S1 Teknik TelekomunikasiFakultas Teknik Elektro

2016/2017

Boole Algebra and Logic Series

CLO1-Week3-Simplify Boolean

Expression Using Algebra Axiom

• Understand the Axiom of Boolean Algebra

• Knowing the Boolean Algebra Theorem

• Understand the Concept of Duality in Boolean Algebra

• Understand how to solve Boolean Algebra Expression

by Thruth Table

• Understand how to simplify Boolean Expression by

Using Boolean Algebra

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Outline

Why should be “Simple”?

• Digital computers contain circuits that implement Boolean functions.

• The simpler that we can make a Boolean function, the smaller the circuit that will result.

– Simpler circuits are cheaper to build, consume less power, and run faster than complex circuits.

• With this in mind, we always want to reduce our Boolean functions to their simplest form.

• There are a number of Boolean identities that help us to do this.

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Axiom in Boolean Algebra

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1a. 0 0 = 01b. 1 + 1 = 1

2a. 1 1 = 12b. 0 + 0 = 0

3a. 0 1 = 1 0 = 03b. 1 + 0 = 0 + 1 = 1

4a. If x = 0, then = 14b. If x = 1, then = 0

x_

x_

Single Variable Theorem

• Null Law

• Identity Law

• Idempotent Law

• Inverse Law

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5a. x 0 = 05b. x + 1 = 1

6a. x 1 = x6b. x + 0 = x

7a. x x = x7b. x + x = x

8a. x = 08b. x + = 1

9. = x

x_

x_

x

__

Duality Concept

• Duality of Boolean Expression is obtained byreplacing the AND operator with the equivalentof an OR operator, and the operator OR byequivalent operators AND, bit '0' with theequivalent bit '1' and bit '1' to bits '0‘

• This principle will be useful in the manipulation ofBoolean algebra in a logic circuit simplification

• To simplify the Boolean Expression, you need touse Boolean Algebra Law and also itsCharacteristic

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Two or Three Variable-Characteristic in Algebra

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10a. x y = y x10b. x + y = y + x

11a. x (y z) = (x y) z11b. x + (y + z) = (x + y) + z

12a. x (y + z) = x y + x z12b. x + (y z) = (x + y) (x + z)

13a. x + x y = x13b. x (x + y) = x

Commutative}

}

}

}

Associative

Distributive

Absorptive

Two or Three Variable-Characteristic in Algebra

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14a. x y + x = x

14b. (x + y) . (x + ) = x

15a.

15b.

Combining

Theorem of DeMorgan}

}y

y

x . y x= + y

x + y x y= .

X = X'

X = (X')'} Symbol Equivalence

For Example

By Using Boolean Theorem, Proof that:

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1. X • Y + X • Y' = X

2. X + X • Y = X

Examples: Proof Theorem of Algebra Boole By Using Axiom and Single Variable Theorem:

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1. Proof Theorem : X • Y + X • Y' = XX • Y + X •Y' = X • (Y + Y')

X • (Y + Y') = X • (1)

X • (1) = X

2. Proof Theorem : X + X • Y = XX + X • Y = X • 1 + X • Y

X • 1 + X • Y = X • (1 + Y)

X • (1 + Y) = X • (1)

X • (1) = X

Distributive :

Complement :

Identity :

Identity

Distributive

Identity

Identity

THEOREM of DeMorgan

(X + Y)' = X' • Y'

(X • Y)' = X' + Y'

NOR Gate equivalent with

AND Gate with complemented input

NAND Gate equivalent withOR Gate with complemented input

Example:

Z = A' B' C + A' B C + A B' C + A B C'

Z' = (A + B + C') • (A + B' + C') • (A' + B + C') • (A' + B' + C)

DeMorgan theorem can be used to convertDeMorgan theorem can be used to convert

statement AND/OR be a statement OR/AND

X 0 0 1 1

Y 0 1 0 1

X 1 1 0 0

Y 1 0 1 0

X + Y 1 0 0 0

X•Y 1 0 0 0

X 0 0 1 1

Y 0 1 0 1

X 1 1 0 0

Y 1 0 1 0

X + Y 1 1 1 0

X•Y 1 1 1 0

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A Truth Table can be expressed as a Boolean function

A truth table can be expressed in two forms that are equivalent Boolean function

The functions of the equation obtained from a truth table called a canonical form.

Sum of Products Form

Also called disjunctive normal form, an expansionPart minterm truth table

F = A' B C + A B' C' + A B' C + A B C' + A B C

0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

F' = A' B' C' + A' B' C + A' B C'

F 0 0 0 1 1 1 1 1

F 1 1 1 0 0 0 0 0

C 0 1 0 1 0 1 0 1

B 0 0 1 1 0 0 1 1

A 0 0 0 0 1 1 1 1

TABEL KEBENARAN (TRUTH TABLE)

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Sum of Products

F in form of SoP :

F(A,B,C) = Sm(3,4,5,6,7)

= m3

+ m4

+ m5

+ m6

+ m7= A' B C + A B' C' + A B' C

+ A B C' + A B C

Simplify by Using Boolean Algebra Theorem:

F = A B' (C + C') + A' B C + A B (C' + C)

= A B' + A' B C + A B

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

= A + A' B C

= A + B C

Realization Results SimplificationSoP expression

B

C

A

F

A B C = m 1 A B C = m 2 A B C = m 3 A B C = m 4 A B C = m 5 A B C = m 6 A B C = m 7

A

0 0 0 0 1 1 1 1

B

0 0 1 1 0 0 1 1

C

0 1 0 1 0 1 0 1

Minterms

A B C = m 0

CANONICAL SUM of PRODUCT (SoP)

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Product of Sums / Conjunctive Normal Form / Maxterm Expansion

Boolean function PoS readings based on the truth table:

Maxterm form :

Define rows in the truth tableWith F = 0.'0' in the input field is a notationPut without complement.'1' on the input field notationPut the complement.

F(A,B,C) = PM(0,1,2)

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

F’(A,B,C) = PM(3,4,5,6,7)

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

A 0 0 0 0 1 1 1 1

B 0 0 1 1 0 0 1 1

C 0 1 0 1 0 1 0 1

Maxterms A + B + C = M 0 A + B + C = M 1 A + B + C = M 2 A + B + C = M 3 A + B + C = M 4 A + B + C = M 5 A + B + C = M 6 A + B + C = M 7

PERNYATAAN CANONICAL PRODUCT of SUM (PoS)

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Canonical Sum of Products

Minimized Sum of Products

Canonical Products of Sums

A

B

F 2

F 3

F 1

C

COMPARISON OF IMPLEMENTATION

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16

End of Chapter 3

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