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CS 140 Lecture 6

CS 140 Lecture 6. Professor CK Cheng UC San Diego. Part I. Combinational Logic Implementation K-map Quine-McCluskey. Quine-McCluskey Method Given two sets of F R D find min sum of products. Exploit the adjacency to find prime implicants.

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CS 140 Lecture 6

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  1. CS 140 Lecture 6 Professor CK Cheng UC San Diego

  2. Part I. Combinational Logic • Implementation • K-map • Quine-McCluskey

  3. Quine-McCluskey Method • Given two sets of F R D • find min sum of products. • Exploit the adjacency to find prime implicants. • Find essential primes among primes via prime implicant chart.

  4. Id a b c d f (a,b,c,d) 0 0 0 0 0 1 1 0 0 0 1 1 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 0 5 0 1 0 1 0 6 0 1 1 0 0 7 0 1 1 1 0 8 1 0 0 0 1 9 1 0 0 1 - 10 1 0 1 0 - 11 1 0 1 1 0 12 1 1 0 0 0 13 1 1 0 1 0 14 1 1 1 0 1 15 1 1 1 1 0 Example Given f(a,b,c,d) F = Sm(0,1,2,8,14) D = Sm(9,10)

  5. Corresponding4-variable K-map b 0 4 12 8 1 0 0 1 1 5 13 9 1 0 0 - d 3 7 15 11 0 0 0 0 c 2 6 14 10 1 0 1 - a f (a, b, c, d) = b’c’ + b’d’ + acd’

  6. Quine-McCluskey Approach • Draw truth table that only include F and D. Order by • The number of ones. Divide into groups. a 0 0 0 1 1 1 1 c 0 0 1 0 0 1 1 Id 0 1 2 8 9 10 14 b 0 0 0 0 0 0 1 d 0 1 0 0 1 0 0 f 1 1 1 1 - - 1 I II III IV Each group has an equal number of ones in the input combination. Each minterm can only be adjacent to the minterms in the next group.

  7. Continue again. Pair up rows from adjacent regions if they differ by exactly one bit. Put a dash where they differ. a 0 0 - - - 1 1 1 c 0 - 0 0 1 0 - 1 (0,1) (0,2) (0,8) (1,9) (2,10) (8,9) (8,10) (10,14) b 0 0 0 0 0 0 0 - d - 0 0 1 0 - 0 0 I&II II&III III&IV

  8. We stop when we can no longer combine rows. Make sure that we cover the entire onset. Primes Sm(0,1,8,9) Sm(0,2,8,10) Sm(10,14) a - - 1 c 0 - 1 b 0 0 - d - 0 0 (0,1,8,9) (0,2,8,10) (10,14) Draw the prime implicant chart: primes versus the minterms in on-set. Circle essential primes. m8 X X m0 X X m1 X m2 X m14 X Sm(0,1,8,9) Sm(0,2,8,10) Sm(10,14) f(a,b,c,d) = Sm(0,1,8,9) + Sm(0,2,8,10) + Sm(10,14) f (a,b,c,d) = b’c’ + b’d’ + acd’

  9. Id a b c d f (a,b,c,d) 0 0 0 0 0 1 1 0 0 0 1 0 2 0 0 1 0 1 3 0 0 1 1 0 4 0 1 0 0 1 5 0 1 0 1 0 6 0 1 1 0 0 7 0 1 1 1 1 8 1 0 0 0 1 9 1 0 0 1 - 10 1 0 1 0 0 11 1 0 1 1 0 12 1 1 0 0 - 13 1 1 0 1 0 14 1 1 1 0 0 15 1 1 1 1 1 Another example Given f(a,b,c,d) w/ F = Sm(0,2,4,7,8,15) D = Sm(9,12)

  10. a 0 0 0 1 1 1 0 1 c 0 1 0 0 0 0 1 1 a 0 0 - - 1 1 - c - 0 0 0 0 0 1 Id 0 2 4 8 9 12 7 15 b 0 0 1 0 0 1 1 1 d 0 0 0 0 1 0 1 1 (0,2) (0,4) (0,8) (4,12) (8,9) (8,12) (7,15) b 0 - 0 1 0 - 1 d 0 0 0 0 - 0 1 I II III IV V (0,4,8,12) - - 0 0 m7 X m15 X m0 X X m2 X m4 X m8 X Sm(0,2) Sm(7,15) Sm(0,4,8,12) f(a,b,c,d) = Sm(0,2) + Sm(7,15) + Sm(0,4,8,12) f (a,b,c,d) = a’b’d + bcd + c’d’

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