WI12-hw2-sol

WI12-hw2-sol - Hussain Al-Asaad 4. PROBLEM 2.43 00 01 11 10...

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UC Davis 1 Hussain Al-Asaad U NIVERSITY OF C ALIFORNIA —D AVIS D EPARTMENT OF E LECTRICAL OMPUTER E NGINEERING EEC180A—D IGITAL S YSTEMS I Winter 2012 P ROBLEM S ET 2—S OLUTION 1. PROBLEM 2.17 (PART D ) = (Rule 8D) (Rule 5D) (Rule 1) (Rule 6) (Rule 9D) The original representation of the function has 8 literals. However, the simplified result has 1 literal, which corre- sponds to saving 7 literals. 2. PROBLEM 2.25 (PART B ) 3. PROBLEM 2.27 (PART C ) f X Y Z   X Y + X Y Z + + X Y Z + + = f X Y Z   X Y + X Y + ZZ + X Y + X Y + 0 + X Y + X Y + Y X + Y X + Y 00 01 11 10 YZ WX 1 1 1 1 1 1 0 1 0 0 0 0 1 1 0 1 00 01 11 10 The representation of the function uses 3 literals. f W X Y Z    M 1 3 7 9 11 15     = f W X Y Z    Z XY + = 00 01 11 10 CD AB x 0 0 0 x 0 1 0 1 0 0 1 1 x 1 0 00 01 11 10 f A B C D   D A C + =
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Unformatted text preview: Hussain Al-Asaad 4. PROBLEM 2.43 00 01 11 10 CD AB 1 1 1 1 1 1 00 01 11 10 Input A B C D Output F 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 1 1 0 1 1 1 1 (a) Truth Table (b) (c) (d) No simplification possible using K-Map f A B C D m 3 5 6 9 10 12 = f A B C D M 0 1 2 4 7 8 11 13 14 15 = f A B C D ABCD ABCD ABCD ABCD ABCD ABCD + + + + + =...
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This note was uploaded on 02/21/2012 for the course EEC 180A taught by Professor Redinbo during the Spring '08 term at UC Davis.

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WI12-hw2-sol - Hussain Al-Asaad 4. PROBLEM 2.43 00 01 11 10...

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