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

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”
{(1,0), Var, (NOT, AND, OR), Thms}
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

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

4. Basic Functions (계속) AND OR NOT 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

5. Basic Functions (계속)

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

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

8. Laws of Boolean Algebra
Commutative Law
the order of literals does not matter
A + B = B + A
A B = B A

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

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

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

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

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

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

15. Rules of Boolean Algebra (계속)
A•1 =A
ANDing anything with 1 will yield the anything A A X
X=A1=A

16. Rules of Boolean Algebra (계속)
A+A = A
ORing with itself will give the same result A X
A=A+A =A

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

18. Rules of Boolean Algebra (계속)
A•A = A
ANDing with itself will give the same result A X
A=AA=A

19. Rules of Boolean Algebra (계속)
A•A’ =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

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

21. Rules of Boolean Algebra (계속)
A + AB = A A B X

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

23. Rules of Boolean Algebra (계속)
(A + B)(A + C) = A + BC A B C X Y

24. DeMorgan’s Theorems 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

26. Look at (A +B +C + D)’ = A’ • B’ • C’ • D’

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’)

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.

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)

30. Truth Table from Logic Circuit
Input
Output A B C D
A(B+CD) 1
A(B+CD)=m11+m12+m13+m14+m15 =(11,12,13,14,15)

31. Simplification Using Boolean Algebra
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

32. A’BC+AB’C’+A’B’C’+AB’C+ABC
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’

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}

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

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 = = = (0,1,2,3,10,11,13)

36. Product-of-Sums Form
Implementation of a POS Expression
OR-AND logic

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)

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)
POS=(1)(4)(6) = (1, 4, 6) (=(A+B+C’)(A’+B+C)(A’+B’+C))

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 1
A’B’C
AB’C’
ABC

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
A+B+C 1
A+B’+C
A+B’+C’
A’+B+C’
A’+B’+C

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 1

42. Boolean Expression
Truth Table
Logic Diagram

43. Simplification methods
Karnaugh Map
Simplification methods
Boolean algebra(algebraic method)
Karnaugh map(map method))
XY+XY=X(Y+Y)=X

45. Three- and Four-input Kanaugh maps
Gray code

48. F(X,Y,Z)=m(0,1,2,6) =(XY+YZ)=X’Y’ + YZ’

49. Example) F(X,Y,Z)=m(2,3,4,5) =XY+XY

50. Four-Variable Map 16 minterms : m0 ~ m15 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

53. F(W, X,Y,Z)=m(0,2,7,8,9,10,11) = WX’ + X’Z’ + W’XYZ

54. Karnaugh Map SOP Minimization
Mapping a Standard SOP Expression

55. Ex 4-21) Ex 4-22)

56. Mapping a Nonstandard SOP Expression
Numerical Expression of a Nonstandard Product Term
Ex 4-23) A’+AB’+ABC’
A’ AB’ ABC’
010
011

57. Ex 4-24) B’C’+AB’+ABC’+AB’CD’+A’B’C’D+AB’CD

58. Karnaugh Map Simplification of SOP Expressions
Group 2n 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

59. Ex 4-26) F=B+A’C+AC’D

60. 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

61. AB’C+A’BC+A’B’C+A’B’C’+AB’C’
Ex 4-28) Minimize the following expression
AB’C+A’BC+A’B’C+A’B’C’+AB’C’
Sol) B’+A’C

62. 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

63. Mapping Directly from a Truth Table

64. 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

65. Karnaugh Map POS Minimization
Use the Duality Principle
F(A,A, , + , 1,0)  F*(A,A, + , ,0,1)
SOP  POS

66. Ex 4-30) (A’+B’+C+D)(A’+B+C’+D’)(A+B+C’+D) (A’+B’+C’+D’)(A+B+C’+D’)
Sol)

67. 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)

68. (1+0+0+0)(0+0+0+0)(0+0+1+0)(1+0+0+1)(0+1+0+0)(1+1+0+0)
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)
( )( )( )( )( )( )
F=(C+D)(A’+B+C)(A+B+D)

69. 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)

71. Five/Six –Variable K-Maps
Five Variable K-Map : {A,B,C,D,E} BC DE
A=0
A=1

72. Six Variable K-Map : {A,B,C,D,E,F} 00 11

73. Ex 4-34)
Sol) A’D’E’+B’C’D’+BCD+ACDE

74. Digital System Application : 7-Segment LED Driver
Seven-Segment LED driver

75. g = m(2,3,4,5,6,8,9) =A+BC’+B’C+CD’ A B C D CD AB 0 1 3 2 4 5 7 6
g = m(2,3,4,5,6,8,9) =A+BC’+B’C+CD’ CD AB

76. Figure 4-59 Karnaugh map minimization of the segment-a logic expression.

77. Figure 4-60 The minimum logic implementation for segment a of the 7-segment display.

78. End of Ch. 4