1 Ch. 4 Boolean Algebra and Logic Simplification Boolean Operations and Expressions Laws and Rules...

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

BzuPages.COM

2Introduction

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), 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

BzuPages.COM

3

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

Basic Functions

BzuPages.COM

4

• ANDZ=X Y or Z=XY Z=1 if and only if X=1 and Y=1, otherwise

Z=0

• ORZ=X + YZ=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

• NOTZ=X orZ=1 if X=0, Z=0 if X=1

Basic Functions (계속 )

BzuPages.COM

5Basic Functions (계속 )

BzuPages.COM

6Boolean 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

BzuPages.COM

7Basic Identities of Boolean Algebra

The relationship between a single variable X, its complement X, and the binary constants 0 and 1

BzuPages.COM

8Laws of Boolean Algebra

•Commutative Lawthe order of literals does not matter–A + B = B + A

–A B = B A

BzuPages.COM

9

•Associative Lawthe 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 )

BzuPages.COM

10

•Distributive Law : A(B + C) = AB + AC

A

B

C

X

YX=Y


BzuPages.COM

11

(A+B)(C+D) = AC + AD + BC + BD

A

BCD

XY X=Y


BzuPages.COM

12

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

Rules of Boolean Algebra

BzuPages.COM

13

A+1=1

ORing with 1 must give a 1 since if any input is 1 an OR gate will give a 1

A

XX=A+1=1

Rules of Boolean Algebra ( 계속 )

BzuPages.COM

14

A•0=0

In math if 0 is multiplied with anything you get 0. If you AND anything with 0 you get 0

A

XX=A0 = 0


BzuPages.COM

15

A•1 =AANDing anything with 1 will yield the anything

A

XX=A1=A

A


BzuPages.COM

16

A+A = A

ORing with itself will give the same result

A

A

X

A=A+A =A


BzuPages.COM

17

A+A’=1

Either A or A’ must be 1 so A + A’ =1

A

A’

XX=+A’=1


BzuPages.COM

18

A•A = A

ANDing with itself will give the same result

A

A

XA=AA=A


BzuPages.COM

19

A•A’ =0

In digital Logic 1’ =0 and 0’ =1, so AA’=0 since one of the inputs must be 0.

A

A’

XX=AA’=0


BzuPages.COM

20

A = (A’)’If you not something twice you are back to the

beginning

A

XX=(A’)’=A


BzuPages.COM

21

A

B

X

A + AB = A


BzuPages.COM

22

A + A’B = A + B If A is 1 the output is 1 If A is 0 the output is B

AB

XY X=Y


BzuPages.COM

23

A

B

C

X

Y

(A + B)(A + C) = A + BC


BzuPages.COM

24

•DeMorgan’s TheoremF(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

BzuPages.COM

25

BzuPages.COM

26

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

BzuPages.COM

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’)

BzuPages.COM

28Boolean 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.

BzuPages.COM

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)

BzuPages.COM

30Truth Table from Logic Circuit

Input OutputA B C D A(B+C

D)0 0 0 0 00 0 0 1 00 0 1 0 00 0 1 1 00 1 0 0 00 1 0 1 00 1 1 0 00 1 1 1 01 0 0 0 01 0 0 1 01 0 1 0 01 0 1 1 11 1 0 0 11 1 0 1 11 1 1 0 11 1 1 1 1

A(B+CD)=m11+m12+m13+m14+m15 =(11,12,13,14,15)

BzuPages.COM

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

BzuPages.COM

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 expressionA’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’

BzuPages.COM

33Standard Forms of Boolean Expressions

•The Sum-of-Products(SOP) FormEx) AB+ABC, ABC+CDE+B’CD’

•The Product-of-Sums(POS) FormEx) (A+B)(A+B+C), (A+B+C)(C+D+E)(B’+C+D’)

•Principle of Duality : SOP POS

•Domain of a Boolean ExpressionThe set of variables contained in the expressionEx) A’B+AB’C : the domain is {A, B, C}

BzuPages.COM

34

• Implementation of a SOP ExpressionAND-OR logic

•Conversion of General Expression to SOP FormA(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

BzuPages.COM

35Standard 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’DSol) 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)

BzuPages.COM

36Product-of-Sums Form

• Implementation of a POS ExpressionOR-AND logic

BzuPages.COM

37Standard 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)

BzuPages.COM

38Converting 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 POSSol) 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))

BzuPages.COM

39Boolean Expressions and Truth Tables

•Converting SOP Expressions to Truth Table FormatEx 4-18) A’B’C+AB’C’+ABC =(1,4,7)

InputsA B C

OutputX

Product Term

0 0 0 00 0 1 1 A’B’C0 1 0 00 1 1 01 0 0 1 AB’C’1 0 1 01 1 0 01 1 1 1 ABC

BzuPages.COM

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)

InputsA B C

OutputX Sum Term

0 0 0 0 A+B+C0 0 1 10 1 0 0 A+B’+C0 1 1 0 A+B’+C’1 0 0 11 0 1 0 A’+B+C’1 1 0 0 A’+B’+C1 1 1 1

BzuPages.COM

41

Ex 4-20) Determine standard SOP and POS from the truth tableSol) (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’)

InputsA B C

OutputX

0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 11 0 1 01 1 0 11 1 1 1

BzuPages.COM

42

Boolean Expression

Truth Table

Logic Diagram

BzuPages.COM

43Karnaugh Map

•Simplification methods–Boolean algebra(algebraic method)–Karnaugh map(map method))

XY+XY=X(Y+Y)=X

BzuPages.COM

44

BzuPages.COM

45

Three- and Four-input Kanaugh mapsGray code

BzuPages.COM

46

BzuPages.COM

47

BzuPages.COM

48

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

BzuPages.COM

49

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

0 1 3 2

4 5 7 6

BzuPages.COM

50

Four-Variable Map16 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

BzuPages.COM

51

BzuPages.COM

52

BzuPages.COM

53

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

BzuPages.COM

54Karnaugh Map SOP Minimization

•Mapping a Standard SOP Expression

BzuPages.COM

55

Ex 4-21) Ex 4-22)

BzuPages.COM

56

•Mapping a Nonstandard SOP Expression– Numerical Expression of a Nonstandard Product Term

Ex 4-23) A’+AB’+ABC’A’ AB’ ABC’000 100 110001 101010011

BzuPages.COM

57

Ex 4-24) B’C’+AB’+ABC’+AB’CD’+A’B’C’D+AB’CDB’C’ AB’ ABC’ AB’CD’ A’B’C’D AB’CD0000 1000 1100 1010 0001 10110001 1001 11011000 10101001 1011

BzuPages.COM

58Karnaugh 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

BzuPages.COM

59

Ex 4-26) F=B+A’C+AC’D

BzuPages.COM

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

BzuPages.COM

61

Ex 4-28) Minimize the following expressionAB’C+A’BC+A’B’C+A’B’C’+AB’C’

Sol) B’+A’C

BzuPages.COM

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

BzuPages.COM

63Mapping Directly from a Truth Table

BzuPages.COM

64Don’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

BzuPages.COM

65Karnaugh Map POS Minimization

•Use the Duality PrincipleF(A,A, , + , 1,0) F*(A,A, + , ,0,1)

SOP POS

BzuPages.COM

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)

BzuPages.COM

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)

BzuPages.COM

68

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)

BzuPages.COM

69Converting 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)

BzuPages.COM

70

BzuPages.COM

71Five/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

00

01

11

10

BCDE A=0

A=1

BzuPages.COM

72

•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

00

01

11

10

CDEF

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

BzuPages.COM

73

Ex 4-34)Sol) A’D’E’+B’C’D’+BCD+ACDE

BzuPages.COM

74Digital System Application : 7-Segment LED Driver

Seven-Segment LED driver

BzuPages.COM

75

A B C D

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

g = m(2,3,4,5,6,8,9) =A+BC’+B’C+CD’CD

AB

BzuPages.COM

76

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

BzuPages.COM

77

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

End of Ch. 4

1 Ch. 4 Boolean Algebra and Logic Simplification Boolean Operations and Expressions Laws and Rules...

Documents

Transcript of 1 Ch. 4 Boolean Algebra and Logic Simplification Boolean Operations and Expressions Laws and Rules...