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),
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.