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Two-Level Simplification

Two-Level Simplification. All Boolean expressions can be represented in two-level forms Sum-of-products Product-of-sums. Canonical S.O.P. form. Canonical forms are very easy to produce Just read them off of a truth table But, they’re not the most efficient representation.

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Two-Level Simplification

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  1. Two-Level Simplification • All Boolean expressions can be represented in two-level forms • Sum-of-products • Product-of-sums Canonical S.O.P. form • Canonical forms are very easy to produce • Just read them off of a truth table • But, they’re not the most efficient representation Reduced S.O.P. form • Reduced two-level forms are more efficient

  2. B A B=1 B=0 A=0 A’B’ A’B AB AB’ A=1 Venn Diagrams Consider a Venn Diagram for 2 sets, A and B A’B’ A’B AB AB’

  3. A B F 0 0 0 0 1 1 F(A,B) 1 0 1 B 0 1 1 1 0 A 0 00 01 1 10 11 Karnaugh maps 2-variable K-map 1 0 Space for A’B’ 1 0 Space for A’B Space for AB’ Space for AB

  4. C CD C 0 1 00 01 11 10 AB AB 000 001 0000 0001 0011 0010 00 00 m1 m1 m2 m0 m0 m3 010 011 0100 0101 0111 0110 01 01 m3 m5 m6 m2 m4 m7 B B C 110 111 1100 1101 1111 1110 11 11 m7 m13 m14 m6 m12 m15 A A 100 101 1000 1001 1011 1010 10 10 m5 m9 m10 m4 m8 m11 D Karnaugh maps K-maps can represent up to four variables easily f(A,B,C) f(A,B,C,D) 3-variable K-map 4-variable K-map Numbering Scheme: 00, 01, 11, 10 Gray Code — only a single bit changes from one number to the next

  5. C CD C 0 1 00 01 11 10 AB AB 00 00 f(A,B,C) 01 01 B B C 11 11 A A 10 10 D 0 0 0 0 0 0 0 0 0 0 0 Filling in a K-map F(A,B,C,D) = ABC’D’ + AB’CD’ + ABC’D + AB’CD + A’BCD F (A,B,C) = A’B’C’ + ABC’ + A’B’C + AB’C 1 1 1 1 1 0 0 1 1 1 0 0 1 4-variable K-map 3-variable K-map

  6. B 0 1 A 0 0 1 1 0 1 B 0 1 A 0 1 1 1 0 0 Finding Combinations with K-maps We can combine A’B and AB F = A’B + AB = B We can combine A’B’ and A’B G = A’B’ + A’B = A’ With Karnaugh maps, adjacent 1’s mean we can combine them

  7. C C 0 1 AB 00 01 B 11 A 10 Adjacencies in the K-map Neighbors Wrap from top to bottom Wrap from left to right

  8. F F C C 1 1 1 1 C C 0 0 1 1 AB AB 1 1 0 0 00 00 0 0 0 0 01 01 B B 0 0 1 1 11 11 A A 10 10 3-variable K-map examples F(C,B,A) = A’BC’ + AB’C+ A’B’ In the K-map, adjacency wraps from left to right and from top to bottom F(C,B,A) = A’C’+ B’C Same function, alternative “circling” Note: Larger circles are better

  9. G C 0 0 C 0 1 AB 1 1 00 1 1 01 B 0 0 11 A 10 3-variable K-map examples We can use the combining theorem on larger units as well. G(A,B,C) = A’BC’ + A’BC + ABC’ + ABC = A’B(C’ + C) + AB(C’ + C) = A’B + AB = B(A’ + A) = B • What can we circle? • Any rectangle that contains all ones • As long as its size is a power of two • 1, 2, 4, 8, 16, ... • No rectangles of 3, 5, 6, ... Find the smallest number of the largest possible rectangles that cover all the 1’s at least once (overlapping circles are allowed)

  10. CD 00 01 11 10 AB F 00 1 1 1 1 m1 m2 m0 m3 01 m5 m6 m4 m7 1 1 1 1 B C 11 m13 m14 m12 m15 A 0 1 0 1 10 m9 m10 m8 m11 1 0 1 1 D 4-variable K-map example F(A,B,C,D) = åm(0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14) Find the smallest number of the largest possible rectangles that cover all the 1’s • Start at upper left corner and search for 1’s: • Circled? – Go to next ‘1’ • Not circled? – Circle largest term that contains this ‘1’ and go to next ‘1’ • Tie? – Skip this square for now and come back to it later... F(A,B,C,D) = A’ + BC’D + CD’ + B’D’ + B’C

  11. B G 0 1 A C 0 0 0 0 0 1 1 1 1 1 0 0 1 CD CD C 0 1 00 00 01 01 11 11 10 10 AB AB AB 1 1 1 1 1 0 0 0 0 0 1 0 1 00 00 00 0 0 0 0 0 1 1 1 1 1 01 01 01 B B B 1 1 1 1 1 0 0 0 0 0 C C 11 11 11 Q A A A 10 10 10 P D D P= A Å B Å C Å D K-maps for XORs and XNORs F = A’B + AB’ = A Å B G = A’B’C + A’BC’ + ABC’ + AB’C = A Å B Å C Q= A Å B Å C Å D

  12. CD 00 01 11 10 AB F 00 m1 m2 m0 m3 1 1 0 0 01 m5 m6 m4 m7 0 0 0 1 B F = A’C + A’BD’ + AD C 11 m13 m14 m12 m15 A 0 0 1 1 10 m9 m10 m8 m11 1 0 1 0 D Product-of-Sums We can circle 0’s to find a sum-of-products for the complement F(A,B,C,D) = Sm(0,1,5,8,10,12,14) F’ = A’C + A’BD’ + AD DeMorgan’s Law F = (A+C’)(A+B’+D)(A’+D’) Product-of-Sums! • Circling 1’s gives S.O.P. for F • Complementing S.O.P. of F gives P.O.S. for F’ • Circling 0’s gives S.O.P. for F’ • Complementing S.O.P. for F’ gives P.O.S. for F

  13. CD CD 00 00 01 01 11 11 10 10 AB AB F F 00 00 0 0 1 1 1 1 0 0 m1 m1 m2 m2 m0 m0 m3 m3 01 01 0 0 1 1 1 1 x x m5 m5 m6 m6 m4 m4 m7 m7 B B C C 11 11 m13 m13 m14 m14 m12 m12 m15 m15 A A x x x x 0 0 0 0 10 10 m9 m9 m10 m10 m8 m8 m11 m11 0 0 1 1 0 0 0 0 D D K-maps and Don’t Cares Invalid Inputs (Don’t Cares) can be treated as 1's or 0's if it is advantageous to do so F(A,B,C,D) = Sm(1,3,5,7,9) + Sd(6,12,13) F = assuming x’s are zero A’D + B’C’D Tie! - Skip and come back F = using don’t cares A’D +C’D By treating this X as a "1", a largerrectangle can be formed

  14. C D F F F A B 1 2 3 AB = CD A 0 0 0 0 0 F1 N B 1 1 AB < CD 0 1 F2 1 1 0 AB > CD C 1 F3 1 1 N 2 D 0 0 0 0 1 =, >, < 0 0 1 1 1 0 1 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 1 0 1 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 0 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 1 0 Example: 2-bit Comparator Will need a 4-variable K-map for each of the 3 output functions DC and BA are two-bitbinary numbers

  15. F2 F1 F3 CD CD CD 00 00 00 01 01 01 11 11 11 10 10 10 AB AB AB 0 0 0 1 0 0 0 0 0 1 1 1 00 00 00 m1 m1 m1 m2 m2 m2 m0 m0 m0 m3 m3 m3 0 0 0 1 1 0 0 0 0 0 1 1 01 01 01 m5 m5 m5 m6 m6 m6 m4 m4 m4 m7 m7 m7 B B B 0 0 1 0 1 1 0 1 0 0 0 0 11 11 11 A A A m13 m13 m13 m14 m14 m14 m12 m12 m12 m15 m15 m15 0 0 0 1 1 1 0 0 0 0 1 0 C C C 10 10 10 m9 m9 m9 m10 m10 m10 m8 m8 m8 m11 m11 m11 D D D K-maps for 2-bit comparator F1 = AB==CD F2 = AB<CD F3 = AB>CD F1 = F2 = A’B’C’D’ + A’BC’D + ABCD + AB’CD’ A’B’D + A’C +B’CD -OR- F3 = BC’D +AC’ +ABD’ F1’ = B’D +A’C +BD’ +AC’ F1 = (B+D’)(A+C’)(B’+D)(A’+C)

  16. N2 N1 A B C D W X Y Z 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 1 0 1 0 1 0 1 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 0 1 0 1 0 X X X X 1 0 1 1 X X X X 1 1 0 0 X X X X 1 1 0 1 X X X X 1 1 1 0 X X X X 1 1 1 1 X X X X BCD Decrement by 1 • BCD – Binary Coded Decimal • Represents ‘0’ through ‘9’ in four bits • Binary patterns for 10-15 are invalid inputs • Decrement by 1 function • N2 = N1 – 1 • 0 – 1 = 9 (rolls over)

  17. Y W Z X CD CD CD CD 00 00 00 00 01 01 01 01 11 11 11 11 10 10 10 10 AB AB AB AB 0 0 0 0 0 1 0 1 0 0 0 0 00 00 00 00 m1 m1 m1 m1 m2 m2 m2 m2 m0 m0 m0 m0 m3 m3 m3 m3 0 1 1 1 1 1 0 0 0 0 0 0 01 01 01 01 m5 m5 m5 m5 m6 m6 m6 m6 m4 m4 m4 m4 m7 m7 m7 m7 B B B B x x x x x x x x x x x x 11 11 11 11 A A A A m13 m13 m13 m13 m14 m14 m14 m14 m12 m12 m12 m12 m15 m15 m15 m15 x x x x x x 0 1 1 1 0 C C C C 0 10 10 10 10 m9 m9 m9 m9 m10 m10 m10 m10 m8 m8 m8 m8 m11 m11 m11 m11 D D D D 0 0 1 1 1 0 1 0 x x x x x x 1 0 BCD Decrement by One W = A’B’C’D’ +AD X = BD +BC +AD’ Y = CD +BC’D’ +AD’ Z = D’

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