I am new to this, but I wanted to ask how it would be possible to convert a dB value into resistance. I've been looking at an isolated analog switch from http://www.ti.com/lit/ds/symlink/ts5a3359.pdf that has an OFF isolation value of -64dB at 2.5v (more on pg 10), and I need help converting this. I'd appreciate any help.
You've got a voltage divider, the series element is a capacitor, and the output (shunt) element is a 50 ohm resistor.
Chances are you will not be loading your analog multiplexor with 50 ohms, but this is how the TI measurements are taken.
The attenuation level of -64dB (lets call it 60dB, for now, to ease the fears of using dB) means 1,000:1 voltage or 1,000,000:1 power.
Thus the crosstalk, thus the capacitances of FET junctions and FET gates and the non-ideal VDD bypassing and the non-ideal real-world package inductances, is 1/1,000 of the input level at 1MHz.
The shunt reactance is that 50 ohm resistor; the series reactance is 1,000X larger at 50,000 ohms at 1MHz ---- and is a capacitor.
Given 1pF at 1GHz is -j159 ohms, or -j159,000 ohms at 1MHz, we easily conclude there is an effective 3pF (to produce -j53,000 ohms at 1MHz).
Thus your crosstalk circuit, or your isolation circuit, looks like this: