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Gate-level Minimization

Gate-level Minimization. The procedure of simplifying Boolean expressions (in 2-4) is difficult since it lacks specific rules to predict the successive steps in the simplification process. Alternative: Karnaugh Map (K-map) Method.

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Gate-level Minimization

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  1. Gate-level Minimization • The procedure of simplifying Boolean expressions (in 2-4) is difficult since it lacks specific rules to predict the successive steps in the simplification process. • Alternative: Karnaugh Map (K-map) Method. K-map method can be seen as a pictorial form of the truth table. y y x m0 m1 m2 m3 x Two-variable map

  2. Two-variable K-MAP y y x x y y y y x x x x

  3. Two-variable K-MAP y y y y x x x x The three squares can be determined from the intersection of variable x in the second row and variable y in the second column.

  4. Three-Variable K-Map How is this map useful? • Any two adjacent squares differ by only one variable. • From the postulates of Boolean algebra, the sum of two minterms in adjacent squares can be simplified to a simple AND term.

  5. Three-Variable K-Map Example 1

  6. Three-Variable K-Map Example 2 Simplify:

  7. Three-Variable K-Map Example 3 Simplify:

  8. Three-Variable K-Map Example 4 Given: (a) Express F in sum of minterms. (b) Find the minimal sum of products using K-Map (a)

  9. Three-Variable K-Map Example 4 (continued)

  10. Three-variable K-Map: Observations • One square represents one minterm  a term of 3 literals • Two adjacent squares  a term of 2 literals • Four adjacent squares  a term of 1 literal • Eight adjacent squares  the function equals to 1

  11. Four-Variable K-Map

  12. Four-Variable K-Map Simplify F(w,x,y,z) = S(0,1,2,4,5,6,8,9,12,13,14) Example 5 1

  13. Four-Variable K-Map Example 6 Simplify F(A,B,C,D) =

  14. Four-variable K-Map: Observations • One square represents one minterm  a term of 4 literals • Two adjacent squares  a term of 3 literals • Four adjacent squares  a term of 2 literal • Eight adjacent squares  a term of 1 literal • sixteen adjacent squares  the function equals to 1

  15. SUM of PRODUCT and PRODUCT OF SUM Simplify the following Boolean function in: (a) sum of products (b) product of sums Combining the one’s: (a) Combining the zero’s: Taking the the complement: (b)

  16. SOP and POS gate implementation PRODUCT OF SUM (POS) SUM OF PRODUCT (SOP)

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