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Recently I had an exam where we were tested on logic circuits. I encountered something on that exam that I had never encountered before. Forgive me for I do not remember the exact problem given and we have not received our grade for it; however I will describe the problem.

The problem had a 3 or 4 inputs. We were told to simplify then draw a logic circuit design for that simplification. However, when I simplified, I ended up eliminating the other inputs and ended up literally with just

A

I had another problem like this as well where there was 4 inputs and when I simplified, I ended up with three. My question is:

What do I do with the eliminated inputs? Do I just not have it on the circuit? How would I draw it?

munchschair
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2 Answers2

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Apparently it not eliminating inputs but the resulted expression is the simplified outcome which you need to think of implementing with logic circuit.

As an example if you have a expression given with 3 inputs namely with the combination of A, B & c, possible literals can be 2^9 = 9 between 000 through 111. Now when you said your simplification lead to just A that mean, when any of those 9 input combinations will result in to value which contain by A.

An example boolean expression simplified to output A truth table is as follows, 

 A B | Output = A 
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 0 0 |     0
 0 1 |     0
 1 0 |     1
 1 1 |     1
Sunil Bojanapally
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Typically an output is a requirement which would not be eliminated, even if it ends up being dependent on a single input. If input A flows through to output Y, just connect A to Y in the diagram. If output Y is always 0 or 1, connect an incoming 0 or 1 to output Y.

On the other hand, inputs are possible, not required, factors in the definition of the problem. Inputs that have no bearing on the output need not be shown in the circuit diagram at all.

kjhughes
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  • This is how I ended up drawing it on the exam but seeing as I had never seen it before, it kind of petrified me heh. Also, there is something about it that bothers me. If the input is not being drawn put in the circuit, it seems like it never entered the equation to begin with. – munchschair Mar 12 '14 at 14:46
  • It's ok. There was a time prior to understanding the true essence of the output that it was unclear whether some input possibility mattered. You worked it through and found that the input did, in fact, not matter. It can therefore be omitted, as could the phase of the moon or any other superfluous factor that does not affect the output. ;-) – kjhughes Mar 12 '14 at 14:58