What will the current division be between these two sources? Assuming the capacitor value is fairly large, the resistor represents 10k of impedance. The bottommost triode (6DJ8) and its cathode resistor represents about 343k of impedance. The formula is a familiar one: 
        Z = (mu + 1)Rk + rp.
The answer then is that the resistor will contribute 34 times more current than the bottommost triode circuit. Consequently, a 1 volt pulse at the topmost triode's grid will give rise to a 0.1 mA increase in current flow (1/10k), which will increase the voltage across the topmost plate resistor by 1 volt. Since the plate resistor is terminated into the power supply the 1 volt increase in voltage pushes the connection to the plate down 1 volt.
   At the same time the bottommost cathode resistor sees a microscopic increase in its current flow equal to 1 / 343k. The resistor that sits in between the triodes, on the other hand, sees the same microscopic increase in current, as it is in the same current path as the cathode resistor. But since the cathode it is connected to is following its grid, the resistor is pulled up with the cathode's movement.
   So to consolidate, if the bottom triode sees a +1 volt pulse at its grid, the output at its plate will be -1 volts, as the top triode's cathode has not moved; the output at the top triode's plate will be -1 volts, as the top triode's plate resistor will see the same increase in current flow. Now, if the top triode sees a +1 volt pulse at its grid, the output at its plate will be -1 volts, as its cathode moving up 1 volt will create a current pulse through the extra resistor, which will also flow through the plate resistor; the output at the bottom triode's plate will be +1 volts, as the top triode's cathode will climb 1 volt. Since the bottom triode's effective impedance is so great, it cannot buck this movement.

   What happens if a +1 volt pulse is applied to both grids at the same time? The quick answer is since the top triode's plate resistor sees twice the current pulse that pulsing one triode would have given rise to, the plate resistor sees twice the voltage increase or -2 volts of swing. And if the pulses are both negative, the top output swings +2 volts in response.
   Now if the top triode sees a +1 volt pulse and the bottom sees a -1 volt pulse, the top output nulls the two countervailing current pulses, in other words, 0 volts. The same hold true if the if the top triode sees a -1 volt pulse and the bottom sees a +1 volt pulse. 

Conclusion
   We have simple adding/subtracting circuit, what do we do with it? Do you remember how bad quadraphonic sound was? Where did the fault lie? Was it a bad idea in general or did the fault lie in the cheap ICs used in the processors? The SQ encoding system was simple enough and many of records of the period used the encoding system, even it they did not advertise the fact. What would those records sound like if decoded and played on an all-tube system? The results just might be worth the effort.

                                 //JRB

pg. 13

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