1 ch. 4 boolean algebra and logic simplification boolean operations and expressions laws and rules...

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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

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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

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

Basic Functions

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• 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 )

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5Basic Functions (계속 )

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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

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7Basic Identities of Boolean Algebra

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

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8Laws of Boolean Algebra

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

–A B = B A

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•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 )

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

A

B

C

X

YX=Y

Laws of Boolean Algebra (계속 )

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

A

BCD

XY X=Y

Laws of Boolean Algebra (계속 )

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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

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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 )

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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

Rules of Boolean Algebra ( 계속 )

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

A

XX=A1=A

A

Rules of Boolean Algebra ( 계속 )

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A+A = A

ORing with itself will give the same result

A

A

X

A=A+A =A

Rules of Boolean Algebra ( 계속 )

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A+A’=1

Either A or A’ must be 1 so A + A’ =1

A

A’

XX=+A’=1

Rules of Boolean Algebra ( 계속 )

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A•A = A

ANDing with itself will give the same result

A

A

XA=AA=A

Rules of Boolean Algebra ( 계속 )

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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

Rules of Boolean Algebra ( 계속 )

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

beginning

A

XX=(A’)’=A

Rules of Boolean Algebra ( 계속 )

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A

B

X

A + AB = A

Rules of Boolean Algebra ( 계속 )

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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

Rules of Boolean Algebra ( 계속 )

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A

B

C

X

Y

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

Rules of Boolean Algebra ( 계속 )

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•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

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

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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’)

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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.

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•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)

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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)

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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

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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’

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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}

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• 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

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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)

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36Product-of-Sums Form

• Implementation of a POS ExpressionOR-AND logic

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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)

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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))

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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

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• 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

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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

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Boolean Expression

Truth Table

Logic Diagram

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43Karnaugh Map

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

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

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Three- and Four-input Kanaugh mapsGray code

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

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

0 1 3 2

4 5 7 6

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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

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

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54Karnaugh Map SOP Minimization

•Mapping a Standard SOP Expression

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Ex 4-21) Ex 4-22)

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•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

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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

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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

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Ex 4-26) F=B+A’C+AC’D

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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

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

Sol) B’+A’C

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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

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63Mapping Directly from a Truth Table

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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

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65Karnaugh Map POS Minimization

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

SOP POS

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Ex 4-30) (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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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)

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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)

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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)

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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

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•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

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Ex 4-34)Sol) A’D’E’+B’C’D’+BCD+ACDE

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74Digital System Application : 7-Segment LED Driver

Seven-Segment LED driver

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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

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

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

End of Ch. 4