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EECS 465: Digital Systems

EECS 465: Digital Systems. Lecture Notes # 2. Two-Level Minimization Using Karnaugh Maps. SHANTANU DUTT. Department of Electrical & Computer Engineering University of Illinois, Chicago Phone: (312) 355-1314: e-mail: dutt@ece.uic.edu URL: http://www.ece.uic.edu/~dutt. 2.

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EECS 465: Digital Systems

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  1. EECS 465: Digital Systems Lecture Notes # 2 Two-Level Minimization Using Karnaugh Maps SHANTANU DUTT Department of Electrical & Computer Engineering University of Illinois, Chicago Phone: (312) 355-1314: e-mail: dutt@ece.uic.edu URL: http://www.ece.uic.edu/~dutt

  2. 2 Concepts in 2-Level Minimization • Defn: An implicant g of a function is a product term (e.g., g=xyz) s.t. when g=1  f=1 • Defn: A minterm is an implicant whose product term representation consists of all n variables of an n-var. function (each in either compl.or uncompl. form) • Defn: Let g, and h be 2 implicants of a function f. g is said to cover h if h=1  g=1 (e.g., g = xz, h = xyz) • Defn: Let G = {g1, .., gk} be a set of implicants of func f, and let h be another implicant of f. Then G is covers h if h=1  g1+ .. + gk =1 • Defn: An SOP or POS expression is also called a 2-level expression • Defn: A 2-level AND-OR (OR-AND) gate realizationof an SOP • (POS) expression is one in which all product terms (OR • terms in POS) in the SOP (POS) expression are implemented • by multiple input AND (OR) gates & the ORing (ANDing) • of the product (OR) terms is realized by a multiple input OR • (AND) gate.

  3. 3 E.g. f1 = AB + BC + ACD --- 2-level. (SOP) f2 = ( B+C )( A+D )( C+D ) --- 2-level (POS) f3 =AB + AC( B+D ) --- not SOP or POS level 1 I/Ps --- not 2-level --- 3 level ( can be realized directly by 3 levels of gates) A B A f3 C B level 2 I/Ps D level 2 gates level 3 gate level 0 I/Ps level 1 gates

  4. 4 A AB + BC + ACD B f1 B C A C Level 2 D Level 1 B (B + C)(A + D)(C + D) C f2 A D C D

  5. 5 • Defn. A gate in a logic circuit is a level-1 gate if all its inputs are primary inputs (i.e., literals, A, A, B, B, etc.) A gate is a level-i gate i > 1 if the highest-level gate whose output feeds the (level-i) gate is a level-(i-1) gate. • NOTE: If a given expr. is not 2-level, we can convert it to a 2-level one using the distributive law. The goal of 2-level minimization is to minimize the number of literals (a literal is a var. in complemented or uncompl. form) in a 2-level logic expr.. This approx. reduces the total # of inputs over all gates in the circuit in a 2-level gate realization.

  6. 6 E.g. f = ABC + ABC + ABC (non-minimized 9 literals) A B (12 gate I/Ps) = Circuit complexity C A f B C A B C f = AB + AC (minimized 4 literals) A f B (6 gate I/Ps) = Circuit complexity A C Note: Minimizing # of literals + # of product terms  (approx) minimizing total # of gate inputs

  7. 7 Multilevel Minimization (Example): f = x1x2x3 + x2x3x4 + x1x5 + x4x5 (2-level minimized) Apply distributive law (factoring) f = x2x3(x1 + x4) + x5(x1 + x4) = (x2x3 + x5)(x1 + x4) -- Multilevel expr. Gate Impl.: x2 x3 f x5 Multilevel circuit x1 x4

  8. 8 Defn. Two implicants (or product terms) of a function f are said to be adjacent (logically) if they have the same literals except in one variable xi which occurs in uncomplemented form (xi) in one implicant and in complemented form (xi) in the other. The 2 implicants are said to differ in xi. E.g. ABC, ABC (adjacent implicants, differ in B) ABD, ABC (not adjacent) NOTE: Adjacent implicants can be combined into one implicant by the combining theorem (ABC + ABC = AC)

  9. 9 Karnaugh Maps 2-variable TT outputs A B Z1 Z2 Binary place-value ordering (Binary ordering) 0 0 0 0 Physically adjacent but not logically adjacent. 0 1 1 1 1 0 0 1 1 1 1 0 logically adjacent but not physically adjacent AB + AB B Another ordering of inputs ( Gray-code ordering) A B Z1 0 0 0 0 1 1 1 1 1 1 0 0 Z1 = B Physically as well as logically adjacent.

  10. 10 n-variable Gray-code ordering Defn. Gi = i-bit Gray-code ordering rev(Gi) = reverse order of Gi Gn = 0 Gn-1 1 rev(Gn-1) G1 = 0 Base. 1 G2 = 0 G1 1 rev(G1) 0 0 0 1 1 1 1 0 A B C 0 0 0 0 0 1 0 1 1 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 = G3 = 0 G2 1 rev(G2) = Convert to a 2-dimensional Gray-code ordered TT x 110 AB C 00 01 11 10 y 0 1 Throw away variable (A) changing across the 2 squares. We thus obtain BC. 1 1 001

  11. 11 Example ( 3-var. K-Map) Consensus = BC AB C Function f: 00 01 11 10 0 1 1 1 1 1 1 f = AB + AB + AC + BC Four Prime Implicants AB, BC, AC, AB formed. Not all these PIs are needed in the expression. Only a minimum set that covers all minterms is needed. Defn. A Prime Implicant (PI) of a function f is an implicant of f that is not covered by any other implicant of f. Defn. An Essential PI is a PI that covers/includes at least one minterm that is not covered by any other PI.

  12. 12 In the above example, AB & AB are essential PIs. These will need to be present in any SOP expr. of the function. To form a minimal set, the PI BC can be used. Thus f = AB + AB + BC is a minimized expression. Larger than 2-minterm PIs can also exist: AB AB E.g., C C 00 01 11 10 00 01 11 10 0 1 1 1 1 1 0 1 1 1 1 1 AB AB B C B 2-minterm implicants (AB, AB in the 1st example above) can be merged if they are logically adjacent to form a 4-minterm implicant and so on.

  13. 13 When an implicant can not be “grown” any further, then it is a PI. Defn. In a K-map, 2 implicants can be logically adjacent if they are symmetric, i.e., if : (1) They are disjoint (no common minterms). (2) They cover the same # of minterms. (3) Each minterm in one implicant is adjacent to a unique minterm in the other implicant. AB More Examples: Both PIs (A, BC) are essential. C 00 01 11 10 AB Redrawn 0 1 00 01 11 10 1 1 1 1 1 C 0 1 1 1 1 1 1 Not symmetric Thus, f = A + BC is minimized. Symmetric ( logically adjacent) 2-minterm implicants : Can be “merged” to form a 4-minterm implicant, which will be a PI.

  14. 14 4-variable K-Map: AB CD 00 01 11 10 00 01 11 10

  15. 15 Example: f = AC + D Instead of starting with 2-minterm implicants & growing them, you can form larger implicants directly by grouping power of 2 (2, 4, 8, etc.) # of minterms so as to form a rectangle or square ( a convex region). AB CD 00 01 11 10 00 1 1 1 1 1 1 01 AC 11 10 1 1 1 1 Convex D Concave AB CD 00 01 11 10 Invalid grouping (region formed is concave) f = AB + AC + BCD 00 01 11 10 1 1 1 1 Valid groupings 1 1 1 1

  16. 16 Another Example: AB CD BD TT: 00 01 11 10 A B C D Z 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 1 1 0 1 1 1 1 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 1 1 00 01 1 1 ABD 1 C 11 10 1 1 1 1 Binary order 1 1 1 1 f = C + BD + ABD

  17. 17 Don’t Cares in K-Maps Many times certain input combinations are invalid for a function (E.g. BCD to 7-segment display functions; see pp. 212-214 in Katz text ) For such combinations, we do not care what the output is, and we put an ‘x’ in the o/p column for the TT AB CD 00 01 11 10 A B C f 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 x 1 0 0 0 1 0 1 1 1 1 0 x 1 1 1 x 0 1 x x 1 0 0 1 1 0 f = BC + AB by considering all x’s as 0s f = B The x’s can, however, be profitably used to make larger PIs & thus further simplify the function. by selectively considering the 010 cell’s x as 1

  18. 18 In a K-map, the ‘x’s can be profitably used to form larger implicants (which have fewer literals) However, when we need to form a minimal PI cover, we need to worry only about covering the minterms ( the 1s), and not the Xs. • Thus as far as x’s are concerned, we can have the cake and eat it too! Example: f(A,B,C,D) = • m(1,3,5,7,9) + • d(2,6,12,13) MSBs AB A B C D 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 CD 00 01 11 10 0 4 12 x 8 1 1 5 1 13 x 9 1 3 1 7 1 15 11 2 x 6 x 14 10 00 01 LSBs CD 11 10 AD f = CD + AD or CD + AC If input comb. 2,6 occur o/p is 0. AC If inputs 2,6 occur o/p = 1. Which to use for lower power?

  19. 19 STEPS IN K-MAP MINIMIZATION (sop EXPR.) STEP1: Form all PIs by including the Xs with the 1s to form larger PIs. (Do not form PIs covering only Xs !) STEP2: Identify all essential PIs by visiting each minterm (1s) & checking if it is covered by only one PI. If so, then that PI is an essential PI. STEP3: Identify all 1s not covered by essential PIs. Select a minimal- cost set of PIs S to cover them. This requires some trial-and error (in fact, this is a hard computational problem). STEP4: Minimal SOP Expr. for function f: f = • essential PIs + • Pi • S Pi PI Pi is in set S

  20. 20 Minimal POS Expr. from a K-Map : METHOD 1: Obtain minimal SOP expression for the complement function f, and complement this SOP expression to get a minimal POS expr. for f (using De-Morgan’s Law) Big M notation Example: F (A,B,C,D) = • M(0,2,4,8,10,11,14,15) • D(6,12,13) Xs of F 0s of F AB F F AB CD CD 00 01 11 10 D 1 1 x 1 0 0 x 0 0 0 1 1 1 x 1 1 0 0 x 0 1 1 x 1 00 01 11 10 00 01 11 10 Complement 0 4 12 8 1 5 13 9 3 1 7 1 15 0 11 0 2 0 6 x 14 0 10 0 AC F =D + AC F = D + AC = D(A +C)

  21. 21 METHOD2 : Direct Method : (1) Obtain all prime implicates(PTs) (largest groups of 0s & Xs of size 2 , i • 0, forming a convex region, that can not be “grown” any further). (2) Identify all essential PTs (those that cover at least one 0 not covered by any other PT). (3) Choose a minimal set T of PTs to cover the 0s not covered by the set of essential PTs. (4) The expression for a PT is an OR term obtained by discarding all changing variables, & keeping variables complemented if they are constant at 1 or uncomplemented if they are constant at 0 & taking the OR (sum) of these literals. (5) Minimal POS Expr.: f = • (essential PTs) • qi qi T

  22. 22 Example: (Direct Method) AB (OR terms) CD 00 01 11 10 0 0 x 0 1 1 x 1 1 1 0 0 0 x 0 0 00 01 11 10 Prime Implicate ( Groups of 0s that cannot be grown any further) Both D and A + C are Essential Prime Implicates. D (A + C) F = D( A + C ) ( Product of Prime Implicates )

  23. 23 5-variable K-map: f(A,B,C,D,E)---Juxtapose two 4-var. K-submaps one for the MSB A=0 and the other for A=1: BC BC DE 00 01 11 10 00 01 11 10 DE 00 01 11 10 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 16 20 28 24 17 21 29 25 19 23 31 27 18 22 30 26 A=0 A=1 4-var. K-submap 4-var. K-submap Adjacencies of a square: Adjacent squares of the 4-var. K-submap plus the “corresponding” or “mirror” square in the other 4-var. K-submap. Example: BC BC 00 01 11 10 DE DE 00 01 11 10 00 01 11 10 16 20 28 x 24 17 21 29 1 25 19 x 23 1 31 x 27 18 22 1 30 1 26 00 01 11 10 0 1 4 12 1 8 1 5 13 1 9 3 1 7 1 15 x 11 2 6 14 x 10 A=0 A=1

  24. 24 6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submaps one for the MSB A=0 and other for A=1 CD CD EF 00 01 11 10 00 01 11 10 EF 00 01 11 10 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 16 20 28 24 17 21 29 25 19 23 31 27 18 22 30 26 5-var. K-submap A=0 B=0 B=1 CD CD 00 01 11 10 EF EF 00 01 11 10 00 01 11 10 00 01 11 10 48 52 60 56 49 53 61 57 51 55 63 59 50 54 62 58 32 36 44 40 33 37 45 41 35 39 47 43 34 38 46 42 5-var. K-submap A=1 B=1 B=0 Adjacencies of a square: Adjacent squares of the 5-var. K-submap plus the “corresponding” or “mirror” square in the other 5-var. K-submap.

  25. 25 6-variable K-map: f(A,B,C,D,E,F,)---Example: CD CD EF 00 01 11 10 00 01 11 10 EF 00 01 11 10 00 01 11 10 0 4 12 8 1 5 13 9 3 7 15 11 2 6 14 10 16 20 28 24 17 21 29 25 19 23 31 27 18 22 30 26 A=0 B=0 B=1 CD CD 00 01 11 10 EF EF 00 01 11 10 00 01 11 10 00 01 11 10 48 52 60 56 49 53 61 57 51 55 63 59 50 54 62 58 32 36 44 40 33 37 45 41 35 39 47 43 34 38 46 42 A=1 B=1 B=0

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