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Sorry if this is me going into teacher mode, but this program will translate into object code that's very similar to the OP's original program: the conditional expressions written with ? and : will generate jumping code just like the conditional statements written with if and else. A nice alternative, though.
With input 5, the original program has input1 = 1 and input2 = 0 (and input3 = 1), so the result is flag = 1. Swanand's program gives flag = 0 in that case. I'm not 100% sure this matters in the OP's situation, but the two programs do behave differently.
I am sure you will find that the cables have simple wires running from one end to the other, so that the voltage reaching your AC adapter is the same as the voltage coming out of the wall. This being the case, there is no difference between one cable and the other, and you can safely use the new cable to replace the old one.
Interesting idea, but (sorry to pull a trump card) that's not part of the circuit analyses I've read -- and not consistent with the 5 μA current labelling R30. I'll give a link in the question to an analysis.
I'm sorry, FakeM, but by "that's not right," I meant, "that's a helpful suggestion, but it's inconsistent with what I understand about the rest of the circuit." And the problem is precisely that the information given on the circuit diagram seems inconsistent. I don't want to build it and then find it can't be tuned!
That's not right. That node is the output of an op amp that sums the control voltages for the oscillator, and is normally a DC voltage that determines the frequency, increasing by 18-21 mV per octave. (18mV would be normal for an exponential pair at room temperature, but Q7 and Q8 are in an oven at about 80C, and 21 mV/oct is the right figure here.)
