K Maps in higher dimensions, Elements of Sequential Circuits - PowerPoint PPT Presentation

K Maps in higher dimensions, Elements of Sequential Circuits (Latches) CSE 140: Components and Design Techniques for Digital Systems Diba Mirza Dept. of Computer Science and Engineering University of California, San Diego 1

Review Some Definitions • Complement: variable with a bar over it A , B , C • Literal: variable or its complement A , A , B , B , C , C • Implicant: product of literals ABC , AC , BC • Implicate: sum of literals ( A+B+C) , ( A+C) , ( B+C) • Minterm: product that includes all input variables ABC , ABC , ABC • Maxterm: sum that includes all input variables (A+B+C) , (A+B+C) , (A+B+C) 2

Minimum Product of Sum Given f (a,b,c) = Σ m (3, 5) + Σ d (0, 4) ab 11 10 00 01 c 0 2 6 4 0 X 0 0 X 1 3 7 5 1 0 1 0 1 Prime Implicates: Essential Primes Implicates: Min exp: f(a,b,c) = 3

Minimum Product of Sum Given f (a,b,c) = Σ m (3, 5) + Σ d(0, 4) ab 11 10 00 01 c 0 2 6 4 0 X 0 0 X 1 3 7 5 1 0 1 0 1 Prime Implicates: Π M (0, 1), Π M (0, 2, 4, 6), Π M (6, 7) Essential Primes Implicates: Π M (0, 1), Π M (0, 2, 4, 6), Π M(6, 7) Min exp: f(a,b,c) = (a+b)(c )(a ’ +b ’ ) 4

Corresponding Circuit Min exp: f(a,b,c) = (a+b)(c )(a ’ +b ’ ) a b f(a,b,c,d) a ’ b ’ c 5

Minimum product of sum: Ex 2 • Reduce the following to a POS form • First find the essential prime implicates ab 11 00 01 10 cd 0 4 12 8 1 0 0 1 00 1 5 13 9 01 1 0 0 X 3 7 15 11 11 0 0 0 0 2 6 14 10 10 1 0 1 X 6

Minimum product of sum: Ex2 • Reduce the following to a POS form • First find the essential prime implicates ab 11 00 01 10 cd 0 4 12 8 1 0 0 1 00 1 5 13 9 01 1 0 0 X 3 7 15 11 11 0 0 0 0 2 6 14 10 10 1 0 1 X 7

Minimum product of sum: Ex 2 • Reduce the following to a POS form • First find the essential prime implicates ab 11 00 01 10 cd 0 4 12 8 1 0 0 1 00 1 5 13 9 01 1 0 0 X 3 7 15 11 11 0 0 0 0 2 6 14 10 10 1 0 1 X 8

Min product of sums: Ex3 Given f (a,b,c,d) = Π M(3, 11, 12, 13, 14). Π D(4, 8, 10) K-map ab 00 01 11 10 cd 0 4 12 8 00 1 5 13 9 01 11 3 7 15 11 2 6 14 10 10 9

Min product of sums: Ex3 Given f (a,b,c,d) = Π M(3, 11, 12, 13, 14). Π D(4, 8, 10) ab 00 01 11 10 cd 0 4 12 8 00 1 X 0 X 1 5 13 9 01 1 1 0 1 d 11 3 7 15 11 0 1 1 0 2 6 14 10 10 1 1 0 X a 10

Prime Implicates: Π M (3,11), Π M (12,13), Π M(10,11), Π M (4,12), Π M (8,10,12,14) PI Q: Which of the following is a non-essential prime implicate? A. Π M(3,11) B. Π M(12,13) C. Π M(10,11) ab D. Π M(8,10,12,14) 00 01 11 10 cd 0 4 12 8 00 1 X 0 X 1 5 13 9 01 1 1 0 1 d 3 7 15 11 11 0 1 1 0 2 6 14 10 10 1 1 0 X 11 a

(V) (25pts) (Karnaugh Map) Use Karnaugh map to simplify function f ( a , b , c ) = Σ m (1, 6) + Σ d (0, 5). List all possible minimal product of sums expres- sions. Show the Boolean expressions. No need for the logic diagram. ab 11 10 00 01 c 0 2 6 4 0 X 0 1 0 1 3 7 5 1 1 0 0 X 12

ab 11 10 00 01 c 0 2 6 4 0 X 0 1 0 1 3 7 5 1 1 0 0 X 13

Five variable K-map a=0 a=1 bc 11 10 00 01 11 10 00 01 de c c 0 4 12 8 16 20 28 24 00 1 5 13 9 17 21 29 25 01 e e 3 7 15 11 19 23 31 27 11 d d 2 6 14 10 18 22 30 26 10 b b a Neighbors of m 5 are: minterms 1, 4, 7, 13, and 21 Neighbors of m 10 are: minterms 2, 8, 11, 14, and 26 14

Reading a Five variable K-map a=0 a=1 bc 11 10 00 01 11 10 00 01 de c c 0 4 12 8 16 20 28 24 00 1 1 1 1 1 1 1 1 5 13 9 17 21 29 25 01 e e 3 7 15 11 19 23 31 27 11 1 1 1 1 1 d d 1 1 1 1 1 1 2 6 14 10 18 22 30 26 10 b b a 15

Six variable K-map ab=(0,0) ab=(0,1) bc de d d 0 4 12 8 16 20 28 24 1 5 13 9 17 21 29 25 f f 3 7 15 11 19 23 31 27 e e 2 6 14 10 18 22 30 26 c c d d 32 36 44 40 48 52 60 56 33 37 45 41 49 53 61 57 a f f 35 39 47 43 51 55 63 59 e e 34 38 46 42 50 54 62 58 c c ab=(1,0) ab=(1,1) b 16

Design example: two-bit comparator A B C D LT EQ GT 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 A N1 LT A B < C D 1 1 1 0 0 B 0 1 0 0 0 0 1 EQ A B = C D 0 1 0 1 0 C N2 GT A B > C D 1 0 1 0 0 D 1 1 1 0 0 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 1 0 1 1 1 0 0 1 1 0 0 0 0 1 block diagram 0 1 0 0 1 and 1 0 0 0 1 truth table 1 1 0 1 0 we'll need a 4-variable Karnaugh map for each of the 3 output functions 17

Design example: two-bit comparator (cont’d) A A A 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 0 0 0 0 1 1 D D D 1 1 0 1 0 0 1 0 0 0 0 0 C C C 1 1 0 0 0 0 0 1 0 0 1 0 B B B K-map for LT K-map for EQ K-map for GT LT = A' B' D + A' C + B' C D EQ = A' B' C' D' + A' B C' D + A B C D + A B' C D’ = (A xnor C) • (B xnor D) GT = B C' D' + A C' + A B D' LT and GT are similar (flip A/C and B/D) 18 Source: Rosing

Design example: 2x2-bit multiplier A2 A1 B2 B1 P8 P4 P2 P1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 A1 P1 0 1 0 0 0 1 A2 P2 1 0 0 0 1 0 B1 P4 1 1 0 0 1 1 1 0 0 0 0 0 0 0 B2 P8 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0 0 0 0 0 0 0 1 0 0 1 1 block diagram 1 0 0 1 1 0 and 1 1 1 0 0 1 truth table 4-variable K-map for each of the 4 output functions 19

Design example: 2x2-bit multiplier (cont’d) A2 A2 K-map for P4 K-map for P8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 B1 B1 0 0 1 0 0 0 0 1 B2 B2 0 0 0 0 0 0 1 1 A1 A1 A2 A2 K-map for P2 K-map for P1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 B1 B1 0 1 0 1 0 1 1 0 B2 B2 0 1 1 0 0 0 0 0 A1 A1 20