gain stage rather than the usual roll-off provided by a first-order filter (bypass cap alone).