8/12/2019 logika11

1/6

1

LOGIKA MATEMATIKA2004/2005

Aljabar Boolean

2

Developed by George Boole in the 1850s

M a t h e m a t i c a l t h e o r y o f l o g i c . Shannon was the first to use Boolean

Algebra to solve problems in electronicc i r c u i t d e s i g n . ( 1 9 3 8 ) .

The behavior of a Digital Circuit can bedescribed in terms of Boolean Algebra.

Boolean algebra provides the operations andthe rules for working with the set{0, 1}.

Boolean Algebra

3

Purpose of Boolean Algebra

Boolean Algebra is thetool that allows logicalideas to be representedin a mathematical way,simplified and analyzed,and then translated intor ea l h a r d wa r e fo rimplementation (usingl o g i c a l g a t e s ) .

Logical idea

Gate - implementation

Mathematicalrepresentation

4

Boolean Constants

In Boolean algebra, there are onlytwo constants, true and false

0VOff0false

+5VOn1true

Voltagelevel

State ofa switch

Binarydigit

Booleanconstant

5

Boolean Variables

Boolean variables are variables thatstore values that are Booleanc o n s t a n t s .

6

Boolean Operator AND

If A and B are Boolean variables (or expressions)then

A AND Bis true if and only if bothA and B are true.

If A and B are Boolean variables (or expressions)then

A AND Bis false if and only if eitherA or B is false or theyreboth false.

We denote the AND operation likemultiplication in ordinary algebra: AB or A.B

8/12/2019 logika11

2/6

7

Boolean Operator OR

If A and B are Boolean variables (or expressions)

then A OR Bis true if and only if at least one of A and B istrue.If A and B are Boolean variables (or expressions)then

A OR Bis false if and only if both A and B are false.

We denote the OR operation like additionin ordinary algebra: A+B

8

Boolean Operator NOT

If A is a Boolean variable (or expression)

thenNOT A

has the opposite value from A.

We denote the NOToperation by putting abar over the variable (or expression) :

_A

9

Truth Table for AND

111

001

010

000

ABBA

10

Truth Table for OR

111

101

110

000

A+BBA

11

Truth Table for NOT

01

10

_

AA

12

Boolean Operations

The complement is denoted by a bar (on the slides, wewill use a minus sign). It is defined by

-0 = 1 and -1 = 0.

The Boolean sum, denoted by + or by OR, has thefollowing values:

1 + 1 = 1, 1 + 0 = 1, 0 + 1 = 1, 0 + 0 = 0

The Boolean product, denoted by or by AND, has thefollowing values:

1 1 = 1, 1 0 = 0, 0 1 = 0, 0 0 = 0

8/12/2019 logika11

3/6

13

The rules of Boolean Algebra are:

AND Operations ()

00 = 0 A0 = 010 = 0 A1 = A01 = 0 AA = A11 = 1 AA' = 0

OR Operations (+)

0+0 = 0 A+0 = A1+0 = 1 A+1 = 10+1 = 1 A+A = A1+1 = 1 A+A' = 1

NOT Operations (')

0' = 1 A'' = A1' = 0

Associative Law

(AB)C = A(BC) = ABC

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

Distributive Law

A(B+C) = (AB) + (AC)

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

Some Theorems

A+(AB) = A

A(A+B) = A

A(A+B) = ABA+(AB) = A+B

A+(AB) = A+BA+(AB) = A+B

ABC = (A+B+C)(ABC) = A+B+C

Commutative LawCommutative Law

AAB = BB = BAAA+B = B+AA+B = B+A

PrecedencePrecedence

AA B = (AB = (A))BBAAB+C = (AB+C = (AB) + CB) + CA+BA+BC = A + (BC = A + (BC)C)

DeMorgan's Theorem

(AB)' = A' + B' (NAND)(A+B)' = A' B' (NOR)

De Morgans theorems

14

Boolean Functions and Expressions

Definition: Let B = {0, 1}. The variable x is

called a Boolean variable if it assumes valuesonly from B.

A function from Bn, the set {(x1, x2, , xn)|xiB, 1 i n} , to B is called a Booleanfunction of degree n.

Boolean functions can be represented usingexpressions made up from the variables andBoolean operations.

15

Boolean Functions and Expressions

Question: How many different Boolean functions of degree 1 arethere?

Solution: There are four of them, F1, F2, F3, and F4:

0

1

F3

1

0

F2

101

100

F4F1x

16

Boolean Functions and Expressions

Question: How many different Boolean functions of degree 2 arethere?

Solution: There are 16 of them, F1, F2, , F16:

1

0

0

0

F2

0

0

0

0

F1

010

101

011

000

F3yx

1

1

1

0

F8

0

1

1

0

F7

0

0

0

1

F9

0

0

1

0

F5

1

1

0

0

F4

1

0

1

0

F6

0

1

0

1

F11

1

0

0

1

F10

0

1

1

1

F12

1

0

1

1

F14

0

0

1

1

F13

1

1

0

1

F15

1

1

1

1

F16

17

Evaluating a Boolean Expression

Unlike ordinary algebra, for a BE, there are

only finitely many possible assignments of

values to the variables; so, theoretically, we

can make a table, called a truth table, that

shows the value of the BE for every possible

set of values of the variables.

For convenience, use

0 = false

1 = true 18

Filling in a Truth Table

If there are N variables, there are 2N

possible combinations of values

Thus, there are 2N rows in the truth table

Fill in the values by counting up from 0 inbinary

8/12/2019 logika11

4/6

19

Truth Table for (X+Y)Z

1110

0001

1101

0011

1

0

0

0

X

1

0

1

0

Z

11

01

00

00

(X+Y)ZY

20

Construct a truth table for

_ ___

E = AB + (A+C)B

Example

21

111

011

101

001

110

010

100

000

CBA

0

0

1

1

0

0

1

1

_

B

_ ___

E =AB + (A+C)B

0

0

1

1

0

0

0

0

_

AB

1

1

1

1

1

0

1

0

A+C

0

0

0

0

0

1

0

1

___

(A+C)

0

0

0

0

0

1

0

0

___

(A+C)B

0

0

1

1

0

1

0

0

E

22

Boolean Functions and Expressions

There is a simple method for deriving aBoolean expression for a function that isdefined by a table. This method is based onminterms.

Definition: A literal is a Boolean variable orits complement. A minterm of the Booleanvariables x1, x2, , xn is a Boolean producty 1y 2 y n , where y i = x i or y i = -x i .

Hence, a minterm is a product of n literals,w ith one l i tera l for each var iab le .

23

Boolean Functions and Expressions

Example: Give a Boolean expression for the Boolean functionF(x, y) as defined by the following table:

110

001

011

000

F(x, y)yx

Possible solution: F(x, y) = (-x).y

24

Boolean Functions and Expressions

Consider F(x,y,z) again:

F(x, y, z) = 1 if and only if:

x = y = z = 0 or

x = y = 0, z = 1 or

x = 1, y = z = 0Therefore,

F(x, y, z) =(-x)(-y)(-z) +(-x)(-y)z +x(-y)(-z)

0

0

1

1

F(x, y, z)

1

0

1

0

z

00

10

10

00

yx

0

0

0

1

1

0

1

0

11

11

01

01

8/12/2019 logika11

5/6

8/12/2019 logika11

6/6

31

Gambarlah circuituntuk menghasilkanoutput sbb :

zyxzyx

zyx

xyx

).(

).(

).(

++

+

+

32

Carilah SPE dari fungsi F(x,y,z) = (x+y)~z,

kemudian gambarkan circuit yang sesuaidengan fungsi tersebut.

Jawab :

33

Carilah SPE dari fungsi boolean berikut ini dangambarkan circuit yang sesuai.

yzxzyxF

zyxzyxF

)(),,(

),,(

+=

++=