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My apologies if this is the wrong section.

I am having trouble trying to determine the purpose of circuit. As in, what would this circuit be useful for, if it was installed in a logic board.

The circuit is as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

I believe that the truth table is as follows:

enter image description here

I will admit that this problem is based on previous exam papers the lecturer has set, but I can't seem to find an answer anywhere.

Any help would be gratefully appreciated.

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  • \$\begingroup\$ Your truth table shows five signals, but there are only four signals in the graph. Do you mean ¬X by X`? If so, there's a typo in your last row. \$\endgroup\$
    – Hearth
    Commented Apr 1, 2019 at 21:32
  • \$\begingroup\$ @Hearth - thanks, hadn't notice the typo. Will fix \$\endgroup\$
    – Clauric
    Commented Apr 1, 2019 at 21:44
  • \$\begingroup\$ Even putting aside the fact that it looks an awful lot like a half adder, there's only 16 ways (including the trivial ones) to map two inputs to two outputs -- given that, every single possible one will have been, or perhaps still is, useful on some board some place. It's only most useful as an instrument of 2's compliment arithmetic because of our racial obsession with adding things. \$\endgroup\$
    – TimWescott
    Commented Apr 1, 2019 at 22:10

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This isn't the solution to your assignment but I will give you a pretty good hint... What is the equivalent circuit for a half adder? (See below)

enter image description here

Picture can be found here... not my picture.

Wow, this sure looks pretty similar to your circuit above (wink wink)... Now look at the truth table... It's not the same as what your diagram has but there is some relationship. You notice that binary addition will give you a sum and a carry bit. Now try to relate a half adder circuit with the circuit that you have provided. What's the difference? How will X,Y,B,D relate? Could it be another binary operation other than binary addition? (The answer is yes but now it's up to you to figure that out.)

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  • \$\begingroup\$ thanks for the information. I had thought it was related to a half added, but wasn't sure. \$\endgroup\$
    – Clauric
    Commented Apr 1, 2019 at 21:46
  • \$\begingroup\$ @Clauric You're welcome :) \$\endgroup\$
    – user103380
    Commented Apr 1, 2019 at 21:49

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