# Gate Level Minimization

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1 Gate Level Minimization By Dr. M. Hebaishy Digital Logic Design Ch-

2 Simplifying Boolean Equations Example : Y = AB + AB Example 2: = B (A + A) T8 = B () T5 = B T Y = A(AB + ABC) = A (AB ( + C ) ) T8 = A(AB () ) T2 = A(AB) T = (AA)B T7 = AB T3 Digital Logic Design Ch-2

3 Simplify the following Boolean functions to a minimum number of literals. Digital Logic Design Ch-3

4 Simplify the following Boolean functions to a minimum number of literals. Y=AB+A(A+C)+B(A+C) Solution Y=AB+AA+AC+AB+BC Y=AB+A+AC+BA+BC Y=AB+A+AC+BC Y=A(B++C)+BC Y=A +BC Y=A+BC T (A.A=A) T (A+A=A) T (A+=A) T (A.=A) Before Simplification Y=AB+A(A+C)+B(A+C) After Simplification Y=A+BC Digital Logic Design Ch-4

5 Simplify the following Boolean functions to a minimum number of literals. Solution Digital Logic Design Ch-5

6 quiz Simplify the following Boolean functions to a minimum number of literals. Digital Logic Design Ch-6

7 DeMorgan s Theorem Y = AB = A + B A B A B Y Y Y = A + B = A B A B A B Y Y Digital Logic Design Ch-7

8 Properties of NAND Gate Basic Gates Representation Using NAND Gate NOT Gate AND Gate OR Gate NOR Digital Logic Design Ch-8

9 Properties of NOR Gate Basic Gates Representation Using NOR Gate NOT Gate AND Gate OR Gate NAND Digital Logic Design Ch-9

10 - Simplify The Following Boolean Expressions Using De Morgan Theories? Solution Implementation of Boolean Function using NAND Gate 2- Simplify The Following Boolean Expressions Using De Morgan Theories? Implementation of Boolean Function using NOR Gate Digital Logic Design Ch-

11 Apply De Morgan Theories Implementation using NAND-NOR Gates Y [(AB)C] [(DE)F] = [(AB)C] [(DE)F] = (AB)C+(DE)F = (A+B)C+(D+E)F Y [(A B) C] [(D E) F] = [(A B) C] [(D E) F] = [(A B) C] [(D E) F] = [(AB) C] [(DE) F] Digital Logic Design Ch-

12 quiz Apply De Morgan Theories on the following Boolean Functions? Digital Logic Design Ch-2

13 Conversion between Canonical Forms Sum of minterms: there are 2 n minterms and 2 2n combinations of function with n Boolean variables. Example: express F = A+BC' as a sum of minterms. F = A+B'C = A (B+B') + B'C = AB +AB' + B'C = AB(C+C') + AB'(C+C') + (A+A')B'C = ABC+ABC'+AB'C+AB'C'+A'B'C F = A'B'C +AB'C' +AB'C+ABC'+ ABC = m + m 4 +m 5 + m 6 + m 7 F(A, B, C) = S(, 4, 5, 6, 7) or, built the truth table first Digital Logic Design Ch-3

14 Conversion between Canonical Forms Product of Maxterms Product of maxterms: using distributive law to expand. x + yz = (x + y)(x + z) = (x+y+zz')(x+z+yy') = (x+y+z)(x+y+z')(x+y'+z) Example : express F = xy + x'z as a product of maxterms. F = xy + x'z = (xy + x')(xy +z) = (x+x')(y+x')(x+z)(y+z) = (x'+y)(x+z)(y+z) x'+y = x' + y + zz' = (x'+y+z)(x'+y+z') F = (x+y+z)(x+y'+z)(x'+y+z)(x'+y+z') = M M 2 M 4 M 5 F(x, y, z) = P(, 2, 4, 5) Digital Logic Design Ch-4

15 Conversion between Canonical Forms The complement of a function expressed as the sum of minterms equals the sum of minterms missing from the original function. F(A, B, C) = S(, 4, 5, 6, 7) Thus, F'(A, B, C) = S(, 2, 3) By DeMorgan's theorem F(A, B, C) = P(, 2, 3) F'(A, B, C) =P (, 4, 5, 6, 7) m j ' = M j Sum of minterms = product of maxterms Interchange the symbols S and P and list those numbers missing from the original form» S of 's» P of 's Digital Logic Design Ch-5

16 Example F = xy + x z F(x, y, z) = S(, 3, 6, 7) F(x, y, z) = P (, 2, 4, 6) Digital Logic Design Ch-6

17 Standard Forms Canonical forms are very seldom the ones with the least number of literals. Standard forms: the terms that form the function may obtain one, two, or any number of literals. Sum of products: F = y' + xy+ x'yz' Product of sums: F 2 = x(y'+z)(x'+y+z') F 3 = A'B'CD+ABC'D' Digital Logic Design Ch-7

18 Two-level implementation Implementation Multi-level implementation Digital Logic Design Ch-8

19 Simplification using Karnaugh-map Types of k-maps K-Map for 2- Variables K-Map for 3-Variables K-Map for 4-Variables Digital Logic Design Ch-9

20 Karnaugh Maps (K-Maps) Boolean expressions can be minimized by combining terms K-maps minimize equations graphically PA + PA = P A B C Y Y C AB Y AB C ABC ABC ABC ABC ABC ABC ABC ABC Digital Logic Design Ch-2

21 K-Map 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 Prime implicant : implicant corresponding to the largest circle in a K-map Digital Logic Design Ch-2

22 K-Map Rules Every must be circled at least once Each circle must span a power of 2 (i.e., 2, 4) squares in each direction Each circle must be as large as possible A circle may wrap around the edges A don't care (X) is circled only if it helps minimize the equation Digital Logic Design Ch-22

23 2-Inputs K-Map Use k-map to simplify the following truth table Inputs Output A B Y Y A. B A. B A. B B B A A A B After Simplification Y A B Digital Logic Design Ch-23

24 A two-variable map Four minterms x' = row ; x = row y' = column ; y = column A truth table in square diagram Fig. (a): xy = m 3 Fig. 2(b): x+y = x'y+xy'+xy ) Two-variable Map = m +m 2 +m 3 2) Representation of functions in the map Digital Logic Design Ch-24

25 3-Inputs K-Map Y AB C ABC ABC ABC ABC ABC ABC ABC ABC A Truth Table K-Map Y B C Y AB C Y = AB + BC Digital Logic Design Ch-25

26 3-Inputs K-Map A three-variable map Eight minterms The Gray code sequence Any two adjacent squares in the map differ by only on variable» Primed in one square and unprimed in the other» e.g., m 5 and m 7 can be simplified» m 5 + m 7 = xy'z + xyz = xz (y'+y) = xz Three-variable Map Digital Logic Design Ch-26

27 3-Inputs K-Map m and m 2 (m 4 and m 6 ) are adjacent m + m 2 = x'y'z' + x'yz' = x'z' (y'+y) = x'z' m 4 + m 6 = xy'z' + xyz' = xz' (y'+y) = xz' Digital Logic Design Ch-27

28 3-Inputs K-Map Inputs Output A. B C Y A. B. C A. B. C A. B. C A. B. C A. B. C Y C A B Digital Logic Design Ch-28

29 3-Inputs K-Map Example: simplify the Boolean function F(x, y, z) = S(2, 3, 4, 5) F(x, y, z) = S(2, 3, 4, 5) = x'y + xy' Map for Example F(x, y, z) = Σ(2, 3, 4, 5) = x'y + xy' Digital Logic Design Ch-29

30 3-Inputs K-Map Example : simplify F(x, y, z) = S(3, 4, 6, 7) F(x, y, z) = S(3, 4, 6, 7) = yz+ xz' Map for Example ; F(x, y, z) = Σ(3, 4, 6, 7) = yz + xz' Digital Logic Design Ch-3

31 3-Inputs K-Map Consider four adjacent squares 2, 4, and 8 squares m +m 2 +m 4 +m 6 = x'y'z'+x'yz'+xy'z'+xyz' = x'z'(y'+y) +xz'(y'+y) = x'z' + xz = z' m +m 3 +m 5 +m 7 = x'y'z+x'yz+xy'z+xyz =x'z(y'+y) + xz(y'+y) =x'z + xz = z Digital Logic Design Ch-3

32 3-Inputs K-Map Example 3.3: simplify F(x, y, z) = S(, 2, 4, 5, 6) F(x, y, z) = S(, 2, 4, 5, 6) = z'+ xy' F(x, y, z) = Σ(, 2, 4, 5, 6) = z' +xy' Digital Logic Design Ch-32

33 3-Inputs K-Map Example : let F = A'C + A'B + AB'C + BC a) Express it in sum of minterms. b) Find the minimal sum of products expression. Ans: F(A, B, C) S(, 2, 3, 5, 7) = C + A'B Map for Example 3.4, A'C + A'B + AB'C + BC = C + A'B Digital Logic Design Ch-33

34 4-Inputs K-Map The map 6 minterms Combinations of 2, 4, 8, and 6 adjacent squares Four-variable Map Digital Logic Design Ch-34

35 4-Inputs K-Map Example : simplify F(w, x, y, z) = S(,, 2, 4, 5, 6, 8, 9, 2, 3, 4) F = y'+w'z'+xz' Map for Example; F(w, x, y, z) = Σ(,, 2, 4, 5, 6, 8, 9, 2, 3, 4) = y' + w' z' +xz' Digital Logic Design Ch-35

36 4-Inputs K-Map Example : simplify F = A B C + B CD + A B C D + AB C Map for Example ; A B C + B CD + A B C D + AB C = B D + B C +A CD Digital Logic Design Ch-36

37 4-Inputs K-Map A B C D Y Y CD AB Y = AC + ABD + ABC + BD Digital Logic Design Ch-37

38 4-Inputs K-Map A B C D Y X X X X X X X Y AB CD X X X X X Y = A + BD + C X X Digital Logic Design Ch-38

39 4-Inputs K-Map INPUTS OUTPUT Digital Logic Design Ch-39

40 4-Input K-Map Before Simplification Before Simplification Y ABCD ABCD ABCD ABCD Y ABCD ABCD ABCD ABCD +ABCD+ABCD+ABCD+ABCD + ABCD+ABCD+ABCD After Simplification After Simplification +ABCD+ABCD+ABCD+ABCD + ABCD+ABCD+ABCD Y ABC AD AD AB Y AC D BC Digital Logic Design Ch-4

41 4-Inputs K-Map Before Simplification Before Simplification After Simplification After Simplification Digital Logic Design Ch-4

42 Example: implement F( x, y, z) (,2,3,4,5,7) F( x, y, z) xy x y z Digital Logic Design Ch-42

43 Mano - K-Map Quiz - Simplify the following Boolean functions, using Karnaugh maps: (a) * F (x, y, z) = (2, 3, 6, 7) (b) * F (A, B, C, D) = (4, 6, 7, 5) (c) * F (A, B, C, D) = (3, 7,, 3, 4, 5) (d) * F (w, x, y, z) = (2, 3, 2, 3, 4, 5) (e) F (w, x, y, z) = (, 2, 3, 4, 5) (f) F (w, x, y, z) = (8,, 2, 3, 4) 2- Simplify the following Boolean functions, using four-variable maps: (a) * F (w, x, y, z) = (, 4, 5, 6, 2, 4, 5) (b) F (A, B, C, D) = (2, 3, 6, 7, 2, 3, 4) (c) F (w, x, y, z) = (, 3, 4, 5, 6, 7, 9,, 3, 5) (d) * F (A, B, C, D) = (, 2, 4, 5, 6, 7, 8,, 3, 5) 3- Simplify the following Boolean expressions, using four-variable maps: (a) * A B C D + AC D + B CD + A BCD + BC D (b) * x z + w xy + w (x y + xy ) (c) A B C D + AB D + A BC + ABCD + AB C (d) A B C D + BC D + A C D + A BCD + ACD 4- Simplify the following Boolean expressions, using four-variable maps: (a) * w z + xz + x y + wx z (b) AD + B C D + BCD + BC D (c) * AB C + B C D + BCD+ ACD + A B C + A BC D (d) wxy + xz + wx z + w x Digital Logic Design Ch-43

44 Multilevel NAND Circuits Boolean function implementation AND-OR logic NAND-NAND logic» AND AND + inverter» OR: inverter + OR = NAND» For every bubble that is not compensated by another small circle along the same line, insert an inverter. Implementing F = A(CD + B) + BC Digital Logic Design Ch-44

45 NAND Implementation Implementing F = (AB +A B)(C+ D ) Digital Logic Design Ch-45

46 NOR Implementation NOR function is the dual of NAND function. The NOR gate is also universal. Logic Operation with NOR Gates Digital Logic Design Ch-46

47 NOR Implementation Example: F = (AB +A B)(C + D ) Implementing F = (AB +A B)(C + D ) with NOR gates Digital Logic Design Ch-47

48 General Quiz - Use K-Map to Simplify the following Boolean Functions F ABCD ABCD ABCD ABCD ABCD ABCD F ABCD ABCD ABCD ABCD ABCD ABCD ABCD F ABCD ABCD ABCD ABCD ABCD F ABCD ABCD ABCD ABCD ABCD ABCD ABCD ABCD 2-Simplify the following Boolean equations using Boolean theorems. Check for correctness using a truth table or K- map. 3-Simplify each of the following Boolean equations. Sketch a reasonably simple combinational circuit implementing the simplified equation. Digital Logic Design Ch-48

49 Quiz 4- Write Boolean equations for the circuit in Figure : 5- Minimize the Boolean equations from Exercise 4 and sketch an improved circuit with the same function? Digital Logic Design Ch-49

50 Quiz 6- Write a Boolean equation in sum-of-products canonical form for each of the truth tables in Figure - Write a Boolean equation in product-of-sums canonical form for the truth tables in Figure? - Minimize each of the Boolean equations from Exercise Digital Logic Design Ch-5

51 Quiz 7- Using De Morgan equivalent gates and bubble pushing methods, redraw the circuit in Figure so that you can find the Boolean equation by inspection. Write the Boolean equation. Digital Logic Design Ch-5

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