Tutorial 3 Solution - = AB + A + C + AD (adjacency) = A + C...

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EE 2000 Logic Circuit Design, Semester A, 2008/09 Tutorial 3 Solution Week 4 (22 nd , 24 th , 25 th September, 2008) Level 1: Review Questions Question 3: Implement the function f(A, B, C) = AC + BC’ by using NAND operators only. Answer: ) ( ) ( ) , , ( C B AC C B AC C B AC C B A f = + = + = f B C A Question 9: What is the logic which controls a staircase light associated with two switches A and B located at bottom and top of staircase respectively? Answer: XNOR (or XOR, depends on the design) Question 42: If a three-variable switching function is expressed as the product of maxterms by f(A, B, C) = Π M(0, 3, 5, 6) Answer: c) Σ m (1, 2, 4, 7) Level 2: Problems Question 13: a) Simplify the logic function f = AB + AC’ + C + AD + AB’C + ABC b) Simplify Boolean function F = A’C + A’B + AB’C + BC Answer: a) f = AB + AC’ + C + AD + AB’C + ABC = (AB + ABC) + AC’ + C + AD + (AB’C + ABC) (idempotency) = AB + AC’ + C + AD + AC (absorption, adjacency)
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Unformatted text preview: = AB + A + C + AD (adjacency) = A + C (absorption) b) F = A’C + A’B + AB’C + BC = A’B + (A’ + AB’ + B)C (distributivity) = A’B + (A’ + B’ + B)C (redundancy) = A’B + (A’ + 1)C (complementation) = A’B + (1)C (null) = A’B + C (identity) Level 3: Challenge Exercise Question I: Proof that (x + y) ♁ (x + z) = x’(y z) Answer: (x + y) (x + z) = (x + y )’(x + z) + (x + y)(x + z)’ = (x’y’)(x + z) + (x + y)(x’z’) (DeMorgan) = x’y’z + x’z’y (complementation) = x’ (y’z + z’y) (distributivity) = x’(y z) Question II: How to implement the following expression with 2-input NAND gates only? (a) DE ABC + (b) E D ABC + + Answer: A ) ( ) ) ( ( ) ( ) ( DE C AB DE ABC DE ABC DE ABC = = + = + B C D E ) ( ) ( ) ) ( ( ) ( ) ( ) ) ( ( ) ( ) ( ) ( E D C AB E D C AB E D ABC E D ABC E D ABC = = = + + = + + A B C D E...
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Tutorial 3 Solution - = AB + A + C + AD (adjacency) = A + C...

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