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CS 140 Lecture 4 Combinational Logic: K-Map

This lecture, presented by Professor CK Cheng from the CSE Department at UC San Diego, covers the fundamentals of combinational logic design using Karnaugh Maps (K-maps). The session highlights the implementation of 4-variable K-maps, how to derive Boolean expressions, and the significance of essential prime implicants in simplifying logic functions. In addition, examples are provided to illustrate the drawing of K-maps and using them to find minimal sums of products while addressing don't care conditions. This is a vital skill for students learning digital logic design.

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CS 140 Lecture 4 Combinational Logic: K-Map

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  1. CS 140 Lecture 4Combinational Logic: K-Map Professor CK Cheng CSE Dept. UC San Diego

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

  3. Id a b c d f (a,b,c,d) 0 0 0 0 0 0 1 0 0 0 1 0 2 0 0 1 0 1 3 0 0 1 1 1 4 0 1 0 0 0 5 0 1 0 1 0 6 0 1 1 0 1 7 0 1 1 1 1 8 1 0 0 0 0 9 1 0 0 1 0 10 1 0 1 0 1 11 1 0 1 1 1 12 1 1 0 0 0 13 1 1 0 1 0 14 1 1 1 0 1 15 1 1 1 1 1 4-variable K-maps

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

  5. 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 Another example w/ 4 bits:

  6. 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’

  7. Boolean Expression K-Map Variable xi and its compliment xi’ Two half planes Rxi, Rxi’  Product term P (PXi* e.g. b’c’) Intersect of Rxi* for all i in P (Rb’ intersect Rc’)  U Each minterm  1-cell Two minterms are adjacent iff they differ by one and only one variable, eg:abc’d, abc’d’ The two 1-cells are neighbors  Each minterm has n adjacent minterms Each 1-cell has n neighbors 

  8. Procedure Input: Two sets of F R D • Draw K-map. • Expand all terms in F to their largest sizes (prime implicants). • Choose the essential prime implicants. • Try all combinations to find the minimal sum of products. (This is the most difficult step)

  9. 4-input K-map

  10. 4-input K-map

  11. K-maps with Don’t Cares

  12. K-maps with Don’t Cares

  13. Example Given F = Sm (0, 1, 2, 8, 14) D = Sm (9, 10) 1. Draw 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

  14. 2. Prime Implicants: Largest rectangles that intersect On Set but not Off Set that correspond to product terms. Sm (0, 1, 2, 9), Sm (0, 2, 8, 10), Sm (10, 14) 3. Essential Primes: Prime implicants covering elements in F that are not covered by any other primes. Sm (0, 1, 8, 9), Sm (0, 2, 8, 10), Sm (10, 14) 4. Min exp: Sm (0, 1, 8, 9) + Sm (0, 2, 8, 10) + Sm (10, 14) f(a,b,c,d) = b’c’ + b’d’+ acd’

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

  16. 2. Prime Implicants: Largest rectangles that intersect On Set but not Off Set that correspond to product terms. E.g. Sm (0, 4), Sm (0, 1), Sm (1, 3), Sm (3, 11), Sm (14, 15), Sm (11, 15), Sm (13, 15) 3. Essential Primes: Prime implicants covering elements in F that are not covered by any other primes. E.g. Sm (0, 4), Sm (14, 15) 4. Min exp: Sm (0, 4), Sm (14, 15), ( Sm (3, 11) or Sm (1,3) ) f(a,b,c,d) = a’c’d’+ abc+ b’cd (or a’b’d)

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