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Fraf asked Logic gates and boolean algebra?
Hi. See pic: http://www.flickr.com/photos/67251822@N08/7163176942/in/photostream My textbook is silent on this one so I need help.
And got the following answer:
Ugh...I haven't done one of these in many years. Used to be good at it, though... The required expression is Y = AB(A+B). This uses AND and OR gates which according to the answers we aren't allowed to use. But what if we transform the whole thing using DeMorgan's law? Y = A' + B' + (A'B') Negating flips the values of all the variables, changes every AND to an OR and every OR to an AND. So count the operations, now. to get Y we need... a) An inverter to produce A' (a 2-input NAND or NOR gate can do this; just connect A to both inputs) b) An inverter to produce B' (again, a 2-input NAND or NOR can do this) c) A 3-input OR gate (they won't give us OR gates, but we can use a NOR gate and then invert the output with a NOR or NAND) d) An AND gate to produce the A'B'. (we can't have AND gates, but we can use a NAND and then invert the output with a NOR or NAND). So counting up elements from the above list, Y requires a) 1 (NAND or NOR) b) 1 (NAND or NOR) c) 1 NOR + 1 (NAND or NOR) d) 1 NAND + 1(NAND or NOR) Total:1 NOR, 1 NAND, 4 of whichever. So I believe (5) is correct; any combination of six gates with at least one NAND and at least one NOR is fine. (As mentioned above, this expression is stupidly easy if you simplify it. AB(A+B) = AAB + ABB = AB + AB = AB...but they aren't letting us do that.)