bharat

When I first saw the looseelectron's reply "Capacitor needs to be big enough to pass the lowest frequency signal of interest, into the chip without significant attenuation associated with reactive voltage division between the capacitor and the inpout impedance of the chip."
I wasn't very sure about it. Therefore I am elaborating his point in my words for sake of others: High Pass RC filter (C coupling and 50 ohms termination resistor), will have pole freq = 1/(2*pi*Rterm*Ccoup). The response of RC filter would be, from that pole freq to infinity  the gain would be 0 dB and below pole freq there will be attenuation of <0dB. Therefore one should pick the value of Ccoup such that, the pole freq should be less than or equal to the minimum freq of of the data. After 8b/10b encoding, the input data can be represented as band pass filter ( because minimum and maximum number of transitions are fixed), the minimum freq of that band pass filter should coincide with the pole freq of the HP filter. We would never want that data frequency should see any attenuation. Obviously, if one goes for 64b/66b, the band pass filter will have much lower min freq as compared to 8b/10b encoding; therefore to accommodate 64b/66b encoding, one needs to use larger Ccoup to bring down pole frequency to lower values.
Regards, Bharat
