1.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-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 Reduced 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 two-level forms are more efficient

2.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-2 Venn Diagrams Consider a Venn Diagram for 2 sets, A and B AB’ A’B AB A’B’ B=0 B=1 A=0 A=1 A B A’B’ AB’ AB A’B

3.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-3 Karnaugh maps 2-variable K-map A 0 0 1 1 B 0 1 0 1 F 0 1 1 0 0 1 1 0 B A 01 0 1 00 10 01 11 F(A,B) Space for A’B Space for AB Space for A’B’ Space for AB’

4.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-4 Karnaugh maps K-maps can represent up to four variables easily 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 A f(A,B,C) f(A,B,C,D) AB C 00 01 11 10 01 000001 010011 110111 100101 B AB CD 00011110 00 01 11 10 00000001 01000101 11001101 10001001 00110010 01110110 11111110 10111010 A B C D C m0m0 m1m1 m2m2 m3m3 m6m6 m7m7 m4m4 m5m5 m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10

5.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-5 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 3-variable K-map 4-variable K-map 11 1 1 00 0 0 11 11 1 000 0 0 00 0 00 0 f(A,B,C) AB C 00 01 11 10 01 B C A AB CD 00011110 00 01 11 10 A B C D

6.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-6 Finding Combinations with K-maps F = A’B + AB = B G = A’B’ + A’B = A’ We can combine A’B and AB We can combine A’B’ and A’B With Karnaugh maps, adjacent 1’s mean we can combine them B A 01 01 01 0 1 B A 01 11 00 0 1

7.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-7 Adjacencies in the K-map Wrap from left to right Wrap from top to bottom Neighbors AB C 00 01 11 10 01 B C A

8.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-8 3-variable K-map examples F(C,B,A) = A’C’ + B’C In the K-map, adjacency wraps from left to right and from top to bottom F(C,B,A) = A’BC’ + AB’C + A’B’ Same function, alternative “circling” Note: Larger circles are better 0 1 F AB C 00 01 11 10 01 B C A 1 1 1 00 0 0 1 F AB C 00 01 11 10 01 B C A 1 1 1 00 0

9.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-9 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) G AB C 00 01 11 10 01 B C A 1 00 00 1 11

10.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-10 4-variable K-map example 00 0 1 1 111 111 11 111 F(A,B,C,D) =  m(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) = AB CD 00011110 00 01 11 10 A B C D F m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 A’+ BC’D+ CD’+ B’D’+ B’C

11.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-11 K-maps for XORs and XNORs B A 01 01 1 0 0 1 F = A’B + AB’ = A  B G = A’B’C + A’BC’ + ABC’ + AB’C = A  B  C P= A  B  C  D Q= A  B  C  D G AB C 00 01 11 10 01 B C A 0 1 1 0 01 1 0 AB CD 00011110 00 01 11 10 A B C D 0 1 1 0 01 1 0 0 1 1 0 01 1 0 Q AB CD 00011110 00 01 11 10 A B C D 0 1 1 0 01 1 0 0 1 1 0 01 1 0 P

12.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-12 Product-of-Sums F(A,B,C,D) =  m(0,1,5,8,10,12,14) We can circle 0’s to find a sum-of-products for the complement 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 Product-of-Sums! F’ = AB CD 00011110 00 01 11 10 A B C D F m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 0 0 1 1 1 1 00 0 0 00 0 1 1 1 A’C+ A’BD’+ AD F = A’C + A’BD’ + AD F = (A+C’)(A+B’+D)(A’+D’) DeMorgan’s Law

13.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-13 K-maps and Don’t Cares F(A,B,C,D) =  m (1,3,5,7,9) +  d (6,12,13) Invalid Inputs (Don’t Cares) can be treated as 1's or 0's if it is advantageous to do so By treating this X as a "1", a larger rectangle can be formed F = assuming x’s are zero Tie! - Skip and come back AB CD 00011110 00 01 11 10 A B C D F m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 0 1 1 1 x 0 01 0 x 10 x 0 0 0 A’D + B’C’D AB CD 00011110 00 01 11 10 A B C D F m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 0 1 1 1 x 0 01 0 x 10 x 0 0 0 F = using don’t cares A’D + C’D

14.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-14 Example: 2-bit Comparator Will need a 4- variable K-map for each of the 3 output functions 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0 =, >, < AB = CD AB < CD AB > CD A B C D N 1 N 2 F1F1 F2F2 F3F3 DC and BA are two-bit binary numbers F 1 F 2 F 3 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 B 0 1 0 1 A 0 0 1 1

15.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-15 K-maps for 2-bit comparator AB CD 00011110 00 01 11 10 A B C D F1F1 m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 AB CD 00011110 00 01 11 10 A B C D F2F2 m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 AB CD 00011110 00 01 11 10 A B C D F3F3 m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 F 1 = AB==CDF 2 = AB<CDF 3 = AB>CD 1 1 1 1 0 0 0 0 00 00 0 00 0 1 1 11 1 1 00 0 0 00 00 00 1 1 1 11 1 0 0 0 000 00 00 F 1 =F 2 = F 3 = A’B’C’D’+ A’BC’D+ ABCD+ AB’CD’A’B’D+ A’C+ B’CD BC’D+ AC’+ ABD’ F 1 ’ = B’D+ A’C+ BD’+ AC’ F 1 = (B+D’)(A+C’)(B’+D)(A’+C) -OR-

16.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-16 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) ABCDWXYZ000010010001000000100001001100100100001101010100011001010111011010000111100110001010XXXX1011XXXX1100XXXX1101XXXX1110XXXX1111XXXXABCDWXYZ000010010001000000100001001100100100001101010100011001010111011010000111100110001010XXXX1011XXXX1100XXXX1101XXXX1110XXXX1111XXXX N1 N2

17.  Seattle Pacific University EE 1210 - Logic System DesignKMaps-17 BCD Decrement by One AB CD 00011110 00 01 11 10 A B C D W m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 AB CD 00011110 00 01 11 10 A B C D X m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 AB CD 00011110 00 01 11 10 A B C D Y m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 AB CD 00011110 00 01 11 10 A B C D Z m0m0 m1m1 m4m4 m5m5 m 12 m 13 m8m8 m9m9 m3m3 m2m2 m7m7 m6m6 m 15 m 14 m 11 m 10 x x x x x x 1 1 0 0 0 00 00 0 x x x x x x 1 11 1 0 0 000 0 x x x x x x 1 11 1 0 0 0 0 00 x x x x x x 1 1 1 1 1 0 0 00 0 W = X = A’B’C’D’+ AD Y = Z = BD+ BC+ AD’ CD+ BC’D’+ AD’ D’