2. 1 Ch. 4 Boolean Algebra and Logic Simplification Boolean Operations and Expressions Laws and Rules of Boolean Algebra Boolean Analysis of Logic Circuits Simplification Using Boolean Algebra Standard Forms of Boolean Expressions Truth Table and Karnaugh Map Programmable Logic: PALs and GALs Boolean Expressions with VHDL

3. Information Security Lab. 2 Introduction Boolean Algebra George Boole(English mathematician), 1854 “An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and Probabilities” Boolean Algebra {(1,0), (NOT, AND, OR} Mathematical tool to expression and analyze digital (logic) circuits Claude Shannon, the first to apply Boole’s work, 1938 – “A Symbolic Analysis of Relay and Switching Circuits” at MIT This chapter covers Boolean algebra, Boolean expression and its evaluation and simplification, and VHDL program

4. Information Security Lab. 3 Boolean functions : NOT, AND, OR, exclusive OR(XOR) : odd function exclusive NOR(XNOR) : even function(equivalence) Basic Functions

5. Information Security Lab. 4 AND Z=X  Y or Z=XY Z=1 if and only if X=1 and Y=1, otherwise Z=0 OR Z=X + Y Z=1 if X=1 or if Y=1, or both X=1and Y=1. Z=0 if and only if X=0 and Y=0 NOT Z=X or Z=1 if X=0, Z=0 if X=1 Basic Functions ( 계속 )

6. Information Security Lab. 5 Basic Functions ( 계속 )

7. Information Security Lab. 6 Boolean Operations and Expressions Boolean Addition – Logical OR operation Ex 4-1) Determine the values of A, B, C, and D that make the sum term A+B’+C+D’ Sol) all literals must be ‘0’ for the sum term to be ‘0’ A+B’+C+D’=0+1’+0+1’=0  A=0, B=1, C=0, and D=1 Boolean Multiplication – Logical AND operation Ex 4-2) Determine the values of A, B, C, and D for AB’CD’=1 Sol) all literals must be ‘1’ for the product term to be ‘1’ AB’CD’=10’10’=1  A=1, B=0, C=1, and D=0

8. Information Security Lab. 7 Basic Identities of Boolean Algebra The relationship between a single variable X, its complement X, and the binary constants 0 and 1

9. Information Security Lab. 8 Laws of Boolean Algebra Commutative Law the order of literals does not matter – A + B = B + A – A B = B A

10. Information Security Lab. 9 Associative Law the grouping of literals does not matter – A + (B + C) = (A + B) + C (=A+B+C) – A(BC) = (AB)C (=ABC) Laws of Boolean Algebra ( 계속 )

11. Information Security Lab. 10 Distributive Law : A(B + C) = AB + AC ABCXYABCXY X=Y Laws of Boolean Algebra ( 계속 )

12. Information Security Lab. 11 (A+B)(C+D) = AC + AD + BC + BD ABCDXYABCDXY X=Y Laws of Boolean Algebra ( 계속 )

13. Information Security Lab. 12 A+0=A In math if you add 0 you have changed nothing in Boolean Algebra ORing with 0 changes nothing AXAX X=A+0 =A Rules of Boolean Algebra

14. Information Security Lab. 13 A+1=1 ORing with 1 must give a 1 since if any input is 1 an OR gate will give a 1 AXAX X=A+1 =1 Rules of Boolean Algebra ( 계속 )

15. Information Security Lab. 14 A0=0 In math if 0 is multiplied with anything you get 0. If you AND anything with 0 you get 0 AXAX X=A0 = 0 Rules of Boolean Algebra ( 계속 )

16. Information Security Lab. 15 A1 =A ANDing anything with 1 will yield the anything AXAX X=A1 =A A Rules of Boolean Algebra ( 계속 )

17. Information Security Lab. 16 A+A = A ORing with itself will give the same result AAXAAX A=A+A =A Rules of Boolean Algebra ( 계속 )

18. Information Security Lab. 17 A+A’=1 Either A or A’ must be 1 so A + A’ =1 A A’ X X=+A’ =1 Rules of Boolean Algebra ( 계속 )

19. Information Security Lab. 18 AA = A ANDing with itself will give the same result AAXAAX A=AA =A Rules of Boolean Algebra ( 계속 )

20. Information Security Lab. 19 AA’ =0 In digital Logic 1’ =0 and 0’ =1, so AA’=0 since one of the inputs must be 0. A A’ X X=AA’ =0 Rules of Boolean Algebra ( 계속 )

21. Information Security Lab. 20 A = (A’)’ If you not something twice you are back to the beginning AXAX X=(A’)’ =A Rules of Boolean Algebra ( 계속 )

22. Information Security Lab. 21 ABXABX A + AB = A Rules of Boolean Algebra ( 계속 )

23. Information Security Lab. 22 A + A’B = A + B If A is 1 the output is 1 If A is 0 the output is B ABXYABXY X=Y Rules of Boolean Algebra ( 계속 )

24. Information Security Lab. 23 ABCXYABCXY (A + B)(A + C) = A + BC Rules of Boolean Algebra ( 계속 )

25. Information Security Lab. 24 DeMorgan’s Theorem F(A,A, , +, 1,0) = F(A, A, +, ,0,1) – (A B)’ = A’ + B’ and (A + B)’ = A’ B’ – DeMorgan ’ s theorem will help to simplify digital circuits using NORs and NANDs his theorem states DeMorgan’s Theorems

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27. Information Security Lab. 26 Look at (A +B +C + D)’ = A’ B’ C’ D’

28. Information Security Lab. 27 Ex 4-3) Apply DeMorgan’s theorems to (XYZ)’ and (X+Y+Z)’ Sol) (XYZ)’=X’+Y’+Z’ and (X+Y+Z)’=X’Y’Z’ Ex 4-5) Apply DeMorgan’s theorems to (a) ((A+B+C)D)’ (b) (ABC+DEF)’ (c) (AB’+C’D+EF)’ Sol) (a) ((A+B+C)D)’= (A+B+C)’+D’=A’B’C’+D’ (b) (ABC+DEF)’=(ABC)’(DEF)’=(A’+B’+C’)(D’+E’+F’) (c) (AB’+C’D+EF)’=(AB’)’(C’D)’(EF)’=(A’+B)(C+D’)(E’+F’)

29. Information Security Lab. 28 Boolean Analysis of Logic Circuits Boolean Expression for a Logic Circuit Figure 4-16 A logic circuit showing the development of the Boolean expression for the output.

30. Information Security Lab. 29 Constructing a Truth Table for a Logic Circuit – Convert the expression into the min-terms containing all the input literals – Get the numbers from the min-terms – Putting ‘1’s in the rows corresponding to the min- terms and ‘0’s in the remains Ex) A(B+CD)=AB(C+C’) (D+D’) +A(B+B’)CD =ABC(D+D’) +ABC’(D+D’) +ABCD+AB’CD =ABCD+ABCD’+ABC’D+ABC’D’ +ABCD+AB’CD =ABCD+ABCD’+ABC’D+ABC’D’ +AB’CD =m11+m12+m13+m14+m15=  (11,12,13,14,15)

31. Information Security Lab. 30 Truth Table from Logic Circuit InputOutput ABCDA(B+CD) 00000 00010 00100 00110 01000 01010 01100 01110 10000 10010 10100 10111 11001 11011 11101 11111 A(B+CD)=m11+m12+m13+m14+m15 =  (11,12,13,14,15)

32. Information Security Lab. 31 Ex 4-8) Using Boolean algebra, simplify this expression AB+A(B+C)+B(B+C) Sol) AB+AB+AC+BB+BC =B(1+A+A+C)+AC=B+AC Simplification Using Boolean Algebra

33. Information Security Lab. 32 Ex 4-9) Simplify the following Boolean expression (AB’(C+BD)+A’B’)C Sol) (AB’C+AB’BD+A’B’)C=AB’CC+A’B’C=(A+A’)B’C=B’C Ex 4-10) Simplify the following Boolean expression A’BC+AB’C’+A’B’C’+AB’C+ABC Sol) (A+A’)BC+(A+A’)B’C’+AB’C=BC+B’C’+AB’C =BC+B’(C’+AC)=BC+B’(C’+A)=BC+B’C’+AB’ Ex 4-11) Simplify the following Boolean expression (AB +AC)’+A’B’C Sol) (AB)’(AC)’+A’B’C=(A’+B’)(A’+C’)+A’B’C=A’+A’B’ +A’C’+B’C+A’B’C =A’(1+B’+C’+B’C)+B’C=A’+B’C’

34. Information Security Lab. 33 Standard Forms of Boolean Expressions The Sum-of-Products(SOP) Form Ex) AB+ABC, ABC+CDE+B’CD’ The Product-of-Sums(POS) Form Ex) (A+B)(A+B+C), (A+B+C)(C+D+E)(B’+C+D’) Principle of Duality : SOP  POS Domain of a Boolean Expression The set of variables contained in the expression Ex) A’B+AB’C : the domain is {A, B, C}

35. Information Security Lab. 34 Implementation of a SOP Expression AND-OR logic Conversion of General Expression to SOP Form A(B+CD)=AB +ACD Ex 4-12) Convert each of the following expressions to SOP form: (a) AB+B(CD+EF) (b) (A+B)(B+C+D) Sol) (a) AB+B(CD+EF)=AB+BCD+BEF (b) (A+B)(B+C+D)=AB+AC+AD+ BB+BC+BD =B(1+A+C+D)+ AC+AD=B+AC+AD

36. Information Security Lab. 35 Standard SOP Form (Canonical SOP Form) – For all the missing variables, apply (x+x’)=1 to the AND terms of the expression – List all the min-terms in forms of the complete set of variables in ascending order Ex 4-13) Convert the following expression into standard SOP form: AB’C+A’B’+ABC’D Sol) domain={A,B,C,D}, AB’C(D’+D)+A’B’(C’+C)(D’+D)+ABC’D =AB’CD’+AB’CD+A’B’C’D’+A’B’C’D+A’B’CD’+A’B’CD+ABC’D =1010+1011+0000+0001+0010+0011+1101 =0+1+2+3+10+11+13 =  (0,1,2,3,10,11,13)

37. Information Security Lab. 36 Product-of-Sums Form Implementation of a POS Expression OR-AND logic

38. Information Security Lab. 37 Standard POS Form (Canonical POS Form) – For all the missing variables, apply (x’x)=0 to the OR terms of the expression – List all the max-terms in forms of the complete set of variables in ascending order Ex 4-15) Convert the following expression into standard POS form: (A+B’+C)(B’+C+D’)(A+B’+C’+D) Sol) domain={A,B,C,D}, (A+B’+C)(B’+C+D’)(A+B’+C’+D) =(A+B’+C+D’D)(A’A+B’+C+D’)(A+B’+C’+D) =(A+B’+C+D’)(A+B’+C+D)(A’+B’+C+D’)(A+B’+C+D’)(A +B’+C’+D)=(0100) )(0101)(0110)(1101)=  (4,5,6,13)

39. Information Security Lab. 38 Converting Standard SOP to Standard POS Step 1. Evaluate each product term in the SOP expression. Determine the binary numbers that represent the product terms Step 2. Determine all of the binary numbers not included in the evaluation in Step 1 Step 3. Write in equivalent sum term for each binary number Step 2 and expression in POS form Ex 4-17) Convert the following SOP to POS Sol) SOP= A’B’C’+A’BC’+A’BC+AB’C+ABC=0+2+3+5+7 =  (0,2,3,5,7) POS=(1)(4)(6) =  (1, 4, 6) (=(A+B+C’)(A’+B+C)(A’+B’+C))

40. Information Security Lab. 39 Boolean Expressions and Truth Tables Converting SOP Expressions to Truth Table Format Ex 4-18) A’B’C+AB’C’+ABC =  (1,4,7) Inputs A B C Output X Product Term 0 0 00 0 0 11A’B’C 0 1 00 0 1 10 1 0 01AB’C’ 1 0 10 1 1 00 1 1 11ABC

41. Information Security Lab. 40 Converting POS Expressions to Truth Table Format Ex 4-19) (A+B+C)(A+B’+C)(A+B’+C’)(A’+B+C’)(A’+B’+C) = (000)(010)(011)(101)(110) =  (0,2,3,5,6) Inputs A B C Output X Sum Term 0 0 00A+B+C 0 0 11 0 1 00A+B’+C 0 1 10A+B’+C’ 1 0 01 1 0 10A’+B+C’ 1 1 00A’+B’+C 1 1 11

42. Information Security Lab. 41 Ex 4-20) Determine standard SOP and POS from the truth table Sol) (a) Standard SOP F=A’BC+AB’C’+ABC’+ABC (b) Standard POS F=(A+B+C)(A+B+C’)(A+B’+C) (A’+B+C’) Inputs A B C Output X 0 0 00 0 0 10 0 1 00 0 1 11 1 0 01 1 0 10 1 1 01 1 1 11

43. Information Security Lab. 42 Boolean Expression Truth Table Logic Diagram

44. Information Security Lab. 43 Karnaugh Map Simplification methods – Boolean algebra(algebraic method) – Karnaugh map(map method)) – Quine-McCluskey(tabular method) XY+XY=X(Y+Y)=X

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46. Information Security Lab. 45 Three- and Four-input Kanaugh maps Gray code

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49. Information Security Lab. 48 Gray code sequence generation

50. Information Security Lab. 49 F(X,Y,Z)=  m(0,1,2,6) =(XY+YZ)=X’Y’ + YZ’

51. Information Security Lab. 50 Example) F(X,Y,Z)=  m(2,3,4,5) =XY+XY 0 1 3 2 4 5 7 6

52. Information Security Lab. 51 Example) F(X,Y,Z)=  m(0,2,4,6) = XZ+XZ =Z(X+X)=Z

53. Information Security Lab. 52 Four-Variable Map 16 minterms : m 0 ~ m 15 Rectangle group – 2-squares(minterms) : 3-literals product term – 4-squares : 2-literals product term – 8-squares : 1-literals product term – 16-squares : logic 1

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56. Information Security Lab. 55 F(W, X,Y,Z)=  m(0,2,7,8,9,10,11) = WX’ + X’Z’ + W’XYZ

57. Information Security Lab. 56 Karnaugh Map SOP Minimization Mapping a Standard SOP Expression

58. Information Security Lab. 57 Ex 4-21) Ex 4-22)

59. Information Security Lab. 58 Mapping a Nonstandard SOP Expression – Numerical Expression of a Nonstandard Product Term Ex 4-23) A’+AB’+ABC’ A’AB’ABC’ 000 100110 001 101 010 011

60. Information Security Lab. 59 Ex 4-24) B’C’+AB’+ABC’+AB’CD’+A’B’C’D+AB’CD B’C’ AB’ ABC’ AB’CD’ A’B’C’D AB’CD 0000 1000 1100 1010 0001 1011 0001 1001 1101 1000 1010 1001 1011

61. Information Security Lab. 60 Karnaugh Map Simplification of SOP Expressions Group 2 n adjacent cells including the largest possible number of 1s in a rectangle or square shape, 1<=n Get the groups containing all 1s on the map for the expression Determine the minimum SOP expression form map

62. Information Security Lab. 61 Ex 4-26) F=B+A’C+AC’D

63. Information Security Lab. 62 Ex 4-27) (a) AB+BC+A’B’C’ (b) B’+AC+A’C’ (c) A’C’+A’B+AB’D (d) D’+BC’+AB’C

64. Information Security Lab. 63 Ex 4-28) Minimize the following expression AB’C+A’BC+A’B’C+A’B’C’+AB’C’ Sol) B’+A’C

65. Information Security Lab. 64 Ex 4-29) Minimize the following expression B’C’D’+A’BC’D’+ABC’D’+A’B’CD+AB’CD+A’B’CD’+A’BCD’ +ABCD’+AB’CD’ Sol) D’+B’C

66. Information Security Lab. 65 Mapping Directly from a Truth Table

67. Information Security Lab. 66 Don’t Care Conditions it really does not matter since they will never occur(its output is either ‘0’ or ‘1’) The don’t care terms can be used to advantage on the Karnaugh map

68. Information Security Lab. 67 Karnaugh Map POS Minimization Use the Duality Principle F(A,A, , +, 1,0)  F * (A,A, +, ,0,1) SOP  POS

69. Information Security Lab. 68 Ex 4-30) (A’+B’+C+D)(A’+B+C’+D’)(A+B+C’+D) (A’+B’+C’+D’)(A+B+C’+D’) Sol)

70. Information Security Lab. 69 Ex 4-31) (A+B+C)(A+B+C’)(A+B’+C)(A+B’+C’)(A’+B’+C) Sol) (0+0+0)(0+0+1)(0+1+0)(0+1+1)(1+1+0)=A(B’+C) AC+AB’=A(B’+C)

71. Information Security Lab. 70 Ex 4-32) (B+C+D)(A+B+C’+D)(A’+B+C+D’)(A+B’+C+D)(A’+B’+C+D) Sol) (B+C+D)=(A’A+B+C+D)=(A’+B+C+D)(A+B+C+D) (1+0+0+0)(0+0+0+0)(0+0+1+0)(1+0+0+1)(0+1+0+0)(1+1+0+0) F=(C+D)(A’+B+C)(A+B+D)

72. Information Security Lab. 71 Converting Between POS and SOP Using the K-map Ex 4-33) (A’+B’+C+D)(A+B’+C+D)(A+B+C+D’)(A+B+C’+D’) (A’+B+C+D’)(A+B+C’+D) Sol)

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74. Information Security Lab. 73 Five/Six –Variable K-Maps Five Variable K-Map : {A,B,C,D,E} 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 16 17 19 18 20 21 23 22 28 29 31 30 24 25 27 26 00 01 11 10 BC DE A=0 A=1

75. Information Security Lab. 74 Six Variable K-Map : {A,B,C,D,E,F} 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 16 17 19 18 20 21 23 22 28 29 31 30 24 25 27 26 00 01 11 10 CD EF 00 10 01 11 AB 32 33 35 34 36 37 39 38 44 45 47 46 40 41 43 42 48 49 51 50 52 53 55 54 60 61 62 63 56 57 59 58

76. Information Security Lab. 75 Ex 4-34) Sol) A’D’E’+B’C’D’+BCD+ACDE