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Chapter 3

Chapter 3. Gate-level Minimization. Simplification Using Prime Implicants. A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2.

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Chapter 3

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  1. Chapter 3 Gate-level Minimization

  2. Simplification Using Prime Implicants • A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2. • A prime implicant is called an Essential Prime Implicant if it is the only prime implicant that covers (includes) one or more minterms. • Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map.

  3. 1

  4. CD C B B D D B C 1 1 1 1 1 1 1 1 BD BD 1 1 B B 1 1 1 1 A A 1 1 1 1 1 1 1 1 A B D D AD Minterms covered by single prime implicant Example of Prime Implicants • Find ALL Prime Implicants ESSENTIAL Prime Implicants C

  5. Y X W Z Prime Implicant Practice • Find all prime implicants for: F (W,X,Y,Z) = (0,2,3,8,9,10,11,12,13,14,15) 1 1 1 1 1 1 1 1 1 1 1 Prime implicants are: W, X‘Y, and X‘Z‘ Note that all of these prime implicants are essential.

  6. Y X W Z Another Example • Find all prime implicants for: G(W,X,Y,Z) = (0,2,3,4,7,12,13,14,15) • Hint: There are seven prime implicants! 1 1 1 1 1 1 1 1 1 Prime Implicants are WX, XY’Z', W’Y’Z', W’X’Z', W’X’Y, W’YZ, XYZ. There is only one essential prime implicant: WX.

  7. 3-4 Five-Variable Map

  8. Example 3-7

  9. 3-5 Product-of-Sums Simplification E.g. Simplify the following Boolean function in product-of-sums form F(A,B,C,D)= m(0,1,2,5,8,9,10) • Mark with 1’s the minterms of F. • Mark the remaining squares with 0’s. These represent F’. • Find the simplified F’.F’ = AB + CD + BD’ • Complement 3 to obtain a simplified F in product-of-sums formF = (A’ + B’)(C’ + D’) (B’ +D)

  10. C B A D Product-of-Sums Simplification Simplify :F= (0,1,2,5,8,9,10)in Product-of-Sums Form • Mark with 0’s the Maxterms of F • Combine 0’s to obtain a simplified F’ in SOP. • Complement 2 to obtain a simplified F in POS. F’ = AB + CD + BD’ F = (A’+B’)(C’+D’)(B’+D)

  11. Example 3-8 Simply Boolean function F(A, B, C, D) =(0, 1, 2,5, 8, 9, 10) in (a) sum of products and (b) product of sums (b) 1. Obtain simplified complemented function: (a) F = B’D’ F’ = AB + CD+BD’ + B’C‘ 2. Applying DeMorgan’s theorem to obtain F + A’C’D F = (A’ + B’) (C’+D’) (B’ + D)

  12. Gate Implementation for Example 3-8

  13. F(x, y, z) = (1, 3, 4, 6) = (0, 2, 5, 7) F(x, y, z) = (1, 3, 4, 6) = (0, 2, 5,7) F = x’z + xz‘ F’ = xz + x’z’  F = (x’ + z’)(x + z)

  14. 3.6 Don't Cares Condition • Sometimes a function table or map contains entries for which it is known: • the input values for the minterm will never occur, or • The output value for the minterm is not used • In these cases, the output value need not be defined • Instead, the output value is defined as a “don't care” • By placing “don't cares” ( an “x” entry) in the function table or map, the cost of the logic circuit may be lowered. • Example 1: A logic function having the binary codes for the BCD digits as its inputs. Only the codes for 0 through 9 are used. The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”

  15. Example 3-9 product of sums?

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