1. CS 140 Lecture 5 Professor CK Cheng 10/10/02

2. Part I. Combinational Logic 1.Spec 2.Implementation K-map: Sum of products Product of sums

3. Implicant: A product term tat covers at least an element in F and has no intersect with R. Prime Implicants: Largest rectangles that intersect On Set but not Off Set that correspond to product terms. Essential Primes: Prime implicants covering elements in F that are not covered by any other primes.

4. Example Given F =  m (3, 5), D =  m (0, 4) 0 1 3 2 4 5 7 6 c a b - 0 1 1 - 1 0 0 Primes:  m (3),  m (4, 5) Essential Primes:  m (3),  m (4, 5) Min exp: f(a,b,c) = a’bc + ab’

5. 5 variable K-map 0 1 3 2 d a c b 4 5 7 6 12 13 15 14 8 9 11 10 16 17 19 18 d a c b e 20 21 23 22 28 29 31 30 24 25 27 26 Neighbors of 5 are: 1, 4, 13, 7, and 21 Neighbors of 10 are: 2, 8, 10,14, and 26

6. 6 variable K-map d a c b d a c d a c b 48 49 51 50 d a c e 52 53 55 54 60 60 63 62 56 57 59 58 f 32 33 35 34 36 37 39 38 44 45 47 46 40 41 43 42 b b 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 16 17 19 18 20 21 23 22 28 29 31 30 24 25 27 26

7. Min product of sums Given F =  m (3, 5), D =  m (0, 4) 0 1 3 2 4 5 7 6 c a b - 0 1 0 - 1 0 0 Prime Implicates:  M (0,1),  M (0,2,4,6),  M (6,7) Essential Primes Implicates:  M (0,1),  M (0,2,4,6),  M (6,7) Min exp: f(a,b,c) = (a+b)(c )(a’+b’)

8. Corresponding Circuit a b a’ b’ c f(a,b,c,d)

9. Another min product of sums example Given R =  m (3, 11, 12, 13, 14) D =  m (4, 8, 10) K-map d a c b 1 1 0 1 - 0 1 1 0 0 1 0 - 1 0 - 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

10. Prime Implicates:  M (3,11),  M (12,13),  M(10,11),  M (4,12),  M (8,10,12,14) Essential Primes:  M (8,10,12,14),  M (3,11),  M(12,13)