hw2solv3 - a) (3 points) f1 = ab’ + bc’d + efg b) (3...

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1 ECE 2030F: Introduction to Computer Engineering Fall 2007 Homework 2 Solution (Total 50 Points) 1. (10 points) Here is an example: F = A’B’D + A’BC’ + ABD’ + ACD = A’C’D + BC’D’ + ABC + B’CD All product terms are prime implicants, but none of them are essential. 2. (10 points) a) y = (((ab’)’cd)’e’)’ = (a’ + b)cd + e = a’cd + bcd + e b) y = ((abc)’ (d’e) ’ (fg’)’)’ = abc + d’e + fg’
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2 3. (10 points)
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Unformatted text preview: a) (3 points) f1 = ab’ + bc’d + efg b) (3 points) f2 = a[b’(c + e) + cd] a b c d e f2 c) (4 points) f3 = c’(ab + d) + d’(ab+e) 3 4. (10 points) y = ab’ + bc’d + efg = (a’+ b)’ + (b’+ c + d’)’ + (e’+ f’ + g’)’ = (((a’+ b)’ + (b’+ c + d’)’ + (e’+ f’ + g’)’) ’)’ 5. (10 pints) y = a[b’(c + e) + cd] = a b’c + ab’e + acd = ((a b’c)’ (ab’e)’ (acd)’)’ Use only two levels of logic:...
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This note was uploaded on 06/17/2008 for the course ECE 2030 taught by Professor Wolf during the Fall '07 term at Georgia Institute of Technology.

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hw2solv3 - a) (3 points) f1 = ab’ + bc’d + efg b) (3...

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