A normal Cathode Follower does not have a pair of resistors wrapped around its input and output. Resistors R3, R4 were added to better balanced the circuit's output impedance.
 For a 6DJ8, the result would be 5 mA greater conduction, as 0.5 V x 0.01 A/V = 5 mA.      Moving to the top triode, the 1 volt pulse at the output will be relayed through the resistor string of R4 and R3 to the grid of the top triode. As resistors R1 and R2 also define a voltage divider, once again only 0.5 volts of the original 1 volt will present itself to the grid. But the cathode has not remained at a fixed voltage as was the case in the previous example, but instead has moved up +1 volt with the connection of the battery. Thus the cathode voltage must be subtracted from the change in grid voltage, in this case 0.5 V - 1 V = -0.5 V. Then this negative going signal is multiplied against the transconductance (Gm) of the triode, which much as in the previous case, if its cathode resistor is not bypassed, equals    (mu +1)/(rp + (mu + 1)Rk), otherwise just    (mu +1)/rp.    For a 6DJ8, the result would be 5.15 mA less conduction, as -0.5 V x 0.0103 A/V = -5.15 mA. If the mu of the triode is large, we can ignore the difference, as it will be less than the variation between triodes anyway. But if the mu is small, say 2, as it is for the 6AS7, then resistor R4 should be made (mu +1)/mu times larger in value.
 Broskie Cathode Follower
 Once again, the easiest path to understanding what the output impedance of a circuit is to imagine connecting a 1 volt battery to the output (or any portion of the circuit to determine the impedance at that point) and then to calculate the resulting current flow into or out of the battery and then taking the inverse of the current as the output impedance. If we calculate the output for the bottom half of this circuit and then do the same for the top half, we will have, once we parallel the two results, the output impedance for the entire circuit. Starting with the bottom half first, the 1 volt pulse at the output will be relayed through the resistor string of R2 and R1 to the grid of the bottom triode. As resistor R1 and R2 define a voltage divider, only 0.5 volts of the original 1 volt will present itself to the grid. This 0.5 volt signal is then multiplied against the transconductance (Gm) of the triode, which, if its cathode resistor is not bypassed, equals     mu/(rp + (mu + 1)Rk), otherwise just     mu/rp.
 Why the difference in Gm from the top section to bottom? When a signal is applied to the cathode rather than the grid, we have more than just a change in the grid-to-cathode voltage, we have a change in cathode-to-plate voltage as well. Do not forget that the triode, unlike the transistor or MOSFET, has rp, a change in cathode-to-plate voltage will mean a change in the flow of current through the triode. So, effectively, the cathode has an amplification factor of mu + 1 and a transconductance figure of (mu + 1)/rp, while the plate has an amplification factor of 1/mu and a transconductance figure of 1/rp.
 pg. 10